How many unpaired electrons are present in each of the following?

(a) [CoF6]3 - (high spin)

(b) [Mn (CN) 6]3 - (low spin)

(c) [Mn (CN) 6]4 - (low spin)

(d) [MnCl6]4 - (high spin)

(e) [RhCl6]3 - (low spin)

Answer: A. 4 unpaired electrons

B. Zero unpaired electrons

C. 1 unpaired electron

D. 5 unpaired electrons

E. Zero unpaired electrons

Explanation:

In A, Oxidation state of Co is + 3 and Electronic configuration is [Ar]3d6

F is weak field ligand, causes no pairing of Electrons hence it has 4 unpaired electrons&2 are paired in t2g orbitals (dXY)

In B, Mn is in + 3 with electronic configuration 3d4&CN is a strong field ligand hence causes pairing of Electrons hence it results 0 unpaired electron

In C, Mn is in + 2 with electronic configuration 3d5 sinceCN is a strong field ligand hence it leaves one unpaired electron

In D, Mn is in+2 with 3d5& five unpaired electrons since cl is a weak field ligand causes no pairing.

In E, Rh is in + 3, with d6 configuration and it is a low spin complex hence pairing of Electrons involved. So it leaves zero unpaired electrons.