Question

Which of the following species has magnetic moment value of 3.87 B.M. ?
a)Fe³⁺ b)Cr²⁺
c)Co²⁺ d)Au³⁺
Explain your answer.

Answer 1

Magnetic moment = μ = $\sqrt{n(n+2)}or\left[n\right(n+2){]}^{\frac{1}{2}}$
n= number of unpaired electrons
Given magnetic moment= 3.87 BM
​Hence, $$\begin{gathered} 3.87=\sqrt{n(n+2)} \\ squaringbothsides, \\ 14.97=n(n+2) \\ Lettakethe14.97=15 \\ So,15=n(n+2) \\ hencen=3 \end{gathered}$$
So, number of unpaired electrons= 3
Electronic configurations:
Atomic number of Fe = 26, Fe³⁺=23 = 1s²2s²2p⁶3s²3p⁶4s⁰3d^​5 number of unpaired electron = 5
Atomic number of Cr= 24, Cr²⁺= 22= 1s²2s²2p⁶3s²3p⁶4s⁰3d⁴ number of unpaired electrons= 4
Atomic number of Co= 27, Co²⁺ = 25= 1s²2s²2p⁶3s²3p⁶4s⁰3d⁷ number of unpaired electrons= 3
Hence, Co²⁺ will have magnetic moment=3.87 B.M.