In the brown ring complex $[Fe (H X_{2} O) X_{5} (NO)] X^{2 +}$ , treat the nitrosyl ligand as $NO X^{+}$ , so that iron is in the +1 state, and assume the water ligands give a high-spin arrangement. The spin-only magnetic moment of the complex ion is

Answer: B

With the ligands

5 \times 0

and

NO X^{+}

(+1) and overall charge +2, the iron is

Fe X^{+}

(

3 d^{7}

, since Fe is

3 d^{6} 4 s^{2}

). In a high-spin arrangement the

d^{7}

ion has 3 unpaired electrons.

μ = n (n + 2) = 15 = 3.87 BM

. The values 4.90 and 5.92 correspond to

d^{6}

and

d^{5}