Which of the following pair of transition metal ions have same value…

The correct answer is Option (2) → Ni²⁺ and Ti²⁺

The spin-only magnetic moment depends only on the number of unpaired electrons, n:

    \( \displaystyle \mu=\sqrt{n(n+2)}\ \mathrm{BM} \)

So two ions have the same spin-only moment when they have the same number of unpaired electrons. For a 3d ion, the 4s electrons are removed first, then 3d electrons, and the d electrons are placed singly in the five d orbitals before pairing.

Ni (Z = 28) is [Ar]3d⁸4s²; removing the two 4s electrons gives Ni²⁺: 3d⁸. Five electrons go in singly and three pair up, leaving 2 unpaired.

Ti (Z = 22) is [Ar]3d²4s²; removing the two 4s electrons gives Ti²⁺: 3d², with 2 unpaired electrons.

    \( \displaystyle \mu=\sqrt{2(2+2)}=\sqrt{8}=2.83\ \mathrm{BM}\ \text{for both} \)

So Ni²⁺ and Ti²⁺ have the same spin-only magnetic moment.

Why the other options are incorrect

• Option (1) Fe²⁺ and V²⁺ — Incorrect. Fe (Z = 26) [Ar]3d⁶4s² gives \( \ce{Fe^{2+}} \): 3d⁶, 4 unpaired (4.90 BM); V (Z = 23) [Ar]3d³4s² gives \( \ce{V^{2+}} \): 3d³, 3 unpaired (3.87 BM).

• Option (3) Ti³⁺ and Co²⁺ — Incorrect. Ti (Z = 22) [Ar]3d²4s² loses both 4s electrons and one 3d electron to give \( \ce{Ti^{3+}} \): 3d¹, 1 unpaired (1.73 BM); Co (Z = 27) [Ar]3d⁷4s² gives \( \ce{Co^{2+}} \): 3d⁷, 3 unpaired (3.87 BM).

• Option (4) Mn²⁺ and Cr²⁺ — Incorrect. Mn (Z = 25) [Ar]3d⁵4s² gives \( \ce{Mn^{2+}} \): 3d⁵, 5 unpaired (5.92 BM); Cr (Z = 24) [Ar]3d⁵4s¹ loses the 4s electron and one 3d electron to give \( \ce{Cr^{2+}} \): 3d⁴, 4 unpaired (4.90 BM).