The calculated spin only magnetic moments (BM) of the anionic and cationic species of $\left[\mathrm{Fe}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]_{2}$ and $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$, respectively, are:

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Coordination Compounds

The calculated spin only magnetic moments (BM) of the anionic and cationic species of

[Fe (H_{2} O)_{6}]_{2}

and

[Fe (CN)_{6}]

, respectively, are:

Answer: (C)

Both complexes hold iron in the

+ 2

state, a

3 d^{6}

configuration, so the whole difference comes from the ligand field strength. In the anionic

[Fe (CN) X_{6}] X^{4 -}

, cyanide is a strong-field ligand that forces all six d-electrons to pair up as

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

; with no unpaired electrons,

μ = n (n + 2) = 0 BM

. In the cationic

[Fe (H X_{2} O) X_{6}] X^{2 +}

, water is a weak-field ligand and leaves the configuration

t_{2 g}^{4} e_{g}^{2}

with four unpaired electrons, giving

μ = 4 (4 + 2) = 24 \approx 4.9 BM

. Taking them in the order asked, anionic then cationic, the spin-only moments are

0

and

4.9 BM

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Single correct option