Coordination Compounds - Result Question 59
####66. $[Cr(H_2 O)_6] Cl_3($ at no. of $Cr=24)$ has a magnetic moment of $3.83 B$. M. The correct distribution of $3 d$ electrons in the chromium of the complex is
(a) $3 d _{x y}^{1},(3 d _{x^{2}-y^{2}})^{1}, 3 d _{y z}^{1}$
[2006]
(b) $3 d _{x y}{ }^{1}, 3 d _{y z}{ }^{1}, 3 d _{x z}{ }^{1}$
(c) $3 d _{x y}{ }^{1}, 3 d _{y z}{ }^{1}, 3 d_z^{1} 2$
(d) $(3 d_x^{2}-y^{2})^{1}, 3 d_z^{2}, 3 d _{x z}{ }^{1}$
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Solution:
(b) $\mu=\sqrt{n(n+2)}$
$3.83=\sqrt{n(n+2)}$
on solving $n=3$
As per magnetic moment, it has three unpaired electron.
$Cr^{3+}$ will have configuration as $=$
$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{3}$
So $3 d _{x y}{ }^{1} 3 d _{y z}{ }^{1} 3 d _{x z}{ }^{1}$
