A) \[S{{F}_{4}}\] and \[Xe{{F}_{4}}\]
B) \[SO_{3}^{2-}\] and \[NO_{3}^{-}\]
C) \[B{{F}_{3}}\] and \[N{{F}_{3}}\]
D) \[BrO_{3}^{-}\] and \[Xe{{O}_{3}}\]
Correct Answer: D
Solution :
| [a] \[S{{F}_{4}}=\] irregular tetrahedron (\[s{{p}^{3}}d\], one lone pair) \[Xe{{F}_{4}}\]= square planar (\[s{{p}^{3}}{{d}^{2}}\], two lone pair) |
| [b] \[SO_{3}^{2-}=\] pyramidal (\[s{{p}^{3}}\] one lone pair) \[NO_{3}^{-}=\] trigonal planar \[(s{{p}^{2}})\] |
| [c] \[B{{F}_{3}}=\]trigonal planar \[(s{{p}^{2}})\] \[N{{F}_{3}}=\] pyramidal \[(s{{p}^{3}})\] |
| [d] \[BrO_{3}^{-}=\] pyramidal (\[s{{p}^{3}},\] one lone pair) \[Xe{{O}_{3}}=\] pyramidal (\[s{{p}^{3}}\], one lone pair) |
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