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.You need to login to perform this action.
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