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CUET · CHEMISTRY · PYQ PAPER 2023

The oxidation state, geometry and number of unpaired electrons in \([Cr(NH_3)_6]^{3+}\) is :

  1. A +3, Octahedral, 2
  2. B +0, Octahedral, 2
  3. C +3, distorted octahedral, 0
  4. D +3, Octahedral, 3
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Answer & Solution

Correct Answer

(D) +3, Octahedral, 3

Step-by-step Solution

Detailed explanation

Oxidation state of Cr: \(x + 6(0) = +3 \implies x = +3\) Coordination number = 6 \(\implies\) Octahedral geometry Cr: \([Ar] 3d^5 4s^1\) \(Cr^{3+}\): \([Ar] 3d^3\) For \(d^3\) in octahedral field: \(t_{2g}^3 e_g^0\) Number of unpaired electrons = 3
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