question_answer

The number of unpaired electrons in the  square planar \[{{[Pt{{(CN)}_{4}}]}^{2-}}\]ion is

A)  2             

B)  1

C)  0             

D)  3

Correct Answer: C

Solution :

 The electronic configuration of \[Pt=[Xe]4{{f}^{14}},5{{d}^{9}},6{{s}^{1}}\] \[\therefore \] \[P{{t}^{2+}}=[Xe]\,4{{f}^{14}},5{{d}^{8}},6{{s}^{0}}\] \[{{[Pt{{(CN)}_{4}}]}^{2-}}=[Xe]\,4{{f}^{14}}\] \[\therefore \]No unpaired electron is present in \[{{[Pt{{(CN)}_{4}}]}^{2-}}\]ion.