$\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}ight)_{6}ight] \mathrm{Cl}_{3}$ (atomic number of…

$\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}ight)_{6}ight] \mathrm{Cl}_{3}$ (atomic number of $\mathrm{Cr}=24$) has a magnetic moment of $3.83 \mathrm{BM}$. The correct distribution of $3 d$-elections in the chromium present in the complex is
  1. $3 d_{x y}^{1}, 3 d_{y z}^{1}, 3 d_{z x}^{1}$
  2. $3 d_{x y}^{1}, 3 d_{y z}^{1}, 3 d_{z^{2}}^{1}$
  3. $3 d^{1}{ }_{\left(x^{2}-y^{2}ight)}, 3 d_{z^{2}}^{1}, 3 d_{z x}^{1}$
  4. $3 d_{x y}^{1}, 3 d_{\left(x^{2}-y^{2}ight)}^{1}, 3 d^{1}{ }_{x z}$

Solution

$\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}ight)_{6}ight] \mathrm{Cl}_{3}$ (at, no. of $\mathrm{Cr}=24$ ) has a magnetic moment of $3.83$ B.M. The correct distribution of 3 d electrons of chromium in the complex is $3 \mathrm{~d}_{\mathrm{xy}}^{1}, 3 \mathrm{~d}_{\mathrm{yz} z}^{1}, 3 \mathrm{~d}_{\mathrm{xz}}^{1}$.
The magnetic moment of $3.83$ B.M corresponds to 3 unpaired electrons.
$
\begin{array}{l}
3.83=\sqrt{\mathrm{n}(\mathrm{n}+2)} \\
\mathrm{n}=3
\end{array}
$
All these electrons are present in $\mathrm{t}_{2 \mathrm{~g}}$ orbitals. These correspond to $3 \mathrm{~d}_{\mathrm{xy}}^{1}, 3 \mathrm{~d}_{\mathrm{yz}}^{1}, 3 \mathrm{~d}_{\mathrm{xz}}^{1}$. *

Asked in: JEE-TOPICTESTS-CHEMISTRY

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