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.