A) \[NH_{4}^{+}\]and \[NH_{2}^{-}\]
B) \[CH_{3}^{-}\]and \[CH_{3}^{+}\]
C) \[SO_{4}^{2-},PO_{4}^{3-}\]and \[[BF_{4}^{-}]\]
D) \[NH_{4}^{+}\]and \[N{{H}_{3}}\]
Correct Answer: C
Solution :
Hybridizations on a molecule can be determined as follows\[X=SA+\frac{1}{2}(G-V)X=SA+\frac{1}{2}(G-V+a)\]for simple molecule for polyatomic anions \[X=SA+\frac{1}{2}(G-V+C)\] for polyatomic cation| Value of X | Hybrid state | No. of SA | No. of L | Shape | Type of molecule and example |
| 2 | sp | 2 | 0 | Linear | \[A{{B}_{2}}:Be{{F}_{2}},\] \[C{{O}_{2}},C{{S}_{2}},NO_{2}^{+}\] |
| 3 | \[s{{p}^{2}}\] | 3 | 0 | Trigonal planar | \[A{{B}_{3}}:B{{F}_{3}},\] \[AlC{{l}_{3}},S{{O}_{3}},\] \[NO_{3}^{-}CO_{3}^{2-}\] |
| 2 | 1 | Bent | \[A{{B}_{2}}L:S{{O}_{2}},\] \[PbC{{l}_{2}},SnC{{l}_{2}},\] \[NO_{2}^{-}\] | ||
| 4 | \[S{{P}^{3}}\] | 4 | 0 | Tetrahedral | \[A{{B}_{4}}:C{{H}_{4}},\] \[SiC{{l}_{4}},SO_{4}^{2-},\] \[ClO_{4}^{-}\] |
| 3 | 1 | Pyramidal | \[A{{B}_{3}}L:N{{H}_{3}},\] \[N{{F}_{3}},P{{H}_{3}},PC{{l}_{5}},\] \[{{H}_{3}}{{O}^{+}}\] | ||
| 2 | 2 | Bent | \[A{{B}_{2}}{{L}_{2}}:{{H}_{2}}O,\] \[NH_{2}^{-}\] | ||
| 4 | 1 | See saw | \[A{{B}_{4}}L:S{{F}_{4}},\] \[Te{{F}_{4}}\] | ||
| 5 | \[s{{p}^{3}}d\] | 3 | 2 | T shape | \[A{{B}_{2}}{{L}_{2}}:Cl{{F}_{3}},\] \[B{{F}_{3}}\] |
| 2 | 3 | Linear | \[A{{B}_{2}}{{L}_{3}}:Xe{{F}_{2}},\] \[I_{3}^{-},Br_{3}^{-}\] | ||
| 6 | \[s{{p}^{3}}{{d}^{2}}\] | 5 | 1 | Square pyramid | \[A{{B}_{5}}L:I{{F}_{5}},\] \[Br{{F}_{5}}\] |
| 4 | 2 | Square planar | \[A{{B}_{4}}{{L}_{2}}:Xe{{F}_{4}}\] |
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