The spin magnetic moment of cobalt in the compound named Mercuric tet…

The correct answer is Option (3) → \(\sqrt{15}\)

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

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

Mercuric tetrathiocyanatocobaltate(II) is \( \ce{Hg[Co(SCN)4]} \). Mercuric means \( \ce{Hg^{2+}} \), so the complex ion is \( \ce{[Co(SCN)4]^{2-}} \), and with \( \ce{SCN^{-}} \) = −1:

    \( x + 4(-1) = -2 \Rightarrow x = +2 \)

Co (Z = 27) is [Ar] 3d⁷ 4s²; removing the two 4s electrons gives \( \ce{Co^{2+}} \): [Ar] 3d⁷. Filling the five 3d orbitals by Hund’s rule (\( \ce{SCN^{-}} \) is a weak field ligand and the complex is tetrahedral, so no extra pairing occurs):

    3d⁷: [↑↓] [↑↓] [↑] [↑] [↑] → 3 unpaired electrons

    \( \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\ \mathrm{BM} \)

NCERT (2026-27 edition):
Magnetic Properties – “… For the compounds of the first series of transition metals, the contribution of the orbital angular momentum is effectively quenched and hence is of no significance. 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).”