Match the LIST-I with LIST-II

List-I Electronic configuration of tetrahedral metal ion		List-II Crystal Field Stabilization Energy ($$\Delta_t$$)
A.	d$$^2$$	I.	-0.6
B.	d$$^4$$	II.	-0.8
C.	d$$^6$$	III.	-1.2
D.	d$$^8$$	IV.	-0.4

Choose the correct answer from the options given below:

For tetrahedral complexes, the (e) orbitals are stabilized by (0.6\Delta_t) and the (t_2) orbitals are destabilized by (0.4\Delta_t). Therefore, the Crystal Field Stabilization Energy (CFSE) is given by

$$\text{CFSE}=[n(e)(-0.6)+n(t_2)(+0.4)]\Delta_t.$$

For a (d^2) configuration, the electron distribution is (e^2t_2^0). Hence,

$$\text{CFSE}=2(-0.6)\Delta_t+0(+0.4)\Delta_t=-1.2\Delta_t,$$

which corresponds to III.

For a (d^4) configuration, the electron distribution is (e^2t_2^2). Thus,

$$\text{CFSE}=2(-0.6)\Delta_t+2(+0.4)\Delta_t=-1.2\Delta_t+0.8\Delta_t=-0.4\Delta_t,$$

which corresponds to IV.

For a (d^6) configuration, the electron distribution is (e^3t_2^3). Therefore,

$$\text{CFSE}=3(-0.6)\Delta_t+3(+0.4)\Delta_t=-1.8\Delta_t+1.2\Delta_t=-0.6\Delta_t,$$

which corresponds to I.

For a (d^8) configuration, the electron distribution is (e^4t_2^4). Hence,

$$\text{CFSE}=4(-0.6)\Delta_t+4(+0.4)\Delta_t=-2.4\Delta_t+1.6\Delta_t=-0.8\Delta_t,$$

which corresponds to II.

Thus, the correct matching is

$$A\rightarrow III,\qquad B\rightarrow IV,\qquad C\rightarrow I,\qquad D\rightarrow II.$$

Hence, the correct sequence is A-III, B-IV, C-I, D-II.