The correct order of bond angle is:

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Question

(B) In $PX_3$ molecules,the central atom $P$ is $sp^3$ hybridized with one lone pair.As the electronegativity of the surrounding halogen atom decrease

Vedclass · Accepted Answer

$PF_3 < PCl_3 < PBr_3 < PI_3$

Vedclass · Answer

(B) In $PX_3$ molecules,the central atom $P$ is $sp^3$ hybridized with one lone pair.As the electronegativity of the surrounding halogen atom decreases $(F > Cl > Br > I)$,the bond pair electrons move further away from the central atom.According to the $VSEPR$ theory,as the size of the surrounding atom increases,the bond pair-bond pair repulsion decreases,but the bond angle increases due to the decrease in electronegativity of the surrounding atoms.Therefore,the correct order of bond angle is $PI_3 > PBr_3 > PCl_3 > PF_3$.

List-$I$ (Compound)	List-$II$ (Shape/geometry)
$A$. $NH_3$	$I$. Trigonal Pyramidal
$B$. $BrF_5$	$IV$. Square Pyramidal
$C$. $XeF_4$	$II$. Square Planar
$D$. $SF_6$	$III$. Octahedral

Similar Questions

The correct order of $H-N-H$ bond angles of ammonia,ammonium ion,and amide ion is:

In which of the following molecules/ions are all the bond angles not equal?

Which of the following pairs has the same structure?

Match List-$I$ with List-$II$.

List-$I$ (Compound) List-$II$ (Shape/geometry)

$A$. $NH_3$ $I$. Trigonal Pyramidal

$B$. $BrF_5$ $IV$. Square Pyramidal

$C$. $XeF_4$ $II$. Square Planar

$D$. $SF_6$ $III$. Octahedral

Choose the correct answer from the options given below:

The compound$(s)$ with $2$ lone pairs of electrons on the central atom is(are):
$(A)$ $BrF_5$
$(B)$ $ClF_3$
$(C)$ $XeF_4$
$(D)$ $SF_4$

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