Solutions 2

Q1:

We are asked to normalize \( |\psi \rangle = |\psi_A \rangle + \lambda|\psi_B\rangle \),

Givens: \( \langle \psi_A|\psi_A \rangle = \langle \psi_B|\psi_B \rangle = 1\) and \( \langle \psi_A|\psi_B \rangle = \langle \psi_B|\psi_A \rangle = S \)

The normalization constant N, would be given by:

\( N = \frac{1}{\sqrt{\langle \psi | \psi \rangle}} \)

Since \(\langle \psi | \psi \rangle = ( \langle \psi_A | + \lambda \langle \psi_B| ) ( |\psi_A \rangle + \lambda|\psi_B \rangle) = \langle \psi_A|\psi_A \rangle + \lambda \langle \psi_A| \psi_B \rangle + \lambda \langle \psi_B| \psi_A \rangle + \lambda^2 \langle \psi_B|\psi_B \rangle = 1 + 2\lambda S + \lambda^2 \) ,

\( N = \frac{1}{\sqrt{1 + 2\lambda S + \lambda^2 }}\)

Q2:

The \(Cl−Sn−Cl\) bond angle of 90° implies the central tin atom is essentially unhybridized. It uses its p-orbitals to bond to chlorine.

Q3:

Given :

\[|\psi_A \rangle = N \left( 0.61 | 2p_z \rangle + 0.79 | 2p_x \rangle - 0.5 | 2s \rangle \right)\]

\[|\psi_B \rangle = N \left( 0.61 | 2p_z \rangle - 0.79 | 2p_x \rangle - 0.5 | 2s \rangle \right)\]

a). Bond angle relates to "s character" by the following relations: \( \%s = \frac{\gamma^2}{1+\gamma^2} \) and \( \cos(\theta) = -\gamma^2 \).

Rearranging we have \( \theta = \arccos(-\gamma^2) = \arccos(-\frac{\%s}{1-\%s}) \)

\( \%s\) from part (d) is found to be .2 so:

\( \theta = \arccos(-\frac{.2}{1-.2}) = \arccos(-.25) = 1.3181 \) rad \( = 104.4775° \)

b). This is consistent with the experimentally observed value of 104.5°.

c). The bond angles would probably be closer to 90° due to orthogonality.

d). "s character" is given by the probability of being in the s state: \( \frac{\langle s| \psi_A\rangle^2}{ \langle \psi_A| \psi_A \rangle} = \frac{\langle s| \psi_B\rangle^2}{ \langle \psi_B| \psi_B \rangle} = \frac{N^2.5^2}{N^2(.61^2 + .79^2 + .5^2)} = .2006\)

"p character" is the probability of being in a p state: 1- P(s state) \(= 1 - .2006 = .7994 \)

e). The \( |2p_y \rangle \) is not mixed in because the bond lies in the xz plane and has no x component.

Q4:

Valence bond theory only uses combinations of atomic orbitals from the same atoms to form new orbitals, which are always localized around single atoms. Molecular orbital theory combines atomic orbitals of different atoms to form new molecular orbitals which are delocalized over different atoms.

Q5:

Demonstrate that each of the three \(sp^2\) orbitals satisfies the orthonormality criteria for wavefunctions. This means showing that each \(|sp^3 \rangle \) meets the criteria: \( \langle sp^3_i | sp^3_j \rangle = \delta_{ij} \)

Designating the three sp³ orbitals as:

\( | 1 \rangle = \frac{1}{\sqrt{3}} | s \rangle + \sqrt{\frac{2}{3}} | p_z \rangle \)

\( |2 \rangle = \frac{1}{\sqrt{3}} | s \rangle + \sqrt{\frac{1}{2}} | p_x \rangle + \sqrt{\frac{1}{6}} | p_z \rangle \)

\( |3 \rangle = \frac{1}{\sqrt{3}} | s \rangle - \sqrt{\frac{1}{2}} | p_x \rangle + \sqrt{\frac{1}{6}} | p_z \rangle \)

Normality:

\( \langle 1 | 1 \rangle = (\frac{1}{\sqrt{3}} \langle s| + \sqrt{\frac{2}{3}} \langle p_z|) (\sqrt{\frac{1}{3}} | s \rangle + \sqrt{\frac{2}{3}} | p_z \rangle) = \frac{1}{3}( \langle s |s \rangle + \sqrt{2} \langle s|p_z \rangle + \sqrt{2} \langle p_z| s \rangle + 2 \langle p_z|p_z \rangle) = \frac{1}{3}(1 + 0 + 0 + 2) = 1 \)

For the longer wavefunctions, only surviving terms in the dot product are included:

\(\langle 2|2 \rangle = (\frac{1}{\sqrt{3}} \langle s| + \sqrt{\frac{1}{2}} \langle p_x | + \sqrt{\frac{1}{6}} \langle p_z | ) (\frac{1}{\sqrt{3}} | s \rangle + \sqrt{\frac{1}{2}} | p_x \rangle + \sqrt{\frac{1}{6}} | p_z \rangle) = \frac{1}{3} \langle s|s \rangle + \frac{1}{2} \langle p_x|p_x \rangle + \frac{1}{6} \langle p_z|p_z \rangle = \frac{1}{3} + \frac{1}{2} + \frac{1}{6}=1 \)

\(\langle 3|3 \rangle = (\frac{1}{\sqrt{3}} \langle s| + \sqrt{\frac{1}{2}} \langle p_x | - \sqrt{\frac{1}{6}} \langle p_z | ) (\frac{1}{\sqrt{3}} | s \rangle + \sqrt{\frac{1}{2}} | p_x \rangle - \sqrt{\frac{1}{6}} | p_z \rangle) = \frac{1}{3} \langle s|s \rangle + \frac{1}{2} \langle p_x|p_x \rangle + \frac{1}{6} \langle p_z|p_z \rangle = \frac{1}{3} + \frac{1}{2} + \frac{1}{6}=1 \)

Orthogonality:

\( \langle 1 | 2 \rangle = \langle 2|1 \rangle = (\frac{1}{\sqrt{3}} \langle s| + \sqrt{\frac{2}{3}} \langle p_z|)\sqrt{ (\frac{1}{6} }| s \rangle + \sqrt{\frac{1}{2}} | p_x \rangle = \frac{1}{3} \langle s|s \rangle - \sqrt{\frac{2}{18}} \langle p_z| p_z\rangle = ( \frac{1}{3} -\frac{1}{3}) = 0 \)

\( \langle 1 | 3 \rangle = \langle 3|1 \rangle = (\frac{1}{\sqrt{3}} \langle s| + \sqrt{\frac{2}{3}} \langle p_z|)\sqrt{ (\frac{1}{6} }| s \rangle - \sqrt{\frac{1}{2}} | p_x \rangle = \frac{1}{3} \langle s|s \rangle - \sqrt{\frac{2}{18}} \langle p_z| p_z\rangle = ( \frac{1}{3} -\frac{1}{3}) = 0 \)

\( \langle 2 |3 \rangle = \langle 3|2 \rangle = ( \langle s| + \sqrt{\frac{1}{2}} \langle p_x | + \sqrt{\frac{1}{6}} \langle p_z | ) (\frac{1}{\sqrt{3}} | s \rangle + \sqrt{\frac{1}{2}} | p_x \rangle - \sqrt{\frac{1}{6}} | p_z \rangle) = \frac{1}{3} \langle s|s \rangle - \frac{1}{2} \langle p_x|p_x \rangle + \frac{1}{6} \langle p_z|p_z \rangle = (\frac{1}{3} - \frac{1}{2} + \frac{1}{6}) = 0 \)

Q6:

Draw the molecular orbitals, fill in according to the valence electrons and then bond order = \(\frac{1}{2}\) (bonding electrons - antibonding electrons). Paramagnetic species are those that have unpaired electrons.

Molecule	Bond Order	Paramagnetic
\(C^+_2\)	1.5	yes
\(C_2\)	2	no
\(C^-_2\)	2.5	yes
\(N^+_2\)	2.5	yes
\(N_2\)	3	no
\(N^-_2\)	2.5	yes
\(O^+_2\)	2.5	yes
\(O_2\)	2	yes
\(O^-_2\)	1.5	yes

Q7:

1. six

2. With only 6 valence electrons, the lowest three orbitals on the Walsh diagram are occupied and hence we would predict a bent geometry for \(NH_2^+\).

3. four

4. With 4 valence electrons, the lowest two orbitals on the Walsh diagram are occupied and hence we would predict a linear geometry for \(BH_2^+\).