Arrange the following in the decreasing order of their magnetic moment values:
A. Sc
B. V
C. Mn
D. Fe
E. Ni
Choose the correct answer from the option given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → C > D > B > E > A
The magnetic moment rises with the number of unpaired electrons, \( \mu = \sqrt{n(n+2)} \) BM. The configuration of each atom is written from its atomic number and the five 3d orbitals are filled singly first (Hund’s rule):
A. Sc (Z = 21): [Ar] 3d1 4s2 ⇒ 1 unpaired electron, μ = √3 = 1.73 BM — last (fifth).
B. V (Z = 23): [Ar] 3d3 4s2 ⇒ 3 unpaired electrons, μ = √15 = 3.87 BM — third.
C. Mn (Z = 25): [Ar] 3d5 4s2 ⇒ 5 unpaired electrons (half-filled 3d), μ = √35 = 5.92 BM — first.
D. Fe (Z = 26): [Ar] 3d6 4s2 ⇒ one pair + 4 unpaired electrons, μ = √24 = 4.90 BM — second.
E. Ni (Z = 28): [Ar] 3d8 4s2 ⇒ three pairs + 2 unpaired electrons, μ = √8 = 2.84 BM — fourth.