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X-ray generation
X-rays are a form of high-energy electromagnetic radiation. We have talked about ways to generate EM
radiation before: think back to semiconductors. The visible light emitted from LEDs has energy on the order of a few $\mathrm{eV}$s. X-rays are much more energetic! They can range from hundreds to hundreds of thousands of $\mathrm{eV}$s. Therefore, to generate $\mathrm{x}$-rays, we need a much higher energy differential than the band gap of even an insulator.
To achieve such a massive $\Delta E$, we turn to another old friend: interatomic transitions! When we first went quantum, we talked about Bohr's model of quantization in the hydrogen atom. Though the Bohr model really only applies to the hydrogen atom, atoms with higher $\mathrm{Z}$ are roughly proportional to the Bohr model prediction by a factor of $(Z-1)^2$. Though the ionization energy of hydrogen was only $13.1 \mathrm{eV}$, with the intra-atomic energy transitions even lower in energy, larger elements can have energy states separated by massive amounts of energy: enough to produce $\mathrm{x}$-rays.
To actually generate $\mathrm{x}$-rays, we need to excite an electron between two atomic energy levels within an atom. For historical reasons, the atomic transitions that produce $\mathrm{x}$-rays are denoted in Siegbahn notation: the familiar our $\mathrm{n}=1, \mathrm{n}=2$, and $\mathrm{n}=3$ shells are called $\mathrm{K}, \mathrm{L}$, and $\mathrm{M}$, respectively.
The $\mathrm{x}$-rays that are emitted are named by a letter with a subscript: the letter tells us which shell the electron ended in, and the subscript tells us which shell the electron started in. For example, $K_\alpha$ radiation ends in the $\mathrm{K}$ shell $(\mathrm{n}=1)$, and the $\alpha$ tells us it started one shell higher, in the $\mathrm{n}=2$, or $\mathrm{L}$ shell. Similarly, $K_\beta$ ends in the $\mathrm{n}=1$ shell, but comes from two shells up $(\beta)$, or the $\mathrm{M}$ shell. Don't worry about the subshells; we'll just focus on transitions between principal energy levels. These $\mathrm{x}$-rays are called characteristic $\mathrm{x}$-rays, because their character (properties) are atom-specific.
Finally, we need to consider how this energy transition is excited. This is done with high energy electrons: the electrons are accelerated by applying high voltages. When they crash into a heavy element, $\mathrm{x}$-rays are produced (if the electrons are energetic enough). The act of high-energy electrons colliding into a solid is enough to produce some $\mathrm{x}$-rays: these are called Bremsstrahlung.
When the electrons collide with the solid target, they rapidly decelerate as they interact with the charged nuclei. Above all else, energy must be conserved: the excess kinetic energy dissipated through deceleration is emitted as a photon. These photons have a continuous spectrum, as shown in the plot above. However, there is a hard lower bound on the wavelength that can be emitted: since energy and wavelength are inversely proportional, this lower bound is determined by the energy of the incoming electrons (you can also think of it as an upper bound on energy/frequency).
In addition to the Bremsstrahlung, characteristic $\mathrm{x}$-rays are emitted. They show up at specific wavelengths in the $\mathrm{x}$-ray spectrum. We can determine which intraatomic transition the peaks correspond to by referencing the diagram above, and recalling that the energy levels get closer together as $\mathrm{n}$ increases.
The $K_{\beta}$ transition is larger in energy than $K_{\alpha}$, so it occurs at a lower wavelength. The $L_{\alpha}$ would be the lowest energy (highest wavelength) of the three, and would appear to the right on this plot.
Diffraction and Bragg's Law
Next time we’ll talk about what we can do with $\mathrm{x}$-rays, but for now, we need to brush up on Bragg’s law. The basic idea is that when light that is incident on a periodic structure satisfies the Bragg condition, it scatters coherently. The Bragg condition gives the angle at which coherent scattering occurs as a function of the wavelength of the incident light and the periodicity of the lattice:
$n\lambda = 2d\sin \theta \nonumber$
This equation is the condition for constructive interference, and $\mathrm{n}$ is an integer ($n=1$ dominates for most lattices), $\lambda$ is the wavelength of the $\mathrm{x}$-rays, $\theta$ is an experimental knob, and $d$ is the interplanar spacing, $d_{hkl}$.
From the figure above, we can see that when the Bragg condition is fulfilled, what is actually happening is that the extra path length traveled by neighboring photons that scatter coherently is exactly $d\sin \theta$ as it comes in and again as it comes out. Therefore the incoming and outgoing waves constructively interfere, and light registers as bouncing off the crystal. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/06%3A_Recitations/6.15%3A_X-ray_Generation_Diffraction_and_Bragg%27s_Law.txt |
X-ray diffraction
Last time we discussed how to generate $\mathrm{x}$-rays; now, let’s talk about one application that plays a big role in materials characterization. X-ray diffraction is a tool that allows us to probe the crystal structure of a sample. Let’s walk through the process:
First, we need to generate $\mathrm{x}$-rays: this process is shown on the left. As we discussed, by shooting high energy electrons at a target made of heavy atoms (copper is a common choice), both characteristic $\mathrm{x}$-rays and Bremsstrahlung are generated. Then, by bombarding an unknown sample with the $\mathrm{x}$-rays, and measuring the scattered light as a function of incident angle, we can probe the interplanar spacing that exists within the crystal. However, recall that the Bragg condition is wavelength-dependent: to really do this effectively, we only want to keep a narrow band of wavelengths. This is accomplished by employing a filter, removing everything except the $\mathrm{K}_{\alpha}$ $\mathrm{x}$-rays, for example.
After all that, what is really measured in an XRD experiment is the intensity of scattered $\mathrm{x}$-rays as a function of incident angle. A typical XRD spectrum looks something like this:
Plotted here is intensity/counts of $\mathrm{x}$-rays on a detector as a function of $2\theta$, or twice the incident angle. It is helpful to utilize the following table to translate an XRD plot to a crystal structure/element:
$\begin{array}{c|c|c|c|c|c|c} \text { Peak # } & 2 \theta & \theta & \sin ^2 \theta & \text { normalized } & \text { integer } & \text { plane } \ \hline & \text { from } & \leftarrow \times \dfrac{1}{2} & \text { calculator } & \text { divide all } & \text { multiply } & \ & \text { graph } & & & \leftarrow \text { by } & \leftarrow \text { by } & h^2+k^2+l^2= \ & & & & \begin{array}{c} \leftarrow \ \text { smallest of } \ \text { an integer } \end{array} & \text { (often } 2 \text { or } 3 \text { ) } & \leftarrow \end{array}$
Selection rules
As we discussed in lecture, we can apply selection rules to determine what kind of crystal structure a sample has. For each of the Bravais lattices, selection rules tell us which planes have coherent scattering and which do not:
Cubic crystal Allowed planes ($\mathrm{hkl}$) Forbidden planes($\mathrm{hkl}$)
$\mathrm{SC}$ any $\mathrm{h,k,l}$ none
$\mathrm{BCC}$ $\mathrm{h}+\mathrm{k}+\mathrm{l} =$ even number $\mathrm{h}+\mathrm{k}+\mathrm{l} =$ odd number
$\mathrm{FCC}$
$\mathrm{h,k,l}$ all even
$\mathrm{h,k,l}$ all odd
$\mathrm{h,k,l}$ mixed even and odd
Example: Use the following XRD spectrum to determine the element that was investigated. It is either $\mathrm{BCC}$ or $\mathrm{FCC}$. Copper $K_{\alpha}$ radiation was used, which has a wavelength of 1.54 Angstroms.
$\begin{array}{c|c} \text { Peak # } & 2 \theta \ \hline \hline 1 & 44.51 \ 2 & 51.90 \ 3 & 76.45 \ 4 & 93.02 \ 5 & 98.50 \end{array}$
We are given a plot and the values of $2\theta$, so next let’s construct the chart as described above. The first two columns are given, and we can quickly calculate the next two:
$\begin{array}{c|c|c|c|c|c|c} \text { Peak # } & 2 \theta & \theta & \sin ^2 \theta & \text { normalized } & \text { integer } & \text { plane } \ \hline 1 & 44.51 & 22.25 & 0.1434 & & & \ 2 & 51.90 & 25.95 & 0.1915 & & & \ 3 & 76.45 & 38.23 & 0.3829 & & & \ 4 & 93.02 & 46.51 & 0.5263 & & & \ 5 & 98.50 & 49.25 & 0.5739 & & & \end{array}$
For the next column, normalized values, we simply divide all of the values in the $\sin ^2 \theta$ column by the $s m a l l e s t$ value in the $\sin ^2 \theta$ column, $0.1434$.
$\begin{array}{c|c|c|c|c|c|c} \text { Peak # } & 2 \theta & \theta & \sin ^2 \theta & \text { normalized } & \text { integer } & \text { plane } \ \hline 1 & 44.51 & 22.25 & 0.1434 & 1 & & \ 2 & 51.90 & 25.95 & 0.1915 & 1.33 & & \ 3 & 76.45 & 38.23 & 0.3829 & 2.66 & & \ 4 & 93.02 & 46.51 & 0.5263 & 3.65 & & \ 5 & 98.50 & 49.25 & 0.5739 & 4 & & \end{array}$
Looking at the values in the normalized column, they seem to be approximately factors of $\frac{1}{3}$. To create integers, let's try multiplying each by 3 .
$\begin{array}{c|c|c|c|c|c|c} \text { Peak # } & 2 \theta & \theta & \sin ^2 \theta & \text { normalized } & \text { integer } & \text { plane } \ \hline 1 & 44.51 & 22.25 & 0.1434 & 1 & 3 & \ 2 & 51.90 & 25.95 & 0.1915 & 1.33 & 4 & \ 3 & 76.45 & 38.23 & 0.3829 & 2.66 & 8 & \ 4 & 93.02 & 46.51 & 0.5263 & 3.65 & 11 & \ 5 & 98.50 & 49.25 & 0.5739 & 4 & 12 & \end{array}$
Great, this created integers! The integers in this column should be the sum of $h^2, k^2$, and $l^2$ that form a particular plane. Let's try to match some planes:
$\begin{array}{c|c|c|c|c|c|c} \text { Peak # } & 2 \theta & \theta & \sin ^2 \theta & \text { normalized } & \text { integer } & \text { plane } \ \hline 1 & 44.51 & 22.25 & 0.1434 & 1 & 3 & (111) \ 2 & 51.90 & 25.95 & 0.1915 & 1.33 & 4 & (200) \ 3 & 76.45 & 38.23 & 0.3829 & 2.66 & 8 & (220) \ 4 & 93.02 & 46.51 & 0.5263 & 3.65 & 11 & (311) \ 5 & 98.50 & 49.25 & 0.5739 & 4 & 12 & (222) \end{array}$
It takes a bit of time to develop intuition as to how to combine the squares of three numbers to form integers, but it comes with practice. For now, you can double check that these ($\mathrm{hkl}$) planes really do generate the integers in the neighboring column. Also note that any member of a planar family could generate the integer in question: it's ok to choose any plane within that family! For example, the (200), (020), and (002) planes all satisfy $h^2+k^2+l^2=4$, so any would work there.
Next, we should check the selection rules. We see that for each of the planes, the values of $\mathrm{h}, \mathrm{k}$, and $\mathrm{l}$ are either all odd or all even. Further the (111) plane is forbidden for BCC crystals, since $1+1+1=3$, which is odd. We can conclude that this must be an FCC structure!
Finally, we need to figure out which FCC element it is. We can combine some old principles to figure this out. We know the Bragg condition is
$\lambda=2 d_{h k l} \sin \theta$
where $\theta$ corresponds to one of the peaks above, which signify constructive interference. Recall also that the interplanar spacing is given by
$d_{h k l}=\dfrac{a}{\sqrt{h^2+k^2+l^2}}$
By plugging in one of the entries from the table above, we can figure out what the lattice constant is. Let's use $\theta=25.95$ degrees and the (200) plane:
\begin{gathered}
a=d_{h k l} \sqrt{h^2+k^2+l^2} \
a=\dfrac{\lambda}{2 \sin \theta} \sqrt{h^2+k^2+l^2} \
a=\dfrac{1.54 \AA}{2 * \sin \left(25.95^{\circ}\right)} \sqrt{2^2+0^2+0^2}=3.52 \AA
\end{gathered}
Looking at the periodic table, an FCC element with lattice parameter around $3.52 \AA$ is nickel! | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/06%3A_Recitations/6.16%3A_X-ray_Diffraction_and_Selection_Rules.txt |
Point defects
Point defects are local, so-called zero-dimensional defects in a lattice. They consist of an atom and its immediate surroundings, no more than a few atomic layers away. In 3.091, we’ll talk about three types of defects that occur in covalent solids, and two that occur in ionic solids. As a reminder, the individual atoms in a covalent solid have no charge relative to each other. The three defects are pictured below:
A self-interstitial occurs in crystals with atoms that are reasonably small compared to their spacing: it is sometimes possible for an atom to squeeze in between the atoms that are in their correct lattice position. A substitutional defect occurs when an atom in the homogeneous crystal is replaced by something else- think Ge in Si, for example. Finally, a vacancy occurs when one of the atoms in the crystal is missing from its standard lattice site. Remember also - though these are shown in 2D here, the real crystals they occur in are 3D Bravais lattices!
In an ionic crystal, all of the atoms have a relative charge: usually they are arranged to alternate, so the material retains both local and global charge neutrality. Therefore, it is important to account for charge neutrality when thinking about defects. Two ionic defects are shown below.
A Schottky defect occurs when there are two simultaneous vacancies – one anion and one cation – in close proximity. A Frenkel defect occurs when a cation vacancy neighbors an interstitial cation. Realistically, this occurs when a cation leaves its spot in the crystal lattice and becomes an interstitial. Recall that cations have a much smaller atomic radius than anions: it is much more practical for this sort of atomic migration to occur with a small cation than a huge anion!
Many, many defects occur in even the purest of crystals. In fact, they are entropically favorable! They can interact with each other, and even move around through the lattice in a thermally-activated process. In 3.091, it is sufficient to be able to identify them.
Example: How many oxygen vacancies are generated if you dope $\mathrm{ZrO}_2$ with $0.5 \mathrm{~g}$ of $\mathrm{Sc}_2 \mathrm{O}_3$?
Answer
In zirconia, each zinc is stoichiometrically matched with two oxygen atoms. Therefore, the charge on the zirconium atom must be $\mathrm{Zr}^{4+}$. In scandium oxide, there are two scandium ions per three oxygen ions, so it must be coordinated as $\mathrm{Sc}^{3+}$. If we dope zirconia with scandium oxide, the cations will substitute for the cations, and the oxygen will just go on oxygen sites. But each $\mathrm{Zr}^{4+}$ that is replaced by $\mathrm{Sc}^{3+}$ leaves a local charge imbalance in the lattice of one missing positive charge. This can be counteracted with oxygen vacancies in the lattice, effectively creating missing negative charge to balance things out. For every $2 \mathrm{Sc}$ $^{3+}$ atoms added, one $\mathrm{V}_O^{\prime \prime}$ must be generated in order to maintain charge neutrality.
\begin{gathered}
0.5 \mathrm{~g} \mathrm{Sc}_2\mathrm{O}_3 \left(\dfrac{1 \mathrm{~mol}\mathrm{~Sc}_2\mathrm{O}_3}{137.9\mathrm{~g}}\right) = 0.0036 \mathrm{~mol}\mathrm{Sc}_2\mathrm{O}_3 \
0.0072 \mathrm{~mol} \mathrm{~Sc}^{3+} \operatorname{added} \times \frac{1 \mathrm{~mol} \mathrm{~V}_O^{\prime \prime}}{2 \mathrm{~mol} \mathrm{~Sc}^{3+}}=0.0036 \mathrm{~mol} \mathrm{~V}_O^{\prime \prime} \
\# \text{ oxygen vacancies }=\left(0.0036 \mathrm{~mol} \mathrm{~V}_O^{\prime \prime}\right)\left(6.022 \times 10^{23} \dfrac{\text { vacancies }}{\text { mol vacancies }}\right)=2.18 \times 10^{21}
\end{gathered}
Arrhenius-like vacancy activation
Arrhenius provided a mathematical description for thermally activated processes which relates the rate at which a process occurs to temperature. The dependence is exponential and parameterized by energy: you can think of it as an "activation energy" to achieve a state.
$x \propto e^{-\frac{E_a}{k_B T}}$
Here, $\mathrm{E}$ is the relevant activation energy in $[\mathrm{J}]$ or $[\mathrm{eV}], k_B$ is the Boltzmann constant (with units $[\mathrm{J} / \mathrm{k}]$ or $[\mathrm{eV} / \mathrm{k}]$ to match the energy), and $\mathrm{T}$ is temperature in Kelvin. If we call the constant of proportionality A, we can write this equation in a convenient linear form by taking the log of both sides:
\begin{gathered}
\ln (x)=\ln \left(A e^{-\frac{E}{k_B T}}\right)=\ln A-\dfrac{E}{k_B T} \
\ln x=-\dfrac{E}{k_B} \dfrac{1}{T}+\ln A
\end{gathered}
Here we have a convenient slope-intercept form that allows us to read off the exponential rate and constant from a simple plot.
We'll come back to Arrhenius later in the course, but for now we will apply this exponential relationship to calculating the thermally-activated process of vacancy creation. Let's call the rate of vacancy formation $r_{form}$ and the rate of vacancy annihilation $r_{destroy}$. We can describe these rates as Arrhenius processes:
$r_{\text {form }}=A_{\text {form }} e^{-\frac{E_{a, \text { form }}}{k_B T}} \quad \quad r_{\text {destroy }}=A_{\text {destroy }} e^{-\frac{E_{a, \text { destroy }}}{k_B T}}$
In equilibrium, the flux of vacancies that form must be equal to the flux of vacancies are destroyed. The flux of vacancies formed is simply the number of sites available to form a vacancy times the rate at which vacancies form, and the flux of vacancies destroyed is the number of sites already having a vacancy times the rate at which vacancies are destroyed. Let's say our crystal has $\mathrm{N}$ atoms and $N_V$ vacancies, then in equilibrium:
\begin{gathered}
N * r_{\text {form }}=N_V * r_{\text {destroy }} \
\dfrac{N_V}{N}=\dfrac{r_{\text {form }}}{r_{\text {destroy }}}=\dfrac{A_{\text {form }}}{A_{\text {destroy }}} e^{-\frac{E_{\text {form }}-E_{\text {destroy }}}{k_B T^T}}
\end{gathered}
We can wrap the difference in formation and annihilation activation energies into a $\Delta E$ that we'll call $E_V$. Then finally, the fraction of vacancies as a function of temperature is
$\dfrac{N_V}{N}=\dfrac{A_{\text {form }}}{A_{\text {destroy }}} e^{-\frac{E_V}{k_B T}}$
Example: At room temperature $(295 \mathrm{k})$, a piece of Si has 3 vacancies per 10 million atoms. When it is heated to $400 \mathrm{k}$, there are 15 vacancies per 10 million atoms. What is $E_V$?
Answer
Let’s start up by setting up the equations for the fraction of vacancies at each temperature:
$\dfrac{N_{V, 295}}{N}=\dfrac{A_{\text {form }}}{A_{\text {destroy }}} e^{-\frac{E_V}{295 * k_B}} \quad \dfrac{N_{V, 400}}{N}=\dfrac{A_{\text {form }}}{A_{\text {destroy }}} e^{-\frac{E_V}{295 * k_B}}$
Here we essentially have two equations with two unknowns: one is $E_V$, the quantity we are looking for, and the other is $N \frac{A_f \text { orm }}{A_d \text { estroy }}$, which we can treat as a constant. Here we are making an assumption that $\frac{A_f \text { orm }}{A_d \text { estroy }}$ is temperature independent. We can divide the two equations to get rid of the constant we don't care about, and then solve for $E_V$:
\begin{gathered}
\frac{\dfrac{N_{V, 295}}{N}}{\frac{N_{V, 400}}{N}}=\dfrac{\frac{A_{\text {form }}}{A_{\text {destroy }}}}{\frac{A_{\text {form }}}{A_{\text {destroy }}}} \dfrac{e^{-\frac{E_V}{295 k_B}}}{e^{-\frac{E_V}{40 T}}} \
\dfrac{N_{V, 295}}{N_{V, 400}}=e^{\frac{E_V}{k_B}\left(\frac{1}{400}-\frac{1}{295}\right)}=\dfrac{3}{15}
\end{gathered}
Taking the log of both sides
\begin{gathered}
\ln \left(\dfrac{3}{15}\right)=\dfrac{E_V}{8.617 \times 10^{-5} \mathrm{eV} / \mathrm{k}}\left(\dfrac{1}{400}-\dfrac{1}{295}\right) \
E_V=0.16 \mathrm{eV}
\end{gathered} | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/06%3A_Recitations/6.17%3A_Point_Defects_and_Arrhenius-like_Vacancy_Activation.txt |
Line defects
In addition to point defects, crystals can have defects that cover larger areas and involve the coordination of more atoms than just one. Point defects are called 0D, or zero dimensional; in keeping with this nomenclature, line defects can be thought of as 1D defects (involving a line of atoms), and planar defects can be thought of as 2D defects (involving a plane). One important line defect is called a dislocation. A dislocation amounts to an extra line of atoms which persists part of the way through the crystal: the termination within the crystal is noted with a $\mathrm{T}$ shape as below:
The bulk of the crystal maintains its equilibrium structure, well-aligned with the lattice. However, you can see that the region around the origin of the dislocation has been forced out of its equilibrium position in the lattice. The spacing between atoms is different around the end of the dislocation; the bonds are disrupted. It takes energy to generate a dislocation, but it also takes energy for it to move in the crystal. Therefore, even though they are not the most energetically favorable state, they are meta-stable in equilibrium once generated.
Mechanics and stress–strain curves
One of the most informative experiments to explore the mechanical properties of a material is generating a stress–strain curve. As we discussed in lecture, this is usually done with an Instron; the basic principle is to pull (or push) on a material in a carefully controlled way, and to measure its response.
Example: Label the plot with the following features of stress-strain curves: elastic deformation, plastic deformation, yield stress, elastic modulus, yield point, fracture
Answer
Elastic deformation refers to the linear response of a material to applied strain. This typically happens for low amounts of applied stress in crystalline materials. The slope of the stress-strain curve in the elastic region is called the elastic modulus. Plastic deformation refers to nonlinear strain response that is irreversible. The transition between elastic and plastic deformation is the yield point, and the stress at which the material yields is the yield stress. When the material is not able to stretch any more, it breaks: this is fracture. The relevant regimes are labeled below:
Slip
Even failure in crystalline materials is systematic: when significant force is applied to a material, enough to break bonds, the material can yield. Except in cases of high impact, this generally happens along slip planes. Slip planes are the most densely-packed planes in a lattice. Since the bond density within a slip plane is as large as it gets in the crystal, the out-of-plane density is lower: this means that fewer bonds must be broken for rows of atoms to move past each other as the material yields. As it yields, you can think of a planar face dividing sections of material forced in different directions. The directions that slip happens along are called slip directions, and with similar rationale, they are the most closely-packed directions in the crystal.
6.19: Glasses and Cooling Curves
Glasses
Sometimes, instead of possessing the carefully arranged crystalline bonds we’ve been talking about, materials end up partially or completely jumbled. In particular, elements that have flexibility in the rotational orientation of bonds can form these disordered structures, like long, interlinked chains frozen in a random orientation. We refer to these amorphous structures as glasses. In $3.091$ we’ll focus primarily on silica: a glass based on silicon dioxide.
The scale from order to disorder is a spectrum. Single crystals have near-perfect order: everything is oriented the same way, and there is long-range order in the crystalline lattice. Then there are polycrystalline materials, which possess multiple crystalline regions arranged in grains. Near the grain boundaries, there are defects; the sudden change in crystal orientation can add local strain. Polycrystalline materials have short-range order, on the scale of the grains, but they lack the homogeniety of single crystals. At the other end of the spectrum are amorphous materials. These glassy materials lack any semblance of order: the complete rotational freedom of the silicon bond provides little constraint on the arrangement of bonds.
As it turns out, it is possible to control whether a melt forms a crystal or a glass by carefully engineering the cooling process. Although silica glasses are able to form, the regular structure of the crystalline lattice is still the most energetically favorable structure. In a crystal, the atoms are optimally arranged to minimize energy by balancing the Coulombic attraction between protons and electrons with the repulsion of like charges.
Cooling curves
In summary, the primary knob we can turn to determine whether a melt ends up as a crystal or a glass is processing. One metric that allows us to keep track of the affect various processing steps have affected the melt is the molar volume. You can think of molar volume as a proxy for disorder. For the optimally-arranged crystal, the atoms are minimally spaced, and the molar volume is minimized. The more disordered the material becomes, the more space each atom takes up, and the larger the molar volume is! By measuring the molar volume as a function of temperature, we can compare how ordered or disordered different processing methods are.
Example: Label the plot with the following features of cooling curves: liquid, supercooled liquid, $\mathrm{T}_m, \mathrm{T}_g$, the crystalline and glassy regimes, and the curves with the fastest and slowest cooling rates
Answer
Liquid is the state of matter at high temperatures, when the material forms a melt. It has a higher slope than the solid regions because the coefficient of thermal expansion $\left(\frac{\partial \bar{V}}{\partial T}\right)$ is generally higher for liquids than solids. The melting point, $\mathrm{T}_m$, can be determined by looking for a discontinuity in the plot: during the phase transition from liquid to solid involves a big change in molar volume at constant temperature. Correspondingly, the crystalline regime is the low temperature solid that follows the phase transition. Supercooled liquid is any liquid that remains at temperatures below the melting point- it has the same slope as the liquid above the melting point. The points in the supercooled liquid regime where the slope changes are the glass transition temperatures, $\mathrm{T}_g$. Unlike the crystalline phase transition, the glass transition is continuous (but not smooth): you can think of the glass transition as just freezing the viscous liquid melt in place, as is. Remember that depending on the specifics of the processing, glasses with different $\mathrm{T}_g$ and different molar volumes can form. Below $\mathrm{T}_g$, the glass becomes a solid: this is the glassy regime. Finally, it takes a long time atoms to rearrange into ordered structures, so crystals are processed with the slowest cooling rates. The glasses with the highest molar volumes are the most disordered: these melts have to freeze rapidly, and they have the fastest cooling rates.
Example: Which of the following samples will yield a glass with higher molar volume?
$\begin{array}{lll}\text { melt is very viscous } & \text { OR } & \text { melt is very fluid } \ \text { melt is cooled rapidly } & \text { OR } & \text { melt is cooled slowly } \ \text{equilibrium crystal structure is complex} & \text{ OR } & \text{equilibrium crystal structure is simple} \end{array}$
Answer
A viscous melt does not flow easily: it is more difficult for the atoms to move around than in a very fluid melt. Therefore, with all other factors held constant, the viscous melt should produce a glass with higher molar volume. Following the rationale from the previous section, the melt that is cooled rapidly will yield a glass with higher molar volume compared to a melt that is cooled slowly. Finally, if the equilibrium crystal structure formed by a melt is very complicated, it takes time for all of the atoms to make it to their optimal lattice sites. With all other factors held constant, the melt with a more complex equilibrium crystal structure will yield a higher molar volume glass than the melt with a simple equilibrium crystal structure. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/06%3A_Recitations/6.18%3A_Line_Defects_Mechanics_and_Stress-strain_Curves_and_Slip.txt |
Glass formers
In $3.091$, we focus on silica glass. The chemical composition of silica is $\mathrm{SiO}_2$: you are now well equipped to check this compound is charge-neutral. Although the charges are balanced, if you were to draw a Lewis dot diagram for $\mathrm{SiO}_2$, you might notice that there are lots of lone pairs of electrons that aren't involved in bonding. You may also recall that silicon, with its four valence electrons, typically forms tetrahedrally coordinated (or $\mathrm{sp}^3$ ) bonds. What ends up happening when liquid silica solidifies is that oxygen atoms are shared between silicon atoms, acting like bridges: these are referred to as bridging oxygens. If we were to flatten crystalline silica to represent it in 2D, it might look something like this (not a true projection):
Here, the dotted lines represent half-bonds that would continue on throughout $3 \mathrm{D}$ space. Each oxygen atom is a bridging oxygen, because it is shared between adjacent silicon atoms. Though the stoichiometry is $\mathrm{SiO}_2$, you can think of the tetrahedral structure as silicate groups $\left(\left[\mathrm{SiO}_4\right]^{4-}\right)$ where the oxygen ions are then shared with neighboring $\mathrm{Si}$ ions, yielding a charge-neutral compound.
$\mathrm{Si}+4\left(\dfrac{1}{2} \mathrm{O}\right)=\mathrm{Si}+2 \mathrm{O}=\mathrm{SiO}_2$
Although I've shown a crystalline representation above, the same concept applies for glasses: the difference is that in a glass, the arrangement of the bonds is not perfectly tetrahedral. In lecture and in Recitation 19 , we we talked about various processing methods that yield more disordered (glassy) or more ordered (crystalline) materials. Generally, when a glass forms instead of a crystal, the bonds do not have enough time (or energy) to become arranged in neat tetrahedrons: instead, they are rotated, stretched, and the long range structure is generally amorphous. The chemistry, however, remains the same: for silica glass, there are bridging oxygen atoms that connect neighboring silicon atoms, and a 2:1 ratio of oxygen atoms to silicon atoms. The resulting glass is like a large network of haphazardly-arranged $\mathrm{Si}-\mathrm{O}$ bonds: for this reason, silica is called a network former.
Network modifiers
In contrast to a network former, there can also be network modifiers: elemental defects introduced into the melt that disrupt the highly connected silica network. Some examples include $\mathrm{Na}_2 \mathrm{O}, \mathrm{CaO}_2$ and $\mathrm{Al}_2 \mathrm{O}_3$. These ionic oxides dissociate inside the glass, and the strongly electropositive cations have an affinity for the electronegative oxygen atoms. Here's an example using $\mathrm{Na}_2 \mathrm{O}$:
Let's focus in on one bridging oxygen. If we think of it like a chemical reaction, we can add a number of units of $\mathrm{Na}_2 \mathrm{O}$.
Each $\mathrm{Na}_2 \mathrm{O}$ in the melt dissociates into two sodium atoms and an oxygen atom. One of the covalent bonds connecting the bridging oxygen atom is broken by the sodium atoms. Of course, mass is conserved, and the oxygen from $\mathrm{Na}_2 \mathrm{O}$ can form a new covalent bond with the $\mathrm{Si}$ whose bond was broken. In the final state, the $\mathrm{Na}^{+}$ atoms form ionic bonds with the newly nonbridging oxygen atoms. The network of silica has been split.
For other ionic salts, you can think of each $\mathrm{O}_2^{2-}$ that is added as breaking one bond, but creating two nonbridging oxygen atoms. The more network modifier that is added, the more this process happens, and the shorter the continuous chains of silica get! As the chains get shorter and shorter, they can slide past each other more easily during cooling. It's harder for them to get tangled. Therefore, the more network modifier is added, the lower the free volume of the glass that results (holding all other factors constant), because the shorter chains can pack more efficiently than the jumbled mess of pure silica glass.
Example: Sketch a glass cooling curve comparing a glass with a lot of network modifier added to a glass with little network modifier added, assuming the cooling rate is the same for both.
Answer
Note that the glass with more network modifier has a lower $\mathrm{T}_g$! | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/06%3A_Recitations/6.20%3A_Glass_Formers_and_Network_Modifiers.txt |
Reaction kinetics
Not every reaction is the the same: early on, we talked about stoichiometry and how chemical equations give information about what is reacting and in what ratios. We learned how to determine which reactants would be limiting, and which would be in excess. The rate of a reaction is the change in concentration as a function of change in time. Reactants are consumed, so their rate is negative, and products are formed, so their rate is positive.
To determine the rate of a reaction experimentally, we simply measure and plot the concentration as a function of time. The order of a reaction refers to the relationship between concentration and rate. If the reaction rate is independent of the concentration of a reagent, the reaction is $0^{t h}$ order in that reagent. If the reaction rate doubles when the concentration of a reagent is doubled, the reaction is $1^{s t}$ order in that reagent. Finally, if the reaction rate quadruples when the concentration of a reagent is doubled, the reaction is $2^{\text {nd }}$ order in that reagent. Note that order is defined with respect to a specific reagent (usually for each reactant), not for a reaction as a whole!
When there are multiple reactants, it can be difficult to determine which reactant is causing the rate to change in what way. The best way to isolate the effect of changing concentration is to modulate the concentration of each reactant in turn, while holding the others constant.
Say we have a reaction of the following form:
$a \mathrm{~A}+\mathrm{bB} \rightarrow \mathrm{cC}+\mathrm{dD} \nonumber$
Conservation of mass allows us to write a generalized rate law, since we know that the rate at which the reactants disappear must equal the rate at which products appear:
$\text { rate }=-\dfrac{1}{a} \dfrac{d[A]}{d t}=-\dfrac{1}{b} \dfrac{d[B]}{d t}=\dfrac{1}{c} \dfrac{d[C]}{d t}=\dfrac{1}{d} \dfrac{d[D]}{d t} \nonumber$
Say the dependence of rate on concentration is as follows:
$\begin{array}{l|c|c|c} & {[\mathrm{A}]} & {[\mathrm{B}]} & \text { rate } \ \hline \hline \text { Experiment 1 } & 2 \mathrm{M} & 1 \mathrm{M} & 140 \mathrm{mM} / \mathrm{s} \ \text { Experiment 2 } & 2 \mathrm{M} & 4 \mathrm{M} & 560 \mathrm{mM} / \mathrm{s} \ \text { Experiment 3 } & 1 \mathrm{M} & 1 \mathrm{M} & 35 \mathrm{mM} / \mathrm{s} \end{array}$
Notice that if we compare Experiments 1 and 2, the concentration of $\mathrm{B}$ quadruples, while the concentration if $\mathrm{A}$ is kept constant, and the rate also quadruples. This reaction must be first order in $\mathrm{B}$! Comparing Experiments 1 and 3, the concentration of $\mathrm{A}$ halves, the concentration of $\mathrm{B}$ is constant, and the rate decreases by a factor of 4 . Since the reaction increases with the square of the concentration of $\mathrm{A}$, this reaction must be second order in $\mathrm{A}$. The general rate law can be expressed as
$r=k[A]^2[B]$
Here, the constant k is called the rate constant. The overall order of the reaction is given by the sum of the rates of the individual components. In our example above, the overall reaction rate is 3.
Rate laws and rate constants
The rate law for a $0^{t h}$ order reaction is
$r=k[A]^0=k \nonumber$
The units of the rate constant must therefore be units of rate, or $\mathrm{M} / \mathrm{s}$. The integrated rate law can be found by integrating the rate law with respect to time, recalling that rate is change in concentration per unit time. For $0^{\text {th }}$ order, the integrated rate law is
$[A]=[A]_0-k t \nonumber$
A similar analysis can be performed for first and second order reactions; the characteristics are summarized below:
$\begin{array}{c|c|c|c} & 0^{t h} \text { order } & 1^{s t} \text { order } & 2^{\text {nd }} \text { order } \ \hline \text { rate law } & \mathrm{r}=\mathrm{k} & \mathrm{r}=\mathrm{k}[\mathrm{A}] & \mathrm{r}=\mathrm{k}[\mathrm{A}]^2 \ \text { units of } \mathrm{k} & M s^{-1} & s^{-1} & M^{-1} s^{-1} \ \text { integrated rate law } & {[A]=[A]_0-k t} & \ln [A]=\ln [A]_0-k t & \frac{1}{[A]}=\frac{1}{[A]_0}+k t \end{array}$
If you’ve had some diff eq, try to produce the integrated rate laws for each order for the rate laws given! You just need to separate variables and choose appropriate integration limits.
Example: You measure the rate at which reactants are consumed for the following reaction. Determine the overall order of the reaction and write the general rate law. What are the units of the rate constant $k$?
$2 \mathrm{NO}(g)+2 \mathrm{H}_2(g) \rightarrow 2 \mathrm{~N}_2(g)+2 \mathrm{H}_2 \mathrm{O}(g)$
$\begin{array}{c|c|c|c} \text { Trial } & {[\mathrm{NO}]} & {\left[\mathrm{H}_2\right]} & \text { rate } \ \hline \hline 1 & 0.1 \mathrm{M} & 0.1 \mathrm{M} & 1.2 \mathrm{mM} / \mathrm{s} \ 2 & 0.1 \mathrm{M} & 0.2 \mathrm{M} & 2.4 \mathrm{mM} / \mathrm{s} \ 3 & 0.3 \mathrm{M} & 0.1 \mathrm{M} & 10.8 \mathrm{mM} / \mathrm{s} \end{array}$
Answer
Comparing trial 2 to trial 1, the rate doubles while the concentration of $\mathrm{H}_2$ doubles, so the reaction must be first order in $\left[\mathrm{H}_2\right]$. Comparing trial 3 to trial 1 , the concentration of $\mathrm{NO}$ triples and the rate increases by a factor of $9\left(=3^2\right)$. The reaction must therefore be second order in $[\mathrm{NO}]$.
The overall rate law for this reaction is
$\text { rate }=-\dfrac{d[N O]}{d t}=-\dfrac{d\left[H_2\right]}{d t}=k[N O]^2\left[H_2\right] \nonumber$
The overall order of the reaction is the sum of the orders of the constituent reactants: in this case, $2+1=3$, so the overall reaction order is 3 . Finally, both sides of the rate equation must have the same units: the rate has units of $\mathrm{M} / \mathrm{s}$. We must choose the units of $\mathrm{k}$ such that the right hand side matches. The concentrations contribute $\mathrm{M}^3$; therefore, the units of $\mathrm{k}$ must be $\mathrm{M}^{-2} \mathrm{~s}^{-1}$ to match. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/06%3A_Recitations/6.21%3A_Reaction_Kinetics_Rate_Laws_and_Rate_Constants.txt |
Equilibrium
Last time when we discussed reaction rates, we focused on the rate at which the forward reaction progresses– the rate at which reactants become products. Of course, reactions can progress in both the forward and reverse direction, and it is frequently thermodynamically favorable for this to happen. A reaction reaches a dynamic equilibrium when both the forward and reverse reaction are proceeding at the same rate.
The equillibrium constant, $\mathrm{K}_{e q}$, is a ratio between the reverse reaction and the forward reaction in equilibrium. For a reaction
$a[A]+b[B] \rightarrow c[C]+d[D] \nonumber$
the equilibrium constant is
$K_{e q}=\dfrac{[C]^c[D]^d}{[A]^a[B]^b} \quad \quad \text { (use when system is in equilibrium) } \nonumber$
Each reaction has an equilibrium constant, and here, the superscripts are the stoichiometric coefficients (unlike the rate constant!). When the system is not in equilibrium, such as when the system is perturbed or the reactants are first mixed, the ratio of the rate constants is usually represented by a $\mathrm{Q}$.
$Q=\dfrac{[C]^c[D]^d}{[A]^a[B]^b} \quad \text { (use when system is not in equilibrium) } \nonumber$
By comparing $\mathrm{Q}$ and $\mathrm{K}_{e q}$, the progress of the reaction can be determined. The reaction has reached equilibrium when $\mathrm{Q}=\mathrm{K}_{e q}$. If $\mathrm{Q}<\mathrm{K}_{e q}$, the reaction proceeds to produce more products. If $\mathrm{Q}>\mathrm{K}_{e q}$, the reaction proceeds to produce more reactants.
Example: Write an expression for the equilibrium constant for the following reaction:
$2 \mathrm{NO}+2 \mathrm{H}_2 \rightarrow \mathrm{N}_2+2 \mathrm{H}_2 \mathrm{O}$
Answer
We can read off the expression for the equilibrium constant from the balanced equation.
$K_{e q}=\dfrac{\left[N_2\right]\left[H_2 O\right]^2}{[N O]^2\left[H_2\right]^2} \nonumber$
Solubility product
The solubility product is the equilibrium constant for the dissolution of a solid in a solution. The solid that is dissolving is called the solute, and the liquid that is doing the dissolving is called the solvent. For a solubility product, the solid isn’t included: you can think of the concentration of the solid as being constant throughout the reaction.
Example: Find the solubility product of calcium chloride dissolving in water.
Answer
The equation governing the dissolution of calcium chloride is just the splitting of the compound into constituent ions:
$\mathrm{CaCl}_2(s) \rightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) \nonumber$
To write the solubility product, we will ignore the solid part and focus only on the ions in solution, noted (aq) for aqueous. Therefore, the solubility product can be expressed as
$K_{s p}=\left[C a^{2+}\right]\left[C l^{-}\right]^2 \nonumber$
Common ion effect
The presence of additional ions in solution can have an impact on the solubility of a solid. In particular, if two solids that share an ion are both dissolved in solution, the equilibrium solubility shifts. The solubility of a salt is usually lower in the presence of another soluble salt which shares a common ion.
Example: What happens if we add silver chloride to an $0.2 \mathrm{M}$ solution of calcium chloride? The solubility product if silver chloride is $1.77 \times 10^{-10}$.
Answer
First, we need to write a balanced equation for the dissolution of $\mathrm{AgCl}$:
$A g C l(s) \rightarrow A g^{+}(a q)+C l^{-}(a q) \nonumber$
Correspondingly, the expression for the solubility product is
$K_{s p}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right] \nonumber$
We want to solve for the concentration of $\mathrm{AgCl}$ in solution which already contains calcium ions: one handy tool for this process is an ICE table, which stands for initial, change, equilibrium. When setting up an ICE table, we consider the species that is actively dissolving, not the species that already exists in solution.
The initial concentration of ions in solution is determined by the calcium chloride solution: we know it's an $0.2 \mathrm{M}$ solution. Let's call the amount of $\mathrm{AgCl}$ that dissolves $x$. From our balanced equation above, we know that there are the same number of moles of $\mathrm{Ag}$ and $\mathrm{Cl}$ ions in solution. We can set up the ICE table for $\mathrm{AgCl}$ as follows:
$\begin{array}{c|c|c|c} & {[\mathrm{AgCl}]_s} & {\left[\mathrm{Ag}^{+}\right](\mathrm{aq})} & {\left[\mathrm{Cl}^{-}\right](\mathrm{aq})} \ \hline \hline \text { Initial } & \text { solid } & 0 \mathrm{M} & 0.2 \mathrm{M} \ \text { Change } & - & +x \mathrm{M} & +x \mathrm{M} \ \text { Final } & - & x \mathrm{M} & 0.2+x \mathrm{M} \end{array}$
At every point as the reaction is proceeding, the equation governing the solubility product must be satisfied. Therefore we can solve for the final concentration of $\mathrm{AgCl}$ in solution as follows:
\begin{gathered}
K_{s p}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]=x(0.2+x)=1.77 \times 10^{-10} \
x^2+0.2 x-1.77 \times 10^{-10}=0
\end{gathered}
Here, we can either directly solve the quadratic equation and keep only the physical solutions (positive concentration), or we can linearize and solve (since $\mathrm{x}$ is small). Either way, we come up with
$x=8.85 \times 10^{-10} M \nonumber$
Therefore, the concentration of of $\mathrm{AgCl}$ that will dissolve in $0.2 \mathrm{M} \mathrm{CaCl}_2$ is $8.85 \times 10^{-10} \mathrm{M}$. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/06%3A_Recitations/6.22%3A_Equilibrium_Solubility_Product_and_Common_Ion_Effect.txt |
Acids and bases
In lecture, we talked about Brønsted-Lowry acids and bases as well as Lewis acids and bases. The BrønstedLowry definition of acids and bases is more narrow: a Brønsted-Lowry acid is a proton donor, while a Brønsted-Lowry base is a proton acceptor: to be clear here, the proton is a hydrogen ion, $\mathrm{H}^{+}$. More generally, a Lewis acid is an electron acceptor, while a Lewis base is an electron donor. Water is amphoteric: it can be either an acid or a base. Water serves as the solvent for many acid-base reactions. Consider adding a generic acid to water:
$H A(a q)+H_2 O(l) \rightarrow H_3 O^{+}(a q)+A^{-}(a q) \nonumber$
Similarly, consider adding a generic base to water:
$B^{-}(a q)+H_2 O(l) \rightarrow H B(a q)+O^{-}(a q) \nonumber$
These serve as prototypical acid-base reactions: we can identify conjugate acid-base pairs to relate species that lost/gained a proton (or electron) on either side of the reaction.
Example: Identify the conjugate acid-base pairs in the following reaction:
$\mathrm{CH}_3\mathrm{CO}_2\mathrm{H} (aq) + \mathrm{NH}_3 (aq) \rightarrow \mathrm{CH}_3\mathrm{CO}_2^{-} + \mathrm{NH}_4^{+}$
Answer
Dissociation
In the same way we wrote a general equilibrium constant, we can write an acid/base specific equivalent. The equilibrium constant for our acid dissociation reaction above would be
$K_{e q}=\dfrac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{A}^{-}\right]}{\left[\mathrm{H}_2 \mathrm{O}\right][\mathrm{HA}]}$
However, since the water is acting as a solvent and is present in excess, it has a constant concentration: we therefore define the acid constant as
$K_a=K_{e q}\left[H_2 O\right]=\dfrac{\left[H_3 O^{+}\right]\left[A^{-}\right]}{[H A]}$
Similarly, the base dissocation constant is
$K_b=\dfrac{[H B]\left[O H^{-}\right]}{[B-]}$
The stronger the acid or base, the more readily it dissolves in solution, and the greater the magnitude of $K_a$ or $K_b$. There are only a few strong acids- acids that fully dissociate into their constituent ions. These include $\mathrm{HCl}, \mathrm{HBr}, \mathrm{HI}, \mathrm{HNO}_3, \mathrm{HClO}_3, \mathrm{HCLO}_4$, and $\mathrm{H}_2 \mathrm{SO}_4$. The strong bases are mostly alkali salts, including $\mathrm{LiOH}, \mathrm{NaOH}, \mathrm{KOH}, \mathrm{RbOH}, \mathrm{CsOH}, \mathrm{Ca}(\mathrm{OH})_2, \mathrm{Sr}(\mathrm{OH})_2, \mathrm{Ba}(\mathrm{OH})_2$. Of course, the strength of an acid or base is related to its bond energy: polar acids with big differences in electronegativity have a low bond energy rapidly dissociate when placed in water, for example.
Example: Write an expression for the dissociation and for the acid or base dissociation constants for these acids $\left(\mathrm{HF}, \mathrm{CH}_3 \mathrm{CO}_2 \mathrm{H}\right)$ and bases $\left(\mathrm{NH}_3, \mathrm{NaOH}\right)$.
Answer
Hydrofluoric acid, $\mathrm{HF}$, is a strong acid. In water, the proton dissociates:
$H F(a q)+H_2 O(l) \rightarrow H_3 O^{+}(a q)+F^{-}(a q) \nonumber$
Its acid dissociation constant is
$K_a=\dfrac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{F}^{-}\right]}{[H F]} \nonumber$
We can take a similar approach for acetic acid, a weak acid.
$\mathrm{CH}_3 \mathrm{CO}_2 \mathrm{H}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{CH}_3 \mathrm{CO}_2^{-}(a q) \nonumber$
The acid dissociation constant is
$K_a=\dfrac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{CH}_3 \mathrm{CO}_2^{-}\right]}{\left[\mathrm{CH}_3 \mathrm{CO}_2 \mathrm{H}\right]} \nonumber$
For ammonia, the base gains a proton when placed in water:
$\mathrm{NH}_3(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{NH}_4^{+}(a q)+\mathrm{OH}^{-}(a q) \nonumber$
The base dissociation constant for ammonia is
$K_b=\dfrac{\left[\mathrm{NH}_4^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_3\right]} \nonumber$
Finally, sodium hydroxide fully dissociates:
$\mathrm{NaOH}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{Na}^{+}(a q)+\mathrm{OH}^{-}(a q) \nonumber$
The base dissociation constant is
$K_b=\dfrac{[N a+]\left[\mathrm{OH}^{-}\right]}{[N a O H]} \nonumber$
Example: Which of these acids is the strongest? sulfurous acid $\left(\mathrm{H}_2 \mathrm{SO}_3 ; \mathrm{K}_a=1.54 \times 10^{-2}\right)$, phosphoric $\operatorname{acid}\left(\mathrm{H}_2 \mathrm{PO}_4^{-} ; \mathrm{K}_a=6.23 \times 10^{-8}\right)$, citric acid $\left(\mathrm{H}_3 \mathrm{C}_6 \mathrm{H}_5 \mathrm{O}_7 ; \mathrm{K}_a=8.4 \times 10^{-4}\right)$
Answer
Stronger acids have larger acid dissociation constants, so sulfurous acid is stronger than citric acid and phosphoric acid. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/06%3A_Recitations/6.23%3A_Acids_Bases_and_Dissociation.txt |
pH and pOH
The concentration of ions in solution can be quite high- and can differ from solution to solution by many orders of magnitude. For this reason, it is useful to refer to a logarithmic scale. We will use an operator $\mathrm{p}$ defined to be the negative $\log$ (base 10) of the value it precedes: for example, $\mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right]$.
The $\mathrm{pH}$ of a solution is a measure of how acidic it is: a $\mathrm{pH}$ of 7 indicates the solution is neutral, while lower values are acidic and higher values are basic. The compliment to $\mathrm{pH}$ is $\mathrm{pOH}$, a measure of the concentration of hydroxide ions: $\mathrm{pOH}=-\log _{10}\left[\mathrm{OH}^{-}\right]$. A high $\mathrm{pOH}$ corresponds to an acidic solution, while a low $\mathrm{pOH}$ indicates a basic solution. $\mathrm{pH}$ and $\mathrm{pOH}$ are related as
$p H+p O H=14$
Example: The $\mathrm{K}_a$ for nitrous acid, $\mathrm{HNO}_2$, is $4.0 \times 10^{-4}$. Determine the $\mathrm{pH}$ and $\mathrm{pOH}$ of $0.1 \mathrm{M}$ nitrous acid.
Answer
First, let's write an expression for the dissociation of nitrous acid:
$\mathrm{HNO}_2(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{NO}_2^{-}(a q)+\mathrm{H}_3 \mathrm{O}^{+}(a q) \nonumber$
Then, we can write an expression for the acid dissociation constant:
$K_a=\dfrac{\left[\mathrm{NO}_2^{-}\right]\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}{\left[\mathrm{HNO}_2\right]} \nonumber$
To find the $\mathrm{pH}$, we need to determine the concentration of $\mathrm{H}^{+}$ ion (equivalently $\mathrm{H}_3 \mathrm{O}^{+}$). The initial concentration of nitrous acid is $0.1 \mathrm{M}$: we can set up an ICE table to solve for the final concentrations:
$\begin{array}{c|c|c|c} & {\left[\mathrm{HNO}_2\right]} & {\left[\mathrm{NO}_2^{-}\right]} & {\left[\mathrm{H}_3 \mathrm{O}^{+}\right]} \ \hline \hline \text { Initial } & 0.1 & 0 \mathrm{M} & 0 \mathrm{M} \ \text { Change } & -x \mathrm{M} & +x \mathrm{M} & +x \mathrm{M} \ \text { Final } & 0.1-x \mathrm{M} & x \mathrm{M} & x \mathrm{M} \end{array}$
Plugging in the final values to the expression for $\mathrm{K}_a$, we can solve for $x$ to find the concentration of hydrogen ions:
\begin{gathered}
K_a=\dfrac{\left[\mathrm{NO}_2^{-}\right]\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}{\left[\mathrm{HNO}_2\right]}=\frac{x^2}{0.1-x}=4.0 \times 10^{-4} \
x^2=4.0 \times 10^{-5}-x\left(4.0 \times 10^{-4}\right) \
x^2+x\left(4.0 \times 10^{-4}\right)-4.0 \times 10^{-5}=0 \
{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=x \approx 0.006}
\end{gathered}
Finally, we can solve for $\mathrm{pH}$ :
$p H=-\log \left[H^{+}\right]=-\log \left[H_3 O^{+}\right]=-\log (0.006)=2.21 \nonumber$
The value of $\mathrm{pOH}$ follows using the relation that $\mathrm{pH}+\mathrm{pOH}=14$ :
$p O H=14-p H=11.79 \nonumber$
Example: What is the $\mathrm{pH}$ of $0.04 \mathrm{M} \mathrm{HI}$, which is a strong acid?
Answer
The key phrase to solving this problem is 'strong acid.' When we have a strong acid, we don't need an acid dissociation constant because we know it fully dissociates! Here, the concentration of hydrogen ion simply equals the initial concentration of acid:
$p H=-\log \left[H^{+}\right]=-\log (0.04)=1.4 \nonumber$
$\mathbf{p K}_a$ and $\mathbf{p K}_b$
In the same way that it is convenient to express the concentration of hydronium or hydroxide ions as a logarithmic value, it can be useful to express the acid and base dissociation constants as $\mathrm{pK}_a$ and $\mathrm{pK}_b$, since they can again differ by many orders of magnitude between species. The definition of $\mathrm{p}$ as an operator remains the same, so $\mathrm{pK}_a=-\log _{10}\left(\mathrm{~K}_a\right)$ and $\mathrm{pK}_b=-\log _{10}\left(\mathrm{~K}_b\right)$.
Example: For an $0.2 \mathrm{M}$ solution of pyridine $\left(\mathrm{C}_5 \mathrm{H}_5 \mathrm{~N}\right)$, a base with a $\mathrm{pH}$ of $9.28$, determine the $\mathrm{pK}_b$.
Answer
First, let's write a dissociation equation for pyridine:
$C_5 H_5 N(a q)+H_2 O(l) \rightarrow C_5 H_5 N^{+}(a q)+O H^{-} \nonumber$
Then, we can convert the $\mathrm{pH}$ to a $\mathrm{pOH}$, since pyridine is a base:
$p O H=14-p H=14-9.28=4.72 \nonumber$
Next, we can convert the $\mathrm{pOH}$ into a concentration of hydroxide ions:
\begin{gathered}
p O H=-\log _{10}\left[O H^{-}\right] \
{\left[O H^{-}\right]=10^{-p O H}=10^{-4.72}=1.89 \times 10^{-5}}
\end{gathered}
Then, we can write an expression for the base dissociation constant:
$K_b=\dfrac{\left[C_5 H_5 \mathrm{NH}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[C_5 \mathrm{H}_5 \mathrm{~N}\right]} \nonumber$
And write our ICE table: Plugging these final values into the expression from above for $\mathrm{K}_b$:
$\begin{array}{c|c|c|c} & {\left[\mathrm{C}_5 \mathrm{H}_5 \mathrm{~N}\right]} & {\left[\mathrm{C}_5 \mathrm{H}_5 \mathrm{NH}^{+}\right]} & {\left[\mathrm{OH}^{-}\right]} \ \hline \hline \text { Initial } & 0.2 & 0 \mathrm{M} & 0 \mathrm{M} \ \text { Change } & -x \mathrm{M} & +x \mathrm{M} & +x \mathrm{M} \ \text { Final } & 0.2-x \mathrm{M} & x \mathrm{M} & x \mathrm{M} \end{array}$
$K_b=\dfrac{x^2}{0.2-x} \nonumber$
Above, we determined the value of $x=\left[\mathrm{OH}^{-}\right]=1.89 \times 10^{-5}$, so we can simply plug in to solve for $\mathrm{K}_b$ :
$K_b=\dfrac{\left(1.89 \times 10^{-5}\right)^2}{0.2-1.89 \times 10^{-5}}=1.8 \times 10^{-9} \nonumber$
Finally,
$p K_b=-\log \left(K_b\right)=8.74 \nonumber$ | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/06%3A_Recitations/6.24%3A_pH_and_pOH_pKa_and_pKb.txt |
Polymers
A polymer is a long molecule comprised of many (poly) iterations of a molecular unit cell (mer). There are many knobs available to turn to design a polymer with just the right properties. In 3.091, we’ll focus on two types of polymerization reaction: radical and condensation.
Radical polymerization
One mechanism to achieve polymerization is radical polymerization, a chain reaction that is started with the introduction of an initiator with a free radical. The initiator is shown below as $\mathrm{R}$; more important than its chemistry is the free radical, a highly reactive single electron. The radical polymerization of polyethylene is shown below:
The free radical introduces an extra electron to the monomer, breaking the double bond between the carbon atoms and propagating through to react with another nearby monomer (shown as a half-bond in the polymer above). As long as there is monomer near the end of the chain, the reaction will proceed. Of course, as with any reaction, it is critical that mass is conserved. We can double check that this is the case by counting electrons. The polyethylene monomer has 12 electrons, and with the radical, the system has 13 electrons. In the polymerized picture, there are still 13 electrons.
Radical polymerization works for all sorts of steric groups, where steric refers to the spatial arrangement of atoms in a molecule or side group. For polyethylene, the steric side groups are simply hydrogen atoms; in general, there are limitless options for side groups. Below, the generic reaction shows that what is really important for a radical polymerization to take place is the presence of a carbon=carbon double bond. The
leftmost part of the figure shows a generalized representation of a polymer: there are groups on the end (symbolized as $\mathrm{R}$, but they don’t necessarily need to be the same), and $\mathrm{n}$ repeat units that make up the backbone of the polymer.
Condensation polymerization
Condensation polymerization is a type of step-growth reaction that occurs between monomer units with end groups that react to form water as a by-product. The monomers that react could all be the same, or they could be different: this distinction impacts the size of the repeat unit in the polymer that results.
Here, the polymerization reaction mechanism is that the hydroxyl group on the end of one monomer reacts with the hydrogen that terminates the other monomer. Each time this reaction happens, the resulting polymer grows. The polymerization can be halted in many ways; of course, when all of the monomer is consumed or there aren’t monomers near the end of the chain to react, it stops growing. In this case, the hydroxyl group on one end of the monomer comes from the hydroxyl-terminated monomer. The other end could end up being either an $\mathrm{OH}$ group or a $\mathrm{H}$ group, depending on what the terminal monomer is- here, it’s drawn assuming an even number of monomers reacted. For each of $2n$ monomers that add on to the chain, one water molecule is formed as a by-product: this lends the name condensation reaction. Alternatively, a similar process could take place if all of the monomer units are terminated with hydroxyl groups:
The same caveats apply here: there is nothing special about having two different monomers; it could just as easily be a single monomer or many more. The key difference in this case is that there is an extra oxygen on the end of one of the monomers when the water condenses off: this integrates into the backbone as shown, and results in a polymer that is terminated by a hydroxyl group at both ends.
These examples are simply meant as illustrations. With so much flexibility as to the choice of monomer groups, mixtures, and reaction conditions, there is really a whole world of properties that can be engineered.
Polymer properties
The thesis of many efforts in materials science is that structure dictates properties, and polymers are no exception. We have to look at both the chemistry of the monomers and the specific polymerization and processing conditions to understand the electronic and molecular structure, which in turn provide insights as to the micro- and macroscopic properties of the material that results.
Monomer composition: the size and composition of the side groups can play a big role in how tightly a polymer packs. If there are big groups (with a lot of steric bulk), the backbones of the polymer chains can’t sit very close together, resulting in a lower degree of crystallinity.
IMFs: just as the size of the side groups plays a role, the composition can affect how tightly bound neighboring chains are to each other. For example, chains of polyethylene, with just $\mathrm{H}$ for side groups, slide across each other much easier than chains with side groups that are polar or can form hydrogen bonds.
Backbone structure: depending on how the specific reaction is run, the resulting polymer chains can end up being either linear or branched. Branched polymers pack less densely: you can think of branched polymers as a tangled mess of tree branches, while linear polymers are branches that have been cut into straight pieces and stacked.
Chain length: shorter chains can more easily slide past each other and move around, while long chains get tangled. It’s harder to pull apart a polymer comprised of long chains. If long polymer chains are a tangled mess of cooked spaghetti, short chains are like macaroni. The chain length that results from a given polymerization reaction is called the degree of polymerization, and it is often advantageous to try and engineer the resulting distribution of chain lengths to be as narrow as possible.
Tacticity: the tacticity of a polymer refers to how the side groups are arranged: if all on the same side, it is isotactic. If the groups alternate positions, the polymer is syndiotactic, and if they are randomly arranged, it is atactic. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/06%3A_Recitations/6.25%3A_Polymers_Radical_Polymerization_Condensation_Polymerization_and_Polymer_Properties.txt |
Steady state diffusion
Diffusion refers to the net movement of a species down a concentration gradient, from an area of high concentration to an area of low concentration. Importantly, diffusion can take place in any phase of matter, including in solids! Steady state (time independent) diffusion is described by Fick’s first law:
$J=-D \dfrac{d C}{d x}$
Here, $\mathrm{J}$ is the diffusion flux: the rate at which an amount of a substance passes through a surface area. The diffusion flux has units of $\frac{\text { amount }}{\text { areaxtime }}$, or $\frac{\mathrm{mol}}{\mathrm{m}^2 s}$. $\mathrm{D}$ is the diffusion coefficient, which is sometimes called diffusivity. It depends on the specific circumstances the diffusion is occurring in, including what materials are involved and the state of the surrounding environment. It has units of $\frac{m^2}{s}$. Finally, here $\mathrm{C}$ is concentration ( $M$ or $\frac{\mathrm{mol}}{\text { volume }}$, so $\frac{d C}{d x}$ is the change in concentration with respect to change in position. Therefore, Fick's first law tells us how concentration change flows over the region between two different concentrations.
Example: If you heat steel in a carbon rich atmosphere, carbon atoms will diffuse into the steel, hardening it. Calculate the diffusivity of a steel plate if the flux of carbon atoms through the plate is $2 \times 10^{-6} \frac{g}{m^2 s}$, the concentration $2 \mathrm{~mm}$ under the surface is $300 \frac{g}{m^3}$, and the concentration $5 \mathrm{~mm}$ under the surface is $100 \frac{g^3}{m^3}$.
Answer
We can rearrange Fick’s First law and plug in:
\begin{gathered}
D=-J\left(\dfrac{d C}{d x}\right)^{-1}=-J\left(\dfrac{x_2-x_1}{C_2-C_1}\right)=-2 \times 10^{-6} \dfrac{g}{m^2 s}\left(\dfrac{\left(2 \times 10^{-3}-5 \times 10^{-3}\right) m}{\left(3 \times 10^2-1 \times 10^2\right) \frac{g}{m^3}}\right) \
D=3 \times 10^{-11} \dfrac{m^2}{\mathrm{~s}}
\end{gathered}
Where we have approximated the concentration gradient as being linear.
Diffusion coefficient
Diffusion is a thermally-activated process, and the temperature dependence is reflected in the diffusion coefficient. The diffusion coefficient (diffusivity) is also described by an Arrhenius relationship! Picture an atom diffusing through a crystalline solid: it has to move through the lattice. One way this can happen is, if the atom is sufficiently small, it can travel hop from interstitial site to interstitial site. Larger atoms can diffuse, too, but it’s a more energy-consuming process. One mechanism for this to happen is if a vacancy is first created in the lattice, and then the diffusing atom moves into the now-vacant lattice site. As vacancies diffuse through the lattice, other species can follow. Of course, there are more moving parts with this process, and more energy is required.
The energy required for diffusion to occur can be thought of as an activation energy. The diffusion coefficient is
$D=D_0 e^{-\dfrac{E_a}{k_B T}}$
Here, $D_0$ is the maximum value of diffusivity, $E_a$ is the activation energy, $k_B$ is the Boltzmann constant, and $T$ is temperature. By describing the diffusion coefficient with an Arrhenius relation, recall that we are saying that at a given temperature, $T$, the ratio of the thermal energy given by $k_B T$ to the activation energy $E_a$ sets the value of the diffusion coefficient, and it is maximized at infinite temperature.
Fick’s second law
If the concentration profile varies with respect to time, the steady-state assumption no longer holds, and instead Fick’s second law is used instead:
$\dfrac{\partial C}{\partial t}=D \dfrac{\partial^2 C}{\partial x^2}$
Solutions to Fick’s second law are of the form
$\dfrac{C(x, t)-C_0}{C_s-C_0}=1-\operatorname{erf}\left(\dfrac{x}{2 \sqrt{D t}}\right)$
Here, $C_s$ is the concentration of the source, $C_0$ is the initial concentration, $C(x, t)$ is the expression for concentration as a function of position and time, $\mathrm{D}$ is the diffusion coefficient, and erf is the error function. | textbooks/chem/Inorganic_Chemistry/Introduction_to_Solid_State_Chemistry/06%3A_Recitations/6.26%3A_Steady_State_Diffusion_Diffusion_Coefficient_Ficks_Second_Law.txt |
All matter has intrinsic wave properties. These are described mathematically by the Schrödinger Equations and it's solutions. The wavenature of electrons and other fundamental principles (eg charge and momentum) together produce the wave mechanics of electron. The effects of electron wave mechanics are far reaching, responsible for such phenomena as electricity, emission and absorption, and bonding and hybridization.
Background
Wave-Nature of Matter
Accurate explanations of atomic natural physical chemical phenomena are dependent on energy quantization. The realization of this fundamental characteristic of matter was developed through treatment of a couple well known experiments, notably Max Planck's explanation of black body radiation and Einstein's explanation of the photoelectric effect. The conclusions of energy quantization were consolidated by Louis De Broglie as
$\Lambda = \dfrac{h}{p} \nonumber$
and, by rearrangement:
$f= \dfrac{E}{h} \nonumber$
On Waves
Quantum mechanically, an electron can be described by a wave function oscillating in space and time that has mean values equal to the expectation values of observables corresponding to given operators. According to the Born interpretation of quantum mechanics, the complex conjugate of this wavefunction correlates to the electron's positional probability density.
Electrons are fermions. They are charged particles. When they are confined by a potential to a limited space they display harmonics analogous to those of other wavelike phenomena. This occurs most notable in atoms and molecules. The hydrogen atom proves the most simple atomic example. The three dimensional harmonics of an electron bound within the potential energy well of a proton results in what are conventionally called orbitals. Orbitals are commonly depicted as contours of some percent of the complex conjugate of the electron's approximate wave function, though realistically without external potentials they diffuse infinitely.
A bound electron occupies higher harmonics of the bound state with increasing energy. Energy can only be increased in specific quanta as demanded for the wave function to exist. The discrete energy levels of higher harmonics correspond to higher orbitals.
Electrons can gain energy to exist in a higher orbital. When this process occurs via interaction with electromagnetic radiation, it is referred to as absorption. Similarly, the regression of an electron into a lower energy orbital results in the release of electromagnetic radiation, and is referred to as emission. Because of the quantized energy levels demanded by a bound system, electrons in a molecule or atom can only absorb or emit light at specific frequencies, which depend on the properties of the system.
Certain materials have energy level spacing such that excitation by an energy source can produce a greater number of electrons in an excited state than in the ground state. This is known as population inversion. When this happens for a transition which releases light upon relaxation, light of a specific nature is produced that has great practical importance. This light is monochromatic, and can be channeled back and forth through the medium (gain medium) and allowed only to disperse through a very narrow slit to produce monochromatic, directional, coherent light source. The apparatus is called a LASER, which is an acronym for light amplification by stimulated emission of radiation.
The inherent charge of electron incites movement of the particle in accordance to the forces described by coulomb's law. Rotational motion of a charged particle produces an electric field. The potential of an electron attraction to positive charge can be used to store energy in chemical form. The motion of electrons by batteries or other sources through conductive media, such as copper wire can be facilitated to do work. Computer. Light.
Electronic absorption and emission within roughly 350 to 750 nm produces radiation that is in the visible spectrum. the sky is blue because of the interaction of light from the sun with electrons in atoms in the atmosphere in what's known as scattering. Scattering of this type occurs to the inverse cube of wavelength so light with shorter wavelength (blue in the visible) is scattered much more than other wavelengths. The other light passes through the atmosphere or something. maybe i have this backwards.
electrons can tunnel due to their wave nature. quantum mechanical tunneling is where a particle goes somewhere that is classically impossible, meaning that it simply did not have enough energy to to past a potential barrier, ut it did. We experts in science call this quantum weirdness. There are a lot of electrons, but perhaps not more than there are stupid people in the world. This page needs revision.
Most basic wave to satisfy boundary conditions (which are...) is
$A \sin(n\pi x/L) \nonumber$
The superposition principle allows for the fourier theorem which allows an infinite number of such waves to be combined to form any any curve that obeys the requirements of a bound system.
Such a wave provides an accurate (nonrelativistic) description of the electron. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/01%3A_Basic_Concepts-_Atoms/1.05%3A_An_Introduction_to_Wave_Mechanics/1.5A%3A_Wave_Mechanics_of_Electrons.txt |
Introduction
There are four quantum numbers ($n$, $l$, $m_{l}$, $m_{s}$). No two electrons in the same atom can have the same four quantum numbers. Each quantum number describes a different aspect of the electron and its orbital. These numbers are obtained from the solution of the Schrödinger Equation for atoms in spherical coordinates.
n, the principal quantum number
$n$tells you about the size of the orbital. It is related to how far the electron is from the atom. It is also related to the energy of the electron. $n$can be any positive integer number. An orbital has $n-1$radial nodes as well, which describe a radius at which the probability of finding the electron is 0.
l, azimuthal quantum number
lt tells you about the angular momentum of the electron in the orbital. It defines the shape of the orbital. lt can be any integer between (and including) 0 and ($n-1$). Each orbital has $l$planar nodes.
Some examples of orbitals with different $l$:
$l=0$is an s orbital, with no planar nodes.
$l=1$gives a p orbital, which has one planar node.
ml, magnetic quantum number
$m_{l}$ tells you about the angular momentum projected on to the z axis. It tells you of the orientation of the orbital. It can be any integer between $-l$ and $l$.
$l=1$ has 3 different (but degenerate) $m_{l}$ possible values:-1,0 or 1.
$p_{x}$
$p_{y}$
$p_{z}$
ms spin projection quantum number
$m_{s}$tells you about the spin of the electron. An electron is a fermion, a type of quantum particle which is only allowed to have $m_{s}$equal to -1/2 or 1/2.
Pauli Exclusion Principle
The Pauli Exclusion principle states that no two identical fermions can share the same quantum state simultaneously. An electron, being a fermion, obeys this principle, so no two electrons in the same atom can have the same quantum numbers. Each orbital has space for 2 electrons, with $m_{s}$ values with opposite sign. These electrons are considered spin-paired.
Energies of the Orbitals
The energies of the orbitals depend solely on $n$ and $l$ so there are degenerate states that result from various $m_{l}$ values. For example, there are 3 p orbitals, all of which are equal in energy and therefore degenerate. Here is a table of possible quantum number arrangements and the corresponding orbitals from $n=1$ to $n=5$:
$n$ $l$ $m_{l}$ $m_{s}$ Orbital
$1$ $0$ $0$ $\pm \frac{1}{2}$ $1s$
$2$ $0$ $0$ $\pm \frac{1}{2}$ $2s$
$2$ $1$ $\pm 1$ or $0$ $\pm \frac{1}{2}$ $2p_{x}, 2p_{y} \; or \; 2p_{z}$
$3$ $0$ $0$ $\pm \frac{1}{2}$ $3s$
$3$ $1$ $\pm 1$ or $0$ $\pm \frac{1}{2}$ $3p_{x}, 3p_{y} \; or \; 3p_{z}$
$3$ $2$ $\pm 2, \pm 1$ or $0$ $\pm \frac{1}{2}$ $3d_{z^2}, 3d_{x^2-y^2}, 3d_{xy},$$3d_{xz}\; or \; 3d_{yz}$
$4$ $0$ $0$ $\pm \frac{1}{2}$ $4s$
$4$ $1$ $\pm 1$ or $0$ $\pm \frac{1}{2}$ $4p_{x}, 4p_{y} \; or \; 4p_{z}$
$4$ $2$ $\pm 2, \pm 1$ or $0$ $\pm \frac{1}{2}$ $4d_{z^2}, 4d_{x^2-y^2}, 4d_{xy},$$4d_{xz} \; or \; 4d_{yz}$
$4$ $3$ $\pm 3, \pm 2, \pm 1$ or $0$ $\pm \frac{1}{2}$ $4f_{x^3}, 4f_{y^3}, 4f_{z^3},$$4f_{x (z^2-y^2)}, 4f_{y (z^2-x^2)},$$4f_{z (x^2-y^2)}or \; 4f_{xyz}$
$5$ $0$ $0$ $\pm \frac{1}{2}$ $5s$
$5$ $1$ $\pm 1$or $0$ $\pm \frac{1}{2}$ $5p_{x}, 5p_{y} \; or \; 5p_{z}$
$5$ $2$ $\pm 2, \pm 1$ or $0$ $\pm \frac{1}{2}$ $5d_{z^2}, 5d_{x^2-y^2}, 5d_{xy}, 5d_{xz} \; or \; 5d_{yz}$
$5$ $3$ $\pm 3, \pm 2, \pm 1$ or $0$ $\pm \frac{1}{2}$ $5f_{x^3}, 5f_{y^3}, 5f_{z^3}, 5f_{x (z^2-y^2)}, 5f_{y (z^2-x^2)}, 5f_{z (x^2-y^2)} or \; 5f_{xyz}$
$5$ $4$ $\pm 4, \pm 3, \pm 2, \pm 1$ or $0$ $\pm \frac{1}{2}$ $5g_{z^4}, 5g_{z^3 x}, 5g_{z^3 y}, 5g_{z^{2} xy}, 5g_{z^2 (x^2-y^2)}, 5g_{zx^3}, 5g_{zy^3}, 5g_{xy (x^2-y^2)} \; or 5g_{x^4+y^4}$
Problems
1. How many electrons can have the quantum numbers $n=2$ and $l=0$? List the acceptable values for $m_{l}$ and $m_{s}$ for these values.
2. How many electrons can have the quantum numbers $n=2$ and $l=2$?
3. Which set of orbitals do the quantum numbers $n=3$ and $l=2$describe? How many radial and planar nodes do these orbitals have?
Solutions
1. There may be 2 electrons with these quantum numbers, one with $m_{l}=0$ and $m_{s}= \frac{1}{2}$ and the other with $m_{l}=0$ and $m_{s}= -\frac{1}{2}$.
2. This is not a permitted value of $l$for $n=2$, since the greatest acceptable value is given by $l=n-1$.
3. This represents the 3d orbitals. They have 2 planar orbitals and 0 radial nodes.
• Bryn Ellison
1.6F: Size of Orbitals
Figure 1. General Trend of how the radius change within the Periodic Table
Contributors and Attributions
• Name #1 here (if anonymous, you can avoid this) with university affiliation
1.6H: The ground state of Hydrogen
Hydrogen is the simplest atoms, which only contains an electron and a proton. The ground state of hydrogen is the lowest allowed energy level and has zero angular momentum. However, it is the most stable state in which a single electron occupied the 1s atomic orbital.
Contributors and Attributions
• Name #1 here (if anonymous, you can avoid this) with university affiliation | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/01%3A_Basic_Concepts-_Atoms/1.06%3A_Atomic_Orbitals/1.6A%3A_Electronic_Quantum_Numbers.txt |
Learning Objective
• To quantify the shielding effect experienced by atomic electrons.
We have previously described the concepts of electron shielding, orbital penetration and effective nuclear charge, but we did so in a qualitative manner. In this section, we explore one model for quantitatively estimating the impact of electron shielding, and then use that to calculate the effective nuclear charge experienced by an electron in an atom. The model we will use is known as Slater's Rules (J.C. Slater, Phys Rev 1930, 36, 57).
Slater's Rules
The general principle behind Slater's Rule is that the actual charge felt by an electron is equal to what you'd expect the charge to be from a certain number of protons, but minus a certain amount of charge from other electrons. Slater's rules allow you to estimate the effective nuclear charge $Z_{eff}$ from the real number of protons in the nucleus and the effective shielding of electrons in each orbital "shell" (e.g., to compare the effective nuclear charge and shielding 3d and 4s in transition metals). Slater's rules are fairly simple and produce fairly accurate predictions of things like the electron configurations and ionization energies.
Slater's Rules
• Step 1: Write the electron configuration of the atom in the following form:
(1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f) (5s, 5p) . . .
• Step 2: Identify the electron of interest, and ignore all electrons in higher groups (to the right in the list from Step 1). These do not shield electrons in lower groups
• Step 3: Slater's Rules is now broken into two cases:
• the shielding experienced by an s- or p- electron,
• electrons within same group shield 0.35, except the 1s which shield 0.30
• electrons within the n-1 group shield 0.85
• electrons within the n-2 or lower groups shield 1.00
• the shielding experienced by nd or nf valence electrons
• electrons within same group shield 0.35
• electrons within the lower groups shield 1.00
These rules are summarized in Figure $1$ and Table $1$.
Shielding happens when electrons in lower valence shells (or the same valence shell) provide a repulsive force to valence electrons, thereby "negating" some of the attractive force from the positive nucleus. Electrons really close to the atom (n-2 or lower) pretty much just look like protons, so they completely negate. As electrons get closer to the electron of interest, some more complex interactions happen that reduce this shielding.
Table $1$: Slater's Rules for calculating shieldings
Group Other electrons in the same group Electrons in group(s) with principal quantum number n and azimuthal quantum number < l Electrons in group(s) with principal quantum number n-1 Electrons in all group(s) with principal quantum number < n-1
[1s] 0.30 - - -
[ns,np] 0.35 - 0.85 1
[nd] or [nf] 0.35 1 1 1
The shielding numbers in Table $1$ were derived semi-empirically (i.e., derived from experiments) as opposed to theoretical calculations. This is because quantum mechanics makes calculating shielding effects quite difficult, which is outside the scope of this Module.
Calculating S
Sum together the contributions as described in the appropriate rule above to obtain an estimate of the shielding constant, $S$, which is found by totaling the screening by all electrons except the one in question.
$S = \sum n_i S_i \label{2.6.0}$
where
• $n_i$ is the number of electrons in a specific shell and subshell and
• $S_i$ is the shielding of the electrons subject to Slater's rules (Table $1$)
Example $1$: The Shielding of 3p Electrons of Nitrogen Atoms
What is the shielding constant experienced by a 2p electron in the nitrogen atom?
Given: Nitrogen (N)
Asked for: $S$, the shielding constant, for a 2p electron (Equation \ref{2.6.0})
Strategy:
1. Determine the electron configuration of nitrogen, then write it in the appropriate form.
2. Use the appropriate Slater Rule to calculate the shielding constant for the electron.
Solution A N: 1s2 2s2 2p3
N: (1s2)(2s2,2p3)
Solution B
$S[2p] = \underbrace{0.85(2)}_{\text{the 1s electrons}} + \underbrace{0.35(4)}_{\text{the 2s and 2p electrons}} = 3.10\nonumber$
As Table $1$ indicates,
• the 1s electrons shield the other 2p electron to 0.85 "charges".
• the 2s and 2p electrons shield the other 2p electron equally at 0.35 "charges".
Exercise $1$: The Shielding of valence p Electrons of Bromine Atoms
What is the shielding constant experienced by a valence p-electron in the bromine atom?
Answer
$S = 2+8+8 \times 0.85 + 10 + 4 \times 0.35 = 28.20 \nonumber$
Example $2$: The Shielding of 3d Electrons of Bromine Atoms
What is the shielding constant experienced by a 3d electron in the bromine atom?
Given: Bromine (Br)
Asked for: S, the shielding constant, for a 3d electron
Strategy:
1. Determine the electron configuration of bromine, then write it in the appropriate form.
2. Use the appropriate Slater Rule to calculate the shielding constant for the electron.
Solution A Br: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5
Br: (1s2)(2s2,2p6)(3s2,3p6)(3d10)(4s2,4p5)
Ignore the group to the right of the 3d electrons. These do not contribute to the shielding constant.
Solution B S[3d] = 1.00(18) + 0.35(9) = 21.15
Exercise $2$: The Shielding of 3d Electrons of Copper Atoms
What is the shielding constant experienced by a valence d-electron in the copper atom?
Answer
S = 21.15
Calculating Zeff
One set of estimates for the effective nuclear charge ($Z_{eff}$) was presented in Figure 2.5.1. Previously, we described $Z_{eff}$ as being less than the actual nuclear charge ($Z$) because of the repulsive interaction between core and valence electrons. We can quantitatively represent this difference between $Z$ and $Z_{eff}$ as follows:
$S=Z-Z_{eff} \label{2.6.1}$
Rearranging this formula to solve for $Z_{eff}$ we obtain:
$Z_{eff}=Z-S \label{2.6.2}$
We can then substitute the shielding constant obtained using Equation $\ref{2.6.2}$ to calculate an estimate of $Z_{eff}$ for the corresponding atomic electron.
Example $3$: The Effective Charge of p Electrons of Boron Atoms
What is the effective nuclear charge experienced by a valence p- electron in boron?
Given: Boron (B)
Asked for: $Z_{eff}$ for a valence p- electron
Strategy:
1. Determine the electron configuration of boron and identify the electron of interest.
2. Use the appropriate Slater Rule to calculate the shielding constant for the electron.
3. Use the Periodic Table to determine the actual nuclear charge for boron.
4. Determine the effective nuclear constant.
Solution:
A B: 1s2 2s2 2p1 . The valence p- electron in boron resides in the 2p subshell.
B: (1s2)(2s2,2p1)
B S[2p] = 1.00(0) + 0.85(2) + 0.35(2) = 2.40
C Z = 5
D Using Equation \ref{2.6.2}, $Z_{eff} = 2.60$
Exercise $3$
What is the effective nuclear charge experienced by a valence d-electron in copper?
Answer
$Z_{eff} = 7.85$
Summary
Slater's Rules can be used as a model of shielding. This permits us to quantify both the amount of shielding experienced by an electron and the resulting effective nuclear charge. Others performed better optimizations of $Z_{eff}$ using variational Hartree-Fock methods. For example, Clementi and Raimondi published "Atomic Screening Constants from SCF Functions." J Chem Phys (1963) 38, 2686–2689. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/01%3A_Basic_Concepts-_Atoms/1.07%3A_Many_Electron_Atoms/1.7C%3A_Penetration_and_Shielding.txt |
For the most part, the introduction to this chapter is about why inorganic chemistry tends to be more complicated to understand and learn than organic:
• Number of elements
• Variation of properties (size, DHIP, DHEA, c) even among similar ones
• Covalency, electrovalency and multiple valencies
• Physical state, molecular aggregation and stability range (temperature) and solubility
• Air and water stability
• Coordination number and respect for the octet rule
• s, pp-pp, pp-dp and dp-dp bonding possibilities
Much order can be introduced into this potential chaos by relating the properties of the elements to their position in the periodic table (or, as was originally done by Mendeleev et al, arranging the elements into the periodic table according to their properties.
Heavier Elements
There is little chemistry to learn about for many of the transuranium elements especially the newest (104 - 108) because of their extremely short half-lives. In some cases only one or two atoms have been made!
The rest of the chapter is divided into secions A and B which, respectively, describe the pure elements themselves and then the general group trends and non-trends.
Monatomic Elements
Includes all the noble gases and many metal vapours. The noble gases do not form bonds to each other because of their closed valence shell configuration, and the metals have their own kind of bonding which does not work in the gas phase (see below).
Diatomic Elements
These are H2, the simplest of all neutral molecules, the halogens which also have single bonds, and N2 and O2 which have multiple pp-pp bonding.
Discrete (Small) Polyatomic Molecules: Sn and Se8, P4.
Since pp-pp bonding is not a very useful option for elements below the first short period because of core-core repulsion at useful p-bonding distances, P, S and Se can form clusters (or chains - see the next group).
Sulphur, which can form two single bonds to itself, comes in various ring sizes up to S20, S8 being the most stable ("orthorhombic" sulphur).
White phosphorus, which can form three, comes as P4 with P-P at 2.21 Å, which is more or less normal. A particularly strained P-P, i.e. "bent" would be longer. Nevertheless, P4 is the most reactive form of phosphorus, for example, spontaneously inflaming in air.
Boron and carbon, capable of forming 3 and 4 bonds, respectively, could have been included here too, but see below.
Elements with Extended Structures
Boron is a somewhat anomalous case because it is electron deficient. The allotropic forms of boron are all hased on the icosahedral B12 unit. In this unit, each boron has 5 nearest neighbours, and since the units are weakly bonded to form more extended 3-frameworks, the borons that will have more remote neighbours as well. There are nowhere near enough electrons and atomic orbitals to do this using localized 2-electron - 2-centre Lewis type bonds. As usual molecular orbital theory comes to the rescue, but the B12 cage alone is a rather complex system to use as an example. The topic of the boranes (boron hydrides), to be covered later, might help shed some light on how the molecular orbitals might be set up. The allotropes of crystalline boron are the most complex of the structures of the elements, and their study has been futher complicated by the presence of small (but stoichiometric) amounts of carbon or nitrogen leading to some false ones.
The basic icosohedral B12 unit
Depending on the number of bonds that they can form the following elements can form chain, sheet, or three dimensional structures:
C P S
Si As Se
Ge Sb Te
Sn Bi
"Plastic" sulphur (Sn) has a rubbery texture and is thermally unstable at room temperature with respect to conversion to S8. Selenium also forms chains, but infinite spirals in an ordered (crystalline) structure. Tellurium is similar and the solid has semiconductor propeties.
Crystalline black phosphorus has a "double sheet" structure. This structure is shared by its cogeners (members of the same group) but they are metals. Red phosporus, the other common form, contains chains of phosphorus tetrahedra.)
Graphite has 2-dimensional sheets of trigonal sp2 carbon atom arranged in aromatic six membered rings. Because of the extended delocalized orbitals, graphite is a conductor, especially parallel to the sheets.
Diamond is a 3-dimensional lattice of tetrahedral sp3 carbon atoms arranged in saturated six-membered rings all in the chair form. Diamond is an insulator.
A number of additional carbon structures have been discovered quite recently. Buckmeisterfullerene C60, the simplest of the "bucky balls" is shown on the right. It was discovcerd in soot. (Its shape, like an international soccer ball, is an icosahedron with all 10 corners cut off to 1/3 of the way along each edge which generates 10 new pentagonal faces and converts the old triangular ones to hexagons.) There are a number of elongated spheroidal structures and tubes capped by hemispheres known all made by cooling carbon vapour from and arc or plasma discharge. Chemists are still trying to make them do something useful!
Heats of formation for carbon forms are: Graphite, 0 ; diamond, 2.9; and C60, 38.1 kJ mol-1.
Metals
The structures of many metals are based on hexagonal or cubic close packing (Chapter 4 section 7), or the less efficient body-centered cubic (Fig 8-6). Figure 8-7 shows the distribution of these structures.
Metals are sometimes described as a regular array of cations immersed in a sea of valence electrons which are completely delocalized and free to roam through the entire mass. A simplified theoretical picture of this follows:
In the molecule Li2 the bond results from overlap of the 2s orbitals each of which contains 2 electrons. The diagram below shows several representations of this:
The part of the diagram on the left shows the energy levels at the observed internuclear distance marked with a vertical dotted line on the right hand diagram. A crude representation of the two molecular orbitals is also shown. (The internal spherical nodes that the 2s orbitals possess are omitted.)
The next four diagrams show orbital energies (and orbitals) for hypothetical linear molecules, Li3, Li4, Li5 and Li¥.
Variation of orbital enery with internuclear separation for a hypothetical linear Li3 molecule
.
Variation of orbital enery with internuclear separation for a hypothetical linear Li4 molecule.
Variation of orbital enery with internuclear separation for a hypothetical linear Li5 molecule.
At the left is the diagram for the hypothetical Li¥. The infinite number of morse curves merge together to give the "band" shown.
Because lithium has the 2s1 valence shell configuration, only half of this band will be filled, and electrons can easily move into the empty orbitals accounting for the metallic conduction observed. The occupied levels/orbitals are shown by the shaded area in the diagram below:
The situation described above is not real. Real crystals are three-dimensional, and in addition to the molecular orbitals formed by the atomic s-orbitals, there will be another set formed by overlap of the atomic p-orbitals. The text figures 8.8 and 8.9 are a more realistic representation.
Cohesive Energies of the Metals
See figure 8-10. Note that the main group metals have relatively low enthalpies of atomization correlating roughly with the number of valence electrons. Maxima occur at the half-filled shell after which antibonding parts of the conduction bands begin to be populated.
Hydrogen: 1s1
With an intermediate electronegativity and one valence electron, hydrogen might range from H+ to H- in its compounds:
• The proton (radius = 1.5x10-13 cm) cannot exist in condensed states (liquid or solids) because it exerts too powerful a polarizing influence on molecules near to itself. Therefore, it is always complexed, for example as H3O+.
• Covalent compounds are the normal situation.
• When combined with very electropositive elements e.g. Na, K, Ca it behaves as H- (hydride ion) and is a very powerful reducing agent.
Helium (1s2) and the noble gases (ns2np6)
• The energy necessary to make a valence state is not compensated by bond formation above krypton which forms a few flourides. The rest of the group have fairly extensive chemistry with fluorine, chlorine and oxygen.
(The first noble gas compound, originally thought to be Xe2+[PtF6]2- by analogy with [O2]2+[PtF6]2-, was made by Bartlett in 1962. Prior to that they had been thought to be inert.)
Elements of the First Short Period (Li - F)
They are poor representatives of their groups. (The second short period elements are much better for this.) The reason is connected to their small size and lack of (empty) d-orbitals which limits their coordination number to 4 in the absence of multicentre bonding.
• To achieve their normal valencies, Be, B and C "use" a valence state where one of the 1s electrons is promoted to the p-subshell.
• "Electron deficient" Li, Be and B compounds are Lewis acids. This involves a change in hybridization. The extent to which the coordination number can be increased is limited by negative charge build-up on the acceptor.
• Li, Be and B also tend to be involved in molecules featuring 3-centre, 2 e- bonding as a way to reduce their electron deficiency.
• N, O and (perhaps) F compounds can be Lewis bases because of the available electon pairs on these atoms. The sharing of more than one pair is opposed by positive charge build-up on the donor atom, and sometimes steric factors.
• Li+ exists in rather few compounds because of its small size. Covalent and multicenter bonding is common e.g. the lithium alkyls.
• Be2+ would be even more polarizing than Li+ so no compounds containing it exist. Covalency is the norm e.g. BeF2 and [Be(H2O)4]2+, which hydrolyses extensively to BeOH containing species.
• for Li, Be and B there is a marked similarity to the diagonally related Mg, Al and C. Although the valencies differ, the chemical behaviour such as covalency of halides or coordination chemistry tends to be somewhat similar. For example, there are close parallels between the lithium alkyls and the magnesium alkyls or between beryllium halides and the aluminium halides.
• For B, C and N, there are no (long-lived) compounds in which the elements are cationic. Covalency dominates the chemistry of these elements.
• There are compounds containing anions e.g. C22- (acetylide) and N3-, which is a major product when Mg burns in air.
• The chemistry of O and F is dominated by anion formation and covalency. Remember, O2- is too strong a base to exist in water though solid compounds are known. Fluoride, F- is only a weak base and so known in solids and aqueous solution.
• C, N and O commonly participate in multiple bonding by the use of p-orbitals. N2 is a rather unreactive molecule because of the great strength of the triple bond. O2 is much more reactive, partly because it is a bi-radical. F2 is very reactive because the filled p-orbitals perpendicular to the bond repel each other.
Covalent Bonds
Covered as part of the previous section.
Elements of the Second Short Period (Na to Ar)
These elements are much more representative of the chemistry of the elements below them in their groups. This specially true for the non-metalic elements:
• The elements of this period and those below it do not tend to use their p-orbitals for p-bonding. Rather the empty d-orbitals are used if p-bonding occcurs.
For example, aside from special sterically encumbered cases, there are few compounds containing Si=Si, P=P or S=S bonds. An example of an exception is (CH3)3C-P=P-C(CH3). pp-pp-bonding is not stable with respect to addition reactions: a kinetic effect.
• Multiple bonding that does occur usually involves a first short period element such as oxygen or nitrogen. Examples include Cl3PO, SO2 or ClO2. Such molecules may contain pp-dp-bonding, and the octet rule is often violated. On the other hand, there is no analogue in silicon chemistry of O=C=O. SiO2 is a network solid although the Si-O-Si groupings are linear which does hint and p-interactions.
• The d-obitals can be invoked to explain valence states where the coordination number exceeds 4 (and the octet is exceeded also.) Examples include PCl5, SF6 and IF7. The possibility of higher coordination numbers lead also to different reactivity, e.g. CCl4 which is inert to hydrolysis and SiCl4.
• The metals differ from the first memeber of their group, e.g. sodium does not form covalent compounds, aluminum is a cation-forming metal quite unlike boron, and the coordination numbers are often 6 rather than 4 for these second row elements.
The Remainder of the Non-Transition Elements
This section contains a group-by-group summary of the properties characterizing each group and indicating, again, how the top member of each group differs. The following is list of key properties appropriate for comparisions:
1. Metallic character.
2. Properties of the oxides:
1. Ionic vs covalent
2. Acidic vs basic
3. Properties of the halides:
1. Ionic vs molecular
2. Ease of hydrolysis
4. General trends in electrovalence and covalence.
5. Trends in structure:
1. Coordination numbers for discrete species
2. Aggregation in the solid state to increase coordination number
6. Properties of the hydrides.
7. Tendency to catenate.
8. The relative importance of pp-pp vs pp-dp vs dp-dp bonding.
9. The general strength of covalent bonding.
10. The relative importance of the lower valent states vs the hight valent states.
Group IA or 1
See text Table 8.1. The elements have chemistry dominated by the ionic 1+ state. The behaviour going down a group is well-behaved. The following decrease down a group:
1. melting points and heats of sublimation.
2. lattice energies (mostly).
3. Effective hydrated radii and hydration energies.
4. Ionization Enthalpies.
Group IIA or 2 and IIB or 12
See text Table 8.2. For group IIA (Mg, Ca, Sr, Ba and Ra) which all form the 2+ ionic state there are again well-behaved trends including those mentioned for group IA and adding:
1. Solubility of the sulphates and carbonates.
2. Thermal instability of the carbonates and nitrates.
which decrease going down the group. Group IIB metals (Zn, Cd and Hg) have two s-electrons outside filled d-subshells. In Zn and Cd the d-electrons do a somewhat poorer screening job than the closed shells of the group IIA so that the 2+ ions (which are more polarizing) are more inclined to form complex ions with NH3, halides and CN-.
Mercuric ion, Hg2+, where the f-subshell also has been filled, is quite different in that it is even more inclined to form complexes. The oxidation potential for mercury is positive, whereas the others are all negative, which means that mercury metal is rather easily produced in its reactions. It also readily forms the mercurous ion, Hg22+.
Group IIIA or 3 and IIIB or 13
See text Table 8.3. The text includes the group IIIA elements and the rest of the lanthanides among the non-transition metals because of the relative simplicity of their chemistry. They all behave as quite electropositive elements and form the 3+ ion. Scandium behaves as if it falls somewhere between aluminum and gallium in terms of its covalency. The lanthanides are so similar to each another that separation is difficult. (Cation exchange chromatography is used.)
All the group IIIB elements form compounds containing the 3+ ion, but there is a marked tendency towards covalency especially for aluminum e.g. its alkyls are volatile liquids and its chloride sublimes.
Group IIIB is the first group to clearly demonstrate the "inert pair effect". On descending the group, there is a tendency for the 1+ ions to become more and more stable. Tl+ is more stable than Tl3+ and its chemistry is complicated by its redox behaviour.
(The unusual stability of Hg0 could also be ascribed to the inert pair effect.)
Group IVB or 14
See text Table 8.4. This is the group in which the property differences between the first member, carbon, and the rest is perhaps most marked. Knowing the chemistry of carbon is almost useless for predicting the chemistry of the remaining elements! The chemistry of carbon is dominated by its extraordinary ability to catenate. While silicon can form analogues to the hydrocarbons, multiple bonding is not viable, and the Si-O bond is the important bond in silicon chemistry.
The chemistry of the group IVB elements in their IV oxidation state is predominantly covalent: SiCl4 is a liquid as is Pb(C2H5)4, the infamous anti-knock gasoline additive.
The covalent bonds generally weaken on descending the group.
The divalent state of carbon (in carbenes) is very reactive and it comes in both singlet and triplet forms. Silicon has virtually no chemistry in this state.
For the rest of the group, the II state becomes successively more stable going down the group. The "inert pair" is usually stereoactive in both molecular and solid state aggregated compounds. The M(II) compounds can never be considered truly ionic. The trend is probably associated with decreasing covalent bond strengths compared to the energy needed to obtain the valence state because the ionization potential differences are small:
GeCl2 + Cl2 GeCl4 (Rapid at 25 oC)
SnCl2 + Cl2 SnCl4 (Slow at 25 oC)
PbCl2 + Cl2 PbCl4 (Does not go except at higher temperature and pressure)
PbBr4 and PbI4 cannot be made.
Group VB or 15
See text Table 8.5. The oxidation states V and III are normal for this group. Phosphorus compounds are all covalent, but the tendency towards ionic behaviour increases down the group especially for the III state, but, the only species that can be considered as predominantly cations are SbO+and BiO+ and Bi3+ in its fluoride.
Group VIB or 16
See text Table 8.6. This group is relatively well-behaved so that its features can be tablulated:
1. The 2- ions exist in salts with electropositive elements.
2. There are also anions containing one covalent bond e.g. RS- or HS-.
3. There are covalent compounds of type (for example) R2E or X2E (R is organic, X is a halogen and E is the group VIB element).
4. There are monocations of type R3E+
5. There are anionic ompounds where the group VI element is in the IV or VI state, and with coordination numbers 4, 5 and 6 e.g. SCl4, SeF5-, TeF6. There are stereoactive lone pairs to worry about!
6. The hydrides, H2E decrease in stability down the group.
7. The elements themselves become gradually more metallic.
8. The tendency to form anionic complexes increases e.g. SeBr62-, TeBr62- and even PoI62-.
Group VIIB or 17
See text Table 8.7. With the elements of group IA, this group is the most well-behaved. The properties progress nicely down the group:
1. The elements all form th X- ion and compounds with one covalent bond e.g. RX or HX.
2. Interhalogens e.g. ClF have properties somewhat intermediate between the parent dihalogens, but they are polar in the expected sense.
3. The higher oxidation states are II, V and VII.
4. There are no simple cationic ions X+ but Br2+, I2+, Cl3+, Br3+ and several larger iodine cations have been prepared.
The Transition Elements of the d and f-Blocks
Not covered in Chem 242. (Winter 1999) | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/01%3A_Basic_Concepts-_Atoms/1.08%3A_The_Periodic_Table.txt |
Ground state electron configurations are the foundation for understanding molecular bonding, properties, and structures. From the electrons in an atom, to the differing orbitals and hybridization, the ground state electron configuration sheds light on many different atomic properties. Fundamentally, understanding electron configuration leads to an understanding of the periodic table.
Introduction
In 1913, Niels Bohr proposed that electrons could orbit an atom at a certain distance without collapsing into the atom, and that each orbit distance had its own energy level. He proposed that each orbital’s angular momentum, M, was equal to a multiple, n, of Plank’s constant, h, divided by 2π. This gives the equation:
M = nħ where ħ= h/2π and n= 1,2,3,4
This model proposed the Bohr atom, which shows circular orbits surrounding the nucleus.
Orbitals
In addition to having different energy levels, orbitals also have different shapes and orientations, and each can be occupied by two electrons. For each principal quantum number, n, there is one s orbital, three p orbitals, five d orbitals and seven f orbitals. Therefore, an s orbital can hold two electrons, a p orbital can hold six electrons, a d orbital can hold ten electrons, and an f orbital can hold 14 electrons.
Ground State Electron Configuration
Quantum numbers
There are four quantum numbers n, l, ml, and ms. The principal quantum number n is a positive integer (1,2,3,4) and it represents the energy of the orbital. The angular momentum quantum number l, is from 0 to n – 1. The l values of 0, 1, 2, and 3 correspond to the s, p, d and f orbitals, respectively. The magnetic quantum number ml ranges from –l to +l. This quantum number dictates the orbital orientation, such as px, py, or pz. The quantum spin number ms, is either +1/2 or -1/2 and it dictates the electron spin.
Aufbau Principle
The Aufbau principle states that electrons must fill lowest energy shells first.
Following the model, electrons fill the 1s orbital with two electrons, then the 2s with two electrons, then the 2p with six electrons, then the 3s with two electrons, etc.
There are some exceptions to the Aufbau Principle. This occurs mainly with electrons in the d orbital where extra stability is obtained from a half filled or fully filled d orbital. Therefore, if there are 4 electrons, or 9 electrons in the d orbital, it will move one electron from the s orbital below it to fill the extra space.
Example \(1\): Chromium
The electron configuration of chromium
Solution
Cr's electron configuration, following the model would be: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^23d^4\0, but instead it is \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^13d^5\), because there is extra stability gained from the half-filled d orbital.
Hund's Rule
Hund’s rule states that when filled sub-levels other than s orbital, electrons must not be spin paired in the orbitals until each orbital contains one electron, and no orbital can have two electrons with the same spin (ms).
Pauli Exclusion Principle
Pauli Exclusion Principle states that no two electrons can have the same quantum numbers. An orbital can only hold 0, 1, or 2 electrons. They must have opposite spins if there are 2 electrons in the orbittal.
Periodic Trend
Valence electron shells in the periodic table follow a trend. This can be referred to as the s block, the p block, the d block and the f block (lanthanides and actinides) meaning that, in its ground state, an element in a certain "block" will have its valence electrons in the s, p, d, or f orbitals depending.
How to Write Ground State Electron Configurations
Basics
Electron configurations are written using the principal quantum number n, followed by the orbital (s, p, d, or f) with the total number of electrons written as a superscript. Example: 1s2 For writing ground state electron configurations, a few main steps should be followed.
1. Find the amount of electrons in the atom. Example: Na: 11 e- Na+: 10 e-
2. Fill orbitals following the model until all electrons have been accounted for.
Example: Na: 11 e- 1s2 2s2 2p6 3s1 or Na+: 1s2 2s2 2p6
1. After that, it is important to check for a nearly half-filled or filled d orbital (d4 or d9) and adjust accordingly by removing an electron from the s orbital beneath it.
Example \(2\): Chromium
What is the electronic configuration of chromium
Solution
Cr: 1s2 2s2 2p6 3s2 3p64s23d4 half filled orbital, s orbital beneath it
1s2 2s2 2p6 3s2 3p6 4s13d5
Shorthand
Because writing the entire electron configuration can become cumbersome, there is a shorthand option. It is done by using the symbol of the noble gas in the period above the element to represent the electron configuration before it.
Example: Na: [Ne] 3s1
Problems
1. Write the expanded and shortened ground state electron configuration for Cl.
2. Write the expanded and shortened ground state electron configuration for Cr.
3. Write the expanded and shortened ground state electron configuration for Cu.
4. Write the expanded and shortened ground state electron configuration for Co2+
5. Write the ground state electron configuration for P3-
Solution 1. Expanded: 1s2 2s2 2p6 3s2 3p5
Shorthand: [Ne] 3s2 3p5
Solution 2. Expanded: 1s2 2s2 2p6 3s2 3p6 4s13d5
Shorthand: [Ar] 4s13d5
Solution 3. Expanded: 1s2 2s2 2p6 3s2 3p6 4s13d10
Shorthand: [Ar] 4s13d10
Solution 4. Expanded: 1s2 2s2 2p6 3s2 3p6 4s23d5
Shorthand: [Ar} 4s23d5
Solution 5. 1s2 2s2 2p6 3s2 3p6 | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/01%3A_Basic_Concepts-_Atoms/1.09%3A_The_Aufbau_Principle/1.9A%3A_Ground_State_Electronic_Configurations.txt |
The electrons of an atom are typically divided into two categories: valence and core electrons. Valence electrons occupy the outermost shell or highest energy level of an atom while core electrons are those occupying the innermost shell or lowest energy levels. This difference greatly influences the role of the two types of electrons in a chemical reaction. Generally, valence electrons can participate in the formation of chemical bonding, but core electrons cannot. While core electrons are not involved in bonding, they influence the chemical reactivity of an atom.
The electron configuration of a oxygen atom is
$\ce{O}: \,1s^22s^22p^4 \label{1}$
which may be shorted
$\ce{O}:\, [He]2s^2 2p^4 \label{2}$
where the $[He]$ stands for the configuration of helium ($1s^2$). Similarly, the configuration of calcium with 20 electrons can be written
$\ce{Ca}:\, [Ar]4s^2 \label{3}$
where the $[Ar]$ stands for the configuration of argon ($1s^22s^22p^6 3s^2 3p^6$). Electronic configurations that are the same as noble gases are very stable since they have a full octet (except helium with a full 1s orbital).
The $1s$ electrons in oxygen do not participate in bonding (i.e., chemistry) and are called core electrons. The valence electrons (i.e., the $2s^22p^4$ part) are valence electrons, which do participate in the making and breaking of bonds. Similarly, in calcium (Equation $\ref{3}$), the electrons in the argon-like closed shell are the core electrons and the the two electrons in the 4s orbital are valence electrons.
Example $1$: Cobalt
What are the core and valence electrons in cobalt?
Solution
Start by writing the electron configuration of cobalt with 27 electrons:
$1s^22s^22p^63s^23p^64s^23d^7 \nonumber$
However, argon has the electronic structure $1s^22s^22p^23s^23p^6$, so we can rewrite the configuration as
$[Ar]4s^23d^7 \nonumber$
The two electrons in the $4s$ orbital and the seven electrons in the $3d$ are the valence electrons: all others are core electrons.
The periodicity of valance electrons can be seen in the Periodic Table. Basically, the periodicity is only applied to the main group elements, while in transition metals, rules are complex.
The core electrons remain the same in the increase of group numbers in the main group elements. On the other hand, the valance electrons increase by one from left to right of a main period, and remain the same down the column of a main group. This evolution gives periodical change in property of a period, and similar chemical property of a group, which is called periodical trend. The number of valence electrons in a main period is the same as its group number. The table below shows this rule clearly.
Under construction
Figure 1: 1A + 2A are metals. 3A to 8A are non-metals.
However, this periodicity cannot be applied to the transition group, which is more complicated than that of the main group. Although the outermost electrons can be easily determined, the apparent valence electrons considered in chemical reactivity are complex and fluctuated. Electrons going into d sublevel can play either a role of valence electrons or shielding electrons. So there is not always a certain number of apparent valence electrons. The number of apparent valence electrons for the first transition metal period is shown in the table below.
Under construction
Figure 2: Valence electrons for transition metals.
Relationship with Chemical Reactivity
The chemical reactivity of an atom is mainly determined by valence electrons. Atoms which have a complete shell of valence electrons tend to be chemically inert. Atoms with one or two valence electrons are highly reactive. This phenomenon can be explained by Hund's rule, which states that orbitals that are empty, half-full, or full are more stable than those that are not. For example, Ne is chemically inert because it has two valence electrons that fill its outermost shell which makes it stable compared to atoms such as Al, which has three valence electrons, but its valence electrons does not fill its outermost shell.
Although core electrons do not take part in chemical bonding, they play a role in determining the chemical reactivity of an atom. This influence is generally due to the effect it has on valence electrons. The effect can be observed from the gradual change of chemical reactivity in a group. As you go down a group, more shells are occupied by electrons, which increases the size of the atom. The more core electron shells an atom has, the larger the size of the atom, and the farther the valence electrons are from the nucleus, thus the valence electrons will experience less effective nuclear charge and will be easily lost. For example, $\ce{Na}$ and $\ce{K}$ can both react with water, but K has a more radical reaction because it has more shells of core electrons which makes the valence electron in its outermost orbital much easier to lose than the valence electron of Na.
References
1. Miessler, Gary L., and Donald A. Tarr. Inorganic Chemistry. Upper Saddle River, NJ: Pearson Prentice Hall, 2010. Print.
2. Brown, Ian David. The Chemical Bond in Inorganic Chemistry the Bond Valence Model. Oxford: Oxford UP, 2006. Print. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/01%3A_Basic_Concepts-_Atoms/1.09%3A_The_Aufbau_Principle/1.9B%3A_Valence_and_Core_Electrons.txt |
Example of IE1 of Magnesium: Mg(g) -> Mg+(g) + e- I1 = 738 kJ/mol
IE1 stands for the first ionization energy: the energy the atom requires to expel the first electron from its orbital. Similarly, the second ionization energy, will be the energy needed to expel the second electron.
Mg+(g) -> Mg2+(g) + e- I2 = 1,451 kJ/mol
However, IE2 of Magnesium will be larger than that of IE1 because it is not energetically favorable to separate an electron from a positively charged ion.
The general pattern of the ionization energy as they are in regard to the period table is that the IE increases across a period, and decreases down a group. Because it requires more energy to remove an electron from a stable atom, the noble gases are usually associated with the highest IE1. Because their valence shells are already filled and stablized, they will require much more energy to disrupt that stability. The first electron that is expelled is the most loosely held to the atom.
On the other hand, the group 1 elements are usually associated with the lowest IE1. Since only one electron occupies the valence shell of these atoms, it will be more energetically favorable for them to lose the electron in order to achieve a full orbital shell.
However, there are few exceptions. The IE1 decreases when crossing from element in group 15 to the element in group 16. The group 15 has half-filled electronic configuration ns2 np3. This type of configuration is very stable; it’s hard to remove electron from valence shell. Therefore, element in group 15 requires greater value of IE1 than group 16. Another exception is that going from Be (group 2) to B (group 13), the IE1 decreases because Be has the filled shell 2s2 which is more stable than the electronic configuration of B 2s2 2p1. Hence, Be will require more IE1 than B. Similarly, the IE1 decreases when going from elements in group 12 to group 13
Electron Affinities
Electron affinity, often abbreviated as EA, is the energy released when an electron is added to a valence shell of the atom.
F(g) + e- -> F-(g) EA = -328 kJ/mol
[When an electron is added to an atom, energy is given off. This process is exothermic. ]
Atoms like the noble gases will not gain an electron because they are already in their most stable state with a full shell. Atoms like F will most likely gain an electron because when a free electron is added to the outer shell of fluorine, it will have obtained a full shell. Generally, atoms increasing across a period will increase in EA also.
Exothermic vs endothermic process
O(g) + e- -> O-(g) EA1= -141.0 kJ/mol
O-(g) + e- -> O2-(g) EA2 = +744kJ/mol
When an electron is added to an atom, the energy change is exothermic because of the attraction of the electron to the nucleus. However, in the case of EA2 where the electron is added to an anion, the repulsion between the anion and this newly added electron will overwhelm the attraction of the electron to the nucleus. Therefore, this process will be endothermic, as opposed to EA1.
Periodic Trend
The general trend of IE and EA along a periodic table.
Outside Links
• This is not meant for references used for constructing the module, but as secondary and unvetted information available at other site
• Link to outside sources. Wikipedia entries should probably be referenced here. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/01%3A_Basic_Concepts-_Atoms/1.10%3A_Ionization_Energies_and_Electron_Affinities.txt |
Lewis Structures are very similar to electron dot diagrams except for the fact that shared electrons between atoms are shown as lines. Lone pairs of electrons themselves are usually represented by dots around the atoms. These diagrams are helpful because they allow us to predict the shape of the molecule and see the positions of each atom relative to the others.
2.2D: Valence Bond model of bonding in (F 2) (O 2) and (N
The bonding in simple homonuclear molecules such as F2, O2, and N2 is quite similar in principle, but have fundamental differences in structure that can be described easily using Valence Bond models. The three molecules are comprised of atoms that are adjacent to one another in their period, and will thus have similar size and mass. The atoms do reside in different columns of the table though, and as such will have fundamentally different characteristics such as valence electron count and bond order in their respective diatomic molecules in which the elements are found naturally.
2.3D: MO Applied to (F 2) and (O 2)
Molecular orbitals (MO) are constructed from atomic orbitals. In O2 and F2, there is a crossover of the sigma and the pi ortbials: the relative energies of the sigma orbitals drop below that of the pi orbitals'. Information from the MO diagram justify O2's stability and show that it's bonding order is 2. The LUMO (lowest unoccupied molecular orbital) and HOMO (highest occupied molecular orbital) of difluoride's MO diagram help explain why the molecule is very stable - the diagram also tells us that the bond order is 1.
• Dieu Huynh
2.3b: MO theory of bonding in H
Note that there is a nodal plane in the anti-bonding MO.
Bond order
Bond order = 1/2 (#e- in bonding MO - #e- in antibonding MO)
For H2, bond order = 1/2 (2-0) = 1, which means H2has only one bond. The antibonding orbital is empty. Thus, H2 is a stable molecule.
Again, in the MO, there is no unpaired electron, so H2 is diamagnetic.
Problems
1. What does the MO of H2+ look like? What is its bond order? What is its magnetic property? Explain.
2. What does the MO of H2- look like? What is its bond order? What is its magnetic property? Explain.
3. Which one is the most stable: H2, H2+, or H2-? Why?
4. When a hydrogen atom accepts an electron, it becomes a hydride H-. Theoretically would it be possible to form a molecule from two hydrides, that is to form H22-? Why?
Answers:
1-
Bond order = 1/2 (1-0) = 1/2
Paramagnetic because it has one unpaired e- in the σ(1s) orbital.
2-
Bond order = 1/2 (2-1) = 1/2
Paramagnetic because it has one unpaired e- in the σ*(1s) orbital.
3- H2 is the most stable because it has the highest bond order (1), in comparison with the bond orders (1/2) of H2+ and H2-.
4- Theoretically it would not be possible to form a molecule from two hydrides because the anti-bonding and bonding orbitals would cancel each other out. So, the bond order is zero. Because the antibonding ortibal is filled, it destabilizes the structure, making the "molecule" H22- very non-stable.
Bond order = 1/2 (2-2) = 0 ---> no bond formation. Thus, this molecule doesn't exist. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/02%3A_Basic_Concepts-_Molecules/2.01%3A_Bonding_Models_-_An_Introduction/2.1B%3A_Lewis_Structures.txt |
Introduction
Trends across a period follow from the increasing number of protons in the nucleus and the decrease in radius. Both contributions can be explained by the change in effective nuclear charge.
Trends down a group follow from the increasing number of electron shells and the increased distance of the outer electrons from the nucleus. The major factor is the increasing size.
The properties of an element are largely determined by their electronic configurations, giving rise to recurring patterns or periodic behaviour. Examples are shown in the diagrams below including ionization energy, electron affinity, electronegativity and atomic radii. It is this periodicity of properties, manifestations of which were noticed well before the underlying theory was known, that led to the establishment of the periodic law (the properties of the elements recur at varying intervals) and the development of the first periodic tables. The modern periodic table is a tabular arrangement of the chemical elements, organized on the basis of their atomic number (number of protons in the nucleus), electronic configurations, and recurring chemical properties.
In the RSC Tutorial Chemistry Text on Main Group Chemistry, it notes that "When an element forms a chemical compound, electrons can be considered to be either lost, gained or shared with other atoms. These tendencies can be assessed by the parameters of ionization energy (IE), electron affinity (EA) and electronegativity (E). Prediction of bond types as either ionic or covalent then allows prediction of the chemical and physical properties of chemical substances."
So how are these parameters defined and how do they vary with atomic number?
Effective nuclear charge
The concept of the effective nuclear charge (often symbolized as Zeff or Z*) relates to the net positive charge experienced by an electron in a multi-electron atom. The term "effective" is used because the shielding effect of negatively charged inner electrons prevents higher orbital electrons from experiencing the full nuclear charge due to the repelling effect of the lower inner-layer electrons.
In an atom with one electron, that electron experiences the full charge of the positive nucleus. In this case, the effective nuclear charge can be calculated from Coulomb's law.
However, in an atom with many electrons the outer electrons are simultaneously attracted to the positive nucleus and repelled by the inner negatively charged electrons. The effective nuclear charge on such an electron is given by the following equation:
Zeff = Z - S
where
Z is the number of protons in the nucleus (atomic number), and
S is the shielding calculated from the electrons between the nucleus and the electron in question. A systematic method for determining this is given by "Slater's rules".
These can be summarised as follows:
Arrange the electrons into a sequence of groups in order of increasing principal quantum number n, and for equal n in order of increasing azimuthal quantum number l, except that s- and p- orbitals are kept together.
[1s] [2s,2p] [3s,3p] [3d] [4s,4p] [4d] [4f] [5s, 5p] [5d] etc.
Any electron higher in the sequence to the electron under consideration contributes nothing to the shielding, S, and is ignored.
For an electron in an ns or np orbital:
0.35 comes from each other electron within the same group except for the [1s] group, where the other electron contributes only 0.30.
0.85 for each electron with principal quantum number n one less than that of the group, i.e (n-1) shell
1.00 for each electron with principal quantum number two or more less, i.e (n-2) etc. shell
For an electron in an nd or nf orbital:
0.35 comes from each other electron within the same group
1.00 for each electron "closer" to the atom than the group. This includes electrons with the same principal quantum number but in s or p orbitals.
In tabular form, the rules are summarized as:
Group Other electrons in the same group Electrons in group(s) with principal quantum number n and azimuthal quantum number < l Electrons in group(s) with principal quantum number n-1 Electrons in all group(s) with principal quantum number < n-1
[1s] 0.30 - - -
[ns, np] 0.35 - 0.85 1
[nd] or [nf] 0.35 1 1 1
Example 1.
Consider a sodium cation, Na+, a fluorine anion, F-, and a neutral neon atom, Ne. Each has 10 electrons, 1s2 2s2 2p6 so the shielding from the 1s and 2s/2p electrons is 2 * 0.85 + 7 * 0.35 = 4.15 but the effective nuclear charge varies because each has a different atomic number:
Zeff F- = 9 - 4.15 = 4.85
Zeff Ne = 10 - 4.15 = 5.85
Zeff Na+ = 11 - 4.15 = 6.85
So the sodium cation has the largest effective nuclear charge, and can be expected to have the smallest radius.
Example 2.
Predict whether K would be more energetically stable with a configuration of
1s2 2s2 2p6 3s2 3p6 4s1 or 1s2 2s2 2p6 3s2 3p6 3d1
For K, Z=19 and considering the 4s electron then the screening constant S can be calculated from:
S= (8 * 0.85) + (10 * 1.0) = 16.8
Zeff = 19 - 16.8 = 2.2
For the 3d calculation of S:
S= (18 * 1.0) = 18
Zeff = 19 - 18 = 1
Accordingly, an electron in the 4s (as opposed to the 3d) orbital would come under the influence of a greater effective nuclear charge in the ground state of potassium and so will be the orbital that is occupied.
Ionization Energy
The Ionization Energy (IEn) of an element is defined as the internal energy change associated with the removal of an electron from the gaseous atom, E, in its ground state, i.e. at 0 K. The first IE is therefore the energy required for the reaction:
E(g) → E+(g) + e- energy required = IE
This energy change is generally considered equivalent to the enthalpy change at 298 K (ΔH298 K). Estimates of the error suggest < 10 kJmol-1 which when compared to typical IE values often in their thousands, is insignificant.
The diagrams above show the variation in the values of the 1st five IE's as a function of Z up to Nd (60).
Features to note for IE1 are:
• the values associated with the noble gases are the highest in each period
• the values associated with the group 1 elements are generally the lowest in each period, (group 2 elements for the 2nd IE and group 3 for the 3rd IE etc.)
• the gradual increase in values across a given period (applies to IE2-5 as well)
• the drop in values on going from an element in group 15 to its neighbour in group 16 e.g. for N-O, P-S, As-Se
• the drop in values on going from an element in group 2 or 12 to its neighbour in group 13, e.g. for Be-B, Mg-Al and Zn-Ga, Cd-In
• the rather similar values for a given row of d-block elements
These observations can be accounted for in terms of the effective nuclear charge since the further away from the positively charged nucleus that a negatively charged electron is located, the less strongly that electron is attracted to the nucleus and the more easily it can be removed. So, as the atomic radius decreases from left to right across the Period the 1st Ionization Energy increases.
Electron Affinity
The electron affinity(EA) of an element E is defined as minus the internal energy change associated with the gain of an electron by a gaseous atom, at 0 K :
$E_{(g)} + e^- → E^-_{(g)}$
Unlike ionization energies, which are always positive for a neutral atom because energy is required to remove an electron, electron affinities can be positive (energy is released when an electron is added), negative (energy must be added to the system to produce an anion), or zero (the process is energetically neutral).
Chlorine has the most positive electron affinity of any element, which means that more energy is released when an electron is added to a gaseous chlorine atom than to an atom of any other element, EA= 348.6 kJmol-1 and the Group 17 elements have the largest values overall. The addition of a second electron to an element is expected to be much less favored since there will be repulsion between the negatively charged electron and the overall negatively charged anion. For example, for O the values are:
$O_{(g)} + e^ \rightarrow O^-_{(g)} \;\;\;\; EA = +141\; kJmol^{-1}$
$O^-_{(g)} + e^- → O^{2-}){(g)} \;\;\;\; EA = -798\; kJmol^{-1}$
Electronegativity
The concept of Electronegativity originated with Linus Pauling in the 1930's and was defined as "the power of an atom in a molecule to attract electrons to itself".
The values proposed by Pauling were calculated based on differences in bond dissociation enthalpy values found when comparing homo-diatomic molecues with hetero-diatomic molecules. For example , the bond energy of chlorine monofluoride, ClF, is about 255 kJ mol-1 which is significantly greater than for either of the two homo-nuclear species Cl2 and F2 (242 and 153 kJ mol-1 respectively). Pauling attributed this to an electrostatic attraction between the partially charged atoms in the heternuclear species. That is the excess bond energy came from an ionic contribution to the bond.
The method of calculating the Pauling values is:
D(XY) = [D(XX).D(YY)]1/2 + 96.48 * (χY - χX)2 where the 96.48 factor means D values are in kJ mol-1
In Housecroft and Sharpe the average, rather than geometric mean is used, and this is rearranged to give:
ΔD = D(XY) - {½ * [D(XX) - D(YY)] } = (χY - χX)2 = (Δχ)2
or Δχ = √(ΔD)
As only differences in electronegativity were defined, it was necessary to choose an arbitrary reference point in order to construct a scale. Hydrogen was chosen as the reference, since it formed covalent bonds with a large variety of elements: its electronegativity was fixed at 2.20.
The Mulliken scale was calculated by taking the average of the Ionization Energy and the Electron Affinity (when both were given in units of eV).
χM = ½ (IE1 + EA1) where both IE1 and EA1 are in eV
A variant of this (2006) that converts the values to roughly the Pauling scale is:
χM = 0.00197 * (IE1 + EA1) + 0.19 where IE1 and EA1 are now given in kJ mol-1
The plots above indicate that while the absolute values are different, the trends are quite similar and the 2 curves are comparable when scaled appropriately.
The Allred-Rochow scale considered that electronegativity was related to the charge experienced by an electron on the "surface" of an atom: the higher the charge per unit area of atomic surface the greater the tendency of that atom to attract electrons. Their scale was dependent on Zeff and inversely proportional to the square of the covalent radius, rcov.
χAR = (3590 * Zeff / r2cov) + 0.744 where rcov is in pm
The values range between 0 and 10. Once again a good correlation to the Pauling scale was found and this applies as well to other Electronegativity scales.
Atomic Radius
Have a look at an interactive visual display (JSmol) showing the periodic table of elements with atomic and ionic radii.
The atomic radius of a chemical element is a measure of the size of its atoms, usually the mean or typical distance from the center of the nucleus to the boundary of the surrounding cloud of electrons. Since the boundary is not a well-defined physical entity, there are various non-equivalent definitions of atomic radius. Three widely used definitions of atomic radius are Van der Waals radius, ionic radius, and covalent radius.
Covalent radius is defined as half the covalent bond length when the two atoms bonded are homonuclear (½ X-X bond).
van der Waals radius is defined as half of the internuclear separation of two non-bonded atoms of the same element on their closest possible approach.
It is not possible to measure the sizes of both metallic and nonmetallic elements using a single technique and method. To get values for comparison, theoretical quantum mechanical functions have been used instead to calculate atomic radii.
In the periodic table, atomic radii decrease from left to right across a period and increase from top to bottom down the groups. As a result of these two trends, the largest atoms are found in the lower left corner of the periodic table, and the smallest are found in the upper right corner.
The radius increases sharply between the noble gas at the end of each period and the alkali metal at the beginning of the next period. These trends of the atomic radii (and of various other chemical and physical properties of the elements) can be explained by the electron shell theory of the atom; they provided important evidence for the development and confirmation of quantum theory. The atomic radii decrease across the Periodic Table because as the atomic number increases, the number of protons increases across the period, but the extra electrons are only added to the same quantum shell. Therefore, the effective nuclear charge towards the outermost electrons increases, drawing the outermost electrons closer. As a result, the electron cloud contracts and the atomic radii decreases.
The Lanthanide Contraction
The chart on the right above can be used to explain why Zirconium and Hafnium are two of the hardest elements in the Periodic Table to separate. In addition why the teaching of Transition Metal Chemistry is often covered in 2 courses since the properties of the first row elements are expected to be quite different to those of the second and third row.
Considering that the size of Gallium is smaller than Aluminium suggests that the 3d contraction is having an impact as well.
Ionic Radius
Although neither atoms nor ions have sharp boundaries, they are sometimes treated as if they were hard spheres with radii such that the sum of ionic radii of the cation and anion gives the distance between the ions in a crystal lattice.
Ions may be larger or smaller than the neutral atom, depending on the ion's charge. When an atom loses an electron to form a cation, the lost electron no longer contributes to shielding the other electrons from the charge of the nucleus; consequently, the other electrons are more strongly attracted to the nucleus, and the radius of the atom gets smaller. Similarly, when an electron is added to an atom, forming an anion, the added electron shields the other electrons from the nucleus, with the result that the size of the atom increases. Typical values range from 50 pm to over 220 pm.
Atomic and Ionic Radius Retrieved 24 November 2014
Return to the course outline or move on to Lecture 2: Polarizing power and polarizability, Anomalous behavior of row 2 elements.
2.4B: Isoelectronic Molecules
Isoelectronic Species Periodic Trends
The observation that isoelectronic species are usually isostructural, first made by Penny and Southerland in 1936, known as the isoelectronic principle (Geoff). Table 1 shows an example of isostructural isoelectronic species periodic trends. All of these molecules are octahedral and isoelectronic within their periods.
Isostructural Isoelectronic Species in Differeing Groups and Periods
Table 1 Group 13 Group 14 Group 15 Group 16 Group 17
Period 3 AlF63- SiF62- PF6- SF6 ClF6+
Period 4 GaF63- GeF62- AsF6- SeF6 BrF6+
Period 5 InF63- SnF62- SbF6- TeF6 IF6+
Other interesting trends appear in the periodic table including:
• Isoelectronic matricies--all isoelectronic species in a matrix defined by total electrons and valence electrons vary by progression in group number. For example, a 14 electron/10 valence electron diatomic matrix would have molecules such as CN-, CO, and N2.
• Isoelectronic arrays--an atom is replaced with another which alters the charge and continues the isoelectronic relationship. For example: BeF42-, BF4-, and CF4.
Problems
1) Why is it important to understand the term "isoelectronic" and its trends throughout the periodic table?
2) Which isoelectronic species has the largest radius?
a. Mg2+ b.N3- c. O2- d. F-
3) Name 3 isostructural molecules
Answers:
1) Isoelectronic can help predict chemical reactions and interactions between molecules.
2) N3-
3) AlF63-, SiF62-, PF6- | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/02%3A_Basic_Concepts-_Molecules/2.04%3A_The_Octet_Rule_and_Isoelectronic_Species/2.4.D%3A_Periodic_Trends.txt |
There are a few different 'types' of electronegativity which differ only in their definitions and the system by which they assign values for electronegativity. For example, there is Mulliken electronegativity which is defined as "the average of the ionization energy and electron affinity of an atom"3, which as we will see, differs slightly from Pauling's definition of electronegativity.
2.05: Electronegativity Values
Linus Pauling described electronegativity as “the power of an atom in a molecule to attract electrons to itself.”1 Basically, the electronegativity of an atom is a relative value of that atom's ability to attract election density toward itself when it bonds to another atom. The higher the electronegativity of an element, the more that atom will attempt to pull electrons towards itself and away from any atom it bonds to. The main properties of an atom dictate it's electronegativity are it's atomic number as well as its atomic radius. The trend for electronegativity is to increase as you move from left to right and bottom to top across the periodic table. This means that the most electronegative atom is Fluorine and the least electronegative is Francium.
There are a few different 'types' of electronegativity which differ only in their definitions and the system by which they assign values for electronegativity. For example, Mulliken electronegativity defines electronegativity as the "the average of the ionization energy and electron affinity of an atom."3 As we will see, this definition differs slightly from Pauling's definition of electronegativity.
Pauling Electronegativity
Linus Pauling was the original scientist to describe the phenomena of electronegativity. The best way to describe his method is to look at a hypothetical molecule that we will call XY. By comparing the measured X-Y bond energy with the theoretical X-Y bond energy (computed as the average of the X-X bond energy and the Y-Y bond energy), we can describe the relative affinities of these two atoms with respect to each other.
$Δ\text{Bond Energies} = (X-Y)_{measured} – (X-Y)_{expected} \nonumber$
If the electonegativities of X and Y are the same, then we would expect the measured bond energy to equal the theoretical (expected) bond energy and therefore the Δ bond energies would be zero. If the electronegativities of these atoms are not the same, we would see a polar molecule where one atom would start to pull electron density toward itself, causing it to become partially negative.
By doing some careful experiments and calculations, Pauling came up with a slightly more sophisticated equation for the relative electronegativities of two atoms in a molecule:
$EN(X) - EN(Y) = 0.102 \sqrt{Δ}.\nonumber$
In that equation, the factor 0.102 is simply a conversion factor between kJ and eV to keep the units consistent with bond energies.
By assigning a value of 4.0 to Fluorine (the most electronegative element), Pauling was able to set up relative values for all of the elements. This was when he first noticed the trend that the electronegativity of an atom was determined by it's position on the periodic table and that the electronegativity tended to increase as you moved left to right and bottom to top along the table. The range of values for Pauling's scale of electronegativity ranges from Fluorine (most electronegative = 4.0) to Francium (least electronegative = 0.7). 2 Furthermore, if the electronegativity difference between two atoms is very large, then the bond type tends to be more ionic, however if the difference in electronegativity is small then it is a nonpolar covalent bond.
Exercise $1$
Explain the difference between Electronegativity and Electron Affinity
Exercise $2$
Predict the order or increasing electronegativity from the following elements
1. F, Li, C, O
2. Te, Cl, S, Se
3. Cs, At, Tl, I | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/02%3A_Basic_Concepts-_Molecules/2.05%3A_Electronegativity_Values/2.5A%3A_Pauling_Electronegativity_Values.txt |
A method for estimating electronegativity was developed by Robert Mulliken (1896–1986; Nobel Prize in Chemistry 1966) who noticed that elements with large first ionization energies tend to have very negative electron affinities and gain electrons in chemical reactions. Conversely, elements with small first ionization energies tend to have slightly negative (or even positive) electron affinities and lose electrons in chemical reactions. Mulliken recognized that an atom’s tendency to gain or lose electrons could therefore be described quantitatively by the average of the values of its first ionization energy and the absolute value of its electron affinity.
Robert S. Mulliken proposed that the arithmetic mean of the first ionization energy ($E_{I_1}$) and the electron affinity ($E_{ea}$) should be a measure of the tendency of an atom to attract electrons. As this definition is not dependent on an arbitrary relative scale, it has also been termed absolute electronegativity. Using our definition of electron affinity, we can write Mulliken’s original expression for electronegativity as follows:Mulliken’s definition used the magnitude of the ionization energy and the electron affinity. By definition, the magnitude of a quantity is a positive number. Our definition of electron affinity produces negative values for the electron affinity for most elements, so vertical lines indicating absolute value are needed in Equation $\ref{1}$ to make sure that we are adding two positive numbers in the numerator.
$\chi = \dfrac{|E_{I_1} + E_{ea}|}{2} \label{1}$
Elements with a large first ionization energy and a very negative electron affinity have a large positive value in the numerator of Equation $\ref{1}$, so their electronegativity is high. Elements with a small first ionization energy and a small electron affinity have a small positive value for the numerator in Equation $\ref{1}$, so they have a low electronegativity. Inserting the appropriate data into Equation $\ref{1}$ gives a Mulliken electronegativity value for fluorine of 1004.6 kJ/mol. To compare Mulliken’s electronegativity values with those obtained by Pauling, Mulliken’s values are divided by 252.4 kJ/mol, which gives Pauling’s value (3.98).
However, it is more usual to use a linear transformation to transform these absolute values into values that resemble the more familiar Pauling values. For ionization energies and electron affinities in electronvolts:
$\chi_{Mulliken} = 0.187 (E_{I_1}+E_{ea})+0.17 \label{2}$
and for energies in kJ/mol,
$\chi_{Mulliken} = (1.97 \times 10^{-3})(E_{I_1}+E_{ea})+0.19 \label{3}$
The Mulliken electronegativity can only be calculated for an element for which the electron affinity is known, fifty-seven elements as of 2006. The Mulliken electronegativity of an atom is sometimes said to be the negative of the chemical potential. By inserting the energetic definitions of the ionization potential and electron affinity into the Mulliken electronegativity, it is possible to show that the Mulliken chemical potential is a finite difference approximation of the electronic energy with respect to the number of electrons., i.e.,
$\mu_{Mulliken}= -\chi_{Mulliken} = -\dfrac{E_{I_1} + E_{ea}}{2} \label{4}$
All electronegativity scales give essentially the same results for one element relative to another. Even though the Mulliken scale is based on the properties of individual atoms and the Pauling scale is based on the properties of atoms in molecules, they both apparently measure the same basic property of an element. In the following discussion, we will focus on the relationship between electronegativity and the tendency of atoms to form positive or negative ions. We will therefore be implicitly using the Mulliken definition of electronegativity. Because of the parallels between the Mulliken and Pauling definitions, however, the conclusions are likely to apply to atoms in molecules as well.
Significance
Despite being developed from a very different set of principles than Pauling Electronegativity, which is based on bond dissociation energies, there is a good correlation between Mullikin and Pauling Electronegativities for the atoms, as shown in the plot below.
Although Pauling electronegativities are usually what are found in textbooks, the Mullikin electronegativity more intuitively corresponds to the "ability of an atom to draw electrons toward itself in bonding," and is probably a better indicator of that property. However, because of the good correlation between the two scales, using the Pauling scale is sufficient for most purposes. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/02%3A_Basic_Concepts-_Molecules/2.05%3A_Electronegativity_Values/2.5B%3A_Mulliken_Electronegativity_Values.txt |
Allred-Rochow Electronegativity is a measure that determines the values of the electrostatic force exerted by the effective nuclear charge on the valence electrons. The value of the effective nuclear charges is estimated from Slater's rules. The higher charge, the more likely it will attract electrons. Although, Slater's rule are partly empirical. So the Allred-Rochow electronegativity is no more rigid than the Pauling Electronegativity.
Electronegativity
Pauling established Electronegativity as the "power" of an atom in a molecule to attract electron to itself. It is a measure of the atom's ability to attract electron to itself while the electron is still attached to another atom. The higher the values, the more likely that atom can pull electron from another atom and into itself. Electronegativity correlates with bond polarity, ionization energy, electron affinity, effective nuclear charge, and atomic size.
Table 1: Pauling Electronegativity Values
H
2.1
Li
1.0
Be
1.6
B
2.0
C
2.50
N
3.0
O
3.5
F
4.0
Na
0.9
Mg
1.3
Al
1.6
Si
1.9
P
2.2
S
2.5
Cl
3.0
K
0.8
Ca
1.3
Sc
1.4
Ti
1.5
V
1.6
Cr
1.7
Mn
1.6
Fe
1.8
Co
1.9
Ni
1.9
Cu
1.9
Zn
1.7
Ga
1.6
Ge
2.0
As
2.2
Se
2.6
Br
2.8
Rb
0.8
Sr
1.0
Y
1.2
Zr
1.3
Nb
1.6
Mo
2.2
Te
2.1
Ru
2.2
Rh
2.3
Pd
2.2
Ag
1.9
Cd
1.7
In
1.8
Sn
2.0
Sb
2.1
Te
2.1
I
2.7
Cs
0.8
Ba
0.9
La
1.1
Hf
1.3
Ta
1.5
W
1.7
Re
1.9
Os
2.2
Ir
2.2
Pt
2.2
Au
2.4
Hg
1.9
Tl
2.0
Pb
2.3
Bi
2.0
Po
2.0
At
2.2
The periodic trend for electronegativity generally increases from left to right and decreases as it go down the group. The exception are Hydrogen and the noble gases because the noble gases are content with their filled outermost shells, and hydrogen cannot bear to lose a valence electron unlike the rest of the group 1 metals. The elements in the halogen group usually have the highest electronegativity values because they only need to attract one valence electron to complete the octet in their outer shell. Whereas the group 1 elements except for Hydrogen, are willing to give up their only valence electron so they can fulfill having a complete, filled outer shell.
Slater's rules
Slater's rules are rules that provides the values for the effective nuclear charge concept, or $Z_{eff}$. These rules are based on experimental data for electron promotion and ionization energies, and $Z_{eff}$ is determined from this equation:
$Z_{eff} = Z - S \label{A}$
where
• $Z$ is the nuclear charge,
• $Z_{eff}$ is the effective nuclear charge, and
• $S$ is the shielding constant
Through this equation, this tells us that electron may get reduced nuclear charge due to high shielding. Allred and Rochow used $Z_{eff}$ because it is accurate due to the involvement of shielding that prevents electron to reach its true nuclear charge: $Z$. When an atom with filled s-shell attracts electrons, those electrons will go to the unfilled p-orbital. Since the electrons have the same negative charge, they will not only repel each other, but also repel the electrons from the filled s-shell. This creates a shielding effect where the inner core electrons will shield the outer core electrons from the nucleus. Not only would the outer core electrons experience effective nuclear charge, but it will make them easily removed from the outer shell. Thus, It is easier for outer electrons to penetrate the p shell, which has little likelihood of being near the nuclear, rather than the s shell. Consider this, each of the outer electron in the (ns, np) group contributes S = 0.35, S = 0.85 in the (n - 1) shell, and S = 1.00 in the (n - 2) or lower shells.
Example 1: Slater's Rules
What is the $Z_{eff}$ for the 4s electrons in Ca.
Solution
Since $\ce{Ca}$ has atomic number of 20, $Z = 20$.
Then, we find the electron configuration for Ca, which is 1s22s22p63s23p64s2.
Now we got that, we can use Slater's rules:
\begin{align*} Z_{eff} &= Z - S \[4pt] &= 20 - ((8\times 0.85) + (10 \times 1.00)) \[4pt] &= 3.2 \end{align*}
So, Ca has a $Z_{eff}$ of 3.2.
Allred-Rochow Electronegativity
Allred and Rochow were two chemists who came up with the Allred-Rochow Electronegativity values by taking the electrostatic force exerted by effective nuclear charge, Zeff, on the valence electron. To do so, they came up with an equation:
$\chi^{AR} = \left(\dfrac{3590 \times Z_{eff}}{r^2_{cov}}\right) + 0.744 \label{1}$
At the time, the values for the covalent radius, $r_{cov}$, were inaccurate. Allred and Rochow added certain perimeters so that it would more closely correspond to Pauling's electronegativity scale.
Table 2: Allred-Rochow Electronegativity Values
H
2.20
Li
0.97
Be
1.47
B
2.01
C
2.50
N
3.07
O
3.50
F
4.10
Na
1.01
Mg
1.23
Al
1.47
Si
1.74
P
2.06
S
2.44
Cl
2.83
K
0.91
Ca
1.04
Sc
1.20
Ti
1.32
V
1.45
Cr
1.56
Mn
1.60
Fe
1.64
Co
1.70
Ni
1.75
Cu
1.75
Zn
1.66
Ga
1.82
Ge
2.02
As
2.20
Se
2.48
Br
2.74
Rb
0.89
Sr
0.99
Y
1.11
Zr
1.22
Nb
1.23
Mo
1.30
Te
1.36
Ru
1.42
Rh
1.45
Pd
1.35
Ag
1.42
Cd
1.46
In
1.49
Sn
1.72
Sb
1.82
Te
2.01
I
2.21
Cs
0.86
Ba
0.97
La
1.08
Hf
1.23
Ta
1.33
W
1.40
Re
1.46
Os
1.52
Ir
1.55
Pt
1.44
Au
1.42
Hg
1.44
Tl
1.44
Pb
1.55
Bi
1.67
Po
1.76
At
1.90
In this table, the electronegativities increases from left to right just like Pauling's scale because the $Z$ is increasing. As we go down the group, it decreases because of the larger atomic size that increases the distance between the electrons and nucleus.
Problems
1. From lowest to highest, order the elements in terms of Zeff: Ni, Cu, Zn, Ga, Ge
2. Using the equations above, find the Zeff for the Br by using Slater's rules.
3. Using the equations above, Find the Xar for Br. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/02%3A_Basic_Concepts-_Molecules/2.05%3A_Electronegativity_Values/2.5C%3A_Allred-Rochow_Electronegativity_Values.txt |
Molecular geometry and coordinates
Consider a diatomic molecule AB. Imagine fixing this molecule at a very specific spatial location, as shown below:
To locate the molecule so specifically, we would need to give the $x$, $y$, and $z$ coordinates of each of its atoms, i.e.,
\begin{align*}(x_A ,y_A ,z_A) &= r_A\ (x_B ,y_B ,z_B) &= r_B\end{align*}
which is a total of 6 numbers.
However, we note that the molecule looks the same, no matter where in space it is located. This is called translational invariance, and it implies that we can give only the coordinates of one of the atoms relative to the other, which is equivalent to giving the vector difference between $r_B$ and $r_A$ (or vice-versa), which we will call the vector $r$:
$r=r_B -r_A=(x,y,z)$
which is only 3 numbers. This is the same as arbitrarily placing atom A at the origin of our $xyz$ coordinate system.
We also note that the spatial orientation of the molecule is arbitrary, since the molecule looks the same at any viewing angle. For a diatomic, its orientation can be specified by giving two angles: the angle it makes with the z-axis and the angle of its projection onto the $xy$ plane with the x-axis. The choice of these angles is arbitrary. This leaves only 1 number left, which is the distance between A and B, called the molecule's bond length.
\begin{align*}r &= |r|=\sqrt{x^2 +y^2 +z^2}\ &= |r_b - r_A| = \sqrt{(x_B -x_A)^2 +(y_B -y_A)^2 +(z_B -z_A)^2}\end{align*}
This is an internal degree of freedom and is the only important number we need to give in order to convey the geometry of the diatomic.
In spite of this simplification, it is often necessary to specify all of the coordinates of the atoms in a molecule. Molecular modeling packages, which are becoming increasingly important in chemical research, require a full set of coordinates for each atom as input. Similarly, molecular data banks, such as the protein data bank (PDB) will give molecular structures as files of $x$, $y$, and $z$ coordinates. Thus, being able to determine a set of coordinates given only bond lengths and bond angles, and conversely being able to determine bond lengths and angles from a set of coordinates is an extremely important skill. A few examples of how to do this will be illustrated below.
Example 1
The diatomic $AB$. How do we determine a set of coordinates for $AB$ given only its bond length $r$. Since its absolute location in space and its orientation are arbitrary, any set of coordinates that reproduces the correct bond length will suffice. Thus, since a diatomic is linear, we may place it along one of the axes of our coordinate system with one of the atoms at the origin:
Now we see that the coordinates of atom $A$ will simply be
$r_A =(0,0,0)$
and the coordinates of atom $B$, since $B$ lies on the x-axis a distance $r$ away from $A$, will be
$r_B =(r,0,0)$
Clearly, this set of coordinates reproduces the correct bond length:
$|r_B -r_A|=\sqrt{r^2 +0^2 +0^2}=r$
Thus, for the molecule $HCl$, whose bond length is $r=1.284 \ \stackrel{\circ}{A}$, a set of coordinates could be
\begin{align*}r_H &= (0,0,0)\ r_Cl &= (1.284,0,0)\end{align*}
in $\stackrel{\circ}{A}$. This is just one possibility. We could have chosen either atom to be at the origin (or anywhere else in space for that matter), and chosen the bond to lie along any axis (or not along any particular axis), as we choose, so long as the correct bond length is reproduced.
Example 2
Water, $H_2 O$: The geometry of water is bent (we will see how to determine this later), with a bond angle of $104.5^\circ$ and an $OH$ bond length of approximately $1.0 \ \stackrel{\circ}{A}$.
To determine a set of coordinates for $H_2 O$, we note that the molecule is planar, so we may choose it to lie in the $xy$ plane. We will place the oxygen at the origin with the hydrogens as shown below:
The coordinates of the oxygen can be written down immediately:
$r_{O}=(0,0,0)$
For each of the hydrogens, note that the y-axis bisects the angle, giving two right triangles. An $OH$ bond forms the hypotenuse of one of these triangles, so that the $x$ and $y$ coordinates will be determined from the sine and cosine of the angle $\theta /2$, as can be shown using simple trigonometry:
\begin{align*}r_{H_1} &= (d_{OH}sin\theta /2,d_{OH}cos\theta /2,0)=(0.7907,0.6122,0)\ r_{H_2} &= (-d_{OH}sin\theta /2,d_{OH}cos\theta /2,0)=(-0.7907,0.6122,0) \end{align*}
To verify that the bond lengths are correctly reproduced, we compute the magnitudes of the vector differences $r_{H_1}-r_O$ and $r_{H_2}-r_O$:
\begin{align*}|r_{H_1}-r_O| &= \sqrt{(0.7907)^2 +(0.6122)^2 +0^2}=1.0 \stackrel{\circ}{A}\ |r_{H_2}-r_O| &= \sqrt{(0.7907)^2 +(0.6122)^2 +0^2}=1.0 \stackrel{\circ}{A}\end{align*}
In order to verify that the bond angle is correct, we note that the angle between two vectors $a$ and $b$ is given by the formula:
$\theta_{ab}=cos^{-1}\left ( \frac{a\cdot b}{|a||b|} \right )$
where $a\cdot b$ is the dot product of $a$ and $b$ defined to be
$a\cdot b=a_x b_x +a_y b_y +a_z b_z$
Thus, the $H_1 -O-H_2$ angle is given by
\begin{align*}\theta_{H_1 - O-H_2} &= cos^{-1}\left ( \frac{(r_{H_1}-r_O)\cdot (r_{H_2}-r_O)}{|r_{H_1}-r_O||r_{H_2}-r_O|} \right )\ &= cos^{-1}\left ( \frac{-(0.7907)^2+(0.6122)^2}{1.0*1.0} \right )\ &= cos^{-1}(-0.2504)=104.5^\circ\end{align*}
Within a group of the periodic table, bond lengths tend to increase with increasing atomic number $Z$. Consider the Group 17 elements:
\begin{align*}& F_2 \;\;\;\; d=141.7 \;pm\ & Cl_2 \;\;\;\; d=199.1 \, pm \ & Br_2 \;\;\;\; d=228.6 \, pm\ & I_2 \;\;\;\; d=266.9 \, pm\end{align*}
which corresponds to an increased valence shell size, hence increased electron-electron repulsion. An important result from experiment, which has been corroborated by theory, is that bond lengths tend not to vary much from molecule to molecule. Thus, a $CH$ bond will have roughly the same value in methane, $CH_4$ as it will in aspirin, $C_9 H_8 O_4$.
Bond dissociation energies were discussed in the last lecture in the context of ionic bonds. There we used the symbol $\Delta E_d$ measured in $kJ/mol$. This measures the energy required to break a mole of a particular kind of bond. A similar periodic trend exists for bond dissociation energies. Consider the hydrogen halides:
\begin{align*} & HF \;\;\;\; \Delta E_d =565 \ kJ/mol \;\;\;\; d= 0.926 \ \, pm\ & HCl \;\;\;\; \Delta E_d =429 \ kJ/mol \;\;\;\; d= 128.4 \ \, pm\ & HBr \;\;\;\; \Delta E_d =363 \ kJ/mol \;\;\;\; d= 142.4 \ \, pm\ & HI \;\;\;\; \Delta E_d =295 \ kJ/mol \;\;\;\; d= 162.0 \ \, pm \end{align*}
Thus, as bond lengths increase with increasing $Z$, there is a corresponding decrease in the bond dissociation energy.
$CC$ bonds are an exception to the the rule of constancy of bond lengths across different molecules. Because $CC$ bonds can be single, double, or triple bonds, some differences can occur. For example, consider the $CC$ bond in the molecules ethane $(C_2 H_6)$, ethylene $(C_2 H_4)$ and acetylene $(C_2 H_2)$:
\begin{align*} & C_2 H_6 \;\;\;\; (single)\;\;\;\; d=1.536 \ \stackrel{\circ}{A}\;\;\;\; \Delta E_d=345 \ kJ/mol\ & C_2 H_4 \;\;\;\; (double)\;\;\;\; d=133.7 \, pm\;\;\;\; \Delta E_d=612 \ kJ/mol\ & C_2 H_2 \;\;\;\; (tirple)\;\;\;\; d=126.4 \, pm\;\;\;\; \Delta E_d=809 \ kJ/mol\end{align*}
The greater the bond order, i.e., number of shared electron pairs, the greater the dissociation energy. The same will be true for any kind of bond that can come in such different flavors'', e.g., $NN$ bonds, $OO$ bonds, $NO$ bonds, $CO$ bonds, etc.
Polar covalent bonds
Most real chemical bonds in nature are neither truly covalent nor truly ionic. Only homonuclear bonds are truly covalent, and nearly perfect ionic bonds can form between group I and group VII elements, for example, KF. Generally, however, bonds are partially covalent and partially ionic, meaning that there is partial transfer of electrons between atoms and partial sharing of electrons.
In order to quantify how much ionic character (and how much covalent character) a bond possesses, electronegativity differences between the atoms in the bond can be used. We have already seen one method for estimating atomic electronegativities, the Mulliken method. In 1936, Linus Pauling came up with another method that forms the basis of our understanding of electronegativity today.
Pauling's method
Recall the Mulliken's method was based on the arithmetic average of the first ionization energy $IE_1$ and the electron affinity $EA$. Both of these energies are properties of individual atoms, hence this method is appealing in its simplicity. However, there is no information about bonding in the Mulliken method. Pauling's method includes such information, and hence is a more effective approach.
To see how the Pauling method works, consider a diatomic $AB$, which is polar covalent. Let $\Delta E_{AA}$ and $\Delta E_{BB}$ be the dissociation energies of the diatomics $A_2$ and $B_2$, respectively. Since $A_2$ and $B_2$ are purely covalent bonds, these two dissociation energies can be used to estimate the pure covalent contribution to the bond $AB$. Pauling proposed the geometric mean of $\Delta E_{AA}$ and $\Delta E_{BB}$, this being more sensitive to large differences between these energies than the arithmetic average:
$pure \ covalent \ contribution=\sqrt{\Delta E_{AA} \Delta E_{BB}}$
If $\Delta E_{AB}$ is the true bond dissociation energy, then the difference
$\Delta E_{AB}-\sqrt{\Delta E_{AA} \Delta E_{BB}}$
is a measure of the ionic contribution. Let us define this difference to be $\Delta$:
$\Delta =\Delta E_{AB}-\sqrt{\Delta E_{AA} \Delta E_{BB}}$
Then Pauling defined the electronegativity difference $\chi_A -\chi_B$ between atoms $A$ and $B$ to be
$\chi_A -\chi_B = 0.102 \sqrt{\Delta}$
where $\Delta$ is measured in $kJ/mol$, and the constant $0.102$ has units $mol^{1/2} /kJ^{1/2}$, so that the electronegativity difference is dimensionless. Thus, with some extra input information, he was able to generate a table of atomic electronegativities that are still used today (Table A2).
To use the electronegativities to estimate degree of ionic character, simply compute the absolute value of the difference for the two atoms in the bond. As an example, consider again the hydrogen halides:
\begin{align*} & HF \;\;\;\; |\chi_F -\chi_H|=1.78\ & HCl \;\;\;\; |\chi_{Cl} -\chi_H|=0.96\ & HBr \;\;\;\; |\chi_{Br} -\chi_H|=0.76\ & HI \;\;\;\; |\chi_I -\chi_H|=0.46\end{align*}
As the electronegativity difference decreases, so does the ionic character of the bond. Hence its covalent character increases.
Electric dipole moment
In a nearly perfect ionic bond, such as $KF$, where electron transfer is almost complete, representing the molecule as
$K^+ F^-$
is a very good approximation, since the charge on the potassium will be approximately $1e$ and the charge on the fluorine will be approximately $-1e$. For a polar covalent bond, such as $HF$, in which only partial charge transfer occurs, a more accurate representation would be
$H^{+\delta}F^{-\delta}$
where $\delta$, expressed in units of $e$, is known as a partial charge. It suggests that a fraction of an electron is transferred, although the reality is that there is simply a little more electron density on the more electronegative atom and a little less on the electropositive atom.
How much charge is actually transferred can be quantified by studying the electric dipole moment of the bond, which is a quantity that can be measured experimentally. The electric dipole moment of an assembly of charges $Q_1 ,Q_2 ,...,Q_N$ having positions$r_1 ,r_2 ,...,r_N$ is defined to be
$\mu =Q_1 r_1 +Q_2 r_2+...+ Q_N r_N=\sum_{i=1}^{N}Q_i r_i$
Thus, for a diatomic with charges $Q_1 =Q=\delta e$ and $Q_2 =-Q =-\delta e$ on atoms 1 and 2, respectively, the dipole moment, according to the definition, would be
\begin{align*}\mu &= Q_1 r_1 +Q_2 r_2\ &= Qr_1 -Qr_2\ &=Q(r_1 -r_2)\end{align*}
Hence, the magnitude of the dipole moment is
$\mu = |\mu|=Q|r_1 -r_2|=QR$
where $R$ is the bond length. As an example, consider $HF$, which has a partial charge on $H$ of $0.41 e$, which means $\delta =0.41$, and a bond length of $0.926 \ \stackrel{\circ}{A}$. Thus, the magnitude of the dipole moment is
$|\mu|=0.41(1.602*10^{-19}C)(0.926*10^{-10}m)=6.08*10^{-30}C\cdot m$
Thus, the units of the dipole moment are Coulomb meters. However, as this example makes clear, this is a very large unit and awkward to work with for molecules. A more convenient unit is the Debye $(D)$, defined to be
$1D=3.336*10^{-30}\; \text{Coulomb} \cdot \text{meters}$
$1 \ D$ is actually the dipole moment of two charges $+e$ and $-e$ separated by a distance of $0.208 \stackrel{\circ}{A}$. Thus, for a diatomic with partial charges $+\delta$ and $-\delta$, the dipole moment in $D$ is given by
$\mu (D)=\frac{\delta *R(\stackrel{\circ}{A})}{0.2082 \ \stackrel{\circ}{A}D^{-1}}$
and the percent ionic character is defined in terms of the partial charge $\delta$ by
$percent \ ionic \ character=100\% *\delta$
As an example, consider $HF$ again, for which $\delta = 0.41$. The bond length is $R=0.926 \ \stackrel{\circ}{A}$. Thus, its dipole moment will be
$\mu (D)=\frac{0.41*0.926 \stackrel{\circ}{A}}{0.2082 \ \stackrel{\circ}{A}D^{-1}}=1.82D$
and its percent ionic character is $41\%$.
Experimental importance of the dipole moment
The electric dipole moment lies at the heart of a widely used experimental method for probing the vibrational dynamics of a system. If a system is exposed to a monochromatic electromagnetic field from a laser, then the electric dipole moment couples to the electric field component $E(r,t)$ in such a way that the energy is
${\cal E} = -{\mu}\cdot {\bf E}({\bf r},t)$
In general, the electric field is a function of space and time having the general wave form
${\bf E}({\bf r},t) = {\bf E}_0\cos\left({\bf k}\cdot {\bf r}- \omega t\right)$
where $\omega$ is the frequency of the field $\omega=2\pi c/\lambda$, with $c$ the speed of light and $\lambda$ the wavelength, and $k$ is called the wave vector, $|k|=2\pi /\lambda$, and the direction of $k$ is the direction of wave propagation (this will be covered in more detail next semester). In most experiments, the wavelength is long enough compared to the size of the system studied that one can take the electric field to be spatially constant and consider only the time dependence. In this case,
${\cal E}\approx -\mu \cdot E_0 cos(\omega t)$
Thus, the electric field varies as a simple cosine function at a single frequency $\omega$.
The importance of the coupling between the dipole moment and the time-dependent electric field is that the frequency of the field can be varied over a range of natural frequencies in a given chemical system. Thus, chemical bonds vibrate at a particular natural frequency, three-atom bending modes have their characteristic frequency, etc. What one seeks in this experiment is a report'' of the natural frequencies in the system, since from such a report, one can often tell one local chemical environment from another.
By sweeping through a range of frequencies, the coupling of the field to the dipole moment suggests that the local charge distribution will respond to the oscillations of the field at the field frequency. Thus, if the field frequency is tuned'' to be that of a bond stretch, the charge distribution in the bond will be stimulated and report on the frequency of the bond, etc. At each frequency, the intensity $I$ of the response can be measured, and a plot of $I$ vs. $\omega$ is produced. Such a plot is called an infrared spectrum. The figure below shows the infrared spectrum for liquid water (left) and for 13 M (blue) and 1 M (red) KOH solutions (right).
In the left panel, the solid curve is the water spectrum obtained from a computer simulation, while the dashed curve is the experimentally obtained spectrum. On the right, the red and blue curves are from computer simulations, while the inset at the upper right is the experimentally measured spectrum. The peaks in the spectra occur at particular vibrational frequencies in the system. The water spectrum shows very distinct bands, while the spectrum of the KOH solutions shows both bands and continuum regions. The latter arise from the fact that protons can be transferred from water to hydroxide. As the proton moves across a hydrogen bond between water and the hydroxide ion, the vibrations in the bond sweep through a range of frequencies as the proton is transferred, giving rise to the continuum. This feature in the infrared spectra of solutions of strong acids and bases is known as Zundel polarization. More information on how we compute these spectra and how the computer simulation are performed can be found in the following research papers: | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/02%3A_Basic_Concepts-_Molecules/2.06%3A_Dipole_Moments/2.6B%3A_Molecular_Dipole_Moments.txt |
Trigonal Bipyramidal Species are those that have a central p-block atom and are attached to 5 other atoms. It is classified as a EX5 molecule, where E stands for the central atom, and X stands for the atoms that are attached. It makes sense, then, to classify this molecule a 5-coordinate system. The VSEPR (valence shell electron pair repulsion) model is the what helps us identify how a molecule may or may not be trigonal bipyramidal.
Trigonal Bipyramidal Species
The central atom of the triognal bipyramidal species is bonded to 5 separate molecules. This means that the central atom must have an extended valence shell, in order to all for the 5 bonds to occur. The central atom is thus typically a p-block atom.
There are two different ways to classify the 5 X atoms, either as axial or equitorial. It is important to understand the difference between the two, so that we can reference them in future discussion.
The axial atoms are the ones above and below the central atom. These are the points of the 'pyramids' the molecule makes. (NOTE: Bipyramidal means two pyramids. If you look at a picture you should be able to see the pyramids relative to the top half and the bottom half of the central atom." The equatorial atoms are the ones that lie in the horizontal plane of the central atom. (Think of the equator relative to the Earth.)
Let us continue this discussion with the assumption that the EX5 is composed of just atoms (and not lone pairs). The bond angles between the equatorial atoms is 120 degrees. The bond angles between the axial and equitorial atoms is 90 degrees. The bond angles between the two axial atoms is 180 degrees. The angles formed by these atoms is the most stable conformation that can be maintained by the trigonal bipyramidal species.
Other Common Shapes for 5 Coordinate Molecules
As we have already discussed, when the central atom has 5 bonding pairs of electrons (those electrons being shared between two atoms), the shape of the molecule is trigonal bipyramidal. This section gives a little bit of information about how the shape of a molecule is affected when not all of the pairs of electrons are nonbonding.
Before that, however, it is important to discuss how bond angles are affected by lone pairs. According to VSEPR, electron-electron repulsion decreases in the following manner:
lone pair - lone pair > lone pair - bonding pair > bonding pair - bonding pair.
This means that presence of lone pairs will increase electron-electron repulsion, causing a change in the bond angles.
The first alternate shape for a 5 coordinate molecule, then, is where one of the bonding pairs of electrons is replaced with a lone pair. The best place for the lone pair to be is equatorial to the central atom. This allows for the least amount of electron-electron repulsion possible. This shape is called Seesaw.
The next possible shape is where there are two lone pairs on the central atom. Once again, the best place for these two molecules is equatorial to the central atom, for the least amount of electron repulsion to occur. This molecule is called T-shaped.
The last alternate shape for a 5 coordinate system is when there are three lone pairs. These are all equatorial to the central atom to establish the lowest energy conformation. This molecule is considered to be Linear.
Stereoisomers
Due to the fact that there are two possible types of X atoms (axial or equatorial) stereoisomers of trigonal bipyramidal species (with more than one type of A atom) are possible. If there are two different X atoms, then those atoms have the option of being cis or trans to each other. Meaning if one is axial, then the other can be equatorial (cis) or also axial (trans). Stereoisomers occur when two (or more) molecules with the same molecular formula have different molecular shapes.
Outside Links
1. The Relationship Between the Number of Regions of Valence Electrons
and the Molecular Geometry Around an Atom. Tutor-Page.com. 05 Nov. 2010.
<www.tutor-pages.com/Chemistry..._Nonpolar.html>
2. Trigonal Bipyramidal Molecular Geometry. Wikipedia. 05 Nov. 2010. <http://en.Wikipedia.org/wiki/Trigona...cular_geometry>
3. Molecular Shape. Spark Notes. 05 Nov. 2010. <http://img.sparknotes.com/content/te...17002_0423.gif>
Problems
1. Explain why a trigonal bipyramidal molecule takes on the shape that it does.
2. What are the other possible shapes a 5-coordinate molecule can have? Give examples if possible.
3. What is a stereoisomer? Give an example using a trigonal bipyramidal molecule.
2.9.02: Trigonal Bipyramidal Structures
Trigonal Bipyramidal electron configuration is part of the VSEPR model of an atom with 5 electron pairs.
Introduction
Trigonal Bipyramidal structures are electronic configurations of molecules. VSEPR gives us a handful of parent shapes. The one with 5 sets of bonded or non-bonded electrons.
Molecular Geometry of the Trigonal Bipyramidal Structures
Number of Lone pairs Geometry Bond Angles
0 Trigonal Bipyramidal 90 and 120
1 Seesaw 90 and 120
2 T-Shaped 90
3 Linear 180
The order of most repulsion to least repulsion among bonding and lone pair electrons are:
Lone pair-Lone pair > Lone pair-Bond pair > Bond pair-Bond pair
To decide where to place lone pairs on the parent Trigonal Bipyramidal structure, we must place lone pairs far away from each other and bond pairs. The seesaw shape maximizes the bond angles of the single lone pair and the other atoms in the molecule. The lone pair is in an equatorial position offering 120 and 90 degree bond angles, compared to only 90 degree bond angles if placed at the axial position.
The T shaped structure minimizes the remaining bond pair-bond pair angles at 90 degrees and maximizes the lone pair-lone pair bond angle at 120.
The linear structure does the same by making sure that the lone pairs are kept at a maximum of 120 degree bond angles. As a result, all are placed in the equatorial position, leaving room at the axial postion for the atoms. Thus, the bond angles of the atoms are 180 degrees from each other.
Stereoisomers
Since there are two types of atoms on a Trigonal Bipyramidal structure, axial and equatorial, there are different Stereoisomers that could arise depending on the substituents attached. For example, if there is 4 X atoms and 1 Y atom attached to the central atom, Y could either be in an equatorial position or in an axial position. If there are 3 X atoms and 2 Y atoms, then one Y atom could be placed in a equatorial position and the other in an axial position, or both Y atoms could be placed in the same position.
Problems
1. How would you arrange SeCl4?
2. How would you arrange ArCl2?
3. How would you arrange ICl3?
Contributors and Attributions
• Ian Marton (UC Davis) | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/02%3A_Basic_Concepts-_Molecules/2.09%3A_Molecular_Shape_-_Stereoisomerism/2.9.01%3A_Trigonal_Bipyramidal_Species.txt |
In inorganic chemistry, an octahedron is classified by its molecular geometry in which its distict shape is described as having six atoms, groups of atoms or electron pairs symmetrically arranged around one central atom, defining the vertices of an octahedron. The prefix octa, which means eight, comes from the fact that the molecule has eight symmetrical faces. All atoms are 90 degrees apart from one another, and 180 degrees apart from the atom, directly across and opposite from it. In regards to identifying each species, we will be looking at three separate unique shapes with different numbers of bond pairs and lone pairs. This allows us to distinguish and classifiy the octahedrals based on the following shapes: octahedral, square pyramidal, and square planar.
Introduction
To be able to understand and distinguish the difference between the three types of octahedral species and how they differ from one molecule to the next, it is essential to try to visualize shapes geometrically and in 3D. A simple comprehension of geometry is required to be able to imagine molecules in 3D, as well as having basic background knowledge of the concept of bonding pairs and lone pairs. This allows one to recognize and see the difference in the molecular design for each individual molecule. We will begin by describing the design of an octahedral and then continue on to the next two molecules.
An octahedral is best described as a central atom symmetrically arranged by six other atoms. What makes this molecule different from other species is the fact that it is surrounded by six, either the same or different, atoms. There are six bonding pairs in this molecule and no lone electron pairs. The molecule below has no lone pairs of electrons surrounding it, thus allowing it to have a distinct shape. In a three dimensional sense, we may think of a x, y, and z coordinate plane having both its positive and negative coordinate systems. Another way of looking at it would be in the sense that all the faces of the molecule are present; through this reference, it resembles what would be a three dimensional prism. All the atoms are spread apart 90 degrees from each other and 180 from the atom directly across and opposite from it. The reson for the spacing is due to the molecule arranging itself in the most stable form possible, limiting the bond-pair to bond-pair interaction. Here is basic, but clear example of what an octahedral looks like:
Octahedral (6 bond pairs and 0 electron pairs)
The next molecule that we will examine is known as a square pyramidal. This molecule has a lot of the same characteristics as that of an octahedral in the sense it consist of a central atom that is still symmetrically surrounded by six other atoms. What makes this molecule different from the previous molecule is the fact that this molecule does not consist of only bond-pair atoms surrounding it. There is one pair of electrons that has taken the place of one of the atoms and because these electrons are now present, it gives the molecule a distict new look. There are five bonding pairs and one electron pair. The atoms have to arrange themselves in the most stable form possible, not only limiting the bond-pair to bond-pair interaction, but also limiting the bond-pair to electron-pair interaction. The easiest way to visualize what this molecule looks like to visualize the x, y, and z coordinate plane again, but this time remove what would be considered the negative y coordinate axis and put a pair of lone pair electrons in its place. The molecule is still considered apart of the octahedral species because it still satisfies the 6 atom requirement, but in terms of its shape, the electrons effect the shape. This allows it to have its new shape. If you actually exclude those electrons and lay the molecule on the surface, you can see that it looks like a three dimensional pyramid with a square base. Again all the atoms and electron pair are 90 degrees apart from each other and 180 from the atom directly across and opposite from it. Here is what a square pyramidal would look like:
Square Pyramidal (5 bond pairs and 1 electron pair)
The last of the octahedral species is known as a square planar. This molecule resembles both of the previous molecules, but more similarly resembles a square pyramidal. It still has many of the characteristics of a square pyramidal, but what makes it different is that rather than having only one pair of electrons replacing the position of an atom, there are two pairs of electrons that are replacing the position of two atoms. To visualize what this molecule looks like, we refer back to the x, y, and z coordinate system, the only difference is this time we are taking away the entire y coordinate, and replacing it with electrons on what would be the positive y coordinate axis as well as placing a pair of electrons in what would be considered the negative y coordinate axis. The reason for this arrangement goes back to having the molecule arrange itself in the most stable form possible limiting interactions between bond-pair to bond-pair, bond-pair to electron-pair, and electron-pair to electron-pair. If you try visualizing what this would look like, it almost resembles a three-dimensional "X" with two pairs of lone electrons. Because the lone pairs of electrons are still present, that allows this molecule to still be considered an octahedral due to the fact that it still meets the requirements of being surrounded by 6 atoms or groups. In regards to its shape the electron pairs cause repulsion, thus allowing it to have its new shape. The atoms and electrons are still 90 degrees apart from eachother and 180 degrees from the atom directly across and opposite from it. Here is what a square planar would look like:
Square Planar (4 bond pairs and 2 electron pairs)
Heading #2
Rename to desired sub-topic. You can delete the header for this section and place your own related to the topic. Remember to hyperlink your module to other modules via the link button on the editor toolbar.
Outside Links
• Sample octahedral image adapted from Wikipedia key word octahedral geometry:en.Wikipedia.org/wiki/Octahedral_molecular_geometry
• Sample square planar image adapted from Wikipedia key word square planar geometry:en.Wikipedia.org/wiki/Square_planar_molecular_geometry
• Sample square pyramidal image adapted from Wikipedia key word square pyramidal geometry:en.Wikipedia.org/.../Square_pyramidal_molecular_geometry
Problems
1. What causes the three different octahedral species to arrange the way they do? What conditions must be met?
2. Can two seperate electron-pair stand at 90 degrees apart from eachother? Why?
3. Give one example of a molecules that would fall into the category of a octehedral, square pyramidal, and square planar.
Answers
1. The molecules take the arrangment they do due to trying to arrange themselves in the most stable structure possible limiting the interaction between bond-pair and electron-pair interaction. As long as these conditions can be met, it is possible for the structure to not only exist, but remain stable.
2. This again goes back to satisfying the conditions of keeping the molecule as stable as possible by limiting lone-pair to lone -pair interaction as well as same sign interaction. Because electrons hold the same kind of charge, they can not be near eachother due to same charge repulsion and so they need to be as far away as possible from eachother so that the molecule may be stable.
3. Molecules that would fall into the category of triganol planar based on their molecular geometry would be SF6, a molecule that falls into the category of a square pyramidal would be BrF5 and one molecule that would fall into a category of a square planar would be [AuCl2]-.
Contributors and Attributions
• Name #1 here (if anonymous, you can avoid this) with university affiliation | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/02%3A_Basic_Concepts-_Molecules/2.09%3A_Molecular_Shape_-_Stereoisomerism/2.9B%3A_Octahedral_Species.txt |
Coordination numbers from 2 to 6 and the associated coordination complexes are very common, especially in transition metal chemistry. These complexes can take on a variety of geometries and each individual complex will have unique properties based on its substituents and their arrangement spatially. Though there is no set number of ligands, ions, or bound atoms required for a species to be considered "highly coordinated" versus general coordination, when the coordination number exceeds seven it is generally considered a high coordination number. High coordination numbers are of particular interest in solution phase and inorganic chemistry.
Introduction
The coordination number of a specified atom in a chemical species is the "number of other atoms directly linked to that specified atom."[1] Coordination itself is "the formation of a covalent bond, the two shared electrons of which have come from only one of the two parts of the molecular entity linked by it, as in the reaction of a Lewis acid and a Lewis base to form a Lewis adduct; alternatively, the bonding formed in this way."[2] Coordination number is a measure of one of two things:
1. In a molecule, the coordination number is simply the number of bound atoms to that molecule.
2. For an ion the coordination number is the number of ions that are either associated to, or can associate with a given ionic species.
3. Coordination requires a sigma bond be present or available to be present, and so a pi bond does not contribute to the coordination number of a species.
Lower Coordination Numbers
Following are a few examples of more commonly encountered coordination numbers
Coordination Number 2
Coordination numbers of two are more common in organic molecules, or in non transition-metal containing species and relatively rare in transition metal containing species. Coordination numbers of two result in a linear or collinear spacial arrangement.
Carbon Dioxide, Coordination Number 2 [Ag(NH3)2]+, Coordination Number 2
Coordination Number 6
Coordination number 6 is the most common coordination number, especially among transition metal containing species.[4]
Generally, the octahedron is the preferred coordination geometry. Other possible coordination geometries exist for coordination number six, such as trigonal prismatic.
[Co(H2O)6)]3+, Coordination number 6
Higher Coordination
Higher coordination numbers; 7,8, 9, and above exist though they are less common.
Coordination Number 7
Seven coordinated atoms or molecules results in three main geometries:
• Pentagonal bipyramid
• Capped octahedron
• Capped trigonal prism
Coordination Number 8
Eight coordinated atoms or molecules results in two main geometries:
• Square antiprism
• Dodecahedron
Coordination Number 9
The only structure yet identified for coordination number 9 systems is the tricrapt trigonal prism.
Two examples of these structures are [ReH9]2- and [TcH9]2- with the representative geometry shown below.
[TcH9]2-, Coordination number 9
Higher Coordination Numbers
Species exist with 10, 11, and 12 coordination numbers.
• Coordination numbers 10 and 11 are unique to complexes involving lanthanides and actinides. [4]
• Coordination number 12 has a structure that is involved in boron chemistry, the icosahedron.
• Coordination number 15 is the highest reported coordination number currently, being described for [PbHe15]2+[3]
Problems
1. What is the coordination number of the nitrate ion NO3-?
2. What is the coordination number around Co in this complex? How many different ligands are around Cobalt?
3. What coordination number do you expect [Th(NO3)6]2- to have? (Draw out a rough structure, remember lone pairs are required for coordination. What is the likely geometry for this?
Answers
1. Coordination number 3. Pi Bonds do not contribute to coordination number.
2. Coordination number 6. There is one ligand around Cobalt; EDTA is hexadentate.
3. Coordination number 12; Each NO3- ligand is bidentate. The likely geometry is icosahedral like other 12 coordination number species.
Contributors and Attributions
• Brian Atwood (UC Davis) | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/02%3A_Basic_Concepts-_Molecules/2.09%3A_Molecular_Shape_-_Stereoisomerism/2.9D%3A_High_Coordination_Numbers.txt |
Symmetry is important in the study of chemistry. It can be studied on both the molecular level and the crystalline level, of which the former will be discussed in this book.
A molecule is symmetrical if it stays in an indistinguishable configuration after some movement. For example, the molecule \(BH_{3}\) in 3.1 can be rotated about an axis perpendicular to the screen and passes through the boron atom. If the rotation angle is 120°, the new configuration is indistinguishable from the original form.
3.1
The study of symmetry has many applications, some of which will be introduced in this chapter. Also in this chapter, some of the terminology and definitions used in group theory, the mathematical representation of symmetry, will be introduced.
3.02: Symmetry Operations and Elements
Figure $1$: An example of a symmetry operation is a 180° rotation where the resulting position is indistinguishable from the original. A 180° rotation is called a C2 operation; the axis of rotation is the symmetry element. (CC-BY-NC-SA; Kathryn Haas)
Introduction
The symmetry of a molecule consists of symmetry operations and symmetry elements. A symmetry operation is an operation that is performed to a molecule which leaves it indistinguishable and superimposable on the original position. Symmetry operations are performed with respect to symmetry elements (points, lines, or planes).
An example of a symmetry operation is a 180° rotation of a water molecule in which the resulting position of the molecule is indistinguishable from the original position (see Figure $1$). In this example, the symmetry operation is the rotation and the symmetry element is the axis of rotation.
There are five types of symmetry operations including identity, reflection, inversion, proper rotation, and improper rotation. The improper rotation is the sum of a rotation followed by a reflection. The symmetry elements that correspond to the five types of symmetry operations are listed in Table $1$.
Table $1$: Table of elements and operations
Element Operation Symbol
Identity identity E
Proper axis rotation by (360/n)o Cn
Symmetry plane reflection in the plane σ
Inversion center inversion of a point at (x,y,z) to (-x,-y,-z) i
Improper axis rotation by (360/n)o, followed by reflection in the plane perpendicular to the rotation axis Sn
Symmetry Operations and Elements
Identity (E)
All molecules have the identity element. The identity operation is doing nothing to the molecule (it doesn't rotate, reflect, or invert...it just is).
Proper Rotation and Proper Axis (Cn)
A "proper" rotation is just a simple rotation operation about an axis. The symbol for any proper rotation or proper axis is C(360/n), where n is the degree of rotation. Thus, a 180° rotation is a C2 rotation around a C2 axis, and a 120° rotation is a C3 rotation about a C3 axis.
PRINCIPLE AXIS: The principle axis of a molecule is the highest order proper rotation axis. For example, if a molecule had C2 and C4 axes, the C4 is the principle axis.
Reflection and Symmetry Planes (σ)
Symmetry planes are mirror planes within the molecule. A reflection operation occurs with respect to a plane of symmetry. There are three classes of symmetry elements:
1. σh (horizontal): horizontal planes are perpendicular to principal axis
2. σv (vertical): vertical planes are parallel to the principal axis
3. σd (dihedral): dihedral planes are parallel to the principle axis and bisecting two C2' axes
Inversion and Inversion Center (i)
The inversion operation requires a point of symmetry (a center of symmetry within a molecule). In other words, a point at the center of the molecule that can transform (x,y,z) into (-x,-y,-z) coordinate. Structures of tetrahedron, triangles, and pentagons lack an inversion center.
Improper rotation (Sn)
Improper rotation is a combination of a rotation with respect to an axis of rotation (Cn), followed by a reflection through a plane perpendicular to that Cn axis. In short, and Sn operation is equivalent to Cn followed by $\sigma_h$.
Problems
1. Water molecule H2O was used as an example and was mentioned that water was rotated 180 degree around an axis bisecting the oxygen, then the molecule was superimposable on the original water molecule, how about CO2? Is it going to be like the water molecule since CO2 also has 2 atoms of oxygen.
Of course not, because every molecule has different molecular shape. To recognize the symmetry of any molecule the structure and the molecular shape of that molecule should be defined. The water molecule is bent but CO2 is not, and if CO2 is rotated 360 degree around the axis bisecting the C atom then it can be superimposed on the original molecule. We then see the symmetry for the CO2.
2. Why should all of the five symmetry elements be done on a molecule in order to find the point group the molecule belongs to, why is performing only one or two of the symmetry elements not enough for recognizing the point group?
One or two of the symmetry elements will not be able to tell us everything about the molecule's symmetry since those one or two properties do not tell us everything about the molecule. Also, while different molecules may have one or two symmetrical properties in common, the five properties will not be the same for all molecules.
3. What does the symbol Cn stand for and what does n represent? Why is it important to identify n?
C is the axis of rotation and n is the order of the axis.
4. How are the character tables helpful?
The character table tells us about all the operational elements performed on the molecule and indicates if we have forgotten to perform any of the symmetry elements. The tables serve as a checklist because all the operational elements should be done on the molecule in order to find the point group of the molecule.
5. Why is important to find symmetry in molecules?
Symmetry tells us about bounding for that molecular bonding. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/03%3A_Introduction_to_Molecular_Symmetry/3.01%3A_Introduction.txt |
Sometimes, new symmetry operations form by carrying out two or more simpler operations successively to result in an indistinguishable configuration. For example, the improper rotation axis that was introduced in the last section results from rotation about an axis of rotation (Cn) followed by reflection about a plane perpendicular to that axis ($\sigma$h).
$S_{n}=C_{n}\times\sigma_{h} \nonumber$
Note also that Cn in this case need not be a symmetry element for the molecule. For example, staggered ethane has an S6 symmetry element although it does not have a C6. Another example of a symmetry element that results from a combination of two different elements is the inversion center (i), which results from a C2 rotation followed by a reflection about a plane perpendicular to that axis ($\sigma$h).
$i=C_{2}\times\sigma_{h} \nonumber$
Note that an inversion center is a special case of an improper rotation axis because it results when, in the first equation, $n=2$. That is,
$i=S_{2} \nonumber$
In fact, any symmetry operation can be carried out multiple time in a row. For example, when BH3 is rotated twice by 120°, the two-step operation can be symboled by $C_{3}^{2}$. When the C3 operation is performed a third time, the molecule returns to its original configuration; i.e,
$C_{3}^{3}=E \nonumber$
In general,
$C_{n}^{n}=E \nonumber$
3.04: Point Groups
Point groups are used to describe molecular symmetries and are a condensed representation of the symmetry elements a molecule may posses. This includes both bond and orbital symmetry. Knowing molecular symmetry allows for a greater understanding of molecular structure and can help to predict many molecular properties.
Introduction
Point groups are a quick and easy way to gain knowledge of a molecule. They not only contain a molecule's symmetry elements, but also give rise to a character table, which is a complete set of irreducible representations for a point group. A molecule's point group can be determined by either elucidating each symmetry element contained in a molecule or by properly using the Schreiber chart (see below).
Point groups usually consist of (but are not limited to) the following elements:
• E - The identity operator. This operation leaves a molecule completely unchanged and exists for mathematical purposes.
• Cn - The Cn proper axis of rotation is a 360/n° rotation that when performed leaves a molecule the same. A proper rotation with the highest value of n is known as the major axis of rotation.
• σ - The mirror plane. The mirror plane can be described as a plane which produces a reflection of part of a molecule that is unnoticeable and can be labeled as either σh , σv , σd.
• i - The inversion center. A molecule has a center of inversion if, when inverted, the molecule is unchanged.
See the section on symmetry elements for a more thorough explanation of each.
Each point group is associated with a specific combination of symmetry elements
Each point group has it's own combination of symmetry elements. Listed below are some of the many point groups and their respective symmetry elements, according to category, followed by a representative example.
Non axial groups
C1: E C1: E, i
Cn groups
C2: E, C2 (notice the major axis of rotation is the point group) C3: E, C3, C32
H2O2 C2
Dn groups
D2: E C2(z), C2(y), C2(x) D3: E, 2C3, 3C2
Cnv groups
C2v: E, C2, σv(xz), σv'(yz) C3v: E, 2C3, 3σv
H2O C2v
Cnh groups
C2h: E, C2, i, σh C3h: E, C3, C32, σh, S3, S33
B(OH)3
Dnh groups
D2h: E
C2H4 D2h
How to determine a molecules point group
A molecule's point group can be determined by calculating all the symmetry elements of a molecule and matching them to a respective point group. This process, however, is greatly simplified when the Schreiber chart is used:
Problems
1. Determine the point group of BH3 by calculating all it's symmetry elements then use the chart and determine which method is faster.
2. Determine the point groups of BH3 and NH3. Why is there a difference?
3. What is the point group of PPh3?
4. Determine the point groups of CO2 and H2O and then compare them.
5. Propose a molecule with no symmetry. What is it's point group? | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/03%3A_Introduction_to_Molecular_Symmetry/3.03%3A_Successive_Operations.txt |
Although the method of assigning a point group to a molecule depends on some knowledge of the symmetry elements the molecule has, it does not require the consideration of all elements. For example, the molecule $CO_{2}$ can be assigned the point group $D_{\infty h}$ using the scheme in section 3.4 by merely knowing that the molecule is linear and that it has the a center io inversion $i$. The point group assignment in this case did not require knowing that the molecule also has $\sigma_{h}$, for example. This type of knowledge is gained by examining character tables.
Definition
A character table is a table that contains the symmetry information of the molecule. This information can be used to analyze the molecule's behavior in many application, among which is spectroscopy. Each point group has its own character table. The following table is the character table of the point group $C_{2v}$:
Table 1: The character table of the point group $C_{2v}$.
Understanding Character Tables
Symbols under the first column of the character tables
A (Mulliken Symbol) (singly degenerate or one dimensional) symmetric with respect to rotation of the principle axis
B (Mulliken Symbol) (singly degenerate or one dimensional) anti-symmetric with respect to rotation of the principle axis
E (Mulliken Symbol) (doubly degenerate or two dimensional)
T (Mulliken Symbol) (thirdly degenerate or three dimensional )
Subscript 1 symmetric with respect to the Cnprinciple axis, if no perpendicular axis, then it is with respect to σv
Subscript 2 anti-symmetric with respect to the Cnprinciple axis, if no perpendicular axis, then it is with respect to σv
Subscript g symmetric with respect to the inverse
subscript u
anti-symmetric with respect to the inverse
prime symmetric with respect to
double prime anti-symmetric with respect to
Symbols in the first row of the character tables
E describes the degeneracy of the row (A and B= 1) (E=2) (T=3)
Cn 2pi/n= number of turns in one circle on the main axis without changing the look of a molecule (rotation of the molecule)
Cn' 2π/n= number of turns in one circle perpendicular to the main axis, without changing the structure of the molecule
Cn" 2π/n= number of turns in one circle perpendicular to the Cn' and the main axis, without changing the structure
σ' reflection of the molecule perpendicular to the other sigma
σv (vertical) reflection of the molecule vertically compared to the horizontal highest fold axis.
σh or d (horizontal) reflection of the molecule horizontally compared to the horizontal highest fold axis.
i Inversion of the molecule from the center
Sn rotation of 2π/n and then reflected in a plane perpendicular to rotation axis.
#Cn the # stands for the number of irreducible representation for the Cn
the # stands for the number irreducible representations for the sigmas.
the number in superscript in the same rotation there is another rotation, for instance Oh has 3C2=C42
other useful definitions
(Rx,Ry) the ( , ) means they are the same and can be counted once.
x2+y2, z2 without ( , ) means they are different and can be counted twice.
3.06: Significance of Recognizing Symmetry Elements
In the forgoing sections of this chapter, the necessary tools to assign a point group to a molecule and to be able to read a character table were introduced. A good question at this point is: What is the importance of these tools in inorganic chemistry?
In fact, there are many applications in the study of inorganic chemistry where these symmetry tools become handy. In general, these applications belong to one of the following categories:
• Understanding the spectroscopic properties of moelcules (introduced in section 3.7);
• Determining the chirality of molecular species (introduced in section 3.8);
• Constructing molecular and hybrid orbitals (Chapter 5). | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/03%3A_Introduction_to_Molecular_Symmetry/3.05%3A_Character_Tables_-_An_Introduction.txt |
All molecules are constantly vibrating, and can absorb energy from an incoming photon to increase their vibrations. The two types of vibrational spectroscopy are infrared spectroscopy and Raman spectroscopy. Vibrational spectroscopy is the science of measuring exactly which wavelengths of light are absorbed by a molecule. This technique could be used identify an unknown molecule by comparing its absorption to that of other molecules. Or vibrational spectroscopy could be used to gain further understanding of the physical properties of a known molecule.
Introduction
Molecular vibration can be modeled by balls attached by springs. Displacing an atom from its most stable position requires energy proportional to the displacement. IR spectroscopy can be used to characterize a molecule is the energy of its vibrations falls in the infrared range. Alternatively, higher energy light can be absorbed, the re-emitted at a different wavelength and/or in a different direction; this leads to Raman spectroscopy. So a molecule can absorb radiation to change bond lengths or positions with respect to the other atoms in the molecule.3
From the number, frequency, and intensity of these absorptions or emissions we can gain insight into the composition of the sample being measured. The number of different absorptions is indicative of the number of different atoms and shape of the molecule. The frequency or wavelength absorbed is indicative of the energy of the bonds and vibrations. And the intensity of the absorptions is related to the concentration of the analyte.1
Vibrational Modes
The different possible vibrations are called vibrational modes. Vibrational modes are determined by all the different ways the atoms in the molecule can move with respect to eachother, called the vibrational degrees of freedom. Vibrational degrees of freedom differ from the total degrees of freedom in that translation (movement through space) and rotation do not contribute to the vibrational degrees of freedom.4
To find the number of vibrational modes one must first know the point group of the molecule. From there, the point group's character table will list all of the possible symmetry operations for the molecule. Each symmetry operation will leave some atoms in the molecule in place and/or move other atoms in the molecule. Count up the number of atoms that do not move for each symmetry operation, and multiply that number by the symmetry operation's contribution. This gives the total representation of atomic motion. From there, the vibrational modes can be found by reducing the total representation, according to the equation:
n=1/h*ΣXR*XI*N
where n is the number of modes with that symmetry, h is the total number of symmetry operations, XI is the number of irreducible representations (the value calculated above), XR is the number of reducible representations (the entry in the character table), and N is the number of identical symmetry operations. Finally, subtracting the rotational modes and translational modes from the reduced representation gives the number of vibrational modes.6
Three of the vibrational modes for the C2V point group. Used with permission from http://en.Wikipedia.org/wiki/Infrared_spectroscopy
Infrared Spectroscopy
A vibrational mode will be observed in an infrared spectrum if it leads to a change in the molecular dipole moment. Compared to Raman spectroscopy, the infrared photon is completely absorbed and its energy is transfered to the vibration of the molecule, not re-emitted.4
Different bonds have different energies associated with them, and require different amounts of energy to stretch or bend. In general, the stronger the bond, the more energy required to deform it. So very weak bonds will only be deformed by low energy radiation, and strong bonds will only be deformed by high energy radiation. This leads to characteristic frequencies where only certain vibrations are absorbed. For example, C-H bonds are typically the only bonds observed in the range from 2960-2850 cm-1. So if an absorption is present at that frequency, it can be assumed that it is due to a C-H bond.1
Raman Spectroscopy
A vibrational mode will be observed in Raman spectroscopy if it leads to a change in the polarizability of the electron cloud of a molecule.4
If a photon has energy that is significantly higher than the energy of the vibrational states, it may either be deflected without any change in energy, or it may interact with the molecule and either take energy from it or give energy to it.2
When the photon is absorbed and re-emitted in a different direction, it is called Rayleigh scattering, and this is strongly dependent on the wavelength of the incoming light. When there is a change in the energy of the photon, it is called Stokes scattering or anti-Stokes scattering, depending on whether energy is absorbed or lost by the molecule. This change in the behavior of the incoming photons can be measured, and will provide information about the concentration and chemical properties of the analyte.3
Diagram used with permission from http://en.Wikipedia.org/wiki/Raman_spectroscopy
Problems
1. Why is it that the symmetric stretch of carbon dioxide (CO2) is IR inactive, while the symmetric stretch of carbonyl sulfide (COS) is IR active?
A: Because the symmetric stretch of CO2 does not lead to a change in the dipole moment of the molecule, but the symmetric stretch of COS does.
2. How many vibrational modes are IR and Raman active for C20?
A: 17 IR active modes, 29 Raman Active modes.
3. Are there any differences between the vibrational spectrums of H2O, H2S, and H2Se?
A: Although they have the same number of atoms, the same point group, and the same number of vibrational modes, the frequencies at which the vibrational modes occur will be different.
3.7C: Vibrational Spectroscopy of Linear and Bent triatomic Molecules
Vibrational spectroscopy, AKA Infrared (IR) Spectroscopy, is a highly useful and beneficial tool for determining structures and functional groups contained in compounds. Although used for many different structure types of polyatomic molecules, this particular Module is dedicated to triatomic molecules with bent and linear spacial molecular geometry.
Introduction
Triatomic molecules are molecules that contain three atoms. The atoms in triatomic molecules can all be the same, as in I3-, all be different, as in HCN, or can be a mix like CO2. Examples include H2O, which is a bent and has a bond angle of 109o, and a linear triatomic molecule such as CO2. All bent tri-atomic molecules belong to the point group C2v, while all the liner tri-atomic molecules with an inversion center belong to the Dinfinity-h point group; Those without an inversion center belong to the point group Cinfinity-v. This gives these two structures very different infrared spectrum even though they have the same number of atoms.
Linear molecules
Using VESPR theory there are several ways to achieve a linear structure. Tri-atomic molecules where the central atom is using ALL of its electrons in the bonds with the surrounding molecules, or in other words the central atom does not have any lone pairs surrounding it, will give rise to a linear molecule. Examples include \(CO_2\) and \(BeH_3\). This electronic configuration gives the central atom a sp hybridization. Triatomic molecules where the central atom does not use all of its electron pairs in the bonds between the other two atoms will, under certain circumstances, also give rise to a linear species. When the central atom is surrounded by three, or four lone pairs in addition to the two elements already attached will also give rise to a linear molecule. Examples include \(KrF_2\).
Bent Molecules
There are also several ways to give rise to bent molecules using VESPR theory. A tri-atomic molecule with one, or two lone pair on the central atom will also give rise to bent species. One lone pair examples include SO2, and two lone pair examples include H2O
Contributors and Attributions
• David Phinney (UC Davis)
3.08: Chiral Molecules
Introduction
Around the year 1847, the French scientist Louis Pasteur provided an explanation for the optical activity of tartaric acid salts. when he carried out a particular reaction, Pasteur observed that two types of crystals precipitated. Patiently and carefully using tweezers, Pasteur was able to separate the two types of crystals. Pasteur noticed that the types rotated the plane polarized by the same amount but in different directions. These two compounds are called enantiomers.
What are Enantiomers?
Two compounds are enantiomers if they are non-superimposable mirror images of each other. As was mentioned, enantiomers are characterized by their ability to rotate plane-polarized light. They also have the same physical properties (e.g., melting point, etc.) relative to each other. As a result, they are also referred to as being optically active. When it comes to symmetry, there are some general rules of thumb that help determine whether a molecule is chiral or achiral. This can be very useful because sometimes molecules can have relatively complicated structures and geometries that knowing whether or not they are chiral becomes a daunting task. The goal, as a result, is to determine the point group of the molecule and the symmetry elements associated with it, then inferring the chirality of the molecule.
Using Symmetry to Determine Chirality
For a molecule to be chiral, it must lack:
1. Center of inversion $i$ and a plane of symmetry $\sigma$.
2. An improper rotation axis (rotation-reflection axis) $S_{n}$.
However, since, by definition, an improper rotation axis is a rotation about an certain axis followed by reflection about a plane perpendicular to that axis, and an inversion center is simply $S_{2}$, the absence of an improper axis requires, in most cases, that absence of both a plane of symmetry and an inversion center. As a result, it suffices, in most cases, to check for improper rotation axes to determine whether a molecule is chiral or not.
As a result of the previous discussion, there are a few classes of point groups that lack an improper axis. Those classes are $C_{1}$, $C_{n}$, and $D_{n}$. Cis-dichlorobis(ethylenediamine)cobalt (III) has two enantiomers that are chiral (figure 1), but the trans compound is achiral. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/03%3A_Introduction_to_Molecular_Symmetry/3.07%3A_Vibrational_Spectroscopy/3.7A%3A_Vibrational_Spectroscopy.txt |
Guystav Kirchoff and Robert Bunsen first used atomic absorption spectroscopy—along with atomic emission—in 1859 and 1860 as a means for identify atoms in flames and hot gases. Although atomic emission continued to develop as an analytical technique, progress in atomic absorption languished for almost a century. Modern atomic absorption spectroscopy has its beginnings in 1955 as a result of the independent work of A. C. Walsh and C. T. J. Alkemade.13 Commercial instruments were in place by the early 1960s, and the importance of atomic absorption as an analytical technique was soon evident.
10.4.1 Instrumentation
Atomic absorption spectrophotometers use the same single-beam or double-beam optics described earlier for molecular absorption spectrophotometers (see Figure 10.26 and Figure 10.27). There is, however, an important additional need in atomic absorption spectroscopy—we must covert the analyte into free atoms. In most cases our analyte is in solution form. If our sample is a solid, then we must bring it into solution before the analysis. When analyzing a lake sediment for Cu, Zn, and Fe, for example, we bring the analytes into solution as Cu2+, Zn2+, and Fe3+ by extracting them with a suitable reagent. For this reason, only the introduction of solution samples is considered in this text.
What reagent we choose to use depends on our research goals. If we need to know the total amount of metal in the sediment, then we might use a microwave digestion using a mixture of concentrated acids, such as HNO3, HCl, and HF. This destroys the sediment’s matrix and brings everything into solution. On the other hand, if our interest is biologically available metals, we might extract the sample under milder conditions, such as a dilute solution of HCl or CH3COOH at room temperature.
Atomization
The process of converting an analyte to a free gaseous atom is called atomization. Converting an aqueous analyte into a free atom requires that we strip away the solvent, volatilize the analytes, and, if necessary, dissociate the analyte into free atoms. Desolvating an aqueous solution of CuCl2, for example, leaves us with solid particulates of CuCl2. Converting the particulate CuCl2 to gas phases atoms of Cu and Cl requires thermal energy.
$\ce{CuCl}_{2(aq)} \rightarrow \ce{CuCl}_{2(s)} \rightarrow \ce{Cu}_{(g)} + \ce{2Cl}_{(g)}$
There are two common atomization methods: flame atomization and electrothermal atomization, although a few elements are atomized using other methods.
Flame Atomizer
Figure 10.42 shows a typical flame atomization assembly with close-up views of several key components. In the unit shown here, the aqueous sample is drawn into the assembly by passing a high-pressure stream of compressed air past the end of a capillary tube immersed in the sample. When the sample exits the nebulizer it strikes a glass impact bead, converting it into a fine aerosol mist within the spray chamber. The aerosol mist is swept through the spray chamber by the combustion gases—compressed air and acetylene in this case—to the burner head where the flame’s thermal energy desolvates the aerosol mist to a dry aerosol of small, solid particles. The flame’s thermal energy then volatilizes the particles, producing a vapor consisting of molecular species, ionic species, and free atoms.
Compressed air is one of the two gases whose combustion produces the flame.
Figure 10.42 Flame atomization assembly with expanded views of (a) the burner head showing the burner slot where the flame is located; (b) the nebulizer’s impact bead; and (c) the interior of the spray chamber. Although the unit shown here is from an older instrument, the basic components of a modern flame AA spectrometer are the same.
Burner. The slot burner in Figure 10.42a provides a long optical pathlength and a stable flame. Because absorbance increases linearly with the path length, a long path length provides greater sensitivity. A stable flame minimizes uncertainty due to fluctuations in the flame.
The burner is mounted on an adjustable stage that allows the entire assembly to move horizontally and vertically. Horizontal adjustments ensure that the flame is aligned with the instrument’s optical path. Vertical adjustments adjust the height within the flame from which absorbance is monitored. This is important because two competing processes affect the concentration of free atoms in the flame. The more time the analyte spends in the flame the greater the atomization efficiency; thus, the production of free atoms increases with height. On the other hand, a longer residence time allows more opportunity for the free atoms to combine with oxygen to form a molecular oxide. For an easily oxidized metal, such as Cr, the concentration of free atoms is greatest just above the burner head. For metals, such as Ag, which are difficult to oxidize, the concentration of free atoms increases steadily with height (Figure 10.43). Other atoms show concentration profiles that maximize at a characteristic height.
Figure 10.43 Absorbance versus height profiles for Ag and Cr in flame atomic absorption spectroscopy.
Flame. The flame’s temperature, which affects the efficiency of atomization, depends on the fuel–oxidant mixture, several examples of which are listed in Table 10.9. Of these, the air–acetylene and the nitrous oxide–acetylene flames are the most popular. Normally the fuel and oxidant are mixed in an approximately stoichiometric ratio; however, a fuel-rich mixture may be necessary for easily oxidized analytes.
Table 10.9 Fuels and Oxidants Used for Flame Combustion
fuel
oxidant
temperature range (oC)
natural gas
air
1700–1900
hydrogen
air
2000–2100
acetylene
air
2100–2400
acetylene
nitrous oxide
2600–2800
acetylene
oxygen
3050–3150
Figure 10.44 shows a cross-section through the flame, looking down the source radiation’s optical path. The primary combustion zone is usually rich in gas combustion products that emit radiation, limiting is usefulness for atomic absorption. The interzonal region generally is rich in free atoms and provides the best location for measuring atomic absorption. The hottest part of the flame is typically 2–3 cm above the primary combustion zone. As atoms approach the flame’s secondary combustion zone, the decrease in temperature allows for formation of stable molecular species.
Figure 10.44 Profile of typical flame using a slot burner. The relative size of each zone depends on many factors, including the choice of fuel and oxidant, and their relative proportions.
Sample Introduction. The most common means for introducing samples into a flame atomizer is a continuous aspiration in which the sample flows through the burner while we monitor the absorbance. Continuous aspiration is sample intensive, typically requiring from 2–5 mL of sample.
Flame microsampling allows us to introduce a discrete sample of fixed volume, and is useful when we have a limited amount of sample or when the sample’s matrix is incompatible with the flame atomizer. For example, continuously aspirating a sample that has a high concentration of dissolved solids—sea water, for example, comes to mind—may build-up a solid deposit on the burner head that obstructs the flame and that lowers the absorbance. Flame microsampling is accomplished using a micropipet to place 50–250 μL of sample in a Teflon funnel connected to the nebulizer, or by dipping the nebulizer tubing into the sample for a short time. Dip sampling is usually accomplished with an automatic sampler. The signal for flame microsampling is a transitory peak whose height or area is proportional to the amount of analyte that is injected.
Advantages and Disadvantages of Flame Atomization. The principal advantage of flame atomization is the reproducibility with which the sample is introduced into the spectrophotometer. A significant disadvantage to flame atomizers is that the efficiency of atomization may be quite poor. There are two reasons for poor atomization efficiency. First, the majority of the aerosol droplets produced during nebulization are too large to be carried to the flame by the combustion gases. Consequently, as much as 95% of the sample never reaches the flame. A second reason for poor atomization efficiency is that the large volume of combustion gases significantly dilutes the sample. Together, these contributions to the efficiency of atomization reduce sensitivity because the analyte’s concentration in the flame may be a factor of 2.5 × 10–6 less than that in solution.14 This is the reason for the waste line shown at the bottom of the spray chamber in Figure 10.42.
Electrothermal Atomizers
A significant improvement in sensitivity is achieved by using the resistive heating of a graphite tube in place of a flame. A typical electrothermal atomizer, also known as a graphite furnace, consists of a cylindrical graphite tube approximately 1–3 cm in length and 3–8 mm in diameter. As shown in Figure 10.45, the graphite tube is housed in an sealed assembly that has optically transparent windows at each end. A continuous stream of an inert gas is passed through the furnace, protecting the graphite tube from oxidation and removing the gaseous products produced during atomization. A power supply is used to pass a current through the graphite tube, resulting in resistive heating.
Figure 10.45: Diagram showing a cross-section of an electrothermal analyzer.
Samples of between 5–50 μL are injected into the graphite tube through a small hole at the top of the tube. Atomization is achieved in three stages. In the first stage the sample is dried to a solid residue using a current that raises the temperature of the graphite tube to about 110oC. In the second stage, which is called ashing, the temperature is increased to between 350–1200oC. At these temperatures any organic material in the sample is converted to CO2 and H2O, and volatile inorganic materials are vaporized. These gases are removed by the inert gas flow. In the final stage the sample is atomized by rapidly increasing the temperature to between 2000–3000oC. The result is a transient absorbance peak whose height or area is proportional to the absolute amount of analyte injected into the graphite tube. Together, the three stages take approximately 45–90 s, with most of this time used for drying and ashing the sample.
Electrothermal atomization provides a significant improvement in sensitivity by trapping the gaseous analyte in the small volume within the graphite tube. The analyte’s concentration in the resulting vapor phase may be as much as 1000× greater than in a flame atomization.15 This improvement in sensitivity—and the resulting improvement in detection limits—is offset by a significant decrease in precision. Atomization efficiency is strongly influenced by the sample’s contact with the graphite tube, which is difficult to control reproducibly.
Miscellaneous Atomization Methods
A few elements may be atomized by a chemical reaction that produces a volatile product. Elements such as As, Se, Sb, Bi, Ge, Sn, Te, and Pb, for example, form volatile hydrides when reacted with NaBH4 in acid. An inert gas carries the volatile hydrides to either a flame or to a heated quartz observation tube situated in the optical path. Mercury is determined by the cold-vapor method in which it is reduced to elemental mercury with SnCl2. The volatile Hg is carried by an inert gas to an unheated observation tube situated in the instrument’s optical path.
10.4.2 Quantitative Applications
Atomic absorption is widely used for the analysis of trace metals in a variety of sample matrices. Using Zn as an example, atomic absorption methods have been developed for its determination in samples as diverse as water and wastewater, air, blood, urine, muscle tissue, hair, milk, breakfast cereals, shampoos, alloys, industrial plating baths, gasoline, oil, sediments, and rocks.
Developing a quantitative atomic absorption method requires several considerations, including choosing a method of atomization, selecting the wavelength and slit width, preparing the sample for analysis, minimizing spectral and chemical interferences, and selecting a method of standardization. Each of these topics is considered in this section.
Developing a Quantitative Method
Flame or Electrothermal Atomization? The most important factor in choosing a method of atomization is the analyte’s concentration. Because of its greater sensitivity, it takes less analyte to achieve a given absorbance when using electrothermal atomization. Table 10.10, which compares the amount of analyte needed to achieve an absorbance of 0.20 when using flame atomization and electrothermal atomization, is useful when selecting an atomization method. For example, flame atomization is the method of choice if our samples contain 1–10 mg Zn2+/L, but electrothermal atomization is the best choice for samples containing 1–10 μg Zn2+/L.
Table 10.10: Concentration of Analyte Yielding an Absorbance of 0.20
Concentration (mg/L)a
element
flame atomization
electrothermal atomization
Ag
1.5
0.0035
Al
40
0.015
As
40b
0.050
Ca
0.8
0.003
Cd
0.6
0.001
Co
2.5
0.021
Cr
2.5
0.0075
Cu
1.5
0.012
Fe
2.5
0.006
Hg
70b
0.52
Mg
0.15
0.00075
Mn
1
0.003
Na
0.3
0.00023
Ni
2
0.024
Pb
5
0.080
Pt
70
0.29
Sn
50b
0.023
Zn
0.3
0.00071
a Source: Varian Cookbook, SpectraAA Software Version 4.00 Pro.
b As: 10 mg/L by hydride vaporization; Hg: 11.5 mg/L by cold-vapor; and Sn:18 mg/L by hydride vaporization
Selecting the Wavelength and Slit Width. The source for atomic absorption is a hollow cathode lamp consisting of a cathode and anode enclosed within a glass tube filled with a low pressure of Ne or Ar (Figure 10.46). Applying a potential across the electrodes ionizes the filler gas. The positively charged gas ions collide with the negatively charged cathode, sputtering atoms from the cathode’s surface. Some of the sputtered atoms are in the excited state and emit radiation characteristic of the metal(s) from which the cathode was manufactured. By fashioning the cathode from the metallic analyte, a hollow cathode lamp provides emission lines that correspond to the analyte’s absorption spectrum.
Because atomic absorption lines are narrow, we need to use a line source instead of a continuum source (compare, for example, Figure 10.18 with Figure 10.20). The effective bandwidth when using a continuum source is roughly 1000× larger than an atomic absorption line; thus, PT P0, %T 100, and A 0. Because a hollow cathode lamp is a line source, PT and P0 have different values giving a %T < 100 and A > 0.
Figure 10.46: Photo of a typical multielemental hollow cathode lamp. The cathode in this lamp is fashioned from an alloy containing Co, Cr, Cu, Fe, Mn, and Ni, and is surrounded by a glass shield to isolate it from the anode. The lamp is filled with Ne gas. Also shown is the process leading to atomic emission. See the text for an explanation.
Each element in a hollow cathode lamp provides several atomic emission lines that we can use for atomic absorption. Usually the wavelength that provides the best sensitivity is the one we choose to use, although a less sensitive wavelength may be more appropriate for a larger concentration of analyte. For the Cr hollow cathode lamp in Table 10.11, for example, the best sensitivity is obtained using a wavelength of 357.9 nm.
Another consideration is the intensity of the emission line. If several emission lines meet our need for sensitivity, we may wish to use the emission line with the largest relative P0 because there is less uncertainty in measuring P0 and PT. When analyzing samples containing 10 mg Cr/L, for example, the first three wavelengths in Table 10.11 provide an appropriate sensitivity. The wavelengths of 425.5 nm and 429.0 nm, however, have a greater P0 and will provide less uncertainty in the measured absorbance.
Table 10.11 Atomic Emission Lines for a Cr Hollow Cathode Lamp
wavelength (nm)
slit width (nm)
mg Cr/L giving A = 0.20
P0 (relative)
357.9
0.2
2.5
40
425.4
0.2
12
85
429.0
0.5
20
100
520.5
0.2
1500
15
520.8
0.2
500
20
The emission spectrum from a hollow cathode lamp includes, besides emission lines for the analyte, additional emission lines for impurities present in the metallic cathode and from the filler gas. These additional lines are a source of stray radiation that leads to an instrumental deviation from Beer’s law. The monochromator’s slit width is set as wide as possible, improving the throughput of radiation, while, at the same time, being narrow enough to eliminate the stray radiation.
Preparing the Sample. Flame and electrothermal atomization require that the sample be in solution. Solid samples are brought into solution by dissolving in an appropriate solvent. If the sample is not soluble it may be digested, either on a hot-plate or by microwave, using HNO3, H2SO4, or HClO4. Alternatively, we can extract the analyte using a Soxhlet extractor. Liquid samples may be analyzed directly or extracted if the matrix is incompatible with the method of atomization. A serum sample, for instance, is difficult to aspirate when using flame atomization and may produce an unacceptably high background absorbance when using electrothermal atomization. A liquid–liquid extraction using an organic solvent and a chelating agent is frequently used to concentrate analytes. Dilute solutions of Cd2+, Co2+, Cu2+, Fe3+, Pb2+, Ni2+, and Zn2+, for example, can be concentrated by extracting with a solution of ammonium pyrrolidine dithiocarbamate in methyl isobutyl ketone.See Chapter 7 to review different methods for preparing samples for analysis.
Minimizing Spectral Interference. A spectral interference occurs when an analyte’s absorption line overlaps with an interferent’s absorption line or band. Because they are so narrow, the overlap of two atomic absorption lines is seldom a problem. On the other hand, a molecule’s broad absorption band or the scattering of source radiation is a potentially serious spectral interference.
An important consideration when using a flame as an atomization source is its effect on the measured absorbance. Among the products of combustion are molecular species that exhibit broad absorption bands and particulates that scatter radiation from the source. If we fail to compensate for these spectral interference, then the intensity of transmitted radiation decreases. The result is an apparent increase in the sample’s absorbance. Fortunately, absorption and scattering of radiation by the flame are corrected by analyzing a blank.
Spectral interferences also occur when components of the sample’s matrix other than the analyte react to form molecular species, such as oxides and hydroxides. The resulting absorption and scattering constitutes the sample’s background and may present a significant problem, particularly at wavelengths below 300 nm where the scattering of radiation becomes more important. If we know the composition of the sample’s matrix, then we can prepare our samples using an identical matrix. In this case the background absorption is the same for both the samples and standards. Alternatively, if the background is due to a known matrix component, then we can add that component in excess to all samples and standards so that the contribution of the naturally occurring interferent is insignificant. Finally, many interferences due to the sample’s matrix can be eliminated by increasing the atomization temperature. For example, by switching to a higher temperature flame it may be possible to prevent the formation of interfering oxides and hydroxides.
If the identity of the matrix interference is unknown, or if it is not possible to adjust the flame or furnace conditions to eliminate the interference, then we must find another method to compensate for the background interference. Several methods have been developed to compensate for matrix interferences, and most atomic absorption spectrophotometers include one or more of these methods.
One of the most common methods for background correction is to use a continuum source, such as a D2 lamp. Because a D2 lamp is a continuum source, absorbance of its radiation by the analyte’s narrow absorption line is negligible. Only the background, therefore, absorbs radiation from the D2 lamp. Both the analyte and the background, on the other hand, absorb the hollow cathode’s radiation. Subtracting the absorbance for the D2 lamp from that for the hollow cathode lamp gives a corrected absorbance that compensates for the background interference. Although this method of background correction may be quite effective, it does assume that the background absorbance is constant over the range of wavelengths passed by the monochromator. If this is not true, subtracting the two absorbances may underestimate or overestimate the background.
Other methods of background correction have been developed, including Zeeman effect background correction and Smith–Hieftje background correction, both of which are included in some commercially available atomic absorption spectrophotometers. Consult the chapter’s additional resources for additional information.
Minimizing Chemical Interferences. The quantitative analysis of some elements is complicated by chemical interferences occurring during atomization. The two most common chemical interferences are the formation of nonvolatile compounds containing the analyte and ionization of the analyte.
One example of the formation of nonvolatile compounds is the effect of PO43– or Al3+ on the flame atomic absorption analysis of Ca2+. In one study, for example, adding 100 ppm Al3+ to a solution of 5 ppm Ca2+ decreased the calcium ion’s absorbance from 0.50 to 0.14, while adding 500 ppm PO43– to a similar solution of Ca2+ decreased the absorbance from 0.50 to 0.38. These interferences were attributed to the formation of nonvolatile particles of Ca3(PO4)2 and an Al–Ca–O oxide.16
When using flame atomization, we can minimize the formation of nonvolatile compounds by increasing the flame’s temperature, either by changing the fuel-to-oxidant ratio or by switching to a different combination of fuel and oxidant. Another approach is to add a releasing agent or a protecting agent to the samples. A releasing agent is a species that reacts with the interferent, releasing the analyte during atomization. Adding Sr2+ or La3+ to solutions of Ca2+, for example, minimizes the effect of PO43– and Al3+ by reacting in place of the analyte. Thus, adding 2000 ppm SrCl2 to the Ca2+/PO43– and Ca2+/Al3+ mixtures described in the previous paragraph increased the absorbance to 0.48. A protecting agent reacts with the analyte to form a stable volatile complex. Adding 1% w/w EDTA to the Ca2+/PO43– solution described in the previous paragraph increased the absorbance to 0.52.
Ionization interferences occur when thermal energy from the flame or the electrothermal atomizer is sufficient to ionize the analyte
$\ce{M}_{(g)} \overset{\Delta}{\rightleftharpoons} \ce{M}^+_{(g)} + e^− \tag{10.24}$
where M is the analyte. Because the absorption spectra for M and M+ are different, the position of the equilibrium in reaction 10.24 affects absorbance at wavelengths where M absorbs. To limit ionization we add a high concentration of an ionization suppressor, which is simply a species that ionizes more easily than the analyte. If the concentration of the ionization suppressor is sufficient, then the increased concentration of electrons in the flame pushes reaction 10.24 to the left, preventing the analyte’s ionization. Potassium and cesium are frequently used as an ionization suppressor because of their low ionization energy.
Standardizing the Method. Because Beer’s law also applies to atomic absorption, we might expect atomic absorption calibration curves to be linear. In practice, however, most atomic absorption calibration curves are nonlinear, or linear for only a limited range of concentrations. Nonlinearity in atomic absorption is a consequence of instrumental limitations, including stray radiation from the hollow cathode lamp and the variation in molar absorptivity across the absorption line. Accurate quantitative work, therefore, often requires a suitable means for computing the calibration curve from a set of standards.
Most instruments include several different algorithms for computing the calibration curve. The instrument in my lab, for example, includes five algorithms. Three of the algorithms fit absorbance data using linear, quadratic, or cubic polynomial functions of the analyte’s concentration. It also includes two algorithms that fit the concentrations of the standards to quadratic functions of the absorbance.
When possible, a quantitative analysis is best conducted using external standards. Unfortunately, matrix interferences are a frequent problem, particularly when using electrothermal atomization. For this reason the method of standard additions is often used. One limitation to this method of standardization, however, is the requirement that there be a linear relationship between absorbance and concentration.
The best way to appreciate the theoretical and practical details discussed in this section is to carefully examine a typical analytical method. Although each method is unique, the following description of the determination of Cu and Zn in biological tissues provides an instructive example of a typical procedure. The description here is based on Bhattacharya, S. K.; Goodwin, T. G.; Crawford, A. J. Anal. Lett. 1984, 17, 1567–1593, and Crawford, A. J.; Bhattacharya, S. K. Varian Instruments at Work, Number AA–46, April 1985.
Representative Method 10.2: Determination of Cu and Zn in Tissue Samples
Description of Method
Copper and zinc are isolated from tissue samples by digesting the sample with HNO3 after first removing any fatty tissue. The concentration of copper and zinc in the supernatant are determined by atomic absorption using an air-acetylene flame.
Procedure
Tissue samples are obtained by a muscle needle biopsy and dried for 24–30 h at 105oC to remove all traces of moisture. The fatty tissue in the dried samples is removed by extracting overnight with anhydrous ether. After removing the ether, the sample is dried to obtain the fat-free dry tissue weight (FFDT). The sample is digested at 68oC for 20–24 h using 3 mL of 0.75 M HNO3. After centrifuging at 2500 rpm for 10 minutes, the supernatant is transferred to a 5-mL volumetric flask. The digestion is repeated two more times, for 2–4 hours each, using 0.9-mL aliquots of 0.75 M HNO3. These supernatants are added to the 5-mL volumetric flask, which is diluted to volume with 0.75 M HNO3. The concentrations of Cu and Zn in the diluted supernatant are determined by flame atomic absorption spectroscopy using an air-acetylene flame and external standards. Copper is analyzed at a wavelength of 324.8 nm with a slit width of 0.5 nm, and zinc is analyzed at 213.9 nm with a slit width of 1.0 nm. Background correction using a D2 lamp is necessary for zinc. Results are reported as mg of Cu or Zn per gram of FFDT.
Questions
1. Describe the appropriate matrix for the external standards and for the blank?
The matrix for the standards and the blank should match the matrix of the samples; thus, an appropriate matrix is 0.75 M HNO3. Any interferences from other components of the sample matrix are minimized by background correction.
2. Why is a background correction necessary for the analysis of Zn, but not for the analysis of Cu?
Background correction compensates for background absorption and scattering due to interferents in the sample. Such interferences are most severe when using a wavelength less than 300 nm. This is the case for Zn, but not for Cu.
3. A Cu hollow cathode lamp has several emission lines. Explain why this method uses the line at 324.8 nm.
wavelength
(nm)
slit width (nm)
mg Cu/L for
A = 0.20
P0 (relative)
217.9
0.2
15
3
218.2
0.2
15
3
222.6
0.2
60
5
244.2
0.2
400
15
249.2
0.5
200
24
324.8
0.5
1.5
100
327.4
0.5
3
87
With 1.5 mg Cu/L giving an absorbance of 0.20, the emission line at 324.8 nm has the best sensitivity. In addition, it is the most intense emission line, which decreases the uncertainty in the measured absorbance.
Example 10.10
To evaluate the method described in Representative Method 10.2, a series of external standard is prepared and analyzed, providing the results shown here.17
μg Cu/mL absorbance μg Cu/mL absorbance
0.000
0.100
0.200
0.300
0.400
0.000
0.006
0.013
0.020
0.026
0.500
0.600
0.700
1.000
0.033
0.039
0.046
0.066
A bovine liver standard reference material was used to evaluate the method’s accuracy. After drying and extracting the sample, a 11.23-mg FFDT tissue sample gives an absorbance of 0.023. Report the amount of copper in the sample as μg Cu/g FFDT.
Solution
Linear regression of absorbance versus the concentration of Cu in the standards gives a calibration curve with the following equation.
$A = \mathrm{−0.0002 + 0.0661 × \dfrac{g\: Cu}{mL}}$
Substituting the sample’s absorbance into the calibration equation gives the concentration of copper as 0.351 μg/mL. The concentration of copper in the tissue sample, therefore, is
$\mathrm{\dfrac{\dfrac{0.351\: g\: Cu}{mL} × 5.000\: mL}{0.01123\: g\: sample} = 156\: g\: Cu/g\: FFDT}$
10.4.3 - Evaluation of Atomic Absorption Spectroscopy
Scale of Operation
Atomic absorption spectroscopy is ideally suited for the analysis of trace and ultratrace analytes, particularly when using electrothermal atomization. For minor and major analyte, sample can be diluted before the analysis. Most analyses use a macro or a meso sample. The small volume requirement for electrothermal atomization or flame microsampling, however, makes practical the analysis micro and ultramicro samples.
See Figure 3.5 to review the meaning of macro and meso for describing samples, and the meaning of major, minor, and ultratrace for describing analytes.
Accuracy
If spectral and chemical interferences are minimized, an accuracy of 0.5–5% is routinely attainable. When the calibration curve is nonlinear, accuracy may be improved by using a pair of standards whose absorbances closely bracket the sample’s absorbance and assuming that the change in absorbance is linear over this limited concentration range. Determinate errors for electrothermal atomization are often greater than that obtained with flame atomization due to more serious matrix interferences.
Precision
For absorbance values greater than 0.1–0.2, the relative standard deviation for atomic absorption is 0.3–1% for flame atomization and 1–5% for electrothermal atomization. The principle limitation is the variation in the concentration of free analyte atoms resulting from variations in the rate of aspiration, nebulization, and atomization when using a flame atomizer, and the consistency of injecting samples when using electrothermal atomization.
Sensitivity
The sensitivity of a flame atomic absorption analysis is influenced strongly by the flame’s composition and by the position in the flame from which we monitor the absorbance. Normally the sensitivity of an analysis is optimized by aspirating a standard solution of the analyte and adjusting operating conditions, such as the fuel-to-oxidant ratio, the nebulizer flow rate, and the height of the burner, to give the greatest absorbance. With electrothermal atomization, sensitivity is influenced by the drying and ashing stages that precede atomization. The temperature and time used for each stage must be optimized for each type of sample.
See Chapter 14 for several strategies for optimizing experiments.
Sensitivity is also influenced by the sample’s matrix. We have already noted, for example, that sensitivity can be decreased by chemical interferences. An increase in sensitivity may be realized by adding a low molecular weight alcohol, ester, or ketone to the solution, or by using an organic solvent.
Selectivity
Due to the narrow width of absorption lines, atomic absorption provides excellent selectivity. Atomic absorption can be used for the analysis of over 60 elements at concentrations at or below the level of μg/L.
Time, Cost, and Equipment
The analysis time when using flame atomization is short, with sample throughputs of 250–350 determinations per hour when using a fully automated system. Electrothermal atomization requires substantially more time per analysis, with maximum sample throughputs of 20–30 determinations per hour. The cost of a new instrument ranges from between $10,000–$50,000 for flame atomization, and from $18,000–$70,000 for electrothermal atomization. The more expensive instruments in each price range include double-beam optics, automatic samplers, and can be programmed for multielemental analysis by allowing the wavelength and hollow cathode lamp to be changed automatically. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/04%3A_Experimental_Techniques/4.03%3A_Elemental_Analysis/4.3B%3A_Atomic_Absorption_Spectroscopy_%28AAS%29.txt |
Magnetism results from the circular motion of charged particles. This property is demonstrated on a macroscopic scale by making an electromagnet from a coil of wire and a battery. Electrons moving through the coil produce a magnetic field (Figure $1$), which can be thought of as originating from a magnetic dipole or a bar magnet.
Magnetism results from the circular motion of charged particles.
Electrons in atoms also are moving charges with angular momentum so they too produce a magnetic dipole, which is why some materials are magnetic. A magnetic dipole interacts with an applied magnetic field, and the energy of this interaction is given by the scalar product of the magnetic dipole moment, and the magnetic field, $\vec{B}$.
$E_B = - \vec{\mu} _m \cdot \vec{B} \label {8.4.1}$
Magnets are acted on by forces and torques when placed within an external applied magnetic field (Figure $2$). In a uniform external field, a magnet experiences no net force, but a net torque. The torque tries to align the magnetic moment ($\vec{\mu} _m$ of the magnet with the external field $\vec{B}$. The magnetic moment of a magnet points from its south pole to its north pole.
In a non-uniform magnetic field a current loop, and therefore a magnet, experiences a net force, which tries to pull an aligned dipole into regions where the magnitude of the magnetic field is larger and push an anti-aligned dipole into regions where magnitude the magnetic field is smaller.
Quantum Effects
As expected, the quantum picture is different. Pieter Zeeman was one of the first to observe the splittings of spectral lines in a magnetic field caused by this interaction. Consequently, such splittings are known as the Zeeman effect. Let’s now use our current knowledge to predict what the Zeeman effect for the 2p to 1s transition in hydrogen would look like, and then compare this prediction with a more complete theory. To understand the Zeeman effect, which uses a magnetic field to remove the degeneracy of different angular momentum states, we need to examine how an electron in a hydrogen atom interacts with an external magnetic field, $\vec{B}$. Since magnetism results from the circular motion of charged particles, we should look for a relationship between the angular momentum $\vec{L}$ and the magnetic dipole moment $\vec{\mu} _m$.
The relationship between the magnetic dipole moment $\vec{\mu} _m$ (also referred to simply as the magnetic moment) and the angular momentum $\vec{L}$ of a particle with mass m and charge $q$ is given by
$\vec{\mu} _m = \dfrac {q}{2m} \vec{L} \label {8.4.2}$
For an electron, this equation becomes
$\vec{\mu} _m = - \dfrac {e}{2m_e} \vec{L} \label {8.4.3}$
where the specific charge and mass of the electron have been substituted for $q$ and $m$. The magnetic moment for the electron is a vector pointing in the direction opposite to $\vec{L}$, both of which classically are perpendicular to the plane of the rotational motion.
Exercise $1$
Will an electron in the ground state of hydrogen have a magnetic moment? Why or why not?
The relationship between the angular momentum of a particle and its magnetic moment is commonly expressed as a ratio, called the gyromagnetic ratio, $\gamma$. Gyro is Greek for turn so gyromagnetic simply relates turning (angular momentum) to magnetism. Now you also know why the Greek sandwiches made with meat cut from a spit turning over a fire are called gyros.
$\gamma = \dfrac {\mu _m}{L} = \dfrac {q}{2m} \label {8.4.4}$
In the specific case of an electron,
$\gamma _e = - \dfrac {e}{2m_e} \label {8.4.5}$
Exercise $2$
Calculate the magnitude of the gyromagnetic ratio for an electron.
To determine the energy of a hydrogen atom in a magnetic field we need to include the operator form of the hydrogen atom Hamiltonian. The Hamiltonian always consists of all the energy terms that are relevant to the problem at hand.
$\hat {H} = \hat {H} ^0 + \hat {H} _m \label {8.4.6}$
where $\hat {H} ^0$ is the Hamiltonian operator in the absence of the field and $\hat {H} _m$ is written using the operator forms of Equations $\ref{8.4.1}$ and $\ref{8.4.3}$),
$\hat {H}_m = - \hat {\mu} \cdot \vec{B} = \dfrac {e}{2m_e} \hat {L} \cdot B \label {8.4.7}$
The scalar product
$\hat {L} \cdot \vec{B} = \hat {L}_x B_x + \hat {L}_y B_y + \hat {L}_z B_z \label {8.4.8}$
simplifies if the z-axis is defined as the direction of the external field because then $B_x$ and $B_y$ are automatically 0, and Equation \ref{8.4.6} becomes
$\hat {H} = \hat {H}^0 + \dfrac {eB_z}{2m_e} \hat {L} _z \label {8.4.9}$
where $B_z$ is the magnitude of the magnetic field, which is along the z-axis.
We now can ask, “What is the effect of a magnetic field on the energy of the hydrogen atom orbitals?” To answer this question, we will not solve the Schrödinger equation again; we simply calculate the expectation value of the energy, $\left \langle E \right \rangle$, using the existing hydrogen atom wavefunctions and the new Hamiltonian operator.
$\left \langle E \right \rangle = \left \langle \hat {H}^0 \right \rangle + \dfrac {eB_z}{2m_e} \left \langle \hat {L} _z \right \rangle \label {8.4.10}$
where
$\left \langle \hat {H}^0 \right \rangle = \int \psi ^*_{n,l,m_l} \hat {H}^0 \psi _{n,l,m_l} d \tau = E_n \label {8.4.11}$
and
$\left \langle \hat {L}_z \right \rangle = \int \psi ^*_{n,l,m_l} \hat {L}_z \psi _{n,l,m_l} d \tau = m_l \hbar \label {8.4.12}$
Exercise $3$
Show that the expectation value $\left \langle \hat {L}_z \right \rangle = m_l \hbar$.
The expectation value approach provides an exact result in this case because the hydrogen atom wavefunctions are eigenfunctions of both $\hat {H} ^0$ and $\hat {L}_z$. If the wavefunctions were not eigenfunctions of the operator associated with the magnetic field, then this approach would provide a first-order estimate of the energy. First and higher order estimates of the energy are part of a general approach to developing approximate solutions to the Schrödinger equation. This approach, called perturbation theory, is discussed in the next chapter.
The expectation value calculated for the total energy in this case is the sum of the energy in the absence of the field, $E_n$, plus the Zeeman energy, $\dfrac {e \hbar B_z m_l}{2m_e}$
\begin{align} \left \langle E \right \rangle &= E_n + \dfrac {e \hbar B_z m_l}{2m_e} \[4pt] &= E_n + \mu _B B_z m_l \label {8.4.13} \end{align}
The factor
$\dfrac {e \hbar}{2m_e} = - \gamma _e \hbar = \mu _B \label {8.4.14}$
defines the constant $\mu _B$, called the Bohr magneton, which is taken to be the fundamental magnetic moment. It has units of $9.2732 \times 10^{-21}$ erg/Gauss or $9.2732 \times 10^{-24}$ Joule/Tesla. This factor will help you to relate magnetic fields, measured in Gauss or Tesla, to energies, measured in ergs or Joules, for any particle with a charge and mass the same as an electron.
Equation \ref{8.4.13} shows that the $m_l$ quantum number degeneracy of the hydrogen atom is removed by the magnetic field. For example, the three states $\psi _{211}$, $\psi _{21-1}$, and $\psi _{210}$, which are degenerate in zero field, have different energies in a magnetic field, as shown in Figure $3$.
The $m_l = 0$ state, for which the component of angular momentum and hence also the magnetic moment in the external field direction is zero, experiences no interaction with the magnetic field. The $m_l = +1$ state, for which the angular momentum in the z-direction is +ħ and the magnetic moment is in the opposite direction, against the field, experiences a raising of energy in the presence of a field. Maintaining the magnetic dipole against the external field direction is like holding a small bar magnet with its poles aligned exactly opposite to the poles of a large magnet (Figure $5$). It is a higher energy situation than when the magnetic moments are aligned with each other.
Exercise $4$
Carry out the steps going from Equation $\ref{8.4.10}$ to Equation $\ref{8.4.13}$.
Exercise $5$
Consider the effect of changing the magnetic field on the magnitude of the Zeeman splitting. Sketch a diagram where the magnetic field strength is on the x-axis and the energy of the three 2p orbitals is on the y-axis to show the trend in splitting magnitudes with increasing magnetic field. Be quantitative, calculate and plot the exact numerical values using a software package of your choice.
Exercise $6$
Based on your calculations in Exercise $2$ sketch a luminescence spectrum for the hydrogen atom in the n = 2 level in a magnetic field of 1 Tesla. Provide the numerical value for each of the transition energies. Use cm-1 or electron volts for the energy units. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/04%3A_Experimental_Techniques/4.09%3A_Electron_Paramagnetic_Resonance_%28EPR%29_Spectroscopy/4.9B%3A_The_Zeeman_Electronic_Effect.txt |
Mössbauer spectroscopy is a versatile technique used to study nuclear structure with the absorption and re-emission of gamma rays, part of the electromagnetic spectrum. The technique uses a combination of the Mössbauer effect and Doppler shifts to probe the hyperfine transitions between the excited and ground states of the nucleus. Mössbauer spectroscopy requires the use of solids or crystals which have a probability to absorb the photon in a recoilless manner, many isotopes exhibit Mössbauer characteristics but the most commonly studied isotope is 57Fe.
Introduction
Rudolf L. Mössbauer became a physics student at Technical University in Munich at the age of 20. After passing his intermediate exams Mössbauer began working on his thesis and doctorate work in 1955, while working as an assistant lecturer at Institute for Mathematics. In 1958 at the age of 28 Mössbauer graduated, and also showed experimental evidence for recoilless resonant absorption in the nucleus, later to be called the Mössbauer Effect. In 1961 Mössbauer was awarded the Nobel Prize in physics and, under the urging of Richard Feynman, accepted the position of Professor of Physics at the California Institute of Technology.
Mössbauer Effect
The recoil energy associated with absorption or emission of a photon can be described by the conservation of momentum. In it we find that the recoil energy depends inversely on the mass of the system. For a gas the mass of the single nucleus is small compared to a solid. The solid or crystal absorbs the energy as phonons, quantized vibration states of the solid, but there is a probability that no phonons are created and the whole lattice acts as the mass, resulting in a recoilless emission of the gamma ray. The new radiation is at the proper energy to excite the next ground state nucleus. The probability of recoilless events increases with decreasing transition energy.
$P_R = P_{\gamma} \nonumber$
$P^2_\gamma = P^2_{\gamma} \nonumber$
$2 M E_R = \dfrac{E^2_{\gamma}}{c^2} \nonumber$
$E_R = \dfrac{E^2_\gamma}{2M{c^2}} \nonumber$
Doppler Effect
The Doppler shift describes the change in frequency due to a moving source and a moving observer. $f$ is the frequency measured at the observer, $v$ is the velocity of the wave so for our case this is the speed of light $c$, $v_r$ is the velocity of the observer, $v_s$ is the velocity of the source which is positive when heading away from the observer, and $f_0$ is the initial frequency.
$f = {\left (\dfrac{v+v_r}{v+v_s}\right)} f_0 \nonumber$
$f = {\left (\dfrac{c}{c+v_s}\right)} f_0 \nonumber$
In the case where the source is moving toward a stationary observer the perceived frequency is higher. For the opposite situation where the source travels away from the observer frequencies recorded at the observer will be of lower compared to the initial wave. The energy of a photon is related to the product of Planck's constant and the frequency of the electromagnetic radiation. Thus for increasing frequencies the corresponding energy also increase, and the same is true in the reverse case where frequencies decrease and therefore energy decreases.
$E = \dfrac{hc}{\lambda} = hv \nonumber$
The energy differences between hyperfine states are minimal (fractions of an eV) and the energy variation is achieved by the moving the source toward and away from the sample in an oscillating manner, commonly at a velocity of a few mm/s. The transmittance is then plotted against the velocity of the source and a peak is seen at the energy corresponding to the resonance energy.
In the above spectrum the emission and absorption are both estimated by the Lorentzian distribution.
Mössbauer Isotopes
By far the most common isotopes studied using Mössbauer spectroscopy is 57Fe, but many other isotopes have also displayed a Mössbauer spectrum. Two criteria for functionality are
1. The excited state is of very low energy, resulting in a small change in energy between ground and excited state. This is because gamma rays at higher energy are not absorbed in a recoil free manner, meaning resonance only occurs for gamma rays of low energy.
2. The resolution of Mössbauer spectroscopy depends upon the lifetime of the excited state. The longer the excited state lasts the better the image.
Both conditions are met by 57Fe and it is thus used extensively in Mössbauer spectroscopy. In the figure to the right the red colored boxes of the periodic table of elements indicate all elements that have isotopes visible using the Mössbauer technique.
Hyperfine Interactions
Mössbauer spectroscopy allows the researcher to probe structural elements of the nucleus in several ways, termed isomer shift, quadrupole interactions, and magnetic splitting. These are each explained by the following sections as individual graphs, but in practice Mössbauer spectrum are likely to contain a combination of all effects.
Isomer Shift
An isomeric shift occurs when non identical atoms play the role of source and absorber, thus the radius of the source, $R_s$, is different that of the absorber, $R_a$, and the same holds that the electron density of each species is different. The Coulombic interactions affects the ground and excited state differently leading to a energy difference that is not the same for the two species. This is best illustrated with the equation:
$R_A \neq R_S \nonumber$
$\rho_S \neq \rho_S \nonumber$
$E_A \neq E_S \nonumber$
$\delta = E_A-E_S = \dfrac{2}{3}nZ{e^2}{(\rho_A - \rho_S)}(R^2_{es} - R^2_{gs}) \nonumber$
Where delta represents the change in energy necessary to excite the absorber, which is seen as a shift from the Doppler speed 0 to V1. The isomer shift depends directly on the s-electrons and can be influenced by the shielding p, d, f electrons. From the measured delta shift there is information about the valance state of the absorbing atom
The energy level diagram for $\delta$ shift shows the change in source velocity due to different sources used. The shift may be either positive or negative.
Quadrupole Interaction
The Hamiltonian for quadrupole interaction using ${}^{57}Fe$ nuclear excited state is given by
$H_Q = \dfrac{eQV_{ZZ}}{12}[3I^2_Z-I(I+1) + \eta(I^2_X-I^2_y)] \nonumber$
where the nuclear excited states are split into two degenerate doublets in the absence of magnetic interactions. For the asymmetry parameter $\eta = 0$ doublets are labeled with magnetic quantum numbers $m_{es} = \pm 3/2$ and $m_{es} = \pm 1/2$, where the $m_{es} = \pm 3/2$ doublet has the higher energy. The energy difference between the doublets is thus
$\Delta{EQ} = \dfrac{eQV_{zz}}{2}\sqrt{1+\dfrac{\eta^2}{3}} \nonumber$
The energy diagram and corresponding spectrum can be seen as
Magnetic Splitting
Magnetic splitting of seen in Mössbauer spectroscopy can be seen because the nuclear spin moment undergoes dipolar interactions with the magnetic field
$E(m_I) = -g_n{\beta_n}{B_{eff}}m_I \tag{14}$
where $g_n$ is the nuclear g-factor and $\beta_n$ is the nuclear magneton. In the absence of quadrupole interactions the Hamiltonian splits into equally spaced energy levels of
The allowed gamma stimulated transitions of nuclear excitation follows the magnetic dipole transition selection rule:
$\Delta I = 1 \nonumber$
and
$\Delta m_I = 0, \pm 1 \nonumber$
where $m_I$ is the magnetic quantum number and the direction of $\beta$ defines the nuclear quantization axis. If we assume $g$ and $A$ are isotropic (direction independent) where $g_x = g_y = g_z$ and $B$ is actually a combination of the applied and internal magnetic fields:
$H = g\beta{S}\centerdot{B}+AS\centerdot{I} - g_n\beta_nB\centerdot{I} \nonumber$
The electronic Zeeman term is far larger then the nuclear Zeeman term, meaning the electronic term dominates the equation so $S$ is approximated by $\langle S \rangle$ and
$\langle S_z\rangle = m_s = \pm \dfrac{1}{2} \nonumber$
and
$\langle S_x \rangle = \langle S_y \rangle \approx 0 \nonumber$
$H_n = A \langle S \rangle \centerdot{I} - g_n\beta_nB\centerdot{I} \nonumber$
Pulling out a $-g_n\beta_n$ followed by $I$ leaves
$H_n = -g_n\beta_n \left( -\dfrac{A \langle S \rangle}{g_n\beta_n} + B\right){I} \nonumber$
Substituting the internal magnetic field with
$B_{int} = -\dfrac{A \langle S \rangle }{g_n\beta_n} \nonumber$
results in a combined magnetic field term involving both the applied magnetic field and the internal magnetic field
$H_n = -g_n\beta_n(B_{int} + B)\centerdot{I} \nonumber$
which is simplified by using the effective magnetic field $B_{eff}$
$H_n = -g_n\beta_nB_{eff}\centerdot{I} \nonumber$
Problems
1. The magnetic splitting of $m_I = 0$ intensity of transition is related to $sin^2(\theta)$ of the angle between the incoming gamma ray and the effective magnetic field. When is the intensity of transition at max?
2. Why is it important for the sample to be in solid or crystalline state?
3. What case will result in a delta shift of 0. 00 mm/s?
4. Why is the Doppler effect important to Mössbauer spectroscopy?
5. Why are both the emission and absorption distributions the same? (both estimated with Lorentzian functions)
4.10B: What Can Isomer Shift Data Tell Us
Among the drawbacks of the technique are the limited number of gamma ray sources and the requirement that samples be solid in order to eliminate the recoil of the nucleus. Mössbauer spectroscopy is unique in its sensitivity to subtle changes in the chemical environment of the nucleus including oxidation state changes, the effect of different ligands on a particular atom, and the magnetic environment of the sample.
As an analytical tool Mössbauer spectroscopy has been especially useful in the field of geology for identifying the composition of iron-containing specimens including meteors and moon rocks. In situ data collection of Mössbauer spectra has also been carried out on iron rich rocks on Mars.[10]
In another application, Mössbauer spectroscopy is used to characterize phase transformations in iron catalysts, e.g., those used for Fischer–Tropsch synthesis. While initially consisting of hematite (Fe2O3), these catalysts transform into a mixture of magnetite (Fe3O4) and several iron carbides. The formation of carbides appears to improve catalytic activity, however it can also lead to the mechanical break-up and attrition of the catalyst particles, which can cause difficulties in the final separation of catalyst from reaction products.[11]
Mössbauer spectroscopy has also been used to determine the relative concentration change in the oxidation state of antimony (Sb) during the selective oxidation of olefins. During calcination all the Sb ions in an antimony-containing tin dioxide catalyst transform into the +5 oxidation state. Following the catalytic reaction, almost all Sb ions revert from the +5 to the +3 oxidation state. A significant change in the chemical environment surrounding the antimony nucleus occurs during the oxidation state change which can easily be monitored as an isomer shift in the Mössbauer spectrum.[12]
This technique has also been used to observe the second-order transverse Doppler effect predicted by the theory of relativity, because of very high energy resolution.[13]
• Wikipedia | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/04%3A_Experimental_Techniques/4.10%3A_Mossbauer_Spectroscopy/4.10A%3A_The_Technique_of_Mossbauer_Spectroscopy.txt |
Learning Objectives
• Demonstrate how photoelectron spectroscopy can be used to resolve the absolute energies of molecular orbitals.
Photoelectron spectroscopy (PES) utilizes photo-ionization and analysis of the kinetic energy distribution of the emitted photoelectrons to study the composition and electronic state of the surface region of a sample.
• X-ray Photoelectron Spectroscopy (XPS) uses soft x-rays (with a photon energy of 200-2000 eV) to examine electrons in core-levels.
• Ultraviolet Photoelectron Spectroscopy (UPS) using vacuum UV radiation (with a photon energy of 10-45 eV) to examine electrons in valence levels.
Both photoelectron spectroscopies are based upon a single photon in/electron out process. The energy of a photon of all types of electromagnetic radiation is given by the Planck–Einstein relation:
$E = h \nu \label{5.3.1}$
where $h$ is Planck constant and $\nu$ is the frequency (Hz) of the radiation. UPS is a powerful technique to exam molecular electron structure since we are interested in the molecular orbitals from polyatomic molecules (especially the valence orbitals) and is the topic of this page.
Photoelectron spectroscopy uses monochromatic sources of radiation (i.e. photons of fixed energy). In UPS the photon interacts with valence levels of the molecule or solid, leading to ionization by removal of one of these valence electrons. The kinetic energy distribution of the emitted photoelectrons (i.e. the number of emitted photoelectrons as a function of their kinetic energy) can be measured using any appropriate electron energy analyzer and a photoelectron spectrum can thus be recorded.
The process of photoionization can be considered in several ways. One way is to look at the overall process for a species $A$:
$A + \text{photon} \rightarrow A^+ + e^- \label{5.3.2}$
Conservation of energy then requires that (after using Equation $\ref{5.3.1}$):
$E(A) + h\nu = E(A^+ ) + E(e^-) \label{5.3.3}$
Since the free electron's energy is present solely as kinetic energy ($KE$) (i.e., there is no internal energy in a free electron)
$E(e^-) = KE \nonumber$
Equation $\ref{5.3.3}$ can then be rearranged to give the following expression for the KE of the photoelectron:
$KE = h\nu - \left[ E(A^+ ) - E(A) \right] \label{5.3.4}$
The final term in brackets represents the difference in energy between the ionized and neutral species and is generally called the vertical ionization energy ($IE$) of the ejected electron; this then leads to the following commonly quoted equations:
$KE = h\nu - IE \label{5.3.5}$
or
$IE= h\nu - KE \label{Big}$
The vertical ionization energy is a direct measure of the energy required to just remove the electron concerned from its initial level to the vacuum level (i.e., a free electron). Photoelectron spectroscopy measures the relative energies of the ground and excited positive ion states that are obtained by removal of single electrons from the neutral molecule.
Note
Equation \ref{5.3.5} may look familiar to you as it the same equation Einstein used to describe the photoelectric effect except the vertical ionization energy ($IE$) is substituted for workfunction $\Phi$. Both vertical ionization energy and workfunctions are metrics for the binding energy of an electron in the sample.
At a fundamental level, ionization energies are well-defined thermodynamic quantities related to the heats of protonation, oxidation/reduction chemistry, and ionic and covalent bond energies. Ionization energies are closely related to the concepts of electronegativity, electron-richness, and the general reactivity of molecules. The energies and other characteristic features of the ionization bands observed in photoelectron spectroscopy provide some of the molecular orbitals detailed and specific quantitative information regarding the electronic structure and bonding in molecules.
Ionization is explicitly defined in terms of transitions between the ground state of a molecule and ion states as shown in Equation $\ref{Big}$ and as illustrated in the Figure 10.4.2 . Nonetheless, the information obtained from photoelectron spectroscopy is typically discussed in terms of the electronic structure and bonding in the ground states of neutral molecules, with ionization of electrons occurring from bonding molecular orbitals, lone pairs, antibonding molecular orbitals, or atomic cores. These descriptions reflect the relationship of ionization energies to the molecular orbital model of electronic structure.
Ionization energies are directly related to the energies of molecular orbitals by Koopmans' theorem, which states that the negative of the eigenvalue of an occupied orbital from a Hartree-Fock calculation is equal to the vertical ionization energy to the ion state formed by removal of an electron from that orbital (Figure 10.4.3 ), provided the distributions of the remaining electrons do not change (i.e., frozen).
$I_j = - \epsilon_j \label{Koopman}$
There are many limitations to Koopmans' theorem, but in a first order approximation each ionization of a molecule can be considered as removal of an electron from an individual orbital. The ionization energies can then be considered as measures of orbital stabilities, and shifts can be interpreted in terms of orbital stabilizations or destabilizations due to electron distributions and bonding. Koopmans' theorem is implicated whenever an orbital picture is involved, but is not necessary when the focus is on the total electronic states of the positive ions.
Koopmans' Theorem
Koopmans' theorem argues that the negative of the eigenvalue of an occupied orbital from a Hartree-Fock calculation is equal to the vertical ionization energy to the ion state formed by removal of an electron from that orbital.
Several different ionization energies can be defined, depending on the degree of vibrational excitation of the cations. In general, the following two types of ionization energies are considered (Figure 10.4.4 ):
• Adiabatic ionization energy corresponds to the ionization energy associated with this transition $M(X, v” = 0) + h\nu \rightarrow M^+(x, v’ = 0) + e^- \nonumber$ Adiabatic ionization energy that is, the minimum energy required to eject an electron from a molecule in its ground vibrational state and transform it into a cation in the lowest vibrational level of an electronic state x of the cation.
• Vertical ionization energy corresponds to the ionization energy associated with this transition $M(X, v” = 0) + h\nu \rightarrow M^+(x, v’ = n) + e^- \nonumber$ where, the value n of the vibrational quantum number v’ corresponds to the vibrational level whose wavefunction gives the largest overlap with the v” = 0 wavefunction. This is the most probable transition and usually corresponds to the vertical transition where the internuclear separations of the ionic state are similar to those of the ground state.
The geometry of an ion may be different from the neutral molecule. The measured ionzation energy in a PES experiment can refer to the vertical ionization energy, in which case the ion is in the same geometry as the neutral, or to the adiabatic ionzaiton energy, in which case the ion is in its lowest energy, relaxed geometry (mostly the former though). This is illustrated in the Figure 10.4.4 . For a diatomic the only geometry change possible is the bond length. The figure shows an ion with a slightly longer bond length than the neutral. The harmonic potential energy surfaces are shown in green (neutral) and red (ion) with vibrational energy levels. The vertical ionzation energy is always greater than the adiabatic ionzation energy.
Differing Ionization Energies
You have been exposed to three metrics of ionization energies already, which are similar, but with distinct differences:
• The ionization energy (also called adiabatic ionization energy) is the lowest energy required to effect the removal of an electron from a molecule or atom, and corresponds to the transition from the lowest electronic, vibrational and rotational level of the isolated molecule to the lowest electronic, vibrational and rotational level of the isolated ion.
• The binding energy (also called vertical ionization energy) is the energy change corresponding to an ionization reaction leading to formation of the ion in a configuration which is the same as that of the equilibrium geometry of the ground state neutral molecule.
• The workfunction is the minimum energy needed to remove an electron from a (bulk) solid to a point in the vacuum.
Example 10.4.1 : Molecular Hydrogen
As you remember, the molecular orbital description of hydrogen involves two $|1s \rangle$ atomic orbitals generating a bonding $1\sigma_g$ and antibonding $2\sigma_u^*$ molecular orbitals. The two electrons that are responsible for the $\ce{H_2}$ bond are occupied in the $1\sigma_g$.
The PES spectrum has a single band that corresponds to the ionization of a g electron. The multiple peaks are due to electrons ejecting from a range of stimulated vibrational energy levels. When extensive vibrational structure is resolved in a PES molecular orbital, then the removal of an electron from that molecular orbital induces a significant change in the bonding (in this case an increase in the bond length since the bond order has been reduced).
Example 10.4.2 : Molecular Nitrogen
Diatomic nitrogen is more complex than hydrogen since multiple molecular orbitals are occupied. Four molecular orbitals are occupied (the two $1\pi_u$ orbitals are both occupied). The UV photoelectron spectrum of $N_2$, has three bands corresponding to $3σ_g$, $1π_u$ and $2σ_u$ occupied molecular orbitals. Both $3σ_g$ and $2σ_u$ are weakly bonding and antibonding. The $1\sigma_g$ orbital is not resolved in this spectrum since the incident light $h\nu$ used did not have sufficient energy to ionize electrons in that deeply stabilized molecular orbital.
Note that extensive vibrational structure for the $1π_u$ band indicates that the removal of an electron from this molecular orbital causes a significant change in the bonding.
Hydrogen Chloride
The molecular energy level diagram for $\ce{HCl}$ is reproduced in Figure 10.4.5
Important aspects of molecular orbital diagram in Figure 10.4.5 :
• The H 1s energy lies well above the Cl 2s and 2p atomic orbitals;
• The valence electron configuration can be written 24;
• The H 1s orbital contributes only to the σ molecular orbitals, as does one of the Cl 2p orbitals (hence the lines in Figure 10.4.5 connecting these atomic orbitals and the and molecular orbitals);
• The remaining Cl 2p orbitals (ie those perpendicular to the bond axis) are unaffected by bonding, and these form the molecular orbitals;
• The orbitals are nonbonding - they are not affected energetically by the interaction between the atoms, and are hence neither bonding nor antibonding;
• The orbital is weakly bonding, and largely Cl 2p;
• The 3σ* orbital is antibonding, and primarily of H 1s character;
Figure 10.4.6 shows the analogous MO diagram and photoelectron spectrum for $\ce{HCl}$. The spectrum has two bands corresponding to non-bonding 1p (or $1\pi$) molecular orbitals (with negligible vibrational structure) and the 3s bonding molecular orbital (vibrational structure).
The higher energy (more stabilized) core molecular orbitals are not observed since the incident photon energy $h\nu$ is below their ionization energies.
Water
In the simplified valence bond theory perspective of the water molecule, the oxygen atom form four $sp^3$ hybrid orbitals. Two of these are occupied by the two lone pairs on the oxygen atom, while the other two are used for bonding. Within the molecular orbital picture, the electronic configuration of the $\ce{H_2O^{+}}$ molecule is $(1a_1)^2 (2a_1)^2 (1b_2)^2 (3a_1)^2 (1b_1)^2$ where the symbols $a_1$, $b_2$ and $b_1$ are orbital labels based on molecular symmetry that will be discussed later (Figure 10.4.7 ). Within Koopmans' theorem:
• The energy of the 1b1 HOMO corresponds to the ionization energy to form the $\ce{H_2O^{+}}$ ion in its ground state $(1a_1)^2 (2a_1)^2 (1b_2)^2 (3a_1)^2 (1b_1)^1$.
• The energy of the second-highest molecular orbitals $3a_1$ refers to the ion in the excited state $(1a_1)^2 (2a_1)^2 (1b_2)^2 (3a_1)^1 (1b_1)^2$.
The Hartree–Fock orbital energies (with sign changed) of these orbitals are tabulated below and compared to the experimental ionization energies.
Molecular orbital Hartree–Fock orbital Energies (eV) Experimental Ionization Energies (eV)
2a1 36.7 32.2
1b2 19.5 18.5
3a1 15.9 14.7
1b1 13.8 12.6
As explained above, the deviations between orbital energy and ionization energy is small and due to the effects of orbital relaxation as well as differences in electron correlation energy between the molecular and the various ionized states.
The molecular orbital perspective has the lone pair in different orbitals (one in a non-bonding orbital ($1b_1$ and one in the bonding orbitals). We tern to the photoelectron spectroscopy to help identify which theory is more accurate (i.e., describes reality better). The photoelectron spectrum of water in Figure 10.4.6 can be interpreted as having three major peaks with some fine structure arises from vibrational energy changes. The light source used in this experiment is not sufficiently energetic to ionize electrons from the lowest lying molecular orbitals.
If water was formed two identical O-H bonds and two lone pairs on the oxygen atom line valence bond theory predicts, then the PES in Figure 10.4.8 would have two (degenerate) peaks, one for the two bonds and one for the two lone pairs. The photoelectron spectrum clearly shows three peaks in the positions expected for the molecular orbitals in Figure 10.4.8 .
If the molecular orbitals in Figure 10.4.7 represent the real electronic structure, how do we view the bonding? These molecular orbitals are delocalized and bare little relationship to the familiar 2-center bonds used in valence bond theory. For example, the $2a_1$ $1b_1$ and $3a_1$ molecular orbitals all have contributions from all three atoms, they are really 3-centered molecular orbitals. The bonds however can be thought of as representing a build up of the total electron density which loosely put is a total of all the orbital contributions. Despite this, we keep the ideas of hybridization and 2-center bonds because they are useful NOT because they represent reality
Summary
A photoelecton spectrum can show the relative energies of occupied molecular orbitals by ionization. (i.e. ejection of an electron). A photoelectron spectrum can also be used to determine energy spacing between vibrational levels of a given electronic state. Each orbital energy band has a structure showing ionization to different vibrational levels.
Contributors and Attributions
• Roger Nix (Queen Mary, University of London) | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/04%3A_Experimental_Techniques/4.12%3A_Photoelectron_Spectroscopy_%28PES_UPS_XPS_ESCA%29.txt |
In a modern ab initio electronic structure calculation on a closed shell molecule, the electronic Hamiltonian is used with a single determinant wavefunction. This wavefunction, $\psi$, is constructed from molecular orbitals, $\psi$ that are written as linear combinations of contracted Gaussian basis functions, $\varphi$
$\varphi _j = \sum \limits _k c_{jk} \psi _k \label {10.69}$
The contracted Gaussian functions are composed from primitive Gaussian functions to match Slater-type orbitals (STOs). The exponential parameters in the STOs are optimized by calculations on small molecules using the nonlinear variational method and then those values are used with other molecules. The problem is to calculate the electronic energy from
$E = \dfrac { \displaystyle \int \psi ^* \hat {H} \psi d \tau }{\displaystyle \int \psi ^* \psi d \tau} \label {10.70}$
and find the optimum coefficients $c_{jk}$ for each molecular orbital in Equation $\ref{10.69}$ by using the Self Consistent Field Method and the Linear Variational Method to minimize the energy as was described in the previous chapter for the case of atoms.
To obtain the total energy of the molecule, we need to add the internuclear repulsion to the electronic energy calculated by this procedure. The total energy of the molecule can be calculated for different geometries (i.e. bond lengths and angles) to find the minimum energy configuration. Also, the total energies of possible transition states can be calculated to find the lowest energy pathway to products in chemical reactions.
$V_{rs} = \sum \limits _{r=1}^{N-1} \sum \limits _{s=r+1}^{N} \dfrac {Z_r Z_s}{r_{rs}} \label {10.71}$
Exercise $1$
For a molecule with three nuclei, show that the sums in Equation $\ref{10.71}$ correctly include all the pairwise potential energy terms without including any twice.
As we improve the basis set used in calculations by adding more and better functions, we expect to get better and better energies. The variational principle says an approximate energy is an upper bound to the exact energy, so the lowest energy that we calculate is the most accurate. At some point, the improvements in the energy will be very slight. This limiting energy is the lowest that can be obtained with a single determinant wavefunction. This limit is called the Hartree-Fock limit, the energy is the Hartree-Fock energy, the molecular orbitals producing this limit are called Hartree-Fock orbitals, and the determinant is the Hartree-Fock wavefunction.
Exercise $2$
Write a one-sentence definition of the Hartree-Fock wavefunction that captures all the essential features of this function.
Restricted vs. Unrestricted Hartree-Fock
You may encounter the terms restricted and unrestricted Hartree-Fock. The above discussion pertains to a restricted HF calculation. In a restricted HF calculation, electrons with $\alpha$ spin are restricted or constrained to occupy the same spatial orbitals as electrons with $\beta$ spin. This constraint is removed in an unrestricted calculation. For example, the spin orbital for electron 1 could be $\psi _A (r_1) \alpha (1)$, and the spin orbital for electron 2 in a molecule could be $\psi _B (r_2) \beta (2)$, where both the spatial molecular orbital and the spin function differ for the two electrons. Such spin orbitals are called unrestricted. If both electrons are constrained to have the same spatial orbital, e.g. $\psi _A (r_1) \alpha (1)$ and $\psi _A (r_2) \beta (2)$, then the spin orbital is said to be restricted. While unrestricted spin orbitals can provide a better description of the electrons, twice as many spatial orbitals are needed, so the demands of the calculation are much higher. Using unrestricted orbitals is particular beneficial when a molecule contains an odd number of electrons because there are more electrons in one spin state than in the other.
Carbon Dioxide
Now consider the results of a self-consistent field calculation for carbon monoxide, $\ce{CO}$. It is well known that carbon monoxide is a poison that acts by binding to the iron in hemoglobin and preventing oxygen from binding. As a result, oxygen is not transported by the blood to cells. Which end of carbon monoxide, carbon or oxygen, do you think binds to iron by donating electrons? We all know that oxygen is more electron-rich than carbon (8 vs 6 electrons) and more electronegative. A reasonable answer to this question therefore is oxygen, but experimentally it is carbon that binds to iron.
A quantum mechanical calculation done by Winifred M. Huo, published in J. Chem. Phys. 43, 624 (1965), provides an explanation for this counter-intuitive result. The basis set used in the calculation consisted of 10 functions: the ls, 2s, 2px, 2py, and 2pz atomic orbitals of C and O. Ten molecular orbitals (mo’s) were defined as linear combinations of the ten atomic orbitals, which are written as
$\psi _k = \sum \limits _{j=1}^{10} C_{kj} \varphi _j \label {10.72}$
where $k$ identifies the mo and $j$ identifies the atomic orbital basis function. The ground state wavefunction $\psi$ is written as the Slater Determinant of the five lowest energy molecular orbitals $\psi _k$. Equation $\ref{10.73}$ gives the energy of the ground state,
$E = \dfrac {\left \langle \psi |\hat {H} | \psi \right \rangle}{\left \langle \psi | \psi \right \rangle} \label {10.73}$
where the denominator accounts for the normalization requirement. The coefficients $C_{kj}$ in the linear combination are determined by the variational method to minimize the energy. The solution of this problem gives the following equations for the molecular orbitals. Only the largest terms have been retained here. These functions are listed and discussed in order of increasing energy.
• $1s \approx 0.94 1s_o$. The 1 says this is the first $\sigma$ orbital. The $\sigma$ says it is symmetric with respect to reflection in the plane of the molecule. The large coefficient, 0.94, means this is essentially the 1s atomic orbital of oxygen. The oxygen 1s orbital should have a lower energy than that of carbon because the positive charge on the oxygen nucleus is greater.
• $2s \approx 0.92 1s_c$. This orbital is essentially the 1s atomic orbital of carbon. Both the $1\sigma$ and $2 \sigma$ are “nonbonding” orbitals since they are localized on a particular atom and do not directly determine the charge density between atoms.
• $3s \approx (0.72 2s_o + 0.18 2p_{zo}) + (0.28 2s_c + 0.16 2p_{zc})$. This orbital is a “bonding” molecular orbital because the electrons are delocalized over C and O in a way that enhances the charge density between the atoms. The 3 means this is the third $\sigma$ orbital. This orbital also illustrates the concept of hybridization. One can say the 2s and 2p orbitals on each atom are hybridized and the molecular orbital is formed from these hybrids although the calculation just obtains the linear combination of the four orbitals directly without the à priori introduction of hybridization. In other words, hybridization just falls out of the calculation. The hybridization in this bonding LCAO increases the amplitude of the function in the region of space between the two atoms and decreases it in the region of space outside of the bonding region of the atoms.
• $4s \approx (0.37 2s_c + 0.1 2p_{zc}) + (0.54 2p_{zo} - 0.43 2s_{0})$. This molecular orbital also can be thought of as being a hybrid formed from atomic orbitals. The hybridization of oxygen atomic orbitals, because of the negative coefficient with 2sO, decreases the electron density between the nuclei and enhances electron density on the side of oxygen facing away from the carbon atom. If we follow how this function varies along the internuclear axis, we see that near carbon the function is positive whereas near oxygen it is negative or possibly small and positive. This change means there must be a node between the two nuclei or at the oxygen nucleus. Because of the node, the electron density between the two nuclei is low so the electrons in this orbital do not serve to shield the two positive nuclei from each other. This orbital therefore is called an “antibonding” mo and the electrons assigned to it are called antibonding electrons. This orbital is the antibonding partner to the $3 \sigma$ orbital.
• $1\pi \approx 0.32 2p_{xc} + 0.44 2p_{xo} \text {and} 2\pi \approx 0.32 2p_{yc} + 0.44 2p_{yo}$. These two orbitals are degenerate and correspond to bonding orbitals made up from the px and py atomic orbitals from each atom. These orbitals are degenerate because the x and y directions are equivalent in this molecule. $\pi$ tells us that these orbitals are antisymmetric with respect to reflection in a plane containing the nuclei.
• $5\sigma \approx 0.38 2_{sC} - 0.38 2_{pC} - 0.29 2p_{zO}$. This orbital is the sp hybrid of the carbon atomic orbitals. The negative coefficient for 2pC puts the largest amplitude on the side of carbon away from oxygen. There is no node between the atoms. We conclude this is a nonbonding orbital with the nonbonding electrons on carbon. This is not a “bonding” orbital because the electron density between the nuclei is lowered by hybridization. It also is not an antibonding orbital because there is no node between the nuclei. When carbon monoxide binds to Fe in hemoglobin, the bond is made between the C and the Fe. This bond involves the donation of the $5\sigma$ nonbonding electrons on C to empty d orbitals on Fe. Thus mo theory allows us to understand why the C end of the molecule is involved in this electron donation when we might naively expect O to be more electron-rich and capable of donating electrons to iron.
Exercise $3$
Summarize how Quantum Mechanics is used to describe bonding and the electronic structure of molecules.
Exercise $4$
Construct an energy level diagram for CO that shows both the atomic orbitals and the molecular orbitals. Show which atomic orbitals contribute to each molecular orbital by drawing lines to connect the mo’s to the ao’s. Label the molecular orbitals in a way that reveals their symmetry. Use this energy level diagram to explain why it is the carbon end of the molecule that binds to hemoglobin rather than the oxygen end. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/04%3A_Experimental_Techniques/4.13%3A_Computational_Methods/4.13A%3A_Hartree-Fock_Theory.txt |
Learning Objectives
• Demonstrate how Hückel's theory approximates the full molecular orbital picture of molecules by treating the $\sigma$-bonding and $\pi$-bonding networks independently.
Molecular orbital theory has been very successfully applied to large conjugated systems, especially those containing chains of carbon atoms with alternating single and double bonds. An approximation introduced by Hückel in 1931 considers only the delocalized p electrons moving in a framework of $\pi$-bonds. This is, in fact, a more sophisticated version of a free-electron model.
The simplest hydrocarbon to consider that exhibits $\pi$ bonding is ethylene (ethene), which is made up of four hydrogen atoms and two carbon atoms. Experimentally, we know that the H–C–H and H–C–C angles in ethylene are approximately 120°. This angle suggests that the carbon atoms are sp2 hybridized, which means that a singly occupied sp2 orbital on one carbon overlaps with a singly occupied s orbital on each H and a singly occupied sp2 lobe on the other C. Thus each carbon forms a set of three $\sigma$ bonds: two C–H (sp2 + s) and one C–C (sp2 + sp2) (part (a) of Figure 10.5.1 ).
The Hückel approximation is used to determine the energies and shapes of the $\pi$ molecular orbitals in conjugated systems. Within the Hückel approximation, the covalent bonding in these hydrocarbones can be separated into two independent "frameworks": the $\sigma$-bonding framework and the the $\sigma$-bonding framework. The wavefunctions used to describe the bonding orbitals in each framework results from different combinations of atomic orbitals. The method limits itself to addressing conjugated hydrocarbons and specifically only $\pi$ electron molecular orbitals are included because these determine the general properties of these molecules; the sigma electrons are ignored. This is referred to as sigma-pi separability and is justified by the orthogonality of $\sigma$ and $\pi$ orbitals in planar molecules. For this reason, the Hückel method is limited to planar systems. Hückel approximation assumes that the electrons in the $\pi$ bonds “feel” an electrostatic potential due to the entire $\sigma$-bonding framework in the molecule (i.e. it focuses only on the formation of $\pi$ bonds, given that the $\sigma$ bonding framework has already been formed).
Conjugated Systems
A conjugated system has a region of overlapping p-orbitals, bridging the interjacent single bonds, that allow a delocalization of $\pi$ electrons across all the adjacent aligned p-orbitals. These $\pi$ electrons do not belong to a single bond or atom, but rather to a group of atoms.
Ethylene
Before considering the Hückel treatment for ethylene, it is beneficial to review the general bonding picture of the molecule. Bonding in ethylene involves the $sp^2$ hybridization of the $2s$, $2p_x$, and $2p_y$ atomic orbitals on each carbon atom; leaving the $2p_z$ orbitals untouched (Figure 10.5.2 ).
The use of hybrid orbitals in the molecular orbital approach describe here is merely a convenience and not invoking valence bond theory (directly). An identical description can be extracted using exclusively atomic orbitals on carbon, but the interpretation of the resulting wavefunctions is less intuitive. For example, the ith molecular orbital can be described via hybrid orbitals
$| \psi_1\rangle = c_1 | sp^2_1 \rangle + c_2 | 1s_a \rangle \nonumber \nonumber$
or via atomic orbitals.
$| \psi_1\rangle = a_1 | 2s \rangle + a_1 | 2p_x \rangle + a_1 | 2p_y \rangle + a_4| 1s_a \rangle \nonumber \nonumber$
where $\{a_i\}$ and $\{c_i\}$ are coefficients of the expansion. Either describe will work and both are identical approaches since
$| sp^2_1 \rangle = b_1 | 2s \rangle + b_1 | 2p_x \rangle + b_1 | 2p_y \rangle \nonumber \nonumber$
where $\{c_i\}$ are coefficients describing the hybridized orbital.
The bonding occurs via the mixing of the electrons in the $sp^2$ hybrid orbitals on carbon and the electrons in the $1s$ atomic orbitals of the four hydrogen atoms (Figure 10.5.1 ; left) resulting in the $\sigma$-bonding framework. The $\pi$-bonding framework results from the unhybridized $2p_z$ orbitals (Figure 10.5.2 ; right). The independence of these two frameworks is demonstrated in the resulting molecular orbital diagram in Figure 10.5.3 ; Hückel theory is concerned only with describing the molecular orbitals and energies of the $\pi$ bonding framework.
Hückel treatment is concerned only with describing the molecular orbitals and energies of the $\pi$ bonding framework.
Since Hückel theory is a special consideration of molecular orbital theory, the molecular orbitals $| \psi_i \rangle$ can be described as a linear combination of the $2p_z$ atomic orbitals $\phi$ at carbon with their corresponding $\{c_i\}$ coefficients:
$| \psi_i \rangle =c_1 | \phi_{1} \rangle +c_2 | \phi_2 \rangle \label{LCAO}$
This equation is substituted in the Schrödinger equation:
$\hat{H} | \psi_i \rangle =E_i | \psi_i \rangle \nonumber$
with $\hat{H}$ the Hamiltonian and $E_i$ the energy corresponding to the molecular orbital to give:
$\hat{H} c_{1} | \phi _{1} \rangle +\hat{H} c_{2} | \phi _{2} \rangle =E c_{1} | \phi _{1} \rangle +E c_{2} | \phi _{2} \rangle \label{SEq}$
If Equation $\ref{SEq}$ is multiplied by $\langle \phi _{1}|$ (and integrated), then
$c_1(H_{11} - ES_{11}) + c_2(H_{12} - ES_{12}) = 0 \label{Eq1}$
where $H_{ij}$ are the Hamiltonian matrix elements (see note below)
$H_{ij} = \langle \phi_i | \hat{H} | \phi_j \rangle = \int \phi _{i}H\phi _{j}\mathrm {d} v\nonumber$
and $S_{ij}$ are the overlap integrals.
$S_{ij}= \langle \phi_i | \phi_j \rangle = \int \phi _{i}\phi _{j}\mathrm {d} v\nonumber$
If Equation $\ref{SEq}$ is multiplied by $\langle \phi _{2} |$ (and integrated), then
$c_1(H_{21} - ES_{21}) + c_2(H_{22} - ES_{22}) = 0 \label{Eq2}$
Both Equations $\ref{Eq1}$ and $\ref{Eq2}$ can better represented in matrix notation,
${\begin{bmatrix}c_{1}(H_{11}-ES_{11})+c_{2}(H_{12}-ES_{12})\c_{1}(H_{21}-ES_{21})+c_{2}(H_{22}-ES_{22})\\end{bmatrix}}=0\nonumber$
or more simply as a product of matrices.
$\begin{bmatrix} H_{11} - ES_{11} & H_{12} - ES_{12} \ H_{21} - ES_{21} & H_{22} - ES_{22} \ \end{bmatrix} \times \begin{bmatrix} c_1 \ c_2 \ \end{bmatrix}= 0 \label{master}$
All diagonal Hamiltonian integrals $H_{ii}$ are called Coulomb integrals and those of type $H_{ij}$ are called resonance integrals. Both integrals are negative and the resonance integrals determines the strength of the bonding interactions. The equations described by Equation $\ref{master}$ are called the secular equations and will also have the trivial solution of
$c_1 = c_2 = 0 \nonumber$
Within linear algebra, the secular equations in Equation $\ref{master}$ will also have a non-trivial solution, if and only if, the secular determinant is zero
$\left| \begin{array} {cc} H_{11} - ES_{11} & H_{12} - ES_{12} \ H_{21} - ES_{21} & H_{22} - ES_{22} \ \end{array}\right| = 0 \label{SecDet}$
or in shorthand notation
$\text{det}(H -ES) =0\nonumber$
Everything in Equation $\ref{SecDet}$ is a known number except $E$. Since the secular determinant for ethylene is a $2 \times 2$ matrix, finding $E$, requires solving a quadratic equation (after expanding the determinant)
$( H_{11} - ES_{11} ) ( H_{22} - ES_{22} ) - ( H_{21} - ES_{21} )( H_{12} - ES_{12} ) = 0\nonumber$
There will be two values of $E$ which satisfy this equation and they are the molecular orbital energies. For ethylene, one will be the bonding energy and the other the antibonding energy for the $\pi$-orbitals formed by the combination of the two carbon $2p_z$ orbitals (Equation $\ref{LCAO}$). However, if more than two $| \phi \rangle$ atomic orbitals were used, e.g., in a bigger molecule, then more energies would be estimated by solving the secular determinant.
Solving the secular determinant is simplified within Hückel method via the following four assumptions:
1. All overlap integrals $S_{ij}$ are set equal to zero. This is quite reasonable since the $\pi-$ orbitals are directed perpendicular to the direction of their bonds (Figure 10.5.1 ). This assumption is often call neglect of differential overlap (NDO).
2. All resonance integrals $H_{ij}$ between non-neighboring atoms are set equal to zero.
3. All resonance integrals $H_{ij}$ between neighboring atoms are equal and set to $\beta$.
4. All coulomb integrals $H_{ii}$ are set equal to $\alpha$.
These assumptions are mathematically expressed as
$H_{11}=H_{22}=\alpha\nonumber$
$H_{12}=H_{21}=\beta\nonumber$
Assumptions 1 means that the overlap integral between the two atomic orbitals is 0
$S_{11}=S_{22}=1\nonumber$
$S_{12}=S_{21}=0\nonumber$
Matrix Representation of the Hamiltonian
The Coulomb integrals
$H_{ii}= \langle \phi _i|H| \phi _i \rangle \nonumber \nonumber$
and resonance integrals.
$H_{ij}= \langle \phi _i|H| \phi _j \rangle \,\,\, (i \neq i) \nonumber \nonumber$
are often described within the matrix representation of the Hamiltonian (specifically within the $| \phi \rangle$ basis):
$\hat{H} = \begin{bmatrix} H_{11} & H_{12} \ H_{21} & H_{22} \end {bmatrix} \nonumber \nonumber$
or within the Hückel assumptions
$\hat{H} = \begin{bmatrix} \alpha & \beta \ \beta & \alpha \end {bmatrix} \nonumber \nonumber$
The Hückel assumptions reduces Equation $\ref{master}$ in two homogeneous equations:
$\begin{bmatrix} \alpha - E & \beta \ \beta & \alpha - E \ \end{bmatrix} \times \begin{bmatrix} c_1 \ c_2 \ \end{bmatrix}= 0 \label{Eq12}$
if Equation $\ref{Eq12}$ is divided by $\beta$:
$\begin{bmatrix} \dfrac{\alpha - E}{\beta} & 1 \ 1 & \dfrac{\alpha - E}{\beta} \ \end{bmatrix} \times \begin{bmatrix} c_1 \ c_2 \ \end{bmatrix}= 0\nonumber$
and then a new variable $x$ is defined
$x = \dfrac {\alpha -E}{\beta} \label{new}$
then Equation $\ref{Eq12}$ simplifies to
$\begin{bmatrix} x & 1 \ 1 & x \ \end{bmatrix} \times \begin{bmatrix} c_1 \ c_2 \ \end{bmatrix}= 0 \label{seceq}$
The trivial solution gives both wavefunction coefficients equal to zero and the other (non-trivial) solution is determined by solving the secular determinant
$\begin{vmatrix}x&1\1&x\\end{vmatrix}=0\nonumber$
which when expanded is $x^{2}-1=0$ so $x=\pm 1$.
Knowing that $E=\alpha -x\beta$ from Equation $\ref{new}$, the energy levels can be found to be
$E=\alpha -\pm 1\times \beta \nonumber$
or
$E=\alpha \mp \beta \nonumber$
Since $\beta$ is negative, the two energies are ordered (Figure 10.5.4 )
• For $\pi_1$: $E_1 =\alpha + \beta$
• For $\pi_2$: $E_2 =\alpha - \beta$
To extract the coefficients attributed to these energies, the corresponding $x$ values can be substituted back into the Secular Equations (Equation $\ref{seceq}$). For the lower energy state ($x=-1$)
$\begin{bmatrix} -1 & 1 \ 1 & -1 \ \end{bmatrix} \times \begin{bmatrix} c_1 \ c_2 \ \end{bmatrix}= 0 \nonumber$
This gives $c_1=c_2$ and the molecular orbitals attributed to this energy is then (based off of Equation $\ref{LCAO}$):
$|\psi_1 \rangle = N_1 (\phi_1 \rangle + | \phi_2 \rangle ) \label{HOMO}$
where $N_1$ is the normalization constant for this molecular orbital; this is the bonding molecular orbital.
For the higher energy molecular orbital ($x=-1$) and then
$\begin{bmatrix} 1 & 1 \ 1 & 1 \ \end{bmatrix} \times \begin{bmatrix} c_1 \ c_2 \ \end{bmatrix}= 0 \nonumber$
This gives $c_1=-c_2$ and the molecular orbitals attributed to this energy is then (based off of Equation $\ref{LCAO}$):
$\psi_2 \rangle = N_2 (\phi_1 \rangle - | \phi_2 \rangle ) \label{LUMO}$
where $N_2$ is the normalization constant for this molecular orbital; this is the anti-bonding molecular orbital.
The normalization constants for both molecular orbitals can obtained via the standard normalization approach (i.e., $\langle \psi_i | \psi_i \rangle =1$) to obtain
$N_1 = N_2 = \dfrac{1}{\sqrt{2}}\nonumber$
These molecular orbitals form the $\pi$-bonding framework and since each carbon contributes one electron to this framework, only the lowest molecular orbital ($| \psi_1 \rangle$) is occupied (Figure 10.5.5 ) in the ground state. The corresponding electron configuration is then $\pi_1^2$.
HOMO and LUMO are acronyms for highest occupied molecular orbital and lowest unoccupied molecular orbital, respectively and are often referred to as frontier orbitals. The energy difference between the HOMO and LUMO is termed the HOMO–LUMO gap.
$\psi ={\$dfrac {1}{\sqrt {2}}}(\phi _{1}+\phi _{2})\,} The 3-D calculated $\pi$ molecular orbitals are shown in Figure 10.5.6 .
Limitations of Hückel Theory
Hückel theory was developed in the 1930's when computers were unavailable and a simple mathematical approaches were very important for understanding experiment. Although the assumptions in Hückel theory are drastic they enabled the early calculations of molecular orbitals to be performed with mechanical calculators or by hand. Hückel Theory can be extended to address other types of atoms in conjugated molecules (e.g., nitrogen and oxygen). Moreover, it can be extended to also treat $\sigma$ orbitals and this "Extended Hückel Theory" is still used today. Despite the utility of Hückel Theory, it is highly qualitative and we should remember the limitations of Hückel Theory:
• Hückel Theory is very approximate
• Hückel Theory cannot calculate energies accurately (electron-electron repulsion is not calculated)
• Hückel Theory typically overestimates predicted dipole moments
Hückel Theory is best used to provide simplified models for understanding chemistry and for a detailed understanding modern ab initio molecular methods discussed in Chapter 11 are needed. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/04%3A_Experimental_Techniques/4.13%3A_Computational_Methods/4.13C%3A_Huckel_MO_Theory.txt |
Summary of applicable formulae
1) Spin-Only magnetic moment
μs.o. = √{4S(S+1)} B.M.
2) For A and E ground terms
μeff = μs.o. (1-α λ /Δ) B.M.
Do not expect Temperature dependence.
3) For T ground terms with orbital angular momentum contribution
μS+L = √{4S(S+1) + L(L+1)} B.M.
T terms generally show marked Temperature dependence.
The examples that follow are arranged showing the experimentally observed values, the theoretical "spin-only" value and possible variations expected.
A number of the examples involve "alums" where the central Transition Metal ion can be considered to be octahedrally coordinated by water molecules.
d1
VCl4
V(IV) tetrahedral
``` 80K 300K μs.o. /B.M.
1.6 1.6 1.73
```
2E ground term - hence don't expect Temperature dependence and small variation from spin-only value can be accounted for by equation 2) above. For less than a half-filled d shell, the sign of λ is positive so the effect on μ should be that μeff < μs.o.
VCl62-
V(IV) octahedral
``` 80K 300K μs.o. /B.M.
1.4 1.8 1.73
```
2T2g ground term - hence do expect Temperature dependence and large variation from spin-only value may be observed at low temperatures.
Since there is a direct orbital angular momentum contribution we should calculate μS+L from equation 3) above.
For a full S+L contribution this would give μS+L = 3 B.M. which is clearly much higher than the 1.8 B.M. found at 300K. So, μs.o. < μobs < μS+L
showing that the magnetic moment is partially quenched.
d2
V3+ in (NH4)V(SO4)2.12H2O (an alum)
V(III) octahedral
``` 80K 300K μs.o. /B.M.
2.7 2.7 2.83
```
3T1g ground term - hence do expect Temperature dependence and large variation from spin-only value may be observed at low temperatures.
Since there is a direct orbital angular momentum contribution we should calculate μS+L from equation 3) above.
For a full S+L contribution this would give μS+L = √(20) = 4.47 B.M. which is clearly much higher than the 2.7 B.M. found at 300K.
So, μobs < μs.o. < μS+L
showing that the magnetic moment is significantly quenched.
In this case, there is no observed Temperature variation between 80 and 300K and it may require much lower temperatures to see the effect?
d3
Cr3+ in KCr(SO4)2.12H2O (an alum)
Cr(III) octahedral
``` 80K 300K μs.o. /B.M.
3.8 3.8 3.87
```
4A2g ground term - hence don't expect Temperature dependence and small variation from spin-only value can be accounted for by equation 2) above. For less than a half-filled d shell the sign of λ is positive so the effect on μ should be that μeff < μs.o.
d4
CrSO4.6H2O
Cr(II) octahedral
``` 80K 300K μs.o. /B.M.
4.8 4.8 4.9
```
5Eg ground term - hence don't expect Temperature dependence and small variation from spin-only value can be accounted for by equation 2) above. For less than a half-filled d shell the sign of λ is positive so the effect on μ should be that μeff < μs.o.
K3Mn(CN)6
Mn(III) low-spin octahedral
``` 80K 300K μs.o. /B.M.
3.1 3.2 2.83
```
3T1g ground term - hence do expect Temperature dependence and large variation from spin-only value may be observed at low temperatures.
Since there is a direct orbital angular momentum contribution we should calculate μS+L from equation 3) above.
For a full S+L contribution this would give μS+L = √(20) = 4.47 B.M. which is clearly much higher than the 3.2 B.M. found at 300K.
So, μs.o. < μobs < μS+L
showing that the magnetic moment is partially quenched.
In this case, there is a small Temperature variation observed between 80 and 300K.
d5
K2Mn(SO4)2.6H2O (an alum)
Mn(II) high-spin octahedral
``` 80K 300K μs.o. /B.M.
5.9 5.9 5.92
```
6A1g ground term - hence do not expect Temperature dependence and L=0 so no orbital contribution possible.
Expect μeff = μs.o.
K3Fe(CN)6
Fe(III) low-spin octahedral
``` 80K 300K μs.o. /B.M.
2.2 2.4 1.73
```
2T2g ground term - hence do expect Temperature dependence and large variation from spin-only value may be observed at low temperatures.
Since there is a direct orbital angular momentum contribution we should calculate μS+L from equation 3) above.
For a full S+L contribution this would give μS+L = √(9) = 3 B.M. which is clearly much higher than the 2.4 B.M. found at 300K.
So, μs.o. < μobs < μS+L
showing that the magnetic moment is partially quenched.
d6
Fe2+ in (NH4)2Fe(SO4)2.6H2O (an alum)
Fe(II) high-spin octahedral
``` 80K 300K μs.o. /B.M.
5.4 5.5 4.9
```
5T2g ground term - hence do expect Temperature dependence and large variation from spin-only value may be observed at low temperatures.
Since there is a direct orbital angular momentum contribution we should calculate μS+L from equation 3) above.
For a full S+L contribution this would give μS+L = √(30) = 5.48 B.M. which is close to the 5.5 B.M. found at 300K.
So, μs.o. < μobs ~ μS+L
showing that the magnetic moment is not showing much quenching.
d7
Cs2CoCl4
Co(II) tetrahedral
``` 80K 300K μs.o. /B.M.
4.5 4.6 3.87
```
4A2 ground term - hence don't expect Temperature dependence and small variation from spin-only value can be accounted for by equation 2) above. For more than a half-filled d shell the sign of λ is negative so the effect on μ should be that μeff > μs.o.
The observed values are somewhat bigger than expected for the small (0.2 B.M.) variation due to equation 2) so other factors must be affecting the magnetic moment. These effects will not be covered in this course!
Co2+ in (NH4)2Co(SO4)2.6H2O (an alum)
Co(II) high-spin octahedral
``` 80K 300K μs.o. /B.M.
4.6 5.1 3.88
```
4T1g ground term - hence do expect Temperature dependence and large variation from spin-only value may be observed at low temperatures.
Since there is a direct orbital angular momentum contribution we should calculate μS+L from equation 3) above.
For a full S+L contribution this would give μS+L = √(27) = 5.2 B.M. which is close to the 5.1 B.M. found at 300K.
So, μs.o. < μobs ~ μS+L
showing that the magnetic moment is not showing much quenching.
d8
Ni2+ in (NH4)2Ni(SO4)2.6H2O (an alum)
Ni(II) octahedral
``` 80K 300K μs.o. /B.M.
3.3 3.3 2.83
```
3A2g ground term - hence don't expect Temperature dependence and small variation from spin-only value can be accounted for by equation 2) above. For more than a half-filled d shell the sign of λ is negative so the effect on μ should be that μeff > μs.o.
The observed values are somewhat bigger than expected for the small (0.2 B.M.) variation due to equation 2) so other factors must be affecting the magnetic moment. These effects will not be covered in this course!
(Et4N)2NiCl4
Ni(II) tetrahedral
``` 80K 300K μs.o. /B.M.
3.2 3.8 2.83
```
3T2 ground term - hence do expect Temperature dependence and large variation from spin-only value may be observed at low temperatures.
Since there is a direct orbital angular momentum contribution we should calculate μS+L from equation 3) above.
For a full S+L contribution this would give μS+L = √(20) = 4.47 B.M. which is higher than the 3.8 B.M. found at 300K.
So, μs.o. < μobs < μS+L
showing that the magnetic moment is partially quenched.
d9
Cu2+ in (NH4)2Cu(SO4)2.6H2O (an alum)
Cu(II) octahedral
``` 80K 300K μs.o. /B.M.
1.9 1.9 1.73
```
2Eg ground term - hence don't expect Temperature dependence and small variation from spin-only value can be accounted for by equation 2) above. For more than a half-filled d shell the sign of λ is negative so the effect on μ should be that μeff > μs.o.
Contributors and Attributions
• The Department of Chemistry, University of the West Indies) | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/04%3A_Experimental_Techniques/4.14%3A_Magnetism/4.14.01%3A_example_magnetic_moment_data_and_their_interpretation.txt |
oct d3,d8 tet d2,d7 ←-----------------------------------→ oct d2,d7 tet d3,d8
Three peaks are predicted in their electronic spectra, namely:
• 4T24A2 ν1
• 4T1(F) ← 4A2 ν2
• 4T1(P) ← 4A2 ν3 ** the band observed in the visible region
The energy of the third transitions is approximately ν3 = 6/5 Δ + 15B (where B is the Racah parameter and ignoring configuration interactions). For Co(II) tetrahedral complexes B has generally been found to be about 750 cm-1. Hence Δ for the three complexes above can be calculated to be roughly, Δ = (ν3 - 15x750) x 5/6, or:
Copyr2Cl2 (16260-15x750) x 5/6 = 4175 cm-1
Copyr2Br2 (15870-15x750) x 5/6 = 3850 cm-1
Copyr2I2 (14925-15x750) x 5/6 = 3063 cm-1
For three unpaired electrons the spin only magnetic moment is predicted to be 3.87 BM. Using a value for the free ion spin orbit coupling constant (λ) of -172 cm-1 then a better approximation of the magnetic moment can be obtained by using the third formula above. This would give a value of:
• Copyr2Cl2 3.87*(1+ 688/4175) = 4.49 B.M. found 4.42 BM
• Copyr2Br2 3.87*(1+ 688/3850) = 4.57 B.M. found 4.50 BM
• Copyr2I2 3.87*(1+ 688/3063) = 4.76 B.M. found 4.48 BM
In the case of the series;
CoI42-, CoBr42-, CoCl42-, Co(NCS)42-
the magnetic moments have been recorded as
4.77, 4.65, 4.59, 4.40 BM
showing even more clearly the inverse effect of the spectrochemical series on the magnetic moment.
4.14.03: Magnetic Moments 1
Magnetic Moments
Lecture 4. CHEM1902 Coordination Chemistry
Magnetic moments are often used in conjunction with electronic spectra to gain information about the oxidation number and stereochemistry of the central metal ion in coordination complexes.
A common laboratory procedure for the determination of the magnetic moment for a complex is the Gouy method which involves weighing a sample of the complex in the presence and absence of a magnetic field and observing the difference in weight. A template is provided for the calculations involved.
For first row transition metal ions in the free ion state, i.e. isolated ions in a vacuum, all 5 of the 3d orbitals are degenerate.
A simple crystal field theory approach to the bonding in these ions assumes that when they form octahedral complexes, the energy of the d orbitals are no longer degenerate but are split such that two orbitals, the dx2-y2 and the dz2 (eg subset) are at higher energy than the dxy, dxz, dyz orbitals (the t2g subset).
For octahedral ions with between 4 and 7 d electrons, this gives rise to 2 possible arrangements called either high spin/weak field or low spin/strong field respectively. The energy gap is dependent on the position of the coordinated ligands in the SPECTROCHEMICAL SERIES.
See an interactive JAVA applet for examples.
Note: For CHEM1902, we assume that all Co(III), d6 complexes are octahedral and LOW spin, i.e. t2g6.
In tetrahedral complexes, the energy levels of the orbitals are again split, such that the energy of two orbitals, the dx2-y2 and the dz2 (e subset) are now at lower energy (more favoured) than the remaining three dxy, dxz, dyz (the t2 subset) which are destabilised.
Tetrahedral complexes are ALL high spin since the difference between the 2 subsets of energies of the orbitals is much smaller than is found in octahedral complexes.
The usual relationship quoted between them is: Δtet ≈ 4/9 Δoct.
Square planar complexes are less commmon than tetrahedral and for CHEM1902 we will assume that the only ions forming square planar complexes are d8 e.g. Ni(II), Pd(II), Pt(II), etc. d8 can therefore be either square planar or tetrahedral. As with octahedral complexes, the energy gap between the dxy and dx2-y2 is Δoct and these d8 systems are all considered strong field / low spin complexes hence they are all diamagnetic, μ=0 Bohr Magneton (B.M.).
The formula used to calculate the spin-only magnetic moment can be written in two forms; the first based on the number of unpaired electrons, n, and the second based on the total electron spin quantum number, S. Since for each unpaired electron, n=1 and S=1/2 then the two formulae are clearly related and the answer obtained must be identical.
Return to Coordination Chemistry Course Outline.
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Copyright © 1995-2015 by Robert John Lancashire, all rights reserved.
Prof. Robert J. Lancashire,
The Department of Chemistry, University of the West Indies,
Mona Campus, Kingston 7, Jamaica.
Created Dec 1995. Links checked and/or last modified 23rd March 2015.
URL http://wwwchem.uwimona.edu.jm/spectra/MagMom.html | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/04%3A_Experimental_Techniques/4.14%3A_Magnetism/4.14.02%3A_Magnetic_Moments.txt |
The GOUY Method
Perhaps the simplest technique for measuring the magnetic susceptibility of metal complexes is the Gouy Method. From a classical description of magnetism, Lenz's Law (around 1834) can be written as
$\dfrac{B}{H} = 1 + 4\pi \dfrac{I}{H}$
or
$\dfrac{B}{H} = 1 + 4\pi \kappa \label{1}$
where
• B/H is called the magnetic permeability of the material and
• κ is the magnetic susceptibility per unit volume, (I/H).
The determination of a magnetic susceptibility depends on the measurement of B/H. Experimentally the Gouy method involves measuring the force on the sample by a magnetic field and is dependent on the tendency of a sample to concentrate a magnetic field within itself.
At any given point, dx, of the sample, the force is given by:
$dF=μ°H κdV (dH)/dx \label{2}$
where
• μ° is the permeability of a vacuum (=1 when using c.g.s. units)
• H is the magnitude of the magnetic field at point, dx,
• dV is the volume of the sample at point dx,
• κ is the magnetic susceptibility per unit volume.
The sample is uniformly packed into a glass tube (Gouy tube) each end of which is at a constant field strength. This is attained by using a tube that is packed to a certain height (say 10 cm) and the tube is suspended between the poles of a magnet such that the bottom of the sample is in the center of the field (a region where a uniform field strength can be readily obtained) whilst the top of the sample is out of the field, i.e. $H=0$. By integrating the above equation, the total force on the sample can be given as:
$F= 1/2 μ° A κ (H^2-H°^2) \label{3}$
and since $H^o =0$ at the top of the sample then
$F= 1/2 μ°A \kappa H^2 \label{4}$
where $A$ is the cross sectional area of the sample.
The force is measured by the apparent change in mass when the magnetic field is switched on, or
$F=g\delta w =1/2 μ°A \kappa H^2 \label{5}$
where $\delta w$ is the apparent change in mass, and g is the acceleration due to gravity.
An allowance needs to be made for the tube, since it will have its own magnetic properties as a result of the air within the tube (which is displaced from the tube when the sample is introduced) and also from the materials used in its construction. Equation $\ref{5}$ becomes:
$g δw'=1/2 A μ^o (κ-κ') H^2$
where
• δw'=δw + δ
• δ is a constant allowing for the magnetic properties of the empty tube
• κ' is the volume susceptibility of the displaced air.
This leads to:
κ= (2gδw')/(μ°AH2) + κ'
Converting from volume susceptibility to gram susceptibility (χg) leads to:
χg = κ/ρ =κ.V/W
where ρ is the density of the sample so that
χg = β δw' / W + κ'V/W
or
χg = (α + β δw') / W
where
• α is a constant allowing for the air displaced by the sample,
• β is a constant that is dependent on the field strength, =2gV/(μ°AH2)
• W is the weight of the sample used.
Written more simply then:
χg cal = β δw' / Wcal (+ α/(Wcal) \]
the last expression is usually negligible.
β is then obtained and from this
χg sam = β δw' / Wsam (+ α/Wsam)
the χg sample can be obtained, again the factor for the susceptibility of air is usually negligible.
To accurately determine the gram magnetic susceptibility of a sample, it is necessary to predetermine the value of the constants α, β and δ. Since these constants are dependent on the amount of sample placed in the tube, the tube itself and the magnetic field strength, it should be emphasized that each experimenter must determine these constants for their particular configuration. That is, results obtained with one tube are not transferable to other Gouy tubes.
The field strength is determined by the current supplied to the electromagnet. In order to ensure a constant magnetic field strength from one measurement to the next, always set the current to the same value. Note that the magnet may display hysteresis effects so that if you do go beyond the 5 Amp value it may take some time to reestablish itself, after you have decreased the power.
Determination of the Constants
Select a tube and piece of nichrome wire to make an assembly which will allow the tube to be suspended from the analytical balance so that the bottom of the tube is aligned halfway between the polefaces of the magnet and the top of the sample is above the magnet and hence subject to essentially zero field, H=0.
1) delta, δ
Adjust the zero setting on the balance, then suspend the empty tube from the balance and weigh it (W1). Set the field to the required strength and reweigh the tube (W2). The force on the tube, δ, therefore is:
δ = W2 - W1
this will normally be negative since the tubes are generally diamagnetic and pushed out of the field, ie. weigh less.
2) alpha, α
Fill the tube to the required height with water and weigh it (check the zero first), this will give W3. Assuming the density of water at this temperature is 1.00 g cm-3 this gives the volume of water (and also that of the sample).
vol. =(W3-W1)/1.00 where the weight changes should be expressed in g.
α=κ'.V
α= 0.029 x (W3-W1) in 10-6 c.g.s. units,
where 0.029 is the volume susceptibility of air /cm3. For strongly paramagnetic samples this correction is generally insignificant.
3) beta, β
The determination of β requires the use of a compound whose magnetic properties have been well established. Common calibrants include HgCo(SCN)4 and [Ni(en)3]S2O3. Since the magnetic properties are often temperature dependent, the susceptibility of the calibrant must be calculated for the temperature at which the sample is measured.
Record the temperature, T1. Fill the tube to the required height with the calibrant (in this case either HgCo(SCN)4 or [Nien3]S2O3 and weigh it with the field off (W4) and with the field on (W5).
For HgCo(SCN)4 the following relationship can be used:
χg = 4985 / (T+10) in 10-6 c.g.s units at temperature T, while the corresponding relationship for [Ni(en)3]S2O3 is:
χg = 3172 / T in 10-6 c.g.s units at temperature T
Using this χg then
β = (χgW - α)/ δw'
where δw' = (W5 - W4) - δ in mg
and W= (W4 - W1) in g
Determination of the Magnetic Susceptibility of your sample
Once α and β are known, then χm' can be determined for the sample in question. Fill the tube to the required height with your sample and weigh it with the field off (W6) and with the field on (W7). From this calculate:
χg = (α + βδw') / W
where δw' = (W7 - W6) - δ in mg
and W = (W6 - W1) in g
To convert from χg to χm the molar mass must be accurately known, since:
χm = χg x R.M.M.
The final correction is for the diamagnetism of the sample
χ'm= χm + χmdia
where χmdia is the susceptibility arising from the diamagnetic properties of the electron pairs (and therefore not a property of the unpaired electrons) and must be allowed for. The values for χmdia have been well documented (Pascal's constants) for different atoms and ions and a selection of them are tabulated.
To summarize, the overall procedure is:
Weigh the empty tube - magnet off/on W1/W2
Weigh the tube with water - magnet off W3
Weigh the tube with calibrant - magnet off/on W4/W5
Weigh the tube with your sample - magnet off/on W6/W7
Record the temperature(s) of calibrant/sample T1/T2 in K
Calculate the Molar Mass of your sample M.M.
Estimate the total diamagnetic correction for your sample D.C.
Calculate the magnetic moment using:
&delta = (W2 - W1) in mg
α = 0.029 x (W3 - W1) in 10-6 c.g.s. units
β = [(χm{Calibrant}) (W4 - W1) - α]/ [(W5 - W4) - δ] at temperature T1
χm {Sample} = [α + β {(W7 - W6) - δ}]/(W6 - W1) at temperature T2
χ'm = ( χm x R.M.M.) + χm dia
also χ'm = μ ° μb2 N/3k. μeff2/T
where μb is the Bohr Magneton, N is Avogadro's number and k is the Boltzmann constant. Hence,
μeff = √ (3k/μ° μb 2N). √ (χ'm T2) B.M.
or μeff = 2.828 √ ( χ'm T2 x 10-6) at temperature T2
where the 10-6 that has been ignored in these expressions is finally included.
(Determination of the magnetic moment using the Gouy method has been simplified by the use of an on-line template or spreadsheet.)
Contributors and Attributions
• The Department of Chemistry, University of the West Indies) | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/04%3A_Experimental_Techniques/4.14%3A_Magnetism/4.14.04%3A_Magnetic_Susceptibility_Measurements.txt |
The geometrical shape and the inherent physical/chemical properties seen in molecules can be attributed to atomic and molecular orbitals. Features of molecular structure can be explained by taking into consideration (1) how orbitals interact within a single atom to form hybrid atomic orbitals and (2) how atomic orbitals between different atoms interact, giving rise to molecular orbitals. This module will serve as a reminder of the fundamental concepts of bonding as they relate to molecular structure, as well as an investigation into the complexities of hybridized atomic orbitals
The Localized Electron Bonding Model
What is a chemical bond? The most elementary way to understand the bonding between two or more atoms is to make use of Lewis dot structures; a structure drawn where each valence electron surrounding an atom is symbolized by a dot and each bonding electron is symbolized by a pair of dots or a dash (Wade p.7). In accordance with the observations of G. N. Lewis, atoms having or sharing 8 valence electrons will tend to be in an arrangement of lowest energy (ACS p.11). A more sophisticated approach to understanding bonding is to consider how atomic orbitals within individual atoms interact with one another forming molecular orbitals and subsequent covalent bonding. A complete understanding of the subject involves knowledge of quantum mechanics, wave functions and molecular symmetry. This module will focus on how atomic orbitals on an individual atom can give rise to hybrid atomic orbitals, which define the geometry of the bond, and how hybrid atomic atoms of different atoms come together to form molecular orbitals, which define the geometry of the molecule.
Why do chemical bonds form? The stability of a covalent bond, the bond formed when atomic orbitals of separate atoms interact to form molecular orbitals, results from there being a large amount of electron density in the region of space between the two nuclei (Wade p.44). This region is known as the bonding region and here the electrons are close to both nuclei, subsequently lowering the overall energy.
The Heisenberg Uncertainty Principle
The Heisenberg Uncertainty Principle rationalizes the inability of chemists and physicists alike to simultaneously determine both the position and momentum of an atomic particle in space. In lieu of this, theoretical chemists had to develop methods of calculating the position of an electron in terms of probabilities rather than assigning electrons fixed location about a nucleus (Barret p.2). These calculations are collectively known as quantum mechanics, with the Schrödinger wave equation (ĤΨ = iħ d/dtΨ) being the quantum mechanic calculation of greatest interest to orbital symmetry. Solving the Schrödinger wave equation gives the shapes of atomic orbitals represented graphically as the probability of finding an electron residing in a region of space about an atom.
Orbitals and Hybridization
Valence Bond theory describes the formation of a chemical bond in terms of overlapping between atomic orbitals. The 1s orbital of hydrogen, for example, can overlap in-phase and combine constructively to form a molecular orbital called a sigma bond.
Furthermore, a 1s orbital is analogous to the fundamental vibration of a guitar string. The wave function is seemingly positive and negative simultaneously. The effect of squaring the wave function gives the distribution of electron density, which can be used to graphically represent the spherical symmetry of an s orbital.
Atomic orbitals can interact to form new molecular orbitals but let us first consider that orbitals within an individual atom can interact amongst themselves giving rise to hybridized atomic orbitals. Yielding a sp hybrid orbital with an electron density predominately concentrated toward one side of the atom. Molecular orbital theory dictates that the number of hybrid orbitals produced must equal the sum of the orbitals that underwent hybridization and be it that we started with one s orbital and one p orbital (for a total of two orbitals) we must finish with a total of two hybrid sp orbitals.
The final result of this hybridization is a pair of directional sp hybrid orbitals pointed in opposite directions, providing enough electron density in the bonding regions to provoke a sigma bond to both the left and the right of the atom. These 2 sp hybrid orbitals generate a bond angle of 180˚, creating a bond formation with linear geometry. Lastly, the degree of orbital hybridization is governed by the number of attachments (ligands) found on a central atom, lone pairs of electrons included. Table 1.1 provides a summary of orbital hybridization wherein the number of ligands attached to a central molecule correlates to the molecules geometry.
Table 1.1 Summary of hybridization
# of attachments
Hybridization
Geometry
angle(s)
2
sp
Linear
180°
3
sp2
Trigonal planar
120°
4
sp3
Tetrahedral
109.5°
5
sp3d
Trigonal bipyramidal
120° and 90°
6
sp3d2
Octahedral
90°
sp Hybridization
As discussed, molecular orbitals form as a result of constructive & destructive wave overlap of atomic orbitals between different atoms as well as the potential for atomic orbitals contained within an atom can combine amongst themselves giving rise to hybrid atomic orbitals. It becomes prudent then to consider the spatial orientation of atomic orbitals during the interaction of orbitals on different atoms in the formation of chemical bonds.
The chemical bonding of compounds with triple bonds, such as alkynes, can be expounded by sp hybridization. Inspection of the electron configuration of carbon reveals that the electrons in the 2s orbital mix with only one of the three available p orbitals. This results in two hybrid sp orbitals and two unaltered p orbitals. C2H2, for instance, is held together then by the overlap of adjacent/approaching sp-sp hybrid orbitals on each carbon atom. The bond that ultimately forms is a sigma bond complemented by additional pi bonds formed by p-p orbital overlap; triple bonds are actually composed of two different types of bonds, sigma and pi. Each carbon also bonds to a hydrogen by means of a sigma bond formed this time by s-sp orbital overlap.
Outside Links
• Interactive Molecular Structure & Bonding at www2.chemistry.msu.edu:80/fac...Jml/intro3.htm
Problems
1. Using the space provided please draw a) an s orbital & all three appropriately labeled p orbitals and b) the product(s) of their hybridization.
a)
b)
2. The constructive overlap between orbitals of Hydrogen forming sigma bonds was discussed. Please describe the sigma* anti-bonding orbital that results from destructive overlap.
3. Please a) draw both the Lewis Dot and VSEPR structure of CO2 labeling the hybridization and bond angle b) draw the orbitals that overlap during bond formation of CO2 c) identify all the symmetry elements.
a) Structure of CO2
b) Orbitals
c) Symmetry Elements
Contributors and Attributions
• Carter, James C., B.S. Environmental Toxicology | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/05%3A_Bonding_in_Polyatomic_Molecules/5.02%3A_Valence_Bond_Theory_-_Hybridization_of_Atomic_Orbitals/5.2B%3A_sp_Hybridization.txt |
The sp2 hybridization is the mixing of one s and two p atomic orbitals, which involves the promotion of one electron in the s orbital to one of the 2p atomic orbitals. The combination of these atomic orbitals creates three new hybrid orbitals equal in energy-level. The hybrid orbitals are higher in energy than the s orbital but lower in energy than the p orbitals, but they are closer in energy to the p orbitals. The new set of formed hybrid orbitals creates trigonal structures, creating a molecular geometry of 120 degrees.
Contributors and Attributions
• Abel Silva, Michael Dai (Yicong)
5.2D: sp3 Hybridization
Introduction
The term “sp3 hybridization” refers to the mixing character of one 2s-orbital and three 2p-orbitals to create four hybrid orbitals with similar characteristics. In order for an atom to be sp3 hybridized, it must have an s orbital and three p orbitals.
From wave function to the visual representation:
Four equivalent sp3 hybrid orbitals, resulting from the combination of one s atomic orbital and three p atomic orbitals, can then describe by four new wave functions (equations 1 – 4)
ψ(sp3) = 0.5 ( ψ2s + ψ2px + ψ2py + ψ2pz) (1)
ψ(sp3) = 0.5 ( ψ2s + ψ2px - ψ2py - ψ2pz) (2)
ψ(sp3) = 0.5 ( ψ2s - ψ2px - ψ2py + ψ2pz) (3)
ψ(sp3) = 0.5 ( ψ2s - ψ2px + ψ2py - ψ2pz) (4)
Plotting any of these four wave functions gives a picture representation of a sp3 orbital. Each hybrid orbital consists of a large lobe and a small lobe, pointing in two opposite direction (figure 1).
The Bond Angle is 109.5o:
When the graphs of the four wave functions are combined, the resulting picture shows the tetrahedral arrangement of the four sp3 hybrid orbitals around the central atom. Because of the tetrahedral molecular geometry, the calculate bond angles between 1 and 2, 1 and 3, 1 and 4, 2 and 3, 2 and 4, and 3 and 4 approximately equal 109.5o (figure 2).
The Energy level and election population:
All four sp3 hybrid orbitals are delocalized—they occupy the same energy level; however, they are higher in energy than the 2s orbital and lower in energy than the 2p orbital (figure 3).
Just like any other atomic orbital, each sp3 hybrid orbital can house 2 elections.
S-character and the stability of the anion:
Each sp3 orbital has 1 part of s-character to 3 parts of p-character. In other words, it has 25% s-character and 75% p-character. Since the s orbital is closer to the nucleus and thus lower in energy than the p orbital, the electrons of sp3 hybridized species are held farther from the nucleus than those in sp2 (33% s-character) and sp (50% s-character) hybridized species. The closer the electrons are to the nucleus, the more stable they are. Therefore, when bearing the negative charge, sp3 species are less stable than sp2 and sp species. Put differently, sp3 species are less likely to get deprotonated (leaving a pair of electron behind).
Hybridization and bond length/bond strength:
The greater the s-character, the closer the electrons are held to the nucleus, the shorter the bond, and the stronger the bond. Thus, sp3 hybridized atoms form longer and weaker bonds than those of sp2 and sp hybridized.
Problems
1. Which of the (*) carbons is/are sp3 hybridized
2. Draw the energy diagram for the orbitals of sp3 hybridzied carbon and nitrogen. Then fill in the correct number of electron.
3. Indicate the hybridization of oxygen in each molecule
4. Which nitrogen atom(s) is/are sp3 hybridized
5. Describe the bonding scheme of CH4.
Answers:
1. a and b
2. Just like the energy diagram in fig.3.
For carbon, each sp3 orbital has 1 electron. For nitrogen, the first sp3 orbital has 2 electrons, then one electron for each of the remaining three
3. All of them (Don't for get the elctron pairs)
4. a and d
5. Carbon has four half-filled sp3 hybrid orbitals. Each orbital overlaps with a partially filled 1s atomic orbital of hydrogen to form 4 sigma bonds. To visualize, hydrogen atoms are placed at the four corner of the tetrahedron.
Quynh Nhu Nguyen | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/05%3A_Bonding_in_Polyatomic_Molecules/5.02%3A_Valence_Bond_Theory_-_Hybridization_of_Atomic_Orbitals/5.2C%3A_sp2_Hybridization.txt |
When Valence Bond Theory is applied to the multiple bonding in polyatomic molecules, there are two types of bonds that can form: homonuclear and heteronuclear. Homonuclear bonds occur between multiples of the same element and heteronuclear bonds occur between different elements. Homonuclear bonds can occur on a diatomic scale all the way up to large polymers made completely of carbon atoms. However, most polyatomic atoms involve heteronuclear bonds. When considering bonding in polyatomic molecules for VB theory, it is easiest to use a mixture of Lewis structures and Atomic Orbital Hybridization to explain.
Bonding In Molecules
When trying to figure out the bonding of a molecule, it is a good idea to look at the electronic configuration of the atoms (Housecroft, 120). This allows you to know how many valence electrons you have to work with. Then the number of bonds attached to each atom should be determined using the lewis structure of the molecule. After the lewis structure has been determined, find the hybridization of each atom in the molecule. With the determined hybridization, a picture of the molecule should be easy to create.
Example: Ethyne
Ethyne, or acetylene as it is commonly referred to as, has a lewis structure that shows that there is a triple bond between the two carbon atoms. Using valence orbital hybridization, sp hybridization is determined for triple bonded molecules, which means that carbon has two free 2p orbitals that form 2 pi bonds. Using this info you would draw the two carbon atoms with sp bonds connecting them and the two pi orbitals from each atom bonding with each other. Each hydrogen atom has its s orbital is overlapped with each carbon atom's second sp bond . The molecule overall would have a linear structure.
Outside Links
• This is not meant for references used for constructing the module, but as secondary and unvetted information available at other site
• Link to outside sources. Wikipedia entries should probably be referenced here.
Problems
1. Provide the lewis structure and hybridization of the carbonyl carbon in formaldehyde.
2. Provide the hybridization of S in SH2 and a picture of the hybrid orbitals.
3. Provide the lewis structure and a drawing of the hybrid orbitals in CHF3.
4. Provide the electronic configuration, hybridization, and lewis structure of PF3.
5. Provide the electronic configuration and a drawing of the hybrid orbitals of CO2.
Answers
1. sp2
2. sp3
3.
4. P=[Ne] 3s2 3p3 ; F=[He] 2s2 2p5 ; sp3
5. C= [He] 2s2 2p2 ; O=[He] 2s2 2p4 ;
• Name #1 here (if anonymous, you can avoid this) with university affiliation
5.3C: (HCN)
HCN, hydrogen cyanide, is a volatile and poisnous compound with distinguished bitter odor. It is linear molecule with a triple bond between C and N atom and has bond angle of 180 degrees. It can be found in fruits that have pits due to the fact that they contain small amounts of cyanohydrins which slowly releases hydrogen cyanide. Also it can be found in exhaust of vehicles and burning nitrogen-containig plastics.
Introduction
The valence bond theory can be explained by overlapping of atomic orbitals which electrons are localized in the reigion to form chemical bonds. However, when you utilize this approach to explain chemical structure of the molecule, you must aware that there are various atomic orbitals for bonding which will significantly influence the structure of the molecule.
Definition of Valence Bond Theory
The Valence Bond thoery simply explains the bond formation just like lewis dot structure, but instead it explains the bonding in terms of covalent bond by quantum mechanics. According to this theory, bond will form when
1) An orbital of one atom occupy another atom's orbital, known as overlap.
2) number of electrons in both orbital is adds up to no more than two.
Just like forming a molecule with lewis dot structure, bonds between atoms complete when two electrons share same orbital together.
Bond strength depends on the the amount of overlap since electrons are attracted to nuclei of both atoms, more electrons will pull more nuceli thus increase bond strength. However, two orbitals can not contian more than two atoms due to the maximum capacity it can hold.
Also, because known atomic geometry can not be able to have effective overlap, atomic orbitals combine with each other and reconfigure themselves into a different configuration. This process is called hybrdization.
This formation of new hybrid orbital is possible by combining several types of orbitals (s,p,d and etc).
Describe HCN molecular bond by using Valence Bond Theory
In HCN molecule, the C atom includes sp-hybridized orbital, since it will combine with only two other atoms to form HCN. One of the sp-hybrid orbitals of carbon atom overlaps with the 1s orbital of H atom, while the other sp-hybrid orabital mixes with one of the nitrogen's atom's three atomic p orbitals which were unhybridized. Because px orbital of C and N will form sigma bond, this leaves with two N atom p-orbitals which form two mutually perpendicular pi bonds to the two atomic p orbitals on the C atom. HCN thus has one single and one triple bond. The latter consists of a sigma bond from the overlap of a C atom sp hybrid orbital with a N atom p orbital, and two mutually perpendicular pi bonds are formed from parallel atomic p orbitals of carbon and nitrogen atoms.
Outside Links
1. Shaik, Sason S., and Philippe C. Hiberty. A Chemist's Guide to Valence Bond Theory. Hoboken, NJ: Wiley-Interscience, 2008. Print.
2. "Bonding and Hybridization." Department of Chemistry & Biochemistry @ Boise State University. Web. 05 Nov. 2010. <chemistry.boisestate.edu/peop...rganic/bonding and hybridization/bonding_hybridization.htm>.
Problems
1. What type of bond is present in the HCN molecular orbitals?
2. What theory is necessary to explain the formation of hybridized orbitals?
3. Explain why HCN is linear.
Answers
1. 1 sigma bond between H and C atoms. 1 sigma bond and 2 pi bond is present between C and N atoms.
2. Valence bond thoery as wells as hybridization. Lewis dot structure can be used to get the basic idea of the structure.
3. Because of the 2 pi bonds and 1 sigma bond formed by the hybridization of 2px, 2py, and 2pz between C and N atoms, this 2p overlap makes the bond stronger and shorter therefore the bond between C and N is linear. Also, based on the property of atoms to be on the position at smallest strichinderance as possible, H atom will be as far away from C atom, which will result in the linear structure.
5.3D: (BF 3)
The Valence Bond Theory is usually represented by the Lewis dot models. Boron is an unusual molecule because it does not follow the octet rule by having eight valence electrons around the boron atom. BF3 has single bonds between the boron atom and the fluorine atoms and contains no double bonds and an empty p orbital (figure 3). This is not predicted by the valence bond theory because it does not allow for any empty orbitals.
Valence bond theory and how it fails
figure 1
Valence Bond Theory states that when a half filled orbital, in contact with another half filled orbital, will hybridize to form a more stable bonding orbital. According to the Valence Bond Theory, the lone pair of one of the fluorine atoms should overlap to form a pi bond with the boron atom through a double bond in order to complete an octet on the boron atom (figure 2), but this does not occur.
There are two major reasons why there is no pi bond between any of the F atoms and B atoms: electronegativity and separation of charge. If one were to write out formal charges of the molecule without the pi bond, they will find that the formal charges on all of the atoms are 0. The boron shares it's 3 valence electrons with the fluorine creating a total of 6 electrons shared with the fluorine, making it's formal charge 0. Each fluorine shares one of it's valence electrons creating a total of 2 shared electrons with boron and has 6 more electrons to themselves, making their formal charges also 0 (figure 1).
figure 2
If one of the fluorines were to form a pi bond with the boron, then the fluorine, which is the most electronegative element, will have an oxidized state and a formal charge of +1 and, consequently the boron atom would have a formal charge of -1. Although this still leaves the BF3 molecule to have an overall formal charge of 0, the most stable form would be BF3 with no pi bonds because each of the atom's formal charge is 0. The most stable state is to leave the boron with an empty p orbital.
The empty p orbital
figure 3
BF3 is a planar molecule because it does not have a lone pair, which makes it have a trigonal planar geometry. BF3 is considered a Lewis acid because it accepts electrons at its empty p orbital.
Examples of reactivity of the empty p orbital
The most common reactions with involving BF3 is to form complexes with ethers.
figure 4
Other molecules like BF3
Carbenes are molecules that are used for different organic synthesis reactions. They typically involve a carbon with a lone pair and 2 bonded halides (like bromide). It also includes this empty p orbital and is fairly reactive.
figure 5
Matthew Shaltes | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/05%3A_Bonding_in_Polyatomic_Molecules/5.03%3A_Valence_Bond_Theory_-_Multiple_Bonding_in_Polyatomic_Molecules/5.3A%3A_Introduction.txt |
Water is a triatomic species, with one oxygen atom for every two hydrogen atoms per molecule. Through a combination of polarized bonds between the oxygen and hydrogen atoms, and a bent molecular geometry that gives the molecule an overall dipole moment, water posesses several unique chemical and physical properties that are unlike any other species of its size.
Molecular Structure
The classic model for the water molecule (H2O) is a central oxygen atom bonded to two hydrogen atoms on either side. Unlike other triatomic species that have a linear shape, the bonds on a water molecule are tilted at a slight angle, due to the presence of lone-pair electrons on the oxygen atom. Through VSEPR theory, it is shown that H2O has a steric number of 4, with two lone pairs and two bond pairs, making its molecular geometry "bent". As a result, the angle between the two oxygen-hydrogen bonds are approximately 104.5°, which is slightly smaller than the angle typically found in tetrahadral-shaped molecules (see above image). This is due to the particularly strong repulsions by the lone-pair electrons on the oxygen atom, which pushes the hydrogens closer together than usual. Each of the oxygen-hydrogen bonds measure approximately 95.84pm in length. This bent geometry due to the presence of lone pair electrons makes water different from typical linear triatomic species, both on an atomic and a macroscopic level.
Physical and Chemical Properties
Through its combination of bent geometry and polarized bonds, H2O has unique physical properties that are uncharacteristic of most other small molecules. As a result of oxygen being more eletronegative than hydrogen, the oxygen-hydrogen bonds have an unequal sharing of electrons, with a majority of the negative charge going to the oxygen atom, and leaving the positive chage to the hydrogens. With the bent structure of H2O, this creates a dipole on the overall molecule, with the oxygen end being negatively charged, and the hydrogen end being positively charged. As a polar molecule with hydrogen atoms, H2O can undergo hydrogen bonding with neighboring H2O atoms, in which the oxygen atoms are weakly bonded to hydrogen atoms from other molecules, due to the differences in electrical charge (see image above). This intermolecular attraction makes water a relatively stable substance, and gives it physical properties such as a relatively high boiling point (100°C) and melting point (0°C). Compare this to other molecules of this size, but are nonpolar, such as methane (CH4), with a boiling point of -161.6 °C and melting point of -182.5 °C.
Problems
1) Would you expect BeH2 to also be a bent triatomic molecule?
- No, because Be lacks the presence of lone pair electrons, which would give the molecule a linear geometry.
2) What makes water highly unusual in its solid phase, compared to its liquid phase?
- Unlike most substances, wherein the solid state is denser than the liquid state, water is less dense in the solid state, meaning its volume expands when freezing.
3) Would you expect ethanol to mix with water? Why or why not?
- Yes, because both substances are polar, the two liquids are miscible, and will mix to form a single homogeneous solution. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/05%3A_Bonding_in_Polyatomic_Molecules/5.04%3A_MO_Theory_-_The_Ligand_Group_Orbital_%28LGO%29_Approach_and_Application_to_Triatomic_Molecules/5.4D%3A_A_Bent_Triato.txt |
Symmetry Labels
E''
2
-1
0
-2
1
0
(Rx,Ry)
(xz,yz)
FIGURE2: Character table for the the point group D3h
B atom in BH3:
+s-orbital: with the shape of the sphere, its function is x2+y2+z2. Therefore, 2s orbital hasa1' symmetry
+p-orbital: has 3 orbitals , px, py, pz. Therefore, 2pz orbital has a2" symmetry
2px and 2py orbital are degenerate and have e' symmetry
3 Hydrogen atoms in BH3: (Ligand group orbitals)
a. Symmetry labels of LGOs:
-With the symmetry operations of BH3 above, we can determine how many LGO unmoved by creating the following table:
D3h
E
2C3
3C2
σh
2S3
v
LGO
3
0
1
3
0
1
-Next, with these values we can apply the following formula to identify the symmetry labels of the Ligan group orbitals
a= 1/h ∑[(N).Xr(R).Xi(R)]
h: the total number of coeficients of symmetry operation
N: the coeficient of the each symmetry operation
Xr(R): the character of the reducible representation corresponding to the R (values that just found in the LGO row
Xi(R): the character of the irreducible representation corresponding to the R (from the character table)
Calculation:
A1'= 1/12 [(1)(3)(1)+(2)(0)(1)+(3)(1)(1)+(1)(3)(1)+(2)(0)(1)+(3)(1)(1)] = 1 A1'
A2'= 1/12 [(1)(3)(1)+(2)(0)(1)+(3)(1)(-1)+(1)(3)(1)+(2)(0)(1)+(3)(1)(-1)]= 0 A2'
E'= 1/12 [(1)(3)(2)+(2)(0)(-1)+(3)(1)(0)+(1)(3)(2)+(2)(0)(-1)+(3)(1)(0)]= 1E'
A1''= 1/12 [(1)(3)(1)+(2)(0)(1)+(3)(1)(1)+(1)(3)(-1)+(2)(0)(-1)+(3)(1)(-1)]= 0 A1''
A2''= 1/12 [(1)(3)(1)+(2)(0)(1)+(3)(1)(-1)+(1)(3)(-1)+(2)(0)(-1)+(3)(1)(1)] = 0 A2''
E''= 1/12 [(1)(3)(2)+(2)(0)(-1)+(3)(1)(0)+(1)(3)(-2)+(2)(0)(1)+(3)(1)(0)] = 0 E ''
Thus, LGO's symmetry labels are a1' + e'
b. How to determine the shapes of the LGOS?
-LGO's symmetry labels are a1' + e'. There are 3 LGOs that can be made out of these symmetry labels,one LGO is from a1' and two LGOs are from e' due to doubly degenerate. In order to determine the shape of each LGO, we would use the wavefunctions.
-Three hydrogens in BH3 are assigned with Ψ1, Ψ2, Ψ3. Now lets look at how each Ψ is affected by the symmetry operations of the D3h and their results are completed in the following table:
D3h
E
C3
C23
C2
C2
C2’’
σh
S3
S23
σv
σv
σv’’
Ψ1
Ψ1
Ψ2
Ψ3
Ψ1
Ψ3
Ψ2
Ψ1
Ψ2
Ψ3
Ψ1
Ψ3
Ψ2
a1’
1
1
1
1
1
1
1
1
1
1
1
1
LGO1
Ψ1
Ψ2
Ψ3
Ψ1
Ψ3
Ψ2
Ψ1
Ψ2
Ψ3
Ψ1
Ψ3
Ψ2
Ψ (a1') = 4Ψ1+4Ψ2+4Ψ3
= 4(Ψ1+Ψ2+Ψ3)
Ψ(a1')= 1/√3 (Ψ1+Ψ2+Ψ3)
-The shape of the LGO1 is
D3h
E
C3
C23
C2
C2
C2’’
σh
S3
S23
σv
σv
σv’’
Ψ1
Ψ1
Ψ2
Ψ3
Ψ1
Ψ3
Ψ2
Ψ1
Ψ2
Ψ3
Ψ1
Ψ3
Ψ2
e’
2
-1
-1
0
0
0
2
-1
-1
0
0
0
LGO1
2Ψ1
-Ψ2
-Ψ3
0
0
0
2Ψ1
-Ψ2
-Ψ3
0
0
0
Ψ(e') = 4 (Ψ1)-2 (Ψ2)-2 (Ψ3)
= 2[ 2(Ψ1)-Ψ2-Ψ3]
Ψ(e') = 1/ √6 (2 Ψ1-Ψ2-Ψ3)
-The shape of the LGO2 is
-Noticed that in the LGO2, we have 1 nodal plane which is the horizonal line between the positive charge and negative charge. Therefore, the LGO3 (doubly degenerate with e') would also 1 nodal plane and its wavefunction would be Ψ(e') = 1/√2 (Ψ2-Ψ3). The shape of the LGO3 is
MO diagram
-As we can see in this diagram, the energy level of 3 LGOs are higher than the 2s orbital and below the 2 p orbital dued to the electronegativy of both Boron and Hydrogen. Hydrogen has higher electronegativity than boron, therefore hydrogen would have lower energy level in the MO diagram.
-In addition, B has 3 electrons in the valence electrons and 3 hydrogens have total 3 electrons. Therefore, the total number of electrons filled in orbitals are 6. With all of the informations above about symmetry labels of B atom and the 3 LGOs, we now construct the MO diagram of BH3. Noticed that, the bonding formation only happens to atoms that have the same symmetry labels. 2s orbital and LGO(1) would contribute 1 electron to give 2 spin pairs electrons at the a1' energy level. 2px and 2py orbitals would bond to the LGO(2) and LGO(3), which give 2 spin pairs electrons at the e' energy level.
FIGURE 3: MO diagram for the formation of BH3
-there is a detailed explaination that performed in this video
Problems
1. What are the non-bonding orbital in this BH3? '
2. In the above MO diagram, why does e' have 2 lines energy levels compared to a1' has only 1 line energy level?
3. What is the bond order of BH3 in this MO diagram?
4. Assume that we have d orbitals in this BH3, what are symmetry labels that d-orbitals have based on the character table?
5. Explain why a1' has the lowest energy level in the MO diagram.
Answers
1. a2''
2. because e' is doubly degenerate molecule orbital.
3. B.O = 3
4. a1' + e' + e''
5. a1' has no node, so it is stable and has lowest energy compare to e'.
5.5B: (NH 3)
C3v E 2C3 v Linear Rotations Quadratic
A1 1 1 1 z x2+y2,z2
A2 1 1 -1 Rz
E 2 -1 0 (x,y)(Rx, Ry) (x2-y2, xy)(xz,yz)
Contributors and Attributions
• Name #1 here (if anonymous, you can avoid this) with university affiliation
• Name #2 here with university affiliation
Tags below. If no tags exist, then add two new ones: "Vet1" and the level of the module content (e.g. "Fundamental"). See FAQ for more details.
5.5C: (CH 4)
This module seeks to explain the bonding of the 4 Hydrogen atoms to the 1 Carbon atom in the molecule CH4 (methane),using the molecular orbital theory. Molecular orbital theory describes orbitals that are formed with the interaction of the atomic orbitals of given atoms. These orbitals are spread out over the entire molecule and electrons fill these orbitals in accordance with the aufbau principle.
Introduction
Various concepts explain the molecular orbital theory in the bonding in methane, including character tables, symmetry, LGOs (ligand group orbital approach), and a qualitative MO diagram.
The Symmetry of CH4
• CH4 belongs to the Td point group and contains: 8C3 axes, 3C2 axes, 6S4 axes, and a dihedral plane of symmetry. Using the character table for the Td point group,
Character Table for Td Point group
Td E 8C3 3C2 6S4 6sigmad
A1 1 1 1 1 1
A2 1 1 1 -1 -1
E 2 -1 2 0 0
T1 3 0 -1 1 -1
T2 3 0 -1 -1 1
Molecular Orbital Diagram for CH4
Rename to desired sub-topic. You can delete the header for this section and place your own related to the topic. Remember to hyperlink your module to other modules via the link button on the editor toolbar.
Outside Links
• This is not meant for references used for constructing the module, but as secondary and unvetted information available at other site
• Link to outside sources. Wikipedia entries should probably be referenced here.
Problems
What are ligand group orbitals, and how are they used in MO theory in polyatomic molecules?
How many Vibrational modes and IR/Ramen stretches are there in CH4?
What are the major differences between VB theory and MO theory applied to polyatomic molecules?
Contributors and Attributions
• Name #1 here (if anonymous, you can avoid this) with university affiliation | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/05%3A_Bonding_in_Polyatomic_Molecules/5.05%3A_MO_Theory_Applied_to_Polyatomic_Molecules_%28BH_3%29_%28NH_3%29_and_%28CH_4%29/5.5A%3A_%28BH_3%29.txt |
Molecular Orbital (MO) theory is used by scientists to understand bonding in molecules. Carbon dioxide is a linear, centrosymmetric molecule with D∞h symmetry. Surrounded by two oxygens, carbon is the central atom. MO therory predicts π bond formation resulting from the interaction of C 2Px and y atomic orbitals with O fragment LGO(Ligand Group Orbitals).
Introduction
Carbon dioxide is a well studied molecule. Carbon dioxide is a covalent compound comprised of three atoms, carbon surrounded by two oxygens. Both carbon and oxygen contain p orbitals that are able to interact based on symmetry compatability. Valence bond (VB) theory predicts four bonds for carbon and two for each oxygen. The bond order of each carbon-oxygen bond is 2. Carbon dioxide contains two double bonds. Each double bond is comprised of one sigma bond and one π bond. Carbon dioxide is an important reagent used in industry, is a cental aspect of our global carbon cycle and is the basis for climate change. Lastly, Molecular Orbital (MO) theory is a better tool to use than VB theory because it does not assume localization of electrons. MO theory is based on Linear Combinations of Atomic Orbitals (LCAO).
MO Theory
Mo theory is used to predict bonding, anti-bonding and non-bonding orbitals. Anti-bonding orbitals are always higher in energy than corresonding bonding orbitals. Anti-bonding orbitals are usually indicated with an *. For example, σ*=sigma anti-bonding and σ=sigma bonding. Bonding can be thought of as atomic orbitals being in phase (constructive) and anti-bonding as being out of phase (destructive). Mo diagrams are graphical representations of orbital interactions based on symmetry compatability. Mo diagrams are based on individual atom electron configurations. For example, B=1s22s22p1. Boron has three valence electrons (highest n=principle quantum number in nS or nP) and 2 core electrons. Diatomic Boron (B2) has an MO diagram:
http://www.meta-synthesis.com/webbook/39_diatomics/diatomics.html
Notice that each individual B atom has 3 valence(n=2) electrons. When combined to form molecular orbitals the bond order is 1: BO=1/2(Bonding-Anti).
Carbon Dioxide MO diagram
The carbon dioxide MO diagram is based on a C atom and an O--O ligand fragment. Carbon has 2S and 2Px,y,z orbitals and the O--O fragment has 2S and 2Px,y,z orbitals that are involved in the formation of molecular orbitals. Since CO2 has D∞h symmetry the central atom's orbital symmetry lables can be obtained from the corresponding point group table: 2S=σg, 2Pzu and 2Px,yu. The LGO symmetry lables can be calculated using the point group table as well: Γσ=2σg + 2σu and Γπ=2πg + 2πu. The MO diagram for CO2 is more complicated than the diagram for B2. The follwing diagram fails to label orbital symmetries but the LGO 2Px,y particpate in the formation of π double bonds. The 2πg orbiatls are nonbonding because the C 2Px,y atomic orbitals are πu. The LGO 2Pz orbitals are involved in σ bonds.
http://cnx.org/content/m32935/latest/
Ater the formation of σ bonds resulting from C 2S and 2Pz electrons with O 2S and 2Pz orbitals the remaining C 2Px and 2Py orbitals interact with the O LGO fragment. Qualitatively, πu(2Px,y) is dipected by the MO diagram as dumbbells overlaping in phase and πg(2Px,y)* as overlapping out of phase.
Outside Links
• This is not meant for references used for constructing the module, but as secondary and unvetted information available at other site
• Link to outside sources. Wikipedia entries should probably be referenced here.
Problems
What does the g subscipt stand for in σg?
answer: The g implies symmetric with respect to inversion throught the center of the molecule.
Is carbon dioxide centrosymmetric? Does it have an inversion center?
answer: Yes carbon dioxide in centrosymmetric and includes an inversion center.
Contributors and Attributions
• Name #1 here (if anonymous, you can avoid this) with university affiliation | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/05%3A_Bonding_in_Polyatomic_Molecules/5.07%3A_MO_Theory_-_Learning_to_Use_the_Theory_Objectively/5.7A%3A_%28pi_%29-Bonding_in_%28CO_2%29.txt |
Crystalline solids
• The atoms, molecules or ions pack together in an ordered arrangement
• Such solids typically have flat surfaces, with unique angles between faces and unique
• 3-dimensional shape
• Examples of crystalline solids include diamonds, and quartz crystals
Amorphous solids
• No ordered structure to the particles of the solid
• No well defined faces, angles or shapes
• Often are mixtures of molecules which do not stack together well, or large flexible molecules
• Examples would include glass and rubber
1.01: The Unit Cell
The Unit Cell refers to a part of a simple crystal lattice, a repetitive unit of solid, brick-like structures with opposite faces, and equivalent edge points. In 1850, Auguste Bravais proved that crystals could be split into fourteen unit cells. Although there are several types of unit cells found in cubic lattices, we will be discussing the basic ones: Simple Cubic, Body-centered Cubic, and Face-centered Cubic. If any atom recrystalizes, it will eventually become the original lattice. Crystallization refers the purification processes of molecular or structures;.
Introduction
The Unit Cell contains seven crystal systems and fourteen crystal lattices. These unit cells are given types and titles of symmetries, but we will be focusing on cubic unit cells. One of the most commonly known unit cells is rock salt NaCl (Sodium Chloride), an octahedral geometric unit cell. The whole lattice can be reproduced when the unit cell is duplicated in a three dimensional structure. These unit cells are imperative for quite a few metals and ionic solids crystallize into these cubic structures. Calculating with unit cells is a simple task because edge-lengths of the cell are equal along with all 90⁰ angles.
Simple Cubic Unit Cells
Simple Cubic unit cells indicate when lattice points are only at the corners. They are the simplest (hence the title) repetitive unit cell. The lattice points at the corners make it easier for metals, ions, or molecules to be found within the crystalline structure. This phenomena is rare due to the low packing of density, but the closed packed directions give the cube shape.
Since the edges of each unit cell are equidistant, each unit cell is identical. In order to be labeled as a "Simple Cubic" unit cell, each eight cornered same particle must at each of the eight corners. This unit cell only contains one atom. Its packing efficiency is about 52%.
To packing efficiency, we multiply eight corners by one-eighth (for only one-eighth of the atom is part of each unit cell), giving us one atom.
8 Corners of a given atom x 1/8 of the given atom's unit cell = 1 atom
To calculate edge length in terms of r the equation is as follows:
2r
An example of a Simple Cubic unit cell is Polonium.
Body-centered Cubic Unit Cells
Body-centered Cubic (BCC) unit cells indicate where the lattice points appear not only at the corners but in the center of the unit cell as well. The atoms touch one another along the cube's diagonal crossing, but the atoms don't touch the edge of the cube. All atoms are identical. This type of unit cell is more common than that of the Simple Cubic unit cell due to tightly packed atoms. Its packing efficiency is about 68% compared to the Simple Cubic unit cell's 52%.
This unit cells contains two atoms.
To determine this, we multiply the previous eight corners by one-eighth and add one for the additional lattice point in the center.
(8 Corners of a given atom x 1/8 of the given atom's unit cell) + 1 additional lattice point = 2 atoms)
To calculate edge length in terms of r the equation is as follows:
$\dfrac{4r}{\sqrt{3}}$
Some examples of BCCs are Iron, Chromium, and Potassium.
It is a common mistake for CsCl to be considered bcc, but it is not. Instead, it is non-closed packed.
Face-centered Cubic Unit Cells
Face-centered Cubic (FCC) unit cells indicate where the lattice points are at both corners and on each face of the cell.
This is a more common type of unit cell since the atoms are more tightly packed than that of a Simple Cubic unit cell. Like the BCC, the atoms don't touch the edge of the cube, but rather the atoms touch diagonal to each face. Its packing efficiency is the highest with a percentage of 74%. Atoms touch one another along the face diagonals. All atoms are identical.
This unit cell contains four atoms.
To determine this, we take the equation from the aforementioned Simple Cubic unit cell and add to the parenthesized six faces of the unit cell multiplied by one-half (due to the lattice points on each face of the cubic cell).
(8 corners of a given atom x 1/8 of the given atom's unit cell) + (6 faces x 1/2 contribution) = 4 atoms)
To calculate edge length in terms of r the equation is as follows:
$2 \sqrt{2}r$
Problems
1. What is the edge length of the atom Polonium if its radius is 167 pm?
2. What type of unit cell is Caesium Chloride as seen in the picture. Briefly explain your answer. Diagram------------------>
3. Give two other examples (none of which is shown above) of a Face-Centered Cubic Structure metal. Briefly explain your reasonings.
4. Silver crystallizes with a FCC; the raidus of the atom is 160 pm. What is the density of the solid silver in grams per cubic centimeters?
Answers
1. Polonium is a Simple Cubic unit cell, so the equation for the edge length is 2r. Therefore, 2(167pm) = 334pm
2. Caesium Chloride is a non-closed packed unit cell. It is common for one to mistake this as a body-centered cubic, but it is not. The reason for this is because the ions do not touch one another. Also, in order to be considered BCC, all the atoms must be the same. Since the middle atome is different than the corner atoms, this is not a BCC.
3. Two examples of a FCC cubic structure metals are Lead and Aluminum. Each contains four atoms, six of which run diagonally on each face.
4. We approach this problem by first finding the mass of the unit cell. Mass of Silver is 107.87 g/mol, thus we divide by Avagadro's number 6.022 x 10-22.
We end up with 1.79 x 10-22 g/atom. Next we find the mass of the unit cell by multiplying the number of atoms in the unit cell by the mass of each atom (1.79 x 10-22 g/atom)(4) = 7.167 x 10-22 grams.
Next we find the edge length by:
${2}\sqrt{2}*{160pm}$
Which equals 4.525 x 10-10 meters.
Now we find the volume which equals the edge length to the third power. We convert meters into centimeters by dividing the edge length by 1 cm/10-2m to the third power. (4.525 x 10-10 m x 1cm/10-2m = 9.265 x 10-23 cubic centimeters.)
Finally, we find the density by mass divided by volume. So, 7.167 x 10-22 grams/9.265 x 10-23 cubic centimeters = 7.74 g/cm3
Contributors and Attributions
• Sonika Sidher--UC Davis | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/06%3A_Structures_and_Energetics_of_Metallic_and_Ionic_solids/6.01%3A_Introduction.txt |
Hydrogen and Fluorine, as well as Chlorine, Bromine and Iodine, exist as diatomic molecules. Both diatomic Hydrogen and diatomic Florine exist as gasses at room temperature and exhibit extremely low melting and boiling points. If the temperature of the system is lowered below the respective melting points of Hydrogen or Fluorine, molecular motion is reduced and a crystalline lattice is formed, creating a solid.
Solid State Hydrogen and Fluorine
Elemental Hydrogen and Elemental Fluorine, H2 and F2 respectively, exist as gases at room temperature. Though the melting point of H2 is 20.4 K (-252.75 °C), a temperature of 14.0 K (-259.15 °C) is required for the solidification of H2.3,4 As the solid structure is formed, the diatomic Hydrogen molecules adopt a hexagonal close packing (hcp) structure.4 Fluorine exhibits similar behaviour. The melting point of Fluorine is 53.54K (-219.61 °C), at which point the diatomic Fluorine adopts a cubic close packed crystalline arrangement (ccp), instead of a hexagonal close packed arrangement.3,4
Ability to use the sphere model of Crystalline structure for H2 and F2 solids
When solid, both dihydrogen and difluorine are small enough to allow for rotation within the solid structure (the radius of Fluorine is about equivalent to that of Hydrogen). This rotation occurs about a central axis midway between the two distal atoms of the molecule, creating two equal radii from the middle of the H-H or F-F bond to the outer boundary of each H or F atom. This rotation occurs 360° in the x, y and z directions of the central axis, as well as all manner of combinations of these three variables. The rotation of the molecule in all directions creates what can be though of as a spherical shell, the boundary of which is created by the distal atoms.4
Some common uses for solid hydrogen include gamma-ray ablation cages for elemental analysis, and a solid state cage in which to look at hydrogen bonding within molecules. 1,2
Outside Links
• This is not meant for references used for constructing the module, but as secondary and unvetted information available at other site
• Link to outside sources. Wikipedia entries should probably be referenced here.
Problems
1. Why can F2 and H2 be considered similar molecules and thus exhibit similar behaviour in rotation?
2. What is the different between a cubic close packing structure and hexagonal close packing?
3. Why do difluorine and dihydrogen exhibit different packing structures.
Contributors and Attributions
• Name #1 here (if anonymous, you can avoid this) with university affiliation
6.3C: Solid Metallic Ele
A metal (from Greek μέταλλον métallon, "mine, quarry, metal") is a material (an element, compound, or alloy) that is typically hard, opaque, shiny, and has good electrical and thermal conductivity. Metals are generally malleable - that is, they can be hammered or pressed permanently out of shape without breaking or cracking - as well as fusible (able to be fused or melted) and ductile (able to be drawn out into a thin wire). About 91 of the 118 elements in the periodic table are metals (some elements appear in both metallic and non-metallic forms).
Atoms of metals readily lose their outer shell electrons, resulting in a free flowing cloud of electrons within their otherwise solid arrangement. This provides the ability of metallic substances to easily transmit heat and electricity. While this flow of electrons occurs, the solid characteristic of the metal is produced by electrostatic interactions between each atom and the electron cloud. This type of bond is called a metallic bond.
Cubic and hexagonal close packing.
Crystalline solids consist of repeating patterns of its components in three dimensions (a crystal lattice) and can be represented by drawing the structure of the smallest identical units that, when stacked together, form the crystal. This basic repeating unit is called a unit cell.
Many metals adopt close packed structures i.e. cubic close packed (face centred cubic) and hexagonal close packed structures. A simple model for both of these is to assume that the metal atoms are spherical and are packed together in the most efficient way (close packing or closest packing). For closest packing, every atom has 12 equidistant nearest neighbours, and therefore a coordination number of 12. If the close packed structures are considered as being built of layers of spheres then the difference between hexagonal close packing and cubic close packed is how each layer is positioned relative to others. It can be envisaged that for a regular buildup of layers:
• hexagonal close packing has alternate layers positioned directly above/below each other, A,B,A,B,...
• cubic close packed (face centered cubic) has every third layer directly above/below each other, A,B,C,A,B,C,...
Body centred cubic
This is not a close packed structure. Here each metal atom is at the centre of a cube with 8 nearest neighbors, however the 6 atoms at the centres of the adjacent cubes are only approximately 15% further away so the coordination number can therefore be considered to be 14 when these are included. Note that if the body centered cubic unit cell is compressed along one 4 fold axis the structure becomes cubic close packed (face centred cubic).
Cubic, Hexagonal and Body-centred Packing
cubic close packing (ccp)
packing efficiency =74%
CN=12
hexagonal close packing (hcp)
packing efficiency =74%
CN=12
body-centred cubic packing (bcc)
packing efficiency =68%
CN=8
Trends in melting point
Melting points are chosen as a simple measure of the stability or strength of the metallic lattice. Some simple trends can be noted. The transition metals have generally higher melting points than the others. In the alkali metals (Group 1) and alkaline earth metals (Group 2) the melting point decreases as atomic number increases, but in transition metal groups with incomplete d-orbital subshells, the heavier elements have higher melting points. For a given period, the melting points reach a maximum at around Group 6 and then fall with increasing atomic number.
Mercury, caesium and gallium have melting points below 30 °C whereas all the other metals have sufficiently high melting points to be solids at "room temperature". The structures of the metals can be summarised by the table below which shows that most metals crystallise in roughly equal amounts of bcc, hcp and ccp lattices.
Crystal structure of metallic elements in the periodic table
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
H
He
453.69
Li
bcc
1560
Be
hcp
MP (K)
At. Symbol
Lattice type
B C N O F Ne
370.87
Na
bcc
923
Mg
hcp
933.47
Al
ccp
Si P S Cl Ar
336.53
K
bcc
1115
Ca
ccp
1814
Sc
hcp
1941
Ti
hcp
2183
V
bcc
2180
Cr
bcc
1519
Mn
1811
Fe
bcc
1768
Co
hcp
1728
Ni
ccp
1357.8
Cu
ccp
692.68
Zn
hcp
302.91
Ga
Ge As Se Br Kr
312.46
Rb
bcc
1050
Sr
ccp
1799
Y
hcp
2128
Zr
hcp
2750
Nb
bcc
2896
Mo
bcc
2430
Tc
hcp
2607
Ru
hcp
2237
Rh
ccp
1828
Pd
ccp
1235
Ag
ccp
594
Cd
430
In
505
Sn
904
Sb
Te I Xe
301.59
Cs
bcc
1000
Ba
bcc
2506
Hf
hcp
3290
Ta
bcc
3695
W
bcc
3459
Re
hcp
3306
Os
hcp
2719
Ir
ccp
2041.4
Pt
ccp
1337.33
Au
ccp
234.32
Hg
577
Tl
hcp
600.61
Pb
ccp
544.7
Bi
527
Po
At Rn
Group 1: Alkali metals
The alkali metals have their outermost electron in an s-orbital and this electronic configuration results in their characteristic properties. The alkali metals provide the best example of group trends in properties in the periodic table, with elements exhibiting well-characterized homologous behaviour.
The alkali metals have very similar properties: they are all shiny, soft, highly reactive metals at standard temperature and pressure and readily lose their outermost electron to form cations with charge +1. They can all be cut easily with a knife due to their softness, exposing a shiny surface that tarnishes rapidly in air due to oxidation by atmospheric moisture and oxygen. Because of their high reactivity, they must be stored under oil to prevent reaction with air, and are found naturally only in salts and never as the free element. In the modern IUPAC nomenclature, the alkali metals comprise the group 1 elements, excluding hydrogen (H), which is only nominally considered a group 1 element.
Group 2: Alkali Earth Metals
extended structures of Li, Mg, Ca
Lithium -bcc
Magnesium -hcp
Calcium -ccp
Note that Housecroft and Sharpe has Ca and Sr both listed as hexagonal and not cubic (face) close packed lattices. Calcium and Strontium exist in several allotropic forms and the lowest temperature forms (for Ca < 450 °C) are ccp. At high temperatures phase transitions occur to give hexagonal.
Return to the course outline or move on to Lecture 5: Structure of the elements Boron, Carbon and Phosphorus, Sulfur. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/06%3A_Structures_and_Energetics_of_Metallic_and_Ionic_solids/6.03%3A_The_Packing_of_Spheres_Model_Applied_to_the_Structures_of_Elements/6.3B%3A_H_and_F_Solids.txt |
Polymorphism is the ability of a solid material to exist in more than one form or crystal structure.
6.04: Polymorphism in Metals
Polymorphism is when a solid material can exist in more than one form or crystal structure. Polymorphism is very similar to Allotropy, but they should not be confused, polymorphism describes molecules, while Allotropy is used to describe atoms. Polymorphism has many uses whether it was in pharmaceutical, explosives or even food. This process was discovered by Eilhard Mitscherlich in 1821.
Kinetic vs. Thermodynamic Stability
When discussing the concept of Stability it is necessary to distinguish between thermodynamic and kinetic stability.
Here B is at lower energy than A so that ΔG is negative. The reaction should therefore proceed spontaneously and B is the more thermodynamically stable species.
The reaction as shown though has a barrier to the progress of the reaction called the Activation Barrier (Ea) and so the reaction may proceed very slowly. The thermodynamics describes only the starting and ending position of the reaction and not the intermediate or transition state. If the kinetics is slow, A is described as being inert while if it proceeds quickly then A is described as being labile.
The conditions that distinguish them are:
if the reaction takes longer than 1 minute under the conditions of concentration 0.1 M, temperature 25°C, then it is INERT,
if under the same conditions the reaction time < 1 minute, then it is LABILE.
In the lecture on isomerism, we depend on the samples being kinetically stable i.e. inert.
In the lecture on Chelation and Stability we concentrate on thermodynamic stability and look at the changes in free energy, enthalpy and entropy during the reaction. When we consider thermodynamic stability we need to be familiar with 2 formulae:
ΔG = - RT ln(K) or alternatively ΔG = - 2.303RT log10(K) ---(1)
ΔG = ΔH - TΔS ---(2)
The first relates the free energy to the stability constant and the second shows the breakdown into the component enthalpy and entropy terms.
6.05: Metallic Radii
Atomic radius is the radius of an atom which measures the distant from its nucleus to the electron. And metallic radii are the radii of the metallic atoms. It is the size of a metallic atom. However the actually measurement of the radius is very vague because electrons don’t stay at one point, they orbits around.
6.7A: Substitutional Alloys
Alloys are mixtures of metals or a mixture of a metal and another element. An alloy may be a solid solution of metal elements (a homogeneous mixture) or a mixture of metallic phases (a heterogeneous mixture of two or more solutions).
6.07: Alloys and Intermetallic Compounds
When a molten metal is mixed with another substance, there are two mechanisms that can cause an alloy to form: (1) atom exchange or (2) interstitial mechanism. The relative size of each element in the mix plays a primary role in determining which mechanism will occur.
When the atoms are relatively similar in size, the atom exchange method usually happens, where some of the atoms composing the metallic crystals are substituted with atoms of the other constituent. This is called a substitutional alloy. Examples of substitutional alloys include bronze and brass, in which some of the copper atoms are substituted with either tin or zinc atoms.
Why Substitutional Alloys Occur: Bonding
The bonding between two metals is best described as a combination of metallic electron "sharing" and covalent bonding, one can't occur without the other and the proportion of one to the other changes depending on the constituents involved. Metals share there electrons throughout there structure, this flow of electrons is the reason behind many of the characteristics associated with metals, including their ability to act as conductors. The different amount and strength of covalent bonds can change depending on the different specific metals involved and how they are mixed. The covalent bonding is what is responsible for the crystal structure as well as the melting point and various other physical properties.
As the similarities between the electron structure of the metals involved in the alloy increase, the metallic characteristics of the alloy decrease. Pure metals are useful but their applications are often limited to each individual metal's properties. Alloys allow metal mixtures that have increased resistance to oxidation, increased strength, conductivity, and melting point; Essentially any property can be manipulated by adjusting alloy concentrations. An example could be Brass Door fixtures, they are strong and resist corrosion better then pure zinc or copper, the two major metals that constitute a brass alloy. The combination also has a low melting point allowing it to be easily cast into many different shapes and sizes.(1) There are many other aspects of substitutional alloys that could be explored in depth, but the basic concept is the idea that each individual metal in an alloy give the final product its chemical and physical properties.
Substitutional alloys played an important role in the development of human society and culture as we know it today. The Bronze age itself is named after the Substitutional alloy consisting of tin in a metallic solution of copper. Ancient bronzes are very impure, or even mislabeled, containing large amounts of zinc and arsenic as well as lots of impurities. These many substitutional alloys allowed for stronger tools and weapons, they allowed for increased productivity in the workshop as well as on the battlefield. The need for raw materials like tin and copper for the production of bronze also spurred an increase in trade, since their ores are rarely found together. The current chemical understanding of substitutional alloys would not be so in depth if it weren't for the usefulness of the alloys to humans.
Summary
An alloy is a mixture of metals that has bulk metallic properties different from those of its constituent elements. Alloys can be formed by substituting one metal atom for another of similar size in the lattice (substitutional alloys), by inserting smaller atoms into holes in the metal lattice (interstitial alloys), or by a combination of both. Although the elemental composition of most alloys can vary over wide ranges, certain metals combine in only fixed proportions to form intermetallic compo
Problems
1. Are substitutional metal alloys naturally occurring on earths surface?
2. What are two characteristics of a metal required for a substitutional alloy to form?
3. Can Oxygen or Nitrogen be a part of the crystal structure of a substitutional alloy?
Solutions
1. No, the oxidizing nature of the earths atmosphere, as well as the need for specific and concentrated metals keeps these from being found naturally occurring.
2. Similar radii and similar electronegativity.
3. Only metallic elements can form the necessary metallic bonds that allow alloys to form. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/06%3A_Structures_and_Energetics_of_Metallic_and_Ionic_solids/6.04%3A_Polymorphism_in_Metals/6.4A%3A_Polymorphism_-_Phase_Changes_in_the_Solid_State.txt |
Figure 6.7B.1: Different atomic mechanisms of alloy formation, showing pure metal, substitutional, and interstitial structures. from Wikipedia.
With the interstitial mechanism, one atom is usually much smaller than the other, so cannot successfully replace an atom in the crystals of the base metal. The smaller atoms become trapped in the spaces between the atoms in the crystal matrix, called the interstices. This is referred to as an interstitial alloy. Steel is an example of an interstitial alloy, because the very small carbon atoms fit into interstices of the iron matrix. Stainless steel is an example of a combination of interstitial and substitutional alloys, because the carbon atoms fit into the interstices, but some of the iron atoms are replaced with nickel and chromium atoms.
6.7C: Intermetallic Compounds
Intermetallic compounds are solid phases containing two or more metallic elements, with optionally one or more non-metallic elements, whose crystal structure differs from that of the other constituents. Under this definition the following are included
• The definition of a metal is taken to include:
• the so-called post-transition metals, i.e. aluminium, gallium, indium, thallium, tin and lead
• some, if not all, of the metalloids, e.g. silicon, germanium, arsenic, antimony and tellurium.
• Wikipedia
6.8A: Electrical Conductivity and Resistivity
• 6.8A: Electrical Conductivity and Resistivity
Electrical resistivity and conductivity is an important property for materials. Different materials have different conductivity and resistivity. Electrical conductivity is based on electrical transport properties. These can be measured with multiple techniques by using a variety of instruments. If electricity easily flows through a material, that material has high conductivity. Some materials that have high conductivity include copper and aluminum.
• 6.8B: Band Theory of Metals and Insulators
Band Theory was developed with some help from the knowledge gained during the quantum revolution in science. In 1928, Felix Bloch had the idea to take the quantum theory and apply it to solids. In 1927, Walter Heitler and Fritz London discovered bands- very closely spaced orbitals with not much difference in energy.
• 6.8C: The Fermi Level
• 6.8D: Band Theory of Semiconductors
6.08: Bonding in Metals and Semicondoctors
Electrical resistivity and conductivity is an important property for materials. Different materials have different conductivity and resistivity. Electrical conductivity is based on electrical transport properties. These can be measured with multiple techniques by using a variety of instruments. If electricity easily flows through a material, that material has high conductivity. Some materials that have high conductivity include copper and aluminum. Electrical conductivity is the measure of how easily electricity flows through a material.
Conductivity vs Resistivity
Conductivity and resistivity are inversely proportional to each other. When conductivity is low, resistivity is high. When resistivity is low, conductivity is high. The equation is as follows:
$\rho = \dfrac{1}{\sigma}$
where
• Resistivity is represented by $\rho$ and is measured in Ohm-meters ($Ωm$),
• Conductivity is represented by $\sigma$ and is measured in Siemens ($1/Ωm$).
Since conductivity is the measure of how easily electricity flows, electrical resistivity measures how much a material resists the flow of electricity.
Electrical Transport Properties
Simply put, electricity is the movement of electrons through a material. As electrons move through a material, it comes into contact with atoms in the material. Collisions slow the electrons down. Each collision increases the resistivity of the material. The easier the electrons continue through a material, the fewer collisions that take place and the higher the conductivity.
When temperature increases, the conductivity of metals usually decreases, while the conductivity of semiconductors increases. This of course assumes that the material is homogenous, which is not always the case. You can calculate resistivity using the following equation
$\dfrac{E}{J} = ρ$
As you already read, ρ is the symbol for resistivity. $E$ is the electric field and has units of Volts per meter (V/m). J is the current density and has units of amps per meter squared (A/m2). The electric field is calculated by dividing the Voltage by the length, l, that voltage is applied.
$E=\dfrac{V}{l}$
The current density is calculated by the equation below
$J=\dfrac{I}{A}$
I is the current and is divided by the cross sectional area, A, over which the current flows.
Resistivity vs Resistance
Resistivity and resistance are two different things. Resistivity does not depend on size or shape. Resistance, however, does. You can calculate resistance with the equation below.
$R=\dfrac{V}{I}$
R refers to resistance and is measured in Ω. $V$ is the voltage and is measured in volts. I measures the current and its unit is amps (A).
References
1. Electrical Conductivity and Resistivity, Heaney, Michael, Electrical Measurement, Signal Processing, and Displays. Jul 2003
2. Levy, Peter M., and Shufeng Zhang. "Electrical Conductivity of Magnetic Multilayered Structures." Physical Review Letters 65.13 (1990): 1643-646. Print.
Problems
1. What is the current density of a material with a resistivity of 12.Ωm and an electric field of 64.V/m?
2. If the voltage of 6V is passed through a substance with a radius of 2m and a length of 3m, what is the electric field?
3. What is the electric field of a material when the current is equal to 25A, the resistance is measured to be 78Ω, the current density equals 24A/m2, and the length the current flows is 100m?
4. A material has a voltage of 150V and width of 24m. The material also has a current of 62A and travels a distance of 5m. What is the conductivity?
5. A metal originally has an electron colliding with every fifth atom and increases from a temperature of 6K to 100K. A semiconductor originally has an electron colliding with every fifth atom and increases from a temperature of 6K to 100K. What material will have a greater resistivity? Why?
Answers to Problems:
1. E/J = ρ ---> J=E/ρ = 64V/m /12Ωm = 5.33A/m2
2. E=V/l = 6V/3m = 2V/m
3. E=V/l
V=IR ---> E=IR/l = 25A x 78Ω/100m = 19.5V/m
4. E/J = ρ
E=V/l
J=I/A ---> ρ=(V/l)/(I/A) = (150V/5m)/(62A/(24m x 5m) = 58Ωm
ρ = 1/σ ---> 1/ρ = σ = 1/58Ωm
5. The material that has the greatest resistivity is the metal because as temperature increases metals are more likely to increase in resistivity and semiconductors usually decrease in resistivity as temperature increase.
Contributors and Attributions
• Michael Ford (UCD) and Alexandra Christman (UCD) | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/06%3A_Structures_and_Energetics_of_Metallic_and_Ionic_solids/6.07%3A_Alloys_and_Intermetallic_Compounds/6.7B%3A_Interstitial_Alloys.txt |
Band Theory was developed with some help from the knowledge gained during the quantum revolution in science. In 1928, Felix Bloch had the idea to take the quantum theory and apply it to solids. In 1927, Walter Heitler and Fritz London discovered bands- very closely spaced orbitals with not much difference in energy.
In this image, orbitals are represented by the black horizontal lines, and they are being filled with an increasing number of electrons as their amount increases. Eventually, as more orbitals are added, the space in between them decreases to hardly anything, and as a result, a band is formed where the orbitals have been filled.
Different metals will produce different combinations of filled and half filled bands.
Sodium's bands are shown with the rectangles. Filled bands are colored in blue. As you can see, bands may overlap each other (the bands are shown askew to be able to tell the difference between different bands). The lowest unoccupied band is called the conduction band, and the highest occupied band is called the valence band.
Bands will follow a trend as you go across a period:
• In Na, the 3s band is 1/2 full.
• In Mg, the 3s band is full.
• In Al, the 3s band is full and the 3p ban is 1/2 full... and so on.
The probability of finding an electron in the conduction band is shown by the equation:
\[ P= \dfrac{1}{e^{ \Delta E/RT}+1} \]
The ∆E in the equation stands for the change in energy or energy gap. t stands for the temperature, and R is a bonding constant. That equation and this table below show how the bigger difference in energy is, or gap, between the valence band and the conduction band, the less likely electrons are to be found in the conduction band. This is because they cannot be excited enough to make the jump up to the conduction band.
ELEMENT ∆E(kJ/mol) of energy gap # of electrons/cm^3 in conduction band insulator, or conductor?
C (diamond) 524 (big band gap) 10-27 insulator
Si 117 (smaller band gap, but not a full conductor) 109 semiconductor
Ge 66 (smaller band gap, but still not a full conductor) 1013 semiconductor
Conductors, Insulators and Semiconductors
A. Conductors
Metals are conductors. There is no band gap between their valence and conduction bands, since they overlap. There is a continuous availability of electrons in these closely spaced orbitals.
B. Insulators
In insulators, the band gap between the valence band the the conduction band is so large that electrons cannot make the energy jump from the valence band to the conduction band.
C. Semiconductors
Semiconductors have a small energy gap between the valence band and the conduction band. Electrons can make the jump up to the conduction band, but not with the same ease as they do in conductors.
There are two different kinds of semiconductors: intrinsic and extrinsic.
i. Intrinsic Semiconductors
An intrinsic semiconductor is a semiconductor in its pure state. For every electron that jumps into the conduction band, the missing electron will generate a hole that can move freely in the valence band. The number of holes will equal the number of electrons that have jumped.
ii. Extrinsic Semiconductors
In extrinsic semiconductors, the band gap is controlled by purposefully adding small impurities to the material. This process is called doping. Doping, or adding impurities to the lattice can change the electrical conductivity of the lattice and therefore vary the efficiency of the semiconductor. In extrinsic semiconductors, the number of holes will not equal the number of electrons jumped. There are two different kinds of extrinsic semiconductors, p-type (positive charge doped) and n-type (negative charge doped).
Problems
1. How do you distinguish between a valence band and a conduction band?
2. Is the energy gap between an insulator smaller or larger than the energy gap between a semiconductor?
3. What two methods bring conductivity to semiconductors?
4. You are more likely to find electrons in a conduction band if the energy gap is smaller/larger? 5. The property of being able to be drawn into a wire is called...
Answers
1. The valence band is the highest band with electrons in it, and the conduction band is the highest band with no electrons in it.
2. Larger
3. Electron transport and hole transport
4. Smaller
5. Ductility
Contributors and Attributions
• Sierra Blair (UCD)
Jim Clark (Chemguide.co.uk) | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/06%3A_Structures_and_Energetics_of_Metallic_and_Ionic_solids/6.08%3A_Bonding_in_Metals_and_Semicondoctors/6.8B%3A_Band_Theory_of_Metals_and_Insulators.txt |
Band Theory was developed with some help from the knowledge gained during the quantum revolution in science. In 1928, Felix Bloch had the idea to take the quantum theory and apply it to solids. In 1927, Walter Heitler and Fritz London discovered bands- very closely spaced orbitals with not much difference in energy.
In this image, orbitals are represented by the black horizontal lines, and they are being filled with an increasing number of electrons as their amount increases. Eventually, as more orbitals are added, the space in between them decreases to hardly anything, and as a result, a band is formed where the orbitals have been filled.
Different metals will produce different combinations of filled and half filled bands.
Sodium's bands are shown with the rectangles. Filled bands are colored in blue. As you can see, bands may overlap each other (the bands are shown askew to be able to tell the difference between different bands). The lowest unoccupied band is called the conduction band, and the highest occupied band is called the valence band.
Bands will follow a trend as you go across a period:
• In Na, the 3s band is 1/2 full.
• In Mg, the 3s band is full.
• In Al, the 3s band is full and the 3p ban is 1/2 full... and so on.
The probability of finding an electron in the conduction band is shown by the equation:
\[ P= \dfrac{1}{e^{ \Delta E/RT}+1} \]
The ∆E in the equation stands for the change in energy or energy gap. t stands for the temperature, and R is a bonding constant. That equation and this table below show how the bigger difference in energy is, or gap, between the valence band and the conduction band, the less likely electrons are to be found in the conduction band. This is because they cannot be excited enough to make the jump up to the conduction band.
ELEMENT ∆E(kJ/mol) of energy gap # of electrons/cm^3 in conduction band insulator, or conductor?
C (diamond) 524 (big band gap) 10-27 insulator
Si 117 (smaller band gap, but not a full conductor) 109 semiconductor
Ge 66 (smaller band gap, but still not a full conductor) 1013 semiconductor
Conductors, Insulators and Semiconductors
A. Conductors
Metals are conductors. There is no band gap between their valence and conduction bands, since they overlap. There is a continuous availability of electrons in these closely spaced orbitals.
B. Insulators
In insulators, the band gap between the valence band the the conduction band is so large that electrons cannot make the energy jump from the valence band to the conduction band.
C. Semiconductors
Semiconductors have a small energy gap between the valence band and the conduction band. Electrons can make the jump up to the conduction band, but not with the same ease as they do in conductors.
There are two different kinds of semiconductors: intrinsic and extrinsic.
i. Intrinsic Semiconductors
An intrinsic semiconductor is a semiconductor in its pure state. For every electron that jumps into the conduction band, the missing electron will generate a hole that can move freely in the valence band. The number of holes will equal the number of electrons that have jumped.
ii. Extrinsic Semiconductors
In extrinsic semiconductors, the band gap is controlled by purposefully adding small impurities to the material. This process is called doping. Doping, or adding impurities to the lattice can change the electrical conductivity of the lattice and therefore vary the efficiency of the semiconductor. In extrinsic semiconductors, the number of holes will not equal the number of electrons jumped. There are two different kinds of extrinsic semiconductors, p-type (positive charge doped) and n-type (negative charge doped).
Problems
1. How do you distinguish between a valence band and a conduction band?
2. Is the energy gap between an insulator smaller or larger than the energy gap between a semiconductor?
3. What two methods bring conductivity to semiconductors?
4. You are more likely to find electrons in a conduction band if the energy gap is smaller/larger? 5. The property of being able to be drawn into a wire is called...
Answers
1. The valence band is the highest band with electrons in it, and the conduction band is the highest band with no electrons in it.
2. Larger
3. Electron transport and hole transport
4. Smaller
5. Ductility
Contributors and Attributions
• Sierra Blair (UCD)
Jim Clark (Chemguide.co.uk) | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/06%3A_Structures_and_Energetics_of_Metallic_and_Ionic_solids/6.08%3A_Bonding_in_Metals_and_Semicondoctors/6.8D%3A_Band_Theory_of_Semiconductors.txt |
https://eng.libretexts.org/Core/Mate...Semiconductors
https://eng.libretexts.org/Core/Mate...iconductors_II
6.10A: Ionic Radii
In a crystal lattice, the ionic radius is a measure of the size of the atom's ion.6 When formed, ionic atoms change in size with respect to their orginal atom. Cation radii will decrease and the anion radii will increase in size compared to their neutral atoms. Questions such as: "What methodology is used by chemists to measure ionic radii?" and "Are there any non-experimental ways to estimate the size of ionic radii?" will be answered in this module. Accordingly, there are many ways to determine ionic radii.
Introduction
In the past, after an atom is ionized, X-ray diffraction is used to measure how much the radius of the atom increased or decreased. However, scientists wanted to use another technique, due to the fact, that X-ray diffraction is difficult to distinguish a boundary between two ions. As a result, the hard sphere model can be used.
Hard-Sphere model
The Hard-Sphere model are impenetrable spheres that do not overlap in space.5 The Hard-Sphere model has been tested by well-known scientists; Lande', Pauling and Goldsmidt. The ion radii measured under crystal state of ionic compound which cations and anions are stacking in pattern as shown below.
The Hard-Sphere model can be applied to metallic and ionic compounds such as NaCl, which is shown below.
In general, scientists uses formula of Internuclear distance to test out the radii of ion then compared with the ion radii had done on X-ray diffraction:
Internuclear distance (d) = rcation + ranion 2
*To calculate ion radii, Lande used ionic compound under solid state (ex: NaCl). This will minize the distribution of electrons.
1. Find the radii of anion (r-) atom.
2. Find internuclear distance (d) between anion and cation.
3. Use Internuclear distance formula to find the r+.
Periodic Trends
As described earlier, cations are smaller in size compared to their neutral atoms while anions are larger in size.Cations are smaller than its neutral atoms because the positive nuclear charge, which holds the electrons in closer, exceeds the negative charge when a metal atom loses an electron. On the contrary, anions are larger because the electrons are not held as tightly, repulsions of electrons increase, and the electrons spread out more due to nonmetal atoms gaining an electron. Refer to the outside link to learn more about the periodic trends for ionic radii (
Problems
1. What is the most general formula that used to determine the ion radii for hard sphere model?
2. Find radius for Cacium ion in Calcium Chloride (CaCl2). List out all the steps (numbers are not necessary)
3. Determine which is larger:
a) K+ or Cs+?
b) La3+ or Lu3+?
c) Ca2+ or Zn2+?
Answers
1. Internuclear distance (d) = rcation + ranion
2. Find the radii of anion (r-) atom; Find internuclear distance (d) between anion and cation; Use Internuclear distance formula to find the r+.
3. a.) Cs+ b.) La3+ c.) Ca2+
6.11A: Structure - Rock Salt (NaCl)
Rock salt also known as NaCl is an ionic compound. It occurs naturally as white cubic crystals. The structure of NaCl is formed by repeating the unit cell. It has an organized structure and has a 1:1 ratio of Na:Cl.
Introduction
Rock salt ($\ce{NaCl}$) is an ionic compound that occurs naturally as white crystals. It is extracted from the mineral form halite or evaporation of seawater. The structure of NaCl is formed by repeating the face centered cubic unit cell. It has 1:1 stoichiometry ratio of Na:Cl with a molar mass of 58.4 g/mol. Compounds with the sodium chloride structure include alikali halides and metal oxides and transition-metal compounds. An important role to many important applications is structure and dynamics of water. Some applications include crystallization of proteins and conformational behavior of peptides and nucleic acids.
Structure
Figure $1$ shows how the Na+ and Cl- ions occupy the space. The smaller ions are the Na+ with has an atomic radius of 102 pm, and the larger ions are the Cl- with an atomic radium of 181 pm. Since NaCl are one to one ratio as a compound, the coordination numbers of Na and Cl are equal. The larger green ions represent Cl- and the smaller purple ions represent Na+. However, the structure of this molecule allows their positions to be switched since the coordination numbers are equivalent.
A Unit Cell
The unit cell of $\ce{NaCl}$ consists of $\ce{Na^{+}}$ ions and $\ce{Cl^{-}}$ ions. There are four types of site: unique central position, face site, edge sites and corner site, which are used to determine the number of Na+ ions and Cl- ions in the unit cell of NaCl. When counting the number of ions, a corner site would be shared by 7 other unit cells. Therefore, 1 corner would be 1/8 of an ion. A similar occurrence happens with the face site and the edge sites. For a face site, it is shared by 1 other unit cell and for an edge site, the ion is shared by 3 other unit cells. $\ce{NaCl}$ is a face centered cubic unit cell which has four cations and four anions. This can be shown by counting the number of ions and multiplying them in relation to their position.
• $\ce{Na^{+}}$: $1_{center} + 12_{edge} \times \dfrac{1}{4} = 4\, \text{sodium ions total per cell} \nonumber$
• $\ce{Cl^{-}}$: $4_{face} \times \dfrac{1}{2} + 8_{corner} \times \dfrac{1}{8} = 4\, \text{chloride ions total per cell} \nonumber$
Each ion in this lattice has six of the other kind of ion as its nearest neighbors, and twelve of the same kind of ions as its second nearest neighbors. There are many ionic compounds that assume this structure including all other halides of Na, Li, K and Rb. CsF, AgF, AgCl, BaO, CoO, and SrS are also among many that will form similar structures to NaCl.
References
1. Gao, H.X., L.-M. Peng, and J.M Zuo. "Lattice dynamics and Debye-Waller factors of some compounds with the sodium chloride structure." Acta Crystallographica: Section A (Wiley-Blackwell) 55.6 (1999): 1014. Academic Search Complete. EBSCO. Web.
2. Housecroft, Catherine E., and Alan G. Sharpe. Inorganic Chemistry. 3rd ed. Harlow: Pearson Education, 2008. Print.
3. Jun Soo, Kim, and Yethiraj Arun. "A Diffusive Anomaly of Water in Aqueous Sodium Chloride Solutions at Low Temperatures." Journal of Physical Chemistry B 112.6 (2008): 1729-1735. Academic Search Complete. EBSCO. Web.
• Michael Ford | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/06%3A_Structures_and_Energetics_of_Metallic_and_Ionic_solids/6.09%3A_Semiconductors/6.9A%3A_Intrinsic_Semiconductors.txt |
This page is going to discuss the structure of the molecule cesium chloride (\(\ce{CsCl}\)), which is a white hydroscopic solid with a mass of 168.36 g/mol. Cesium Chloride is a type of unit cell that is commonly mistaken as Body-Centered Cubic. This misconception is easy to make, since there is a center atom in the unit cell, but CsCl is really a non-closed packed structure type.
Introduction
CsCl has a boiling point of 1303 degrees Celsius, a melting point of 646 degrees Celsius, and is very soluble in water. For the most part this molecule is stable, but is not compatible with strong oxidizing agents and strong acids. Although it is not hazardous, one should not prolong their exposure to CsCl. CsCl is an ionic compound that can be prepared by the reaction:
\[\ce{Cs2CO3 + 2HCl -> 2 CsCl + H2O + CO2}\]
CsCl crystallize in a primitive cubic lattice which means the cubic unit cell has nodes only at its corners. The structure of CsCl can be seen as two interpenetrating cubes, one of Cs+ and one of Cl-. The ions are not touching one another. Touching would cause repulsion between the anion and cation. Some may mistake the structure type of CsCl with NaCl, but really the two are different. CsCl is more stable than NaCl, for it produces a more stable crystal and more energy is released.
Anions and cations have similar sizes. Each Cs+ is surrounded by 8 Cl- at the corners of its cube and each Cl- is also surrounded by 8 Cs+ at the corners of its cube. The cations are located at the center of the anions cube and the anions are located at the center of the cations cube. There is one atom in CsCl. To determine this, the following equation is given:
8 Corners of a given atom x 1/8 of the given atom's unit cell = 1 atom
Applications
Cesium chloride is used in centrifugation, a process that uses the centrifugal force to separate mixtures based on their molecular density. It is also used in the preparation of electrically conducting glasses. Radioactive CsCl is used in some types of radiation therapy for cancer patients, although it is blamed for some deaths.
References
1. Carter, C. Barry., and M. Grant. Norton. "Binary Compounds." Ceramic Materials Science and Engineering. New York, NY: Springer, 2007. pg87-88
2. Quéré, Yves. "Stable Structure of Halides." Physics of Materials. Amsterdam: Gordon and Breach Science, 1998.
• Ana Popovich
6.11C: Structure - Fluorite (CaF)
Calcium Fluoride is a solid and forms a cube like structure that is centralized around the calcium molecules. The crystal lattice structure that Calcium Fluoride is also known as the fluorite structure (Figure \(1\)) where the Ca2+ ions are eight-coordinate, being centered in a cube of eight F ions. Each F is coordinated to four Ca2+ in the shape of a tetrahedron.
Calcium Fluoride is Quasilinear
When Calcium Fluoride is in a single molecule it forms a Quasilinear structure. Quasilinear means the molecule resonates between a linear shape and a bent shape.Calcium Fluoride is a polyatomic molecule that contains one calcium molecule and two fluoride molecules. Calcium Fluoride is a quasilinear molecule the bonds are created from the single electrons of calcium and the single electron from fluoride. \(\ce{CaF2}\) has its electrons contained with in the 3d orbitals and are able to move between dyz and dz2 squared. The molecule in linear when they are in the dz2 orbitals the molecule is also the most stable in this shape. When the electrons are in the dyz orbitals the molecule becomes bent. The molecule resonates between these two shapes making it quasilinear. Figures two and three show how the d-orbitals cause the molecule to bend.
Fig # 1
Fig # 2 Fig # 3
6.11D: Structure - Antifluorite
Antifluorite is a mineral with a crystal structure identical with that of fluorite but with the positions of the cations and anions reversed (Figure \(1\)).
The same crystal structure is found in numerous ionic compounds with formula AB2, such as ceria (CeO2), zirconia (cubic ZrO2), uranium dioxide (UO2). In the corresponding anti-structure, called the antifluorite structure, anions and cations are swapped, such as beryllium carbide (Be2C) or lithium oxide (Li2O), potassium sulfate (K2SO4).
Contributors and Attributions
• en.Wikipedia.org/wiki/Anti-structure | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/06%3A_Structures_and_Energetics_of_Metallic_and_Ionic_solids/6.11%3A_Ionic_Lattices/6.11B%3A_Structure_-_Caesium_Chloride_%28CsCl%29.txt |
ZnS has a unique structure type compared to other molecules, having different types of unique structures. ZnS can have a zinc blende structure which is a "diamond-type network" and at a different temperature, ZnS can become the wurtzite structure type which has a hexagonal type symmetry. Structure-wise, the zinc blende structure is more thermodynamically favored, however, because of the wurtzite structures slow construction, both forms of ZnS can be found.
Summary: Zinc blend is a compound that comes in two forms: sphalerite and wurtzite. These are characterized by a 1:1 stoichiometric ratio of Zinc to Sulfur. It maintains a tetrahedral arrangement in both forms.
Introduction
Zinc sulfide (ZnS) is a unique compound that forms two types of crystalline structures. These two polymorphs are wurtzite and zincblende (also known as sphalerite). Wurtzite has a hexagonal structure, while zincblende is cubic. It is characterized by single bonds between each atom and maintenance of a 1:1 zinc to sulfur ratio.
sphalerite:
Site Zn S
Central 4 0
Face 0 6(1/2) = 3
Corner 0 8(1/8) = 1
Total 4 4
Since the number of atoms in a single unit cell of Zn and S is the same, it is consistent with the formula ZnS.
The ionic radius for Zn2+ is 74pm and for S2- is 190pm. Therefore the ratio between cationic and anionic radii in zinc blend is 0.39 (74pm/190 pm) .This suggests a tetrahedral ion arrangement and four nearest neighbors from standard crystal structure prediction tables. Therefore, four sulfur atoms surround each zinc atom and four zinc atoms surround each sulfur atom.2 The coordination number, the number of of electron pairs donated to a metal by its ligands, for both zinc an sulfur is four.1,2 The difference between wurtzite and zincblende lies in the different arrangements of layers of ions.2
Zincblende (Sphalerite)
Zincblende is characterized as a cubic closet packing (ccp), also known as face-centered cubic, structure.1,4 This crystal lattice structure is shown in Figures 1 & 2 below.
Fig. 1. A break down of cubic closest packing. (Author: Maghémite Date: May 5, 2008. Licensed under the Creative Commons Attribution-Share Alike 3.0 Unported, 2.5 Generic, 2.0 Generic and 1.0 Generic license.)
Fig. 2. A representation of ccp structure. (from Public Domain)
Notice how only half of the tetrahedral sites are occupied.
Thermal stability
Density tends to decrease as temperature increases. In this case, since ccp structures are more dense than hcp structures, so a conversion from sphalerite to wurtzite occurs naturally over time at a rate similar to that of diamond to graphite. The sphalerite structure is favored at 298k by 13kJ/mol, but at 1296K the transition to wurtzite occurs.3
Wurtzite
Wurtzite has a hexagonal closest packing structure (hcp), which is characterized by 12 ions in the corners of each unit that create a hexagonal prism (seen in Fig. 3).2 As discussed previously, zincblende slowly transforms to wurtzite due to thermodynamic stability.
Fig 3. HCP structure of wurtzite. (Creator: Alexander Mann Date: 01/14/2006 Licensed under the Creative Commons Attribution-Share Alike 2.0 Germany license)
Calculating density of a crystal structure
Density = Mass of unit cell / volume of unit cell.
where:
Mass of unit cell = Number of atoms in a unit cell x the mass of each atom
volume of unit cell = a3 x 10-30
Contributors and Attributions
• Emma Mele, UC Davis Animal Biology | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/06%3A_Structures_and_Energetics_of_Metallic_and_Ionic_solids/6.11%3A_Ionic_Lattices/6.11E%3A_Structure_-_Zinc_Blende_%28ZnS%29.txt |
Silicon dioxide, \(\ce{SiO2}\), also known as silica is a linear molecule that is formed by one silicon atom and two oxygen atoms with two sets of doubles bonds and 4 single bonds. Because of its main component: glass, silicon dioxide is a very common and important molecule in the construction industry. One of the forms of silicon dioxide is quartz, which is found in sand.
The Structure of Silica
SiO2 is a 3 dimensional structure and comes from the tetrahedral structure, SiO4. Each of the Silicon atoms are connected to each other with an oxygen atom, which creates a "diamond type network". All forms of SiO2 possess a 3 dimensional shape and has a diamond structure. The bonding angle of Si-O-Si, which is the building block of the SiO2 molecule, is 144 degrees. These are called polymorph and in order to be stable, 3 of these polymorph are suppose to exist. This stable unit creates a a temperature for each of the different forms of SiO2. Forms that have (alpha) are at low temperature, while forms with (beta) are at high temperature. The structure of the different forms of SiO2 is important because it gives each of the different forms of SiO2 different characteristics and functions. Commercially, SiO2 is very important in steel, electronic, and semiconductor industries because of its structure, SiO2 is able to undergo rapid temperature changes and still maintain its shape and structure.
Different Forms of Silica and Their Uses
There are many different forms of SiO2, which mainly derived quartz glass. One form that is derived from quartz is beta-cristobalite, which is found in high temperature. Some other forms are beta-quartz, alpha-quartz, beta-tridymite, alpha-tridymite, alpha=cristobalite, and many more. The alpha and beta stands for the temperature range. Alpha molecules have low temperature while beta molecules have high temperature.
6.11I: Structure - Layers ((CdI 2) and (CdCl 2))
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6.11J: Structure - Perovskite ((CaTiO 3))
Some other unique properties that sets perovskites apart and makes them ideal for technological applications include: it is the only crystal structure that is ferroelectric (spontaneous alignment of the electric dipoles caused by interactions between them) not because of an external magnetic field but due to its crystal structure, its ferro-, pyro-, and piezo-electric properties, and structural properties such as durability and chemical flexibility.
Contributors and Attributions
• Matthew Ibbotson: Undergraduate in Chemical Engineering, Senior | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/06%3A_Structures_and_Energetics_of_Metallic_and_Ionic_solids/6.11%3A_Ionic_Lattices/6.11F%3A_Structure_-_-Cristobalite_%28SiO%29.txt |
Born forces are one type of force that acts upon atoms in an ionic lattice. In simplest terms, because ions have some finite size, electron-electron and nucleus-nucleus interactions occur and give rise to repulsion forces and electrostatic potential, both called Born forces.
Introduction
Lattice energy is the energy released when gaseous cations and anions bond to form a solid ionic compound. With the Born-Lande equation one can calculate the lattice energy of a crystalline ionic compound. Born and Lande theorized around the turn of the century, that lattice energy of a crystalline ionic compound could be found by calculating terms of electrostatic potential and a repulsive potential.
$\Delta U = \dfrac{-LA|Z_+|\,|Z_-| e^2}{4\pi\epsilon_o r} \tag{1}$
with
• $L$ is the avagadro's constant (6.022 \times 10^{23}\) and
• '$A$ is the Madelung Constant
The first potential is the force of attraction. It is a negative value because it pulls the two atoms closer together, and the forming of a bond is energetically favorable. The negative value of Avogadro’s number (6.022x1023) times a Madelung Constant (varies) times the absolute value of the charge of the cation times the absolute value of the charge of the anion, times the charge of an electron (1.6022x10-19 C) all over four times pi times the emissivity of space times the ion radius.
$\Delta U = \dfrac{-LB}{r^n} \tag{2}$
with
• $B$ is the repulsion coefficient and
• '$N$ is the Born exponent
The second equation is the repulsive force. It found by multiplying Avogadro’s number (6.022x1023) by a repulsion coefficient, and dividing that by the ionic radius raised to the power of a Born exponent (some number between 5 and 12).
Implications of Born Forces
Attractive forces are affected by the charge of the ions and their radii. Ions with large charges (like Mg7+ or O2-) have greater attractive potential than those with smaller charges (like Na1+ or F1-). Smaller ions (like Li+ or Cl-) also have greater electrostatic potential than larger ions (like I- or Cs+). The Madelung constant is dependent on the crystal structure type. This value is found in tables online or in a text, but in general can be thought of as large with larger cation-cation distances and anion-anion distances. A structure with fluorite geometry has a relatively large A value, whereas rock salt crystals have a much lower A value.
Repulsive forces are mainly determined by the born exponent. The Born exponent is dictated by the electronic configuration of the noble gas in the row above it on the periodic table (a closed shell).
Max Born
Max Born was a German physicist and mathematician who was instrumental in the development of quantum mechanics. He also made contributions to solid-state physics and optics and supervised the work of a number of notable physicists in the 1920s and 30s.
Born won the 1954 Nobel Prize in Physics for his "fundamental research in Quantum Mechanics, especially in the statistical interpretation of the wave function".
Problems
Plug in and cancel SI units to the Born-Lande equation to find the units of lattice energy, electrostatic potential, and repulsion forces. All energies are in units of kJ/mol
1. Looking at the equations for Born forces and a periodic table what do you expect to have a higher lattice energy (most negative enthalpy): NaCl, LiF, or KCl?
2. Looking at the equations for Born forces and a periodic table what do you expect to have a higher lattice energy (most negative enthalpy): NaOH, Al2O3, or Mg(OH)2?
3. What lattice structure type Rutile or CsCl has the greater relative Madelung constant? (You need not look in a table if you know the shapes of these crystal geometries)
4. What are the oxidation states of Barium and Oxygen in the crystalline lattice BaO?
5. Calculate the lattice energy of Sodium Chloride using the Born-Lande equation (find all needed information from tables). Compare this calculated value to experimentally found value of -787kJ/mol.
Solutions
1. LiF
2. Al2O3
3. Rutile
4. Ba2+ and O2-
5. The calculated value is -756kJ/mol.
6.13D: The Born-Lande Equation
The Born-Landé equation is a concept originally formulated in 1918 by the scientists Born and Landé and is used to calculate the lattice energy (measure of the strength of bonds) of a compound. This expression takes into account both the Born interactions as well as the Coulomb attractions.
Introduction
Due to its high simplicity and ease, the Born-Landé equation is commonly used by chemists when solving for lattice energy. This equation proposed by Max Born and Alfred Landé states that lattice energy can be derived from ionic lattice based on electrostatic potential and the potential energy due to repulsion. To solve for the Born-Landé equation, you must have a basic understanding of lattice energy:
• Lattice energy decreases as you go down a group (as atomic radii goes up, lattice energy goes down).
• Going across the periodic table, atomic radii decreases, therefore lattice energy increases.
The Born-Landé equation was derived from these two following equations. the first is the electrostatic potential energy:
$\Delta U = - \dfrac{N_A M\left | Z^+ \right | \left | Z^- \right |e^2}{4\pi\epsilon_o r} \label{1}$
with
• $M_A$ is Avogadro's constant ($6.022 \times 10^{23}$)
• $M$ is the Madelung Constant (a constant that varies for different structures)
• $e$ is the charge of an electron ($1.6022 \times 10^{-19}$ C)
• $Z^+$ is the cation charge
• $Z^-$ is the anion charge
• $\epsilon_o$ is the permittivity of free space
The second equation is the repulsive interaction:
$\Delta U = \dfrac{N_A B}{r^n} \label{2}$
with
• $B$ is the repulsion coefficient and
• $n$ is the Born Exponent (typically ranges between 5-12) that is used to measure how much a solid compresses
These equations combine to form:
$\Delta U (0K) = \dfrac{N_A M\left | Z^+ \right | \left | Z^- \right |e^2}{4\pi\epsilon_or_o} \left ( 1- \dfrac{1}{n} \right) \label{3}$
with
• $r_0$ is the closest ion distance
Calculate Lattice Energy
Lattice energy, based on the equation from above, is dependent on multiple factors. We see that the charge of ions is proportional to the increase in lattice energy. In addition, as ions come into closer contact, lattice energy also increases.
Example $1$
Which compound has the greatest lattice energy?
• AlF3
• NACl
• LiF
• CaCl2
Solution
This question requires basic knowledge of lattice energy. Since F3 gives the compound a +3 positive charge and the Al gives the compound a -1 negative charge, the compound has large electrostatic attraction. The bigger the electrostatic attraction, the greater the lattice energy.
Example $2$
What is the lattice energy of NaCl? (Hint: you must look up the values for the constants for this compound)
Solution
-756 kJ/mol (again, this value is found in a table of constants)
Example $3$
Calculate the lattice energy of NaCl. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/06%3A_Structures_and_Energetics_of_Metallic_and_Ionic_solids/6.13%3A_Lattice_Energy_-_Estimates_from_an_Electrostatic_Model/6.13C%3A_Born_Forces.txt |
There are many factors to be considered such as covalent character and electron-electron interactions in ionic solids. But for simplicity, let us consider the ionic solids as a collection of positive and negative ions. In this simple view, appropriate number of cations and anions come together to form a solid. The positive ions experience both attraction and repulsion from ions of opposite charge and ions of the same charge. The Madelung constant is a property of the crystal structure and depends on the lattice parameters, anion-cation distances, and molecular volume of the crystal.
1D Crystal
Before considering a three-dimensional crystal lattice, we shall discuss the calculation of the energetics of a linear chain of ions of alternate signs (Figure $1$).
Let us select the positive sodium ion in the middle (at $x=0$) as a reference and let $r_0$ be the shortest distance between adjacent ions (the sum of ionic radii). The Coulomb energy of the other ions in this 1D lattice on this sodium atom can be decomposed by proximity (or "shells").
• Nearest Neighbors (first shell): This reference sodium ion has two negative chloride ions as its neighbors on either side at $\pm r_0$ so the Coulombic energy of these interactions is $\underbrace{ \dfrac{-e^2}{4 \pi \epsilon_o r_o}}_{\text{left chloride ion}} + \underbrace{ \dfrac{-e^2}{4 \pi \epsilon_o r_o}}_{\text{right chloride ion}} = - \dfrac{2e^2}{4 \pi \epsilon_o r_o} \label{eq1}$
• Next Nearest Neighbors (second shell): Similarly the repulsive energy due to the next two positive sodium ions at a distance of $2r_0$ is $\underbrace{ \dfrac{+e^2}{4 \pi \epsilon_o (2r_o)}}_{\text{left sodium ion}} + \underbrace{ \dfrac{+e^2}{4 \pi \epsilon_o (2r_o)}}_{\text{right sodium ion}} = + \dfrac{2e^2}{4 \pi \epsilon_o (2r_o)} \label{eq2}$
• Next Next Nearest Neighbors (third shell): The attractive Coulomb energy due to the next two chloride ions neighbors at a distance $3r_0$ is $\underbrace{ \dfrac{-e^2}{4 \pi \epsilon_o (3r_o)}}_{\text{left chloride ion}} + \underbrace{ \dfrac{-e^2}{4 \pi \epsilon_o (3r_o)}}_{\text{right chloride ion}} = - \dfrac{2e^2}{4 \pi \epsilon_o (3r_o)} \label{eq3}$
and so on. Thus the total energy due to all the ions in the linear array is
$E = - \dfrac{2e^2}{4 \pi \epsilon_o r_o} + \dfrac{2e^2}{4 \pi \epsilon_o (2r_o)} - \dfrac{2e^2}{4 \pi \epsilon_o (3r_o)} - \ldots$
or
$E= \dfrac{e^2}{4 \pi \epsilon_o r_o} \left[ 2 \left (1 -\dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \ldots \right) \right] \label{eq6}$
We can use the following Maclaurin expansion
$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}- \frac{x^3}{4} + \cdots$
to simplify the sum in the parenthesis of Equation \ref{eq6} as $\ln (1+ 1)$ to obtain
\begin{align} E &= \dfrac{e^2}{4 \pi \epsilon_o r_o} \left[ 2 \ln 2 \right] \label{eq7} \[4pt] &= \dfrac{e^2}{4 \pi \epsilon_o } M \end{align}
The first factor of Equation \ref{eq7} is the Coulomb energy for a single pair of sodium and chloride ions, while the $2 \ln 2$ factor is the Madelung constant ($M \approx 1.38$) per molecule. The Madelung constant is named after Erwin Medelung and is a geometrical factor that depends on the arrangement of ions in the solid. If the lattice were different (when considering 2D or 3D crystals), then this constant would naturally differ.
3D Crystal
In three dimensions the series does present greater difficulty and it is not possible to sum the series conveniently as in the case of one-dimensional lattice. As an example, let us consider the the $\ce{NaCl}$ crystal. In the following discussion, assume $r$ be the distance between $\ce{Na^{+}}$ and $\ce{Cl^-}$ ions. The nearest neighbors of $\ce{Na^{+}}$ are six $\ce{Cl^-}$ ions at a distance 1r, 12 $\ce{Na^{+}}$ ions at a distance 2r, eight $\ce{Cl^-}$ ions at 3r, six $\ce{Na^{+}}$ ions at 4r, 24 $\ce{Na^{+}}$ ions at 5r, and so on. Thus, the electrostatic potential of a single ion in a crystal by approximating the ions by point charges of the surrounding ions:
$E_{ion-lattice} = \dfrac{Z^2e^2}{4\pi\epsilon_or} M \label{12.5.4}$
For NaCl is a poorly converging series of interaction energies:
$M= \dfrac{6}{1} - \dfrac{12}{2} + \dfrac{8}{3} - \dfrac{6}{4} + \dfrac{24}{5} ... \label{21.5.5}$
with
• $Z$ is the number of charges of the ions, (e.g., 1 for NaCl),
• $e$ is the charge of an electron ($1.6022 \times 10^{-19}\; C$),
• $4\pi \epsilon_o$ is 1.11265x10-10 C2/(J m).
The Madelung constant depends on the structure type and Equation $\ref{21.5.5}$ is applicable only for the sodium chloride (ei.g, rock salt) lattice geometry. Other values for other structural types are given in Table $2$. $A$ is the number of anions coordinated to cation and $C$ is the numbers of cations coordinated to anion.
Table $2$: Madelung Constants
Compound
Crystal Lattice
M
A : C Type
NaCl NaCl 1.74756 6 : 6 Rock salt
CsCl CsCl 1.76267 6 : 6 CsCl type
CaF2 Cubic 2.51939 8 : 4 Fluorite
CdCl2 Hexagonal 2.244
MgF2 Tetragonal 2.381
ZnS (wurtzite) Hexagonal 1.64132
TiO2 (rutile) Tetragonal 2.408 6 : 3 Rutile
bSiO2 Hexagonal 2.2197
Al2O3 Rhombohedral 4.1719 6 : 4 Corundum
A is the number of anions coordinated to cation and C is the numbers of cations coordinated to anion.
There are other factors to consider for the evaluation of lattice energy and the treatment by Max Born and Alfred Lande led to the formula for the evaluation of lattice energy for a mole of crystalline solid. The Born–Landé equation (Equation $\ref{21.5.6}$) is a means of calculating the lattice energy of a crystalline ionic compound and derived from the electrostatic potential of the ionic lattice and a repulsive potential energy term
$U= \dfrac{N_A M Z^2e^2}{4\pi \epsilon_o r} \left( 1 - \dfrac{1}{n} \right) \label{21.5.6}$
where
• $N_A$ is Avogadro constant;
• $M$ is the Madelung constant for the lattice
• $z^+$ is the charge number of cation
• $z^−$ is the charge number of anion
• $e$ is elementary charge, 1.6022×10−19 C
• $ε_0$ is the permittivity of free space
• $r_0$ is the distance to closest ion
• $n$ is the Born exponent that is typically between 5 and 12 and is determined experimentally. $n$ is a number related to the electronic configurations of the ions involved (Table $3$).
Table $1$: $n$ values for select solids
Atom/Molecule n
He 5
Ne 7
Ar 9
Kr 10
Xe 12
LiF 5.9
LiCl 8.0
LiBr 8.7
NaCl 9.1
NaBr 9.5
Example $1$: $\ce{NaCl}$
Estimate the lattice energy for $\ce{NaCl}$.
Solution
Using the values giving in the discussion above, the estimation is given by
\begin{align*} U_{NaCl} &= \dfrac{(6.022 \times 10^{23} /mol) (1.74756 ) (1.6022 \times 10 ^{-19})^2 (1.747558)}{ 4\pi \, (8.854 \times 10^{-12} C^2/m ) (282 \times 10^{-12}\; m)} \left( 1 - \dfrac{1}{9.1} \right) \nonumber \[4pt] &= - 756 \,kJ/mol\nonumber \end{align*} \nonumber
Much more should be considered in order to evaluate the lattice energy accurately, but the above calculation leads you to a good start. When methods to evaluate the energy of crystallization or lattice energy lead to reliable values, these values can be used in the Born-Hable cycle to evaluate other chemical properties, for example the electron affinity, which is really difficult to determine directly by experiment. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/06%3A_Structures_and_Energetics_of_Metallic_and_Ionic_solids/6.13%3A_Lattice_Energy_-_Estimates_from_an_Electrostatic_Model/6.13E%3A_Madelung_Constants.txt |
Ionic solids tend to be very stable compounds. The enthalpies of formation of the ionic molecules cannot alone account for this stability. These compounds have an additional stability due to the lattice energy of the solid structure. However, lattice energy cannot be directly measured. The Born-Haber cycle allows us to understand and determine the lattice energies of ionic solids.
Introduction
This module will introduce the idea of lattice energy, as well as one process that allows us to calculate it: the Born-Haber Cycle. In order to use the Born-Haber Cycle, there are several concepts that we must understand first.
Lattice Energy
Lattice Energy is a type of potential energy that may be defined in two ways. In one definition, the lattice energy is the energy required to break apart an ionic solid and convert its component atoms into gaseous ions. This definition causes the value for the lattice energy to always be positive, since this will always be an endothermic reaction. The other definition says that lattice energy is the reverse process, meaning it is the energy released when gaseous ions bind to form an ionic solid. As implied in the definition, this process will always be exothermic, and thus the value for lattice energy will be negative. Its values are usually expressed with the units kJ/mol.
Lattice Energy is used to explain the stability of ionic solids. Some might expect such an ordered structure to be less stable because the entropy of the system would be low. However, the crystalline structure allows each ion to interact with multiple oppositely charge ions, which causes a highly favorable change in the enthalpy of the system. A lot of energy is released as the oppositely charged ions interact. It is this that causes ionic solids to have such high melting and boiling points. Some require such high temperatures that they decompose before they can reach a melting and/or boiling point.
Born-Haber Cycle
There are several important concept to understand before the Born-Haber Cycle can be applied to determine the lattice energy of an ionic solid; ionization energy, electron affinity, dissociation energy, sublimation energy, heat of formation, and Hess's Law.
• Ionization Energy is the energy required to remove an electron from a neutral atom or an ion. This process always requires an input of energy, and thus will always have a positive value. In general, ionization energy increases across the periodic table from left to right, and decreases from top to bottom. There are some excepts, usually due to the stability of half-filled and completely filled orbitals.
• Electron Affinity is the energy released when an electron is added to a neutral atom or an ion. Usually, energy released would have a negative value, but due to the definition of electron affinity, it is written as a positive value in most tables. Therefore, when used in calculating the lattice energy, we must remember to subtract the electron affinity, not add it. In general, electron affinity increases from left to right across the periodic table and decreases from top to bottom.
• Dissociation energy is the energy required to break apart a compound. The dissociation of a compound is always an endothermic process, meaning it will always require an input of energy. Therefore, the change in energy is always positive. The magnitude of the dissociation energy depends on the electronegativity of the atoms involved.
• Sublimation energy is the energy required to cause a change of phase from solid to gas, bypassing the liquid phase. This is an input of energy, and thus has a positive value. It may also be referred to as the energy of atomization.
• The heat of formation is the change in energy when forming a compound from its elements. This may be positive or negative, depending on the atoms involved and how they interact.
• Hess's Law states that the overall change in energy of a process can be determined by breaking the process down into steps, then adding the changes in energy of each step. The Born-Haber Cycle is essentially Hess's Law applied to an ionic solid.
Using the Born-Haber Cycle
The values used in the Born-Haber Cycle are all predetermined changes in enthalpy for the processes described in the section above. Hess' Law allows us to add or subtract these values, which allows us to determine the lattice energy.
Step 1
Determine the energy of the metal and nonmetal in their elemental forms. (Elements in their natural state have an energy level of zero.) Subtract from this the heat of formation of the ionic solid that would be formed from combining these elements in the appropriate ration. This is the energy of the ionic solid, and will be used at the end of the process to determine the lattice energy.
Step 2
The Born-Haber Cycle requires that the elements involved in the reaction are in their gaseous forms. Add the changes in enthalpy to turn one of the elements into its gaseous state, and then do the same for the other element.
Step 3
Metals exist in nature as single atoms and thus no dissociation energy needs to be added for this element. However, many nonmetals will exist as polyatomic species. For example, Cl exists as Cl2 in its elemental state. The energy required to change Cl2 into 2Cl atoms must be added to the value obtained in Step 2.
Step 4
Both the metal and nonmetal now need to be changed into their ionic forms, as they would exist in the ionic solid. To do this, the ionization energy of the metal will be added to the value from Step 3. Next, the electron affinity of the nonmetal will be subtracted from the previous value. It is subtracted because it is a release of energy associated with the addition of an electron.
*This is a common error due to confusion caused by the definition of electron affinity, so be careful when doing this calculation.
Step 5
Now the metal and nonmetal will be combined to form the ionic solid. This will cause a release of energy, which is called the lattice energy. The value for the lattice energy is the difference between the value from Step 1 and the value from Step 4.
--------------------------------------------------------------------------------------------------------------------------------------------
The diagram below is another representation of the Born-Haber Cycle.
Equation
The Born-Haber Cycle can be reduced to a single equation:
Heat of formation= Heat of atomization+ Dissociation energy+ (sum of Ionization energies)+ (sum of Electron affinities)+ Lattice energy
*Note: In this general equation, the electron affinity is added. However, when plugging in a value, determine whether energy is released (exothermic reaction) or absorbed (endothermic reaction) for each electron affinity. If energy is released, put a negative sign in front of the value; if energy is absorbed, the value should be positive.
Rearrangement to solve for lattice energy gives the equation:
Lattice energy= Heat of formation- Heat of atomization- Dissociation energy- (sum of Ionization energies)- (sum of Electron Affinities)
Problems
1. Define lattice energy, ionization energy, and electron affinity.
2. What is Hess' Law?
3. Find the lattice energy of KF(s).
Note: Values can be found in standard tables.
4. Find the lattice energy of MgCl2(s).
5. Which one of the following has the greatest lattice energy?
1. A) MgO
2. B) NaC
3. C) LiCl
4. D) MgCl2
6. Which one of the following has the greatest Lattice Energy?
1. NaCl
2. CaCl2
3. AlCl3
4. KCl
Solutions
1. Lattice energy: The difference in energy between the expected experimental value for the energy of the ionic solid and the actual value observed. More specifically, this is the energy gap between the energy of the separate gaseous ions and the energy of the ionic solid.
Ionization energy: The energy change associated with the removal of an electron from a neutral atom or ion.
Electron affinity: The release of energy associated with the addition of an electron to a neutral atom or ion.
2. Hess' Law states that the overall energy of a reaction may be determined by breaking down the process into several steps, then adding together the changes in energy of each step.
3. Lattice Energy= [-436.68-89-(0.5*158)-418.8-(-328)] kJ/mol= -695.48 kJ/mol
4. Lattice Energy= [-641.8-146-243-(737.7+1450.6)-(2*-349)] kJ/mol= -2521.1 kJ/mol
5. MgO. It has ions with the largest charge.
6. AlCl3. According to the periodic trends, as the radius of the ion increases, lattice energy decreases. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/06%3A_Structures_and_Energetics_of_Metallic_and_Ionic_solids/6.14%3A_Lattice_Energy_-_The_Born-Haber_Cycle.txt |
Crystal Defects and Solids
Problems
All diagrams must be hand drawn, no photocopies or computer images will be accepted.
1. Nonstoichiometric compounds
2. Stoichiometric titanium (II) oxide and magnesium oxides have densities of 4.93 and 3.58 g cm-3 respectively. X-ray diffraction studies have shown that both have the NaCl structure, with unit cell edges of 416.2 and 421.2 pm respectively. Assume that the presence of vacancies does not affect the cell edge lengths. Estimate the percentage of vacancies in these solids.
3. Explain the Pilling-Bedworth principle regarding the protective behaviour of oxide films of metal (Swaddle: p103-104). From the following densities (d), evaluate the molar volumes (Vo = molar mass/d) of the metals and of the oxides. Suggest which oxides forms protective layers?
```Metal d(metal) Oxide d(oxide) Vo(metal) Vo(oxide) Protective?
----- -------- ----- -------- --------- --------- ----------
Be 1.85 BeO 3.01
Ca 1.54 CaO 3.25-3.38
Al 2.70 Al2O3 3.97
Fe 7.86 FeO 5.7
Fe2O3 5.24
Fe3O4 5.18
Ti 4.50 TiO2 4.26
------------------
FeO, wuestite; Fe2O3, hematite; Fe3O4, magnetite;
TiO2, rutile.
```
What advantages and disadvantages do Al and Ti have in their application as material for tableware? This is an open ended question, and give a common-sense type discussion.
4. y = 1.6 nm after 1 day
y = 3.5 nm after 1 year Estimate c and k. What are the thickness of the oxide layer at t = 5, 10, and 20 years?
What effect does moisture have on the oxidation of iron?
6.17A: Schottky Defect
Lattice structures are not perfect; in fact most of the time they experience defects. Lattice structures (or crystals) are prone to defects especially when their temperature is greater than 0 K [1]. One of these defects is known as the Schottky defect, which occurs when oppositely charged ions vacant their sites [1].
Introduction
Like the human body, lattice structures (most commonly known as crystals) are far from perfection. Our body works hard to keep things proportional but occasionally our right foot is bigger than our left; similarly, crystals may try to arrange it's ions under a strict layout, but occasionally an ion slips to another spot or simply goes missing. Realistically speaking, it should be expected that crystals will depart itself from order (not surprising considering defects occurs at temperature greater than 0 K). There are many ways a crystal can depart itself from order (thus experiences defects); these defects can be grouped in different categories such as Point Defects, Line Defects, Planar Defects, or Volume or Bulk Defects [2]. We will focus on Point Defects, specifically the defect that occurs in ionic crystal structures (i.e. NaCl) called the Schottky Defect.
Point Defects
Lattice structures (or crystals) undergoing point defects experience one of two types:
1. atoms or ions leaving their spot (thus creating vacancies).
2. atoms or ions slipping into the little gaps in between other atoms or ions; those little gaps are known as interstitials--since atoms or ions in the crystals are occupying interstitials, they inherently become (create) interstitials.
By the simplest definition, the Schottky defect is defined by type one, while type two defects are known as the Frenkel defect. The Schottky defect is often visually demonstrated using the following layout of anions and cations:
+ - + - + - + - + - +
- + - + - + - + (vacant) + -
+ - + - + - + - + - +
- + - (vacant) - + - + - + -
+ - + - + - + - + - +
- + - + - + - + - + -
Figure $1$: The positive symbols represents cations (i.e. Na+) and the negative symbol represents anions (i.e. Cl-).
In addition, this layout is applicable only for ionic crystal compounds of the formula MX--layout for ionic crystals with formula MX2 and M2X3 will be discussed later--where M is metal and X is nonmetal. Notice the figure has exactly one cation and one anion vacating their sites; that is what defines a (one) Schottky Defect for a crystal of MX formula--for every cation that vacant its site, the same number of anion will follow suit; essentially the vacant sites come in pairs. This also means the crystal will neither be too positive or too negative because the crystal will always be in equilibrium in respect to the number of anions and cations.
It is possible to approximate the number of Schottky defects (ns) in a MX ionic crystal compound by using the equation:
$N= \exp^{-\dfrac{\Delta H}{2RT}} \label{3}$
where
• $\Delta{H}$ is the enthalpy of defect formation,
• $R$ is the gas constant,
• $T$ is the absolute temperature (in K), and
N can be calculated by:
$N = \dfrac{\text{density of the ionic crystal compound} \times N_A}{\text{molar mass of the ionic crystal compound}} \label{4}$
From Equation $\ref{3}$, it is also possible to calculate the fraction of vacant sites by using the equation:
$\dfrac{n_s}{N} = \exp^{-\dfrac{\Delta H}{2RT}} \label{5}$
Schottky defects for $MX_2$ and $M_2X_3$
As mentioned earlier, a Schottky defect will always result a crystal structure in equilibrium--where no crystal is going to be too positive or too negative; thus in the case of:
• MX2: one Schottky defect equals one cation and two anion vacancy.
• M2X3: one Schottky defect equals two cation and three anion vacancy.
Problems
1. How does an ionic crystal structure maintain electrical neutrality despite undergoing a Schottky defect?
2. How is a Schottky defect defined for a compound with a MX formula? MX2? M2X3?
3. Given that the enthalpy of defect formation for LiCl is 3.39 x 10-19 J and the density of LiCl is 2.068 g/cm-3. Calculate the number of Schottky defect at 873 K.
4. Using the number of Schottky defect solved for question 3, determine the fraction of vacant site for LiCl.
5. If a anion and a cation vacant its site and occupies a space between other anions and cations, is it still a Schottky defect?
Answers
1. For a MX compound: one anion and one cation vacant their sites, so the overall charge will remain balanced. This is the same for MX2 and M2X3 because appropriate numbers of anions and cations vacant their site thus leaving the overall charge neutral.
2. MX compound: one Schottky defect is when one anion and one cation leave their sites. MX2 compound: one Schottky defect is when one anion and two cations leave their sites. M2X3 is when two anions and three cations leave their sites.
6.17B: Frenkel Defect
The Frenkel defect (also known as the Frenkel pair/disorder) is a defect in the lattice crystal where an atom or ion occupies a normally vacant site other than its own. As a result the atom or ion leaves its own lattice site vacant.
The Frenkel Defect in a Molecule
The Frenkel Defect explains a defect in the molecule where an atom or ion (normally the cation) leaves its own lattice site vacant and instead occupies a normally vacant site (Figure $1$). The cation leaves its own lattice site open and places itself between the area of all the other cations and anions. This defect is only possible if the cations are smaller in size when compared to the anions.
The number of Frenkel Defects can be calculated using the equation:
$\sqrt{NN^*}\, e^{\frac{\Delta H}{2RT}} \nonumber$
where $N$ is the number of normally occupied positions, $N^*$ is the number of available positions for the moving ion, the $\Delta H$ of formation is the enthalpy formation of one Frenkel defect, and $R$ is the gas constant. Frenkel defects are intrinsic defects because the existence causes the Gibbs energy of a crystal to decrease, which means it is favorable to occur.
Solids Found with a Frenkel Defect
The crystal lattices are relatively open and the coordination number is low.
Problems
1. What requirements are needed in order for the Frenkel defect to occur in an atom?
2. What are the differences between the Schottky defect and the Frenkel defect?
Answers
1. A low coordination number as well as having the crystal lattices open for the molecule.
2. The Frenkel defect causes an cation to leave its own lattice and go to another, while Sckhotty defect depicts that an equal number of cations and anions must be absent to maintain charge neutraility.
Contributors and Attributions
• Stanley Hsia, UC Davis | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/06%3A_Structures_and_Energetics_of_Metallic_and_Ionic_solids/6.17%3A_Defects_in_Solid_State_Lattices/6.17.E%3A_Defects_in_Solid_State_Lattices_%28Exercises%29.txt |
Acids
An acid (from the Greek oxein then Latin acidus/acére meaning sour) is a chemical substance whose aqueous solutions were characterized by a sour taste, the ability to turn blue litmus red, and the ability to react with bases and certain metals (like calcium) to form salts. Aqueous solutions of acids have a pH smaller than 7. The lower the pH, the higher the acidity and thus the higher the concentration of hydrogen ions in the solution (using the Arrhenius or Brønsted-Lowry definition).
Some notes on acids-bases, pH and the use of logarithms in calculations are available.
There are a number of common definitions for acids, for example, the Arrhenius, Brønsted-Lowry, and the Lewis definition. The Arrhenius definition defines acids as substances which increase the concentration of hydrogen ions (H+), when dissolved in water. The Brønsted-Lowry definition is an expansion of this and defines an acid as a substance which can act as an H+ donor. By this definition, any compound which can be easily deprotonated can be considered an acid. Examples include alcohols and amines which contain O-H or N-H fragments. A Lewis acid is a substance which can accept a pair of electrons to form a covalent bond. Examples of Lewis acids include all metal cations, and electron-deficient molecules such as boron trifluoride and aluminium trichloride.
Common examples of acids include hydrochloric acid (a solution of hydrogen chloride gas in water, this is the acid found in the stomach that activates digestive enzymes), acetic acid (vinegar is a dilute solution, generally under 5%), sulfuric acid (used in wet-cell car batteries), and tartaric acid (a solid used in baking). As these examples show, acids can be solutions or pure substances, and can be derived from solids, liquids, or gases.
HCl(aq) + NaOH(aq) ⇄ NaCl + H2O
HOAc(aq) + NaOH(aq) ⇄ NaOAc + H2O
H2SO4(aq) + 2NaOH(aq) ⇄ Na2SO4 + 2H2O
HO2CCH(OH)CH(OH)CO2H(aq) + 2NaOH(aq) ⇄ Na2Tartrate + 2H2O
Bases
The "modern" concept of a base in chemistry, stems from Guillaume-François Rouelle who in 1754 suggested that a base was a substance which reacted with acids "by giving it a concrete base or solid form" (as a salt). In addition they gave aqueous solutions which were characterized as slippery to the touch, tasted bitter, changed the colour of indicators (e.g., turned red litmus paper blue), and promoted certain chemical reactions (base catalysis). Examples of bases are the hydroxides of the alkali and alkaline earth metals (NaOH, Ca(OH)2, etc.).
For a substance to be classified as an Arrhenius base, it must produce hydroxide ions in solution. In order to do so, Arrhenius believed the base must contain hydroxide in the formula. This made the Arrhenius model limited, as it did not readily explain the basic properties of aqueous solutions of ammonia (NH3.aq, often written as NH4OH to better fit the Arrhenius model) or its organic derivatives (amines). In the more general Brønsted-Lowry acid-base theory, a base is a substance that can accept hydrogen ions (H+). In the Lewis model, a base is an electron pair donor.
7.01: Introduction
There are three primary theories of acid-base chemistry that are often taught together: Arrhenius theory, Brønsted-Lowry theory, and Lewis acid-base theory. Each theory is introduced below.
Arrhenius Definition of Acids and Bases
The Swedish chemist Svante Arrhenius attributed the properties of acidity to hydrogen ions ($\ce{H^{+}}$) in 1884. An Arrhenius acid is a substance that, when added to water, increases the concentration of $\ce{H^{+}}$ ions in the water. Note that chemists often write $\ce{H^{+}(aq)}$ and refer to the hydrogen ion when describing acid-base reactions, but the free hydrogen nucleus does not exist alone in water. It exists in a hydrated form which for simplicity is often written as the hydronium (hydroxonium) ion, $\ce{H3O^{+}}$. Thus, an Arrhenius acid can also be described as a substance that increases the concentration of hydronium ions when added to water. This definition stems from the equilibrium dissociation (self-ionization) of water into hydronium and hydroxide ($\ce{OH^{-}}$) ions:
$\ce{H_2O(l) + H_2O(l) ⇌ H_3O^{+}(aq) + OH^{-}(aq)} \nonumber$
with $K_w$ defined as $\ce{[H^{+}][OH^{-}]}$.
The value of $K_w$ varies with temperature, as shown in the table below where at 25 °C $K_w$ is approximately $1.0 \times 10^{-14}$, i.e. $pK_w= 14$.
Water temperature Kw / 10-14 pKw
0 °C 0.112 14.95
25 °C 1.023 13.99
50 °C 5.495 13.26
75 °C 19.95 12.70
100 °C 56.23 12.25
In pure water the majority of molecules are $\ce{H2O}$, but the molecules are constantly dissociating and re-associating, and at any time a small number of the molecules (about 1 in 107) are hydronium and an equal number are hydroxide. Because the numbers are equal, pure water is neutral (not acidic or basic) and has an electrical conductivity of 5.5 microSiemen, μS/m. For comparison, sea water's conductivity is about one million times higher, 5 S/m (due to the dissolved salt).
Proton vs. Hydron
Although the term proton is often used for $\ce{H^{+}}$, this should really be reserved for $\ce{H}$ (protium) not $\ce{D}$ (deuterium) or $\ce{T}$ (tritium). The more general term, hydron covers all isotopes of hydrogen.
An Arrhenius base, on the other hand, is a substance which increases the concentration of hydroxide ions when dissolved in water, hence decreasing the concentration of hydronium ions.
Upshot
To qualify as an Arrhenius acid, upon the introduction to water, the chemical must either cause, directly or otherwise:
• an increase in the aqueous hydronium concentration, or
• a decrease in the aqueous hydroxide concentration.
Conversely, to qualify as an Arrhenius base, upon the introduction to water, the chemical must either cause, directly or otherwise:
• a decrease in the aqueous hydronium concentration, or
• an increase in the aqueous hydroxide concentration.
The definition is expressed in terms of an equilibrium expression:
$\text{acid} + \text{base} ⇌ \text{conjugate base} + \text{conjugate acid}. \nonumber$
With an acid, $\ce{HA}$, the equation can be written symbolically as:
$\ce{HA + B ⇌ A^{-} + HB^{+}} \nonumber$
The double harpoons sign, $\ce{<=>}$, is used because the reaction can occur in both forward and backward directions. The acid, $\ce{HA}$, can lose a hydron to become its conjugate base, $\ce{A^{-}}$. The base, $\ce{B}$, can accept a hydron to become its conjugate acid, $\ce{HB^{+}}$. Most acid-base reactions are fast so that the components of the reaction are usually in dynamic equilibrium with each other.
Brønsted-Lowry Definition of Acids and Bases
While the Arrhenius concept is useful for describing many reactions, it has limitations. In 1923, chemists Johannes Nicolaus Brønsted and Thomas Martin Lowry independently recognized that acid-base reactions involve the transfer of a hydron. A Brønsted-Lowry acid (or simply Brønsted acid) is a species that donates a hydron to a Brønsted-Lowry base. The Brønsted-Lowry acid-base theory has several advantages over the Arrhenius theory. Consider the following reactions of acetic acid ($\ce{CH3COOH}$):
$\ce{CH_3COOH + H_2O ⇌ CH_3COO^{-} + H_3O^{+}} \nonumber$
$\ce{CH_3COOH + NH_3 ⇌ CH_3COO^{-} + NH_4^{+}} \nonumber$
Both theories easily describe the first reaction: $\ce{CH3COOH}$ acts as an Arrhenius acid because it acts as a source of $\ce{H3O^{+}}$ when dissolved in water, and it acts as a Brønsted acid by donating a hydron to water. In the second example $\ce{CH3COOH}$ undergoes the same transformation, in this case donating a hydron to ammonia ($\ce{NH3}$), but it cannot be described using the Arrhenius definition of an acid because the reaction does not produce hydronium ions.
Upshot
To qualify as an Brønsted-Lowry acid, the chemical must either cause, directly or otherwise:
• donate a proton.
Conversely, to qualify as an Brønsted-Lowry base, the chemical must either cause, directly or otherwise:
• accept a proton.
Lewis Definition of Acids and Bases
A third concept was proposed in 1923 by Gilbert N. Lewis which includes reactions with acid-base characteristics that do not involve a hydron transfer. A Lewis acid is a species that reacts with a Lewis base to form a Lewis adduct. The Lewis acid accepts a pair of electrons from another species; in other words, it is an electron pair acceptor. Brønsted acid-base reactions involve hydron transfer reactions while Lewis acid-base reactions involve electron pair transfers. All Brønsted acids are Lewis acids, but not all Lewis acids are Brønsted acids.
$\ce{BF3 + F^{-} <=> BF4^{-}} \nonumber$
$\ce{NH3 + H^{+} <=> NH4^{+}} \nonumber$
In the first example $\ce{BF3}$ is a Lewis acid since it accepts an electron pair from the fluoride ion. This reaction cannot be described in terms of the Brønsted theory because there is no hydron transfer. The second reaction can be described using either theory. A hydron is transferred from an unspecified Brønsted acid to ammonia, a Brønsted base; alternatively, ammonia acts as a Lewis base and transfers a lone pair of electrons to form a bond with a hydrogen ion.
Upshot
To qualify as an Lewis acid, the chemical must
• accept an electron pair
Conversely, to qualify as an Lewis base, the chemical must:
• donate an electron pair
Other Acid-Base Definitions
Lux-Flood acid-base definition
This acid-base theory was a revival of the oxygen theory of acids and bases, proposed by German chemist Hermann Lux in 1939 and further improved by Håkon Flood circa 1947. It is still used in modern geochemistry and for the electrochemistry of molten salts. This definition describes an acid as an oxide ion ($\ce{O^{2-}}$) acceptor and a base as an oxide ion donor. For example:
$\ce{MgO (base) + CO2 (acid) <=> MgCO3} \nonumber$
$\ce{CaO (base) + SiO2 (acid) <=> CaSiO3}\nonumber$
$\ce{NO3^{-} (base) + S2O7^{2-} (acid) <=> NO2+ + 2 SO4^{2-}}\nonumber$
Usanovich acid-base definition
Mikhail Usanovich developed a general theory that does not restrict acidity to hydrogen-containing compounds, and his approach, published in 1938, was even more general than the Lewis theory. Usanovich's theory can be summarized as defining an acid as anything that accepts negative species, anions or electrons or donates positive ones, cations, and a base as the reverse. This definition could even be applied to the concept of redox reactions (oxidation-reduction) as a special case of acid-base reactions. Some examples of Usanovich acid-base reactions include:
• species exchanged: anion $\ce{O^{2-}}$ $\ce{Na2O (base) + SO3 (acid) → 2Na^{+} + SO4^{2-}} \nonumber$
• species exchanged: anion $\ce{S^{2-}}$ $\ce{3(NH4)2S (base) + Sb2S5 (acid) → 6NH4+ + 2SbS4^{3-}} \nonumber$
• species exchanged: electron $\ce{2Na (base) + Cl2 (acid) → 2Na^{+} + 2Cl^{-} } \nonumber$
A comparison of the above definitions of Acids and Bases shows that the Usanovich concept encompasses all of the others but some feel that because of this it is too general to be useful. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/07%3A_Acids_bases_and_ions_in_aqueous_solution/7.01%3A_Introduction/7.1A%3A_Acid-Base_Theories_and_Concepts.txt |
Water covers 71% of the Earth's surface and is vital for all known forms of life. On Earth, 96.5% of the planet's water is found in seas and oceans, 1.7% in groundwater, 1.7% in glaciers and the ice caps of Antarctica and Greenland, a small fraction in other large water bodies, and 0.001% in the air as vapour, clouds (formed from solid and liquid water particles suspended in air), and precipitation.
Only 2.5% of the Earth's water is freshwater, and 98.8% of that water is in ice and groundwater. Less than 0.3% of all freshwater is in rivers, lakes, and the atmosphere, and an even smaller amount of the Earth's freshwater (0.003%) is contained within biological bodies and manufactured products.
The major chemical and physical properties of water are:
• Water is a liquid at standard temperature and pressure. It is tasteless and odourless. The intrinsic colour of water and ice is a very slight blue hue, although both appear colourless in small quantities. Water vapour is essentially invisible as a gas.
• Water is the only substance occurring naturally in all three phases as solid, liquid, and gas on the Earth's surface
• Water is transparent in the visible electromagnetic spectrum. Thus aquatic plants can live in water because sunlight can reach them. Infrared light is strongly absorbed by the hydrogen-oxygen or OH bonds.
• Since the water molecule is not linear and the oxygen atom has a higher electronegativity than hydrogen atoms, the oxygen atom carries a partial negative charge, whereas the hydrogen atoms have partial positive charges. As a result, water is a polar molecule with an electrical dipole moment.
• Water can form an unusually large number of intermolecular hydrogen bonds (four) for a molecule of its size. These factors lead to strong attractive forces between molecules of water, giving rise to water's high surface tension and capillary forces. The capillary action refers to the tendency of water to move up a narrow tube against the force of gravity. This property is relied upon by all vascular plants, such as trees.
• The boiling point of water (like all other liquids) is dependent on the barometric pressure. For example, on the top of Mount Everest water boils at 68 °C, compared to 100 °C at sea level at a similar latitude (since latitude modifies atmospheric pressure slightly). Conversely, water deep in the ocean near geothermal vents can reach temperatures of hundreds of degrees and remain liquid.
• Water has a high specific heat capacity, 4181.3 J kg-1 K-1, as well as a high heat of vaporization (40.65 kJ mol-1), both result from the extensive hydrogen bonding between its molecules. These two unusual properties allow water to moderate Earth's climate by buffering large fluctuations in temperature.
• Solid ice has a density of 917 kg m-3. The maximum density of liquid water occurs at 3.98 °C where it is 1000 kg m-3.
• Elements that are more electropositive than hydrogen such as lithium, sodium, calcium, potassium and caesium displace hydrogen from water, forming hydroxides. Since hydrogen is a flammable gas, when given off it is dangerous and the reaction of water with the more electropositive of these elements can be violently explosive so they are often stored in oil.
Most known pure substances display simple behaviour when they are cooled, they shrink. Liquids contract as they are cooled because the molecules move slower and they are less able to overcome the attractive intermolecular forces drawing them closer to each other. Once the freezing temperature is reached, the substances solidify, causing them to contract even more because crystalline solids are usually tightly packed.
Water however water has the anomalous property of becoming less dense when it is cooled to its solid form, ice.
When liquid water is cooled, it initially contracts as expected, until a temperature of 3.98 °C is reached (~4 °C). After that, it expands slightly until it reaches the freezing point, and then when it freezes, it expands by approximately 9%.
Just above the freezing point, the water molecules begin to locally arrange into ice-like structures with an extended hydrogen bonded network. This creates some "openness" in the liquid water, accounting for the decrease in its density. This is in opposition to the usual tendency for cooling to increase the density. At 3.98 °C these opposing tendencies cancel out, producing the density maximum.
Since water expands to occupy a 9% greater volume in the form of ice and is less dense, it floats on liquid water, as in icebergs. Fortunately this happens, since in colder climates where water is susceptible to freezing, if it all turned solid during the winter, it would kill all the life within it.
The extended structure of the water molecule in liquid and solid form seen in the models below provides the explanation for the variation of density with temperature.
7.02: Properties of Water
Solvated H+ ions
The hydron (a completely free or "naked" hydrogen atomic nucleus) is far too reactive to exist in isolation and readily hydrates in aqueous solution. The simplest hydrated form of the hydrogen cation, the hydronium (hydroxonium) ion H3O+ (aq), is a key object of Arrhenius' definition of acid. Other "simple" hydrated forms include the Zundel cation H5O2+ which is formed from a hydron and two water molecules, and the Eigen cation H9O4+, formed from a hydronium ion and three water molecules. The hydron itself is crucial in the more general Brønsted-Lowry acid-base theory, which extends the concept of acid-base chemistry beyond aqueous solutions. Both of these complexes represent ideal structures in a more general hydrogen bonded network defect. A freezing-point depression study determined that the mean hydration ion in cold water is on average approximately H3O+(H2O)6: where each hydronium ion is solvated by 6 water molecules. Some hydration structures are quite large: the H3O+.20H2O magic ion number structure (called magic because of its increased stability with respect to hydration structures involving a comparable number of water molecules).
Structure from Lancaster site
In 1806 Theodor Grotthuss proposed a theory of water conductivity. He envisioned the electrolytic reaction as a sort of "bucket line" where each oxygen atom simultaneously passes and receives a single hydrogen atom. It was an astonishing theory to propose at the time, since the water molecule was thought to be OH not H2O and the existence of ions was not fully understood. The theory became known as the Grotthuss mechanism. The transport mechanism is now thought to involve the inter-conversion between the Eigen and Zundel solvation structures, Eigen to Zundel to Eigen (E-Z-E).
7.8A: Amphoteric Behavior
Oxides and Hydroxides
An early classification of substances arose from the differences observed in their solubility in acidic and basic solutions. This led to the classification of oxides and hydroxides as being either acidic or basic. Acidic oxides or hydroxides either reacted with water to produce an acidic solution or were soluble in aqueous base. Basic oxides and hydroxides either reacted with water to produce a basic solution or readily dissolved in aqueous acids. The diagram below shows there is strong correlation between the acidic or basic character of oxides (ExOy) and the position of the element, E, in the periodic table.
Oxides of metallic elements are generally basic oxides, and oxides of nonmetallic elements acidic oxides. Take for example, the reactions with water of calcium oxide, a metallic oxide, and carbon dioxide, a nonmetallic oxide:
CaO(s) + H2O(l) → Ca(OH)2
CO2(g) + H2O(l) → H2CO3(aq) Calcium oxide reacts with water to produce a basic solution of calcium hydroxide, whereas carbon dioxide reacts with water to produce a solution of carbonic acid.
There is a gradual transition from basic oxides to acidic oxides from the lower left to the upper right in the periodic table.
Basicity of the oxides increase with increasing atomic number down a group:
BeO < MgO < CaO < SrO < BaO Note as well that acidity increases with increasing oxidation state of the element:
MnO < Mn2O3 < MnO2 < Mn2O7
in keeping with the increase in covalency. Oxides of intermediate character, called amphoteric oxides, are located along the diagonal line between the two extremes. Amphoteric species are molecules or ions that can react as an acid as well as a base. The word has Greek origins, amphoteroi (άμφότεροι) meaning "both". Many metals (such as copper, zinc, tin, lead, aluminium, and beryllium) form amphoteric oxides or hydroxides. Amphoterism depends on the oxidation state of the oxide.
For example, zinc oxide (ZnO) reacts with both acids and with bases: In acid: ZnO + 2H+ → Zn2+ + H2O
In base: ZnO + 2OH- + H2O→ [Zn(OH)4]2-
This reactivity can be used to separate different cations, such as zinc(II), which dissolves in base, from manganese(II), which does not dissolve in base.
Aluminium hydroxide is another amphoteric species:
As a base (neutralizing an acid): Al(OH)3 + 3HCl → AlCl3 + 3H2O
As an acid (neutralizing a base): Al(OH)3 + NaOH → Na[Al(OH)4] | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/07%3A_Acids_bases_and_ions_in_aqueous_solution/7.02%3A_Properties_of_Water/7.2B%3A_The_Self_Ionization_of_Water.txt |
Learning Objectives
• Recognize common ions from various salts, acids, and bases.
• Calculate concentrations involving common ions.
• Calculate ion concentrations involving chemical equilibrium.
The common-ion effect is used to describe the effect on an equilibrium when one or more species in the reaction is shared with another reaction. This results in a shifitng of the equilibrium properties.
Introduction
The solubility products Ksp's are equilibrium constants in hetergeneous equilibria (i.e., between two different phases). If several salts are present in a system, they all ionize in the solution. If the salts contain a common cation or anion, these salts contribute to the concentration of the common ion. Contributions from all salts must be included in the calculation of concentration of the common ion. For example, a solution containing sodium chloride and potassium chloride will have the following relationship:
$\mathrm{[Na^+] + [K^+] = [Cl^-]} \label{1}\nonumber$
Consideration of charge balance or mass balance or both leads to the same conclusion.
Common Ions
When $\ce{NaCl}$ and $\ce{KCl}$ are dissolved in the same solution, the $\mathrm{ {\color{Green} Cl^-}}$ ions are common to both salts. In a system containing $\ce{NaCl}$ and $\ce{KCl}$, the $\mathrm{ {\color{Green} Cl^-}}$ ions are common ions.
\begin{align*} \ce{NaCl &\rightleftharpoons Na^{+}} + \color{Green} \ce{Cl^{-}}\[4pt] \ce{KCl &\rightleftharpoons K^{+}} + \color{Green} \ce{Cl^{-}} \[4pt] \ce{CaCl_2 &\rightleftharpoons Ca^{2+}} + \color{Green} \ce{2 Cl^{-}}\[4pt] \ce{AlCl_3 &\rightleftharpoons Al^{3+}} + \color{Green} \ce{3 Cl^{-}}\[4pt] \ce{AgCl & \rightleftharpoons Ag^{+}} + \color{Green} \ce{Cl^{-}} \end{align*}
For example, when $\ce{AgCl}$ is dissolved into a solution already containing $\ce{NaCl}$ (actually $\ce{Na+}$ and $\ce{Cl-}$ ions), the $\ce{Cl-}$ ions come from the ionization of both $\ce{AgCl}$ and $\ce{NaCl}$. Thus, $\ce{[Cl- ]}$ differs from $\ce{[Ag+]}$. The following examples show how the concentration of the common ion is calculated.
Example $1$
What are $\ce{[Na+]}$, $\ce{[Cl- ]}$, $\ce{[Ca^2+]}$, and $\ce{[H+]}$ in a solution containing 0.10 M each of $\ce{NaCl}$, $\ce{CaCl2}$, and $\ce{HCl}$?
Solution
Due to the conservation of ions, we have
$\ce{[Na^{+}] = [Ca^{2+}] = [H^{+}] = 0.10\, \ce M}. \nonumber$
but
\begin{align*} \ce{[Cl^{-}]} &= 0.10 \, \ce{(due\: to\: NaCl)}\[4pt] &+ 0.20\, \ce{(due\: to\: CaCl_2)} \[4pt] &+ 0.10\, \ce{(due\: to\: HCl)} \[4pt] &= 0.40\, \ce{M} \end{align*}
Exercise $1$
John poured 10.0 mL of 0.10 M $\ce{NaCl}$, 10.0 mL of 0.10 M $\ce{KOH}$, and 5.0 mL of 0.20 M $\ce{HCl}$ solutions together and then he made the total volume to be 100.0 mL. What is $\ce{[Cl- ]}$ in the final solution?
$\mathrm{[Cl^-] = \dfrac{0.1\: M\times 10\: mL+0.2\: M\times 5.0\: mL}{100.0\: mL} = 0.020\: M}$
Le Chatelier's Principle states that if an equilibrium becomes unbalanced, the reaction will shift to restore the balance. If a common ion is added to a weak acid or weak base equilibrium, then the equilibrium will shift towards the reactants, in this case the weak acid or base.
Example $2$
Consider the lead(II) ion concentration in this saturated solution of $\ce{PbCl2}$. The balanced reaction is
$\ce{ PbCl2 (s) <=> Pb^{2+}(aq) + 2Cl^{-}(aq)} \label{Ex1.1}$
Defining $s$ as the concentration of dissolved lead(II) chloride, then:
$[Pb^{2+}] = s \nonumber$
$[Cl^- ] = 2s\nonumber$
These values can be substituted into the solubility product expression, which can be solved for $s$:
\begin{align*} K_{sp} &= [Pb^{2+}] [Cl^{-}]^2 \[4pt] &= s \times (2s)^2 \[4pt] 1.7 \times 10^{-5} &= 4s^3 \[4pt] s^3 &= \dfrac{1.7 \times 10^{-5}}{4} \[4pt] &= 4.25 \times 10^{-6} \[4pt] s &= \sqrt[3]{4.25 \times 10^{-6}} \[4pt] &= 1.62 \times 10^{-2}\ mol\ dm^{-3} \end{align*}
The concentration of lead(II) ions in the solution is 1.62 x 10-2 M. Consider what happens if sodium chloride is added to this saturated solution. Sodium chloride shares an ion with lead(II) chloride. The chloride ion is common to both of them; this is the origin of the term "common ion effect".
Look at the original equilibrium expression in Equation \ref{Ex1.1}. What happens to that equilibrium if extra chloride ions are added? According to Le Chatelier, the position of equilibrium will shift to counter the change, in this case, by removing the chloride ions by making extra solid lead(II) chloride.
Of course, the concentration of lead(II) ions in the solution is so small that only a tiny proportion of the extra chloride ions can be converted into solid lead(II) chloride. The lead(II) chloride becomes even less soluble, and the concentration of lead(II) ions in the solution decreases. This type of response occurs with any sparingly soluble substance: it is less soluble in a solution which contains any ion which it has in common. This is the common ion effect.
A Simple Example
If an attempt is made to dissolve some lead(II) chloride in some 0.100 M sodium chloride solution instead of in water, what is the equilibrium concentration of the lead(II) ions this time? As before, define s to be the concentration of the lead(II) ions.
$\ce{[Pb^{2+}]} = s \label{2}\nonumber$
The calculations are different from before. This time the concentration of the chloride ions is governed by the concentration of the sodium chloride solution. The number of ions coming from the lead(II) chloride is going to be tiny compared with the 0.100 M coming from the sodium chloride solution.
In calculations like this, it can be assumed that the concentration of the common ion is entirely due to the other solution. This simplifies the calculation.
So we assume:
$\ce{[Cl^{-} ]} = 0.100\; M \label{3}\nonumber$
The rest of the mathematics looks like this:
\begin{align*} K_{sp}& = [Pb^{2+}][Cl^-]^2 \[4pt] & = s \times (0.100)^2 \[4pt] 1.7 \times 10^{-5} & = s \times 0.00100 \end{align*}
therefore:
\begin{align*} s & = \dfrac{1.7 \times 10^{-5}}{0.0100} \[4pt] & = 1.7 \times 10^{-3} \, \text{M} \end{align*}
Finally, compare that value with the simple saturated solution:
Original solution:
$\ce{[Pb^{2+}]} = 0.0162 \, M \label{5}\nonumber$
Solution in 0.100 M $\ce{NaCl}$ solution:
$\ce{[Pb^{2+}]} = 0.0017 \, M \label{6}\nonumber$
The concentration of the lead(II) ions has decreased by a factor of about 10. If more concentrated solutions of sodium chloride are used, the solubility decreases further.
Adding a common ion to a system at equilibrium affects the equilibrium composition, but not the ionization constant.
Common Ion Effect with Weak Acids and Bases
Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.
Example $\PageIndex{3A}$
The common ion effect of $\ce{H3O^{+}}$ on the ionization of acetic acid
The common ion effect suppresses the ionization of a weak base by adding more of an ion that is a product of this equilibrium.
Example $\PageIndex{3B}$
Consider the common ion effect of $\ce{OH^{-}}$ on the ionization of ammonia
Adding the common ion of hydroxide shifts the reaction towards the left to decrease the stress (in accordance with Le Chatelier's Principle), forming more reactants. This decreases the reaction quotient, because the reaction is being pushed towards the left to reach equilibrium. The equilibrium constant, $K_b=1.8 \times 10^{-5}$, does not change. The reaction is put out of balance, or equilibrium.
$Q_a = \dfrac{[\ce{NH_4^{+}}][\ce{OH^{-}}]}{[\ce{NH_3}]} \nonumber$
At first, when more hydroxide is added, the quotient is greater than the equilibrium constant. The reaction then shifts right, causing the denominator to increase, decreasing the reaction quotient and pulling towards equilibrium and causing $Q$ to decrease towards $K$.
Common Ion Effect on Solubility
Adding a common ion decreases solubility, as the reaction shifts toward the left to relieve the stress of the excess product. Adding a common ion to a dissociation reaction causes the equilibrium to shift left, toward the reactants, causing precipitation.
Example $4$
Consider the reaction:
$\ce{ PbCl_2(s) <=> Pb^{2+}(aq) + 2Cl^{-}(aq)} \nonumber$
What happens to the solubility of $\ce{PbCl2(s)}$ when 0.1 M $\ce{NaCl}$ is added?
Solution
$K_{sp}=1.7 \times 10^{-5} \nonumber$
$Q_{sp}= 1.8 \times 10^{-5} \nonumber$
Identify the common ion: $\ce{Cl^{-}}$
Notice: $Q_{sp} > K_{sp}$ The addition of $\ce{NaCl}$ has caused the reaction to shift out of equilibrium because there are more dissociated ions. Typically, solving for the molarities requires the assumption that the solubility of $\ce{PbCl2(s)}$ is equivalent to the concentration of $\ce{Pb^{2+}}$ produced because they are in a 1:1 ratio.
Because $K_{sp}$ for the reaction is $1.7 \times 10^{-5}$, the overall reaction would be
$(s)(2s)^2= 1.7 \times 10^{-5}. \nonumber$
Solving the equation for $s$ gives $s= 1.62 \times 10^{-2}\, \text{M}$. The coefficient on $\ce{Cl^{-}}$ is 2, so it is assumed that twice as much $\ce{Cl^{-}}$ is produced as $\ce{Pb^{2+}}$, hence the '2s.' The solubility equilibrium constant can be used to solve for the molarities of the ions at equilibrium.
The molarity of Cl- added would be 0.1 M because $\ce{Na^{+}}$ and $\ce{Cl^{-}}$ are in a 1:1 ratio in the ionic salt, $\ce{NaCl}$. Therefore, the overall molarity of $\ce{Cl^{-}}$ would be $2s + 0.1$, with $2s$ referring to the contribution of the chloride ion from the dissociation of lead chloride.
\begin{align*} Q_{sp} &= [\ce{Pb^{2+}}][\ce{Cl^{-}}]^2 \[4pt] &= 1.8 \times 10^{-5} \[4pt] &= (s)(2s + 0.1)^2 \[4pt] s &= [Pb^{2+}] \[4pt] &= 1.8 \times 10^{-3} M \[4pt] 2s &= [\ce{Cl^{-}}] \[4pt] &\approx 0.1 M \end{align*}
Notice that the molarity of $\ce{Pb^{2+}}$ is lower when $\ce{NaCl}$ is added. The equilibrium constant remains the same because of the increased concentration of the chloride ion. To simplify the reaction, it can be assumed that $[\ce{Cl^{-}}]$ is approximately 0.1 M since the formation of the chloride ion from the dissociation of lead chloride is so small. The reaction quotient for $\ce{PbCl2(s)}$ is greater than the equilibrium constant because of the added $\ce{Cl^{-}}$. This therefore shift the reaction left towards equilibrium, causing precipitation and lowering the current solubility of the reaction. Overall, the solubility of the reaction decreases with the added sodium chloride.
7.13B: Hard and Soft M
Hard and Soft Acids and Bases, Pearson's HSAB
This theory proposes that soft acids react faster and form stronger bonds with soft bases, whereas hard acids react faster and form stronger bonds with hard bases, all other factors being equal. The classification in the original work was largely based on equilibrium constants for the reaction of two Lewis bases competing for a Lewis acid.
Hard acids and hard bases tend to have the following characteristics:
• small atomic/ionic radius
• high oxidation state
• low polarizability
• high electronegativity (bases)
Examples of hard acids are: H+, light alkali ions (Li through K are considered to have small ionic radii), Ti4+, Cr3+, Cr6+, BF3. Examples of hard bases are: OH-, F-, Cl-, NH3, CH3COO-, CO32-. The affinity of hard acids and hard bases for each other is mainly ionic in nature.
Soft acids and soft bases tend to have the following characteristics:
• large atomic/ionic radius
• low or zero oxidation state bonding
• high polarizability
• low electronegativity
Examples of soft acids are: CH3Hg+, Pt2+, Pd2+, Ag+, Au+, Hg2+, Hg22+, Cd2+, BH3. Examples of soft bases are: H-, R3P, SCN-, I-. The affinity of soft acids and bases for each other is mainly covalent in nature.
HSAB acids and bases This provides a qualitative approach to looking at the reactions of metal ions with various ligands since, from the diagram above, it is expected that whereas Al(III) and Ti(III) would prefer to react with O-species over S-species, the reverse would be predicted for Hg(II). | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/07%3A_Acids_bases_and_ions_in_aqueous_solution/7.10%3A_Common-Ion_Effect.txt |
The simplest possible chemical reaction involves the transfer of one or more electrons between two atoms. In the simple reaction:
\[\ce{2 Na + Cl2 -> 2 NaCl}\]
we assume that sodium transfers an electron to chlorine creating the sodium ion and the chloride ion. We can write the two steps of this reaction as:
\[\ce{Na -> Na+1 + e-1}\]
and
\[\ce{Cl2 + 2 e-1 > 2 Cl-1 }\]
The first of these steps involves the loss of an electron and is called an oxidation reaction, while the second step involves the gain of an electron and is called a reduction reaction. The simple nemonic "Leo goes Ger" or Loss of electrons - oxidation; Gain of electrons - reduction has been used by several generations of students to remember these definitions. (OK! This is Simba, my apologies to Lion King aficionados.) Oxidation and reduction reactions (usually called redox reactions) are important in biochemistry and important families of enzymes are called reductases or oxidases based on their reduction or oxidation catalysis. We recognize ethanol as common drinking alcohol, but most people don't realize that ethanol is generated by fermentation reactions that naturally take place in the intestines. The body always has a small amount of ethanol being transported across the intestine walls. As a result, the body has an enzyme to metabolize ethanol called ethanol oxidase. (You might want to think about this. The body didn't invent ethanol to compensate for the consumption of fermented beverages like beer and wine since there are relatively recent on an evolutionary time scale. Ethanol oxidase is present in the body because it had to be there to compensate for the background ethanol. I would suspect that all mammals have this enzyme, not just people. Other things that are ingested by people for their psychotropic effects do not necessarily have enzymes for their detoxification, thus they stay in the tissues longer and have a greater potential for doing damage. If you follow my dictum of not consuming anything for which you don't have an enzyme, you'll be relatively safe.) Redox chemistry also forms the basis of corrosion chemistry and battery chemistry (electrochemistry). Simple electrochemical reactions If we take a piece of zinc metal and put it into a beaker of HCl, we will quickly notice the formation of bubbles on the surface of the zinc. Were we to leave the zinc in the acid for several minutes, we'd easily observe that the zinc is being dissolved while a gas is being liberated. This process is illustrated below: The reaction taking place here is obviously:
\[Zn + 2 H+1 ------> Zn+2 + H2\]
In this reaction, the hydrogen ions are being reduced while the zinc is being oxidized. A similar reaction may be observed if we place a zinc bar into a solution containing copper sulfate. Here the reaction is:
\[Zn + Cu+2 -----> Cu + Zn+2\]
Again the zinc is being oxidized while the copper is being reduced. Unlike the case of zinc in acid, the reaction will only continue until the copper has formed a film on the surface of the zinc, at which point the reaction stops since zinc ions are no longer able to escape to the solution.
Rules for assigning oxidation numbers:
1. The oxidation number of a free element = 0.
2. The oxidation number of a monatomic ion = charge on the ion.
3. The oxidation number of hydrogen = + 1 and rarely - 1.
4. The oxidation number of oxygen = - 2 and in peroxides - 1.
5. The sum of the oxidation numbers in a polyatomic ion = charge on the ion. Elements in group 1, 2, and aluminum are always as indicated on the periodic table. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/08%3A_Reduction_and_Oxidation/8.1A%3A_Oxidation_and_Reduction.txt |
Rules for assigning oxidation numbers: 1. The oxidation number of a free element = 0. 2. The oxidation number of a monatomic ion = charge on the ion. 3. The oxidation number of hydrogen = + 1 and rarely - 1. 4. The oxidation number of oxygen = - 2 and in peroxides - 1. 5. The sum of the oxidation numbers in a polyatomic ion = charge on the ion. Elements in group 1, 2, and aluminum are always as indicated on the periodic table. K2CO3 The sum of all the oxidation numbers in this formula equal 0. Multiply the subscript by the oxidation number for each element. To calculate O.N. of C K = (2) ( + 1 ) = + 2 O = (3) ( - 2 ) = - 6 therefore, C = (1) ( + 4 ) = + 4 HSO4 - To calculate O.N. of S The sum of all the oxidation numbers in this formula equal -1. Multiply the subscript by the oxidation number for each element. H = (1) ( + 1 ) = + 1 O = (4) ( - 2 ) = - 8 therefore, S = (1) ( + 6 ) = + 6 Calculate O.N. in following compounds: 1. Sb in Sb2O5 2. N in Al(NO3)3 3. P in Mg3(PO4)2 4. S in (NH4)2SO4 5. Cr in CrO4 -2 6. Hg in Hg(ClO4)2 7. B in NaBO3 8. Si in MgSiF6 9. I in IO3 - 10. N in (NH4)2S 11. Mn in MnO4 - 12. Br in BrO3 - 13. Cl in ClO - 14. Cr in Cr2O7 -2 15. Se in H2SeO3 Reducing Agents and Oxidizing Agents Reducing agent - the reactant that gives up electrons. The reducing agent contains the element that is oxidized (looses electrons). If a substance gives up electrons easily, it is said to be a strong reducing agent. Oxidizing agent - the reactant that gains electrons. The oxidizing agent contains the element that is reduced (gains electrons). If a substance gains electrons easily, it is said to be a strong oxidizing agent. Example: Fe2O3 (cr) + 3CO(g) 2Fe(l) + 3CO2 (g) Notice that the oxidation number of C goes from +2 on the left to +4 on the right. The reducing agent is CO, because it contains C, which loses e - . Notice that the oxidation number of Fe goes from +3 on the left to 0 on the right. The oxidizing agent is Fe2O3, because it contains the Fe, which gains e - . Practice Problems: In any Redox equation, at least one particle will gain electrons and at least one particle will lose electrons. This is indicated by a change in the particle's oxidation number from one side of the equation to the other. For each reaction below, draw arrows and show electron numbers as in the example here. The top arrow indicates the element that gains electrons, reduction, and the bottom arrow indicates the element that looses electrons, oxidation. An arrow shows what one atom of each of these elements gaines or looses. 1. Mg + O2 MgO 2. Cl2 + I - Cl - + I2 3. MnO4 - + C2O4 -2 Mn+2 + CO2 4. Cr + NO2 - CrO2 - + N2O2 -2 5. BrO3 - + MnO2 Br - + MnO4 - 6. Fe+2 + MnO4 - Mn+2 + Fe+3 7. Cr + Sn+4 Cr+3 + Sn+2 8. NO3 - + S NO2 + H2SO4 9. IO4 - + II2 10. NO2 + ClO - NO3 - + Cl - Balancing Redox Equations by the Half-reaction Method 1. Decide what is reduced (oxidizing agent) and what is oxidized (reducing agent). Do this by drawing arrows as in the practice problems. 2. Write the reduction half-reaction. The top arrow in step #1 indicates the reduction half-reaction. Show the electrons gained on the reactant side. Balance with respect to atoms / ions. To balance oxygen, add H2O to the side with the least amount of oxygen. THEN: add H + to the other side to balance hydrogen. 3. Write the oxidation half-reaction. The bottom arrow in step #1 indicates the oxidation half-reaction. Show the electrons lost on the product side. Balance with respect to atoms / ions. To balance oxygen, add H2O to the side with the least amount of oxygen. THEN: add H + to the other side to balance hydrogen. 4. The number of electrons gained must equal the number of electrons lost. Find the least common multiple of the electrons gained and lost. In each half-reaction, multiply the electron coefficient by a number to reach the common multiple. Multiply all of the coefficients in the half-reaction by this same number. 5. Add the two half-reactions. Write one equation with all the reactants from the half-reactions on the left and all the products on the right. The order in which you write the particles in the combined equation does not matter. 6. Simplify the equation. Cancel things that are found on both sides of the equation as you did in net ionic equations. Rewrite the final balanced equation. Check to see that electrons, elements, and total charge are balanced. There should be no electrons in the equation at this time. The number of each element should be the same on both sides. It doesn't matter what the charge is as long as it is the same on both sides. Practice Problems: 1. Identify the oxidizing agent and reducing agent in each equation: H2SO4 + 8HI H2S + 4I2 + 4H2O CaC2 + 2H2O C2H2 + Ca(OH)2 Au2S3 + 3H2 2Au + 3H2S Zn + 2HCl H2 + ZnCl2 2. To make working with redox equations easier, we will omit all physical state symbols. However, remember that they should be there. An unbalanced redox equation looks like this: MnO4 - + H2SO3 + H + Mn+2 + HSO4 - + H2O Study how this equation is balanced using the half-reaction method. It is important that you understand what happens in each step. Be prepared to ask questions about this process in class tomorrow.
8.8A: Ellingham Diagrams
It is possible to use plots of the free energy of formation of metal oxides vs. temperature to predict the temperatures at which a metal is stable and the temperatures at which it will spontaneously oxidize. For temperatures at which the free energy of formation of the oxide is positive, the reverse reaction is favored and the oxide will spontaneously decompose to the metal. From evaluation of the thermodynamic data presented in this figure, it can be seen that at 1100oC, Al will oxidize in an environment that has an oxygen partial pressure of 10-32 atm or greater, while chromium will oxidize in an oxygen partial pressure of 10-19 atm or higher. In general, a vacuum environment will be oxidizing to these elements unless a reducing species such as hydrogen is present. If inadequate oxygen is present, a nonprotective oxide film may be formed which could promote alloy depletion and loss of strength. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/08%3A_Reduction_and_Oxidation/8.1B%3A_Oxidation_States.txt |
An inorganic nonaqueous solvent is a solvent other than water, that is not an organic compound. Common examples are liquid ammonia, liquid sulfur dioxide, sulfuryl chloride and sulfuryl chloride fluoride, phosphoryl chloride, dinitrogen tetroxide, antimony trichloride, bromine pentafluoride, hydrogen fluoride, pure sulfuric acid and other inorganic acids. These solvents are used in chemical research and industry for reactions that cannot occur in aqueous solutions or require a special environment.
Alkali Metal–Liquid Ammonia Solutions. Most metals are insoluble in virtually all solvents, but the alkali metals (and the heavier alkaline earth metals) dissolve readily in liquid ammonia to form solvated metal cations and solvated electrons, which give the solution a deep blue color. Image copyrighted by the Klein research group.
09: Non-aqueous Media
The Brønsted theory encompasses any type of solvent that can donate and accept $H^+$ ions, not just aqueous solutions. The strength of an acid or a base varies depending on the solvent. Non-aqueous acid-base chemistry follows similar rules to those developed for acids and bases in water. For example in liquid ammonia, the solvent autodissociates in the reaction:
$2NH_{3(l)} \rightleftharpoons NH_4^+ + NH_2^- \nonumber$
This equilibrium is analogous to the autodissociation of water, but has a smaller equilibrium constant (K ≈ 10-30). It follows by analogy to water that NH4+ is the strongest acid and NH2- is the strongest base that can exist in liquid ammonia. Because ammonia is a basic solvent, it enhances the acidity and suppresses the basicity of substances dissolved in it. For example, the ammonium ion (NH4+) is a weak acid in water (Ka = 6 x 10-10), but it is a strong acid in ammonia. Similarly, acetic acid is weak in water but strong in ammonia. Solvent leveling in fact makes CH3COOH and NH4Cl both strong acids in ammonia, where they have equivalent acid strength.
Strong acids that are leveled in water have different acid strengths in acidic solvents such as HF or anhydrous acetic acid. For example, acid dissociation of HX in acetic acid (CH3COOH) involves protonating the solvent to make its conjugate acid (CH3COOH2+) and the X- anion. Because CH3COOH2+ is a stronger acid that H3O+, the anion X- (which is a spectator in water) can become a weak base in CH3COOH:
$HX + CH_3COOH \rightleftharpoons CH_3COOH_2^+ + X^- \nonumber$
It follows that acidic solvents magnify the Brønsted basicities of substances that cannot accept protons in water. Conversely, basic solvents magnify the acidity of substances that cannot donate a proton to $OH^-$.
The acidity and basicity of non-aqueous solvents is difficult to quantify precisely, but one good relative measure is the Hammett acidity function, $H_o$, which is defined analogously to pH according to the Henderson-Hasselbach equation:
$H_o = pK_a + \log\left(\dfrac{[\text{base}]}{[\text{conjugate acid}]}\right) \nonumber$
For non-aqueous solvents, or for acidic or basic compounds in dissolved in solvents that do not themselves dissociate, Ho is a rough measure of the pH of the solvent or compound in question. Anhydrous HF and H2SO4 have Ho values of approximately -10 and -12 respectively.
9.02: Relative Permittivity
The dielectric constant (symbol: ε) of a solvent is a measure of its polarity. The higher the dielectric constant of a solvent, the more polar it is.
eg:
The dielectric constant of water is higher than that of methanol; water is more polar than methanol. One practical consequence is a covalent solute dissociates into ions to a greater extent in water than in methanol.
equilibrium constant for dissociation of MX = Kdis
One must not confuse the dielectric constant of a solvent with its dipole moment. The dipole moment of a solvent, or of any covalent compound, is a microscopic property, meaning it is a property of the molecule of the compound. In contrast, the dielectric constant of a solvent is a a macroscopic property, meaning it is a property of a pure sample of the solvent. Given below is the scattergram of the dielectric constants of sixteen solvents against their dipole moments, showing that there is no clear correlation between the dielectric constant of a solvent and its dipolar moment.
graph1
9.04: Acid-Base Behaviour in Non-Aqueous Solvents
pH values are at present undefined in aprotic solvents, as the definition of pH assumes presence of hydronium ions. In other solvents, the concentration of the respective solvonium/solvate ions should be used, such as pCl in POCl3.
9.05: Liquid Sulfur Dioxide
Sulfur dioxide is a versatile inert solvent widely used for dissolving highly oxidizing salts. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/09%3A_Non-aqueous_Media/9.01%3A_Introduction_to_Non-aqueous_Media.txt |
Despite low boiling point (-33.4 ºC), easy to handle.
Solubilities, relatively high dielectric constant (ammonia, eo= 26.7 @ -60ºC; water, eo = 82 @ 18 ºC).
Hence, ionic compounds can be soluble but the lower eo compared to water means that salt with highly charged, non-polarisable anions such as carbonates, sulphates, and phosphates are insoluble.
NH3 is more polarizable than H2O, so salts with more polarizable anions are more soluble, hence the solubility trends.
\[\ce{ F- < Cl- < Br- < I-}\]
\[\ce{PO43- < SO42- < OAc- < NO3}\]
specific solvation
NH3 is a better a-donor than H2O and ammine complexes are formed, especially with the later transition (Ni2+, Cu2+) and B metals (Ag+, Zn2+). Hence higher solubilities for compounds of these metals than those of the A-metals.
9.6B: Self-Ionization of Ammonia
Self-ionization of ammonia is much "weaker" than water.
\[\ce{2NH3 <=> NH4+ + NH2-}\]
with \(K \approx 10^{-30}\) @ 223K. Since ammonia is better proton acceptor than water, the ionization of acids is relatively enhanced in liquid ammonia. For example, acetic acid is a strong acid in liquid ammonia. Liquid ammonia will therefore tolerate very strong bases such as \(\ce{C5H5-}\) that would otherwise be hydrolyzed in water.
Ammonia is kinetically stabilized to reduction (but easily oxidized) by many reagents, e.g., the reaction
\[\ce{ Na + NH3 -> NaNH2 + H2(g}}\]
and is very favorable but slow in the absence of a catalyst such as Fe3+.
9.6C: Reactions in Liquid NH
Much of the chemistry in liquid ammonia can be classified by analogy with related reactions in aqueous solutions. Comparison of the physical properties of \(\ce{NH3}\) with those of water shows \(\ce{NH3}\) has the lower melting point, boiling point, density, viscosity, dielectric constant and electrical conductivity; this is due at least in part to the weaker hydrogen bonding in \(\ce{NH3}\) and because such bonding cannot form cross-linked networks, since each \(\ce{NH3}\) molecule has only one lone pair of electrons compared with two for each H2O molecule.
Solvolysis: synthesis of amides
\[\ce{ OPCl3 + 6NH3 -> OP(NH2)3 + 3NH4Cl}\]
\[\ce{SiCl4 + 8NH3 -> Si(NH2)4 + 4NH4Cl}\]
Metatheses reactions: solubility reversals
\[\ce{ RCl + AgNO3 ->[H2O] AgCl(s) + RNO3}\]
\[\ce{ AgCl + KNO3 ->[NH3] KCl(s) + AgNO3}\]
\[\ce{Ba(NO3)2 + 2AgCl ->[NH3] BaCl2(ppt) + 2AgNO3}\]
Sodamide as a base
\[\ce{ Na + NH3 -> NaNH2 + H2(g)}\]
\[\ce{NaNH2 + C5H6 -> NaC5H5 + NH3}\]
\(\ce{NaC_p}\) (useful reagent)
9.6D: Solutions of s-block Metals in Liquid NH
Solvated electron as a reducing agent (Birch reduction), many examples of compounds in very unusual low oxidation states.
$\ce{[Ni(CN)4]2- + Na ->[\ce{NH3}] [Ni(CN)4]4-}$
$\ce{ Fe(CO)5 + Na ->[\ce{NH3}] [Fe(CO)4]2-}$
$\ce{ Mo(CO)6 + Na ->[\ce{NH3}] [Mo(CO)4]4-}$
$\ce{ [Pt(NH3)4]2+ + Na ->[\ce{NH3}] [Pt(NH3)4]}$
Reduction of salts of Group IV and V elements give polyhedral anions, many examples. $\ce{Ge94^{-}}$, $\ce{Sn52^{-}}$, $\ce{Sn93^{-}}$, $\ce{Pb52^{-}}$, $\ce{Bi42^{-}}$, $\ce{P72^{-}}$, and $\ce{As64^{-}}$
Liquid Ammonia Solutions
A remarkable feature of the alkali metals is their ability to dissolve reversibly in liquid ammonia. Just as in their reactions with water, reacting alkali metals with liquid ammonia eventually produces hydrogen gas and the metal salt of the conjugate base of the solvent—in this case, the amide ion (NH2) rather than hydroxide:
$\ce{M(s) + NH_3(l) \rightarrow 1/2 H_2(g) + M^+(am) + NH_2^{-} (am)} \label{21.20}$
where the (am) designation refers to an ammonia solution, analogous to (aq) used to indicate aqueous solutions. Without a catalyst, the reaction in Equation \ref{21.20} tends to be rather slow. In many cases, the alkali metal amide salt ($\ce{MNH2}$) is not very soluble in liquid ammonia and precipitates, but when dissolved, very concentrated solutions of the alkali metal are produced. One mole of Cs metal, for example, will dissolve in as little as 53 mL (40 g) of liquid ammonia. The pure metal is easily recovered when the ammonia evaporates.
Solutions of alkali metals in liquid ammonia are intensely colored and good conductors of electricity due to the presence of solvated electrons (e, $\ce{NH3}$), which are not attached to single atoms. A solvated electron is loosely associated with a cavity in the ammonia solvent that is stabilized by hydrogen bonds. Alkali metal–liquid ammonia solutions of about 3 M or less are deep blue (Figure $2$) and conduct electricity about 10 times better than an aqueous $\ce{NaCl}$ solution because of the high mobility of the solvated electrons. As the concentration of the metal increases above 3 M, the color changes to metallic bronze or gold, and the conductivity increases to a value comparable with that of the pure liquid metals.
The most common sources of the hydride nucleophile are lithium aluminum hydride ($\ce{LiAlH4}$) and sodium borohydride ($\ce{NaBH4}$). Note! The hydride anion is not present during this reaction; rather, these reagents serve as a source of hydride due to the presence of a polar metal-hydrogen bond. Because aluminum is less electronegative than boron, the Al-H bond in $\ce{LiAlH4}$ is more polar, thereby, making $\ce{LiAlH4}$ a stronger reducing agent. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/09%3A_Non-aqueous_Media/9.06%3A_Liquid_Ammonia/9.6A%3A_Physical_Properties.txt |
Hydrofluoric acid is difficult to work with because is corrodes glass and silica giving gaseous \(\ce{SiF4}\) or the \(\ce{[SiF6]2-}\) so it must be stored in PTFE or dry Cu or Monel metalI.
\[\ce{4HF + SiO2 -> SiF4 + 2H2O}\]
Liquid range: 190 to 292.5 K
Relative permittivity 84 at 273 K; 175 at 200 K
Liquid HF undergoes self-ionization: Kself =2 x10-12 at 273 K
\[\ce{ 3HF <=> [H2F]+ + [HF2]-}\]
dihydridofluorine(1+) ion
difluorohydrogenate(1-) ion
Large electronegativity difference between H (xP = 2.2) and F (xP = 4.0) results in the presence of extensive intermolecular hydrogen bonding in the liquid.
• Hydrogen bonded molecules (~7 molecules on average) in liquid phase
• Cyclic (HF)x species are present in the gas phase
Warning
\(\ce{HF(l)}\) causes burns that are slow to heal or even get progressively worse. Exercise extreme caution if using this solvent.
Very few compounds exist which are strong enough F- acceptors to be considered acids. An example is \(\ce{SbF5}\) which leads to pure HF having superacid properties. Pure HF has a dielectric constant of 84 and is a good solvent like water which is surprisingly gentle. It can be used for example to remove Fe from metalloproteins without damaging the apoprotein primary structure.
9.09: Superacids
The term superacid was originally coined by James Bryant Conant in 1927 to describe acids that were stronger than conventional mineral acids.[1] George A. Olah prepared the so-called magic acid, so-named for its ability to attack hydrocarbons, by mixing antimony pentafluoride (SbF5) and fluorosulfonic acid (FSO3H). The name was coined after a candle was placed in a sample of magic acid. The candle dissolved, showing the ability of the acid to protonate hydrocarbons, which under aqueous acidic conditions cannot be protonated.
At 140 °C , FSO3H–SbF5 converts methane into the tertiary-butyl carbocation, a reaction that begins with the protonation of methane:[2]
• $CH_4 + H^+ \rightarrow CH^+_5$
• $CH^+_5 \rightarrow CH^+_3 + H_2$
• $CH^+_3 + 3 CH_4 \rightarrow (CH_3)_3C^+ + 3H_2$
Fluoroantimonic acid, HSbF6, can produce solutions with H0 down to –28.[3] Fluoroantimonic acid is made by combining HF and SbF5. In this system, HF releases its proton (H+) concomitant with the binding of F by antimony pentafluoride, which (as described below) is a Lewis acid. The resulting anion (SbF6) is both a weak nucleophile and an extraordinarily weak base.
Superacids are useful in reactions such as the isomerization of alkanes. Industrially, anhydrous acid-exchanged zeolites, which are superacid catalysts, are used on a massive scale to isomerize hydrocarbons in the processing of crude oil to gasoline. Superbases such as lithium diethylamide (LiNEt2), alkyllithium compounds (RLi), and Grignard reagents (RMgX) useful in a broad range of organic reactions. LiNEt2 deprotonates C-H bonds to generate reactive carbanions. RLi and RMgX are powerful nucleophiles.
The use of superbases in nonaqueous media allows us to rank the acidities (and measure the pKa's) of different classes of molecules. This ranking is particularly important in understanding the reactions of organic molecules. Note that the order of acidities for hydrocarbons is alkynes >> alkenes, aromatics >> alkanes. This ordering has to do with the hybridization of the carbon atom that forms the carbanion. The negatively charged lone pair of the carbanion is stabilized in orbitals that have high s character (e.g., sp vs. sp2 or sp3). This is because s orbitals have finite probability density at the nucleus and "feel" the positive nuclear charge (thereby stabilizing the extra negative charge on carbon) more than p orbitals. Resonance effects also stabilize carbanions. Thus, cyclopentadiene is more acidic than even an alkyne because the negative charge is delocalized over the entire (aromatic) C5H5- ring when the C5H6 is deprotonated.
Table 3.1.1: Carbon acid acidities in pKa in DMSO. Reference acids in bold.
name formula structural formula pKa
Methane CH4 ~ 56
Propene C3H6 ~ 44
Benzene C6H6 ~ 43
Acetylene C2H2 25
Cyclopentadiene C5H6 18
https://en.Wikipedia.org/wiki/Superacid
9.10A: Physical Properties
Bromine trifluoride is an interhalogen compound with the formula \(\ce{BrF3}\). It is a straw-colored liquid with a pungent odor. It is soluble in sulfuric acid but explodes on contact with water and organic compounds. It is a powerful fluorinating agent and an ionizing inorganic solvent. It is used to produce uranium hexafluoride (\(\ce{UF6}\)) in the processing and reprocessing of nuclear fuel.
9.10: Bromine Trifluoride
• Wikipedia
9.11A: Physical Properties of NO
\(\ce{N2O4}\) is a red-brown liquid with a sharp, unpleasant chemical odor. Low-boiling (boiling point 21.15°C) and held as a liquid by compression. Density 1.448 g / cm3. Consists of an equilibrium mixture of brown NO2 (nitrogen dioxide) and colorless N2O4 (dinitrogen tetroxide). It is a good oxidizing and nitrating agent.
\[\ce{Li + N2O4 -> LiNO3 + NO}\]
9.11B: Reactions in NO
https://pubs.acs.org/doi/abs/10.1021/ed034p555
9.12A: Molten Salt Solvent Systems
When a solid salt melts, it forms a solution of the cations and anions. For example, KOH melts at temperatures above 400 °C and dissociates into K+ and OH- ions which can act as a solvent for chemical reactions. Because of the autodissociation of the OH- solvent, water is always present in a molten KOH flux, according to the acid-base equilibrium:
\[2 OH^- \rightleftharpoons H_2O + O^{2-}\]
It follows that in this very basic solvent, water (the conjugate acid of the solvent) is the strongest acid that can exist. The conjugate base of the solvent, O2-, is the strongest base. This autodissociation equilibrium allows for the acidity of a flux to be easily tuned through the addition or boiling off of water. A "wet" flux is more acidic, and can dissolve metal oxides that contain the basic O2- anion. Conversely a "dry" flux is more basic and will cause oxides to precipitate. Molten hydroxide fluxes can thus be used in the synthesis of oxide crystals, such as the perovskite superconductor (K1-XBaXBiO3). Eutectic mixtures of NaOH and KOH are relatively low melting (≈ 200 °C) and can be used as solvents for crystallizing a variety of basic oxides | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/09%3A_Non-aqueous_Media/9.07%3A_Liquid_Hydrogen_Fluoride/9.7A%3A_Physical_Properties.txt |
In 1671, Robert Boyle discovered and described the reaction between iron filings and dilute acids, which resulted in the production of hydrogen gas. In 1766-81, Henry Cavendish was the first to recognize that hydrogen gas was a discrete substance, and that it produced water when burned. He named it "flammable air". In 1783, Antoine Lavoisier gave the element the name hydrogen (from the Greek υδρο- hydro meaning "water" and -γενης genes meaning "creator") when he and Pierre-Simon Laplace reproduced Cavendish's finding that water was produced when hydrogen was burned. Hydrogen was liquefied for the first time by James Dewar in 1898 by using regenerative cooling and his invention, the vacuum flask. He produced solid hydrogen the next year. Deuterium was discovered in December 1931 by Harold Urey, and tritium was prepared in 1934 by Ernest Rutherford, Mark Oliphant, and Paul Harteck. Heavy water, which consists of deuterium in the place of regular hydrogen, was discovered by Urey's group in 1932.
• 10.1: Hydrogen - The Simplest Atom
• 10.2: The \(H^+\) and \(H^-\) Ions
• 10.3: Isotopes of Hydrogen
Hydrogen has three naturally occurring isotopes, denoted 1H, 2H and 3H. Other, highly unstable nuclei (4H to 7H) have been synthesized in the laboratory but are not observed in nature.
• 10.4: Dihydrogen
Hydrogen is a colorless, odorless and tasteless gas that is the most abundant element in the known universe. It is also the lightest (in terms of atomic mass) and the simplest, having only one proton and one electron (and no neutrons in its most common isotope). It is all around us. It is a component of water (H2O), fats, petroleum, table sugar (C6H12O6), ammonia (NH3), and hydrogen peroxide (H2O2)—things essential to life, as we know it.
• 10.5: Polar and Non-Polar E-H Bonds
• 10.6: Hydrogen Bonding
A hydrogen bond is a weak type of force that forms a special type of dipole-dipole attraction which occurs when a hydrogen atom bonded to a strongly electronegative atom exists in the vicinity of another electronegative atom with a lone pair of electrons. These bonds are generally stronger than ordinary dipole-dipole and dispersion forces, but weaker than true covalent and ionic bonds.
• 10.7: Binary Hydrides - Classification and General Properties
10: Hydrogen
• Hydrogen (not carbon) forms more compounds than any other element! The isotopes are:
• 1H
• 2D (0.0156%)
• 3T (Formation: 14N(p,n)14O and then (using the neutron) 14N(n,3H)12C
Abundance: 10-15 - 10-16 %, Decay: 3H(b-)3He t½ = 12.35 y.
• The different name arise because isotope effects, which affect kinetics and equilibria, are especially marked for hydrogen.
• Deuterium is obtained by fractional distillation of water (or hydrogen sulphide?) or electrolysis of water. D2O is used as a moderator in nuclear reacters and as a source of D for chemical studies of all sorts.
• Made in the lab from the action of acid on Zn or Fe and industrially by the catalysed reaction of steam with hydrocarbons, coal and other organic materials to give "synthesis gas, "syngas":
CH4 + H2O CO + H2
H2 can also be made by the "water-gas reaction" and the catalysed "water-gas shift reaction":
H2O + CO CO + H2
CO + H2O CO2 + H2
The hydrogen can be isolated by absorption of the carbon dioxide, and removal of residual CO and CO2. (The "syngas" is an important industrial raw material itself.)
• It, and "syngas" are used in the production of organics via alcohols:
2CO + 4H2 CH3CH2OH + H2O
and the production of ammonia via the Haber process:
N2 + 3H2 2NH3
• Hydrogen has a bond energy of 434.1 kJ mol-1. It will burn in air but reacts explosively with oxygen and some halogens via chain reactions:
Br2 + light 2Br chain initiation
Br + H2 HBr + H chain propagation
H + Br2 HBr + Br
H + HBr H2 + Br
2Br Br2 chain termination
Notice there is no direct H-H bond fission in this processs.
The Bonding of Hydrogen
As mentioned before, hydrogen is found bonded in two covalent situations:
1. The complexed proton, e.g. H3O+ or NH4+, and "conventional" covalent situations, e.g. CH4. The distinction between these two is rather artificial, except that the cations would have a rather more polarized and labile X-H bond, and are prone to transferring the proton to another molecule, i.e. behaviour as an acid.
2. Hydrides such as KH, which contain the H- ion.
It has some special ways of bonding as well:
1. The formation of metallic hydrides, that is, hydrides which have metallic properties as apposed to ionic H- containing materials, for example PdH0.4-0.7.
Note that the density of Pd is = 11.99 g cm-3 or 11.99/106.42 = 0.113 mol cm-3
so in PdH0.7 the density of hydrogen is 0.113 x 0.7 = 0.0789 mol cm-3 or 0.0789 x 1.008 = 0.0790 g cm-3
i.e. more than liquid hydrogen where the density (at -252.78 oC) is 0.07099 g cm-3!
The form of the hydrogen in metals is ambiguous: it tends to migrate towards the negative end of a potential gradient in wires, so it seems proton-like. On the other hand, salt like properties and M-H bond distances in some metal hydrides are more suggestive of H- compounds. (Perhaps only a small part of the hydrogen is cationic in nature.)
2. The formation of covalent hydrogen bridges for example in B2H6 and (CO)5CrHCr(CO)5. These are prototypes for 2e- - 3-centre bonding.
3. A newish class of hydrogen bonds called "agostic" have been identified in certain transition metal compounds which seem to be a sort of frozen intermediate in the catalytic activation of C__H bonds. They come as "open" and "closed":
4. So called "hydrogen bonding" described in the following section.
The Hydrogen Bond
• When hydrogen is bonded to an electronegative element, X, usually F, O, N or Cl, and there is another molecule around with a Lewis base donor atom,Y , a very strong largely electrostatic bond can form: Xd-__Hd+........Y The X__H bond is slightly longer than it would be without the proximity of the Y, and the H........Y "bond" is much longer than a H__Y bond would be. Usually, hydrogen bonding is identified if this distance appreciably less than the sum of the van der Waals' radii for H and Y (e.g. <(1.30 + 1.40) = 2.70 Å for H........O. In many cases the position of the hydrogen must be inferred: in this case O__H should be about 0.37 + 0.70 = 1.07 Å so an O to O distance less than about 2.70 + 1.07 = 3.75 Å would imply a hydrogen bond is present.)
• The text refers to the case of crystalline NaHCO3 where there are 4 O to O distances: 3.12, 3.15 and 3.19 Å which are well over the van der Waals' contact distance, and 2.55 Å which is somewhat less, and tells us where the H is to be found.
• The hydrogen bond can also be identified by a shift in the infra-red stretching frequency to lower wavelengths. It is also broadened and made more intense. For example, free O__H comes at 3500 cm-1 but can be lowered several hundred wave-numbers by H-bonding.
• The table below gives some approximate hyhdrogen bond enthalpies:
Bond Energy kJ mol-1
F__H........F 30
N__H........N 25
O__H........O 25
N__H........F 21
O__H........N 20
C__H........O 11
N__H........O 10
• Hydrogen bonding affects boiling points (and heats of vaporization) producing some anomalies. See text Figure 9-1.
• There are specially stong hydrogen bonds that are 4e- - 3-centre bonds. Examples include [FHF]- with a centered proton and F to F distance of 2.26 Å.
Ice and Water
• There are 9 forms of ice which exist at high presssure, except the normal ice I.
• The normal hexagonal ice structure is depicted in Fig 9-2.
• In liquid water, an ice-like structure persists, but there are additional "interstitial" water molecules, and the whole system is "fluxional". The density of liquid water is greater than that of ice. The density is maximum at 4 oC.
Hydrates and Water Clathrates
Most hydrates are salts containing water in addition to the cations and anions. The water is sometimes in the first coodination sphere of the cations and is sometimes rather more loosely held:
CuSO4.5H2O CuSO4.3H2O CuSO4.H2O CuSO4 anhydrous
The water molecules are successively more difficult to remove, as the sulphate takes their place in the Cu2+ coordination sphere.
Sometimes the water is so tightly bound, a decomposition occurs on heating:
ScCl3.6H2O ScOCl + 2HCl + 5H2O
"Gas hydrates" are an example of a class of "clathrate" compounds:
1. There is a cubic symmetry form that features a 46-water molecule unit cell including six medium sized and two small cages. With the medium-sized cages trapping a "guest" molecule each the formula would be X.(7.67)H2O. All eight cages filled would give X.(5.76)H2O. Clathrates of the first type are known for Ar, Kr Xe, Cl2, SO2 and CH3Cl (among others).
2. Another geometry, also cubic has 136 H2O's in the unit cell and features 8 larger cages and 16 smaller ones. Molcules such as CHCl3 and CH3CH2Cl can be trapped.
3. One other class made with the salts of R4N+ or R3S+ have the anions acting as part of the "host" structure. Examples are: [(C4H9)4N][C6H5CO2].(39.5)H2O or [(C4H9)3S]F.(20)H2O.
Hydrides
Many hydrogen compounds are collectively called hydrides, including many cases where the hydrogen is actually less electronegative than the atom to which it is bound, and which are not individually named hydrides, in addition to those where it is more electronegative, upto the saline hydrides containing H-. In addition, there is the class of metallic hydrides, See Figure 9-4:
Covalent
1. Neutral - Group IVB (14) CH4 (methane), SiH4 (silane) etc
2. Somewhat basic - Group VB (15) NH3, PH3 (phosphine) etc
3. Weakly acidic - Group VIB (16) H2O, H2S (hydrogen sulphide) etc
4. Strongly acidic - Group VIIB (17) HCl, HI (hydrogen halides) etc
5. The boron hydrides (BnHm)
6. Hydride anions eg LiAlH4 (alanate), LiBH4 (borohydride or boranate)
Preparation:
(8LiH + Al2Cl6 2LiAlH4 + 6LiCl)
(2NaH + B2H6 2NaBh4)
Typical Reactions:
2LiAlH4 + 2SiCl4 2SiH4 + 2LiCl + Al2Cl6
I2 + LiBH4 B2H6 + 2NaI + H2
Saline
Some are made by direct reaction: M + H2 MH (Li - Cs)
or MH2 (Mg - Ba)
The hydrides of Li and Be have covalent character, especially BeH2 which is really a polymer. Most saline hydrides react violently with water to give H2
They are also powerful hydride transfer reagents, sometimes useful for making other hydrides:
B(OR)3 + NaH Na[BH(OR)3]
4NaH + TiCl4 Tio + 4NaCl + 2H2
NaH + ROH NaOR + H2
Transition metal
These include the stoichiometric hydrides, for example UH3 and HCo(CO)4 and hydride anions, for example [ReH9]2-.
There are also the non-stoichiometric (interstitial) ones including PdH0.7 and ZrH1.9.
Dihydrogen as a Ligand
Dihydrogen complexes, where the hydrogen molecule is bonded sideways on to the metal, but the H-H bond is largely intact have been discovered only recently. The metal accepts electrons from the H-H s-orbital, and donates electrons back to the H-H s*-orbital. Both types of bonding should lead to weakening and ultimate cleavage of the H-H bond to give normal hydrides, and only very a few special cases short of this. An example is:
W(CO)3P(Pri3)2 + H2 W(CO)3P(Pri3)2(H2)
H-H by neutron diffraction = 0.75 Å, by X-ray diffraction = 0.84 Å and in H2 = 0.74 Å
Contributors and Attributions
Template:ContriBird | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/10%3A_Hydrogen/10.01%3A_Hydrogen_-_The_Simplest_Atom.txt |
Hydrogen has three naturally occurring isotopes, denoted 1H, 2H and 3H. Other, highly unstable nuclei (4H to 7H) have been synthesized in the laboratory but are not observed in nature.
• 1H is the most common hydrogen isotope with an abundance of more than 99.98%. Because the nucleus of this isotope consists of only a single proton, it is given the descriptive, but rarely used formal name of protium.
• 2H, the other stable hydrogen isotope, is known as deuterium and contains one proton and one neutron in its nucleus. Essentially all deuterium in the universe is thought to have been produced at the time of the Big Bang, and has endured since that time. Deuterium is not radioactive, and does not represent a significant toxicity hazard. Water enriched in molecules that include deuterium instead of normal hydrogen is called heavy water. Deuterium and its compounds are used as a non-radioactive label in chemical experiments and in solvents for 1H-NMR spectroscopy. Heavy water is used as a neutron moderator and coolant for nuclear reactors. Deuterium is also a potential fuel for commercial nuclear fusion.
• 3H is known as tritium and contains one proton and two neutrons in its nucleus. It is radioactive, decaying into helium-3 through beta decay with a half-life of 12.32 years. It is sufficiently radioactive that it can be used in luminous paint, making it useful in such things as watches where the glass moderates the amount of radiation getting out. Small amounts of tritium occur naturally because of the interaction of cosmic rays with atmospheric gases; tritium has also been released during nuclear weapons tests. It is used in nuclear fusion reactions, as a tracer in isotope geochemistry, and specialized in self-powered lighting devices. Tritium has been used in chemical and biological labeling experiments as a radiolabel.
Hydrogen is the only element that has different names for its isotopes in common use today. During the early study of radioactivity, various heavy radioactive isotopes were given their own names, but these names are no longer used, except for deuterium and tritium.
nuclide symbol Z(p) N(n) isotopic mass (u) half-life decay mode Daughter Isotope Isotopic composition
1H 1 0 1.00782503207(10) Stable - - 0.999885(70)
2H - D 1 1 2.0141017778(4) Stable - - 0.000115(70)
3H - T 1 2 3.0160492777(25) 12.32(2) y β 3He <1>17 atoms
10.03: Isotopes of Hydrogen
Properties of hydrogen
The difference of mass between isotopes of most elements is only a small fraction of the total mass and so this has very little effect on their properties, this is not the case for hydrogen. Consider chlorine with Z=17, there are 2 stable isotopes 35Cl (75.77%) and 37Cl (24.23%). The increase is therefore 2 in 35 or less than 6%. Deuterium and tritium are about double and triple the mass of protium and show significant physical and chemical differences particularly with regard to those properties related to mass, e.g. rate of diffusion, density, etc.
Some physical properties of the hydrogen isotopes.
isotope MP /K BP /K ΔHdiss /kJmol-1 Interatomic Distance /pm
H2 13.99 20.27 435.99 74.14
D2 18.73 23.67 443.4 74.14
T2 20.62 25.04 446.9 74.14
Differences between H2O and D2O
Property H2O D2O
Melting point /K 273.15 276.97
Boiling point /K 373.15 374.5
Temperature of maximum density /K 277 284.2
Maximum density /g cm3 0.99995 1.1053
Relative permittivity (at 298 K) 78.39 78.06
Kw (at 298 K) 1 *1014 2 * 1015
Symmetric stretch, ν1 /cm-1
(gaseous molecule)
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The kinetic isotope effect (KIE) is a phenomenon associated with isotopically substituted molecules exhibiting different reaction rates. Isotope effects such as KIEs are invaluable tools in both physical and biological sciences and are used to aid in the understanding of reaction kinetics, mechanisms, and solvent effects.
Introduction
Research was first introduced on this topic over 50 years ago and has grown into an enormous field. The scientists behind much of the understanding and development of kinetic isotope effects were Jacob Bigeleisen and Maria Goeppert Mayer who published the first paper on isotope effects [J. Chem. Phys., 15, 261 (1947)]. Kinetic isotope effects specifically explore the change in rate of a reaction due to isotopic substitution.
An element is identified by its symbol, mass number, and atomic number. The atomic number is the number of protons in the nucleus while the mass number is the total number of protons and neutrons in the nucleus. Isotopes are two atoms of the same element that have the same number of protons but different numbers of neutrons. Isotopes are specified by the mass number.
As an example consider the two isotopes of chlorine, you can see that their mass numbers vary, with 35Cl being the most abundant isotope, while their atomic numbers remain the same at 17.
$^{35}Cl \;\text{and}\; ^{37}Cl$
The most common isotope used in light atom isotope effects is hydrogen ($^{1}H$) commonly replaced by its isotope deuterium ($^{2}H$). Note: Hydrogen also has a third isotope, tritium ($^{2}H$). Isotopes commonly used in heavy atom isotope effects include carbon ($^{12}C$, $^{13}C$, nitrogen ($^{14}N$, $^{15}N$), oxygen, sulfur, and bromine. Not all elements exhibit reasonably stable isotopes (i.e. Fluorine, $^{19}F$), but those that due serve as powerful tools in isotope effects.
Potential Energy Surfaces
Understanding potential energy surfaces is important in order to be able to understand why and how isotope effects occur as they do. The harmonic oscillator approximation is used to explain the vibrations of a diatomic molecule. The energies resulting from the quantum mechanic solution for the harmonic oscillator help to define the internuclear potential energy of a diatomic molecule and are
$E_n = \left(n + \dfrac{1}{2}\right)h \nu \label{1}$
where
• n is a positive integer (n=1,2,3...),
• h is Planck's constant and
• $\nu$ is the frequency of vibration.
The Morse potential is an analytic expression that is used as an approximation to the intermolecular potential energy curves:
$V(l) = D_e{\left(1-e^{-\beta(l-l_o)}\right)}^2 \label{2}$
where
• $V(l)$ is the potential energy,
• $D_e$ is the dissociation energy of the molecule,
• $\beta$ is the measure of the curvature of the potential at its minimum,
• $l$ is displacement, and
• $l_o$ is the equilibrium bond length.
The $D_e$, $\beta$, and $l_o$ variables can be looked up in a textbook or CRC handbook.
Below is an example of a Morse potential curve with the zero point vibrational energies of two isotopic molecules (for example R-H and R-D where R is a group/atom that is much heavier than H or D). The y-axis is potential energy and the x axis is internuclear distance. In this figure EDo and EHo correspond to the zero point energies of deuterium and hydrogen. The zero point energy is the lowest possible energy of a system and equates to the ground state energy. Zero point energy is dependent upon the reduced mass of the molecule as will be shown in the next section. The heavier the molecule or atom, the lower the frequency of vibration and the smaller the zero point energy. Lighter molecules or atoms have a greater frequency of vibration and a higher zero point energy. We see this is the figure below where deuterium is heavier than hydrogen and therefore has the lower zero point energy.
This results in different bond dissociation energies for R-D and R-H. The bond dissociation energy for R-D (ED) is greater than the bond dissociation energy of R-H (EH). This difference in energy due to isotopic replacement results in differing rates of reaction, the effect that is measured in kinetic isotope effects. The reaction rate for the conversion of R-D is slower than the reaction rate for the conversion of R-H.
p>
It is important to note that isotope replacement does not change the electronic structure of the molecule or the potential energy surfaces of the reactions the molecule may undergo. Only the rate of the reaction is affected.
Activation Energies
The energy of the vibrational levels of a vibration (i.e., a bond) in a molecule is given by
$E_n = \left(n + \dfrac{1}{2}\right)h \nu \label{3}$
where we assume that the molecule is in its ground state and we can compare zero-point vibrational energies,
$E_o = \left(\dfrac{1}{2}\right)hv \label{4}$
Using the harmonic oscillator approximation the fundamental vibrational frequency is
$\nu = \dfrac{1}{2 \pi} \sqrt{ \dfrac {k}{\mu} } \label{5}$
where
• $k$ is the force constant of the bond and
• $\mu$ is the reduced mass
$\mu = \dfrac{m_1m_2}{m_1+m_2} \label{6}$
The Arrhenius equation is used to determine reaction rates and activation energies and since we are interested in the change in rate of reactions with different isotopes, this equation is very important,
$k = Ae^{-\frac{E_a}{kT}} \label{7}$
where
• $k$ is the reaction rate,
• $E_a$ is the activation energy, and
• $A$ is the Arrhenius constant.
The Arrhenius equation can be used to compare the rates of a reaction with R-H and R-D,
$k_H = A_He^{-\frac{E_a^H}{kT}} \label{8}$
$k_D = A_De^{-\frac{E_a^D}{kT}} \label{9}$
where kH and kD are the rates of reaction associated with R-H and the isotope substituted R-D. We will then assume the Arrhenius constants are equal ($A_H=A_D$). The ratio of the rates of reaction gives an approximation for the isotope effect resulting in:
$\dfrac{k_H}{k_D} = e^{-\frac{E_a^H - E_a^D}{kT}} \label{10}$
By using the relationship that for both R-H and R-D
$E_o = \left(\dfrac{1}{2}\right)h\nu \label{11}$
a substitution can be made resulting in
$\dfrac{k_H}{k_D} = e^{\frac{h(\nu_H - \nu_D)}{2kT}} \label{12}$
The vibrational frequency (Equation 5) can then be substituted for R-H and R-D and the value of the expected isotope effect can be calculated.
$\dfrac{k_H}{k_D} = e^{\dfrac {h \left( \dfrac{k_{RH}}{\mu_{RH}} - \dfrac{k_{RD}}{\mu_{RD}} \right)}{4\pi kT}} \label{13}$
The same general procedure can be followed for any isotope substitution.
In summary, the greater the mass the more energy is needed to break bonds. A heavier isotope forms a stronger bond. The resulting molecule has less of a tendency to dissociate. The increase in energy needed to break the bond results in a slower reaction rate and the observed isotope effect.
Kinetic Isotope Effects
Kinetic Isotope Effects (KIEs) are used to determine reaction mechanisms by determining rate limiting steps and transition states and are commonly measured using NMR to detect isotope location or GC/MS to detect mass changes. In a KIE experiment an atom is replaced by its isotope and the change in rate of the reaction is observed. A very common isotope substitution is when hydrogen is replaced by deuterium. This is known as a deuterium effect and is expressed by the ratio kH/kD (as explained above). Normal KIEs for the deuterium effect are around 1 to 7 or 8. Large effects are seen because the percentage mass change between hydrogen and deuterium is great. Heavy atom isotope effects involve the substitution of carbon, oxygen, nitrogen, sulfur, and bromine, with effects that are much smaller and are usually between 1.02 and 1.10. The difference in KIE magnitude is directly related to the percentage change in mass. Large effects are seen when hydrogen is replaced with deuterium because the percentage mass change is very large (mass is being doubled) while smaller percent mass changes are present when an atom like sulfur is replaced with its isotope (increased by two mass units).
Primary KIEs
Primary kinetic isotope effects are rate changes due to isotopic substitution at a site of bond breaking in the rate determining step of a reaction.
Example
Consider the bromination of acetone: kinetic studies have been performed that show the rate of this reaction is independent of the concentration of bromine. To determine the rate determining step and mechanism of this reaction the substitution of a deuterium for a hydrogen can be made.
When hydrogen was replaced with deuterium in this reaction a $k_H \over k_D$ of 7 was found. Therefore the rate determining step is the tautomerization of acetone and involves the breaking of a C-H bond. Since the breaking of a C-H bond is involved, a substantial isotope effect is expected.
Heavy Atom Isotope Effects
A rule of thumb for heavy atom isotope effects is that the maximum isotopic rate ratio is proportional to the square root of the inverse ratio of isotopic masses.
• Expected: $\dfrac{k_{32}}{k_{34}} = \sqrt{\dfrac{34}{32}}=1.031$
• Experimental: $\dfrac{k_{32}}{k_{34}} = 1.072$
Secondary KIEs
Secondary kinetic isotope effects are rate changes due to isotopic substitutions at a site other than the bond breaking site in the rate determining step of the reaction. These come in three forms: $\alpha$, $\beta$, and $\gamma$ effects.
$\beta$ secondary isotope effects occur when the isotope is substituted at a position next to the bond being broken.
$\ce{(CH3)2CHBr + H2O ->[k_H] (CH3)2CHOH}$
$\ce{(CD3)2CHBr + H2O ->[k_D] (CD3)2CHOH}$
This is thought to be due to hyperconjugation in the transition state. Hyperconjugation involves a transfer of electron density from a sigma bond to an empty p orbital (for more on hyperconjugation see outside links).
Solvent Effects in Reactions
Reactions may be affected by the type of solvent used (for example H2O to D2O or ROH to ROD). There are three main ways solvents effect reactions:
1. The solvent can act as a reactant resulting in a primary isotope effect.
2. Rapid hydrogen exchange can occur between substrate molecules labeled with deuterium and hydrogen atoms in the solvent. Deuterium may change positions in the molecule resulting in a new molecule that is then reacted in the rate determining step of the reaction.
3. The nature of solvent and solute interactions may also change with differing solvents. This could change the energy of the transition state and result in a secondary isotope effects.
References
1. Baldwin, J.E., Gallagher, S.S., Leber, P.A., Raghavan, A.S., Shukla, R.; J. Org. Chem. 2004, 69, 7212-7219 (This is a great paper using kinetic isotope effects to determine a reaction mechanism. It will interest the organic chemistry oriented reader.)
2. Bigeleisen, J., Goeppert, M., J. Chem. Phys. 1947, 15, 261.
3. Chang, R.; Physical Chemistry for the Chemical and Biological Sciences; University Science Books: Sausalito, CA, 2000, pp 480-483.
4. Isaacs, N.; Physical Organic Chemistry; John Wiley & Sons Inc.: New York, NY; 1995, 2nd ed, pp 287-313.
5. March, J., Smith, M.B.; March’s Advanced Organic Chemistry; John Wiley & Sons, Inc.: Hoboken, NJ, 2007; 6th ed.
6. McMurry, J.; Organic Chemistry; Brooks & Cole: Belmont, CA; 2004, 6th ed.
7. McQuarrie, D.; Quantum Chemistry; University Science Books: Sausalito, CA, 2008, 2nd ed.
8. Rouhi, A.; C&EN. 1997, 38-42.
Problems
1. Describe the difference between primary and secondary kinetic isotope effects.
2. Estimate the kN-H/kN-D for a deuterium substitution on nitrogen given that vH=9.3x1013 Hz and the activation energy is equal to 5.31 kJ/mol.
3. Using the 'rule of thumb' for heavy isotope effects, calculate the expected effect for a bromine isotope substitution, 79Br and 81Br.
4. Explain some of the main ways kinetic isotope effects are used.
5. As discussed, the rate-limiting step in the bromination of acetone is the breaking of a carbon-hydrogen bond. Estimate kC-H/KC-D for this reaction at 285 K. (Given: vtildeC-H=3000 cm-1 and vtildeC-D=2100 cm-1)
Solutions
1. Primary isotope effects involve isotopic substitution at the bond being broken in a reaction, while secondary isotope effects involve isotopic substituion on bonds adjacent to the bond being broken.
2. 8.5
3. 1.0126
4. To determine reaction mechanisms, to determine rate limiting steps in reactions, to determine transition states in reactions.
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Heavy Water
Given that the boiling point of D2O is 101.4 °C (compared to 100.0 °C for H2O), evaporation or fractional distillation can be used to increase the concentration of deuterium in a sample of water by the selective removal of the more volatile light water, H2O. Thus bodies of water that have no outlet, such as the Great Salt Lake in Utah, USA and the Dead Sea in the Jordan Rift Valley, which maintain their level solely by evaporation, have significantly higher concentrations of deuterated water than do lakes or seawater with at least one outlet.
Heavy water is 10.6% denser than ordinary water, a difference not immediately obvious since they are otherwise physically and chemically similar. The difference can be observed by freezing a sample and dropping it into normal water, where it sinks. With respect to taste and smell, rats given a choice between distilled normal water and heavy water avoided the heavy water, based on smell, and it may be that they detected a different taste as well.
The difference in weight increases the strength of water's hydrogen-oxygen bonds, and this in turn is sufficient to cause differences that are important to some biochemical reactions. The human body naturally contains deuterium equivalent to about five grams of heavy water, which is harmless. When a large fraction of water (> 50%) in higher organisms is replaced by heavy water, the result is cell dysfunction and death.
In normal water, about 1 molecule in 3,200 is HDO (one hydrogen in 6,400 is in the form of D), and heavy water molecules (D2O) only occur in a proportion of about 1 molecule in 41 million (i.e. one in 6,4002). Thus semiheavy water molecules are far more common than "pure" (homoisotopic) heavy water molecules. Deuterium oxide was initially obtained by the electrolysis of ordinary water over a considerable period of time. This method of production requires a large cascade of stills or electrolysis chambers and consumes large amounts of power, so that chemical methods are generally now preferred. The most important chemical method is the Girdler sulfide process.
In this process, demineralised and deaerated water is trickled through a series of perforated (seive) plates in a tower, while hydrogen sulfide gas (BP -60 °C) flows upward through the perforations. Deuterium migration preferentially takes place from the gas to the liquid water. This "enriched" water from the cold tower (maintained at 32 °C) is fed to the hot tower (at 130 °C) where deuterium transfer takes place from the water to the hydrogen sulfide gas. An appropriate "cascade" setup accomplishes enrichment via the reversible reaction:
\[H_2O +HDS \rightleftharpoons HDO + H_2S\]
The equilibrium constant, K for, this reaction in terms of the concentrations, can be written as:
K = ([HDO][H2S]) / ([H2O][HDS]) or alternatively:
K = ([HDO]/[H2O]) / ([HDS]/[H2S])
If H and D were the same chemically, the equilibrium constant for the reaction would be equal to unity. However, what is found is that K is not equal to unity, and furthermore it is temperature dependent:
at 25 °C, K = 2.37
at 128 °C, K = 1.84
From the above information, at 32 °C, the equilibrium favours the concentration of deuterium in water. However, at around 130 °C, the equilibrium is now relatively more favorable to the concentration of deuterium in the hydrogen sulfide. In other words, the concentration of HDO in H2O is greater than the concentration of HDS in H2S but the relative concentration of HDS in H2S increases with increasing temperature, making it possible to separate D from H.
In the first stage, the gas is enriched from 0.015% deuterium to 0.07%. The second column enriches this to 0.35%, and the third column achieves an enrichment between 10% and 30% deuterium oxide, D2O. Further enrichment to "reactor-grade" heavy water (> 99% D2O) still requires distillation or electrolysis. The production of a single litre of heavy water requires ~340,000 litre of feed water.
In 1934, Norway built the first commercial heavy water plant with a capacity of 12 tonnes per year. From 1940 and throughout World War II, the plant was under German control and the Allies decided to destroy the plant and its heavy water to inhibit German development of nuclear weapons. In late 1942, a planned raid by British airborne troops failed, both gliders crashing. The raiders were killed in the crash or subsequently executed by the Germans. On the night of 27 February 1943 Operation Gunnerside succeeded. Norwegian commandos and local resistance managed to demolish small, but key parts of the electrolytic cells, dumping the accumulated heavy water down the factory drains. Had the German nuclear program followed similar lines of research as the United States Manhattan Project, the heavy water would not have been crucial to obtaining plutonium from a nuclear reactor, but the Germans did not discover the graphite reactor design used by the allies for this purpose. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/10%3A_Hydrogen/10.03%3A_Isotopes_of_Hydrogen/10.3C%3A_Deuterated_Compounds.txt |
Hydrogen is a colorless, odorless and tasteless gas that is the most abundant element in the known universe. It is also the lightest (in terms of atomic mass) and the simplest, having only one proton and one electron (and no neutrons in its most common isotope). It is all around us. It is a component of water (H2O), fats, petroleum, table sugar (C6H12O6), ammonia (NH3), and hydrogen peroxide (H2O2)—things essential to life, as we know it.
10.04: Dihydrogen
In the laboratory, H2 can be prepared by the action of a dilute non-oxidizing acid on a reactive metal such as zinc, with a Kipp's apparatus.
$\ce{Zn + 2H (aq)^{+} \rightleftharpoons Zn (aq) ^{2+} + H2}\nonumber$
Aluminium can produce H2 upon treatment with bases:
$\ce{2Al + 6 H2O + 2 OH^{-} \rightleftharpoons 2 Al(OH)4^{-} + 3 H2}\nonumber$
The electrolysis of water is another simple method of producing hydrogen. A low voltage current is passed through the water, and gaseous dioxygen forms at the anode while gaseous hydrogen forms at the cathode. Typically the cathode is made from platinum or other inert metal when producing hydrogen for storage. If, however, the gas is to be burnt on site, oxygen is desirable to assist the combustion, and so both electrodes would be made from inert metals. (Iron, for instance, would oxidize, and thus decrease the amount of oxygen given off.) The theoretical maximum efficiency (electricity used versus energetic value of hydrogen produced) is in the range 80-94%.
$\ce{2 H2O(l) \rightleftharpoons 2 H2(g) + O2(g)}\nonumber$
In 2007, it was discovered that an alloy of aluminum and gallium in pellet form added to water could be used to generate hydrogen. The process creates alumina, but the expensive gallium, which prevents the formation of an oxide skin on the pellets, can be re-used. This has important potential implications for a hydrogen economy, as hydrogen could be produced on-site without the need of being transported.
Industrial preparation of hydrogen
Steam reforming is a method for producing hydrogen, carbon monoxide or other useful products from hydrocarbon fuels such as natural gas. This is achieved in a processing device called a reformer which reacts steam at high temperature with the fossil fuel.
At high temperatures (700 - 1100 °C) and in the presence of a metal-based catalyst (nickel), steam reacts with methane to yield carbon monoxide and hydrogen.
$\ce{CH4 + H2O → CO + 3 H2}\nonumber$
In order to produce more hydrogen from this mixture, more steam is added and the water gas shift reaction is carried out:
$\ce{CO + H2O → CO2 + H2}\nonumber$
The mixture of CO and H2 is called "synthesis gas or syngas". Syngas is used as an intermediate in producing synthetic petroleum for use as a fuel or lubricant via the Fischer-Tropsch process and previously the Mobil methanol to gasoline process.
Enzymatic route from xylose
In 2013 a low-temperature, 50 °C, atmospheric-pressure, enzyme-driven process to convert xylose into hydrogen with nearly 100% of the theoretical yield was announced. The process employed 13 enzymes, including a novel polyphosphate xylulokinase (XK).
It was noted that: "Approximately 50 million metric tons of dihydrogen are produced annually from nonrenewable natural gas, petroleum, and coal. H2 production from water remains costly. Technologies for generating H2 from less costly biomass, such as microbial fermentation, enzymatic decomposition, gasification, steam reforming, and aqueous phase reforming, all suffer from low product yields."
Applications of Hydrogen
Large quantities of H2 are used by the petroleum and chemical industries. The largest application of H2 is for the processing ("upgrading") of fossil fuels, and in the production of ammonia. The key consumers of H2 in the petrochemical plant include hydrodealkylation, hydrodesulfurization, and hydrocracking. H2 has several other important uses. H2 is used as a hydrogenating agent, particularly in increasing the level of saturation of unsaturated fats and oils (found in items such as margarine), and in the production of methanol. It is similarly the source of hydrogen in the manufacture of hydrochloric acid. H2 is used as a reducing agent of metallic ores.
Nitrogen is a strong limiting nutrient in plant growth. Carbon and oxygen are also critical, but are more easily obtained by plants from soil and air. Even though air is 78% nitrogen, atmospheric nitrogen is nutritionally unavailable because nitrogen molecules are held together by strong triple bonds. Nitrogen must be 'fixed', i.e. converted into some bioavailable form, through natural or man-made processes. It was not until the early 20th century that Fritz Haber developed the first practical process to convert atmospheric nitrogen to ammonia, which is nutritionally available.
Fertilizer generated from ammonia produced by the Haber process is estimated to be responsible for sustaining one-third of the Earth's population. It is estimated that half of the protein within human beings is made of nitrogen that was originally fixed by this process; the remainder was produced by nitrogen fixing bacteria and archaea.
Dozens of chemical plants worldwide produce ammonia, consuming more than 1% of all man-made power. Ammonia production is thus a significant component of the world energy budget. Modern ammonia-producing plants depend on industrial hydrogen production to react with atmospheric nitrogen using a magnetite catalyst or over a promoted Fe catalyst under high pressure (100 standard atmospheres (10,000 kPa)) and temperature (450 °C) to form anhydrous liquid ammonia. This step is known as the ammonia synthesis loop (also referred to as the Haber-Bosch process):
$\ce{3 H2 + N2 \rightleftharpoons 2 NH3} \,\, ΔH = -92.4\, kJ\,mol^{-1}\nonumber$
Nitrogen (N2) is very unreactive because the molecules are held together by strong triple bonds. The Haber process relies on catalysts that accelerate the cleavage of this triple bond.
At room temperature, the equilibrium is strongly in favor of ammonia, but the reaction doesn't proceed at a detectable rate. Thus two opposing considerations are relevant to this synthesis. One possible solution is to raise the temperature, but because the reaction is exothermic, the equilibrium quickly becomes quite unfavourable at atmospheric pressure. Low temperatures are not an option since the catalyst requires a temperature of at least 400 °C to be efficient. By increasing the pressure to around 200 atm the equilibrium concentrations are altered to give a profitable yield.
The reaction scheme, involving the heterogeneous catalyst, is believed to involve the following steps:
1. N2 (g) → N2 (adsorbed)
2. N2 (adsorbed) → 2 N (adsorbed)
3. H2(g) → H2 (adsorbed)
4. H2 (adsorbed) → 2 H (adsorbed)
5. N (adsorbed) + 3 H (adsorbed) → NH3 (adsorbed)
6. NH3 (adsorbed) → NH3 (g)
Reaction 5 actually consists of three steps, forming NH, NH2, and then NH3. Experimental evidence suggests that reaction 2 is the slow, rate-determining step. This is not unexpected given that the bond broken, the nitrogen triple bond, is the strongest of the bonds that must be broken. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/10%3A_Hydrogen/10.04%3A_Dihydrogen/10.4C%3A_Synthesis_and_Uses.txt |
The position of H in the Periodic Table
In some respects, H does not seem to have a perfect position in the Periodic Table and so many designers have it in more than one position, e.g. in Group 1 or Group 17 and even in Group 14.
Ionization energy of hydrogen
Hydrogen has a single outer electron, like the alkali metals, but they all form positive ions quite readily whereas hydrogen has little tendency to do so. Hydrogen often tends to share its electron with nonmetals rather than losing it to them.
The first ionization energies for H, Li, Na and K are 1312, 520.2, 495.8 and 418.8 kJmol-1. The high IE for H (even bigger than for Xe) can be attributed to the very small size of the atom and the strong attractive force between the proton and electron.
H(g) → H+(g) + e ΔH = 1312 kJmol-1
Xe(g) → Xe+(g) + e ΔH = 1170 kJmol-1
The free proton can only be obtained under extreme conditions such as by an electric arc or in a discharge tube and even then only exists for about half a second. H+ can be found in solvated form where the solvation energy provides the energy needed to overcome the very high ionization energy. Examples are in ammonia, alcohol or water with species like NH4+, ROH2+ and H3O4+ being formed.
The hydrated proton (H3O4+) will be covered in Lecture 6 on acid-base chemistry.
Electron affinity of hydrogen
Hydrogen, like the halogens, exists as diatomic molecules and H atoms have electron configurations with one electron short of a filled outer shell hence the idea of placing H in Group 17. However unlike the halogens with large EA values, the EA for hydrogen is quite small. The formation of H- is much less favourable than the formation of a chloride ion, as seen from the thermodynamic profiles below and it is rare whereas halide ions are common and stable. In addition H has a lower electronegativity value than any of the halogens.
Much more energy is required as well to break the H-H bond compared to the Cl-Cl bond where the steps for comparison are: ½H2 (g) → H. (g) ΔH = 218 kJmol-1
H. (g) + e → H- (g) ΔH = -72.8 kJmol-1
so overall for hydrogen
½H2 (g) + e → H- (g) ΔH = +145.2 kJmol-1
and
½Cl2 (g) → Cl. (g) ΔH = 121 kJmol-1
Cl. (g) + e → Cl- (g) ΔH = -348.6 kJmol-1
overall for chlorine
½Cl2 (g) + e → Cl- (g) ΔH = -227.6 kJmol-1 As a result, only the most active elements, whose Ionization Energies are low, can form ionic hydrides, e.g. NaH.
The covalent radius for H is 37 pm and the estimated radius for H- is ~140 pm indicating a substantial increase. This comes about as a result of the interelectronic repulsion when a second electron is added to the 1s atomic orbital. All the alkali metal hydrides crystallize with the NaCl-type structure and are all considered ionic. They are sometimes called "saline" hydrides. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/10%3A_Hydrogen/10.04%3A_Dihydrogen/10.4D%3A_Reactivity.txt |
Hydride reducing agents
LiH and Al2Cl6 gives lithium aluminium hydride (lithal LiAlH4), NaH reacts with B(OCH3)3 to give sodium borohydride (NaBH4). These find wide scope and utility in organic chemistry as reducing agents.
LiAlH4 is commonly used for the reduction of esters and carboxylic acids to primary alcohols; previously this was a difficult conversion that used sodium metal in boiling ethanol (the Bouveault-Blanc reduction). The solid is dangerously reactive toward water, releasing gaseous hydrogen (H2). Some related derivatives have been discussed for hydrogen storage.
NaBH4 is used in large amounts for the production of sodium dithionite from sulfur dioxide: Sodium dithionite is used as a bleaching agent for wood pulp and in the dyeing industry. NaBH4 consists of the tetrahedral BH4- anion in the crystalline form and is found to exist as three polymorphs: α, β and γ. The stable phase at room temperature and pressure is α-NaBH4, which is cubic and adopts an NaCl-type structure. Millions of kilograms are produced annually, far exceeding the production levels of any other hydride reducing agent.
NaBH4 will reduce many organic carbonyls, depending on the precise conditions. Most typically, it is used in the laboratory for converting ketones and aldehydes to alcohols. For example, reduction of acetone (propanone) to give propan-2-ol.
10.6B: Trends in Boiling Points Melting Points and Enthalpies of Vaporization for p-block Binary Hydrides
Hydrogen bonds
A hydrogen bond is the name given to the electrostatic attraction between polar molecules that occurs when a hydrogen (H) atom bound to a highly electronegative atom such as nitrogen (N), oxygen (O) or fluorine (F) experiences attraction to some other nearby highly electronegative atom. The name is something of a misnomer, as it represents a particularly strong dipole-dipole attraction, rather than a typical covalent bond.
The 2011 IUPAC definition specifies that "The hydrogen bond is an attractive interaction between a hydrogen atom from a molecule or a molecular fragment X-H in which X is more electronegative than H, and an atom or a group of atoms in the same or a different molecule, in which there is evidence of bond formation."
These hydrogen-bond attractions can occur between molecules (intermolecular) or within different parts of a single molecule (intramolecular). The hydrogen bond (5 to 30 kJ/mole) is stronger than a van der Waals interaction, but weaker than covalent or ionic bonds. This type of bond can occur in inorganic molecules such as water and in organic molecules like DNA and proteins.
Intermolecular hydrogen bonding is responsible for the high boiling point of water (100 °C) compared to the other group 16 hydrides that have no hydrogen bonds. Intramolecular hydrogen bonding is partly responsible for the secondary and tertiary structures of proteins and nucleic acids. It plays an important role in the structure of polymers, both synthetic and natural.
BP's of MG hydrides with Noble gases for comparison /K
10.6E: Hydrogen Bonding in Biological Systems
Hydrogen bonding in biological systems.
Base pairs, which form between specific nucleobases (also termed nitrogenous bases), are the building blocks of the DNA double helix and contribute to the folded structure of both DNA and RNA. Dictated by specific hydrogen bonding patterns, Watson-Crick base pairs (guanine-cytosine and adenine-thymine) allow the DNA helix to maintain a regular helical structure that is subtly dependent on its nucleotide sequence. The complementary nature of this based-paired structure provides a backup copy of all genetic information encoded within double-stranded DNA. The regular structure and data redundancy provided by the DNA double helix make DNA well suited to the storage of genetic information, while base-pairing between DNA and incoming nucleotides provides the mechanism through which DNA polymerase replicates DNA, and RNA polymerase transcribes DNA into RNA. Many DNA-binding proteins can recognize specific base pairing patterns that identify particular regulatory regions of genes.
Hydrogen bonding in base pairs | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/10%3A_Hydrogen/10.05%3A_Polar_and_Non-Polar_E-H_Bonds.txt |
Binary hydrides are a class of compounds that consist of an element bonded to hydrogen, in which hydrogen acts as the more electronegative species. Free hydride anions exist only under extreme conditions. Instead, most hydride compounds have hydrogen centers with a hydridic character.
10.07: Binary Hydrides - Classification and General Properties
Compounds of Hydrogen
The chemistry of hydrogen depends mainly on four processes:
1. donation of the valency electron to form the hydrogen ion, H+
2. accepting an electron to form the hydride ion H-
3. sharing the electron with a partner atom to form a pair bond (covalent bond) H-H
4. sharing the electron with an ensemble of atoms to form a metallic bond H.
While H2 is not very reactive under standard conditions, it does form compounds with most elements. Hydrogen can form compounds with elements that are more electronegative, such as halogens (e.g., F, Cl, Br, I), or oxygen; in these compounds hydrogen takes on a partial positive charge. When bonded to fluorine, oxygen, or nitrogen, hydrogen can participate in a form of medium-strength noncovalent bonding called hydrogen bonding, which is critical to the stability of many biological molecules. Hydrogen also forms compounds with less electronegative elements, such as the metals and metalloids, in which it takes on a partial negative charge. These compounds are often known as hydrides.
The term "hydride" suggests that the H atom has acquired a negative or anionic character, denoted H-, and is used when hydrogen forms a compound with a more electropositive element. The existence of the hydride anion, suggested by Gilbert N. Lewis in 1916 for group I and II saline hydrides, was demonstrated by Moers in 1920 by the electrolysis of molten lithium hydride (LiH), producing a stoichiometry quantity of hydrogen at the anode.
Although hydrides can be formed with almost all main-group elements, the number and combination of possible compounds varies widely; for example, there are over 100 binary borane hydrides known, but only one binary aluminium hydride. A simple binary indium hydride has not yet been identified, although larger complexes exist.
10.7B: Metallic Hydrides
Metallic hydrides (also known as interstitial hydrides) involve hydrogen bonds with transition metals. One interesting and unique characteristic of these hydrides are that they can be nonstoichiometric, meaning basically that the fraction of H atoms to the metals are not fixed. Nonstoichiometric compounds have a variable composition. The idea and basis for this is that with metal and hydrogen bonding there is a crystal lattice that H atoms can and may fill in between the lattice while some might, and is not a definite ordered filling. Thus it is not a fixed ratio of H atoms to the metals. Even so, metallic hydrides consist of more basic stoichiometric compounds as well.
10.7C: Saline Hydrides
Saline hydrides (also known as ionic hydrides or pseudohalides) are compounds form between hydrogen and the most active metals, especially with the alkali and alkaline-earth metals of group one and two elements. In this group, the hydrogen acts as the hydride ion ($H^-$). They bond with more electropositive metal atoms. Ionic hydrides are usually binary compounds (i.e., only two elements in the compound) and are also insoluble in solutions.
$2A_{(s)} + H_{2(g)} \rightarrow 2AH_{(s)} \tag{3}$
with $A$ as any group 1 metal.
$A_{(s)} + H_{2(g)} \rightarrow AH_{2(s)} \tag{4}$
with $A$ as any group 2 metal.
Ionic hydrides combine vigorously with water to produce hydrogen gas.
Example $1$: Alkali Metal Hydrides
As ionic hydrides, alkali metal hydrides contain the hydride ion $H^-$ as well. They are all very reactive and readily react with various compounds. For example, when an alkali metal reacts with hydrogen gas under heat, an ionic hydride is produced. Alkali metal hydrides also react with water to produce hydrogen gas and a hydroxide salt:
$MH_{(s)} + H_2O_{(l)} \rightarrow MOH_{(aq)} + H_{2(g)}$
The instability of the hydride ion compared to the halide ions can be seen by comparison of the ΔHf for alkali metal hydrides and chlorides.
Cation ΔHf MH
/ kJmol-1
ΔHf MCl
/ kJmol-1
Li -90.5 -409
Na -56.3 -411
K -57.7 -436
Rb -52.3 -430
Cs -54.2 -433
Saline hydrides are formed by the Group 1 and 2 metals when heated with dihydrogen (H2). They are white, high melting point solids that react immediately with protic solvents, for example:
$NaH + H_2O → NaOH + H_2$
(Their moisture sensitivity means that reaction conditions must be water-free.)
Evidence for the ionic nature of these hydrides is:
1. molten salts show ionic conductivity.
2. X-ray crystal data gives reasonable radius ratios expected for ionic compounds.
3. Observed and calculated Lattice Energies (from Born-Haber cycles etc.) are in good agreement (i.e. show little covalency).
NaH is capable of deprotonating a range of even weak Brønsted acids to give the corresponding sodium derivatives.
$NaH + Ph_2PH → Na[PPh_2] + H-2$
Sodium hydride is sold by many chemical suppliers as a mixture of 60% sodium hydride (w/w) in mineral oil. Such a dispersion is safer to handle and weigh than pure NaH. The compound can be used in this form but the pure grey solid can be prepared by rinsing the oil with pentane or tetrahydrofuran, THF, care being taken because the washings will contain traces of NaH that can ignite in air. Reactions involving NaH require an inert atmosphere, such as nitrogen or argon gas. Typically NaH is used as a suspension in THF, a solvent that resists deprotonation but solvates many organosodium compounds. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/10%3A_Hydrogen/10.07%3A_Binary_Hydrides_-_Classification_and_General_Properties/10.7A%3A_Classification.txt |
Molecular hydrides - covalent hydrides and organic compounds
Hydrogen forms a vast number of compounds with carbon, (the hydrocarbons), and an even larger array with heteroatoms that, because of their general association with living things, are called organic compounds. The study of their properties is covered in organic chemistry and their study in the context of living organisms is covered in biochemistry. By some definitions, "organic" compounds are only required to contain carbon. However, most of them also contain hydrogen, and because it is the carbon-hydrogen bond which gives this class of compounds most of its particular chemical characteristics, carbon-hydrogen bonds are required in some definitions of the word "organic" in chemistry. Millions of hydrocarbons are known, and they are usually formed by complicated synthetic pathways, which seldom involve direct reaction with elementary hydrogen.
Most molecular hydrides are volatile and many have simple structures that can be predicted by the VSEPR model. There are a large number of B hydrides known (boranes) and although the simplest BH3 has been found in the gas phase it readily dimerises to give B2H6.
In inorganic chemistry, hydrides can serve as bridging ligands that link two metal centers in a coordination complex. This function is particularly common in group 13 elements, especially in boranes (boron hydrides) and aluminium complexes, as well as in clustered carboranes, (composed of boron, carbon and hydrogen atoms). The bonding of the bridging hydrogens in many of the boranes is explained in terms of 3 centre - 2 electron bonds.
Diborane is a colourless and highly unstable gas at room temperature with a repulsively sweet odour. Diborane mixes well with air, easily forming explosive mixtures. Diborane will ignite spontaneously in moist air at room temperature.
MP = -164.85 °C, BP= -92.5 °C
B-H (terminal) 119 pm, (bridge) 131 pm
Metallic (interstitial) hydrides
Many transition metal elements form metallic (interstitial) hydrides, in which H2 molecules (and H atoms) can occupy the holes in the metal's crystal structure. They are traditionally termed 'compounds', even though they do not strictly conform to the definition of a compound; more closely resembling common alloys such as steel. These systems are usually non-stoichiometric, with variable amounts of hydrogen atoms in the lattice.
Palladium is unique in its ability to reversibly absorb large amounts of H2 or D2 (up to 900 times its own volume of hydrogen, but no other gases, at room temperature) to form palladium hydride. Structural studies show that the absorbed H fits into octahedral holes in the cubic close packed Pd lattice with a non-stoichiometric formula approximating to PdH0.6 for the β-form. This material has been considered as a means to carry hydrogen for vehicular fuel cells. Interstitial hydrides show some promise as a way for safe hydrogen storage. During the last 25 years many interstitial hydrides have been developed that readily absorb and discharge hydrogen at room temperature and atmospheric pressure. At this stage their application is still limited, as they are capable of storing only about 2 weight percent of hydrogen, insufficient for automotive applications.
10.7E: Covalent Hydrides with Extended Structures
Covalent hydrides refer to hydrogen centers that react as hydrides, or those that are nucleophilic. In these substances, the hydride bond, formally, is a covalent bond much like the bond that is made by a proton in a weak acid. This category includes hydrides that exist as discrete molecules, polymers, oligomers, or hydrogen that has been chem-adsorbed to a surface. A particularly important type of covalent hydride is the complex metal hydride, a powerful (reducing) soluble hydride that is commonly used in organic syntheses (for example, sodium borohydride or NaBH4). Transition metal hydrides also include compounds that can be classified as covalent hydrides. Some are even classified as interstitial hydrides and other bridging hydrides. Classical transition metal hydrides feature a single bond between the hydrogen center and the transition metal.
• Wikipedia | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/10%3A_Hydrogen/10.07%3A_Binary_Hydrides_-_Classification_and_General_Properties/10.7D%3A_Molecular_Hydrides_and_Complexes_Derived_from_them.txt |
The alkali metals make up Group 1 of the periodic table. This family consists of the elements lithium, sodium, potassium, rubidium, cesium, and francium (Li, Na, K, Rb, Cs, and Fr, respectively). Group one elements share common characteristics. They are all soft, silver metals. Due to their low ionization energy, these metals have low melting points and are highly reactive. The reactivity of this family increases as you move down the table. Alkali metals are noted for how vigorously they react with water.
11: Group 1 - Alkali Metals
Chemistry 242 - Inorganic Chemistry II Chapter 10 - Lithium, Sodium, Potassium, Rubidium and Caesium
Introduction
• The common metals of this group are sodium (2.6% in the lithosphere, NaCl) and potassium (2.4%, KClMgCl2.6H2O, carnallite).
• Salt mines are considered to be good places to leave radioactive waste because they are not subject to groundwater flow - that is why the salt got left there in the first place!
• NaOH, Na2CO3, Na2SO4, Na3P3O9 and Na4SiO4 are among the top 50 industrial chemicals.
• Potassium salts, notable K2SO4 or KNO3, are an important component of fertilizer.
• Lithium alkyls are important reagents in synthesis.
• Na+ and K+ are very important physiological ions, and Li+ salts are used to treat certain mental disorders.
• Except Li, the chemistry is predominantly that of the M+ ions. There may be a hint of covalentcy in certain chelate complexes.
• Li+ would have a ratio of charge/radius similar to Mg2+ hence certain similarities.
• Other "look alikes" are NH4+ which is similar to K+ and Tl+ which is a bit like Rb+ or Ag+.
Preparation and Properties
• Lithium and sodium are made by electrolysis of the molten salts or low melting eutectics.
• potassium, rubidium and cesium are made by the reaction of sodium vapour with the molten chlorides:
MCl(l) + Na(g) NaCl(l) + M(g)
Their vapours are more volatile so that the equilibrium (which is on the side of the halides of the metals desired!)) is disturbed.
• The metals are all silvery in colour except Cs which is yellow. They can all be cut with a knife, lithium being the hardest.
• A sodium - potassium (77.2%) alloy melts at -10 and is a useful, if dangerous, reducing agent.
• The metals are normally protected from air under oil. Lithium, sodium and potassium can be handled quickly in air, and rubidium and cesium are handled under argon.
• All react with water to produce hydrogen: lithium slowly, sodium vigorously, potassium violently enough to ignite the hydrogen, and the rest explosively.
• The products of combustion in a free supply of air are:
Li Li2O (oxide) 2OH- in water.
Na Na2O2 (peroxide) 2OH- + 2H2O2 in water
M(K, Rb and Cs) MO2 (suboxide) O2 + 2OH- + H2O2 with water
The difference is attributed to lattice energy effects, that is it is the size of the metal ion that makes one of the potential oxides more stable.
• All metals will reat with alcohols to give the alkoxide.
• All dissolve in mercury to give amalgams. The most useful is th liquid amalgam formed when less than about 6% Na is dissolved in mercury. It is a relatively gentle reducing agent compared to pure sodium.
The Metals in Liquid Ammonia
They all dissolve in ammonia to give solutions which are a beautiful royal blue when relatively dilute and take on a metallic bronze appearance when concentrated. The most important equilibria in the more dilute solutions are:
Na(s) Na(NH3) Na+(NH3) + e-)NH3)
2e-(NH3) e2-(NH3)
The electrons, which are responsible for the blue colour, are trapped in 3.0-3.4 Å cavities in the solvent. The solutions have a lower density than the pure solvent as a result.
At high concentrations, metal atoms cluster, and the solutions become quite metallic in properties, thus the appearance and high electrical conductivity.
Solutions of sodium in liquid ammonia are slowly decomposed by light an drapidly by the catalytic effect of transition metal ions such as Fe(III) to give sodamide and hydrogen:
Na(NH3) + NH3 NaNH2(s) + ½H2(g)
For potassium, rubidium an dcesium, whose amides are soluble in liquid ammonia the reaction is reversible under hydrogen pressure:
e-(NH3) + NH3 NH2- + ½H2(NH3) K = 5x104
The alkali metals are also slightly soluble in other amines, THF and glymes but to a much lesser extent.
Compounds
Oxides - see above.
Hydroxides - The most important are NaOH and KOH which are very deliquescent waxy looking solids, usually sold a pellets or flakes. They are very corrosive alkaline compounds which should be handled with care.
Ionic Salts - Salts of virtually all acids are known. They are colourless unless teh anions are coloured, or there are lattice defects. Lithium is different:
1. Li3N is formed slowly from Li and N2 at room temperature and is ruby red.
2. LiOH is a "covalentish" OH bridged polymer which is not very alkaline compared to the rest.
3. LiF is rather insoluble.
4. LiCl and LiBr are quite soluble in a number of polar organic solvents such as alcohols, acetone, ethyl acetate and pyridine (LiCl).
5. Lithium salts are often hydrated, e/g/ LiClO4.3H2O.
Insoluble salts of the others:
Finding insoluble salts for identification and gravimetric analysis was difficult - there are so few! Examples are: NaZn(UO2)(CH3CO2)9.6H2O and K+3[Co(NO2)6]3- (or Rb+ or Cs+).
Hydrates and Complexes in Solutions
Hydrates
See table 10-1. Li+, Na+ and K+ probably have 4 molecules of water in their first (or primary) hydration sphere, while Rb+ and Cs+ probably have 6. The larger the central ions, the smaller the area of ordering of the water around it, so the effective size of the ions decreases going down the group. This is important in understanding the mobility of the ions, for example down an ion exchange column.
The Crown Ethers and Cryptands
The alkali metals are complexed quite strongly by THF and glymes, but the effect becomes really marked for the so-called "crown ethers". Two examples are shown below:
Each of these crown ethers has an affinity fro a particular metal ion, for example, for 18-crown-6, the binding constants are in the order:
Li+ < (Na+,Cs+) < Rb+ < K+ That is, this crown ether likes K+ best.
Li+ is most strongly bound in dicylohexyl-14-crown-4
Na+ "fits" well in benzo-15-crown-5
Rb+ "fits" best in dicyclohexyl-21-crown-7
Cs+ "fits" best in dicyclohexyl-24-crown-8
The stability orders differ depending on the method of comparison (calculation, gas-phase, solution etc): experimetally in solution, it appears that any crown with –CH2CH2– bridges prefers K+ because of the 5-membered chelate ring size rather than the size of the hole on the crown - the macrocycle just puckers up to fit,but solvent effects may also be very important.
The complexes are used to get normally insoluble ionic compounds into organic solutions and can also help produce metal electrides like in ammonia.
Cryptands and Cryptates
Cryptands are polycyclic cages, usually including nitrogen as well as oxygen to get the necessary junctions. Metal ions are encapsulated even more securely inside them leadin gto cryptates.
Some remarkable compounds have been made:
2Na(NEt3) + 2,2,2-crypt [Na(2,2,2-crypt)]+Na-(s)
Bear in mind that:
2Na Na+ + Na- DH = 438 kJ mol-1
This compound is stable up to -10 oC and has a structure similar to [Na(2,2,2-crypt)]+I- which has normal stability. There is also a [Na(2,2,2-crypt)]+e- known.
Encapsulated Metals in Biology
The transport of alkali metal ions, notably through cell walls, is biologically important and involves certain natural macrocyclic compounds such as valinomycin (Fig 10-VII) and nonactin (Fig 10-3).
Organometallic Compounds
Lithium Compounds
The lithium compounds are very important synthetic reagents. They can be made in hydrocarbon solvents (which is how they are sold) by reactions such as:
C2H5Cl + Li C2H5Li + LiCl(S)
C4H9Li + CH3I C4H9I + CH3Li
The pure compounds are air-sensitive low melting solids or liquids and are associated into small aggregates with multicentre bonding e.g. Li4(CH3)4 (Figure 10-4) or Li6(C2H5)6.
Organosodium and Potassium Compounds
The organometallic compounds of sodium and potassium are predominantly ionic. The most important are NaC5H5 (made from Na in l-NH3 and C5H5 monomer: it is usually a Diels-Alder dimer) and NaCºCR.
Other Alkali Metal Compounds
Look at this section independently.
Summary
1. Lithium:
1. Reacts relatively slowly with water or oxygen, but readily forms the nitride.
2. Has a marked tendency towards covalency, notably in its organometallic compounds.
3. Is often hydrated in its "ionic" compounds.
4. The hydroxide is not a strong base.
5. Some salts are not very soluble in water, but do dissolve in donor organic solvents.
2. Sodium, Potassium, Rubidium and Caesium:
1. Are all very reactive with water and oxygen. Nitrides are not stable at room temperature.
2. Compounds are always predominantly ionic.
3. Hydroxides are strong bases.
4. Salts are almost all water soluble.
3. All elements in the group:
1. Form blue reducing solutions in ammonia.
2. Form stable complexes with crown ethers or cryptands which are significantly soluble in organic solvents. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/11%3A_Group_1_-_Alkali_Metals/11.01%3A_Introduction.txt |
Alkaline earth metals make up the second group of the periodic table. This family includes the elements beryllium, magnesium, calcium, strontium, barium, and radium (Be, Mg, Ca, Sr, Ba, and Ra, respectively). Group 2 elements share common characteristics. Each metal is naturally occurring and quite reactive. These metals are silver and soft, much like the alkali metals of Group 1. These metals also react with water, though not as vigorously. Beryllium, interestingly, does not react with water. E
12: Goup 2- Alkaline Earth Metals
• Beryllium is found in the mineral beryl, Be3Al2(SiO3)6. Beryl with around 2.9% Cr3+ substituting for Al3+ is emerald.
• Beryllium compounds are dangerously toxic and can pass through the skin - care!
• The other elements are faily common in a variety of minerals e.g. limestone, CaCO3, dolomite, CaCO3.MgCO3 and carnallite, KCl.MgCl(6H2O). Less common are strontianite, SrCO3, and barytes, BaCO3.
• Radium, the bottom member of the group is radioactive with a half-life (226Ra)of about 1600 years as a a-emitter. It was first isolated from uranium ore, pitchblende, by M. and P. Curie, by laborious frational cyrstallizations.
• Atomic radii are smaller than those group 1, while there are two valence electrons. Therefore, the elements are harder, with higher melting and vaporizing points. The enthalpies are correspondingly higher.
• The ionic radii of the M2+ ions are much smaller than group 1 M+ ions so that hydration and lattice energies compensate for the higher ionization enthalpies needed to reach the M2+ state.
• The beryllium ion is too small to form many ionic compounds, and magnesium ion is also quite small so they are treated separately.
• Calcium, Strontium and Barium are a "well-behaved" group of three elements with a nice gradual change in properties.
Beryllium
Beryllium is obtained by the reduction of BeCl2 by Ca or Mg or the Mg reduction of BeF2. It is a very light metal, unreactive with air or water at ordinary temperatures. It dissolves in strong non-complexing acids (except HNO3 which passivates it) to give the [Be(H2O)4]2+ ion. Be also dissolves in strong aqueous bases like NaOH to evolve hydrogn and yield the beryllate, Be(OH)42- ion. Its amphoteric behaviour is similar to aluminum. Beryllium salts, which usually come hydrated, are acidic in solution due to hydrolysis:
[Be(H2O)]2+ + H2O [Be(H2O)3(OH)]+ + H3O+
Beryllium compounds are fairly covalent and the chemistry is dominated by the Be attempting to obtain an octet. There are two important strategies identified:
1. The simple Beryllium compounds can react with Lewis bases forming, for example, BeCl2(OEt2)2, and complex anions are also formed e.g. BeF42-, and [Be(H2O)4]2+.
2. Polymerization occurs through bridging groups to give chain polymers, e.g. (BeF2)n, (BeCl2)n and (Be(CH3)2)n. In the halides, the bridges are not electron deficient, but in the alkyls, three-centre, two-electron bonds must be invoked. For bulkier groups, for example certain alkoxides, the chain lengths can be reduced, or polymerization can even be inhibited, e.g. (Be(OBut)2)3 and Be(OC6H4(But)2)2 (below): In the gas phase the polymers break down so that the chloride is linear BeCl2. By rapidly cooling the vapour, short aggregates with two or three Be's can be isolated. Short chain anions such as M2[Be4Cl10] (M+ = K+, Rb+, Tl+ or NH4+)have similar structures.
The compounds BeO and BeS have the wurtzite and blende structures respectively (both are known for ZnS - see Chapter 4) but the bonding would have to be considered quite covalent. Be(OH)2 in insoluble in water.
"Basic beryllium acetate", Be4O(CH3CO2)6, which is soluble in nonpolar solvents e.g. benzene, has a tetrahedral arrangement of Be atoms around the oxygen atom, and an acetate bridging each of the six edges of the tetrahedron (Figure 11-2).
Magnesium
Magnesium, a rather important metal, is mostly made from dolomite and/or seawater. Magnesium can be concentrated using both sources as follows: Dolomite is heated to give an intimate mixture, MgO.CaO, and then the magnesium is ion exchanged out using seawater:
MgO.CaO + 2H2O Mg(OH)2.Ca(OH)2
Mg(OH)2.Ca(OH)2 + Mg2+ 2Mg(OH)2 + Ca2+
This works because Mg(OH)2 is much less soluble than Ca(OH)2.
There are three common ways to get the magnesium metal itself:
1. The "calcined" dolomite is be heated with ferrosilicon, and the magnesium distills out:
MgO.CaO + FeSi Mg + ill-defined silicates of Ca and Fe
2. A low-melting mixture of MgCl2, CaCl2 and NaCl is electrolysed. Magnesium is preferentially reduced at the cathode.
3. Magnesium oxide is reduced with coke at 2000 oC which gives an equilibrium mixture of Mg and CO. Rapid cooling separates the metal from the gaseous CO.
Magnesium is a silvery-white metal protected by an oxide coating. It is attacked by acids (which dissolve the oxide film first) with the evolution of H2 (even with HNO3, which usually reacts to give NO or NO2). It will react with water if its surface is amalgamated. It is used in ultra-light alloys and it is the metal at the centre of chlorophyll.
Magnesium behaves somewhere between Be and the rest of the group 2 elements. Its hydroxide is not very water soluble, while the rest are, and it has some pretty covalent organo-compounds, the grignard reagents are the most important.
Calcium, Strontium and Barium
These metals are made by reduction of their halides by sodium in relatively small quantities. They are all softish and silvery when freashly cut, but they react readily with oxygen and water. Calcium is most used to make the hydride, CaH2, a useful reducing and drying agent.
Binary Compounds
Oxides
• They are all white high-melting compounds with the NaCl structure.
• The most important is probably CaO, made from CaCO3 and is a major ingredient in cement.
• All except MgO, which can be fairly inert, produce a soluble, stongly basic hydroxide with water, and the carbonate with CO2. MgO quite is insoluble in water and is only very mildly basic. It is used as a stomach antacid ("milk of magnesia").
Halides
• Anhydrous calcium chloride is an important drying agent, often mixed with a little cobalt chloride which is blue when anhydrous and pink when hydrated and acts as an indicator of the condition of the dessicant. Magnesium Chloride is also hygroscopic. The affinity for water decreases down th egroup so that Sr and Ba halides are normally anyhdrous.
Other Compounds
• Calcium carbide, CaC2, made from CaO and carbon at high temperature can be a source of acetylene.
• The hydrides are all ionic although MgH2 has come covalent character.
Oxo Salts, Ions and Complexes
• All form oxo salts (CO32-, SO42- etc). MgSO4.7H2O is "Epsom salts", a mild laxative, MgCO3 is used as an antacid. CaSO4.½H2O is known as "plaster of Paris" which sets with water to give CaSO4.2H2O, "gypsum". BaSO4 is used as an imaging material to obtain medical X-rays of the intestinal tract.
• They become less soluble down the group.
• In aqueous solution the ions are probably at least 6-coordinate.
• There are some chelate complexes with oxygen ligands, the best known being [Ca(EDTA)]2- and also Capolyphospates which can be used to sequester Ca an dalso in its analysis.
• The strong complexation of Mg by porphyrins is quite unusual since the ligating atoms are all nitrogen.
Summary
1. Beryllium:
1. Forms covalent compounds almost exclusively, even with the most electronegative elements.
2. Does not form Be2+ compounds, but readily achieves a maximum coordination number of 4 by forming complex ions e.g. BeF42- and [Be(H2O)4]2+ with essentially covalent bonding within the complex.
3. Forms covalent organometallic compounds.
4. Oxide and specially the hydroxide are amphoteric. (The hydroxide is water soluble.)
5. The hydride, halides and alkoxides are oligomeric or polymeric with covalent bridges, and are easily cleaved by Lewis bases.
2. Magnesium:
1. Forms ionic compounds with partial covalent character.
2. Forms many compounds containing the uncomplexed Mg2+ ion and tend to be 6-coordinated by the counter ions.
3. Forms important organometallic compounds which are mainly covalent.
4. Oxide and especially the hydroxide are weakly basic. (The hydroxide is not water soluble.)
5. The halides are essentially ionic.
6. The hydride is partly covalent.
3. Calcium, Strontium and Barium:
1. Form only ionic substances.
2. Form basic oxides and strongly basic hydroxides, more and more soluble down the group.
3. Form readily hydrated halide salts, especially towards the top of the group.
4. Hydrides are ionic.
12.10: Diagonal Relationships between Li and Mg and between Be and Al
A Diagonal Relationship is said to exist between certain pairs of diagonally adjacent elements in the second and third periods of the periodic table. These pairs (Li & Mg, Be & Al, B & Si etc.) exhibit similar properties; for example, Boron and Silicon are both semiconductors, form halides that are hydrolyzed in water and have acidic oxides. Such a relationship occurs because crossing and descending the periodic table have opposing effects. On crossing a period of the periodic table, the size of the atoms decreases, and on descending a group the size of the atoms increases. Similarly, on moving along the period the elements become progressively more covalent, less reducing and more electronegative, whereas on descending the group the elements become more ionic, more basic and less electronegative. Thus, on both descending a group and crossing by one element the changes cancel each other out, and elements with similar properties which have similar chemistry are often found - the atomic size, electronegativity, properties of compounds (and so forth) of the diagonal members are similar. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/12%3A_Goup_2-_Alkaline_Earth_Metals/12.01%3A_Introduction.txt |
Group 13 is sometimes referred to as the boron group, named for the first element in the family. These elements are--not surprisingly--located in column 13 of the periodic table. This group includes boron, aluminum, gallium, indium, thallium, and ununtrium (B, Al, Ga, In, Tl, and Uut, respectively). These elements all have three valence electrons. Boron is the only metalloid in this family. The rest of the elements are considered to be poor metals.
Thumbnail: Crystals of 99.999% gallium. (CC-SA-BY 3.0; Foobar).
13: The Group 13 Elements
Introduction
• Sources:
• Ulexite: NaCa[B5O6(OH)6].5H2O
• Borax: Na2[B4O5(OH)4].8H2O
• Colmanite: Ca2[B3O4(OH)3].2H2O
• Kenite: Na2[B4O5(OH)4].2H2O
See the figure for the borate anions in te above.
• There are no ionic B3+ compounds, c.f. Mg2+ and Li+.
• The compounds which are coordinatively unsaturated (e.g.BCl3) are very strong Lewis bases.
• Tetrahedral adducts and anions are common, for example:
F3B:O(C2H5)2 BF4- B(C6H5)4-
• The boron hydrides (boranes) are a class unto themselves together with the carboranes and all the anions. They frequently feature closed and open polyhedra based on fragments of an icosahedron, but not always.
• Boron resembles silicon more than aluminum in some ways:
• B2O3 and B(OH)3 are acidic rather like SiO2 and Si(OH)4 whereas the Al compounds are weakly amphoteric.
• The borates have some features in common with the silicates.
• The halide compounds of B and Si are readily hydrolysed (except BF3) whereas the halide compounds of Al are only partly hydrolysed.
• The B and silicon hydrides are volatile molecular compounds, which inflame in air, while AlH3 is an involatile solid.
Manufacture and Properties of Boron
Boron is quite difficult to isolate, because it is refractory and reactive at high temperature and so it is difficult to contain:
1. B2O3 + 3Mg 2B + MgO (98% - Wash with NaOH, HCl and HF)
2. 2BCl3 + 3Zn 3ZnCl2 + 2B (900 oC)
3. 2BX3 + 3H2 6HX + 2B (Tantalum catalyst)
Boron is rather inert in most forms which contain icosahedral cages. It is attacked by hot oxidizing acids.
Amorphous boron is more reactive, if white hot, and is attacked by NH3 to form boron nitride which is isomorphous with graphite.
Oxygen Compounds of Boron
See Figure 12-1.
The "anhydrous" borates involve the ions: BO33-, B3O63-, (BO2)nn-, and larger aggregates.
The hydrated ones feature: BO3 units which are planar and BO4 unitsd which are tetrahedral and formally have a -ve charge on the boron. The charge on the ion is equal to th number of these latter units. The structures without BO4 units hydrate readily.
Boric Acid
See Figure 12-2 for a summary.
• Note that B2O3 + SiO2 is "pyrex".
• Note that B(OH)3, normally a weak acid, can be considerably strengthened to the point where it can be titrated with NaOH by "chelation" by an organic compound with neighbouring OH groups e.g. glycerol:
• In general, equilibria involve the processes:
B-O-H + H-O-B B-O-B + H2O
B-O-H + O-H- B-O- + H2O
Halides of Boron
Trihalides
BF3 is the most important and is used on an industrial scale. It is a gas boiling at -101 oC.
It is a very strong Lewis acid:
BF3 + :F- BF4-
BF3 + :L F3B:L
Unlike the others, BF3 is only partly hydrolysed:
4BF3 + 6H2O 3H3O+ + 3BF4- + B(OH)3
BF4- + H2O BF3(OH) + HF
(BCl3 + 3H2O B(OH)3 + 3HCl)
In synthetic organic chemistry it is used as follows:
The conversion of ethers or alcohols with acids to esters e.g:
H+ + RCOOH [RCOOH2]+
[RCOOH2]+ + BF3 [RCO}+ + F3B:OH2
[RCO]+ + R'OH RCOOR' + H+
Friedel-Crafts alkylations and acylations:
RX + BF3 R+ + BF3X-
R+ + PhH PhR + H+
H+ + BF3X- BF3 + HX
Fluoroboric acid: "HBF4" is sold as a 40% solution in water. It is a strong acid and source of BF4- ions useful for crystallizations where a coordinating anion is to be avoided.
Reactions of the Trihalides of Boron.
1. Adduct formation - the main thing to remember that the order of acid strength is counter to naive expectations because of p-bonding effects whnich are strongest for B-F bonds and lead to BF3 being least willing to go from sp2 to sp3. hybridized.
2. Halide exchange reactions:
BCl3 + BBr3 BCl2Br + BBr2Cl
The exchange, presumably through a bridged intermediate, is very facile, so pure mixed compounds cannot be obtained.
3. Elimination of halide - covers the various solvolyses in addition to hydrolysis:
BCl3 + 3C2H5OH B(OC2H5)3 + 3HCl
BCl3 + 3NH(C2H5)2 B(N(C2H5)2)3 + 3HCl
This will happen with any solvent with an exchangeable H.
Subhalides of Boron
They have a B:X ratio less than 1:3
• There are halides BF and BCl.
• There are halides B2X4 known for F, Cl, Br and I. Rotation about the B-B bond is easy. Any multiiple bonding tendency involves the boron to halogen bonds as in BX3. (This is in contrast with the aminoboranes - see below.)
• BnXn compounds have polyhedral cages of boron each carrying one X group. The largest range of compounds is known for Cl where n = 4, 8, 9, 10 and 11. B4Cl4 is tetrahedral, B8Cl8 is a triangular dodecahedron, and B9Cl9 is a rectangularly tricapped trigonal prism. Look them up!
The Hydrides of Boron - the Boranes
Table 12-1 lists the hydrides up to B10H14
Figure 12-4 shows some of their structures as perspective drawings. Note that the lines are intended to clarify the shape of the molecule and do not necessarily represent 2e- - 2-centre bonds.
Note also the nomenclature - the prefix gives the number of boron atoms and the number in parentheses the number of hydrogen atoms, e.g. pentaborane(9) is B5H9.
Synthesis
Diborane (b.p. -92.6 oC can be made by several methods:
3NaBH4 + 4BF3 2B2H6 + 3NaBF4
2NaBH4 + I2 B2H6 + 2NaI + H2
BF3 + 6NaH B2H6 + 6NaF
The last is the main industrial method. The higher boranes are made by thermolysis of diborane under various conditions.
Structures
The connectivities in the boranes cannot be explained using 2e- - 2-centre bonding only, that is the molecules are electron deficient. Valence bond theory has been "extended" by designating three types of 2e- - 3-centre bonding in addition to a normal 2e- - 2-centre B-H and B-B bonds:
An example of the use of this scheme for B10H14 is shown below:
Each hydrogen must have one bond ending at it and each boron must have a total of 4 bonds ending at it. If an atom is in the middle of a three-centre bond, the curved line passing through it counts as only one bond.
Thus, for example, B6 has one normal bond to a terminal hydrogen and one normal bond to B2, plus it is at the end of two three-centre bonds through the bridging hydrogens for a total of four bonds i.e eight electrons.
B2 has one normal bond from B6, one normal bond from its terminal hydrogen, is at the end of a "closed" three-centre bond from B1 and B2, and it is in the centre of an "open" three-centre bond from B5 to B7. This is also equivalent to four bonds.
In some cases, more than one "resonance" (canonical) structure can be formulated to account for the observed molecular shape.
Diborane
1. With oxygen (explosive):
B2H6 + 3O2 B2O3 + 3H2O
2. With water:
B2H6 + 3H2O B(OH)3 + 6H2
or alcohols:
B2H6 + 3HOR B(OR)3 + 6H2
3. Substitution reactions:
B2H6 + HCl B2H5Cl + H2
B2H6 + 6Cl2 2BCl3 + 6HCl
4. Cleavage with Lewis bases:
Symmetrical:
B2H6 + N(CH3)3 2H3BN(CH3)3
Unsymmetrical:
B2H6 + 2NH3 [H2B(NH3)2]+[BH4]-
5. Reduction:
2B2H6 + 2Na NaBH4 + NaB3H8
B2H6 + NaBH4 NaB3H8 + H2
5B2H6 + 2NaBH4 Na2B12H12
Pentaborane(9)
This molecule illustrates two general trends: Attack by bases can remove the somewhat acidic bridging hydrogen:
B5H9 + NaH NaB5H8 + H2
The pyramidal B5H9 loses one of its four bridging hydrogens and the resulting ion is "fluxional", that is, the location of the missing bridge is not stationary, and all the atoms in the base of the pyramid (four borons, four terminal hydrogens and three bridging hydrogens) appear equivalent on the time scale of nmr experiments which might otherwise have distinguished them.
Attack by electrophiles can lead to substitution at the apex of the pyramid:
B5H9 + I2 B5H8I(apical) + HI
Decaborane(14)
Once again the bridging hydrogens can be removed by base:
B10H14 + OH- B10H13- + H2O
or converted to terminal hydrogens by reducing agents:
B10H14 + 2Na Na2B10H14
In this reaction, the product has two bridging hydrogens between B1 and B5 and B7 and B8.
Other nucleophiles will add at B6 and B9 with loss of two bridging hydrogens. Again, the two that are left bridge between B1 and B5 and B7 and B8:
B10H14 + 2CH3CN 6,9-(CH3CN)2B10H12 + H2
Electrophiles substitute terminal hydrogens at the bottom of the "basket" in the 1 and 3 or 2 and 4 positions, e.g.:
B10H14 + I2 2,4-I2B10H12 + 2HI
There are two reactions that lead to a closed cage:
B10H14 + 2Et3NBH3 [Et3NH]+2[B12H12]2-
and
(SEt2)2B10H14 + HCCH B10C2H12 + H2 + 2SEt2
Polyhedral Borane Anions and Carboranes
Realize that two carbon atoms can replace two B-'s in a closo-borane anion BnHn2-. Derived molecule or molecule ions are the nido structures, which are missing one vertex relative to the closo structure and the more open arachno structures which are missing two vertices. Figure 12-12 shows structures from B4 to B12. see also Figure 12-8.
Skip the chemistry of these species.
The Tetrahydroborate Ion (BH4-)
This is an important reducing agent, source of H-, and reagent to make other less ionic borohydrides.
NaBH4 is stable in dry air and alkaline aqueous solution. (It will react with water initially but the reaction stops as the concentration of the hydrolysis product, sodium borate, builds up.)
LiBH4 is similar to NaBH4 but more sensitive to water.
Al(BH4)3 is liquid which explodes with air or water. It probably has pairs of hydrogen bridges like diborane.
Zr(BH4)4 is a molecular solid with three bridging hydrogens connecting each boron to the zirconium.
Boron-Nitrogen Compounds
The following are really equivalent representations, but the text uses the right-hand one to indicate a weaker bond:
Amine Boranes
This is the class of amine - BH3 Lewis adducts. They contain the unit shown above. The safest synthesis is:
H3NRCl + LiBH4 H2RN:BH3 + LiCl + H2
Aminoboranes
These have the structure:
The molecules are flat, and rotation about the B—N bond is restricted, therefore the left-hand structure must be is a significant contributor. It is perhaps not correct to represent them as canonical structures since the geometries would be so different. The cleanest synthetic route is, for example:
(CH3)2NH + BCl3 (CH3)2HN:BCl3
(CH3)2HN:BCl3 (CH3)2N:BCl2 + HCl (on heating)
(CH3)2N:BCl3 + 2RMgBr (CH3)2N:BR2 + 2MgClBr
Borazine
This six membered ring can be synthesised by several routes:
Unlike benzene, borazine undergoes addition reactions:
Notice where the Hd+ and the Cld- end up: This illustrates how unrealistic the formal charges on the boron and nitrogen atoms really are!
Like benzene, borazine can form p-complexes with transition metals:
Aluminium, Gallium, Indium and Thallium
Aluminium is the most common metallic element in the crust of the earth, but the common minerals, for example, felspars and micas are rather difficult to process. Aluminum is obtained from bauxite, (Al2O3.nH2O) and cryolite, Na3AlF6 by electrolysis. Gallium and Indium occur in traces in bauxite and all three are found in certain sulphite ores of other metals. While aluminium is obviously the most important, gallium is used in gallium arsenide semiconductors. The elements are all much more metallic than boron, but there are a number of borderline covalent compounds. All are trivalent, but for thallium the univalent state (Tl+) becomes the dominant state as covalent bond strenghts diminish down the group. Some thallium III compounds are thermodynamically unstable:
$\ce{TlX3 -> TlX + X2} \nonumber$
The MX3 compounds (halides and organometallic) are Lewis bases like boron. The strengths vary in the sequence:$\ce{B > Al > Ga > (In ~ Tl)} \nonumber$
The trihalides are not monomers like BX3 but are more or less associated e.g. Al2Cl6 and (AlF3)n.
All give aqua ions [M(H2O)6]3+ in their salts obtained from aqueous solution.
Occurence, Extraction and Properities of the Elements
To obtain aluminum, bauxite is dissolved in sodium hydroxide to give sodium aluminate, NaAl(OH)4. The insolubles which include hydrated iron oxide are filtered off, and the pH adjusted (with CO2)to reprecipitate the aluminum as Al(OH)3.3H2O which is then dehydrated to Al2O3 and dissolved in molten cryolite, Na3AlF6 for electrolysis. The other (less reactive) metals can be obtained by electrolysis of aqueous solutions.
• The metals are all soft and quite reactive.
• Aluminum is "passivated" by a film of oxide which prevents it reacting with oxygen, water and even dilute nitric acid. (If the surface is amalgamated, reaction with water can occur.)
• The metals dissolve in non-oxidizing dilute acids. Aluminum and gallium are amphoteric and will dissolve in sodium hydroxide.
• They react with the halogens and sulphur.
• Thallium reacts slowly because the Tl+ salts which are formed are often insoluble and coat and passivate the metal surface.
The Oxides
• The most important are g-alumina, used as a stationay phase in liquid-solid chromatography and a-alumina, used as a catalyst in petroleum cracking.
• The gemstone, ruby is Al2O3 contaminated with traces of Cr3+ in place of the Al3+ and sapphire has Fe2+, Fe3+ and Ti4+ replacing some of the Al3+.
The Halides
• All the M(III) trihalides are known except the triiodide of thallium. The compound, TlI3 is actually Tl+[I3]-.
• The coodination numbers are 4 to 6 depending on the relative metal and halogen sizes.
• The 4-coordinate compounds are molecular compounds and have lower melting-points.
• They are strong Lewis bases and some can be used as Friedel-Crafts reaction catalysts:
RCOCl + "AlCl3" RCO+ + AlCl4-
The carbocation goes on to attacks the other organic reagent electrophilicly.
The Aqua Ions, Oxo salts and Aqueous Chemistry
The aqua ions all undergo hydrolysis:
$\ce{[M(H2O)6]^{3+} -> [M(H2O)5(OH)]^{2+} + H^{+}(aq)} \nonumber$
Element Ka
Al 1.12x10-5
Ga 2.5x10-3
In 2x10-4
Tl ~7x10-2
Salts of weak acids cannot exist in solution because the anions would be protonated and the hydroxides would precipitate.
The "hydroxides" of aluminum and gallium are amphoteric:
M(OH)3(s) M3+ + 3OH-
M(OH)3(s) MO2- + H+ + H2O
Depending on the conditions, bridging hydroxide is also common:
2[M(H2O)5(OH)]2+ [(H2O)5MOM(H2O)5]4+ + H2O (etc)
Hydroxides
The real hydroxides, by extension of the above, are complicated structures involving bridging OH-, terminal H2O and perhaps [M(OH)4]- for some metals.
Alums
These are the compounds for which aluminum was originally named. They are double salts of formula MM'(SO4)2.12H2O where M+ is usually an alkali metal ion (not Li+) and M'3+ is Al3+ or another trivalent ion. For example, plain "alum" or "potash alum" is the potassium/aluminum salt and "chrome alum" is the potassium/chromium(III) salt. These compounds are characterized by easily grown octahedral crystals. Each metal ion is 6-coordinated by water.
Coordination Compounds
Examples are: [Al(H2O)6]3+, [AlF6]3-, Cl3Al(N(CH3)3)2, [Al(ox)3]3- and Al(8-hydroxyquinolinate)3
Hydrides
The metal hydrides are not very stable except "AlH3" which is an air-sensitive polymeric material. The tetrahydroaluminate ion AlH4- is an important reducing agent and hydride source which usually comes as lithium aluminum hydride. The analogous gallium compound exists. The compounds are very sensitive to hydrolysis which is very exothermic and can be explosive.
The is a series of MH3 Lewis adducts with donor molecules which are generally more stable to, for example, hydrolysis than the parent hydrides.
Lower Valent Compounds
This section is mainly about Tl+ which resembles K+ and Ag+ in its chemistry. This section was not covered in depth in lectures. Skip it.
Summary of the Periodic Trends for the Elements of Group 13
1. Boron
1. Forms no simple B3+ cation.
2. Forms covalent compounds almost exclusively, and polyatomic ions are internally covalently bonded.
3. Has a maximum covalence of 4 corresponding to an octet.
4. The trivalent compounds are usually strong Lewis acids.
5. Its oxide and "hydroxide" are acidic.
6. Forms many polyatomic borates.
7. The trihalides are easily hydrolysed.
8. Forms many hydrides and hydride anions which are polyhedral clusters: the boranes, carboranes and the borane anions. The simplest BH4- is a very important synthetic reagent
2. Aluminum
1. Readily forms the Al3+ ion which is usually coordinated.
2. Much more metallic than boron and forms many ionic compounds.
3. Forms molecular compounds and ionic lattices with cordination numbers from 4 up to 6 and higher.
4. Forms oxides which are chemically and thermally fairly inert.
5. Forms a mainly basic but quite amphoteric hydroxide.
6. Forms partially hydrolysable halides.
7. Forms a polymeric hydride and the AlH4- ion. The latter is important.
3. Gallium, Indium, and Thallium.
1. Readily form M3+ aquo species and have a rich coordination chemistry.
2. Form increasingly stable M+ compounds especially thallium. Covalent bonds successively weaken down the group enhancing this trend.
3. Halides are increasingly aggregated with the increasing size of the metals.
4. Hydrides and hydride ions are not very important or stable. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/13%3A_The_Group_13_Elements/13.01%3A_Introduction_to_Group_13_Elements.txt |
A non-metal can be classified as an element that mostly lacks metallic attributes. Physically, non-metals tend to be highly volatile (easily vaporised), have low elasticity, and are good insulators of heat and electricity; chemically, they tend to have high ionization energy and electronegativity values, and gain or share electrons when they react with other elements or compounds. Seventeen elements are generally classified as nonmetals; most are gases (hydrogen, helium, nitrogen, oxygen, fluorine, neon, chlorine, argon, krypton, xenon and radon); one is a liquid (bromine); and a few are solids (carbon, phosphorus, sulfur, selenium, and iodine).
On moving across the periodic table, nonmetals are seen to adopt structures with progressively fewer nearest neighbours. Polyatomic nonmetals have structures with either three nearest neighbours, as is the case (for example) of carbon (in its standard state of graphite), or two nearest neighbours (for example) in the case of sulfur. Diatomic non-metals, such as hydrogen, have one nearest neighbour, and the monatomic noble gases, such as helium, have none. This gradual fall in the number of nearest neighbours is associated with a reduction in metallic character and an increase in nonmetallic character.
Allotropes are different structural forms of the same element in which changes in the connectivity of the covalent bonding between atoms results in substances with quite different chemical and/or physical properties. The change between allotropic forms is triggered by factors such as pressure, light, and temperature. Therefore the stability of a particular allotrope depends on particular conditions.
If covalent connectivity is the same but packing is different then you have polymorphs (eg. Monoclinic and Rhombic sulfur (S8) are polymorphs not different allotropes. S8 and S12 are different allotropes of S.
Many nonmetals have allotropes (that are less stable than their standard form) with either nonmetallic or metallic properties. Graphite, the standard state of carbon, has a lustrous appearance and is a fairly good electrical conductor. The diamond allotrope of carbon is nonmetallic, being translucent and having relatively poor electrical conductivity.
Catenation is the ability to form element-element bonded molecular networks.
13.04: The Elements
Allotropes of boron
Boron can be prepared in several crystalline and amorphous forms. The best known crystalline forms are α-rhombohedral, β-rhombohedral, and β-tetragonal. Under special circumstances, boron can form α-tetragonal, and γ-orthorhombic allotropes. Two amorphous forms, one a finely divided powder and the other a glassy solid, are also known and a further 14 allotropes have been reported.
Allotropes of Boron
B12
icosahedral unit
α-rhombohedral
β-rhombohedral
γ-orthorhombic
high pressure form
13.10: Metal Borides
In metal borides, the bonding of boron varies depending on the atomic ratio B/M. Diborides have B/M = 2, as in the well-known superconductor MgB2; they crystallize in a hexagonal AlB2-type layered structure. Hexaborides have B/M = 6 and form a three-dimensional boron framework based on a boron octahedron (Fig. 1a). Tetraborides, i.e. B/M = 4, are mixtures of diboride and hexaboride structures. Cuboctahedron (Fig. 1b) is the structural unit of dodecaborides, which have a cubic lattice and B/M = 12. When the composition ratio exceeds 12, boron forms B12 icosahedra (Fig. 1c) which are linked into a three-dimensional boron framework, and the metal atoms reside in the voids of this framework.[1][2][3]
This complex bonding behavior originates from the fact that boron has only three valence electrons; this hinders tetrahedral bonding as in diamond or hexagonal bonding as in graphite. Instead, boron atoms form polyhedra. For example, three boron atoms make up a triangle where they share two electrons to complete the so-called three-center bonding. Boron polyhedra, such as B6 octahedron, B12 cuboctahedron and B12 icosahedron, lack two valence electrons per polyhedron to complete the polyhedron-based framework structure. Metal atoms need to donate two electrons per boron polyhedron to form boron-rich metal borides. Thus, boron compounds are often regarded as electron-deficient solids.[4]
Icosahedral B12 compounds include[2] α-rhombohedral boron (B13C2), β-rhombohedral boron (MeBx, 23≤x), α-tetragonal boron (B48B2C2), β-tetragonal boron (β-AlB12),[5] AlB10 or AlC4B24, YB25, YB50, YB66, NaB15 or MgAlB14, γ-AlB12,[5] BeB3 [6] and SiB6.[7]
Fig. 2. Relationship between the ionic radius of trivalent rare-earth ion and chemical composition of icosahedron-based rare-earth borides.
YB25 and YB50 decompose without melting that hinders their growth as single crystals by the floating zone method. However, addition of a small amount of Si solves this problem and results in single crystals [8] with the stoichiometry of YB41Si1.2.[9] This stabilization technique allowed the synthesis of some other boron-rich rare-earth (RE) borides.
Albert and Hillebrecht reviewed binary and selected ternary boron compounds containing main-group elements, namely, borides of the alkali and alkaline-earth metals, aluminum borides and compounds of boron and the nonmetals C, Si, Ge, N, P, As, O, S and Se.[10] They, however, excluded the described here icosahedron-based rare-earth borides. Note that rare-earth elements have d- and f-electrons that complicates chemical and physical properties of their borides. Werheit et al. reviewed Raman spectra of numerous icosahedron-based boron compounds.[11]
Figure 2 shows a relationship between the ionic radius of trivalent rare-earth ions and the composition of some rare-earth borides. Note that scandium has many unique boron compounds, as shown in figure 2, because of the much smaller ionic radius compared with other rare-earth elements.[3][12]
In understanding the crystal structures of rare-earth borides, it is important to keep in mind the concept of partial site occupancy, that is, some atoms in the described below unit cells can take several possible positions with a given statistical probability. Thus, with the given statistical probability, some of the partial-occupancy sites in such a unit cell are empty, and the remained sites are occupied.[13] | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/13%3A_The_Group_13_Elements/13.04%3A_The_Elements/13.4B%3A_Structures_of_the_Elements.txt |
Group 14 of the periodic table is often referred to as the carbon group. It is located in column 14 of the periodic table and houses the elements carbon, silicon, germanium, tin, lead, and ununquadium (C, Si, Ge, Sn, Pb, and Uuq, respectively). Each element in this group contains four valence electrons. A unique feature of this group is that the elements can form different anions and cations. Carbon forms a 4- anion whereas silicon and germanium form 4+ cations. Tin and lead can even form 2+ cat
14: The Group 14 Elements
Carbon is capable of forming many allotropes in addition to the well known diamond and graphite forms. The physical properties of carbon vary widely with the allotropic form. For example, diamond is highly transparent, but graphite is opaque and black. Diamond is the hardest naturally-occurring material known, while graphite is soft enough to form a streak on paper (hence its name, from the Greek word "γρáφω" which means "to write"). Diamond has a very low electrical conductivity, while graphite is a very good conductor. Under normal conditions, diamond, carbon nanotubes, and graphene have the highest thermal conductivities of all known materials.
All carbon allotropes are solids under normal conditions, with graphite being the most thermodynamically stable form. They are chemically resistant and require high temperature to react even with oxygen. The system of carbon allotropes spans a range of extremes:
Synthetic nanocrystalline diamond is the hardest material known. Graphite is one of the softest materials known.
Diamond is the ultimate abrasive. Graphite is a very good lubricant, displaying superlubricity.
Diamond is an excellent electrical insulator, and has the highest breakdown electric field of any known material. Graphite is a conductor of electricity.
Diamond is the best known naturally occurring thermal conductor Some forms of graphite are used for thermal insulation (i.e. firebreaks and heat shields), but some other forms are good thermal conductors.
Diamond is highly transparent. Graphite is opaque.
Diamond crystallizes in the cubic system. Graphite crystallizes in the hexagonal system.
Amorphous carbon is completely isotropic. Carbon nanotubes are among the most anisotropic materials ever produced.
14.04: Allotropes of Carbon
Covalent Network Solids are giant covalent substances like diamond, graphite and silicon dioxide (silicon(IV) oxide). This page relates the structures of covalent network solids to the physical properties of the substances.
Diamond
Carbon has an electronic arrangement of 2,4. In diamond, each carbon shares electrons with four other carbon atoms - forming four single bonds.
In the diagram some carbon atoms only seem to be forming two bonds (or even one bond), but that's not really the case. We are only showing a small bit of the whole structure. This is a giant covalent structure - it continues on and on in three dimensions. It is not a molecule, because the number of atoms joined up in a real diamond is completely variable - depending on the size of the crystal.
How to draw the structure of diamond
Don't try to be too clever by trying to draw too much of the structure! Learn to draw the diagram given above. Do it in the following stages:
Practice until you can do a reasonable free-hand sketch in about 30 seconds.
Physical Properties of Diamond
• has a very high melting point (almost 4000°C). Very strong carbon-carbon covalent bonds have to be broken throughout the structure before melting occurs.
• is very hard. This is again due to the need to break very strong covalent bonds operating in 3-dimensions.
• doesn't conduct electricity. All the electrons are held tightly between the atoms, and aren't free to move.
• is insoluble in water and organic solvents. There are no possible attractions which could occur between solvent molecules and carbon atoms which could outweigh the attractions between the covalently bound carbon atoms.
Graphite
Graphite has a layer structure which is quite difficult to draw convincingly in three dimensions. The diagram below shows the arrangement of the atoms in each layer, and the way the layers are spaced.
Notice that you cannot really draw the side view of the layers to the same scale as the atoms in the layer without one or other part of the diagram being either very spread out or very squashed. In that case, it is important to give some idea of the distances involved. The distance between the layers is about 2.5 times the distance between the atoms within each layer. The layers, of course, extend over huge numbers of atoms - not just the few shown above.
You might argue that carbon has to form 4 bonds because of its 4 unpaired electrons, whereas in this diagram it only seems to be forming 3 bonds to the neighboring carbons. This diagram is something of a simplification, and shows the arrangement of atoms rather than the bonding.
The Bonding in Graphite
Each carbon atom uses three of its electrons to form simple bonds to its three close neighbors. That leaves a fourth electron in the bonding level. These "spare" electrons in each carbon atom become delocalized over the whole of the sheet of atoms in one layer. They are no longer associated directly with any particular atom or pair of atoms, but are free to wander throughout the whole sheet. The important thing is that the delocalized electrons are free to move anywhere within the sheet - each electron is no longer fixed to a particular carbon atom. There is, however, no direct contact between the delocalized electrons in one sheet and those in the neighboring sheets. The atoms within a sheet are held together by strong covalent bonds - stronger, in fact, than in diamond because of the additional bonding caused by the delocalized electrons.
So what holds the sheets together? In graphite you have the ultimate example of van der Waals dispersion forces. As the delocalized electrons move around in the sheet, very large temporary dipoles can be set up which will induce opposite dipoles in the sheets above and below - and so on throughout the whole graphite crystal.
Graphite has a high melting point, similar to that of diamond. In order to melt graphite, it isn't enough to loosen one sheet from another. You have to break the covalent bonding throughout the whole structure. It has a soft, slippery feel, and is used in pencils and as a dry lubricant for things like locks. You can think of graphite rather like a pack of cards - each card is strong, but the cards will slide over each other, or even fall off the pack altogether. When you use a pencil, sheets are rubbed off and stick to the paper. Graphite has a lower density than diamond. This is because of the relatively large amount of space that is "wasted" between the sheets.
Graphite is insoluble in water and organic solvents - for the same reason that diamond is insoluble. Attractions between solvent molecules and carbon atoms will never be strong enough to overcome the strong covalent bonds in graphite. conducts electricity. The delocalized electrons are free to move throughout the sheets. If a piece of graphite is connected into a circuit, electrons can fall off one end of the sheet and be replaced with new ones at the other end.
Silicon dioxide: SiO2
Silicon dioxide is also known as silica or silicon(IV) oxide has three different crystal forms. The easiest one to remember and draw is based on the diamond structure. Crystalline silicon has the same structure as diamond. To turn it into silicon dioxide, all you need to do is to modify the silicon structure by including some oxygen atoms.
Notice that each silicon atom is bridged to its neighbors by an oxygen atom. Don't forget that this is just a tiny part of a giant structure extending on all 3 dimensions.
Silicon Dioxide has a high melting point - varying depending on what the particular structure is (remember that the structure given is only one of three possible structures), but around 1700°C. Very strong silicon-oxygen covalent bonds have to be broken throughout the structure before melting occurs. Morevoer, it hard due to the need to break the very strong covalent bonds.Silicon Dioxide does not conduct electricity since there aren't any delocalized electrons with all the electrons are held tightly between the atoms, and are not free to move.Silicon Dioxide is insoluble in water and organic solvents. There are no possible attractions which could occur between solvent molecules and the silicon or oxygen atoms which could overcome the covalent bonds in the giant structure.
Introduction
• There are more compounds of carbon than of any other element except hydrogen. Most of carbon chemistry is handled in different courses (organic and biochemistry).
• There are no compounds containing anything like C4+ but some carbides contain a species approximating C4-.
• There are transient ionic species: carbonium ions such as F3C+, and carbanions such as (NC)3C-. Notice the attached groups which would help to stabilize these species.
• Transient divalent carbon occurs in carbenes, R2C:
• The strong C—C single bond and the ability to form double, C=C, and triple, CºC, bonds contributes to the huge versatility of carbon.
The Chemistry and Physical Properties of Diamond Graphite and the Fullerenes
Carbides
These come in three general types:
Ionic carbides are formed by elements of groups 1, 2 and aluminum. The actual for of the carbon varies, for example, aluminum carbide, based on its hydrolysis product seems to contain "C4-" units:
\[\ce{Al4C3 + 6H2O -> 2Al(OH)3 + 3CH4}\]
but calcium carbide seems to contain [CºC]2- units:
\[\ce{CaC + 2H2O -> Ca(OH)2 + HCºH}\]
Interstitial carbides are compounds of the transition metals with metallic properties and the C in tetrahedral holes in the metal atom lattice. The best known example is the extremely hard tungsten carbide, WC, used in cutting tools.
Covalent carbides include B4C3 and SiC (carborundum - an abrasive with a diamond-like structure) | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/14%3A_The_Group_14_Elements/14.04%3A_Allotropes_of_Carbon/14.4A%3A_Graphite_and_Diamond_-_Structure_and_Properties.txt |
Chemistry 242 - Inorganic Chemistry II Chapter 15 - Silicon, Germanium, Tin and Lead
Introduction
• Silicon is th most abundant element after oxygen in the earth's crust occuring mostly as the silicates and quartz (SiO2).
• The other elements are acutally quite rare (~10-3%) but tin an dlead are found in concetrated pockets of their ores (cassiterite, SnO2 and galena, PbS2) from which they are easily obtained, so they have been known and used since antiquity.
• The existence of germanium ('ekasilicon") was predicted by Mendleev when he constructed the first reasobably complete periodic table. It was isolated in 1886 from coal and zinc ores.
• The main uses of tin an dlead are as the metals and as synthetic reagents as their organo compounds. Silicon and germanium are the basis of the semicondutor industry.
Multiple Bonding
Because of the great versatility of carbon, it is interesting to compare the other members of group 14.
• Carbon dioxide is molecular gas whereas silicon dioxide is a network solid (in all its forms several forms).
• Dehydration of alcohols, ROH yields both alkenes and ROR depending on the conditions, but dehydrating R3SiOH yields only R3SiOSiR3.
• Replacement of the hydrogen with tertiary butyl groups allows the isolation of similar compounds of silicon, germanium and tin.
• Silicon can use empty d-orbitals for pp - dp bonding. Examples are planar (H3Si)3N vs pyramidal (H3C)3N and linear H3SiNCO vs H3CNCO which is bent at the nitrogen. In both cases the nitrogen lone pair is delocalized by p-bonding with the silicon(s).
In CR3OCR3 the Si-O-Si angle is around 109 o whereas in R3SiOSiR3 the angles range from 140 to 180 o indicative of Si=O dp - pp bonding.
(CH3)3COH is a very weak acid but (CH3)3SiOH is stronger because the ion (CH3)3SiO- is stabilized by dp - pp bonding.
Stereochemistry
See Table 15-1 for summary information about the tetravalent state:
• All show tetrahedral coordination.
• Five coordinate complexes e.g. MX5- or MnX5-n- can be trigonal bipyramids or rarely square pyramidal if a constraining chelating ligand is used, e.g.[XSi(O2C2H4)2]-.
• 6-coordinatate complexes are normally octahedral.
See table 15-2 for summary information about divalent states.
• Very often, but not always, the lone pair is stereochemically active and influences the molcular shape. F2Pb is bent and so is SiCl3- .
Isolation and Properties of the Elements
• Silicon and germanium can be made by reduction of their dioxides by carbon or calcium carbide in an electric furnace, and then purified further by zone-refining.
• Tin and Lead are obtained by carbon reduction of their oxides or sulphides. I ffuther purification is necessary they can be dissolved in acid and redeposited electrolytically.
• Silicon and germanium are relatively inert but the following reactions occur:
Si + 2X2 SiX4 (X2 is a halogen)
Si + excess OH- silicates
Si + excess HX no reaction except HF will give SiF62-
Germanium is somewhat more reactive and will dissolve in sulphuric or nitric acids.
Tin and lead dissolve in several acids, hot alkalis and also react with halogens.
Hydrides
Compounds MH4 all exist as spontaneously flammable gases and are not very important.
Chlorides
Compounds MCl4 are all colourless liquids except PbCl4 which is yellow. They are hydrolysed easily to give hydrous oxides. In hydrochloric acid the lower members, tin and lead, give MCl62- ions in aqueous solution. The compounds are intermediates in the synthesis of organo compounds for example the infamous tetraethyl lead.
Oxygen Compounds
• Silica comes on three forms: quartz and crystobalite which are both crystalline, and silica glass. The glass has a very low coefficient of expansion and high melting point so it is relatively resistant to heat and sudden temperature changes. It is also transparent to a large part of the ultraviolet spectrum and therefore used in cells for spectrophotometry in that region.
• There are several important oxides of lead: PbO exists in a red from, litharge, and a yellow form, massicot. It is the most used Pb source of lead for synthesis. Pb2O3 which behaves like a mixture of PbO and PbO2 although it is a well defined structure is called "red lead". It is used as an anti-rust coating for steel. Lead dioxide, PbO2 is maroon in colour and has a structure similar to rutile (TiO2). It is one of the electrode materials in lead/acid batteries.
• The oxides vary from acidic for SiO2 to basic for tin and lead.
Complex Compounds
Anionic Complexes
Silicon forms a very stable fluoroanion:
SiO2(s) + 6HF(aq) 2H+(aq) + SiF62-(aq) + 2H2O
The same anion is formed by the incomplete hydrolysis of SiF4:
SF4 + 2H2O SiO2 + SiF62- + 2H+ + 2HF
The other MF62- ions are hydrolysed by bases or even water in the case of the lead complex ion. All the elements give analogous chloroanions, except silicon. The other fairly important anionic complex is obtained with oxalate, [M(ox)3]2-.
Cationic Complexes and Neutral Adducts
Can be formed with chelating uninegative oxygen donor ligands, e.g. [Pb(acac)3]+.
The MX4 compounds are Lewis acids and can form adducts which are sometimes neutral MX4L or MX4L2 molecular compounds but can also be ionic [MX2L2]X2.
Alkoxides, Carboxylates and Oxo Salts
Typical preparative routes are, for alkoxides:
MCl4 + 4ROH + 4(Et)3N M(OR)4 + 4(Et)3NHCl
Note the use of the triethylamine to "remove" the HCl which would be formed in its absence.
Carboxylates can be made by direct reaction:
Pb3O4 + 8CH3COOH Pb(CH3COO)4 + 2Pb(CH3COO)2 + 8H2O
Lead tetraacetate is used as an oxidizing agent in certain organic reactions.
Tin and Lead (IV) salts are hydrated e.g Pb(SO4)2.2H2O and subject to extensive hydrolysis in aqueous solution.
The Divalent State
• This oxidation state becomes more and more stab;le down the group.
• Silicon dihalides are only transient species. GeF2 and GeCl2 can be isolated.
• Tin II fluoride and chloride are well known and useful. (SnF2 is the active ingredient in many fluoridated toothpastes.) In solution, tin II is easily oxidized by air. The sulphates and nitrates are heavily hydrolysed: Sn3(OH)4(NO3)2 and Sn3(OH)2SO4.
• Lead II is the best defined divalent state. Most lead II salts are not very water soluble. The exception are the nitrate and the acetate. The solid halides are always anyhdrous.
Silenes and Other Organic Compounds
It is possible to make some compounds containing a Si=Si or Ge=Ge double bonds.
The earliest attempts tried the reaction:
2R2SiCl2 + Na/K in THF R2Si=SiR2 + 4K/NaCl
Without the bulky R groups cyclic polymers are generated typically with 6 silicons in a ring. By using sufficiently bulky organic groups, it was possible to prepare transient dimers, R2Si=SiR2, or monomers R2Si. The first silene stable enough to isolate was prepared by the photochemical decomposition of (mes)2Si(SiMe3)2:
2(mes)2Si(SiMe3)2 (mes)2Si=Si(mes)2 + Me3Si-SiMe3
Unlike the carbon analogs, the molecule is not perfectly flat. The Si=Si double bond is 9% shorter than a normal Si-Si single bond, c.f. 13% in an olefin. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/14%3A_The_Group_14_Elements/14.05%3A_Structural_and_Chemical_Properties_of_Silicon_Germanium_Tin_and_Lead.txt |
Carbon Monoxide
• A colorless and very dangerously toxic gas - it has no smell - boiling point -190 oC.
• It is thermodynamically unstable with respect to carbon and carbon dioxide, but the equilibrium is only established at high temperature:
$\ce{2CO(g) -> C(s) + CO2(g)}$
• An important primary industrial chemical, involved in several reaction sequences:
$\ce{CH4 + 2H2O -> CO2 + 4H2}$
$\ce{C + H2O -> CO + H2 (+ H2O) "water gas"}$
$\ce{CO + H2O -> CO2 + H2} \tag{"water gas shift" reaction}$
Mixtures of carbon monoxide and hydrogen are called "synthesis gas" or "syngas".
• Formally CO is the anhydride of formic acid, HCCH, but CO does not react at room temperature with water.
• Carbon monoxide is unique in that it is a weak Lewis base (s-donor through carbon) but a very strong p-acceptor. Asa ligand it stabilizes transition metals in low zero or negative oxidation states.
Carbon Dioxide and Carbonic Acid
• Carbon dioxide makes up about 0.03% (300 ppm) of the earth's atmosphere. It is mainly produced from volcanic activity, fermentation of organic matter and fires of al types.
• Solid carbon dioxide sublimes at -78 oC at atmospheric pressure, making it a useful refrigerant ("dry ice").
• Carbonic acid H2CO3 is produced only very slowly when CO2 is dissolved in water. The equilibrium constants usually quoted are:
$\dfrac{[\ce{H^{+}}]\{\ce{HCO3^{-}}\}}{[\ce{H2CO3}]} = 4.16 \times 10^{-7}$
$\dfrac{[\ce{H{+}}][\ce{CO3^{2-}}]} {[\ce{HCO3^{-}}]} = 4.84 \times 10^{-11}$
but the first is incorrect because the real [H2CO3] is much lower than what is calculated based on dissolved CO2. The real constant is probably closer to 2x10-4 more in keeping with a compound with a C=O bond - see Chapter 7 section 12.
14.9B: Silica Silicates and Aluminosilicates
While you are admiring this beautiful picture of faujasite, remember that the oxygen atoms have two unshared electron pairs in addition to the (Al,Si)-O-Si(or Al) bonds. Thus the oxygen atoms are sites to interact with positive site of molecules that passes by these structures. At present over 150 synthetic zeolites & zeotypes and 40 natural zeolites are known. Synthesis of zeolite is a very active field of study.
Aluminosilicates have three major minerals: Andalusite, sillimanite, and kyanite. Zeochem has been developing and manufacturing molecular sieve adsorbents since 1977. Simply put, their adsorbents are used to "screen" out impurities from a variety of applications by attracting and trapping the targeted contaminants. For example, in natural gas processing, molecular sieves are used to remove specific molecules from the gas stream to allow for more efficient downstream processing. Faujasite is a typical zeolite.
Applications of Zeolites?
As you have read above that there are many different kinds of zeolites, each with a definite structure and associate with it are unique properties. In terms of applications, we are assuming zeolites as porous aluminosilicates with large tunnels and cages for a fluid (gas and liquid) to pass through. The applications are based on the interactions between the fluid phase and the atoms or ions of the zeolites. In general terms, zeolites have many applications:
1. As selective and strong adsorbers: remove toxic material, selective concentrate a particular chemical, as Molecular Sieve. This link will be a very good to discuss zeolites. Currently, the site is under construction, but it has a very good framework. Even many deorderants are zeolite type.
2. As selective ion exchangers: for example used in water softener.
3. Superb solid acid catalysts, when the cations are protons H+. As catalysts, their environmental advantages include decreased corrosion, improved handling, decreased environmentally toxic waste and minimal undesirable byproducts.
4. As builder: a material that enhance or protecting the cleaning power of a detergent. Sodium aluminosilicate is an ion exchange builder often used in laundry detergent as a builder. A builder inactive the hardness of water by either keeping calcium ions in solution, by precipitation, or by ion exchange.
123 ppm CaCO3 = 123 g per 106 g of water.
``` 1 mol CaCO3 2 mol H+ 1 mol z-A 1926 g z-A 100
123 g CaCO3 ----------- ----------- ---------- ---------- ---
100 g CaCO3 1 mol CaCO3 12 mol H+ 1 mol z-A 80
```
= 494 g zeolite A
That 80 % of protons of the zeolite A is used means that we require a little more zeolite A than stoichiometric quantities.
DISCUSSION
Zeolites are aluminosilicates with open frames strcutures discussed above. Replacement of each Si atom by an Al atom in silicates results in having an extra negative charge on the frame. These charges must be balanced by trapping positive ions: H+, Na K+, Ca2+, Cu2+ or Mg2+. Water molecules are also trapped in the frame work of zeolites.
In this example, we assume that when we soak the zeolite in water containing Ca2+, and Mg2+ ions, these ions are more attrative to the zeolite than the small, singly charged protons. We further assumed that 80 percent of the protons in zeolite are replaced by other ions.
58.5 g/mol.
``` 1 mol 0.8*12 mol NaCl 100 58.5 g NaCl
10 kg z-A -------- --------------- --- -----------
2.190 kg 1 mol z-A 20 1 mol NaCl
= 12822 g NaCl = 12.8 kg NaCl
```
DISCUSSION
How much salt is required if 60% of the sodium ions are effectively used to replace all the divalent ions?
14.11: Sulfides
Compounds with C-S bonds
Carbon disulphide is perhaps the most important as a solvent and a synthetic reagent. It gives rise to other carbon sulphur compounds such as:
\[\ce{RO^{-}- + CS2 -> ROCS2- (xanthates)}\]
\[\ce{HS^{-} + CS2 -> CS3^{2-} (thiocarbonate)}\]
\[\ce{R2HN + CS2 -> R2NCS2^{-} (dithiocarbamates)}\]
These ions are important ligands for transition metals, and the dithiocarbamtes are sued as agricultural fungicides.
14.12A: Cyanogen and its Derivatives
Compounds with C-N Bonds: Cyanides and Related Compounds
Cyanogen (NºC–CºN) is a poisonous and flammable gas (bp -21 oC. Although its heat of formation is strongly endothermic 297 kJ mol-1 is is fairly stable. Impure cyanogen polymerizes to form "paracyanogen". Cyanogen is prepared by nitrogen dioxide catalyzed oxidation of hydrogen cyanide by oxygen:
\[\ce{2HCN + NO2 -> (CN)2 + NO + H2O}\]
\[\ce{NO + ½O2 -> NO2}\]
It is also formed by oxidation od CN- with Cu2+:
\[\ce{Cu^{2+} + 2CN^{-} -> CuCN + ½(CN)2}\]
Notice the similarity between the above reaction and the one below:
\[\ce{Cu^{2+} + 2I^{-} -> CuI + ½(I)2}\]
The term "pseudo halogen/halide" is often applied to molecules and derived ions such as (CN)2 and CN-. Notice also the existence of HCN which has its parallel in the hydrohalic acids and the reaction of cyanogen with base:
\[\ce{(CN)2 + 2OH^{-} -> CN^{-} + OCN^{-} + H2O}\]
Compare:
\[\ce{(Cl)2 + 2OH^{-} -> Cl^{-} + OCl^{-} + H2O}\]
The reaction of cyanogen with oxygen produces one of the hottest flames known at about 5000 oC.
Hydrogen cyanide boils at 25.6 oC. It is very poisonous and has an odor of almonds which not everyone can smell. It is a very good solvent due to its high dielectric constant, e = 107. It is made on an industrial scale (~300 000 tons in 1980) as follows:
\[\ce{CH4 + 3O2 + 2NH3 -> 2HCN + 6H2O}\]
with Pt/Rh or Pt/Ir catalyst and 800 oC.
or
\[\ce{CH4 + NH3 -> HCN + 3H2}\]
with Pt catalyst and 1200 oC
Cyanides are made industrially via the calcium cyanamide salt by the processes:
\[\ce{CaC2 + N2 ->[1100 degrees] CaNCN + C}\]
\[\ce{CaNCN + C + Na2CO3 -> CaCO3 + 2NaCN}\]
or
\[\ce{NaNH2 + C -> NaCN (500-600 oC)}\]
(NCN2- which yields cyanamide itself, H2NCN, by hydrolysis of the salt, is isoelectronic with CO2.)
Cyanide is important, among other things, as a very good p-acceptor ligand like CO. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/14%3A_The_Group_14_Elements/14.09%3A_Oxides_Oxoacids_and_Hydroxides/14.9A%3A_Oxides_and_Oxoacids_of_Carbon.txt |
The nitrogen family includes the following compounds: nitrogen (N), phosphorus (P), arsenic (As), antimony (Sb), and bismuth (Bi). All Group 15 elements have the electron configuration ns2np3 in their outer shell, where n is the principal quantum number.
15: The Group 15 Elements
• Nitrogen forms a range of different types of compounds:
• In combination with the most electropositive element ionic nitrides (N3-) are formed.
• Also forms anionic species NR2 and NR2-.
• Forms cationic compounds NR4+ for the maximum covalency of four bonds.
• Forms compounds with single, double and triple covalent bonds to itself, carbon and oxygen.
• There are some paramagnetic compounds of nitrogen, i.e. with an odd number of electrons, notably certain oxides.
• In its three-covalent compounds, the following generallities apply:
• Nitrogen is almost always a Lewis donor (except in very special cases, e.g. NF3).
• Although based on a tetrahedral shape with the lone-pair occupying one site, optical isomers of potentially chiral compounds :NRR'R" cannot be isolated because the molecules invert so easily through a planar transition state (like an umbrella turning inside-out).
• There are a few cases where the lone-pair is delocalized into empty orbitals on the attached groups in which nitrogen atom is planar. An example is N(SiMe3)3.
• Probably due to repulsion between the lone pairs, nitrogen-nitrogen single bonds are rather weak, e.g. 160 kJ mol-1 in H2N-NH2 as compared to a carbon-carbon bond dissociation energy of around 350 kJ mol-1 in CH3-CH3.
• The ability to form multiple bonds via pp - pp overlap is one of the things that distinguishes nitrogen from the other members of group 15. Because of the lone pairs, molecules will normally not be linear.
Occurence and Properties of the Element
• Dinitrogen constitutes 78% of the atmosphere. The ratio of the abundances of the two stable isotopes is 14N/15N = 272.0 so 15N at natural abundance is somewhat useful for nmr studies, and when it is enriched it, can be used in tracer studies.
• Nitrogen is obtained by low temperature distillation (fractionation) of air. It will normally still be contaminated with about 30 ppm of oxygen and some argon. It can be prepared pure by thermally decomposing sodium azide, NaN3.
• The NºN bond is very stron with a dissociation energy of 944.7 kJ mol-1. For this reason, many hitrogen compounds are "endothermic" that is they have positive heats of formation.
• Compared to other molecules with triple bonds, nitrogen is very unreactive. It will react with lithium at room temperature, and there are bacteria which are able to activate it by a mechanism that remains unclear.
• At high temperatures it is more reactive:
N2 + 3H2 2NH3
N2 + O2 2NO
N2 + 3Mg Mg3N2
N2 + CaC2 CaN
CN + C
Nitrides
• The ionic ones only occur with the most electropositive elements. Hydrolysis yields ammonia.
• Transition metals tend to form non-stoichiometric compounds.
• The important covalent nitrides are covered in the section concerning the other element
Hydrides
Ammonia
• This topic has been covered to some extent elsewhere.
• The following sequence of conversions is industrially important:
• Ammonia is very soluble in water due to extensive hydrogen bonding. Th eresulting solutions are conventionally called ammonium hydroxide, but:
NH3 + H2O NH4+ + OH Kb = 1.77x10-5
i.e. there is not much ammonium ion in solutions of ammonia!
• Ammonium salts are mostly soluble, and the ammonium ion, r = 1.48 Å, resembles potassium ion, r = 1.33, Å or rubidium ion, r = 1.48 Å.
• Ammonium salts sublime via this type of reaction:
NH4Cl(s) NH3(g) + HCl(g)
NH4NO3(s) NH3(g) + HNO3(g)
But with oxidizing anions, decomposition can occur:
NH4NO3(l) N2O(g) + 2H2O
(NH4)2Cr2O7(s) N2(g) + 4H2O(g) + NH3
Hydrazine
• Hydrazine bp 114 oC is explosive when pure but organo-substitued hydrazines are more stable.
• Preparation:
NH3 + NaOCl NaOH + NH2Cl NH3 + NH2Cl + NaOH N2H4 + NaCl + H2O but the following reaction is catalysed by traces of transition metal ions, notable Cu2+:
NH2Cl + NH2Cl 2NH4Cl + +N2 Therefore, the reaction is perfomed in the presence of gelatin which complexes the metal ions and suppresses their catalytic effect.
• Hydrazine is a "bifunctional" base:
N2H4 + H2O N2H5+ + OH Kb1 = 8.5x10-7
N2H5+ + H2O N2H62+ + OH Kb2 = 8.9x10-15
These equilibria leasd to two sets of hydrazinium salts.
• Hydrazine and the substituted hydrazines are used as reducing agents. In combination with N2O4 as oxidant, alkylhydrizines are used in teh vernier rocket engines on things like the space shuttle. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/15%3A_The_Group_15_Elements/15.04%3A_The_Elements/15.4A%3A_Nitrogen.txt |
• Phosphorus is found extensively non-crystalline phosphate rocks such as apatite, Ca3(PO4)2.CaX2 where X- = F- and/or Cl- and/or OH-.
Arsenic, antimony and bismuth come mainly from sulphide ores such as mispickel, FeAsS, and stibnite, Sb2S3.
• Apart from the hydrides such as phosphine, PH3, the elements are very different to nitrogen.
The differences arise from lower pp - pp bonding effectiveness replaced by use of the d-orbitals for dp - pp bonding, and gradual increasing stability of the lower oxidation states on descending the group:
• compare O=N-OR with P(-O-R)3.
• Compare the oxides of nitrogen with multiple bonding in all of them with P4O6 and P4O10 containing only single bonds.
• Compare nitric acid, HONO2, with pp - pp bonding with phosphoric acid, (HO)3PO, where phosphorus uses a d-orbital to form a p-bond to the oxygen (perhaps).
• Unlike nitrogen, the rest of the elements of group 15 can exceed 4-covalency, e.g. PF5, PF6- by using empty d-orbitals.
• As ligands with electron rich transition metals, all the group 15 elements, except nitrogen, can use their empty d-orbitals to act as p-acceptors (i.e. they are "soft").
The Elements
• Phosphorus, the most reactive can be obtained by carbon reduction of calcium phosphate:
Ca3(PO4)2 + 6SiO2 + 10C P4 + 6CaSiO3 + 10CO
The yellow phosphorus, P4 must be collected under water since it inflames in air.
• The other elements can be obtained by carbon reduction of their oxides.
• Reactivity:
• All the elements will react directly with oxygen especially if heated.
• The products of reaction with oxidizing acids, e.g. nitric acid illustrate the increasing metallic character down the group. The products are: H3PO4, H3AsO4, Sb4O6 and Bi(NO3)3. Note that the products are in the V oxidation state for phosphorus and arsenic, and the III-state for antimony and bismuth. Bismuth produces a nitrate salt.
• Compounds can be formed by direct reaction with other non-metals. One important compound is gallium arsenide, a semiconductor which is particularly heat resistant.
Hydrides
• The hydrides are all gases and increasingly unstable down the group. The classical (Sherlock Holmes style) test for arsenic is to generate AsH3 by reduction and observe the metallic arsenic mirror produced by decomposition of the gas on the glass of the test tube above the reaction mixture.
• The (Lewis) basic properties decrease down the group. There are some phosphonium, PH4+ compounds, but water tends to be a stronger base than PH3 so they tend to decompose in water.
Halides and Oxo Halides
• Refer to Figure 17-1 for some reactions of PCl3 which are typical of the group.
• Note that the solid state structures sometimes differ from the gas phase ones, for example, PCl5 is probably like PF5, a trigonal bipyramidal molecule in the gas phase, but in the solid state it is ionic: [PCl4]+[PCl6]-. PBr5, in the solid state, is [PBr4]+Br-.
• The Oxo trihalides, notably PCl3O, undergo reaction similar to the trihalides.
• SbOCl and BiOCl are obtained when hydrocloric acid solutions containing Sb3+ or Bi3+ are diluted.
• The pentaflourides are all good flouride ion acceptors to give non-coordinating anions such as AsF6- and the corresponding "super acids".
Oxides
• The oxides of the elements in the V state are most stable (relative to III) for phosphorus in addition to being the most acidic.
• "Phosphorus pentoxide" which is actually P4O10 is one of the most powerful dessicating agents known. It reacts with water to produce phosphoric acid, and can remove water from nitric acid, to give dinitrogen pentoxide and from sulphuric acid, itself a powerful dessicant to give sulphur trioxide.
Its structure is a tetrahedron of phosphorus atoms connected by oxygens on th esix edges of the tetrahedron and each carrying a terminal oxygen atom.
• The phosphorus III oxide, P4O6 is formed when phosphorus burns in a limited supply of oxygen. One of its forms is similar to P4O10 except that the terminal oxygen atoms are absent. It hydrolyses to phosphorous acid, H3PO3. Arsenic and Antimony produce similar compounds.
• The oxide and hydroxide of bismuth III, Bi2O3 and Bi(OH)3, obtained by adding base to solutions of Bi3+ salts, are not acidic at all.
Sulphides
• Some are somewhat related to the oxides but the numbers of terminal sulphurs varies more, others are chain or ribbon structures. Skip the details.
The Oxo Acids and their Esters
• The oxo acids have already been covered to some extent in chapter 5.
• Phosphorous acid is obtained by controlled hydrolysis of PCl3 or P4O6. Its formula is best written HP(O)(OH)2 to illustrate its structure with only two acidic hydrogens (on oxygen).
• Similarly, hypophosphorous acid is best written H2P(O)(OH). It is a monobasic acid.
• Phosphite esters, P(OR)3, are related to PX3 compounds. These compounds are easily oxidised to phosphate esters: OP(OR)3.
• The phosphite esters undergo the Michaelis-Arbusov reaction with alkyl halides:
P(OR)3 + R'X [(RO)3PR']+X- (RO)2(R')PO + RX
The dialkyl phosphonates produced are structurally similar to phosphorous acid where R replaces H.
• (Ortho)phosphoric acid, H3PO4, when pure is a solid melting at 42.5 oC. It i ssold as "syrupy phosphoric acid", an 85% solution in water. It is made by hydrolysing P4O10 or treating phosphate rocks with sulphuric acid. Its dehydration to pyrophosphoric acid (HO)2(O)POP(O)(OH)2 is slow.
• Phosphate esters are important in biochemical processes for energy storage and transfer. The mechanism of their hydrolysis has been extensively studied in this context.
Complexes of the Group 15 Elements
• Antimony forms some complexes, particularly with chelating oxy-ligands. One of the oldest know in the complex with tartrate, K2[Sb2(d-C4H2O6)2].3H2O known as "tartar emetic".
• Bismuth should behave much more like a true metal, but the simple aquo ion, [Bi(H2O)6]3+ does not seem to exist. There is a range of extensively hydrolysed clusters ions containing several bismuth III ions bridged by oxygen, and carrying OH groups. Skip the details!
Phosphorus-Nitrogen Compounds
• Six and eight membered ring systems and linear polymers containing ...-N=P-N=P-... chains can be generated by the reaction:
nPCl5 + nNH4Cl (NPCl2)n + 4nHCl (Solvents: C2H2Cl4 or C6H5Cl)
Then the chlorine atoms can be replace by alkoxy, alkyl, or aryl groups by reaction with NaOR or LiR reagents. The resulting compounds can be made into useful fibres or elastomers.
Compounds with Element-Element Double Bonds
• Just as in group 14, therehave been attempts to stabilize compounds with multiple bonds betweemn the group 15 elements and just as in group 14, bulky groups get the job done. Molecules with double bonds between two phosphorus atoms, a phosphorus and an arsenic or two arsenic atoms have been sythesized using groups such as 2,4,6-(Me3C)3C6H2 and (Me3Si)3C have been used. The typical synthetic routes (where E = P or As) are:
2RPCl2 + 2Mg RP=PR + 2MgCl2
or
RECl2 + H2E'R' RE=E'R' + 2HCl (In the presence of base)
Allotropes of phosphorus
Elemental phosphorus exists in a number of different allotropes.
White phosphorus
The most important form of elemental phosphorus from the perspective of applications and the chemical literature is white phosphorus. It consists of tetrahedral P4 molecules, in which each atom is bound to the other three atoms by a single bond. This P4 tetrahedron is also present in liquid and gaseous phosphorus up to the temperature of 800 °C when it starts decomposing to P2 molecules. Solid white phosphorus exists in two forms. At low-temperatures, the β form is stable. At high-temperatures the α form is predominant. These forms differ in terms of the relative orientations of the constituent P4 tetrahedra.
The history of the match is linked to the discovery of the allotropes of phosphorus.
Allotropes of Phosphorus
white
red
violet - Hittorf
black
Phosphorene is an allotrope of phosphorus normally used to designate a single layer of black phosphorus that may be somewhat flattened. Conceptually the structure is similar to the carbon-based graphene, hence the name phosphorene. However phosphorene is a semiconductor, unlike graphene which is a semimetal. Recently a sample that was about 20 layers thick was shown to demonstrate high-speed data communication on nanoscale optical circuits.
White phosphorus is the most reactive, the least stable, the most volatile, the least dense, and the most toxic of the allotropes. White phosphorus gradually changes to red phosphorus. This transformation is accelerated by light and heat, and samples of white phosphorus almost always contain some red phosphorus and accordingly appear yellow. For this reason, white phosphorus that is aged or otherwise impure is sometimes called yellow phosphorus. White phosphorus glows in the dark (when exposed to oxygen) with a very faint tinge of green and blue, is highly flammable and pyrophoric (self-igniting) upon contact with air and is toxic (causing severe liver damage on ingestion). Owing to its pyrophoricity, white phosphorus has been used as an additive in napalm. The odour of combustion of this form has a characteristic garlic smell, and samples are commonly coated with white "phosphorus pentoxide", which consists of P4O10 tetrahedra with oxygen inserted between the phosphorus atoms and at their vertices. White phosphorus is insoluble in water but soluble in carbon disulfide.
Red phosphorus
In 1847 Anton von Schrotter found that sunlight changed white/yellow into red phosphorus, even when moisture and atmospheric oxygen were rigorously excluded. The red product was separated from the residual yellow phosphorus by treatment with carbon disulfide. Red phosphorus was also prepared from the yellow variety by heating it to about 250 °C. in an inert gas. Heating to higher temperatures reconverted the red modification to the yellow one.
Red phosphorus exists as an amorphous network and does not ignite in air at temperatures below 240 °C.
Violet phosphorus
In 1865, Johann Hittorf heated red phosphorus in a sealed tube at 530 °C. The upper part of the tube was kept at 444 °C. Brilliant opaque monoclinic, or rhombohedral, crystals sublimed.
This form is sometimes known as "Hittorf's phosphorus" (or violet or α-metallic phosphorus).
Black phosphorus
Black phosphorus is the thermodynamically stable form of phosphorus at room temperature and pressure. It is obtained by heating white phosphorus under high pressures (12,000 atmospheres). In appearance, properties and structure it is similar to graphite, being black and flaky, a conductor of electricity, and having puckered sheets of linked atoms.
Black phosphorus has an orthorhombic structure and is the least reactive allotrope: a result of its lattice of interlinked six-membered rings. Each atom is bonded to three other atoms. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/15%3A_The_Group_15_Elements/15.04%3A_The_Elements/15.4B%3A_Phosphorus.txt |
Nitrogen Halides
• Most compounds NX3 are known, together with the mixed ones for F and X. Some, for example NI3.nNH3 are explosive.
• N2F2 and N2F4 are known.
• The halogen azides XN3 are also known.
• NF3 is exceptional in that it has no basic properties. Salts of NF4+ such as [NF4]+[BF4] can be prepared and are powerful oxidizing agents.
15.8A: Dinitrogen Monoxide (N 2O)
Dinitrogen monoxide (Nitrous Oxide)
• Nitrous oxide, N=N=O, also known as "laughing gas" is used as an anesthetic. It is prepared by the decomposition of molten ammonium nitrate. See above.
15.8B: Nitrogen Monoxide (NO)
Nitrogen Monoxide (Nitric Oxide)
• Nitric oxide is a colourless paramagnetic gas.
• It is made, for example, by the reduction of concentrated nitric acid by copper, or reduction of nitrates and nitrites:
HNO3 + 3Cu 3Cu(NO3)2 + 4H2O + 2NO
• Its reaction include:
• Oxidation to NO2 by oxygen or to nitric acid by permanganate (used for its analysis).
• Reduction to N2O (e.g. by sulphur dioxide) or NH2OH (e.g. by Cr2+).
• It disproportionates at high temperature to N2O and NO2.
• It is implicated in blood pressure control.
• An odd-electron molecule, the bond order (which is most clearly rationalized using molecular orbital theory) is 2.5, with the odd electron in a p-antibonding orbital. This electron is easily lost to give the NºO+ ion which is isolectronic with CºO or CºN. The bond in NO+ is 0.09 Å shorter than in NO, and the stretching vibration frequency increases from 1840 cm-1 to 2150-2400 cm-1 depending on how the NO+ is behaving as a s-donor or p-acceptor ligand.
Note that dimerization to O=N-N=O would not increase the (total) number of bonds, and turns out to be energetically unrewarding.
• Nitrosonium salts include [NO]+2SO42- implicated in the "lead chamber process" for the manufacture of sulphuric acid.
15.8C: Dinitrogen Trioxide (N 2O 3)
• This blue liquid, ON-NO2, is made from equimolar quantities of NO and NO2. The N-N bond is 1.89 Å long, i.e. very weak.
15.8D: Dinitrogen Tetraoxide (N 2O 4) and Nitrogen Dioxide (NO 2)
Nitrogen Dioxide and Dinitrogen Tetroxide
• These two oxides are almost always present in an equilibrium mixture:
Solid at -11.2 oC (mp)
straw coloured
0.01% 99.99%
Vapour a 21.5 oC (bp)
Brown
0.1% 99.9%
Gas at 140 oC
Dark brown
100% 0%
The N-N bond in N2O4 is longish at 1.75 Å: the adjacent formal positive charges may be partly responsible. Since the unpaired electon in NO2 can be considered (using molecular orbital theory) to reside in an antibonding p-orbital centred largely on the nitrogen, the formation of N2O4 is more favorable than in the other paramagnetic oxides, NO and ClO2. There are two other isomers of N2O4 which are possible: ONONO2 and ONOONO but thse appear to be less stable.
• The most important reactions are:
• With water: 2NO2 (or N2O4) HNO3 + HNO2 (then 3HNO2 HNO3 + NO + H2O)
• Using it, it is possible to prepare anyhdrous nitrates which are not accessible by other routes. As an example, copper will dissolve in liquid N2O4 to give Cu(NO3)2.N2O4 (The other product is NO)
15.8E: Dinitrogen Pentaoxide (N 2O 5)
• An unstable colorless solid made by dehydration of nitric acid by phosphorus pentoxide (P4O10). In the solid state, its structure is that of the nitronium nitrate, NO2+NO3.
15.9B: Nitrous Acid (HNO 2)
Nitrous Acid
• Nitrous acid can be prepared by:
Ba(NO2)2(aq) + H2SO4(aq) BaSO4(s) + HNO2(aq)
• It readily diproportionates to nitric acid and nitic oxide.
15.9C: Nitric Acid (HNO 3) and its Derivatives
The Nitronium Ion
• This ion is present in nitric acid which autoionizes thus:
2HNO3 NO2+ + NO3 + H2O
It is also formed in mixtures of nitric and sulphuric acid:
HNO3 + H2SO4 NO2+ + HSO4 + H2O
• It is the electrophile in organic nitration reactions.
• It can be isolated as its perchlorate [NO2]+[ClO4] and as its pyrosulphate:
HNO3 + SO3 [NO2]+[HS2O7]
• It is linear and isoelectronic with CO2
Summary of Reaction and Interconversions Among the Oxides, Oxo Acids and Oxo Anions of Nitrogen
These species and many of their interconversions are of considerable industrial and environmental importance:
Reactions that do not involve disproportionation Reactions involving disproportionation
(a) 2NO2 + O3 N2O5 + O2
(b) 2NO2 + H2O2 2HNO3
(c) 2NO2 N2O4
(d) N2O4 + xs Cu Cu(NO3)2
(e) 2HNO2 + 2HI I2 + 2NO + 2H2O
(f) Fe2+ + HNO2 + H+ Fe3+ + NO + H2O
(g) 2NO + O2 + 2NO2
(h) 2Cu + NO2 Cu2O + NO
(i) C + NO2 CO2 + ½N2
(j) NO2 + 2H2 ½N2 + 2H2O
(k) 2NO2 + 7H2 2NH3 + 4H2O
(l) N2 + 3H2 2NH3
(m) 4NH3 + 3O2 2N2 + 6H2O
(n) 4NH3 + 5O2 4NO + 6H2O
(o) N2 + O2 2NO
(a') 2NO2 + H2O HNO3 + HNO2
(b') N2O4 NO+ + NO3
(c') N2O4 + 3H2SO4 NO+ + NO2+ + 3HSO4 + H3O+
(d') 3HNO3 HNO2 + 2NO + H2O
(e') 3NO N2O + NO2 | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/15%3A_The_Group_15_Elements/15.07%3A_Halides_Oxohalides_and_Complex_Halides/15.7A%3A_Nitrogen_Halides.txt |
Learning Objectives
In this lecture you will learn the following
• Organoaresnic and organoantimony compounds.
• Preparation and reactivity of pentavlent As and Sb compounds.
Organoelement compounds of group 15
Organic chemistry of non-metal phosphorus, metalloids such as arsine and antimony along with metallic element bismuth is termed as organoelement chemistry. The importance given to organoarsenic compounds earlier due to their medicinal values was waded out after antibiotics were discovered and also their carcinogenic and toxic properties were revealed. Also, the synthetically important organometallic compounds of group 13 and 14 masked the growth of group 15 elements. However, the organoelement compounds of phosphorus, arsenic and antimony find usefulness as ligands in transition metal chemistry due to their σ-donor and π-acceptor abilities which can be readily tuned by simply changing the substituents. These donor properties are very useful in tuning them as ligands to make suitable metal complexes for metal mediated homogeneous catalysis. Although organoelement compounds can be formed in both +3 (trivalent and tricoordinated) and +5(pentavalent and tetra or pentacoordinated) oxidation states, trivalent compounds are important in coordination chemistry.
For organoelement compounds of group 15, the energy of E—C bond decreases in the order, E = P > As > Sb > Bi, and in the same sequence E—C bond polarity increases.
Organometallic compounds of As(V) and Sb(V)
Due to the strong oxidizing nature of pentahalides, the direct alkylation or arylation to generate ER5 is not feasible, but can be prepared in two steps.
A few representative methods of preparation are given below:
Structures and properties
Pentaalkyl or pentaaryl derivatives are moderately thermally stable. On heating above 100°C, they form trivalent compounds as shown below:
Reaction with water,
Pentavalent compounds readily form “tetrahedral onium” cations and “octahedral and hexacoordinatged ate” anions.
In solid state, Ph5As adopts trigonal bipyramidal geometry, whereas Ph5Sb prefers square based pyramidal geometry although the energy difference between the two is marginal.
The salts of the type [R4E]+ adopt tetrahedral geometry, whereas hexacoordinated anions [R6E]-assume octahedral geometry.
Mixed organo-halo compounds of the type RnEX5-n adopt often dimeric structures due to the presence of lone pairs of electrons on X which can readily coordinate to the second molecule. The following structural types can be anticipated.
The thermal stability of RnEX5-n decreases with decreasing ‘n’. Thermal reactions are essentially the reverse reactions of addition reactions used in the preparation of R5E.
Objectives In this lecture you will learn the following
• Preparation of trivalent compounds.
• Mono and bis derivatives.
• Reaction of organo arsenic and antimony compounds.
• Structural features of organolead compounds.
Organometallic compound of As(III) and Sb(III).
Direct synthesis
Mono- derivatives
Bis derivatives:
Reactions of trialkyl derivatives, R3E
The transition metal chemistry of R3E, phosphines, arsines or stibines has been extensively studied because of their distinct donor and acceptor properties. Among them, the phosphines or tertiary phosphines (R3P) are the most valuable ligands in metal mediated homogeneous catalysis. Interestingly, the steric and electronic properties can be readily tuned by changing the substituents on phosphorus atoms. Chapter 16 is fully dedicated to the chemistry of phosphines.
Properties
Trialkyl derivatives are highly air-sensitive liquids with low boiling points and some of them are even pyrophyric. Triphenyl derivatives are solids at room temperature and are moderately stable and oxidizing agents such as KMnO4, H2O2 or TMNO are needed for oxidation to form Ph3E=O.
Cyclic and acyclic derivatives containing E—E bonds
E—E single bonds:
The E—E bond energies suggest that they do not have greater stability and the stability decreases down the group.
The simplest molecules include Ph2P—PPh2, Me2As—AsMe2 prepared by coupling reactions:
The weakness of E—E bonds accounts for many interesting reactions and a few of such reactions are listed below:
Cyclic and polycyclic derivatives can be prepared by employing any of the following methods:
Problems:
1. Confirm that the octahedral structure of [Ph6Bi]- is consistent with VSEPR theory.
Solution:
Octahedral similar to PF6-
5 (Bi valence electrons) + 6 (each Ph ) + 1 (-ve charge) = 12 electrons
i.e. six pairs, octahedral geometry
2. Comment on the stability of BiMe3 and Al2(iBu)6 with respect to their thermal decomposition and give chemical equations for their decomposition.
Solution:
Similar to other heavy p-block elements, Bi—C bonds are weak and readily undergo homolytic cleavage. The resulting methyl radicals will react with other radicals or form ethane
The Al2(iBu)6 dimer readily dissociates. At elevated temperature dissociation is followed by β-hydrogen elimination. This type of elimination is common for organometallic compounds that have alkyl groups with β-hydrogens, can form stable M—H bonds, and can provide a coordination site on the central metal.
The decomposition reaction is:
1. Using a suitable Grignard reagent, how would you prepare (i) MeC(Et)(OH)Ph; (ii) AsPh3.
Solution:
1. Add a Grignard reagent to a C=O bond, then acidify.
Several possibilities, e.g.
Me-C(O)-Et + PhMgBr → Me-C(OMgBr)(Et)(Ph) → MeC(Et)(OH)Ph or Me-C(O)-Ph + EtMgBr →etc
2. AsCl3 + 3PhMgBr → AsPh3 + 3MgBrCl.
Contirbutors
http://nptel.ac.in/courses/104101006/14 | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/15%3A_The_Group_15_Elements/15.14%3A_Sulfides_and_Selenides/15.14B%3A_Arsenic_Antimony_and_Bismuth_Sulfides.txt |
The oxygen family, also called the chalcogens, consists of the elements found in Group 16 of the periodic table and is considered among the main group elements. It consists of the elements oxygen, sulfur, selenium, tellurium and polonium. These can be found in nature in both free and combined states.
16: The Group 16 Elements
Introduction
• Oxygen is another very ubiquitous (found everywhere) element. Compounds are known with every element except helium, neon and krypton.
• Oxygen forms ionic oxides containing O2– with metals.
• Oxygen is frequently bi-covalent, X-O-Y or X=O.
• O will form one covalent bond, eg OH and O+ can form three, eg H3O+. Vary rarely, tetracovalent O2+ is found. An example is Be4O(ac)6 where the O2+ is at the centre of a tetrahedron of Be atoms.
Ionic Oxides
• The formation of an ionic oxide requires a high lattice energy, and low ionization potential of the cation to overcome the highly endothermic O=O bond dissociation energy, 496 kJ mol-1, and electron attachment enthalpy for 2 electrons, 752 kJ mol-1 in total.
Covalent Oxides
• The compounds with non-metals are covalent and can be molecular, e.g. CO2 or network structures, e.g. SiO2.
• The same two examples demonstrate the ability of oxygen to be involved in pp - pp bonding or pp - dp bonding. (The Si-O-Si bonds in SiO2 are nearly linear.)
Acid-Base Properties of Oxides
• The oxygen anions, oxide, O2–, superoxide, O2, and peroxide, O22 are hydrolysed istantly by water:
O2– + H2O 2OH
O2 + H2O O2 + HO2 + OH
O22– + H2O HO2 + OH
• Acidic Oxides: The non-metal oxides in this category are sometimes called "acid anyhdrides":
N2O5 + H2O 2HNO3
SO3 + H2O H2SO4
CO2 + H2O H2CO3
• Sometime neutral water is not enough and a base is needed:
Sb2O5 + 2OH + 5H2O 2Sb(OH)6
• Sometimes an acidic oxide is needed:
Na2O + SiO2 Na2SiO3
Some oxides are amphoteric and others are more inclined to undergo redox reactions than acid base reactions if they are not inert, e.g. MnO2.
Occurence, Isolation and Allotropy
• There are three natural isotopes of oxygen of relative abundance: 16O = 99.759%, 17O = 0.0374% and 18O = 0.2039%
The heavier ones can be enriched by fractionation of liquid oxygen and they are used in tracer studies.
• There are two molecular forms (allotropes), dioxygen, O2 the common form, and trioxygen, ozone, O3.
• Dioxygen is a paramagnetic molcule with 2 unpaired electrons. Molecular orbital theory provides a ready explanation for this: valence bond theory does not.
• Ozone is a paramagnetic bent molecule formed when dioxygen is passed through an electric discharge or irradiated with ultra-violet light. Its sharp odour can be smelled aroung electric motors and some older photcopiers. The liquid is dark blue. Liquid dioxygen is pale blue.
Chemical Properties of Oxygen and Ozone
• Ozone is the more oxidizing of the two.
• Ozone can be measured by the reaction:
O3 + KI + H2O I2 + 2KOH + O2
The iodine is titrated with thiosulphate.
• Ozone is used in place of chlorine for water treatment.
• Oxygen is a common contaminent in organic and aqueous solvents, and it can be very difficult to remove.
Hydrogen Peroxide
• Hydrogen peroxide, H2O2, is a colorless liquid boiling at 152.1 oC.
• It is ~40% more dense than water, and like water it is extensively hydrogen bonded. When pure it can decompose explosively. The most stable rotational conformation has a tortion angle of 96.5o - see Figure18-1.
• It is unstable with respect to water and dioxygen by -97 kJ mol-1.
• It is somewhat more acidic than water:
H2O2 H+ + HO2 k = 1.5x10-12
• There are two principle methods of synthesis:
1. 2HSO4 HO(O)2S-O-O-S(O)2OH + 2e (by electrolysis)
HO(O)2S-O-O-S(O)2OH + H2O H-O-O-S(O)2OH + H2SO4 (fast)
H-O-O-S(O)2OH + H2O H2SO4 + H2O2 (slow)
The H2O2 can be obtained 90-98% pure by distillation.
• Hydrogen peroxide is a good oxidizing agent, fast in base and slower in acid and often acts via a free radical mechanism.
The Peroxides and Superoxides
• Ionic peroxides are formed by the alkali and alkaline earth elements (not Be). They are also oxidizing agents.
• Paramagnetic ionic superoxides are formed by potassium, rubidium and cesium. They are very powerful oxidizing agents.
• Other covalent peroxides include the peroxo acids, e.g. peroxosulphuric and peroxodisulphuric acid, the intermediates in the production of H2O2 shown above, and the dangerously explosive organic peroxides ROOR which form by free radical oxidation of ethers. (They are the reason why ethers shold never be distilled to dryness unless the peroxides have been destroyed immediately beforehand by washing with with FeSO4 or passing the ether over activated alumina. A solution of Fe2+ and SCN is used to test for their presence - look for the formation of the blood red [FeIII(SCN)]2+ ion.)
The Dioxygenyl Radical (O2+)
See notes on Chapter 21.
Dioxygen as a Ligand
Read this section - oxygen binding to transition metals is of interest because of its binding by haemoglobin and myoglobin. Note the structural types.
Oxygen Compounds as Ligands
More transition metal chemistry. Skip it for now.
Oxygen Florides
See notes on Chapter 20.
Chemistry 242 - Inorganic Chemistry II Chapter 19 - Sulphur, Selenium, Tellurium and Polonium
Introduction
These elements differ considerably from oxygen:
1. Their electronegativity is lower.
2. Their covalent bonding is generally weaker.
3. Hydrogen bonding, where it is a possibility is very weak.
• The do not form compounds where pp - pp bonding is needed, but rather use dp - pp bonding especially with oxygen.
• A covalence exceeding 4 is possible by the use of empty d-orbitals. The 6-coordinate geometry is increasingly favoured down the group.
• There are sulphur compounds with very long chains (second only to carbon)
• Tellurium an dpolonium are fairly metallic in their properties.
Occurence an dReactions of the Elements
• Sulphur occurs "native" (i.e. as sulpur) in deposits from which it is extracted with high pressure hot water (Frasch process). It is also obtained from hydrogen sulphide in natural gas and petroleum - if it were left, it cause a pollution problem when the fuels are burned.
2H2S + SO2 3S + H2O
• Selenium and tellurium come from silver and copper smelting flue gases. (These metals come from sulphide ores. Selenium and tellurium tend to be found with sulphur.)
• Sulphur forms a number of allotropes:
• Behaviour on melting:
Solid ~112 oC
(just melted)
160 oC 444.6 oC
(just boiling)
Vapour phase
S8
Yellow
S13.8 Sn (n is maximum)
Dark brown
Sn, (incl S3, S4
Dark red)
S8 going to S2
at higher temperatures
A rubbery material called "plastic sulphur" can be obtaind by quickly cooling molten sulphur.
• There are several crystalline modifications of S8 stable at different temperatures.
• It is possible to isolate other rings sizes from 6 to 20. Engel's sulpur contains S6 rings.
• Sulphur is used extensively to harden synthetic and natural rubbers - "vulcanization". Bridging S2 units are of the things that hold proteins in their correct shapes.
Hydrides
• The main one is hydrogen sulphide, H2S, which smells of rotten eggs and is very much more poisonous than hydrogen cyanide.
• The compounds are all gases whose stability decreases down the group.
• The acidity of the hydrides increases down the group.
• The series of sulphanes, H2Sn where n is 2 to 6 have been characterized. There are higher ones, but their separation is impossible since the chains tend to break. They are synthesized by the reaction:
SnCl2 + 2H2S H2Sn+2 + 2HCl
Halides and Oxohalides of Sulphur
Sulphur Flourides
• Synthesis:
S8 + xs F2 SF6 (+ SF4 + S2F10)
SCl2(l) + 4NaF(s) + SF4(g) + Na2S + 2NaF
• Sulphur tetrafluoride (bp = -30 oC) is quite reactive, for example it is easily hydrolysed:
SF4 + 2H2O SO2 + 4HF
It is used as a selective fluorinating agent:
>C=O is converted to >CF2
-C(O)OH is converted to -CF3
• Sulphur hexafluoride (sublimes at -64 oC) is very inert kinetically (since the hydrolysis is thermodynamically very favorable). Presumably it cannot further expand its coordination sphere to a reaction intermediate.
It is used as a gaseous electrical insulator, much better than air, because of its high dielectric constant and lack of reactivity.
Sulphur Chlorides
• Disulphur Dichloride and Sulphur Dichloride
¼S8(s) + Cl2(g) S2Cl2
(SCl2 is unstable, decomposing slowly to S2Cl2 and chlorine.)
Sulphur dichloride or disulphur dichloride will dissolve more sulphur to form sulphanes upto around S100Cl2.
• Thionyl Chloride
SO2 + PCl5 SOCl2 + POCl3
Thionyl chloride is a liquid (bp = 80 oC) which is rapidly hydrolysed:
SOCl2 + H2O SO2(g) + 2HCl(g)
Because the hydrolysis products are both gases, one of its great uses in inorganic chemistry is in the dehydration of hydrated metal chlorides to produce the anhydrous substance.
• Sulphuryl Chloride
SO2 + Cl2 SO2Cl2 FeCl3 catalyst
It is used as a chlorinating agent in organic synthesis.
Oxides and Oxo Acids
Take note of Table 19-1 in the text - it contains an important summary of the sulphur oxoacids. Know the names and structures.
Sulphur Dioxide
• is a gas boiling at -10 oC which is a useful solvent when liquified and can act as a ligand towards transition metals.
• Although it is formally the anhydride of sulphurous acid, H2SO3 it does not react with water to any significant extent. However, the soluble salts of bisulphite, HSO32–, and sulphite, SO32– are well characterized.
Sulphur Trioxide
• This oxide is the anyhdride of sulphuric acid, and does indeed react with water to produce it.
• It is produced by the oxidation of sulphur dioxide with oxygen catalysed by a heterogeneous catalyst such as V2O5 (the "contact process") or a homogeneous catalyst such as nitric oxide (the "lead chamber process").
Thiosulphate Ion
• This ion is formed by the reaction of sulphur with sulphite:
SO32–(aq) + S(s) S2O32–(aq)
• It is used in iodometric titrations forming the tetrathionate ion:
S2O32– + I2 2I2– + S4O62–
• It is used as "fixer" in photography, specifically, it dissolves silver chloride by complexing the silver ion as [Ag(S2O3)2]3-:
AgCl(s) + 2Na2S2O3(aq) Na3[Ag(S2O3)2] + NaCl
16.04: The Elements
Oxygen is a member of the chalcogen group (Group 16) of the periodic table and is a highly reactive nonmetallic element and oxidizing agent that readily forms compounds (notably oxides) with most elements. By mass, oxygen is the third-most abundant element in the universe, after hydrogen and helium. At standard temperature and pressure, STP, two atoms of the element bind to form dioxygen, O2, a diatomic gas that is colourless, odourless, and tasteless.
Discovery
Oxygen was discovered independently by Carl Wilhelm Scheele, in Uppsala, Sweden in 1773 or earlier, and Joseph Priestley in Wiltshire, UK in 1774, but Priestley is generally given the credit since his work was published first. The name oxygen was coined in 1777 by Antoine Lavoisier, whose experiments with oxygen helped to discredit the then-popular phlogiston theory of combustion and corrosion. Its name derives from the Greek roots οξυζ oxys, "acid", literally "sharp", referring to the sour taste of acids and -γενης genes meaning "creator", because at the time of naming, it was mistakenly thought that all acids required oxygen in their composition.
Allotropes of oxygen
There are 2 main allotropes of oxygen although a third has recently been shown to form under high pressures.
Some allotropes of oxygen
O2 - dioxygen O3 - ozone O8 - red oxygen
Dioxygen - O2
The common allotrope of elemental oxygen is often just called oxygen, O2, but to help distinguish it from the element may be called dioxygen or molecular oxygen. Elemental oxygen is most commonly encountered in this form, as about 21% (by volume) of the Earth's atmosphere is O2, the remainder largely being dinitrogen, N2. At STP, dioxygen is a colourless, odourless gas, in which the two oxygen atoms are chemically bonded to each other giving rise to two unpaired electrons occupying two degenerate molecular orbitals. The electron configuration of O2 molecules in this form, a diradical, indicates that they should be paramagnetic. This is a classic example of where a simple Lewis structure fails to account for the properties and where an MO approach correctly provides the explanation.
The electron configuration (ignoring 1s orbitals) is:
2s)22s*)22p)22p)42p*)2 and from this the Bond Order is found to be ½(2 -2 + 2 + 4 -2 ) = 2 that is, a double bond as shown in the Lewis structure as well. The difference though is that the Lewis structure does not predict the molecule to be paramagnetic. The bond length is 121 pm and the bond energy is 498 kJmol-1.
A video clip showing liquid dioxygen being poured between the faces of a magnet and attracted into the magnet field has been prepared as a Harvard Natural Sciences Lecture Demonstration.
With 2 electrons to be placed in 2 degenerate orbitals, a number of variations are possible and the arrangement above where the 2 electrons are parallel is considered to be the most stable. Note that the spin multiplicity is given by the formula, 2S+1 and so for S=1 from s=½ + ½ then 2S+1 = 3 i.e. a spin triplet.
a) e-s parallel (triplet 3Σg), b) 2 e-s in 1 orbital (singlet 1Δg), c) e-s opposed (singlet 1Σg) Singlet oxygen is the name commonly used for the electronically excited state shown in b) and it is less stable than the normal triplet state a) by 94.7 kJmol-1. In isolation, singlet oxygen can persist for over an hour at room temperature. The other singlet state at 157.8 kJmol-1 shown in c) is very short lived and relaxes quickly to b). Because of differences in their electron shells, singlet and triplet oxygen differ in their chemical properties. Singlet oxygen is highly reactive.
Reactions of triplet dioxygen are restricted by conservation of spin state rules and at ambient temperatures this prevents direct reaction with all but the most reactive substrates, e.g. white phosphorus. At higher temperatures, or in the presence of suitable catalysts, reactions proceed more readily. For instance, most flammable substances are characterised by an autoignition temperature above which they will undergo combustion in air without an external flame or spark.
The energy difference between the ground state and singlet oxygen is 94.7 kJmol-1 which would correspond to a transition in the near-infrared at ~1263 nm. In the isolated molecule, this transition is strictly forbidden by spin, symmetry and parity selection rules, making it one of nature's most forbidden transitions. In other words, direct excitation of ground state oxygen by light to form singlet oxygen is very improbable. As a consequence, singlet oxygen in the gas phase is extremely long lived (72 minutes). Interaction with solvents however, can reduce the lifetime to microseconds or even nanoseconds.
Various methods for the production of singlet oxygen exist. A photochemical method involves the irradiation of normal oxygen gas in the presence of an organic dye as a sensitizer, such as methylene blue. Singlet oxygen can be produced chemically as well. One of the chemical methods is by the reaction of hydrogen peroxide with sodium hypochlorite. This is convenient in small laboratories and for demonstrative purposes:
H2O2 + NaOCl → O2(1Δg) + NaCl + H2O In photosynthesis, singlet oxygen can be produced from the light-harvesting chlorophyll molecules. One of the roles of carotenoids in photosynthetic systems is to prevent damage caused by any singlet oxygen that is produced by either removing excess light energy from chlorophyll molecules, or quenching the singlet oxygen molecules directly.
Biological considerations
Molecular dioxygen is a potentially strong oxidizing agent, based on its position in the electrochemical series. The standard redox potential is:
O2 + 4H+ + 4e- ⇄ 2 H2O E = +1.229 V
which is comparable to potassium dichromate (1.33 V). Nevertheless, reactions of dioxygen with most substrates tend to proceed very slowly at room temperature, in the gas phase or in solution. This relative inertness (unexpected considering it is a diradical) is critical for sustaining life in a dioxygen atmosphere. Perhaps one of the reasons for this stablilty is that the reaction above shows that a 4 electron change is required and this is highly improbable so that it is more likely that a sequence of steps involving 1 electron changes actually takes place.
O2 + e- → O2-* (superoxide)
O2-* + e- + 2H+ → H2O2 (peroxide)
H2O2 + e- + H+ → OH* (hydroxy radical)
OH* + e- + H+ → H2O (water)
The first step, that generates superoxide, has an adverse potential of -0.32 V (ΔG= +30.9 kJmol-1) which imparts some kinetic stability to the molecule. Note that both dioxygen and superoxide ion are free radicals that exhibit paramagnetism. Species like superoxide, peroxide and the hydroxy radical are far too reactive to be allowed to accumulate in living systems and the primary defence of the cell appears to be to destroy them as soon as they are formed. This is done by a variety of enzymes including superoxide dismutase (SOD) and catalases. Parts of the immune system of higher organisms, however, create peroxide, superoxide, and singlet oxygen to destroy invading microbes. Reactive oxygen species also play an important role in the hypersensitive response of plants against pathogen attack.
Another issue with dioxygen is its low solubility in water at room temperature and atmospheric pressure. Dioxygen is more soluble in water than is dinitrogen. Water in equilibrium with air contains approximately 1 molecule of dissolved O2 for every 2 molecules of N2, despite the atmospheric ratio of approximately 1:4. The solubility of oxygen in water is temperature-dependent, and about twice as much (14.6 mgL-1) dissolves at 0 °C than at 20 °C (7.6 mgL-1). At 25 °C and 1 standard atmosphere (101.3 kPa) of air, freshwater contains about 6.04 milliliters (mL) of dioxygen per liter, whereas seawater contains about 4.95 mgL-1. At 5 °C the solubility increases to 9.0 mgL-1 (50% more than at 25 °C) for water and 7.2 mgL-1 (45% more) for sea water.
The oxygen in water is unavailable to mammals so that divers (and diving mammals such as whales and seals) are entirely dependent on the oxygen carried in the air in their lungs or their gas supply. For divers this is complicated since at higher partial pressures oxygen can cause acute toxicity leading to convulsions. Scuba (self-contained underwater breathing apparatus) divers using compressed air are restricted to diving above 30 m. If the diver descends to depths near 100 m they can become unconsciousness from nitrogen narcosis and death usually results. A dive to 30 m for 20 minutes puts the scuba diver at risk of nitrogen narcosis and decompression illness. By contrast, the elephant seal can dive to 1 km for 1 hour without risk of either condition.
Respiratory pigments are capable of fixing dioxygen from the atmosphere, transporting it to the reacting site and then releasing it. In addition they are able to counteract small fluctuations in supply or demand by storing dioxygen.
O2 can bind to a single metal center either "end-on" (η1-) or "side-on" (η2-). In Haemoglobin and Myoglobin it is "end-on".
fully oxygenated Haemoglobin
(with 4 Fe porphins and O2's highlighted)
Iron based proteins like haemoglobin and myglobin as well as copper based proteins like haemocyanin perform this role in mammalian systems and lobsters, etc.
Molecular dioxygen, O2, is essential for cellular respiration in all aerobic organisms. Oxygen is used in mitochondria to help generate adenosine triphosphate (ATP) during oxidative phosphorylation. The reaction for aerobic respiration is essentially the reverse of photosynthesis and is simplified as:
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O ΔH= -2880 kJmol-1 In vertebrates, O2 diffuses through membranes in the lungs and into red blood cells. Haemoglobin binds O2, changing its colour from bluish red to bright red (CO2 is released from another part of haemoglobin through the Bohr effect). Other animals use haemocyanin (molluscs and some arthropods) or haemerythrin (spiders and lobsters). A liter of blood can dissolve 200 cm3 of O2. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/16%3A_The_Group_16_Elements/16.04%3A_The_Elements/16.4A%3A_Dioxygen.txt |
Ozone is colourless or slightly bluish gas (blue when liquified), slightly soluble in water and much more soluble in inert non-polar solvents such as carbon tetrachloride or fluorocarbons, where it forms a blue solution. At 161 K (-112 °C), it condenses to form a dark blue liquid. It is dangerous to allow this liquid to warm to its boiling point, because both concentrated gaseous ozone and liquid ozone can detonate.
Most people can detect about 0.01 μmol/mol of ozone in air where it has a very specific sharp odour somewhat resembling chlorine bleach. Exposure of 0.1 to 1 μmol/mol produces headaches, burning eyes and irritation to the respiratory passages. Even low concentrations of ozone in air are very destructive to organic materials such as latex, plastics and animal lung tissue.
Seaside air was once considered to be healthy because of its "ozone" content. It is now recognised that the smell in reality is released from bacteria following the breakdown of species from rotting seaweed! (dimethyl sulfoxide, DMS). Birds appear to be attracted to the smell since for them it indicates a plankton bloom, and therefore the presence of fish feeding on the marine plants.
Ozone may be formed from O2 by electrical discharges and by action of high energy electromagnetic radiation. Unsuppressed arcing breaks down the chemical bonds of the atmospheric oxygen surrounding the contacts [O2 → 2O]. Free radicals of oxygen in and around the arc recombine to create ozone [O3]. Certain electrical equipment generate significant levels of ozone. This is especially true of devices using high voltages, such as ionic air purifiers, laser printers, photocopiers, tasers and arc welders. Electric motors using brushes can generate ozone from repeated sparking inside the unit. Large motors that use brushes, such as those used by elevators or hydraulic pumps, will generate more ozone than smaller motors. Ozone is similarly formed in the Catatumbo lightning storms phenomenon on the Catatumbo River in Venezuela, which helps to replenish ozone in the upper troposphere. It is the world's largest single natural generator of ozone. The lightning originates from a mass of storm clouds at a height of more than 5 km, and occurs during 140 to 160 nights a year, 10 hours per day and up to 280 times per hour.
From the imgkid gallery (accessed 31 Jan 2015).
Ozone-Oxygen cycle
Ninety percent of the ozone in the atmosphere sits in the stratosphere, the layer of atmosphere between about 10 and 50 kilometers altitude. The ozone layer is mainly found in the lower portion of the stratosphere, from approximately 20 to 30 kilometres (12 to 19 mi) above Earth, though the thickness varies seasonally and geographically. Here the concentration of O3 varies from two to eight ppm which is much larger than the average ozone concentration in the Earth's atmosphere which is only about 0.3 parts per million.
The natural level of ozone in the stratosphere can be considered to be a result of a balance between sunlight that creates ozone and chemical reactions that destroy it.
Creation: photolysis of an oxygen molecule by high energy UV light splits it into two oxygen atoms. The wavelength in the UV needed to achieve this can be estimated from the O=O Bond Energy of 498 kJmol-1.
By dividing this by 6.022 * 1023 bonds/mol the Energy per bond is found to be 8.27 * 10-19 J/bond.
The wavelength required for photolysis can be evaluated from E= hc / λ
λ= hc /E = (6.626 * 10-34 Js x 3 * 108 m/s) / ( 8.27 * 10-19 J) so λ= 240 nm, i.e. visible light cannot break the O=O bond, but UV-C light with energy > 240 nm can break the O=O bond. {UV ranges are: UV-A (400-315 nm), UV-B (315-280 nm), and UV-C (280-200 nm).}
O2 + hν → 2O
Each oxygen atom then rapidly combines with an oxygen molecule to form an ozone molecule:
O + O2 → O3 The ozone-oxygen cycle: the ozone molecules formed by the reaction above absorb radiation again having an appropriate wavelength in the UV. The triatomic ozone molecule becomes diatomic molecular oxygen plus a free oxygen atom: O3 + hν (240-310 nm) → O2 + O
The atomic oxygen produced quickly reacts with another oxygen molecule to reform ozone:
O + O2 → O3 + K.E.
where "K.E." denotes the excess energy of the reaction which is manifested as extra kinetic energy. These two reactions form the ozone-oxygen cycle, in which the chemical energy released when O and O2 combine is converted into kinetic energy of molecular motion. The overall effect is to convert penetrating UV-B light into heat, without any net loss of ozone. This cycle keeps the ozone layer in a stable balance while absorbing 97-99% of the Sun's medium-frequency ultraviolet light (from about 200 nm to 315 nm wavelength) and preventing it from reaching the Earth's surface, to the benefit of both plants and animals. This reaction is one of two major sources of heat in the stratosphere (the other being the kinetic energy released when O2 is photolyzed into O atoms).
Removal: reaction of an oxygen atom with an ozone molecule leads to production of two oxygen molecules:
O3 + O. → 2 O2
and if two oxygen atoms meet, they can react to form one oxygen molecule:
2 O. → O2
The overall amount of ozone in the stratosphere is determined by a balance between production by solar radiation and removal. The removal rate is slow, since the concentration of O atoms is very low. Certain free radicals, the most important being hydroxyl (OH), nitric oxide (NO) and atoms of chlorine (Cl) and bromine (Br), catalyze the recombination reaction, leading to an ozone layer that is thinner than it would be if the catalysts were not present.
Most of the OH and NO are naturally present in the stratosphere, but human activity, especially emissions of chlorofluorocarbons (CFCs) and halons, has greatly increased the Cl and Br concentrations, leading to ozone depletion. Each Cl or Br atom can catalyze tens of thousands of decomposition reactions before it is removed from the stratosphere.
Follow the Ozone Hole Watch at NASA and see the ozone depletion Wikipedia page.
Ozone depletion
Ozone depletion describes two distinct but related phenomena observed since the late 1970s: a steady decline of about 4% per decade in the total volume of ozone in the ozone layer, and a much larger springtime decrease in stratospheric ozone over Earth's polar regions. Note that the Antarctic would be in darkness (no sunshine) for the winter months, sunshine returning in springtime (September). The latter phenomenon is referred to as the ozone hole.
The most important process in depletion is catalytic destruction of ozone by atomic halogens. The main source of these halogen atoms in the stratosphere is photodissociation of man-made halocarbon refrigerants, solvents, propellants, and foam-blowing agents (CFCs, HCFCs, freons, halons), substances that are referred to as ozone-depleting substances (ODS).
In the simplest example of such a cycle, a chlorine atom reacts with an ozone molecule, taking an oxygen atom with it (forming ClO) and leaving a normal oxygen molecule. The chlorine monoxide (ClO) produced can react with a second molecule of ozone to yield another chlorine atom and two molecules of oxygen. These gas-phase reactions can be written such that in the first step a chlorine atom changes an ozone molecule to ordinary oxygen:
Cl. + O3 → ClO + O2
The ClO generated can then destroy a second ozone molecule and recreate the original chlorine atom, which can repeat the first reaction and continue a cycle to destroy ozone.
ClO + O3 → Cl. + 2 O2 In theory, a single chlorine atom could keep on destroying ozone (acting as a catalyst) for up to two years (the time scale for transport back down to the troposphere) were it not for reactions that remove Cl by forming reservoir species such as hydrogen chloride (HCl). In practise, the average chlorine atom reacts with 100,000 ozone molecules before it is removed from the catalytic cycle.
On a per atom basis, bromine is even more efficient than chlorine at destroying ozone, but at present there is much less bromine in the atmosphere. Both chlorine and bromine significantly contribute to the overall depletion of ozone.
The Montreal Protocol on Substances that Deplete the Ozone Layer (a protocol to the Vienna Convention for the Protection of the Ozone Layer) is an international treaty designed to protect the ozone layer by phasing out the production of numerous substances that are responsible for ozone depletion. It was agreed on in 1987, and entered into force on January 1, 1989, with numerous revisions since then.
The bans on the production of CFCs, halons, and other ozone-depleting chemicals such as carbon tetrachloride and trichloroethane have led to the expectation of a recovery of the ozone layer to 1980 values somewhere between 2050 and 2070. This was the estimate given in a summary document of the Scientific Assessment of the Ozone Depletion 2014 published by the UN Environment Programme (UNEP) and the UN World Meterological Organization (WMO).
Among the key findings of the report were that the authors noted that what happens after 2050 will then largely depend on concentrations of CO2, methane and nitrous oxide - the three main long-lived greenhouse gases in the atmosphere.
NASA view of the largest ozone hole observed above the Antarctic on 24 Sep 2006. Ozone is much less stable than the diatomic allotrope O2, breaking down in the lower atmosphere to normal dioxygen. The O - O distances in O3 are 127.2 pm and the O - O - O angle is 116.78°. The bonding can be expressed as a resonance hybrid with a single bond on one side and double bond on the other producing an overall average bond order of 1.5 for each side.
Ozone is a powerful oxidant (far more so than dioxygen) and has many industrial and consumer applications related to oxidation. This same high oxidizing potential, however, causes ozone to damage mucous and respiratory tissues in animals, and also tissues in plants, above concentrations of about 100 ppb. This makes ozone a potent respiratory hazard and pollutant near ground level.
Ozone oxidizes nitric oxide to nitrogen dioxide: NO + O3 → NO2 + O2
This reaction is accompanied by chemiluminescence. The NO2 can be further oxidized:
NO2 + O3 → NO3 + O2
The NO3 formed can react with NO2 to form N2O5.
Ozone does not react with ammonium salts, but it oxidizes ammonia to ammonium nitrate:
2 NH3 + 4 O3 → NH4NO3 + 4 O2 + H2O Ozone reacts with carbon to form carbon dioxide, even at room temperature: C + 2 O3 → CO2 + 2 O2
Red oxygen - O8
On increasing the pressure of dioxygen at room temperature, the "solid oxygen β-phase" undergoes phase transitions to the δ-phase at 9 GPa and the ε-phase at 10 GPa. Its volume decreases significantly, and it changes colour from blue to deep red. The structure of this red ε-phase, determined in 2006, is based on an O8 cluster (see structure above). It was confirmed that this structure is formed up to 96 GPa. The box-like cluster is an unusual conformation first recognised for oxygen, and as yet has not been experimentally or theoretically reported for any other diatomic molecule.
Liquid oxygen is already used as an oxidant in rockets, and it has been speculated that red oxygen could make an even better oxidant, because of its higher energy density. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/16%3A_The_Group_16_Elements/16.04%3A_The_Elements/16.4B%3A_Ozone.txt |
Allotropes of sulfur
No other element forms more solid allotropes than sulfur. At present, about 30 well characterized sulfur allotropes are known of which the most common form found in nature is the greenish-yellow orthorhombic α-sulfur, containing puckered rings of S8.
α-sulfur
When pure it has a greenish-yellow colour (traces of cyclo-S7 in commercially available samples make it appear yellower). It is practically insoluble in water and is a good electrical insulator with poor thermal conductivity. It is quite soluble in carbon disulfide: 35.5 g/100 g solvent at 25 °C. It has a rhombohedral crystal structure. This is the predominant form found in "flowers of sulfur", "roll sulfur" and "milk of sulfur". It contains S8 puckered rings, alternatively called a crown shape. The S-S bond lengths are all 206 pm and the S-S-S angles are 108° with a dihedral angle of 98°. At 95.3 °C, α-sulfur converts to β-sulfur.
β-sulfur
This is a yellow solid with a monoclinic crystal form and is less dense than α-sulfur. Like the α- form it contains puckered S8 rings and only differs from it in the way the rings are packed in the crystal. It is unusual because it is only stable above 95.3 °C, below this it converts to α-sulfur. It can be prepared by crystallising at 100 °C and cooling rapidly to slow down formation of α-sulfur. It has a melting point of about 120 °C and decomposes at around this temperature.
γ-sulfur
This form, first prepared by F.W Muthmann in 1890, is sometimes called "nacreous sulfur" or "mother of pearl sulfur" because of its appearance. It crystallises in pale yellow monoclinic needles. It contains puckered S8 rings like α-sulfur and β-sulfur and only differs from them in the way that these rings are packed. It is the densest form of the three. It can be prepared by slowly cooling molten sulfur that has been heated above 150 °C or by chilling solutions of sulfur in carbon disulfide, ethyl alcohol or hydrocarbons. It is found in nature as the mineral rosickyite.
Some allotropes of Sulfur
S6 - cyclohexasulfur
α-S8
S12 - cyclododecasulfur
S6 - cyclo-hexasulfur
This was first prepared by M.R. Engel in 1891 who reacted HCl with thiosulfate, HS2O3-. Cyclo-S6 is orange-red and forms rhombohedral crystals. It is called ρ-sulfur, ε-sulfur, Engel's sulfur and Aten's sulfur. Another method of preparation involves reacting a polysulfane with sulfur monochloride: H2S4 + S2Cl2 → cyclo-S6 + 2 HCl (dilute solution in diethyl ether)
The sulfur ring in cyclo-S6 has a "chair" conformation, reminiscent of the chair form of cyclohexane. All of the sulfur atoms are equivalent.
Cyclo-dodecasulfur
Thermodynamically, S12 is the second most stable sulfur ring after S8. Therefore, S12 is formed in many chemical reactions in which elemental sulfur is a product. In addition, S12 is a component of liquid sulfur at all temperatures. The same holds for S18 and S20 which are often formed together with S12. Its structure can be visualised as having sulfur atoms in three parallel planes, 3 in the top, 6 in the middle and three in the bottom.
Liquid sulfur after equilibration contains sulfur homocycles of all sizes and some of these can be isolated by quenching, extraction, fractional precipitation and crystallization depending on their differing solubilities.
Cyclo-S12 can be prepared by heating elemental sulfur to about 200 °C for 5-10 min and then allowing the mixture to cool to 140-160 °C within about 15 min. Once the melt has become less viscous, it is poured in as thin a stream as possible into liquid nitrogen in order to quench the equilibrium. Recrystallization of the yellow powder from CS2 allows the isolation of an adduct which slowly loses the solvent to give the cyclo-dodecasulfur.
Note that both B and S form stable E12 species but the structures (and coordination numbers) are quite different. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/16%3A_The_Group_16_Elements/16.04%3A_The_Elements/16.4C%3A_Sulfur_-_Allotropes.txt |
Hydrogen peroxide (H2O2) is the simplest peroxide (a compound with an oxygen-oxygen single bond) and in its pure form is a colourless liquid that is slightly more viscous than water. It is a strong oxidizer and is used as a bleaching agent and disinfectant. For safety reasons it is normally used as an aqueous solution, also colourless. For consumers, it is usually available from pharmacies at 3 and 6 wt% concentrations. The concentrations are sometimes described in terms of the volume of oxygen gas generated; one milliliter of a 20-volume solution generates twenty milliliters of oxygen gas when completely decomposed. For laboratory use, 100 volume, 30 wt% solutions are the most common. Concentrated H2O2, or 'high-test peroxide' is a reactive oxygen species and has been used as a propellant in rocketry. Diluted H2O2 (between 1.9% and 12%) mixed with ammonium hydroxide has been used to bleach human hair. The chemical's bleaching property lends its name to the phrase "Hollywood peroxide blonde". Hydrogen peroxide can be used for tooth whitening and when mixed with baking soda and salt forms a recipe for home-made toothpaste.
Discovery
Hydrogen peroxide was first described in 1818 by Louis Jacques Thénard, who produced it by treating barium peroxide with nitric acid. An improved version of this process used hydrochloric acid, followed by addition of sulfuric acid to precipitate the barium sulfate byproduct. Thénard's process was used from the end of the 19th century until the middle of the 20th century.
Pure hydrogen peroxide was long believed to be unstable as early attempts to separate it from the water, which is present during synthesis, all failed. This instability was due to traces of impurities (transition metals salts) which catalyze the decomposition of the hydrogen peroxide. Pure hydrogen peroxide was first obtained in 1894 - almost 80 years after its discovery - by Richard Wolffenstein, who produced it via vacuum distillation.
Determination of the molecular structure of hydrogen peroxide proved to be very difficult. In 1892 the Italian physical chemist Giacomo Carrara (1864-1925) determined its molecular weight by freezing point depression, which confirmed that its molecular formula was H2O2. At least half a dozen hypothetical molecular structures seemed to be consistent with the available evidence. In 1934, the English mathematical physicist William Penney and the Scottish physicist Gordon Sutherland proposed a molecular structure for hydrogen peroxide which was very similar to the presently accepted one and which subsequent evidence cumulatively proved to be correct.
The structure of hydrogen peroxide
H2O2 - gas phase
(O-O 147.4 pm)
H2O2 - solid phase
(O-O 145.8 pm)
Although the O-O bond is a single bond, the molecule has a relatively high barrier to rotation of 29.45 kJmol-1 (2460 cm-1); for comparison, the rotational barrier for ethane is just 12.5 kJmol-1. The increased barrier is thought to be due to the repulsion between the lone pairs of the adjacent oxygen atoms.
The molecular structures of gaseous and crystalline H2O2 are significantly different. This difference is attributed to the effects of hydrogen bonding, which is absent in the gaseous state.
Preparation via the anthraquinone process
The anthraquinone process is a process for the production of hydrogen peroxide, which was developed by BASF. The industrial production of hydrogen peroxide is based on the reduction of oxygen with hydrogen, by the direct synthesis from the elements. Instead of hydrogen itself, however, a 2-alkyl-anthrahydroquinone (generated from the corresponding 2-alkyl-anthraquinone by catalytic hydrogenation with palladium) is used. Oxygen and the organic phase react to give hydrogen peroxide and the anthraquinone can be recycled.
The hydrogen peroxide is then extracted with water in a second step and separated by fractional distillation. The anthraquinone thus acts as a catalyst with the overall reaction given by the equation:
H2 + O2 → H2O2 If ozone is used instead of oxygen, dihydrogen trioxide, H2O3 can be produced by this method.
Decomposition
Hydrogen peroxide is thermodynamically unstable and decomposes to form water and oxygen with a ΔH of -98.2 kJmol-1 and a ΔS of 70.5 Jmol-1K-1.
2 H2O2 → 2 H2O + O2 The rate of decomposition increases with rising temperature, concentration and pH, with cool, dilute, acidic solutions showing the best stability. Decomposition is catalysed by various compounds, including most transition metals and their compounds (e.g. manganese dioxide, silver, and platinum). Certain metal ions, such as Fe2+ or Ti3+, can cause the decomposition to take a different path, with free radicals such as (HO.) and (HOO.) being formed.
The decomposition of hydrogen peroxide liberates oxygen and heat; this can be dangerous as spilling high concentrations of hydrogen peroxide on an inflammable substance can cause an immediate fire.
Metal oxides, peroxides and superoxides
When the group 1 metals are heated in an excess of air or in O2, the principal products obtained depend on the metal: lithium oxide, Li2O, sodium peroxide, Na2O2, and the superoxides KO2, RbO2 and CsO2.
4Li + O2 → 2Li2O oxide formation
2Na + O2 → Na2O2 peroxide formation
K + O2 → KO2 superoxide formation
The superoxides and peroxides contain the paramagnetic [O2]- (μ ~ 1.73 BM) and diamagnetic [O2]2- ions respectively. See the MO diagram for O2 above for comparison.
All of the monoxides are known, M2O, and the heavier peroxides and superoxides will decompose to form the oxides. They are all ionic and strong bases with the basicity increasing down the group.
O2- + H2O(aq) → 2 OH- (aq) Sodium peroxoborate is a solid peroxygen compound with exceptional storage stability and no shock sensitivity. It is cheap and readily available, being produced mainly as a solid ingredient of domestic washing formulations, in which it acts as sources of H2O2 in solution for stain bleaching. World annual production in 1995 was approximately 750,000 tonnes, and its use in washing powders dates back to 1907 with Henkel's original "Persil" product in Germany.
Sodium peroxoborate has the empirical formula "NaBO3.xH2O". Two forms that are commercially available correspond stoichiometrically to x = 1 or 4, and are known as the "monohydrate" and "tetrahydrate', respectively. Structurally, however, sodium peroxoborate was shown in 1961 to be the disodium salt of a 1,4.-diboratetroxane dianion. Hence, the "monohydrate" really corresponds to the anhydrous salt, and the "tetrahydrate" to a hexahydrated form of it. The monohydrate form dissolves better than the tetrahydrate and has higher heat stability; it is prepared by heating the tetrahydrate. The compound exists as a dimer as shown below.
The reagent offers low toxicity and a long shelf life. Sodium peroxoborate is a useful reagent in synthetic chemistry as a substitute for the unstable, highly concentrated hydrogen peroxide solutions that can pose a significant explosion hazard and are not commercially available.
Persil and the structure of Na2B2(O2)2(OH)4.6H2O
Return to the course outline or move on to Lecture 5: Structure of the elements (Groups 1 and 2 metals, Boron, Carbon and Phosphorus, Sulfur). | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/16%3A_The_Group_16_Elements/16.05%3A_Hydrides/16.5B%3A_Hydrogen_Peroxide_%28H_2O_2%29.txt |
Oxides of fluorine, chlorine, bromine and iodine [1]
In compounds such as the oxides and oxyacids the halogens (except F) occur in high oxidation states. Some properties of oxides are shown below:
Property OF2 O2F2 Cl2O ClO2 I2O5
Physical appearance and general characteristics Colourless (very pale yellow) gas; explosive and toxic Yellow solid below 119 K; decomposes above 223 K Brownish yellow gas; explosive at high concentrations Yellowish gas or liquid; explosive above 15% volume in air white crystalline hygroscopic solid
Melting point /K 49 119 153 213 573 decomp
Boiling point /K 128 210 275 284 -
ΔfH°(298 K) / kJ mol-1 24.7 18 80.3 102.6 -158.1
Dipole moment /D 0.3 1.44 0.78 1.78 -
O-X bond distance /pm 141 157.5 170.0 147.3 180 (term)
195 (bridging)
Oxygen difluoride was first prepared in 1929; it was obtained by the electrolysis of slightly moist molten potassium fluoride and hydrofluoric acid.[2][3] More recently it has been prepared by the reaction of difluorine with 2% aqueous NaOH solution:
$2F_2 + 2NaOH \rightarrow OF_2 + 2NaF + H_2O \tag{1}$
Although OF2 is the most stable of the oxygen fluorides, conditions must be controlled so as to minimize losses due to decomposition via the secondary reaction.
$OF_2 + 2 OH^- \rightarrow O_2 + 2F^- + H_2O \tag{2}$
This reaction occurs slowly with water at room temperature, more rapidly with base and explosively with steam. The shape of the OF2 molecule has been compared with HOF and H2O and the three give bent structures where the bond angles range from 97 to 104.5 °.
The basic shape of O2F2 resembles that of H2O2 except that the internal dihedral angle is smaller (87° cf. to 111°). The very long O-F bonds (157.5 pm) probably explains why the molecule is able to easily dissociate.
There are at least 10 oxides of chlorine known, of which only 2 will be discussed.
Cl2O is best prepared by treating fresh yellow mercuric oxide with Cl2 gas:
$2 Cl_2 + 2 HgO \rightarrow Cl_2O + HgCl_2 \cdot HgO \tag{3} It can also be prepared by reaction of Cl2 gas with moist sodium carbonate, Na2CO3. \[2Cl_2 + 2Na_2CO_3 + H-2O \rightarrow Cl_2O + 2NaCl + 2NaHCO_3 \tag{4} Dichlorine monoxide is very water soluble (a saturated solution at -9.4 °C contains 143.6 g Cl2O per 100 g H2O) and it hydrolyses to hypochlorous acid. \[Cl_2O + H_2O \rightarrow 2HOCl \tag{5}$
Cl2O
Much of the Cl2O is used to prepare hypochlorites, especially Ca(OCl)2 that is used for water treatment of swimming pools.
Chlorine dioxide (ClO2) was mentioned in the first lecture as being an important industrial chemical for bleaching flour and wood pulp and for water treatment. Over 95% of the chlorine dioxide produced in the world today is made from sodium chlorate:
$2NaClO_3 + 2NaCl + 2H_2SO_4 \rightarrow 2ClO_2 + Cl_2 + 2Na_2SO_4 + 2H_2O \tag{6}$
ClO2 was first discovered by Humphry Davy (1778-1829) in 1811 and was the first of the chlorine oxides to have been isolated. The yellow paramagnetic gas explosively decomposes into chlorine and oxygen and the decomposition is initiated by light. Thus, it is never handled in concentrated form, but is almost always used as a dissolved gas in water in a concentration range of 0.5 to 10 grams per liter. Its solubility increases at lower temperatures: it is thus common to use chilled water (5 °C) when storing at concentrations above 3 grams per liter. In many countries, such as the USA, chlorine dioxide gas may not be transported at any concentration and is almost always produced at the application site using a chlorine dioxide generator. Chlorine dioxide is less corrosive than chlorine and superior for the control of legionella bacteria. It is more effective as a disinfectant than chlorine in most circumstances against water borne pathogenic microbes such as viruses, bacteria and protozoa.
The oxidation number of the Cl is considered to be +3 and Pauling originally proposed that resonance structures be used to account for this. An alternative explanation by Brockway extended this to resonance between 3 electron 2 centre bonds. For many years it was claimed that a dimer was not formed and considering that ClO2 is a radical this might have been one mechanism to provide some stability by using up the odd electron in a 2 electron 2 centre bond. In 1992, Rehr and Jansen conducted magnetic susceptibility and single crystal studies at various temperatures and it was found that below about 170 K the sample became diamagnetic and did indeed form a dimer. They noted that to avoid explosions all equipment had to be scrupulously cleaned with aqua regia prior to use and temperatures kept low.
ClO2
The oxides of bromine are less numerous than for chlorine and they are generally unstable at room temperature decomposing to the elements. For example, BrO2 is a pale yellow crystalline solid formed quantitatively by low-temperature ozonolysis of dibromine.
$Br_2 + 4O_3 → 2BrO_2 + 4O_2 \nonumber$
The oxides of iodine are the most stable of the halogens and the first of these to be prepared was I2O5 which was independently discovered by Joseph Louis Gay-Lussac (1778-1850) and H. Davy in 1813 although the structure was not determined until 1970. The white hygroscopic crystals are very soluble in water and commercial I2O5 has been found to consist of HI3O8 or I2O5.HIO3. Likewise crystals of commercial iodic(V) acid, HIO3, were investigated using single-crystal and powder X-ray diffraction and the crystals turned out again to be HI3O8 or HIO3·I2O5.
partially staggered conformation of I2O5
I2O5 is one of only a few chemicals that is capable of oxidizing carbon monoxide rapidly and completely at room temperature:
I2O5 + 5 CO → I2 + 5 CO2
This reaction has been used as the basis of an analytical method for sampling CO in the atmosphere. I2O5 can be used for the oxidation of a number of other small molecules including: NO, C2H4, SO3, H2S.
Oxoacids and their salts
Numerous oxoacids of the halogens exist and chlorine is able to form a complete series of such acids ranging from HOCl to HClO4 where the formal oxidation state is +1,+3,+5 and +7. [1]
Oxoacids of chlorine Oxoacids of bromine Oxoacids of iodine
Hypochlorous acid HOCl, pKa=7.53 Hypobromous acid HOBr, pKa=8.69 Hypoiodous acid HOI, pKa=10.64
Chlorous acid HOClO (HClO2), pKa=2.0
Chloric acid HOClO2 (HClO3), pka=-1.0 Bromic acid HOBrO2 (HBrO3) Iodic acid HOIO2 (HIO3)
Perchloric acid HOClO3 (HClO4), pka=~-8 Perbromic acid HOBrO3 (HBrO4) Periodic acid HOIO3 (HIO4)
Orthoperiodic acid (HO)5IO (H5IO6)
HOF has been isolated by reaction of difluorine gas with ice at 230 K but it does not ionise in water or give stable salts and on warming to room temperature it decomposes to HF.
2 HOF → 2 HF + O2
The hypohalous acids (X=Cl, Br, I) are conveniently prepared by the reaction of the dihalogen with mercury oxide:
2 X2 + 3 HgO + H2O → 2 HOX + Hg3O2X2
The hypohalous salts formed from the heavier halogens are all weak acids with the hypochlorites being important industrial reagents.
Bleaching powder is a mixture of CaCl2, Ca(OH)2 and Ca(OCl)2 and is manufactured by the reaction of dichlorine on Ca(OH)2. The sodium salt is a well known disinfectant and bleaching agent
Each of these hypohalites are unstable and disproportionate at room temperature to the halates and halide ion:
3[OX]- → [XO3]- + 2X-
The equilibrium constants for these reactions vary with X such that for Cl, K=1027 for Br, K=1015 and for I, K=1020. The reaction speed increases down the group so that for OI- the decomposition is very fast.
Chlorous acid, chloric acid and bromic acid cannot be isolated as pure compounds but exist in aqueous solution and are prepared from the barium salts.
Ba(ClO2)2 + H2SO4(aq) → 2 HClO2(aq) + BaSO4(s)
and Ba(XO3)2 + H2SO4 → 2 HXO3 + BaSO4 (X= Cl, Br)
Iodic acid (HIO3) is a stable white solid at room temperature and is prepared from I2 and nitric acid or from I2O5 in water.
I2O5 + H2O → 2HIO3
The structure below shows trigonal pyramidal HIO3 units and these are connected by extensive H-bonding. In aqueous solution it is a reasonably strong acid with pKa ~ 0.77.
Potassium chlorate is a white crystalline substance that is used:
• as a pesticide. In Finland it was sold under the trade name Fegabit.
• as an oxidizing agent,
• to prepare oxygen,
• as a disinfectant,
• in safety matches, and
• in explosives and fireworks.
Potassium chlorate will readily decompose if heated in contact with a catalyst, typically manganese (IV) dioxide (MnO2). Thus, it may be simply placed in a test tube and heated over a burner and this can provide a convenient source of dioxygen for school laboratories. The reaction is as follows:
2 KClO3(s) + heat → 3 O2(g) + 2 KCl(s)
The materials used need to be of high purity since even small amounts of impurities can lead to explosions and it is considered good practice to test a new batch of potassium chlorate with a small reaction first (less than 1 g!) Even commercial oxygen generators have been known to explode and cause problems such as at least one plane crash and a fire on the space station MIR.[10]
Both potassium bromate and iodate are used in volumetric analysis and in particular the iodate is a primary standard being available in high purity and with excellent stablity. In the standardisation of thiosulfate for example potassium iodate is used a source of diiodine.
[IO3]- + 5 I- + 6 H+ → 2 I2 + 3 H2O
Iodate is used in Andrew's titrations for determination of hydrazine and in the determination of I- in SnI4 in the CHEM3101 laboratory programme.
HIO3
H5IO6 orthoperiodic acid
Several different I(VII) oxyacids are known including periodic acid (HIO4) and orthoperiodic acid (H5IO6). In the latter case there is extensive H-bonding present in the solid state resulting in a 3D network. Wherease perchloric acid exists as discrete monomeric units HIO4 is again extensively H-bonded through cis- edge-sharing octahedra.
The reactions involving the various oxyacids in aqueous solution is complicated, as seen by the following equilibria:
[H3IO6]2- + H+ → [IO4]- + 2H2O
2[H3IO6]2- → 2[HIO5]2- + 2H2O
2[HIO5]2- → [H2I2O10]4- → [I2O9]4- + H2O
Perchlorates
Count Friederich von Stadion is credited with the preparation of the first perchlorate compound in 1816 which he obtained by mixing potassium chlorate with concentrated sulfuric acid and allowing it to stand for 24 hours with frequent agitation. The water-insoluble residue was potassium perchlorate. Perchloric acid is the only oxoacid of chlorine that can be isolated and in the vapour phase the Cl-O bonds are not equivalent unless deprotonated when they become the same length.
HClO4 Pure perchloric acid is a colourless mobile, shock-sensitive liquid and at least 6 different hydrates are known and the monohydrate forms an H-bonded crystalline lattice [H3O]+ [ClO4]-
The perchlorate anion is not a good ligand and it is sometimes used for the isolation of metal complexes. Perchlorate salts are often soluble in acetone which can provide a very useful solvent system for reactions. Unfortunately though the solids are often unstable and can lead to violent explosions. I know someone who lost the tip of their finger when attempting to grind up 1 or 2 mg of complex for an IR spectrum!
Perchlorates in groundwater
Texas chemists have shown evidence for a natural atmospheric origin of perchlorate. For many, perchlorate is almost synonymous with the rocket fuel that has found its way into U.S. groundwater and surface water and subsequently into foods such as lettuce and milk. Perchlorate taints drinking water in 35 US states at levels of at least 4 ppb. When groundwaters of a 60,000-sq-mile region in Texas and New Mexico were sampled, > 80% of the wells returned positive and unexpectedly high levels of perchlorate and it was found even in samples that were unlikely ever to have been exposed to synthetic contamination. The perchlorate levels in many of the Texas samples were more than 20 ppb, and some were as high as 60 ppb. However, there were regions such as the southern high plains (Texas Panhandle) where there was no clear historical or current evidence of the extensive presence of rocket fuel or Chilean fertilizer sources. The occurrence of easily measurable concentrations of perchlorate in such places was difficult to understand. In the southern high plains groundwater, perchlorate was better correlated with iodate, known to be of atmospheric origin, compared to any other species. They showed that perchlorate was readily formed by a variety of simulated atmospheric processes. For example, it could be formed from chloride aerosol by electrical discharge and by exposing aqueous chloride to high concentrations of ozone. They further reported that perchlorate was present in many rain and snow samples, strongly suggesting that some perchlorate was formed in the atmosphere and a natural perchlorate background of atmospheric origin existed.[4][5][6]
Although the scientists were fairly certain that the perchlorate in the Texas wells was of natural origin, they still needed more evidence to prove it. Researchers at Louisiana State University (LSU) and Oak Ridge National Laboratory (ORNL) developed a new stable-isotope ratio technique that could provide the missing piece of information.
To distinguish between natural and synthetic perchlorate it is important to recognise that perchlorate is a non-volatile oxyanion and that once formed it does not exchange oxygen atoms with those in the ambient environment. This means its oxygen ratio signature is fixed. So by looking at both the 17O/16O and 18O/16O ratios it is possible to tell whether it is from a natural source or not.
In this approach, all three stable oxygen isotopes are measured after a thermal decomposition method generates O2 from perchlorate crystals. Measuring just 18O/16O won't give you the answer, because 18O/16O ratios for anthropogenic and atmospheric perchlorate overlap. But if you measure 17O/16O at the same time and compare the two ratios, it is possible to determine its origin. Most oxygen-containing compounds on earth have a correlation between 18O/16O and 17O/16O. Any deviation from this relationship is referred to as the 17O anomaly, or Δ17O. The results of studies on Chilean perchlorates from the Atacama desert caliche deposits found that Δ17O was between 4 and 10 whereas for man-made perchlorates the value was generally less than -0.2.[7]
In the human body, perchlorate inhibits the uptake of iodide by the thyroid and thus may lower the amount of thyroid hormone in the body. Insufficient levels of this hormone can cause permanent neurological damage in children.
The changes in perchlorate concentrations in surface water adjacent to a site of fireworks displays from 2004 to 2006 was monitored. Preceding the fireworks displays, perchlorate concentrations in surface water ranged from 0.005 to 0.081 µg/L, with a mean value of 0.043 µg/L. Within 14 h after the fireworks, perchlorate concentrations spiked to values ranging from 24 to 1028 times the mean baseline value. A maximum perchlorate concentration of 44.2 µg/L was determined following the July 4th event in 2006. After the fireworks displays, perchlorate concentrations decreased toward the background level within 20 to 80 days, with the rate of attenuation correlating to surface water temperature.[8]
The Pentagon, NASA, the US Department of Energy, and defense contractors - who could face expensive cleanups of perchlorate contamination - vigorously objected when the EPA proposed a 1-ppb limit for drinking water. The US military suggested a limit of 200 ppb.
Aqueous Solution Chemistry
The diagrams below show the variation of redox potentials with acid and base, It should be noted that the couples ½X2/X- are independent of pH and together with the estimated value for F2 indicate a steady decrease in oxidising strength moving down the group.
F2 (+2.866 V) > Cl2 (+1.358 V) > Br2 (+1.066 V) > I2 (+0.536 V)
The reason the potential for difluorine must be estimated is that it is too strong an oxidant and will oxidise water to O2 so the potential can not be easily measured!
Given that the value listed for dichlorine is also greater than what is needed to oxidise water (+1.229 V) this should happen as well however this is not the case since there is a slow kinetic step in the oxidation of water. Note as well that this is the same reason why aqueous solutions of potassium permanganate can be prepared, although they are not stable over time.
We already mentioned the variation of acid strength of the HX acids and in a similar way the magnitudes of the E° for the half-reactions
½X2 + e- → X-
are best described in terms of bond-energies, electron affinities and hydration energies.
½X2 → X(g) → X-(g) → X-(aq)
Overall this causes Cl2 to be a more powerful oxidant than Br2 or I2, partly because of a more negative enthalpy of formation of the anion but by taking into consideration the strong interaction with the small Cl- ion with water.
The majority of the couples are pH dependent so that an increase in pH causes a dramatic lowering of the E° value. This is expected since if we consider an example where BrO3- is converted to ½Br2 by the reaction:
BrO3- + 6H+ + 5e- → ½Br2 + 3 H2O (E° = +1.495 V)
the equilibrium constant includes a term with [H+]6 so that if the pH is changed from pH=1 to pH=14 then H+ effectively changes by 10-14 and the estimated difference can be seen in E° as:
ΔG° = - RTln(K)
ΔG° = - nE°F
from which just looking at the effect of pH is ~(RT/nF)ln([H+]6)
or roughly (0.0592/5) * 14 * 6
that is an effect of ~+0.99 V
Experimentally the measured E° for
BrO3- + 3H2O + 5e- → ½Br2 + 6 OH- (E° = +0.519 V)
so estimated value is +1.495 - 0.99 = +0.51 V and found is +0.519 V showing excellent agreement.
Redox processes in acid solutions
Redox processes in basic solutions
These diagrams indicate that in acid, periodate and perbromate are more powerful oxidising agents than perchlorate when being reduced to halate ions (I +1.653, Br +1.74, Cl +1.189 V) and that iodate and iodine are much weaker oxidants compared to the other halates (I +1.34, Br +1.470, Cl +1.214 V) or halogens (I +0.535, Br +1.066, Cl +1.358 V).
return to course outline
Contributors and Attributions
• The Department of Chemistry, University of the West Indies)
16.09: Oxoacids and their Salts
6 | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/16%3A_The_Group_16_Elements/16.09%3A_Oxoacids_and_their_Salts/16.9G%3A_Peroxysulfuric_Acids_%28H_2S_2O_8%29_and_%28H_2SO_5%29.txt |
The halogens are located on the left of the noble gases on the periodic table. These five toxic, non-metallic elements make up Group 17 of the periodic table and consist of: fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At). Although astatine is radioactive and only has short-lived isotopes, it behaves similar to iodine and is often included in the halogen group. Because the halogen elements have seven valence electrons, they only require one additional electron to form a full octet. This characteristic makes them more reactive than other non-metal groups.
17: The Group 17 Elements
Chemistry 242 - Inorganic Chemistry II Chapter 20 - The Halogens: Fluorine, Chlorine Bromine, Iodine and Astatine
Introduction
• All elements except helium, neon and krypton fro some sort of halogen compound.
• The structures vary from ionic to covalent.
• Halides are often the source materials for thsynthesis of other substances.
• The halides are often the "generic" compounds used to illustrate the range of oxidation states for the other elements.
• The class of fluorocarbons is an important group of organic chemicals which often have unique properties.
Occurence and Isolation, and Properties of the Elements
Fluorine
• Occurs in fluorspar, CaF2, cryolite, Na3AlF6 and fluorapatie, 3Ca3(PO4)2.Ca(F,Cl)2. It is actually more abundant than chlorine.
• It is made by electrolysis, for example, of a misture of 2 to 3 parts of HF with KF which melts at 70 - 100 oC. It is a pale yellow gas, bp -118 oC.
• It is the most reactive of all the elements and attacks all the others except He, Ne and Ar so it has to be handled in special apparatus: stainless steel and copper because they become coated with a protective fluoride layer and are the materials of choice. (If all traces of HF are removed, fluorine can be handled in glass apparatus also, but this is nearly impossible.)
• Fluorine will also attack many compounds yielding flourides.
• The reason for its great reactivity seems to be connected with the weak F-F bond probably due to strong repulsion between the non-bonding (lone-pair) electrons. (In this respect it is similar to hydrazine and hydrogen peroxide.)
Chlorine
• Occurs mainly as rock salt, NaCl, potassium chloride and magnesium chloride.
• Chlorine is manufactured by the electrolysis of brine. At one time this was done using a mercury cathode, which also produced sodium amalgam, thence sodium hydroxide by hydrolysis. The sodium hydroxide was often recombined with the chlorine to form sodium hypochlorite (bleach) for use in the paper industry. The mercury is now recognized as a major polluent. Now "memberane" cells are used to prevent the electrolytes around the two electodes from mixing.
• Chlorine is a green gas bp -34.6 oC.
Bromine
• It occurs with chlorine and can be obtained by displacement from bromides with chlorine.
• Bromine is a brown liquid, mp -7.2, bp 58.8 oC.
Iodine
• Iodine is found in brines, and as IO3- with deposits of NaNO3.
• It is black solid with a metallic sheen. It sublime at room temperature and its vapour is purple. It also forms purple solution in non-polar solvents.
• The brown solution obtained by dissolving iodine in potassium iodide contains the linear I3- ion.
The intense blue complex with starch used in iodometric titrations contains the I5- ion.
Halides
They can be made by:
1. Direct reaction. In the case of fluorine, particularly, high oxidation states can be reached.
2. Reaction with an oxide.
3. Halogen exchange reactions.
4. By dehydration of hydrated halides.
Molecular Halides
As a class of compounds, the halides illustrate the distinction between ionic solids, network solids and molecular compounds (which can be solids liquids or gases). This section deals with all three, its heading notwithstanding!
• Broadly, the more covalent the halide, the more likely it is to be truly molecular. Examples would be BCl3 or IF3.
• As the compounds become less covalent, halogen bridges are found, leading first to dimers such as Al2Cl6.
• As the covalency continues to decrease, the amount of bridging increases and network solids can be found. An example is (AlF3)n in which all the aluminum is 6-coordinated and all the fluorines are bridging (2-coordinate). The bonding remains quite covalent.
• At the ionic limit one finds salts such as sodium chloride, where the forces between sodium and chloride ions are almost entirely electrostatic. Both ions are 6-coordinate in NaCl.
• The position along the above progression is influenced by factors such as the electronegativity of the components, their size, and the possibility of multiple bonding which might stabilize a monomeric molecular compound such as BF3 (pp - pp) or SF6 (dp - pp).
The particularly high electronegativity of fluorine leads to some special properties in its compounds:
• CF3COOH is a strong acid unlike CH3COOH.
• NF3 and N(CF3)3 are not at all basic, unlike ammonia.
• -CF3 groups are quite inert to nucleophilic attack, unlike -CH3.
• -CF3 mimics a big halogen with an electronegativity around that of -Cl.
Halogen Oxides
Fluorine forms two oxygen compounds:
1. Oxygen Fluoride, OF2 a yellow gas produced by:
F2 + NaOH(aq) OF2
2. Dioxygen difluoride, O2F2 a yellow-orange solid produced by:
F2(g) + O2(g) O2F2 (Electric discharge through the reactants)
Chlorine does not react directly with oxygen, but forms a dangerously exposive paramagnetic oxide ClO2 in the following reaction:
NaClO3 + SO2 + H2SO4 2ClO2 + 2NaHSO4
It is used as a chlorinating agent in organic synthesis well diluted with air!
Iodine forms one important oxide, I2O5, by dehydration of HIO3:
HIO3 I2O5 + H2O (240 oC)
Its use is as a reagent for quantitative analysis for carbon monoxide:
5CO + I2O5 I2 + 5CO2
then the iodine is titrated as triiodide with thiosulphate:
I2 + I- I3-
S2O32- + I3- 3I- + S4O62- (starch as indicator)
Oxo Acids
• The reaction of the halogens with water produces hypohalous acids, HOX, and halide ion in solution:
X2 + H2O HOX + H+ + X-
The pure compounds are unknown except HOF which is none too stable itself.
The solutions can be used to obtain the salts e.g. sodium hypochloride, bleach. The XO- ions disproprortionate further in basic solution, (IO- very rapidly even in neutral solution):
3XO- 2X- + XO3-
• For chlorine the following reactions are important:
• Production of chlorite, ClO2-, and chlorate, ClO3-:
ClO2 + 2OH- ClO2- + ClO3- + H2O
• Perchlorate is produced by electrolytic oxidation of chlorate:
ClO3- + H2O ClO4- + 2H+ + 2e-
• Periodate species are unlike the others due to the large size and low electronegativity of iodine. The following equilibria are important:
H5IO5 H+ + H4IO6- K = 10-3
H4IO6- IO4- + 2H2O k = 29
H4IO6- H3IO62- + H= k = 1x10-7
Periodate is used as an oxidizing agent in organic synthesis.
Interhalogens
• The species are molecules or molecule ions, for example, BrCl, IF5, Br3+, I3-.
• The neutral species will be XX'n where n is odd and 7 or less and X' will be the lighter halogen. The range of isolable compounds is governed by steric effects.
• Most are made by direct reactions.
Organic Compounds of Fluorine
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9
F
18.9984032
The Group 17 elements include: fluorine, chlorine, bromine, iodine and astatine and according to Housecroft and Sharpe [1] their chemistry is "probably better understood than that of any other group of elements, except the alkali metals". They share common properties such as forming singly bonded atoms as in the diatomic molecules (F2, Cl2, Br2, I2, and At2), and forming singly charged anions (F-, Cl-, Br-, I-, and At-).
For this four lecture course, we will omit the chemistry of astatine since it is radioactive and the most stable isotopes of astatine have half-lives of less than a minute. As a consequence of this, only trace quantities (less than 50 ng) of astatine compounds have been investigated, severely limiting characterisation of properties.
Discussions of the chemistry of the elements in Group 17 therefore focus on four elements: F, Cl, Br, and I. In 1825, the Swedish chemist Jöns Jakob Berzelius applied the term halogens (from the Greek hals, "salt," and gennan, "to form or generate") for an element that produces a salt when it forms a compound with a metal. None of the halogens are found naturally in their elemental form. They are invariably found as salts of the halide ions (F-, Cl-, Br-, and I-). [Note that Br was not discovered until 1826, after the term halogen was proposed, so it was added to the list of halogens later].
The mineral fluorspar (also called fluorite) consisting mainly of calcium fluoride, was first described as far back as 1530 by Georgius Agricola. This is a major source of fluorine in the form of fluoride ions as was cryolite (Na3AlF6) until the major deposits in Greenland ran out in 1987.
In nature, chlorine is found primarily as the chloride ion, a component of the salt that is deposited in the earth or dissolved in the oceans - about 1.9% of the mass of seawater is chloride ions. Even higher concentrations of chloride are found in the Great Salt Lake in Utah, which is 9% Cl- ion by weight, the Dead Sea and in underground brine deposits. Most chloride salts are soluble in water, thus, chloride-containing minerals are usually only found in abundance in dry climates or deep underground. Common chloride minerals include halite (sodium chloride), sylvite (potassium chloride), and carnallite (potassium magnesium chloride hexahydrate). Over 2000 naturally-occurring organochlorine compounds are known.
The largest bromine reserve in the United States is located in Columbia and Union County, Arkansas, U.S. China's bromine reserves are located in the Shandong Province and Israel's bromine reserves are contained in the waters of the Dead Sea.
Iodine naturally occurs in the environment chiefly as a dissolved iodide in seawater, although it is also found in some minerals and soils. This element also exists in small amounts in the mineral caliche, found in Chile, between the Andes and the sea. A type of seaweed, kelp, tends to be high in iodine as well.
17
Cl
35.453
35
Br
79.904
53
I
126.90447
85
At
(210)
X Isotopes Earth's crust (ppm) Sea water (ppm) Universe (ppm)
F 100% - 19 950 0.0001 0.4
Cl (75.77%, 35) (24.23%, 37) 130 18000 1
Br (50.69%, 79) (49.31%, 81) 3 67.3 0.007
I 100%, 127 1.4 0.05 0.001
Extraction
Difluorine F2
HF is the starting material for the preparation of most fluorine containing compounds and this is isolated as a byproduct from the production of phosphoric acid. Treatment of minerals such as fluorspar/ fluorite CaF2, (often associated with phosphate-containing minerals) with sulfuric acid gives hydrogen fluoride by the following reaction:
CaF2 + H2SO4 → 2 HF + CaSO4
Difluorine is prepared by the electrolysis of HF. Since F2 is strongly oxidising, this is the only viable method that can be performed on an industrial scale and given the extreme reactivity of the F2, it requires strictly controlled conditions. It is believed that many were blinded or killed during the many years in trying to perfect the process and experiment with the toxic gas.
"Henry Moissan (1852-1907) eventually succeeded in preparing fluorine in 1886 by the electrolysis of a solution of potassium hydrogen difluoride (KHF2) in liquid hydrogen fluoride (HF). The mixture was needed because hydrogen fluoride is a non-conductor. The device was built with platinum/iridium electrodes in a platinum holder and the apparatus was cooled to -50 °C. The result was to completely isolate the hydrogen produced from the negative electrode from the fluorine produced at the positive one. This is essentially still the way fluorine is produced today. For this achievement, he was awarded the Nobel Prize in 1906."[5].
It is recorded that he died shortly after his return from collecting his Nobel Prize and that death was a result of acute appendicitis. It is not known whether his studies on fluorine contributed to his early death.
"The modern electrolysis cell contains a steel or copper cathode, ungraphitized carbon anode, and a Monel metal (Cu/Ni) diaphragm that is perforated below the surface of the electrolyte, but not above it, thus preventing the H2 and F2 products from recombining. As electrolysis proceeds, the HF content of the melt is renewed by adding dry gas from cylinders." [1]
Dichlorine (Cl2)
The Castner-Kellner process, invented jointly by Hamilton Castner and Karl Kellner in the 1890's is the earliest industrial process for the extraction of dichlorine (Cl2).
The Castner-Kellner Chlor-alkali electrolysis cell
A=graphite anode, C=iron cathode, F=fulcrum system, M=mercury cathode, N=NaOH solution, S=NaCl solution, W=slate wall
In the diagram above there are two types of cells shown, one in the centre and two outer cells. They are separated by slate walls (W) that dip into a pool of liquid mercury (M). The outer cells have a solution a sodium chloride (S), a graphite anode (A) and mercury cathode. The inner cell has a solution of sodium hydroxide (N), an iron cathode (C) and here the mercury acts as the anode. The mercury is able to move between the three cells aided by a rocking mechanism (F), but the solutions are kept separated by the slate walls.
The overall processes are given by the following equations:
2Cl- → Cl2 + 2e-
2Na+ + 2e- → 2Na (amalgam)
and
2Na (amalgam) → 2Na+ + 2e-
2H2O + 2e- → 2OH- + H2
As the electrolysis proceeds the concentration of sodium chloride in the outside cells decreases and the concentration of sodium hydroxide in the center cell increases. When this reaches a concentration of about 50% some of this sodium hydroxide solution is removed from center cell to be replaced with water. Sodium chloride is added to the outside cells to replace that used up. Dihydrogen and dichlorine are collected as gases in high purity. An animation of the working of a modern plant can be see at Euro-Chlor, Chlorine Online
The large scale use of mercury has raised numerous environmental issues and many of these industrial plants have been decommissioned and the sites cleaned up, those that remain are expected to wind up operation by 2020. According to Euro Chlor "In 2007, emissions for all mercury cells across Western Europe reached an all-time low of 0.97 grammes per tonne of chlorine capacity."
Apart from the mercury cell process, two others are in operation and these require less electricity and do not use mercury. They are the "membrane cell process" and the diaphragm cell process".
Dibromine (Br2)
Internationally, bromine reserves are found in Arkansas, USA, the waters of the Dead Sea, Israel and Shandong Province, China. In 2007, approximately 556,000 metric tonnes (worth around US\$2.5 billion) of bromine were produced predominantly by these three countries and this equated to a sixfold increase since the 1960s.
The extraction involves treating the bromide-rich brines with chlorine gas then flushing through with air. In this treatment, bromide anions are oxidized to bromine by the chlorine gas.
2 Br- + Cl2 → 2 Cl- + Br2
Diiodine (I2)
Chilean Caliche
In the Atacama Desert in northern Chile, there are vast deposits of a mixture, often referred to as caliche, composed of gypsum, sodium chloride, plus other salts and sand, associated to Salitre ("Chile Saltpeter"). Salitre, in turn, is a composite of sodium nitrate (NaNO3) and potassium nitrate (KNO3) and was an important source of export revenue for Chile until World War I, when both nitrates began to be industrially produced in large quantities in Europe.
These deposits are the largest known natural source of nitrates in the world, containing up to 25% sodium nitrate and 3% potassium nitrate, as well as iodate minerals, sodium chloride, sodium sulfate, and sodium borate (borax). The caliche beds are from 0.2 to 5 meters thick, and they are mined and refined to produce a variety of products, including sodium nitrate (for agriculture or industry uses), potassium nitrate, sodium sulfate, iodine, and iodine derivatives. The discovery of the iodate impurity in the Chilean saltpetre dates from 1840 and led to the large scale extraction process that saw Chile as the largest producer of iodine worldwide in 2007, followed by Tokyo, Japan and Oklahoma, USA where the iodine was recovered from brine solutions.
2IO3- + 5HSO3- → 3HSO4- + 2SO42- + H2O + I2
The Halogens - Properties and Uses
Difluorine (F2), is a highly toxic, colorless gas, and is highly reactive. It is so reactive that it even forms compounds with Kr, Xe, and Rn, elements once considered inert. Due to its high reactivity it is difficult to find a container in which it can be stored. F2 attacks both glass and quartz, for example, and causes most metals to burst into flame. It has to be handled in equipment built from alloys of copper and nickel where even though it still reacts with these alloys, it forms a layer of fluoride salt on the surface that protects the metal from further reaction.
Fluorine is such a powerful oxidizing agent that it can convert other elements into unusually high oxidation numbers, as in AgF2, PtF6, and IF7.
portion of Teflon
Fluorine is used in the manufacture of Teflon or PTFE poly(tetrafluoroethylene), (C2F4)n. Teflon was first prepared in 1938 by accident, since the intended product was a new refrigerant. The first non-stick frying pan was developed in France in 1954 and in the USA in 1961. Another important use is for lining of valves and gaskets that are inert to chemical reactions. This is needed in Uranium enrichment involving UF6.
Natural uranium is 99.284% 238U isotope, with 235U only constituting about 0.711% of its weight. Isotope enrichment of 235U is difficult because the two isotopes have nearly identical chemical properties, and can only be separated gradually using small mass differences. (235U is only 1.26% lighter than 238U). Gaseous diffusion was the first successful technology used to produce enrich uranium by forcing gaseous uranium hexafluoride through semi-permeable membranes. This produced a slight separation between the molecules containing 235U and 238U.
Large amounts of fluorine were consumed each year to make the "Freons" (such as Freon-12, CCl2F2) used in refrigerators. CFCs (chlorofluorocarbons) and HCFCs hydrochlorofluorocarbons) were produced by halogen exchange starting from chlorinated methanes and ethanes. For example, the synthesis of chlorodifluoromethane from chloroform:
CHCl3 + 2HF → HCF2Cl + 2HCl
2010 marks the year when the production of CFC's is to cease in developing countries, having already been phased out in 2000 in developed countries under the Montreal Protocol, since they are deplete stratospheric ozone. Recycling will then be the only source of CFC's. HCFC's are not regarded as suitable replacements, since although they start degrading before reaching the ozone layer, a proportion still reaches the stratosphere and chlorine buildup was found to be higher than originally predicted. HFC's (hydrofluorocarbons) are considered suitable since they have a shorter lifetime in the lower atmosphere. HFO-1234yf, (CH2=CFCF3) or 2,3,3,3-Tetrafluoropropene is proposed as a refrigerant for automobiles. It has a boiling point of -30 °C.
Fluoridation of water supplies was introduced in many countries in the mid-1900's as a preventative measure against dental decay. The fluorides used include: H2SiF6 and Na2SiF6. Due to the possibility of an increased risk of cancer this practise has been reduced but fluoridation of toothpaste (NaF or monofluorophosphates) has increased as an alternative.
Chlorine (Cl2) is a highly toxic gas with a pale yellow-green color and is a very strong oxidizing agent.
Chlorine useage in Europe, 2008 (/kt) [6]
PVC - door and window frames, piping etc, (34.2%)
Isocyanates and Oxygenates - upholstery, insulation, pesticides, (29.1%)
Inorganics - disinfectants, water treatment, (13.7%)
Organics - detergents, herbicides, insecticides, (7.0%)
Chloromethanes - silicon rubbers, decaffeination, paint strippers, cosmetics, (5.9%)
Epichlorohydrin - epoxy resins, printed circuits, sports boats, (5.4%)
Solvents - metal degreasing, adhesives, (4.7%)
PVC is the third most widely used plastic material in the world, after polyethylene and polypropylene. At the global level, demand for PVC exceeds 35 million tonnes per annum and it is in constant growth (+5% on global average), with higher growth rates in the developing countries. PVC is durable, easy to clean, stain resistant, lightweight, corrosion resistant and needs no maintenance. [7]
Chlorine has been used commercially as a disinfectant and as a bleaching agent. Chlorine was first used as a disinfectant for drinking water in the late 19th century as a means of controlling the spread of water-borne diseases such as typhoid, cholera, dysentery and gastro-enteritis. In the USA in 1900, annual deaths from cholera totalled 25,000 but following the introduction of chlorination, this figure had fallen to fewer than 20 by 1960! In 1991, a misinterpretation of US law resulted in the Peruvian government voluntarily suspending chlorination of water supplies. The resulting cholera epidemic spread to neighbouring countries causing an estimated 1 million cases of cholera and more than 10,000 deaths.[6]
Chemical pulp bleaching aims to remove coloured residual lignin from the pulp and increase its brightness, brightness stability and cleanliness while preserving the strength (cellulose integrity) and carbohydrate yield (cellulose and hemicellulose) of the unbleached fibre, with due regard for potential effects on the environment. [8]
World use of bleached pulp, 1990-2007 (/Mt) [9]
ECF - Elemental Chlorine-Free (ECF) pulp, bleached with chlorine dioxide.
TCF - Totally Chlorine-Free pulp, a small 5% niche market.
other - this includes dichlorine
ECF, using ClO2, accounts for 99% of the bleached chemical pulp production in the USA, while dichlorine was essentially phased out in 2001 in compliance with rules from the Environmental Protection Agency. Cl2 was found to contribute to the formation of perchlorodibenzodioxins PCDD and perchlorodibenzofurans PCDF. A number of these are highly toxic, persistent (lasting for years or even decades before degrading into less dangerous forms), highly volatile and liable to accumulate in fatty tissue, hence they are called persistent organic pollutants (POP).
For example, tetrachlorodibenzodioxin and hexachlorodibenzofuran
Some 85% of pharmaceuticals contain or are manufactured using chlorine, including products to treat Aids, allergies, arthritis, cancer, depression, diabetes, heart disease, hypertension, infections, pneumonia and ulcers. An example is the natural antibiotic vancomycin, an effective medicine in fighting hospital Staphylococcus infections.[6]
Bromine (Br2) is a reddish-orange liquid with an unpleasant, choking odor. The name of the element is derived from the Greek stem bromos, "stench." Bromine is used to prepare flame retardants, fire-extinguishing agents, sedatives and insecticides.
There are approximately 75 different brominated flame retardants (BFRs) of which Deca-BDE, TBBPA and HBCD are the three main commercial products.
Deca-BDE 1,2,3,4,5-pentabromo-6-(2,3,4,5,6-pentabromophenoxy)benzene is a highly effective brominated flame retardant which is used to prevent fires in plastics for electrical and electronic equipment as well as in contract textiles.
TBBPA 2,2',6,6'-Tetrabromo-4,4'-isopropylidenediphenol is the largest BFR in terms of production. 70% is used as a reacted flame retardant in polymers like epoxy resins in electrical and electronic equipment and 20% is used as an additive to plastics.
HBCD 1,2,5,6,9,10-hexabromocyclododecane has been used for many years mainly in thermal polystyrene insulation foams and applied in the back-coating of textiles for upholstered furniture. [10]
Marine organisms are the main source of naturally occuring organobromine compounds. Perhaps the most famous example and certainly oldest of these is a dyestuff that has been used by humans since the 18th century BC, Tyrian purple. The dye, Tyrian purple - 6,6'-dibromoindigo was extracted from the the marine gastropods Murex brandaris and the purple silks produced were a status symbol of the Byzantium imperial court.
Iodine is an intensely colored solid with an almost metallic luster. The solid is relatively volatile, and it sublimes when heated to form a violet-colored gas. Iodine has been used for many years as a disinfectant in "tincture of iodine." Iodine compounds are used as catalysts, drugs, and dyes. Silver iodide (AgI) played an important role in the photographic process and in attempts to seed clouds to make rain. Iodide is also added to salt to protect against goiter, an iodine deficiency disease characterized by a swelling of the thyroid gland.
Among dyes that have a high iodine content is erythrosine B (food red-colour additive E127 (Federal Food, Drug and Cosmetic Act, Red #3)
More on the Properties of Halogens
Contributors and Attributions
• The Department of Chemistry, University of the West Indies) | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/17%3A_The_Group_17_Elements/17.02%3A_Occurrence_Extraction_and_Uses/17.2A%3A_Occurrence.txt |
Iodine
Iodine was discovered by French chemist Bernard Courtois in 1811. His father was a manufacturer of saltpeter (a vital part of gunpowder) and at the time of the French Napoleonic Wars, saltpeter was in great demand. Saltpeter produced from French niter beds required sodium carbonate, which could be isolated from seaweed collected on the coasts of Normandy and Brittany. To isolate the sodium carbonate, the seaweed was burned and the ash washed with water. The remaining waste was destroyed by adding sulfuric acid. Courtois once added excessive sulfuric acid and a cloud of purple vapour rose. He noted that the vapour crystallized on cold surfaces, making dark crystals. Courtois suspected that this was a new element but lacked the funds to pursue it further.
Samples of the material reached Humphry Davy and Joseph Louis Gay-Lussac and in early December 1813 both claimed that they had identified a new element. Arguments erupted between them over who had identified iodine first, but both scientists acknowledged Courtois as the first to isolate the element.
Iodine is found on Earth mainly as the highly water-soluble iodide ion I-, concentrated in oceans and brine pools. Like the other halogens, free iodine occurs mainly as a diatomic molecule I2. In the universe and on Earth, iodine's high atomic number makes it a relatively rare element. However, its presence in ocean water has given it a role in biology. It is the heaviest essential element widely utilized by life in biological functions.
Under standard conditions, iodine is a bluish-black solid that sublimes to form a noxious violet-pink gas. It melts at 113.7 °C (386.85 K) and forms compounds with many elements but is less reactive than the other halogens, and has some metallic light reflectance.
Elemental iodine is slightly soluble in water, with one gram dissolving in 3450 ml at 20 °C and 1280 ml at 50 °C; potassium iodide may be added to increase solubility via formation of triiodide ions (I3-). Nonpolar solvents such as hexane and carbon tetrachloride provide a higher solubility.
Iodine normally exists as a diatomic molecule with an I-I bond length of 270 pm, one of the longest single bonds known. The I2 molecules tend to interact via weak London dispersion forces, and this interaction is responsible for the higher melting point compared to more compact halogens, which are also diatomic. Since the atomic size of iodine is larger, its melting point is higher.
The I-I bond is relatively weak, with a bond dissociation energy of 151 kJmol-1, and most bonds to iodine are weaker than for the lighter halides. One consequence of this weak bonding is the relatively high tendency of I2 molecules to dissociate into atomic iodine.
orthorhombic structure of I2
a= 0.72701, b= 0.97934, c= 0.47900 nm
The halogens, Cl2, Br2, and I2 adopt similar orthorhombic structures in which diatomic molecules lie in layers:
Cl a= 0.624 b= 0.826 c= 0.448 nm
Br a= 0.667 b= 0.872 c= 0.448 nm
I a= 0.72701, b= 0.97934, c= 0.47900 nm
17.06: Metal Halides - Structures and Energetics
Simple binary salts
When considering the structures of simple binary salts only a small number of types are important. The simplest are for crystals with formula MX and include NaCl, CsCl and ZnS where the CN of the anion and cation are the same and may be 6, 8 or 4.
Compounds with formulae of MX2 or M2X will have different CN's for the cation and anion. The most important of these are the fluorite and rutile structures.
For the MX case, one reason why a substance might favour one form over another is due to the geometry of the packing of the spheres. What is needed is to be able to maximise the interactions of oppositely-charged ions while at the same time minimise the interactions of similarly-charged ions.
The larger the difference in the sizes of the ions affects the packing of the larger ions around the smaller ions. It is relatively easy to calculate the radius ratio (r+/r-) and from this determine the limits for the various CN's.
For the ZnS type structure with CN 4:4 the radius ratio is predicted to be in the range 0.22 - 0.41 while for the NaCl type structure with CN 6:6 it is predicted that the radius ratio will be within the range of 0.41 - 0.73. For CsCl structures with CN 8:8, the radius ratio is expected to be greater than 0.73. The Table below shows the values for a number of alkali halides. Experimentally it is found that the only examples under normal conditions of temperature and pressure to adopt the CsCl structure are CsCl, CsBr and CsI whereas several other salts were predicted to have this structure based on their radius ratios.
Among the other factors that might influence the final structure is the interaction between ions in addition to the nearest neighbours. However the energies involved would not seem to be sufficient to alter the results as shown by the Madelung constants for NaCl and CsCl which are 1.74756 and 1.76267 respectively.
Radius ratio values for alkali halides
X- / M+ Li Na K Rb Cs
F 0.44 0.70 0.98 0.92 0.81
Cl 0.33 0.52 0.73 0.82 0.93
Br 0.31 0.49 0.68 0.76 0.87
I 0.28 0.44 0.62 0.69 0.78
Transition Metal Halides
The only stable pentahalide is VF5, which is readily hydrolysed and a strong Lewis acid. In the solid state it exists as an infinite chain polymer with cis-bridging fluorides but in the vapour phase it has a trigonal-bipyramid monomeric structure. M.P. 19.5° and B.P. 48.3°C.
Tetrahalides are formed by Ti and V. The Ti tetrahalides are fairly unreactive in redox and halogenation chemistry, unlike the V compounds. VCl4 and VBr4 dissociate spontaneously under ambient conditions to VX3 and X2. They also tend to halogenate organic material.
All trihalides of the elements from Ti to Cr are known. Mn(III) and Co(III) are too oxidising to coexist with any halide except F- under ambient conditions, whereas Ti(III) and V(III) are moderately strongly reducing. Chromium(III) is fairly stable toward both reduction or oxidation. There is a marked tendency toward decreasing ionic character on passing from left to right across the period and from the fluorides to the heavier halides. Ferric chloride and bromide show essentially covalent behaviour such as low MPs and solubility in donor organic solvents.
Many trihalides can be prepared by direct combination of the elements. In those cases where direct combination gives a higher oxidation state, trihalides can be produced by either thermal dissociation, disproportionation of the higher halide or by reduction- for example TiCl3 can be prepared by reduction of TiCl4 with H2 at high temperatures.
All these trihalides adopt structures in which the metal is six-coordinate, either octahedral or distorted octahedral. Many of the lattices are complicated, but can be represented as CrCl3 or BiI3 types. In some cases it becomes even more complicated since some salts exist in more than one form. For example, at low temperature (below 240K) CrCl3 exists in the rhombohedral form mentioned above, but at room temperature it is monoclinic.
All the first row transition metal elements form dihalides with all the halogens, with the exceptions of TiF2 and CuI2. The instability of TiF2 is probably due to easy disproportionation to Ti and TiF3 whereas the oxidising power of Cu2+ (reducing power of I-) explains the lack of the copper salt. Anhydrous dihalides can generally be synthesised by reaction of the pure metal with hydrogen halide or, for labile metal ions, by dehydration of hydrated salts with a covalent halogen compound e.g. SOCl2.
The difluorides commonly have rutile structures, for example MnF2, the dichlorides CdCl2 structures and the diiodides CdI2 structures. Dibromides have either CdCl 2 or CdI2 structures or both. Dihalides are all ionic and typically dissolve in water to give aquo complexes or mixed aquo-halo-complexes. The solutions of Ti(II), V(II) and Cr(II) are very strongly reducing. They react extremely rapidly with O 2, and Ti(II) even rapidly reduces water to liberate hydrogen. Solutions of Fe(II) undergo slow oxidation in air, but in acid or neutral solution Mn(II), Co(II), Ni(II), and Cu(II) are quite stable to oxygen.
CuI adopts the Zinc Blende cubic close packed structure.
Contributors and Attributions
• The Department of Chemistry, University of the West Indies) | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/17%3A_The_Group_17_Elements/17.04%3A_The_Elements/17.4B%3A_Dichlorine_Dibromine_and_Diiodine.txt |
Some selected properties of the hydrogen halides (HX) are given in the table below [1].
Property of HX HF HCl HBr HI
Physical appearance at 298K Colorless gas Colorless gas Colorless gas Colorless gas
Melting point /K 189 159 186 222
Boiling point /K 293 188 207 237.5
Liquid Range /K 104 29 21 15.5
ΔfusH°(mp) / kJ mol-1 4.6 2 2.4 2.9
ΔvapH°(bp) / kJ mol-1 34 16.2 18 19.8
ΔfH°(298 K) / kJ mol-1 -273.3 -92.3 -36.3 +26.5
ΔfG°(298 K) / kJ mol-1 -275.4 -95.3 -53.4 +1.7
Bond dissociation energy / kJ mol-1 570 432 366 298
Bond length /pm 92 127.5 141.5 161
Dipole moment /D 1.83 1.11 0.83 0.45
predicted pKa's 1.4 -9.3 -11.7 -12.4
At room temperature, all of the halogen halides are gases and have sharp, acid smells. They can be prepared by direct combination of the halogens with H2 or by the action of a concentrated acid (nonoxidising for HBr and HI) on metal halides.
\[CaF_{2(s)} + H_2SO_{4(l)} → CaSO_{4(s)} + 2HF_{(g)}\]
The reaction is exothermic so to optimise yields it is done at 473-523 °K for 30-60 minutes. Sulfuric acid is too strong an oxidising agent to be used in the generation of HBr and HI resulting in partial oxidation, for example:
\[2HBr + H_2SO_4 → Br_2+ 2H_2O + SO_2\]
Phosphoric acid can be used instead though:
\[3NaBr_{(s)} + H_3PO_4 → Na_3PO_{4(s)} + 3HBr_{(g)}\]
Aqueous solutions of HX are generally referred to as hydohalic acids and we will look at some chemistry of both the anhydrous hydrogen halides and the hydrohalic acids.
Commercial production of anhydrous Hydrogen Fluoride began in the 1930's and by the 1980's at least 16 countries were involved in generating over 1 million tonnes worldwide. Initially the HF was used in making refrigerants and for synthetic cryalite Na2AlF6 for aluminium production as well as in uranium processing. A small amount is used in glass etching. Hydrofluoric acid attacks glass by reaction with silicon dioxide to form gaseous or water soluble silicon fluorides. The dissolution process proceeds as follows:
\[SiO_2 + 4 HF → SiF_{4 (g)} + 2 H_2O\]
\[SiO_2 + 6 HF → H_2SiF_6 + 2 H_2O\]
This property has been known since the 17th century, even before a general procedure for the preparation of large quantities of hydrofluoric acid had been devised by Scheele in 1771. Small quantities of HF(aq) are shipped in TEFLON or polyethylene-lined containers with which it does not react.
Figure 1: Edged glass from the Bankfield Museum, Halifax, West Yorkshire, England. (53.43'.57".N; 1.51'.48".W.). from Wikipedia.
HF is miscible with water in all proportions and phase diagrams show several distinct species including
• HF.H2O (mp 237.7 °K)
• 2HF.H2O (mp 197.7 °K)
• 4HF.H2O (mp 172.8 °K).
An interesting high-tech application of HF/DF is in High-Energy Lasers as weapons which were first demonstrated in the 1970's and have now reached the level of being mobile and capable of shooting down incoming missiles. The construction of a deuterium fluoride laser resembles a rocket engine. In the combustion chamber, ethylene is burned in nitrogen trifluoride. This reaction produces free excited fluorine radicals. Just after the nozzle, the mixture of helium and hydrogen or deuterium gas is injected to the exhaust stream; the hydrogen or deuterium reacts with the fluorine radicals, producing excited molecules of deuterium or hydrogen fluoride. The excited molecules then undergo stimulated emission in the optical resonator region of the laser.
Deuterium fluoride lasers have found military applications: the MIRACL laser, the Pulsed Energy Projectile and the Tactical High Energy Lasers are of the deuterium fluoride type. The HF laser is somewhat cheaper and operates at 2.7-2.9 µm and this is affected by the atmosphere. The DF laser operates at 3.8 µm and so is considered more useful for terrestrial applications.
Hydrogen chloride
Hydrogen chloride is one of the the largest volume chemicals to be manufactured as either the gas or aqueous acid. Total world production is estimated at 20 Mt/year. When a cheap source of NaCl was available the two-stage Leblanc process, developed during the Industrial Revolution of the 1700's, was used:
\[NaCl + H_2SO_4 → NaHSO_4 + HCl\]
\[NaCl + NaHSO_4 → Na_2SO_4 + HCl\]
Here salt is converted to sodium sulfate, using sulfuric acid, giving hydrogen chloride as by-product. Initially, this gas was released to the air, but the Alkali Act of 1863 in the UK prohibited such release, the manufacturers then absorbed the HCl waste gas in water, producing hydrochloric acid on an industrial scale. Modern production of HCl is still as a byproduct that is recovered from large scale processes but now more likely to be from an Industrial Organic Scheme.
Commercial HCl is available up to about 40%, above which the evaporation rate makes it too unstable. Industrially it is usually sold at 30-35% in strength and its major use is for cleaning iron surfaces, where 18% is used. Pickling or removal of rust (iron oxide scale) from steel involves the following reaction:
\[Fe_2O-3 + Fe + 6 HCl → 3 FeCl_2 + 3 H_2O\]
The HCl is then regenerated via a closed loop reaction scheme that has been developed so that Fe2O3 is recovered as a by-product as well. When high-purity HCl is required, then burning of dihydrogen in dichlorine is used. This is a highly exothermic reaction. High quality HCl is used in the food and pharmaceutical industry as well as for water treatment.
The acid in our stomachs (gastric acid) consists mainly of HCl (around 0.5%), and large quantities of potassium chloride and sodium chloride. The pH is usually around 1-2.
The infrared spectrum of gaseous hydrogen chloride consists of a number of sharp absorption lines grouped around 2886 cm-1 and for deuterium chloride vapor this has shifted to 2090 cm-1. These classic molecular spectra can be analyzed to provide information about both rotational and vibrational energies of the molecules. The absorption lines shown involve transitions from the ground to the first excited vibrational state of HCl/DCL, but also involve changes in the rotational states. The rotational angular momentum changes by 1 during such transitions. The splitting of the lines shows the difference in rotational inertia of the two chlorine isotopes 35Cl (75.5%) and 37Cl (24.5%). Calculation of the bond length for H-35Cl for example gives a value of 131 pm, close to the accepted value of 127 pm.
Both HBr and HI can be produced by direct combination, using platinum catalysts and elevated temperatures. An alternate method for the production of HI is the quantitative reaction of iodine with hydrazine.
\[2I_2 + N_2H_4 → 4HI + N_2\]
Hydrogen Bromide is very water soluble and aqueous HBr becomes saturated at around 69% by weight at room temperature. A constant boiling point mixture exists at about 47.6% and this boils at 124.3 °C. Boiling more dilute solutions will evaporate some of the water first to eventually arrive at a concentration of 47%.
Hydrogen iodide is extremely soluble in water (425 dm-3 in 1 dm-3, which is roughly equivalent to only 4 water molecules per HI molecule). The hydroiodic acid produced is the strongest acid in the series, pka ~-10.
HI(g) + H2O(l) → H3O+(aq) + I -(aq) pKa ~ -10
HBr(g) + H2O(l) → H3O+(aq) + Br-(aq) pKa ~ -9
HCl(g) + H2O(l) → H3O+(aq) + Cl-(aq) pKa ~ -8
Melting and Boiling Points
As with the dihalogens, it is expected that with increasing size and polarizability the MP's and BP's should increase. The position of HF though needs further explanation.
There is extensive H-bonding in HF in all states and this explains why HF has such a high boiling point (293 °K), even higher than the much heavier HI. Solid HF exists as an H-bonded polymer.
Acidic properties
In water, the hydrohalic acids are formed by reaction with the gaseous hydrogen halides and apart from HF these are completely dissociating strong acids. In the case of HF, where there is a short, strong bond, (bond dissociation energy of 570 kJ mol-1) only a weak acid is formed (pKa=3.45). A qualitative explanation for this behavior is related to the tendency of HF to hydrogen-bond and form ion-pair clusters such as [H3O+F-]
\[H_2O + HF → [H_3O^+F^-] → H_3O^+ + F^-\]
The first step is thought to lie well to the right, but the 2nd step perhaps only 15%. The reaction representing the pK is given by:
HX(aq) → H+(aq) + X-(aq)
ΔG° = -RTln(K)
ΔG° = ΔH° - TΔS°
The factors that influence the degree of dissociation can not all be readily measured but can be simplified as the following 6 steps in the Hess cycle. [1]
Step 1 must be predicted since it is based on having undissociated HX(aq), and apart from HF the hydrogen halides are completely dissociated in aqueous solution. Step 2 is the cleavage of the HX bond. Steps 3 and 5 are independent of the halide and are for the ionisation and hydration of the H+ ion. Step 4 is related to the Electron Affinity and step 6 is the hydration of the gaseous X- ion.
A plot of the energies of steps 1,2,4 and 6 shows little variation for the heavier halides so that it appears that considering only the ΔH terms is not sufficient to explain the trend and it is only when the TΔS terms are considered as well that the ΔG° for HF is predicted to be positive whereas the others are all negative. The predicted value for HF is 1.4 but experimentally the value is found to be 3.45, the trend is accurately predicted but there is clearly much room for improvement!
Safety
In contrast to the weak acidicity of dilute aqueous HF solutions, in concentrated hydrogen fluoride solution, F- ions form a [HF2]-(aq) complex by addition to HF molecules. HF molecules remain ionised to compensate the loss of F- ions. More H+ ions are thus formed, making concentrated HF an effectively strong acid. Anhydrous hydrogen fluoride is an extremely strong acid comparable in strength to anhydrous sulfuric acid.
The fact that it is a weak acid in dilute solutions does not mean it is any less of a hazard and the poorly dissociated HF actually penetrates tissue more quickly than typical acids. Symptoms of exposure to hydrofluoric acid may not be immediately evident and unfortunately any delay in treatment can lead to serious consequences. The severe pain that comes on is thought to be caused by dissolution of Ca in the bone to form insoluble CaF2 and it is has been found necessary in some cases to amputate affected limbs.
Interhalogens
Some properties of interhalogen compounds are listed below. They are all prepared by direct combination of the elements although since in some cases more than one product is possible the conditions may vary by altering the temperature and relative proportions. For example under the same conditions difluorine reacts with dichlorine to give ClF with dibromine to give BrF3 but with diiodine to give IF5.[1]
Compound ClF BrF BrCl ICl IBr ClF3 BrF3 IF3 I2Cl6 ClF5 BrF5 IF5 IF7
Appearance at 298K Colorless gas Pale brown gas impure Red solid Black solid Colorless gas Yellow liquid Yellow solid Orange solid Colorless gas Colorless liquid Colorless liquid Colorless gas
Stereochemistry linear linear linear linear linear T-shaped T-shaped T-shaped planar square-based pyramid square-based pyramid square-based pyramid pentagonal bipyramid
Melting point /K 117 ~240 dissoc. 300(a) 313 197 282 245 (dec) 337 (sub) 170 212.5 282.5 278 (sub)
Boiling point /K 173 ~293 ~278 ~373 389 285 399 - - 260 314 373 -
ΔfH°(298 K) /kJ mol-1 -50.3 -58.5 14.6 -23.8 -10.5 -163.2 -300.8 ~-500 -89.3 -255 -458.6 -864.8 -962
Dipole moment for gas-phase molecule /D 0.89 1.42 0.52 1.24 0.73 0.6 1.19 - 0 - 1.51 2.18 0
Bond distances in gas-phase molecules except for IF3 and I2Cl6 / pm 163 176 214 232 248.5 160 (eq), 170 (ax) 172 (eq), 181 (ax) 187 (eq), 198 (ax) 238 (terminal) 268 (bridge) 172 (basal), 162 (apical) 178 (basal), 168 (apical) 187 (basal), 185 (apical) 186 (eq), 179 (ax)
Valence-shell electron-pair repulsion theory, VSEPR
The premise of the VSEPR is that the valence electron pairs surrounding an atom mutually repel each other, and will therefore adopt an arrangement that minimizes this repulsion, thus determining the molecular geometry. The number of electron pairs surrounding an atom, both bonding and nonbonding, is called its steric number. The VSEPR theory thus provides a simple model for predicting the shapes of such species, in particular for main group compounds. The model combines original ideas of Sidgwick and Powell (1940's) with extensions developed by Nyholm and Gillespie (1950's).
• Each valence shell electron pair of the central atom E in a molecule EXn containing E-X single bonds is stereochemically significant, and repulsions between them determine the molecular shape.
• Electron-electron repulsions decrease in the sequence: lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair.
• Where the central atom E is involved in multiple bond formation to atoms X, electron-electron repulsions decrease in the order: triple bond-single bond > double bond-single bond > single bond-single bond.
• Repulsions between the bonding pairs in EXn depend on the difference between the electronegativities of E and X; electron-electron repulsions are less the more the E.X bonding electron density is drawn away from the central atom E.
The VSEPR theory works best for simple halides of the p-block elements, but may also be applied to species with other substituents. However, the model does not take steric factors (i.e. the relative sizes of substituents) into account.+
Steric No. Basic Geometry
0 lone pair
1 lone pair 2 lone pairs 3 lone pairs
2 linear
3 trigonal planar bent
4 tetrahedral trigonal pyramid bent
5 trigonal bipyramid seesaw T-shaped linear
6 octahedral square pyramid square planar
7 pentagonal bipyramid pentagonl pyramid
Structure of the Interhalogens
The structures found for the various interhalogens conform to what would be expected based on the VSEPR model. For XY3 the shape can be described as T-shaped with 2 lone pairs sitting in equatorial positions of a trigonal bipyramid. For XY5 the shape is a square pyramid with the unpaired electrons sitting in an axial position of an octahedral and XY7 is a pentagonal bipyramid.
XY diatomic interhalogens
The interhalogens with formula XY have physical properties intermediate between those of the two parent halogens. The covalent bond between the two atoms has some ionic character, the larger element, X, becoming oxidised and having a partial positive charge. Most combinations of F, Cl, Br and I are known, but not all are stable.
• Chlorine monofluoride (ClF), the lightest interhalogen, is a colorless gas with a boiling point of 173 °K.
• Bromine monofluoride (BrF) has not been obtained pure - it dissociates into the trifluoride and free bromine. Similarly, iodine monofluoride is unstable - iodine reacts with fluorine to form a pentafluoride.
• Iodine monofluoride (IF) is unstable and disproportionates rapidly and irreversibly at room temperature: 5IF → 2I2 + IF5. However, its molecular properties have been determined by spectroscopy: the iodine-fluorine distance is 190.9 pm and the I-F bond dissociation energy is around 277 kJ mol-1. ΔHf° = -95.4 kJ mol-1 and ΔGf° = -117.6 kJ mol-1, both at 298 K.
IF can be generated, by the following reactions:
I2 + F2 → 2IF at -45 °C in CCl3F;
I2 + IF3 → 3IF at -78 °C in CCl3F;
I2 + AgF → IF + AgI at 0 °C.
• Bromine monochloride (BrCl) is an unstable red-brown gas with a boiling point of 5 °C.
• Iodine monochloride (ICl) consists of red transparent crystals which melt at 27.2 °C to form a choking brownish liquid (similar in appearance and weight to bromine). It reacts with HCl to form the strong acid HICl2. The crystal structure of iodine monochloride consists of puckered zig-zag chains, with strong interactions between the chains.
• Iodine monobromide (IBr) is made by direct combination of the elements to form a dark red crystalline solid. It melts at 42 °C and boils at 116 °C to form a partially dissociated vapor.
XY3 interhalogens
• Chlorine trifluoride (ClF3) is a Colorless gas that condenses to a green liquid, and freezes to a white solid. It is made by reacting chlorine with an excess of fluorine at 250° C in a nickel tube. It reacts more violently than fluorine, often explosively. The molecule is planar and T-shaped.
• Bromine trifluoride (BrF3) is a yellow green liquid that conducts electricity - it ionises to form [BrF2+] + [BrF4-]. It reacts with many metals and metal oxides to form similar ionised entities; with some others it forms the metal fluoride plus free bromine and oxygen. It is used in organic chemistry as a fluorinating agent. It has the same molecular shape as chlorine trifluoride.
• Iodine trifluoride (IF3) is a yellow solid which decomposes above -28 °C. It can be synthesised from the elements, but care must be taken to avoid the formation of IF5. F2 attacks I2 to yield IF3 at -45 °C in CCl3F. Alternatively, at low temperatures, the fluorination reaction I2 + 3XeF2 → 2IF3 + 3Xe can be used. Not much is known about iodine trifluoride as it is so unstable.
• Iodine trichloride (ICl3) forms lemon yellow crystals which can be melted under pressure to a brown liquid. It can be made from the elements at low temperature, or from iodine pentoxide and hydrogen chloride. It reacts with many metal chlorides to form tetrachloriodides, and hydrolyses in water. The molecule is a planar dimer, with each iodine atom surrounded by four chlorine atoms. In the melt it is conductive, which may indicate dissociation: \[I_2Cl_6 → ICl_2^+ + ICl_4^-\]
Chlorine trifluoride, ClF3 was first reported in 1931 and it is primarily used for the manufacture of uranium hexafluoride, UF6 as part of nuclear fuel processing and reprocessing, by the reaction:
\[U + 3 ClF_3 → UF_6 + 3 ClF\]
U isotope separation is difficult because the two isotopes have very nearly identical chemical properties, and can only be separated gradually using small mass differences. (235U is only 1.26% lighter than 238U.) A cascade of identical stages produces successively higher concentrations of 235U. Each stage passes a slightly more concentrated product to the next stage and returns a slightly less concentrated residue to the previous stage.
There are currently two generic commercial methods employed internationally for enrichment: gaseous diffusion (referred to as first generation) and gas centrifuge (second generation) which consumes only 6% as much energy as gaseous diffusion. These both make use of the volatility of UF6. ClF3 has been investigated as a high-performance storable oxidizer in rocket propellant systems. Handling concerns, however, prevented its use.
Note: \(ClF_3\) is Hypergolic
Hypergolic means explode on contact with no need for any activator. One observer made the following comment about \(ClF_3\):
It is, of course, extremely toxic, but that's the least of the problem. It is hypergolic* with every known fuel, and so rapidly hypergolic that no ignition delay has ever been measured. It is also hypergolic with such things as cloth, wood, and test engineers, not to mention asbestos, sand, and water � with which it reacts explosively. It can be kept in some of the ordinary structural metals-steel, copper, aluminium, etc.-because of the formation of a thin film of insoluble metal fluoride which protects the bulk of the metal, just as the invisible coat of oxide on aluminium keeps it from burning up in the atmosphere. If, however, this coat is melted or scrubbed off, and has no chance to reform, the operator is confronted with the problem of coping with a metal-fluorine fire. For dealing with this situation, I have always recommended a good pair of running shoes."[3]
It is believed that prior to and during World War II, ClF3 code named N-stoff ("substance N") was being stockpiled in Germany for use as a potential incendiary weapon and poison gas. The plant was captured by the Russians in 1944, but there is no evidence that the gas was actually ever used during the war.
XY5 interhalogens
• Chlorine pentafluoride (ClF5) is a Colorless gas, made by reacting chlorine trifluoride with fluorine at high temperatures and high pressures. It reacts violently with water and most metals and nonmetals.
• Bromine pentafluoride (BrF5) is a Colorless fuming liquid, made by reacting bromine trifluoride with fluorine at 200° C. It is physically stable, but reacts violently with water and most metals and nonmetals.
• Iodine pentafluoride (IF5) is a Colorless liquid, made by reacting iodine pentoxide with fluorine, or iodine with silver fluoride. It is highly reactive, even slowly with glass. It reacts with elements, oxides and carbon halides. The molecule has the form of a tetragonal pyramid.
• Primary amines react with iodine pentafluoride to form nitriles after hydrolysis with water. \[R-CH_2-NH_2 → R-CN\]
XY7 interhalogens [3]
• Iodine heptafluoride (IF7) is a Colorless gas. It is made by reacting the pentafluoride with fluorine. IF7 is chemically inert, having no lone pair of electrons in the valency shell; in this it resembles sulfur hexafluoride. The molecule is a pentagonal bipyramid. This compound is the only interhalogen compound possible where the larger atom is carrying seven of the smaller atoms.
• All attempts to form bromine heptafluoride have met with failure; instead, bromine pentafluoride and fluorine gas are produced.
Contributors and Attributions
• The Department of Chemistry, University of the West Indies) | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/17%3A_The_Group_17_Elements/17.07%3A_Interhalogen_Compounds_and_Polyhalogen_Ions/17.7A%3A_Interhalogen_Compounds.txt |
The noble gases (Group 18) are located in the far right of the periodic table and were previously referred to as the "inert gases" due to the fact that their filled valence shells (octets) make them extremely nonreactive. The noble gases were characterized relatively late compared to other element groups.
18: The Group 18 Elements
Introduction
Until comparatively recently, it was believed that the noble gases "had no chemistry". The observation that dioxygen reacts with platinum hexafluoride lead N. Bartlett to attempt a similar reaction with Xe in the early '60's. His attempt was prompted by his recognition that the first ionization enthalpy of Xe was almost identical to that of O2 (going to O2+).
O2 + PtF6 [O2]+[PtF6]-
Xe + PtF6 "[Xe]+[PtF6]-"
The compound actually formed with xenon was subsequently found to have a more complicated structure, but nevertheless, the way was paved for an exhaustive investigation of the chemistry of the noble gases.
The Chemistry of Xenon
The known compounds with fluorine and oxygen are listed in Table 21-2: A detailed study of this topic is not necessary for chem 242. There are also a few compounds containing a Xe-C bond, for example:
XeF2 + B(C6H5)3 [C6H5Xe]+[F3B(C6H5)]- + other products
Other Noble Gas Chemistry
Krypton forms only the unstable KrF2. Radon probably has a chemistry more extensive than xenon, but because of its radioactivity, has not been very much studied.
18.05: Compounds of Argon Krypton and Radon
It was initially believed that the noble gases could not form compounds due to their full valence shell of electrons that rendered them very chemically stable and unreactive. All noble gases have full s and p outer electron shells (except helium, which has no p sublevel), and so do not form chemical compounds easily. Because of their high ionization energy and almost zero electron affinity, they were not expected to be reactive.
The heavier noble gases have more electron shells than the lighter ones. Hence, the outermost electrons experience a shielding effect from the inner electrons that makes them more easily ionized, since they are less strongly attracted to the positively charged nucleus. This results in an ionization energy low enough to form stable compounds with the most electronegative elements, fluorine and oxygen, and even with less electronegative elements such as nitrogen and carbon under certain circumstances. These compounds are listed in order of decreasing order of the atomic weight of the noble gas, which generally reflects the priority of their discovery, and the breadth of available information for these compounds.
• Wikipedia
18.06: Group 18 Elemental Solids
Once cooled below their respective melting points, the Group 18 elements form an ordered solid. Group 18 elemental solids are useful in providing a place for electrons to become trapped or reactions to take place. Some examples include the use of solid argon to study highly reactive molecules, and solid neon to allow reaction and formation of xenon hydrides.
Contributors and Attributions
• Jordan Boothe, University of California Davis, Pharmaceutical Chemistry
19.2A: d-block Metals Versus Transition Elements
The elements of the second and third rows of the Periodic Table show gradual changes in properties across the table from left to right as expected. Electrons in the outer shells of the atoms of these elements have little shielding effects resulting in an increase in effective nuclear charge due to the addition of protons in the nucleus. Consequently, the effects on atomic properties are: smaller atomic radius, increased first ionization energy, enhanced electronegativity and more nonmetallic character. This trend continues until one reaches calcium (Z=20). There is an abrupt break at this point. The next ten elements called the first transition series are remarkably similar in their physical and chemical properties. This general similarity in properties has been explained in terms of their relatively small difference in effective nuclear charge over the series. This occurs because each additional electron enters the penultimate 3d shell providing an effective shield between the nucleus and the outer 4s shell.
Thus, the transition elements can be defined as those in which the d electron shells are being filled and so we generally ignore Sc and Zn where Sc(III) is d0 and Zn(II) is d10.
Summary of Physical Properties
It is useful, at the beginning, to identify the physical and chemical properties of transition elements which differ from main group elements (s-block) such as Calcium.
Transition elements:
• have large charge/radius ratio;
• are hard and have high densities;
• have high melting and boiling points;
• form compounds which are often paramagnetic;
• show variable oxidation states;
• form coloured ions and compounds;
• form compounds with profound catalytic activity;
• form stable complexes.
The following table summarises some of the physical properties of transition elements:
Element Group density
/g cm-3
m. p.
/ °C
b.p.
/ °C
radius
/ pm
free atom
configuration
ionization energy
/ kJ mol-1
Uses
Sc 3 2.99 1541 2831 164 [Ar] 3d14s2 631
Ti 4 4.50 1660 3287 147 [Ar]3d24s2 658 -engines/aircraft industry-density is 60% of iron
V 5 5.96 1890 3380 135 [Ar]3d34s2 650 -stainless steel, 19% Cr, 9% Ni the rest Fe
Cr 6 7.20 1857 2670 129 [Ar]3d54s1 653 -alloys eg with C steel, the most significant use
Mn 7 7.20 1244 1962 137 [Ar]3d54s2 717 -alloys eg with Cu
Fe 8 7.86 1535 2750 126 [Ar]3d64s2 759 -alloys eg with C steel, the most significant use
Co 9 8.90 1495 2870 125 [Ar]3d74s2 758 -alloys eg with Cr and W for hardened drill bits
Ni 10 8.90 1455 2730 125 [Ar]3d84s2 737 -alloys Fe/Ni armour plating, resists corrosion
Cu 11 8.92 1083 2567 128 [Ar]3d104s1 746 -high electrical conductivity (2nd to Ag), wiring
Zn 12 7.14 420 907 137 [Ar]3d104s2 906
Densities and metallic radii
The transition elements are much denser than the s-block elements and show a gradual increase in density from scandium to copper. This trend in density can be explained by the small and irregular decrease in metallic radii coupled with the relative increase in atomic mass.
Melting and boiling points
The melting points and the molar enthalpies of fusion of the transition metals are both high in comparison to main group elements. This arises from strong metallic bonding in transition metals which occurs due to delocalization of electrons facilitated by the availability of both d and s electrons.
Ionization Energies
In moving across the series of metals from scandium to zinc a small change in the values of the first and second ionization energies is observed. This is due to the build-up of electrons in the immediately underlying d-sub-shells that efficiently shields the 4s electrons from the nucleus and minimizing the increase in effective nuclear charge from element to element. The increases in third and fourth ionization energy values are more rapid. However, the trends in these values show the usual discontinuity half way along the series. The reason is that the five d electrons are all unpaired, in singly occupied orbitals. When the sixth and subsequent electrons enter, the electrons have to share the already occupied orbitals resulting in inter-electron repulsions, which would require less energy to remove an electron. Hence, the third ionization energy curve for the last five elements is identical in shape to the curve for the first five elements, but displaced upwards by about 580 kJ mol-1.
Electronic Configurations
The electronic configuration of the atoms of the first row transition elements are basically the same. It can be seen in the Table above that there is a gradual filling of the 3d orbitals across the series starting from scandium. This filling is, however, not regular, since at chromium and copper the population of 3d orbitals increase by the acquisition of an electron from the 4s shell. This illustrates an important generalization about orbital energies of the first row transition series. At chromium, both the 3d and 4s orbitals are occupied, but neither is completely filled in preference to the other. This suggests that the energies of the 3d and 4s orbitals are relatively close for atoms in this row.
In the case of copper, the 3d level is full, but only one electron occupies the 4s orbital. This suggests that in copper the 3d orbital energy is lower than the 4s orbital. Thus the 3d orbital energy has passed from higher to lower as we move across the period from potassium to zinc. However, the whole question of preference of an atom to adopt a particular electronic configuration is not determined by orbital energy alone. In chromium it can be shown that the 4s orbital energy is still below the 3d which suggests a configuration [Ar] 3d44s2. However due to the effect of electronic repulsion between the outer electrons the actual configuration becomes [Ar]3d54s1 where all the electrons in the outer orbitals are unpaired. It should be remembered that the factors that determine electronic configuration in this period are indeed delicately balanced.
Redox Couple E°/V
Mn2+(aq.)/Mn(s) -1.18
H+(aq.)/H2(g) 0.00
This shows that elemental Mn is a stronger reductant than dihydrogen and hence should be able to displace hydrogen gas from 1 mol dm-3 hydrochloric acid. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/18%3A_The_Group_18_Elements/18.01%3A_Introduction.txt |
To be able to use Crystal Field Theory (CFT) successfully, it is essential that you can determine the electronic configuration of the central metal ion in any complex.
This requires being able to recognise all the entities making up the complex and knowing whether the ligands are neutral or anionic, so that you can determine the oxidation number of the metal ion.
In many cases the oxidation number for first row transition metal ions will be either (II) or (III), but in any case you may find it easier to start with the M(II) from which you can easily add or subtract electrons to get the final electronic configuration.
A simple procedure exists for the M(II) case.
First write out all the first row transition metals with their symbols and atomic numbers:
22 23 24 25 26 27 28 29
Ti V Cr Mn Fe Co Ni Cu
To see the number of electrons in the 3d orbitals then cross off the first 2, hence:
2 3 4 5 6 7 8 9
So, the electronic configuration of Ni(II) is d8 and the electronic configuration of Mn(II) is d5.
What is the electronic configuration of Fe(III)?
Well, using the above scheme, Fe(II) would be d6, by subtracting a further electron to make the ion more positive, the configuration of Fe(III) will be d5.
This simple procedure works fine for first row transition metal ions, but sorry it is no good for 2nd or 3rd row elements!
Note: For all final Chemistry examinations, a Periodic Table is provided in the inside back cover of the examination booklets. A Periodic Table may NOT necessarily be provided for course tests.
Oxidation Numbers and their Relative Stabilities
The IUPAC definition of the oxidation number in a coordination compound is:
the charge a central atom in a coordination entity would bear if all the ligands were removed along with the electron pairs that were shared with the central atom. It is represented by a Roman numeral.
The transition metals show a wide range of oxidation numbers. The reason for this is the closeness of 3d and 4s energy states as discussed above. The Table below summarizes known oxidation numbers of the first row transition elements. The most prevalent oxidation numbers are shown in bold and those in blue are likely to be met in this course.
Known Oxidation Numbers of First Row Transition Elements*
21 22 23 24 25 26 27 28 29 30
Sc Ti V Cr Mn Fe Co Ni Cu Zn
I I I I I I I I
II II II II II II II II II
III III III III III III III III III
IV IV IV IV IV IV IV
V V V V V
VI VI VI
VII
* The oxidation number zero usually assigned to the elemental state has been omitted from the Table. The elements Cr to Co form several metal carbonyl compounds where the metals are considered to have an oxidation number of zero.
A number of important conclusions can be drawn from this Table.
1. There is an increase in the number of oxidation numbers from Sc to Mn. All seven oxidation numbers are exhibited by Mn. The oxidation number of VII represents the formal loss of all seven electrons from 3d and 4s orbitals. In fact all of the elements in the series can utilize all the electrons in their 3d and 4s orbitals.
2. There is a decrease in the number of oxidation states from Mn to Zn.
This is because the pairing of d-electrons occurs after Mn (Hund's rule) which in turn decreases the number of available unpaired electrons and hence, the number of oxidation states.
3. The stability of higher oxidation states decreases in moving from Sc to Zn. Mn(VII) and Fe(VI) are powerful oxidizing agents and the higher oxidation states of Co, Ni and Zn are unknown.
4. The relative stability of the +2 state with respect to higher oxidation states, particularly the +3 state increases in moving from left to right. This is justifiable since it will be increasingly difficult to remove the third electron from the d orbitals.
5. There is a tendency of intermediate oxidation states to disproportionate. For example, Mn(VI) → Mn(IV) + Mn(VII)
Cu(I) → Cu(0) + Cu(II).
6. The lower oxidation numbers are usually found in ionic compounds and higher oxidation numbers tend to be involved in covalent compounds.
The relative stability of oxidation numbers is an extremely important topic in transition metal chemistry and is usually discussed in terms of the standard reduction potential (E°) values. Thermodynamically E° values are equated to ΔG° values by the relationship: ΔG° = -nFE° where n = number of electrons involved and F = Faraday of electricity. Hence, the E° values indicate the possibility of spontaneous change from one oxidation state to the other. This value however, does not give any information about the reaction rate. Predictions regarding the stability of a particular oxidation state of an element can be made from the Tables of Redox values found in any standard text book or online. | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/19%3A_d-Block_Metal_Chemistry_-_General_Considerations/19.02%3A_Ground_State_Electronic_Configurations/19.2B%3A_Electronic_Configurations.txt |
Introduction to the colour and magnetism of 1st row transition metal complexes
Before beginning a more detailed examination of the spectroscopy and magnetism of transition meal complexes, it is worth while reviewing how far a simple CFT approach will take us.
When electromagnetic radiation is absorbed by atoms or molecules it promotes them to an excited state. Microwave and infrared radiation correspond to lower energy quanta and so initiate rotational and vibrational excitation. Visible and UV light have much higher frequencies and can cause excitations characterstic of electronic excitation: the promotion of an electron from one orbital to another. We expect therefore that molecules will absorb light when the energy corresponds to the energy differences between occupied and unoccupied orbitals. For transition metal ions, the simplest case is Ti(III), solutions of which appear violet.
Absorption of light of frequency ~20,000 cm-1 excites the electron from the t2g subset to the eg subset. This is described as a eg ← t2g transition.
Absorption of green light, i.e. transmission of blue and red, gives a purple solution
ν ~20,000 cm-1
λmax ~ 500 nm
E = hν = hc/λ
Δ ~ 240 kJ mol-1
Rough guide to absorbance and colour
Wavelength Absorbed (nm) Frequency (cm-1) Colour of Light Absorbed Colour of Complex
410 24,400 violet lemon-yellow
430 23,300 indigo yellow
480 20,800 blue orange
500 20,000 blue-green red
530 18,900 green purple
560 17,900 lemon-yellow violet
580 17,200 yellow indigo
610 16,400 orange blue
680 14,700 red blue-green
In spectroscopy it is usual to measure either the amount of light that is absorbed or transmitted through the sample. For UV/Vis, absorbance is given by the Beer-Lambert expression:
A = ε c l
where A is the Absorbance
ε is the molar absorbance (extinction coefficient)
c is the concentration
and l is the path length of the cell
The most common (and cheapest) sample cells have a 1 cm path length and since A is unitless then we can see that the units of ε are mol-1 l cm-1. To move this to an acceptable SI set of units requires converting ε to units of m2 mol-1 and this involves a factor of 1/10.
Thus an ε of 5 mol-1 l cm-1 is equivalent to ε of 0.5 m2 mol-1.
Given that the separation between the t2g and eg levels is Δ then whether there is 1 d electron or several d electrons the simple Crystal Field Theory model would suggest that there is only 1 energy gap hence all spectra should consist of 1 peak. That this is not found in practise means that the theory is not sophisticated enough. What is required is an extension of the theory that allows for multi-electron systems where the energy levels are modified to include electron-electron interactions. This can be achieved by looking at the various quantum numbers for each of the electrons involved and using a system called the Russell-Saunders coupling scheme to describe an electronic state that can adequately describe the energy levels available to a group of electrons that includes these interactions.
Contributors and Attributions
• The Department of Chemistry, University of the West Indies)
19.5B: Paramagnetism
Magnetism
In CHEM1902 (C10K) we introduced the formula used to relate the magnetic moment to the number of unpaired spins in a transition metal complex.
μs.o. = √{4S(S+1)} B.M.
During the laboratory session you will carry out a measurement of the magnetic susceptibility which is a measure of the force exerted by the magnetic field on a unit mass of the sample under investigation. This is related to the number of unpaired electrons per unit weight and hence per mole and in the simplest picture we consider that this is solely dependent on the presence of unpaired electrons.
For a Ti(III) complex with 1 unpaired electron this corresponds to:
μs.o. = 2 √ (1/2 (1/2 + 1)) B.M
μs.o. = √ (3) B.M.
μs.o. = 1.73 Bohr Magneton
We will see later that while the spin-only approximation works in many cases, for a more complete analysis it is necessary to consider the contribution made by the orbital motion of the electron as well.
Contributors and Attributions
• The Department of Chemistry, University of the West Indies)
19.06: Electroneutrality Principle
Isomers are classified into 1) stereoisomers which have different spatial orientations, and 2) constitutional isomers where atoms are connected in different orders.
19: d-Block Metal Chemistry - General Considerations
Pauling's principle of electroneutrality states that each atom in a stable substance has a charge close to zero. It was formulated by Linus Pauling in 1948 and later revised. The principle has been used to predict which of a set of molecular resonance structures would be the most significant, to explain the stability of inorganic complexes and to explain the existence of π-bonding in compounds and polyatomic anions containing silicon, phosphorus or sulfur bonded to oxygen; it is still invoked in the context of coordination complexes. However, modern computational techniques indicate many stable compounds have a greater charge distribution than the principle predicts (they contain bonds with greater ionic character).
• Wikipedia
19.7A: The Kepert Model
The Kepert model is a modification of VSEPR theory used to predict the 3-dimensional structures of transitional metal complexes. In the Kepert model, the ligands attached to the metal are considered to repel each other the same way that point charges repel each other in VSEPR theory. Unlike VSEPR theory, the Kepert model does not account for non-bonding electrons. Therefore, the geometry of the coordination complex is independent of the electronic configuration of the metal center. Thus [MLn]m+ has the same coordination geometry as [MLn]m−. The Kepert model cannot explain the formation of square planar complexes or distorted structures.
The Kepert model predicts the following geometries for coordination numbers of 2 through 6:
1. Linear
2. Trigonal planar
3. Tetrahedral
4. Trigonal bipyramidal or Square pyramidal
5. Octahedral | textbooks/chem/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/19%3A_d-Block_Metal_Chemistry_-_General_Considerations/19.05%3A_Characteristic_Properties_-_A_General_Perspective/19.5A%3A_Color.txt |
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