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JEE Advanced levelSingle correct optionDifficulty 3 of 3qa158

In $[Fe (H X_{2} O) X_{5} (NO)] X^{2 +}$ , the brown ring complex, treat NO as $NO X^{+}$ and water as a weak-field ligand (high spin). The oxidation state of Fe and the spin-only magnetic moment in BM are:

A $+ 1$ ; $3.87$
B $+ 3$ ; $5.92$
C $+ 1$ ; $1.73$
D $+ 2$ ; $4.90$

Answer: A

Charge balance:

x + 1 = + 2

gives

x = + 1

, so Fe is

d^{7}

. High spin

t_{2 g}^{5} e_{g}^{2}

has 3 unpaired electrons, so

μ = 3 \times 5 = 3.87 BM

. The value

1.73

is for low-spin

d^{7}

,

4.90

is high-spin

d^{6}

(neutral NO), and

5.92

is

d^{5}

.

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