Match List I with List II

	List I Complex		List II CFSE ($$\Delta_0$$)
A	$$[Cu(NH_3)_6]^{2+}$$	I	$$-0.6$$
B	$$[Ti(H_2O)_6]^{3+}$$	II	$$-2.0$$
C	$$[Fe(CN)_6]^{3-}$$	III	$$-1.2$$
D	$$[NiF_6]^{4-}$$	IV	$$-0.4$$

Choose the correct answer from the options given below:

For octahedral complexes, Crystal Field Stabilization Energy (CFSE) is calculated using:

• Each electron in $$t_{2g}$$​ contributes $$−0.4Δ_{0​}$$
• Each electron in $$e_g$$​ contributes $$+0.6Δ_{0}$$​

A. $$[Cu(NH_{3})_{6}]^{2+}]$$

Since, $$Cu=+2$$ = $$3d^9$$

arrangement $$t^{6}_{2g}e^{3}_{g}$$

CFSE:

$$\left(6\times\ -0.4\right)+\left(3\times\ +0.6\right)$$

$$=-2.4+1.8$$

$$=-0.6\triangle_o$$

So, A$$\longrightarrow\ $$ I

B. $$[Ti(H_{2}O)_{6}]^{3+}]$$

Since, $$Ti=+3$$ = $$3d^1$$

arrangement $$t^{1}_{2g}e^{0}_{g}$$

CFSE:

$$1\times\ \left(-0.4\right)$$

$$=-0.4\triangle_o$$

So, B$$\longrightarrow\ $$ IV

C. $$[Fe(CN)_{6}]^{3-}]$$

Since, $$Fe=+3$$ = $$3d^5$$

CN is strong Field ligand (it means their will be pairing)

arrangement $$t^{5}_{2g}e^{0}_{g}$$

CFSE:

$$5\times\ \left(-0.4\right)$$

$$=-2.0\triangle_o$$

So, C$$\longrightarrow\ $$ II

D. $$[NiF_{6}]^{4-}]$$

Since, $$Ni=+2$$ = $$3d^8$$

arrangement $$t^{6}_{2g}e^{2}_{g}$$

CFSE:

$$\left(6\times\ -0.4\right)+\left(2\times\ +0.6\right)$$

$$=-2.4+1.2$$

$$=-1.2\triangle_o$$

So, D$$\longrightarrow\ $$ III

Therefore, Option B is correct.