WBJEE · Chemistry · Coordination Compounds
In Cu-ammonia complex, the state of hybridization of \(\mathrm{Cu}^{+2}\) is
- A \(\mathrm{sp}^3\)
- B \(\mathrm{d}^3 \mathrm{~s}\)
- C \(\mathrm{sp}^2 \mathrm{f}\)
- D \(\mathrm{dsp}^2\)
Answer & Solution
Correct Answer
(D) \(\mathrm{dsp}^2\)
Step-by-step Solution
Detailed explanation
In \(\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{+}\) \(\mathrm{Cu}^{+2}\) is in a state of \(d s p^2\) hybridization and shape of the complex is square planar. (One \(e^{-}\)is excited from \(3 d\) to \(4 p\) during complex formation)
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