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. |
| [a] In \[S{{F}_{4}},\] \[H=\frac{1}{2}[6+4-0+0]=5\] |
| [b] In \[I_{3}^{-}\] \[H=\frac{1}{2}[7+2+1]=5\] |
| [c] In \[SbCl_{5}^{2-},\] \[H=\frac{1}{2}[5+5+2]=6\] |
| [d] 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, i.e., \[S{{b}^{3}}{{d}^{2}}\]. (The hybridisation of other species is \[S{{b}^{3}}d\]). |
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