Question

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

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

Answer 1

The correct option is A 1.73 B.M

Apply the spin only magnetic moment formula-
$\mu =\sqrt{n(n+2)}B.M$
n = no. of unpaired electrons.
The given complex is $K\left[Fe\right(CN{)}_5\left(NO\right)]$.
Oxidation state of $Fe$ is +3, in presence of strong field ligand, back pairing will occur and 1 electrons will be unpaired.
therefore n=1
$\mu =\sqrt{n(n+2)}$
$\mu =\sqrt{1(1+2)}=\sqrt{3}=1.73B.M.$