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 :
\[S{{F}_{4}}=\] irregular tetrahedron (\[s{{p}^{3}}d\], one lone pair) \[Xe{{F}_{4}}=\] square planar (\[s{{p}^{3}}{{d}^{2}},\] two lone pair) \[SO_{3}^{2-}\] = pyramidal (\[s{{p}^{3}}\] one lone pair) \[NO_{3}^{-}\] = trigonal planar \[(s{{p}^{2}})\] \[B{{F}_{3}}\]= trigonal planar \[(s{{p}^{2}})\] \[N{{F}_{3}}=\] pyramidal \[(s{{p}^{3}})\] \[BrO_{3}^{-}\] = pyramidal (\[s{{p}^{3}}\], one lone pair) \[Xe{{O}_{3}}=\] pyramidal (\[s{{p}^{3}},\] one lone pair)
You need to login to perform this action.
You will be redirected in
3 sec
