The decreasing order of bond angles in BF_3,N | vedclass

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(A) $1$. $I_3^-$: The central $I$ atom is $sp^3d$ hybridized with $3$ lone pairs in the equatorial positions,resulting in a linear geometry with a bon

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$I_3^- > BF_3 > NH_3 > PF_3$

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(A) $1$. $I_3^-$: The central $I$ atom is $sp^3d$ hybridized with $3$ lone pairs in the equatorial positions,resulting in a linear geometry with a bond angle of $180^\circ$.$2$. $BF_3$: The central $B$ atom is $sp^2$ hybridized with no lone pairs,resulting in a trigonal planar geometry with a bond angle of $120^\circ$.$3$. $NH_3$ and $PF_3$: Both have $sp^3$ hybridization with $1$ lone pair,giving them a trigonal pyramidal geometry. The bond angle in $NH_3$ is $\approx 107^\circ$,while in $PF_3$ it is $\approx 96^\circ$. As the electronegativity of the surrounding atoms decreases or the size of the central atom increases,the bond angle decreases. Thus,$NH_3 > PF_3$.$4$. Combining these,the decreasing order of bond angles is $I_3^- (180^\circ) > BF_3 (120^\circ) > NH_3 (107^\circ) > PF_3 (96^\circ)$.

The decreasing order of bond angles in $BF_3$,$NH_3$,$PF_3$ and $I_3^-$ is

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