Fluorine reacts with dilute $NaOH$ and forms a gaseous product $A$. The bond angle in the molecule of $A$ is
- A$104^{\circ} 40^{\prime}$
- B$103^{\circ}$
- C$107^{\circ}$
- D$109^{\circ} 28^{\prime}$
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Similar Questions
What is the molecular shape of $BrF_3$?
Medium
View SolutionSelect the incorrect match regarding molecular geometry:
Statement $-$ Shape $-$ Example
$A$. $SF_4$ $-$ See-saw $-$ $SF_4$
$B$. $BrF_3$ $-$ $T$-shaped $-$ $BrF_3$
$C$. $XeF_4$ $-$ Square planar $-$ $XeF_4$
$D$. $ICl_2^-$ $-$ Linear $-$ $ICl_2^-$
Statement $-$ Shape $-$ Example
$A$. $SF_4$ $-$ See-saw $-$ $SF_4$
$B$. $BrF_3$ $-$ $T$-shaped $-$ $BrF_3$
$C$. $XeF_4$ $-$ Square planar $-$ $XeF_4$
$D$. $ICl_2^-$ $-$ Linear $-$ $ICl_2^-$
Identify the bond angle $O-S-O$ in the $SO_2$ molecule. (in $^{\circ}$)
What is the shape of $ClO_3^-$?
Medium
View SolutionMatch the following:
List-$I$ List-$II$ $A. BrF_5$ $I. AB_4E$,see-saw $B. SF_4$ $II. AB_4E_2$,square planar $C. XeF_4$ $III. AB_5E$,square pyramidal $D. ClF_3$ $IV. AB_3E_2$,$T$-shape
The correct answer is:
| List-$I$ | List-$II$ |
| $A. BrF_5$ | $I. AB_4E$,see-saw |
| $B. SF_4$ | $II. AB_4E_2$,square planar |
| $C. XeF_4$ | $III. AB_5E$,square pyramidal |
| $D. ClF_3$ | $IV. AB_3E_2$,$T$-shape |
The correct answer is:
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