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. $ds{p}^2,+1,1$
• B. $s{p}^3,+1,Zero$
• C. $s{p}^3,+2,1$
• D. $ds{p}^3,+2,Zero$

Answer 1

Answer: (B)

${\left[C{u}^{+1}{\left(CN\right)}_4^{-4}\right]}^{-3}$
$c{u}^{+}$ ground state=CN $\left(-\right)$is strong field ligand, $\Delta$ is high
Hybridisation $=s{p}^3;$oxidation state of Cu = +1
Number of unpaired electron = 0