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.

List-$I$ (Molecule)	List-$II$ (Shape)
$A$. $BrF_5$	$I$. $T$-shape
$B$. $H_2O$	$II$. See-saw
$C$. $ClF_3$	$III$. Bent
$D$. $SF_4$	$IV$. Square pyramidal

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

Similar Questions

Match List-$I$ with List-$II$:

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

$A$. $BrF_5$ $I$. $T$-shape

$B$. $H_2O$ $II$. See-saw

$C$. $ClF_3$ $III$. Bent

$D$. $SF_4$ $IV$. Square pyramidal

Which among the following molecules is $NOT$ linear in shape?

Which of the following is not correctly matched?

The sets of molecules in which the central atom has no lone pair of electrons are:
$i$. $SnCl_2, NH_3, SF_4$
$ii$. $HgCl_2, SO_3, SF_6$
$iii$. $BeCl_2, BF_3, PCl_5$
$iv$. $ClF_3, BrF_5, XeF_6$

Maximum bond angle is present in :-

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