The correct sequence of decreasing bond angle | vedclass

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Question

(A) The bond angle in hydrides of group $15$ elements depends on the electronegativity of the central atom. As the electronegativity of the central at

Vedclass · Accepted Answer

$NH_3 > PH_3 > AsH_3 > SbH_3$

Vedclass · Answer

(A) The bond angle in hydrides of group $15$ elements depends on the electronegativity of the central atom. As the electronegativity of the central atom decreases down the group $(N > P > As > Sb)$,the bond pair electrons move further away from the central atom. This results in a decrease in the bond pair-bond pair repulsion,leading to a decrease in the bond angle. Therefore,the correct order of bond angles is $NH_3 (107.8^{\circ}) > PH_3 (93.6^{\circ}) > AsH_3 (91.8^{\circ}) > SbH_3 (91.3^{\circ})$.

List-$I$ (Molecule)	List-$II$ (Shape)
$A$. $NH_3$	$I$. Square pyramid
$B$. $BrF_5$	$II$. Tetrahedral
$C$. $PCl_5$	$III$. Trigonal pyramidal
$D$. $CH_4$	$IV$. Trigonal bipyramidal

List-$I$ (Molecule)	List-$II$ (Shape)
$A$. $BrF_5$	$I$. $T$-shape
$B$. $H_2O$	$II$. See-saw
$C$. $ClF_3$	$III$. Bent
$D$. $SF_4$	$IV$. Square pyramidal

The correct sequence of decreasing bond angles for the following hydrides is:

Similar Questions

Match List-$I$ with List-$II$:

List-$I$ (Molecule) List-$II$ (Shape)

$A$. $NH_3$ $I$. Square pyramid

$B$. $BrF_5$ $II$. Tetrahedral

$C$. $PCl_5$ $III$. Trigonal pyramidal

$D$. $CH_4$ $IV$. Trigonal bipyramidal

Choose the correct answer from the options below:

The structure of $NH_3$ is:

Match List-$I$ with List-$II$:

List-$I$ (Molecule) List-$II$ (Shape)

$A$. $BrF_5$ $I$. $T$-shape

$B$. $H_2O$ $II$. See-saw

$C$. $ClF_3$ $III$. Bent

$D$. $SF_4$ $IV$. Square pyramidal

Consider the molecules $CH_4$,$NH_3$ and $H_2O$. Which of the given statements is false?

In which of the following is the bond angle maximum?

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