Schrödinger equation and wave function
There are a lot of questions in this one post which is probably why there has not yet been an attempt to answer this post.

1.) What is the Schrödinger wave equation?

The wave equation (i.e. wave function) is a function which contains all the information about a chemical system.  With an appropriate operator, one can extract this information and gain insight into the system.  For example, the Hamiltonian operator will extract the energy of a system from the wave function.  The wave function is an eigenfunction while the energy (in this example) is the eigenvalue of the eigenfunction.  For example, in the Schrödinger equation

$$ \mathcal{H}\Psi = E\Psi$$

$\mathcal{H}$ is the Hamiltonian operator, $\Psi$ is the wave function (eigenfunction) and $E$ is the energy (eigenvalue).  Notice that by operating on $\Psi$ with $\mathcal{H}$ returns $E$ AND $\Psi$, which by definition, makes this an eigenfunction-eigenvalue equation.

2.) What are radial probability functions?

These are probability distributions for the probability of an electron residing some distance from the nucleus of an atom such as the hydrogen atom.  Consider this plot of various radial probability distribution functions of the hydrogen atom.



Here we see that the probability of an electron (in the hydrogen atom) in a 1s orbital is zero at the nucleus (at 0,0) and a maximum a hair shy of $2 a_0$.  Interestingly an electron in a 2p orbital has a small probability of residing close to the nucleus ($&lt; 1a_0$) but is more likely to reside further away from the nucleus than an electron in a 1s orbital.  This makes sense given that a 2p orbital is higher in energy because 'it' is further away from the nucleus than a 1s orbital.

3.) What are nodes in radial and angular functions?

A node is where the function is zero at a particular point.  This simply means in a physical sense that at this point (at the node), the probability of an electron existing at that point is ZERO.  Nodes are a result of the separation of variables technique when solving the appropriate equation(s). When any one component (or variable) is zero then the resulting function will be zero at that point.

4.) What is electron cloud?

Electron 'cloud' is just a nicer away of saying electron density.  Electrons are not strictly particles.  Electrons do not just 'orbit' a nucleus.  Electrons also exhibit wavelike behavior.  (see: double-slit experiment - wave-like behavior and J. J. Thompsons cathode ray tube experiments - particle-like behavior).  The 'cloud' is obtained simply by 'squaring' the wave function.  More technically, its taking the product of the wave function and its complex conjugate.  The 'cloud' is just a probability distribution of where an electron is likely to be (usually given in 3-dimensions).

5.) What is the Hamiltonian operator?

The Hamiltonian operator is the energy operator to the Schrodinger equation.  Because it is Hermitian, it will return a 'real' result (as opposed to a complex result).  There are different versions of the Hamiltonian depending upon what approximations you want to make.  The full Hamiltonian operator can be written as (in terms of atomic units)

$$\mathcal{H} = -\sum\limits_{i=1}^{N}\dfrac{1}{2}\nabla_i^2
              -\sum\limits_{A=1}^{M}\dfrac{1}{2M_A}\nabla_A^2
              -\sum\limits_{i=1}^{N}\sum\limits_{A=1}^{M}\dfrac{Z_A}{r_{iA}}
              +\sum\limits_{i=1}^{N}\sum\limits_{j&gt;i}^{N}\dfrac{1}{r_{ij}}
              +\sum\limits_{A=1}^{M}\sum\limits_{B&gt;A}^{M}\dfrac{Z_A Z_B}{R_{AB}}
$$

Each term on the right hand side of the equation (going from left to right) is defined as


the kinetic energy of the electrons
the kinetic energy of the nuclei
the potential energy of the electron-nuclear attractions
the potential energy of the electron-electron repulsions
the potential energy of the nuclear-nuclear repulsions


We can write the Hamiltonian using a bit more compact notation such as

$$
\mathcal{H} = \hat{T}_{e} + \hat{T}_{N} + \hat{V}_{eN} + \hat{V}_{ee} + \hat{V}_{NN}
$$

where negative signs are assumed for each $\hat{T}$ operator.  The terms denoted here correspond to the respective terms in the equation before this one.

If we take into account the Born-Oppenheimer approximation (the B.O. approximation assumes that the electron motion is orders of magnitude larger than the nuclei they belong to... so much faster that the nuclei are, relatively, standing still with respect to the electrons), we can make two changes to the full Hamiltonian giving us the electronic Hamiltonian.

$$
\mathcal{H} = \hat{T}_{e} + \hat{V}_{eN} + \hat{V}_{ee} +\ \mathrm{constant}
$$

The nuclei are treated as frozen in space when the B.O. approximation is applied for reasons explained above.  Therefore the kinetic energy of the nuclei goes to zero.  The nuclear-nuclear repulsion is now a constant because the nuclei are fixed.  The repulsion will never change.

6.) What is a Laplace operator?

The Laplace operater (or Laplacian) is simply a differential operator of the form

$$
\nabla^2 = \dfrac{\partial^2}{\partial x^2} +  \dfrac{\partial^2}{\partial y^2} +  \dfrac{\partial^2}{\partial z^2}
$$

if you are working with Cartesian coordinates. This is simply a set of second-order derivative operators within the kinetic energy operator.   



While I may have addressed each question, there is a lot more depth to these answers that lie well beyond the scope of this post.  I encourage you to find more detailed and insightful answers in well-established texts.  I highly recommend Szabo &amp; Ostlund for a solid introduction to quantum mechanics.  You can snag a paperback version of this excellent work for about $10.