Question

[Cr(HO)]Cl (atomic number of Cr = 24) has a magnetic moment of 3.83 BM. The correct distribution of 3d-elections in the chromium present in the complex is

• A. $3d^{1}\,_{xy},3d^{1}\,_{yz},3d^{1}\,_{zx}$
• B. $3d^{1}\,_{xy},3d^{1}\,_{yz},3d^{1}\,_{z^{2}}$
• C. $3d^{1}\,_{_{\left(x^{2}\,y^{2}\right)}},3d^{1}\,_{_{z^{2}}},3d^{1}\,_{zx}$
• D. $3d^{1}\,_{xy},3d^{1}\,_{_{\left(x^{2}\,y^{2}\right)}},3d^{1}\,_{xz}$

Answer 1

Answer: (A)

$3d^{1}\,_{xy},3d^{1}\,_{yz},3d^{1}\,_{zx}$

Answer 2

Magnetic moment indicates that there are three unpaired electrons present in chromium. These must be present in lower energy orbitals which are $3d^{1}_{ xy},\,3d^{1}_{ yz}$ and $3d^{1}_{ xz}.$