CUET · CHEMISTRY · PYQ PAPER 2023
The oxidation state, geometry and number of unpaired electrons in \([Cr(NH_3)_6]^{3+}\) is :
- A +3, Octahedral, 2
- B +0, Octahedral, 2
- C +3, distorted octahedral, 0
- D +3, Octahedral, 3
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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