• In the case of A. [Cr(H₂O)₆]^{3 +}: The central metal atom is Cr and oxidation state x=+3 as water molecules are neutral.

Therefore , having a d³ configuration (1s² 2s² 2p⁶ 3s² 3p⁶ 3d³).

Cr has +3 oxidation state with d³ configuration and left with 2 empty d orbitals and 1 empty 4s orbital which in interaction with 3,4p orbitals and attain the hybridisation d²sp³ and the 6 H₂O electrons will occupy these 6 hybridised orbitals. And the hybridisation scheme will be :

The orbital energy diagram of [Cr(H₂O)₆]³⁺is :

Hence, there will be 3 unpaired electrons.

• In case of B. [Co(CN)₄]^2– : The central metal atom is Co and oxidation state x= -4 (-1) – 2=+2,

Therefore, having a d⁷configuration (1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷).Hence, the Co²⁺ is left with 1 3d orbital,1 4s orbital which in interaction with 2, 4p orbitals and attains hybridisation of dsp²and the 4CN^- electrons will occupy these 4 hybridised orbitals and the scheme will be:

The orbital energy diagram will be:

hence, there will be 1 unpaired electron.

• In the case ofC. [Ni(NH₃)₆]²⁺: The central metal atom is Ni and oxidation state x=+2,

Therefore, having a d⁸ configuration (1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸).Hence, the Ni²⁺ is left with 2, 3d orbital,1 4s orbital which in interaction with 2, 4p orbitals and attain hybridisation of sp³d²and the 6, NH₃ electrons will occupy these 6 hybridised orbitals and the scheme will be:

The orbital energy diagram will be:

hence, there will be 2 unpaired electrons.

• In the case of D. [MnF₆]^4–: The central metal atom is Mnand oxidation state x=-6(-1) – 4 =+2

Therefore, having a d⁵configuration (1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵).Hence, the Mn²⁺ is left with 1 4s, 2 of 4d (outer orbital complex) which in interaction with 3, 4p orbitals and attain hybridisation of sp³d²and the 6, F^-electrons will occupy these 6 hybridised orbitalsand the scheme will be:

hence, there will be 5 unpaired electrons as it a high spin complex.

The correct answer will be the option (i)A (3) B (1) C (5) D (2)

Hence, matching the two columns the right combination will be:

A. [Cr(H₂O)₆]³⁺ - 3. d²sp³, 3

B. [Co(CN)₄]^2–-1. dsp², 1.

C. [Ni(NH₃)₆]²⁺-5. sp³d², 2

D. [MnF₆]^4– -2. sp³d², 5