The correct order of H-M-H bond angle is | vedclass

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(B) In the hydrides of group $15$ elements $(NH_3, PH_3, AsH_3, SbH_3, BiH_3)$,the central atom undergoes $sp^3$ hybridization.As we move down the gro

Vedclass · Accepted Answer

$BiH_3 < SbH_3 < AsH_3 < PH_3 < NH_3$

Vedclass · Answer

(B) In the hydrides of group $15$ elements $(NH_3, PH_3, AsH_3, SbH_3, BiH_3)$,the central atom undergoes $sp^3$ hybridization.As we move down the group,the electronegativity of the central atom decreases.According to Drago's rule,as the electronegativity of the central atom decreases,the bond pair-bond pair repulsion decreases,and the bond angle approaches $90^{\circ}$.Therefore,the bond angle decreases as we move down the group: $NH_3 (107.8^{\circ}) > PH_3 (93.6^{\circ}) > AsH_3 (91.8^{\circ}) > SbH_3 (91.3^{\circ}) > BiH_3 (90^{\circ})$.Thus,the correct increasing order is $BiH_3 < SbH_3 < AsH_3 < PH_3 < NH_3$.

List-$I$ (molecule/ion)	List-$II$ (Number of lone pairs of electrons on the central atom)
$(A)$ $XeF_2$	$(I)$ $2$
$(B)$ $XeO_3$	$(II)$ $0$
$(C)$ $XeF_4$	$(III)$ $3$
$(D)$ $PF_6^-$	$(IV)$ $1$

List-$I$ (Shape)	List-$II$ (Molecules)
$A$. $T$-shaped	$I$. $XeF_4$
$B$. Trigonal planar	$II$. $SF_4$
$C$. Square planar	$III$. $ClF_3$
$D$. See-saw	$IV$. $BF_3$

The correct order of $H-M-H$ bond angle is

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