Which of the following is the incorrect order | vedclass

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Question

(C) $1$. In $COF_2$,$COCl_2$,and $COBr_2$,the bond angle increases as the size of the halogen atom increases due to steric repulsion between the large

Vedclass · Accepted Answer

$BF_3 < BCl_3 < BBr_3 < BI_3$

Vedclass · Answer

(C) $1$. In $COF_2$,$COCl_2$,and $COBr_2$,the bond angle increases as the size of the halogen atom increases due to steric repulsion between the larger halogen atoms. Thus,$COF_2 $2$. In $PF_3$,$PCl_3$,$PBr_3$,and $PI_3$,the bond angle increases as the electronegativity of the halogen decreases,reducing the repulsion between bonding pairs and increasing the bond angle. Thus,$PF_3 $3$. In $BF_3$,$BCl_3$,$BBr_3$,and $BI_3$,all are $sp^2$ hybridized with a bond angle of $120^{\circ}$. Therefore,the bond angle is equal for all,making the order $BF_3 = BCl_3 = BBr_3 = BI_3$. The given order $BF_3 $4$. In $PH_3$ and $PCl_3$,the bond angle in $PH_3$ is approximately $93^{\circ}$ and in $PCl_3$ is approximately $100^{\circ}$. Thus,$PH_3 < PCl_3$ is correct.

Which of the following is the incorrect order of bond angle?

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