Question

Find the oxidation state of Cr in the given complex $K_2\left[Cr\right(NO\left)\right(N{H}_3\left)\right(CN{)}_4]$, $\mu$ = 1.73 BM.

• A. + 1
• B. + 2
• C. 3
• D. None of these

Answer 1

The correct option is A + 1

Given, $k_2\left[\mathrm{cr}\right(NO\left)\left(N{H}_3\right)\right(CN{)}_4]$ and magnetic moment $\left(\mu \right)=1.73$.

$\text{We know,}\mu =\sqrt{n(n+2)}=1.73=\sqrt{3}$

$\therefore n=1$ (Number of unpaired electrons)

since, chromium ion has one unpaired electron, so NO must

be unit positive ligand.

Let, oxidation number of $Cr$ is $x$.

$\begin{array}{cc}\Rightarrow x+1& +0-4=-2\\ \therefore & x=1\end{array}$

Thus oxidation state of $Cr$ is +1 .

so, option (A) is correct.