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Question

Chemistry Question on The Valence Shell Electron Pair Repulsion (VSEPR) Theory

The value of the "spin only" magnetic moment for one of the following configurations is 2.84 BM. The correct one is :

A

d5d^5 (in strong ligand field)

B

d3d^3 (in weak as well as in strong fields)

C

d4d^4 (in weak ligand field)

D

d4d^4 (in strong ligand field)

Answer

d4d^4 (in strong ligand field)

Explanation

Solution

(a) d5d^5 in strong field n = unpaired electron = 1 Magnetic moment =n(n+2)BM= \sqrt{n\left(n + 2\right)}BM =3BM=1.73BM= \sqrt{3} BM = 1.73 \,BM (b)d3\left(b\right) d^{3} in strong/weak field n=3n = 3 Magnetic moment =15=3.87BM= \sqrt{15} = 3.87\,BM (c)d4\left(c\right) d^{4} in weak field n=4n = 4 Magnetic moment =24=4.90BM=\sqrt{24} = 4.90\,BM (d)d4\left(d\right) d^{4 } in strong field n=2n = 2 Magnetic moment =8=2.83BM=\sqrt{8} = 2.83 \,BM