draw an electron dot diagram to show the formation of each of the following compounds: magnesium chloride - periodic properties - electron affinity
draw an electron dot diagram to show the formation of each of the following compounds: magnesium chloride [h = 1, c = 6, mg = 12, cl = 17]
magnesium atom loses 2 electrons to attain a stable electronic configuration and becomes a cation. example: a non-metallic atom like chlorine gains 1 electron to attain a stable electronic configuration and becomes an anion. example: cations and anions are oppositely charged particles which attract one another to form an electrovalent bond leading to the formation of an electrovalent compound here magnesium donates one electron each with two chlorine atoms resulting in the formation of magnesium chloride.
AgNO3(aq) + NaCl(s) = AgCl(s) + NaNO3  
n(AgCl) = 0.765 g / M(AgCl)  
M(AgCl) = 108 + 35.5 = 143.5 g/mol  
n(AgCl) = 0.765 g /143.5 g/mol = 0.00533 mol  
n(AgCl) = chloride ions = 0.00533 mol  
C(chloride ions) = n/V  
C(chloride ions) = 0.00533 mol/1l = 0.00533 mol/ L (a)  
m(NaCl)=n(NaCl)*M(NaCl)  
n(NaCl) = 0.00533 mol  
M(NaCl) = 23 + 35.5 = 58.5 g/mol  
m(NaCl) = 0.00533 mol * 58.5 g/mol = 0.3115 g (b)  
  
**Answer:**  
**a) 0.00533 mol/ L**  
**b) 0.3115 g** The physical and chemical properties of elements depend on the number of valence electrons. Elements present in the same group have the same number of valence electrons. Therefore, elements present in the same group have similar physical and chemical properties. <p>concepts and reason</p>
<p>the concept used to solve this problem is the formal charge calculation of the oxygen atoms present in the ozone molecule.</p>
<p>fundamentals</p>
<p>a charge that is assigned to an atom, which is present in the molecule, is represented as a formal charge. in two bonded atoms, a formal charge assumes that the electrons are equally shared between them. the formal charge is calculated using the following formula:</p>
<p>formal charge \(=\mathrm{v}-\mathrm{n}-\frac{\mathrm{b}}{2}\)</p>
<p>here, \(\mathrm{v}\) is the number of valence electrons, \(\mathrm{n}\) is the number of non-bonding electrons and \(\mathrm{b}\) is the number of bonding electrons.</p>
<p> </p>
<p>the structure is as follows:</p>
<p>calculate the formal charge for the circled oxygen in ozone molecule as follows:</p>
<p><img src="https://img.wizedu.com/images/5af6d1f54822b4dfb8fe89d91ef2011c5fe3020815aa4.png" alt="" width="88" /></p>
<p>formal charge \(=\mathrm{v}-\mathrm{n}-\frac{\mathrm{b}}{2}\)</p>
<p>\(\begin{aligned}=& 6-6-\frac{2}{2} \\ &=-1 \end{aligned}\)</p>
<p>explanation | common mistakes | hint for next step</p>
<p>the marked oxygen atom has 6 valence electrons. it contains 6 non-bonding electrons. the marked oxygen atom is bonded with one oxygen atom. therefore, the bonding electron of oxygen is \(2 .\) substitute the values in a formal charge formula to determine the formal charge of oxygen (marked).</p>
<p> </p>
<p>the structure is as follows:</p>
<p><img src="https://img.wizedu.com/images/39493056a8515708b79350d1976e17b35fe30216f0473.png" alt="" width="88" /></p>
<p>calculate the formal charge for the circled oxygen in ozone molecule as follows:</p>
<p>formal charge \(=\mathrm{v}-\mathrm{n}-\frac{\mathrm{b}}{2}\)</p>
<p>\(=6-2-\frac{6}{2}\)</p>
<p>\(=+1\)</p>
<p>explanation | common mistakes | hint for next step</p>
<p>the oxygen atom in the middle has 6 valence electrons and has 2 non-bonding electrons. the central oxygen atom is bonded with 1 double bond and 1 single bond. so, there is a total of 3 bonds containing 6 electrons (bonding electrons). the formal charge of the middle oxygen atom can be determined by substituting the values in a formal charge formula.</p>
<p> </p>
<p>the structure is as follows:</p>
<p><img src="https://img.wizedu.com/images/b1122ce7723aa4ab1f9ae5c9e457299e5fe3022ca77a5.png" alt="" width="88" /></p>
<p>calculate the formal charge for the circled oxygen in ozone molecule as follows:</p>
<p>formal charge \(=\mathrm{v}-\mathrm{n}-\frac{\mathrm{b}}{2}\)</p>
<p>\(=6-4-\frac{4}{2}\)</p>
<p>\(=0\)</p>
<p>the formal charge of each oxygen atom in \(\mathrm{o}\_{3}\) molecule are labeled as follows:</p>
<p><img src="https://img.wizedu.com/images/1de5ee227194f6d8c7fb9fb0e76160f25fe3024117fde.png" alt="" width="120" /></p>
<p>the oxygen atom present on the right side contains 6 valence electrons which have 2 non-bonded electrons. the marked oxygen is bonded with 1 double bond with a central oxygen atom. so, the number of bonding electrons is \(4 .\) the formal charge of the corner oxygen atom (right corner) can be determined by substituting the values in the formal charge formula.</p>
<p>the formal charge of each oxygen atom in \(\mathrm{o}\_{3}\) molecule are labeled as follows:</p>
<p><img src="https://img.wizedu.com/images/1de5ee227194f6d8c7fb9fb0e76160f25fe3024117fde.png" alt="" width="120" /></p> a) van der waals (present, all molecule/atoms contain it)[readmore](b) dipole-dipole (present, c=o creates a partial charge on c and o)(c) hydrogen bonding (not present, h-bonds are only present in compounds with h attached on o, n or f)(d) ion-ion (not present, only appears on compound with full charges) a) he helium (he) is an element by itself. he atom can only exhibit dispersion forces.b) hclh          1a                   1 × 1 e– = 1 e–cl         7a                   1 × 7 e– = 7 e–   total: 8 valence e–[readmore]cl is more electronegative than h, so hcl is a polar covalent compound. hcl exhibits dipole-dipole interaction.c) h2oh          1a                   1 × 2 e– = 2 e–o          6a                   1 × 6 e– = 6 e–   total: 8 valence e– since hydrogen is bonded directly to oxygen.h2o exhibits hydrogen bonding.d) cl2cl         7a                   1 × 7 e– = 14 valence e–composed of the same atom. there's no electronegativity difference.cl2 can only exhibit dispersion forces. we are going to do the following steps:step 1: determine the classification of the compoundstep 2: write the formula of the compound[readmore]step 1: determine the classification of the compoundrecall the definition of the following:ionic compound - usually made up of a metal and a non-metalcovalent compound - composed of more than one nonmetal togetheracid - substance whose molecules yield hydrogen ions (h+) when dissolved in waterhydrate - substance that contains water or its constituent elementsfor cobaltic iodidecobaltic – metaliodide – nonmetalmetal + nonmetal = ionicstep 2: write the formula of the compoundthe rules for naming ionic compounds are as follows:1. the metal keeps its name and is named and written first.– for transition metals, we use roman numerals to denote its charge.2. the nonmetal keeps its base name but has its ending changed to –ide.for cobaltic iodidecobaltic  → transition metal → +2 charge → co2+iodide → group 7a → -1 charge → i-criss-cross the charges and make them the subscript of the other ioncobaltic iodide → coi2answer: cobaltic iodide is an ionic compound with a formula of coi2 hybridisation: d2sp3magnetic character: paramagneticspin nature of complex: low-spin complex Δl =l αΔt =2 •10⁻²•14•10⁻⁶•70=1.96•10⁻⁵m=1.96•10⁻² mm ions are responsible for the conductivity of electricity but in case of covalent compounds there are no free ions. so, covalent compounds are poor conductor of electricity. tagging|||chemistry||metals and non metals||reaction of metals and non metals we’re being asked which of the given compounds is a lewis acid and lewis base. let’s first define a lewis acid and a lewis base.based on the lewis definition:lewis acid is an electron pair acceptor.some characteristics of a lewis acid:• when hydrogen is connected to an electronegative element such as p, o, n, s or halogens▪ hydrogen gains a partially positive charge → makes hydrogen act as a lewis acid• positively charged metal ions[readmore]▪ can accept negative electron pairs → can act as lewis acids• central element that has less than 8 valence electrons around it▪ more likely to accept an electron pair to follow the octet rule▪ can accept electron pair → can act as lewis acidlewis base is an electron pair donor.some characteristics of a lewis base:• compounds that have lone pair(s) on a central element can act as lewis bases. • compounds that possess a negative charge can act as lewis bases. let’s draw the lewis structure of each given compound and check which of them can act as a lewis acid and lewis base.step 1: determine the central atom in this molecule.step 2: calculate the total number of valence electrons present.   step 3: draw the lewis structure for the molecule.a) cs2step 1: determine the central atom in this molecule.c is less electronegative than s → c will be the central atomstep 2: calculate the total number of valence electrons present.                group             valence electronsc         4a                   1 x 4 e‑ = 4 e-s         6a                   2 x 6 e‑ = 12 e-              total: 16 valence e- step 3: draw the lewis structure for the molecule• c has a total of 8 e- electrons around it and has a lone pair→ complete octet▪ can act as a lewis acidb) sh−step 1: determine the central atom in this molecule.only two atom → no central atomstep 2: calculate the total number of valence electrons present.                group             valence electronss         6a                   1 x 6 e‑ = 6 e-h        1a                   1 x 1 e‑ = 1 e-from -1 charge:                     +1 e-              total: 8 valence e- step 3: draw the lewis structure for the molecule▪ has a negative charge▪ can act as a lewis baseanswer: sh- is the lewis base and its lewis structure is cs2 is the lewis acid and its lewis structure is