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(D) To determine the hybridization of the $C$ atom,we count the number of sigma bonds and lone pairs attached to it.$1$. In $HCOOH$ (formic acid),the

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$CH_3CHO$

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(D) To determine the hybridization of the $C$ atom,we count the number of sigma bonds and lone pairs attached to it.$1$. In $HCOOH$ (formic acid),the $C$ atom is bonded to $H$,$OH$,and double-bonded to $O$. It is $sp^2$ hybridized.$2$. In $(H_2N)_2CO$ (urea),the $C$ atom is bonded to two $NH_2$ groups and double-bonded to $O$. It is $sp^2$ hybridized.$3$. In $HCHO$ (formaldehyde),the $C$ atom is bonded to two $H$ atoms and double-bonded to $O$. It is $sp^2$ hybridized.$4$. In $CH_3CHO$ (acetaldehyde),the methyl carbon $(CH_3)$ is bonded to three $H$ atoms and one $C$ atom. Since it has four sigma bonds,it is $sp^3$ hybridized.

	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 which of the following compounds does the $C$ atom use $sp^3$ hybrid orbitals for bonding?

Similar Questions

$H - C \equiv C - C \equiv C - CH_3$
Compare the bond lengths $a$ and $b$.

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 $0$

Among the following compounds,the one that is polar and has a central atom with $sp^2$ hybridization is:

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

In $CO_2$ molecule,the hybridizations of carbon and oxygen atoms are respectively:

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