$[C r (H_{2} O)_{6}] C l_{3}$ (At. no. of cr = 24) has moment of 3.83 B.M. The correct distribution of 3d electrons in the chromium present in the complex is :

3 d_{x y}^{1}, 3 d_{y z}^{1}, 3 d_{z x}^{1}

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3 d_{x y}^{1}, 3 d_{y z}^{1}, 3 d_{z^{2}}^{1}

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3 d_{(x^{2} - y^{2})}^{1}, 3 d_{z^{2}}^{1}, 3 d_{x z}^{1}

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3 d_{x z}^{1}, 3 d_{(x^{2} - y^{2})}^{1}, 3 d_{y z}^{1}

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Solution

The correct option is D $3 d_{x y}^{1}, 3 d_{y z}^{1}, 3 d_{z x}^{1}$
Magnetic moment $μ = \sqrt{n (n + 2)}$

$3.83 = \sqrt{n (n + 2)}$

$\Rightarrow n = 3$

Magnetic moment indicates that there are three unpaired electrons present in chromium. These must be present in lower energy orbitals which are :

$3 d_{x y}, 3 d_{y z}$ and $3 d_{x z}$ .

Hence, the correct option is A.

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Similar questions

[C r (H_{2} O)_{6}] C l_{3}

[C r {(H_{2} O)}_{6}] C l_{3}

C r = 24

[C r (H_{2} O)_{6}] C l_{3}

B . M .

3 d

The number of d electrons in [Cr(H₂O)₆]³⁺(At. no. of Cr = 24) is:

[C r (N H_{3})_{4} B r C l] C l

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