A) both square planar
B) tetrahedral and square planar
C) both tetrahedral
D) None of these
Correct Answer: C
Solution :
[c]\[Ni{{(CO)}_{4}}\] | \[Ni{{(PP{{h}_{3}})}_{2}}C{{l}_{2}}\] | |
\[O.S.\] | \[N{{i}^{0}}\] | \[N{{i}^{2+}}\] |
\[E.C.\] | \[[Ar]3{{d}^{8}}4{{s}^{2}}\] | \[[Ar]3{{d}^{8}}4{{s}^{0}}\] |
Pairing | Pairing of \[{{e}^{-}}\] | No pairing of \[{{e}^{-}}\] |
Hybridization | \[s{{p}^{3}}\] (tetrahedral) | \[s{{p}^{3}}\] (tetrahedral) |
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