Orbitals are one-electron wave functions, whether they refer to electronic motion in an atom (atomic — Physical Chemistry — Electrochemistry Chemistry Question
Atomic and molecular orbitals
Orbitals are one-electron wave functions, whether they refer to electronic motion in an atom (atomic orbitals) or in a molecule (molecular orbitals) or a solid. Each orbital corresponds to a certain probability distribution of finding an electron in different regions of space.
Atomic orbitals
The 1s orbital of hydrogen atom is given by
1sΨ = e^(-r/a0)
where ao is the Bohr radius (ao = 5.3×10–11 m) and r is the radial co-ordinate (distance of a point in space from the centre).
Normalize the given wave function.
Model Answer
1sψ = N e^(-r/a0)
∫ |ψ|^2 dv = 1
4π N^2 ∫ e^(-2r/a0) r^2 dr = 1
N = 1 / √(π a0^3)
1sΨ = 1 / √(π a0^3) e^(-r/a0)
At what distance from the nucleus is the electron most likely to be found?
Model Answer
Probability of finding an electron between r and r+dr =
= 4πr^2 × (1 / π a0^3) e^(-2r/a0) dr
This is a maximum at r = rmax , given by
d/dr (r^2 e^(-2r/a0)) = 0
This gives rmax = a0
The 1s electron is most likely to be found in the neighborhood of r = a0.
The wave functions for 2s, 2pz and 3dz2 states are given below:
2sΨ = (2 - r/a0) e^(-r/2a0)
2p_zΨ = (r/a0) cosθ e^(-r/2a0)
3d_z2Ψ = (r/a0)^2 (3cos^2θ - 1) e^(-r/3a0)
What are the nodal surfaces of these orbitals?
Model Answer
2p_zΨ = 0 at r = 2a0
Nodal surface is a sphere of radius 2a0.
3d_z2Ψ = 0 at 3 cos^2 θ - 1 = 0 i.e. θ = arccos(±1/sqrt(3))
Nodal surfaces are cones with these values of half-angle, one above the xy plane and the other below it.
(Note: all three wave functions vanish as r . At r = 0 1sΨ does not vanish, but the other two wave functions vanish.)
It turns out that the solution of Schrödinger equation for a one-electron atom yields exactly the ‘good old’ formula of Bohr for quantized energies:
En = -(13.6 eV) Z^2 / n^2
where, for convenience, the numerical value of the combination of constants appearing in the formula has been put in units of eV.
It is fun using this formula for a neutral helium atom, but we must exercise some care. In a helium atom, each electron ‘sees’ the nucleus screened by the other electron. That is, the effective charge of the nucleus ‘seen’ by each electron decreases from its bare value Z = 2 to some other value, say, Zeff .
The ionization energy for a helium atom in its ground state is known experimentally to be 24.46 eV. Estimate Zeff .
Model Answer
Each electron in n = 1 shell of helium atom has energy –Zeff^2 × 13.6 eV.
Helium ground state energy = –Zeff^2 × 27.2 eV
Energy of He+ ground state = – 4 × 13.6 = – 54.4 eV
Ionization energy = (– 54.4 + Zeff^2 × 27.2) eV = 24.46 eV
This gives Zeff = 1.70
Molecular orbitals
Molecular orbitals of a hydrogen molecule ion (H2 + ) can be approximately written as linear combinations of atomic orbitals centered around the two nuclei of the molecule. Consider the (unnormalized) molecular orbitals constructed in this manner from the 1s and 2s orbitals of two hydrogen atoms, say, A and B:
Taking the z-axis along the line joining the two nuclei, the orbital contours of 1 and 1 are shown schematically below :
Similar orbital contours (curves on which the value of is constant) can be drawn for 2 and 2. The energies of these wave functions as a function of internuclear distance are shown below schematically:
Identify the bonding and antibonding orbitals. State qualitatively what makes one orbital bonding and another antibonding.
Model Answer
1 and 2 are bonding orbitals
1 and 2 are antibonding orbitals
Bonding orbital:
No nodal surface between the nuclei. Electronic energy has a minimum at a certain internuclear distance. Qualitative reason: electron has considerable probability of being between the nuclei and thus has attractive potential energy due to both the nuclei.
Antibonding orbital:
Nodal surface between the nuclei. Electronic energy decreases monotonically with internuclear distance. Hence bound state is not possible.
Determine the values of the equilibrium internuclear distance Re and the dissociation energy D of the ground state of H2 + .
Model Answer
Re = 1.32×10^-10 m
D = –13.6 – (–15.36) = 1.76 eV
If the molecular ion H2 + is excited to the state 2, to what atomic states will it dissociate?
Model Answer
It will dissociate to a hydrogen atom in 2s state and a bare hydrogen nucleus (proton).
In the following questions, assume that the energy versus internuclear distance graphs for the orbitals of H2 and He2 are similar to the one shown for H2 + .
Explain why the ground state total electron spin of the neutral H2 molecule is zero.
Model Answer
The two electrons occupy the same molecular orbital with the lowest energy. By Pauli’s principle, their spins must be antiparallel. Hence the total electronic spin is zero.
Write down the electronic configuration of the first excited state of H2 molecule. Predict if it will stay bound or dissociate.
Model Answer
In the first excited state of H2, one electron is in 1 (bonding orbital) and the other in 1 (antibonding orbital). It will dissociate into two hydrogen atoms.
It is difficult to obtain He2 in its ground state, but it has been observed in its excited states. Explain how this is possible.
Model Answer
Using the aufbau principle, in the ground state two electrons of He2 are in 1 (bonding orbital) and two in 1 (antibonding orbital). The bond order is ½ (2 – 2) = 0
Therefore, bond in He2 is unstable and difficult to detect.
However, if one or more electrons are elevated from the antibonding orbital to (higher energy) bonding orbitals, the bond order becomes greater than zero. This is why it is possible to observe He2 in excited states.