$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_3$ (at. no. or $\mathrm{Cr}=24$…
$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_3$ (at. no. or $\mathrm{Cr}=24$ ) has a magnetic moment of 3.83 B.M. The correct distribution of $3 d$ electrons in the Chromium of the complex is:
\(3 \mathrm{~d} \mathrm{xy}^1,\left(3 \mathrm{~d} x^2-y^2\right)^1, 3 \mathrm{~d} y z^1\)
\(3 \mathrm{~d} x y^1, 3 \mathrm{~d} y z^1, 3 \mathrm{~d} x z^1\)
\(3 \mathrm{~d} \mathrm{xy}^1, 3 \mathrm{dyz}{ }^1, 3 d z^{2^{\prime}}\)
Solution
Magnetic moment
$\begin{aligned}
(\mu) & =\sqrt{n(n+2)} \mathrm{BM} \\
3.83 & =\sqrt{n(n+2)} \\
3.83 \times 3.83 & =n^2+2 n \\
14.6689 & =n^2+2 n
\end{aligned}$
On solving this, we get $n=3$.
Hence, the number of unpaired electrons in the $d$-subshell of the penultimate shell of chromium is 3 .
So,
$\mathrm{Cr}^{3+}=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^2 3 d^3$
In $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_3$, the oxidation state of $\mathrm{Cr}$ is +3 .
Hence, in $3 d^3$, the distribution of
electrons is $3 d_{x y}^1, 3 d^1, 3 d^1{ }_{x x^*}$
