A) \[S{{F}_{4}}\]
B) \[I_{3}^{-}\]
C) \[SbCl_{5}^{2-}\]
D) \[PC{{l}_{5}}\]
Correct Answer: C
Solution :
Key Idea Molecules having the same number of hybrid orbitals, have same hybridisation and number of hybrid orbitals, \[H=\frac{1}{2}[V+X-C+A]\] where, V = no. of valence electrons of central atom X = no. of monovalent atoms C = charge on cation A = Charge on anion. In \[S{{F}_{4}},\] \[H=\frac{1}{2}[6+4-0+0]=5\] In \[\text{I}_{\text{3}}^{\text{-}}\] \[H=\frac{1}{2}[7+2+1]=5\] In \[SbCl_{5}^{2-},\] \[H=\frac{1}{2}[5+5+2]=6\] In \[PC{{l}_{5}},\] \[H=\frac{1}{2}[5+5+0-0]=5\] Since, only \[SbCl_{5}^{2-}\] has different number of hybrid orbitals (i.e., 6) from the other given species, its hybridisation is different from the others, ie, \[S{{b}^{3}}{{d}^{2}}.\](The hybridisation of other species is \[S{{b}^{3}}d\]).
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