The number of unpaired electron in gaseous ions Mn3+ (Z = 25), V3+ (Z…

The correct answer is Option (1) → 4, 2, 3

For a gaseous transition metal ion, the electrons are removed from the 4s orbital first and then from 3d. The remaining d electrons are placed in the five 3d orbitals by Hund’s rule: singly first, with parallel spins, and paired only after each orbital has one electron. The number of unpaired electrons is read from these boxes.

• Mn³⁺: Mn (Z = 25) has 25 − 18 = 7 electrons beyond the argon core: [Ar] 3d⁵4s². Removing the two 4s electrons and one 3d electron gives \( \ce{Mn^{3+}} \): [Ar] 3d⁴.

      3d: [↑] [↑] [↑] [↑] [  ] → 4 unpaired

• V³⁺: V (Z = 23) has 23 − 18 = 5 electrons beyond the argon core: [Ar] 3d³4s². Removing the two 4s electrons and one 3d electron gives \( \ce{V^{3+}} \): [Ar] 3d².

      3d: [↑] [↑] [  ] [  ] [  ] → 2 unpaired

• Cr³⁺: Cr (Z = 24) has 24 − 18 = 6 electrons beyond the argon core: [Ar] 3d⁵4s¹ (half-filled 3d is extra stable). Removing the 4s electron and two 3d electrons gives \( \ce{Cr^{3+}} \): [Ar] 3d³.

      3d: [↑] [↑] [↑] [  ] [  ] → 3 unpaired

So the numbers of unpaired electrons are 4, 2 and 3.

Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.

NCERT (2026-27 edition):
Exercises – “4.24 Calculate the number of unpaired electrons in the following gaseous ions: Mn³⁺, Cr³⁺, V³⁺ and Ti³⁺. Which one of these is the most stable in aqueous solution?”