Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 27.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 3.87\text{ BM} \)
The spin-only magnetic moment of an ion depends only on the number of unpaired electrons, n:
\( \displaystyle \mu = \sqrt{n(n+2)}\ \mathrm{BM} \)
So the first task is to find the configuration of the ion. The element with atomic number 27 is cobalt: Co (Z = 27) is \( \mathrm{[Ar]\,3d^{7}\,4s^{2}} \). The divalent ion is formed by losing the two 4s electrons, which gives \( \ce{Co^2+} \): \( \mathrm{3d^{7}} \).
In the five 3d orbitals, seven electrons are arranged as five singly filled orbitals with two of them paired, which leaves three unpaired electrons (water is a weak field ligand, so in aqueous solution the electrons stay unpaired as far as possible):
\( \displaystyle \mathrm{3d^{7}}: \text{ electrons per orbital } 2,\ 2,\ 1,\ 1,\ 1 \qquad n = 3 \)
\( \displaystyle \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\ \mathrm{BM} \)
So the magnetic moment is 3.87 BM.
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Magnetic Properties – …
