Question

Match the LIST-I with LIST-II

Choose the correct answer from the options given below:

• A. A-III, B-IV, C-II, D-I
• B. A-III, B-I, C-IV, D-II
• C. A-III, B-IV, C-I, D-II
• D. A-II, B-I, C-IV, D-III

Hint:

In a tetrahedral complex, use the formula CFSE = [-0.6(number of e electrons) + 0.4(number of t2 electrons)] $\Delta_t$.

Answer 1

The Correct Option is C

In tetrahedral coordination complexes, the five d-orbitals split into two sets: a lower energy doubly degenerate set 'e' ($d_{z^2}$ and $d_{x^2-y^2}$) and a higher energy triply degenerate set '$t_2$' ($d_{xy}, d_{yz}, d_{xz}$).
The energy of an electron in the 'e' set is $-0.6 \Delta_t$ relative to the barycenter, while the energy in the '$t_2$' set is $+0.4 \Delta_t$.
The Crystal Field Stabilization Energy (CFSE) is calculated using the formula:
$$CFSE = [(-0.6 \times n_e) + (0.4 \times n_{t2})] \Delta_t$$
Where $n_e$ is the number of electrons in the 'e' orbitals and $n_{t2}$ is the number of electrons in the '$t_2$' orbitals. Tetrahedral complexes are almost always high spin due to the small splitting energy.

Let us calculate for each configuration:
1. For $d^2$: Electrons fill the lower energy orbitals first. Electronic distribution is $e^2 t_2^0$.
$$CFSE = [2 \times (-0.6) + 0 \times 0.4] \Delta_t = -1.2 \Delta_t$$
This matches with III.

2. For $d^4$: In high spin, electrons occupy both levels before pairing. Electronic distribution is $e^2 t_2^2$.
$$CFSE = [2 \times (-0.6) + 2 \times 0.4] \Delta_t = [-1.2 + 0.8] \Delta_t = -0.4 \Delta_t$$
This matches with IV.

3. For $d^6$: Electronic distribution is $e^3 t_2^3$.
$$CFSE = [3 \times (-0.6) + 3 \times 0.4] \Delta_t = [-1.8 + 1.2] \Delta_t = -0.6 \Delta_t$$
This matches with I.

4. For $d^8$: Electronic distribution is $e^4 t_2^4$.
$$CFSE = [4 \times (-0.6) + 4 \times 0.4] \Delta_t = [-2.4 + 1.6] \Delta_t = -0.8 \Delta_t$$
This matches with II.

Comparing the results, the correct match is A-III, B-IV, C-I, D-II.