Past exam of the chemistry A course of the University of Cambridge / 2021 / Paper 1 / A4 / a

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii
- iii a
  - Solution iii
- iv a
  - Solution iv
- v a
  - Solution v

i

Solution

The trigonal pyramidal molecular geometry has C3v point group symmetry: identity, two rotations about the threefold axis, and three vertical mirror reflections. Choose

z

along the threefold axis through nitrogen and the center of the hydrogen triangle.

ii

Solution

In the C3v point group, the nitrogen valence atomic orbitals transform as

2 s : A_{1}, 2 p_{z} : A_{1}, (2 p_{x}, 2 p_{y}) : E .

The

s

orbital is invariant under every symmetry operation;

p_{z}

is also invariant because each operation preserves the

z

coordinate. The other two components mix as a two-dimensional irreducible representation.

iii

Solution

Use the classes

(E, 2 C_{3}, 3 σ_{v})

of the C3v point group. For the three hydrogen

1 s

atomic orbitals, the character of a representation counts orbitals fixed by each symmetry operation. Identity fixes three, a nontrivial threefold rotation fixes none, and a vertical mirror fixes one. Therefore

χ_{H} = (3, 0, 1) .

The character inner product gives multiplicities

n_{A_{1}} = \frac{3 + 3}{6} = 1, n_{A_{2}} = \frac{3 - 3}{6} = 0, n_{E} = \frac{2 \times 3}{6} = 1.

Thus $Γ_{H} = A_{1} \oplus E$ , with the

E

contribution accounting for two independent symmetry orbitals.

iv

Solution

Neglecting intersite orbital overlap, one normalized real set of symmetry-adapted linear combinations is

χ_{A_{1}} χ_{E, 1} χ_{E, 2} = (s_{A} + s_{B} + s_{C}) / 3, = (2 s_{A} - s_{B} - s_{C}) / 6, = (s_{B} - s_{C}) / 2 .

The coefficient vectors are orthonormal. The first has equal phase at all three sites; the second opposes site

A

to

B, C

; the third has zero amplitude at

A

and opposite phases at

B, C

. Any orthonormal rotation of the two

E

functions is equally valid. With nonzero intersite orbital overlap, normalization must instead use the overlap matrix.

Figure 1.
Original coefficient-and-phase sketches of the three hydrogen symmetry orbitals in ammonia
. Details.

v

Solution

A_{1}

symmetry orbital mixes with both nitrogen

2 s

and

2 p_{z}

, giving three

a_{1}

molecular orbitals. The two hydrogen

E

functions mix with

(2 p_{x}, 2 p_{y})

, giving a bonding

e

pair and an antibonding

e

pair. Eight valence electrons fill the lower levels as

(2 a_{1})^{2} (1 e)^{4} (3 a_{1})^{2} .

This uses conventional numbering that reserves

1 a_{1}

for the omitted nitrogen core

1 s

orbital. The low

2 a_{1}

and

1 e

levels provide the principal N–H bonding molecular orbitals; the highest occupied

3 a_{1}

level is the mainly nitrogen lone pair, with some bonding admixture permitted by the pyramidal geometry. Empty

4 a_{1}

and

2 e

levels are antibonding. Symmetry fixes which orbitals can mix, rather than their quantitative energy separations.

Figure 1.
Original qualitative ammonia molecular orbital diagram comparing pyramidal and planar geometries; the core orbital is omitted
. Details.

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