question_answer

\[[Cr{{({{H}_{2}}O)}_{6}}]C{{l}_{3}}\] (at no. of Cr = 24) has a magnetic moment of 3.83 B.M. The correct distribution of 3d electrons in the Chromium of the complex is

A) \[3{{d}_{x{{y}^{1}}}},\] \[{{(3{{d}_{{{x}^{2}}-{{y}^{2}}}})}^{1}},\] \[3{{d}_{y{{z}^{1}}}}\]

B) \[3{{d}_{x{{y}^{1}}}},\] \[3{{d}_{y{{z}^{1}}}},\] \[3{{d}_{x{{z}^{1}}}}\]

C) \[3{{d}_{x{{y}^{1}}}},\] \[3{{d}_{y{{z}^{1}}}},\] \[3{{d}_{d{{z}^{2}}}}\]

D) \[{{(3{{d}_{{{x}^{2}}-{{y}^{2}}}})}^{1}},\] \[3{{d}_{\,{{z}^{\,2}}}},\] \[3{{d}_{x{{z}^{1}}}}\]

Correct Answer: B

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. \[C{{r}^{3+}}\] will have configuration as \[Cr\]  \[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{4}}4{{s}^{2}}\] \[C{{r}^{3+}}1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{3}}\] So \[3{{d}^{1}}_{xy}3{{d}^{1}}_{yz}3d{}_{xz}^{1}\]