In compound X,all the bond angles are exactly | vedclass

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

Question

(C) The compound $X$ is carbon tetrachloride $(CCl_4)$.$CCl_4$ undergoes $sp^3$ hybridization and possesses a perfectly symmetrical tetrahedral geomet

Vedclass · Accepted Answer

Carbon tetrachloride

Vedclass · Answer

(C) The compound $X$ is carbon tetrachloride $(CCl_4)$.$CCl_4$ undergoes $sp^3$ hybridization and possesses a perfectly symmetrical tetrahedral geometry because all four substituents attached to the central carbon atom are identical ($Cl$ atoms).In molecules like chloromethane $(CH_3Cl)$,chloroform $(CHCl_3)$,or iodoform $(CHI_3)$,the presence of different atoms attached to the central carbon leads to unequal bond pairs and lone pair-bond pair repulsions,causing the bond angles to deviate from the ideal tetrahedral angle of $109^o 28'$.

	Molecule	Hybridization	Geometry	Number of lone pairs of electrons on central atom
$I$	$SiH_4$	$sp^3$	tetrahedral	$0$
$II$	$BeCl_2$	$sp^2$	linear	$1$
$III$	$SF_4$	$dsp^3$	square planar	$1$
$IV$	$SnCl_2$	$sp$	linear	$0$

In compound $X$,all the bond angles are exactly $109^o 28'$,$X$ is

Similar Questions

If a molecule $MX_3$ has zero dipole moment,the sigma bonding orbitals used by $M$ are

$BF_3$ is a planar and electron-deficient compound. The hybridization and the number of electrons around the central atom,respectively,are:

The $BCl_3$ molecule is planar. What type of hybridization does the boron atom undergo in this molecule?

The types of hybridisation on the five carbon atoms from left to right in $pent-1-en-4-yne$ are

Which of the following sets is correct based on the table provided below?
Molecule Hybridization Geometry Number of lone pairs of electrons on central atom
$I$ $SiH_4$ $sp^3$ tetrahedral $0$
$II$ $BeCl_2$ $sp^2$ linear $1$
$III$ $SF_4$ $dsp^3$ square planar $1$
$IV$ $SnCl_2$ $sp$ linear $0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module