Consider the following low-spin complexes \(\mathrm{K}_{3}\left[\mathrm{Co}\left(\mathrm{NO}_{2}\right)_{6}\right], \mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right], \mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\), \(\mathrm{Cu}_{2}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) and \(\mathrm{Zn}_{2}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\). The sum of the spin-only magnetic moment values of complexes having yellow colour is ______ B.M. (answer is nearest integer)

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Coordination Compounds

Consider the following low-spin complexes

K_{3} [Co (NO_{2})_{6}], K_{4} [Fe (CN)_{6}], K_{3} [Fe (CN)_{6}]

Cu_{2} [Fe (CN)_{6}]

and

Zn_{2} [Fe (CN)_{6}]

.
The sum of the spin-only magnetic moment values of complexes having yellow colour is ______
B.M. (answer is nearest integer)

Answer: 0

Among the given low-spin complexes,

K_{3} [C o (N O_{2})_{6}]

and

K_{4} [F e (C N)_{6}]

are the yellow ones;

K_{3} [F e (C N)_{6}]

is red,

C u_{2} [F e (C N)_{6}]

is reddish-brown, and

Z n_{2} [F e (C N)_{6}]

is white. In

K_{3} [C o (N O_{2})_{6}]

, cobalt is

C o^{3 +}

(

3 d^{6}

); the strong-field nitro ligand forces a low-spin

t_{2 g}^{6} e_{g}^{0}

arrangement with zero unpaired electrons, so

μ = 0

. In

K_{4} [F e (C N)_{6}]

, iron is

F e^{2 +}

(

3 d^{6}

); the strong-field cyanide ligand likewise gives a low-spin

t_{2 g}^{6} e_{g}^{0}

configuration with zero unpaired electrons, so

μ = 0

. The sum of the spin-only magnetic moments of the yellow complexes is

0 + 0 = 0

B.M.

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Integer answer