Question

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 1

Answer: A. 4 unpaired electrons

B. Zero unpaired electrons

C. 1 unpaired electron

D. 5 unpaired electrons

E. Zero unpaired electrons

Step-by-step 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.