Question: 46. ‘Spin only’ magnetic moment is same for which of the following ions?
A. \mathrm{Ti}^{3+}
B. \mathrm{Cr}^{2+}
C. \mathrm{Mn}^{2+}
D. \mathrm{Fe} e^{2+}
E. \mathrm{Sc}^{3+}
Choose the most appropriate answer from the options given below :
(1) B and D only
(2) A and E only
(3) B and C only
(4) A and D only
Answer: Option (1)
Explanation:
The spin only magnetic moment depends only on the number of unpaired electrons
and is given by the formula:
\mu = \sqrt{n(n+2)} \text{ BM}where n is the number of unpaired electrons.
Electronic configuration and unpaired electrons for each ion:
\mathrm{Ti}^{3+} : Atomic number = 22,
configuration = [\mathrm{Ar}]3d^{1}, unpaired electrons = 1
\mathrm{Cr}^{2+} : Atomic number = 24,
configuration = [\mathrm{Ar}]3d^{4}, unpaired electrons = 4
\mathrm{Mn}^{2+} : Atomic number = 25,
configuration = [\mathrm{Ar}]3d^{5}, unpaired electrons = 5
\mathrm{Fe}^{2+} : Atomic number = 26,
configuration = [\mathrm{Ar}]3d^{6}, unpaired electrons = 4
\mathrm{Sc}^{3+} : Atomic number = 21,
configuration = [\mathrm{Ar}]3d^{0}, unpaired electrons = 0
Both \mathrm{Cr}^{2+} and \mathrm{Fe}^{2+} have 4 unpaired electrons,
so their spin only magnetic moments are equal.
Therefore, the correct answer is Option (1).