Q. 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
WBJEEWBJEE 2016
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
