NEET2006ChemistryCoordination CompoundsActual
[ Cr ( H ₂ O )₆ ] Cl ₃ (at. no. or 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:
Options
- A( (3 ~d x ^2- y ^2 )^1, 3 ~d z ^ 2^1 , 3 dxz )
- B(3 ~d xy ^1, (3 ~d x^2-y^2 )^1, 3 ~d y z^1 )
- C(3 ~d x y^1, 3 ~d y z^1, 3 ~d x z^1 )
- D(3 ~d xy ^1, 3 dyz ^1, 3 d z^ 2^ )
Correct answer
D. (3 ~d xy ^1, 3 dyz ^1, 3 d z^ 2^ )
Step-by-step solution
Magnetic moment aligned ( ) & = n(n+2) BM 3.83 & = n(n+2) 3.83 3.83 & =n^2+2 n 14.6689 & =n^2+2 n aligned On solving this, we get n=3 . Hence, the number of unpaired electrons in the d -subshell of the penultimate shell of chromium is 3 . So, Cr ³⁺=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^2 3 d^3 In [ Cr ( H ₂ O )₆ ] Cl ₃ , the oxidation state of Cr is +3 . Hence, in 3 d^3 , the distribution of electrons is 3 d_ x y ^1, 3 d^1, 3 d^1 _ x x^*