What is the correct order of bond angles (from smallest to largest) in $H_2S, NH_3, BF_3,$ and $SiH_4$?
- A$H_2S < SiH_4 < NH_3 < BF_3$
- B$H_2S < NH_3 < BF_3 < SiH_4$
- C$H_2S < NH_3 < SiH_4 < BF_3$
- D$NH_3 < H_2S < SiH_4 < BF_3$
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View Solution$SbF_5$ reacts with $XeF_4$ and $XeF_6$ to form ionic compounds $[XeF_3]^+ [SbF_6]^-$ and $[XeF_5]^+ [SbF_6]^-$ respectively. What are the molecular shapes of the $[XeF_3]^+$ and $[XeF_5]^+$ ions,respectively?
Match the following molecules/ions in List-$I$ with the number of lone pairs of electrons on the central atom in List-$II$.
List-$I$ (molecule/ion) List-$II$ (Number of lone pairs of electrons on the central atom) $(A)$ $XeF_2$ $(I)$ $2$ $(B)$ $XeO_3$ $(II)$ $0$ $(C)$ $XeF_4$ $(III)$ $3$ $(D)$ $PF_6^-$ $(IV)$ $1$
| List-$I$ (molecule/ion) | List-$II$ (Number of lone pairs of electrons on the central atom) |
|---|---|
| $(A)$ $XeF_2$ | $(I)$ $2$ |
| $(B)$ $XeO_3$ | $(II)$ $0$ |
| $(C)$ $XeF_4$ | $(III)$ $3$ |
| $(D)$ $PF_6^-$ | $(IV)$ $1$ |
Molecular bromine on reaction with three moles of molecular fluorine produces an interhalogen compound. The total number of lone pairs at the central halogen atom is
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