TS EAMCET · Chemistry · Coordination Compounds
The electronic configuration of \(\mathrm{Cr}\) in \(\mathrm{Cr}(\mathrm{CO})_6\) as calculated using crystal field theory is
- A \(t_{2 g}^4 e_g^0\)
- B \(t_{2 g}^3 e_g^1\)
- C \(t_{2 g}^6 e_g^0\)
- D \(t_{2 g}^4 e_g^2\)
Answer & Solution
Correct Answer
(C) \(t_{2 g}^6 e_g^0\)
Step-by-step Solution
Detailed explanation
\(\mathrm{Cr}(\mathrm{CO})_6\) is octahedral geometry and \(\mathrm{CO}\) is strong ligand. Therefore, low spin complex formed. \(\operatorname{Cr}(z=24)[\mathrm{Ar}] 3 d^5 4 s^1 \text { or }[\mathrm{Ar}] 3 d^6 4 s^{\circ}\)…
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