The bond angles in methane (I),ammonia (II),a | vedclass

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(B) $1$. In methane $(CH_4)$,the hybridization is $sp^3$ with no lone pair,resulting in a bond angle of $109.5^\circ$.$2$. In ammonia $(NH_3)$,the hyb

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$II < III < I$

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(B) $1$. In methane $(CH_4)$,the hybridization is $sp^3$ with no lone pair,resulting in a bond angle of $109.5^\circ$.$2$. In ammonia $(NH_3)$,the hybridization is $sp^3$ with one lone pair. Due to lone pair-bond pair repulsion,the bond angle decreases to $107^\circ$.$3$. In trimethylamine $(N(CH_3)_3)$,the hybridization is $sp^3$ with one lone pair. However,the bulky methyl groups cause steric repulsion,which increases the bond angle to approximately $108^\circ$ to $109^\circ$.$4$. Comparing the values: $NH_3$ $(107^\circ)$ $5$. Therefore,the correct order of increasing bond angles is $(II) < (III) < (I)$.

The bond angles in methane $(I)$,ammonia $(II)$,and trimethylamine $(III)$ increase in the order:

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