Question

The magnetic momentum of $K\left[Fe\right(CN{)}_5\left(NO\right)]$ is:

• A. $1.73BM$
• B. $5.47BM$
• C. $6.9BM$
• D. zero

Answer 1

Answer: (A)

$1.73BM$

Apply the formula-

$M.M.=\mu =\sqrt{n(n+2)}$

$n$ = no. of unpaired electrons.

oxidation state of $Fe$ is $+3$, in presence of strong field ligand, back pairing will occur and $1$ electrons will be unpaired.

Put $n=1$ in the formula-

$M.M.=\mu =\sqrt{n(n+2)}$

$M.M=\mu =\sqrt{1(1+2)}=\sqrt{3}=1.73$