Which of the following is the incorrect order | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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-$I$	List-$II$
$a$. See-Saw Shape	$i$. $XeF_4$
$b$. Square Pyramidal	$ii$. $ClF_3$
$c$. $T$-Shape	$iii$. $PbCl_2$
$d$. Bent Shape	$iv$. $SF_4$
	$v$. $BrF_5$

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

Similar Questions

The number of bonding and non-bonding electron pairs on the $N$ atom in $NO_3^-$ are respectively ..........

The shapes of $SOF_4$,$XeO_2F_2$ and $ClF_3$ are respectively:

The compound that has the largest $H-M-H$ bond angle $(M = N, O, S, C)$ is :

Match the following:
List-$I$ List-$II$
$a$. See-Saw Shape $i$. $XeF_4$
$b$. Square Pyramidal $ii$. $ClF_3$
$c$. $T$-Shape $iii$. $PbCl_2$
$d$. Bent Shape $iv$. $SF_4$
$v$. $BrF_5$

The correct answer is

Which of the following is $NOT$ tetrahedral in shape?

Vedclass Test Series

Exam Paper Generator

Online Exam Module