The correct order of bond angle is

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Question

(A) Let us analyze each option:$A$: In $NCl_3$ and $NBr_3$,the central atom is the same $(N)$. As the electronegativity of the surrounding atom decrea

Vedclass · Accepted Answer

$NCl_3 > NBr_3$

Vedclass · Answer

(A) Let us analyze each option:$A$: In $NCl_3$ and $NBr_3$,the central atom is the same $(N)$. As the electronegativity of the surrounding atom decreases $(Cl > Br)$,the bond pair-bond pair repulsion decreases,leading to a smaller bond angle. Thus,$NCl_3 > NBr_3$ is correct.$B$: In hydrides of group $16$ $(H_2O, H_2S, H_2Se, H_2Te)$,as the size of the central atom increases,the bond angle decreases. Thus,$H_2S > H_2Te$ is the correct order.$C$: Both $N_3^-$ and $I_3^-$ are linear molecules with a bond angle of $180^{\circ}$. Thus,$N_3^- = I_3^-$.$D$: In $NO_3^-$ ($sp^2$ hybridized,$120^{\circ}$) and $NO_2^-$ ($sp^2$ hybridized with one lone pair,$ NO_2^-$.Therefore,the correct statement is $A$.

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

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:

Which of the following molecules is symmetrical?

The number of lone pairs of electrons on the central $I$ atom in $I_{3}^{-}$ is $.......$

Which of the following molecules has the smallest bond angle?

Which has a larger bond angle between $PH_3$ and $PH_4^+$? Why?

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