Match List-I with List-II

	LIST-I Coordination Complex		LIST-II Number of unpaired electrons
A.	$$[Cr(CN)_6]^{3-}$$	I.	0
B.	$$[Fe(H_2O)_6]^{2+}$$	II.	3
C.	$$[Co(NH_3)_6]^{3+}$$	III.	2
D.	$$[Ni(NH_3)_6]^{2+}$$	IV.	4

Choose the correct answer from the options given below:

We need to match each coordination complex with its number of unpaired electrons.

A. $$[Cr(CN)_6]^{3-}$$

Chromium in $$[Cr(CN)_6]^{3-}$$: Oxidation state of Cr = +3 (since $$x + 6(-1) = -3$$, so $$x = +3$$).

$$Cr^{3+}$$: Electronic configuration = $$[Ar] 3d^3$$

$$CN^-$$ is a strong field ligand, so electrons pair up in $$t_{2g}$$ orbitals first.

Configuration: $$t_{2g}^3 e_g^0$$ → 3 unpaired electrons → matches II

B. $$[Fe(H_2O)_6]^{2+}$$

Iron in $$[Fe(H_2O)_6]^{2+}$$: Oxidation state of Fe = +2.

$$Fe^{2+}$$: Electronic configuration = $$[Ar] 3d^6$$

$$H_2O$$ is a weak field ligand, so no pairing occurs beyond what is necessary.

Configuration: $$t_{2g}^4 e_g^2$$ (high spin) → 4 unpaired electrons → matches IV

C. $$[Co(NH_3)_6]^{3+}$$

Cobalt in $$[Co(NH_3)_6]^{3+}$$: Oxidation state of Co = +3.

$$Co^{3+}$$: Electronic configuration = $$[Ar] 3d^6$$

$$NH_3$$ is a strong field ligand for $$Co^{3+}$$, so all electrons pair in $$t_{2g}$$.

Configuration: $$t_{2g}^6 e_g^0$$ (low spin) → 0 unpaired electrons → matches I

D. $$[Ni(NH_3)_6]^{2+}$$

Nickel in $$[Ni(NH_3)_6]^{2+}$$: Oxidation state of Ni = +2.

$$Ni^{2+}$$: Electronic configuration = $$[Ar] 3d^8$$

For octahedral $$Ni^{2+}$$ ($$d^8$$), regardless of ligand field strength, the configuration is $$t_{2g}^6 e_g^2$$.

This always gives 2 unpaired electrons → matches III

Therefore, the correct matching is:

A → II, B → IV, C → I, D → III

The correct answer is Option A.