question_answer

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

A)  0                                           

B)  1

C)  4                                           

D)  6

Correct Answer: A

Solution :

                 In \[{{[Pt{{(CN)}_{4}}]}^{2-}}\], \[C{{N}^{-}}\] is a strong field ligand and outer configuration of Pt is \[5{{d}^{9}},6{{s}^{1}}\]. Thus, outer configuration of \[P{{t}^{2+}}\] is \[5{{d}^{8}}\]. \[C{{N}^{-}}\] will cause pairing of electrons. Hence,                 \[{{[Pt{{(CN)}_{4}}]}^{2-}}=\]                                                 \[ds{{p}^{2}}\] hybridisation Therefore, number of unpaired electrons in \[{{[Pt{{(CN)}_{4}}]}^{2-}}\] zero.