In the complex ion (\left\[Cu\left(CN\right)\_{4}\right]^{3-... | Chemistry | SolveeIt | Solveeit

Today, August 23

Question

In the complex ion $\left[Cu\left(CN\right)_{4}\right]^{3-}$ the hybridization state, oxidation state and number of unpaired electrons of copper are respectively

A

$dsp^{2},+1,1$

B

$sp^{3},+1,Zero$

C

$sp^{3},+2,1$

D

$dsp^{3}, +2, Zero$

Answer

$sp^{3},+1,Zero$

Explanation

Solution

$\left[Cu^{+1}\left(CN\right)^{-4}_{4}\right]^{-3}$
$cu^{+}$ ground state=CN $\left(-\right)$ is strong field ligand, $\Delta$ is high
Hybridisation $=sp^{3};$ oxidation state of Cu = +1
Number of unpaired electron = 0