Which statement is correct about CO_2 if the | vedclass

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(B) In $CO_2$,the carbon atom is $sp$ hybridized.If the central carbon atom uses $s$ and $P_x$ orbitals for hybridization,the hybrid orbitals lie alon

Vedclass · Accepted Answer

$i, ii, iii$

Vedclass · Answer

(B) In $CO_2$,the carbon atom is $sp$ hybridized.If the central carbon atom uses $s$ and $P_x$ orbitals for hybridization,the hybrid orbitals lie along the $x$-axis.Consequently,the $\pi$ bonds are formed by the remaining unhybridized $P_y$ and $P_z$ orbitals.This satisfies statement $(i)$.Since the hybrid orbitals are along the $x$-axis,the molecule can lie in any plane containing the $x$-axis,such as the $XY$ plane (statement $ii$) or the $XZ$ plane (statement $iii$).It cannot lie in the $YZ$ plane because the $x$-axis is perpendicular to the $YZ$ plane.Thus,the correct statements are $(i)$,$(ii)$,and $(iii)$.

	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	$0$

Similar Questions

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 $0$

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