question_answer

In Cu-ammonia complex, the state of hybridisation of \[C{{u}^{2+}}\]is

A)  \[s{{p}^{3}}\]                                   

B)  \[{{d}^{3}}s\]

C)  \[s{{p}^{2}}f\]                                 

D)  \[ds{{p}^{2}}\]

Correct Answer: D

Solution :

                 Cu-ammonia complex is \[{{[Cu{{(N{{H}_{3}})}_{4}}]}^{2+}}\] \[\text{N}{{\text{H}}_{\text{3}}}\]being strong field ligand, shifts the unpaired electron to higher level.