A) \[{{\operatorname{H}}_{2}}S < Si{{H}_{4}} < N{{H}_{3}} < B{{F}_{3}}\]
B) \[{{\operatorname{NH}}_{3}} < {{H}_{2}}S < SiH{{ }_{4}}< B{{F}_{3}}\]
C) \[{{\operatorname{H}}_{2}}S < N{{H}_{3}} < Si{{H}_{4}}< B{{F}_{3}}\]
D) \[{{\operatorname{H}}_{2}}S < N{{H}_{3}}< B{{F}_{3}}< Si{{H}_{4}}\]
Correct Answer: C
Solution :
\[{{\operatorname{H}}_{2}}S < N{{N}_{3}} < Si{{H}_{4}} < B{{F}_{3}}\] Bond angle compound \[92{}^\circ 6{}^\circ {{H}_{2}}S\] \[107{}^\circ N{{H}_{3}}\] \[10{{9}^{o}}28'Si{{H}_{4}}\] \[120{}^\circ B{{F}_{3}}\]You need to login to perform this action.
You will be redirected in
3 sec