Match List-I (ion) with List-II (approximate spin-only magnetic momen…

The correct answer is Option (1) → A-r, B-s, C-p, D-q

For a first-series transition metal ion the magnetic moment depends only on the number of unpaired electrons n, through the spin-only formula:

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

Fe, Co and Ni lose their two 4s electrons to form M²⁺ (Cu loses its one 4s electron and one 3d electron), and the d electrons then occupy the five 3d orbitals singly before pairing.

(A) Cu²⁺ — This matches (r) 1.8 BM. Cu (Z = 29) is [Ar]3d¹⁰4s¹, so Cu²⁺ is 3d⁹ with 1 unpaired electron: \( \mu=\sqrt{1\times3}=1.73\ \mathrm{BM} \approx 1.8\ \mathrm{BM} \).

(B) Fe²⁺ — This matches (s) 5.0 BM. Fe (Z = 26) is [Ar]3d⁶4s², so Fe²⁺ is 3d⁶ with 4 unpaired electrons: \( \mu=\sqrt{4\times6}=4.90\ \mathrm{BM} \approx 5.0\ \mathrm{BM} \).

(C) Ni²⁺ — This matches (p) 2.9 BM. Ni (Z = 28) is [Ar]3d⁸4s², so Ni²⁺ is 3d⁸ with 2 unpaired electrons: \( \mu=\sqrt{2\times4}=2.84\ \mathrm{BM} \approx 2.9\ \mathrm{BM} \).

(D) Co²⁺ — This matches (q) 4.0 BM. Co (Z = 27) is [Ar]3d⁷4s², so Co²⁺ is 3d⁷ with 3 unpaired electrons: \( \mu=\sqrt{3\times5}=3.87\ \mathrm{BM} \approx 4.0\ \mathrm{BM} \).

NCERT (2026-27 edition):
4.3.9 Magnetic Properties – “… For these, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the ‘spin-only’ formula, i.e., \( \mu=\sqrt{n(n+2)} \) where n is the number of unpaired electrons and μ is the magnetic moment in units of Bohr magneton (BM). … The magnetic moments calculated from the ‘spin-only’ formula and those derived experimentally for some ions of the first row transition elements are given in Table 4.7.”

Table 4.7: Calculated and Observed Magnetic Moments (BM)