Which of the following is the correct order o | 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-bond pair repulsion decreases. Consequently,the bond angle decreases as we move down the group. Thus,the correct order of decreasing 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)
$A$. $XeF_4$	$I$. Square pyramidal
$B$. $ClF_3$	$II$. Pentagonal bipyramidal
$C$. $BrF_5$	$III$. Square planar
$D$. $IF_7$	$IV$. Bent $T$-shape

List-$I$ (species)	List-$II$ ($O-N-O$ angle)
$A$. $NO_2^+$	$1$. $180^{\circ}$
$B$. $NO_2$	$2$. $132^{\circ}$
$C$. $NO_2^-$	$3$. $115^{\circ}$
$D$. $NO_3^-$	$4$. $120^{\circ}$

Which of the following is the correct order of decreasing bond angles in the given hydrides?

Similar Questions

Match the following:

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

$A$. $XeF_4$ $I$. Square pyramidal

$B$. $ClF_3$ $II$. Pentagonal bipyramidal

$C$. $BrF_5$ $III$. Square planar

$D$. $IF_7$ $IV$. Bent $T$-shape

Discuss the molecular structure of $BrF_3$ based on the $VSEPR$ theory.

The number of species below that have two lone pairs of electrons on their central atom is ..... . (Round off to the nearest integer)
$SF_{4}, BF_{4}^{-}, ClF_{3}, AsF_{3}, PCl_{5}, BrF_{5}, XeF_{4}, SF_{6}$

Match List-$I$ with List-$II$ and select the correct answer:

List-$I$ (species) List-$II$ ($O-N-O$ angle)

$A$. $NO_2^+$ $1$. $180^{\circ}$

$B$. $NO_2$ $2$. $132^{\circ}$

$C$. $NO_2^-$ $3$. $115^{\circ}$

$D$. $NO_3^-$ $4$. $120^{\circ}$

To which type does $ClF_3$ belong among $AB_4$,$AB_4E$,and $AB_3E_2$? Explain.

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