(a) Write t

(a) (i) [FeF₆]^3-

• The electronic configuration of Fe (26) is [Ar] 3d⁶4s². The oxidation state of Fe in [FeF₆]^3- is +3.

• This leads to the electronic configuration of Fe³⁺ becoming [Ar] 3d⁵4s⁰.

• The electrons from the fluorine atoms now overlap with the vacant orbitals of Fe ion.

• Since there are 6 F atoms, 6 orbitals are required. So, one 4s orbital, three 4p orbitals and two 4d orbitals undergo sp³d² hybridisation to form 6 orbitals of same energy.

• Now, since fluorine is a strong ligand, pairing does not take place.

• Thus, the complex [FeF₆]^3- is paramagnetic and has octahedral geometry.

(ii) [Ni (CO)₄]

• The electronic configuration of Ni (28) is [Ar] 3d⁸4s². The oxidation state of Ni in [Ni (CO)₄] is 0.

• Now, since 4 orbitals are required for the overlapping of the electrons of carbonyl group, the two electrons from the 4s orbital shift to the 3d orbital.

• The configuration now becomes [Ar] 3d¹⁰4s⁰.

• One 4s and three 4p orbitals undergo sp³ hybridisation to form 4 orbitals of equal energy.

• The electrons from CO group now overlap with the orbitals to form [Ni (CO)₄] complex.

• Thus, [Ni (CO)₄] is diamagnetic and tetrahedral.

(b)

• On moving across the spectrochemical series we see that CO is a stronger ligand than CN^-.

• Thus, complexes formed with CO group tend to be more stable than those formed by CN^- ligand.

• Another reason is that CO is capable of forming synergic bonds which literally translated to ‘self-strengthening’ bond.