The species having bond angles of 120^{circ} | vedclass

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Question

(C) To determine the species with a bond angle of $120^{\circ}$,we analyze the hybridization and geometry of each molecule:$1$. $ClF_3$: The central a

Vedclass · Accepted Answer

$BCl_3$

Vedclass · Answer

(C) To determine the species with a bond angle of $120^{\circ}$,we analyze the hybridization and geometry of each molecule:$1$. $ClF_3$: The central atom $Cl$ has $7$ valence electrons. It forms $3$ bonds with $F$ atoms and has $2$ lone pairs. The hybridization is $sp^3d$,and the geometry is $T$-shaped with bond angles less than $90^{\circ}$.$2$. $NCl_3$: The central atom $N$ has $5$ valence electrons. It forms $3$ bonds with $Cl$ atoms and has $1$ lone pair. The hybridization is $sp^3$,and the geometry is pyramidal with bond angles approximately $107^{\circ}$.$3$. $BCl_3$: The central atom $B$ has $3$ valence electrons. It forms $3$ bonds with $Cl$ atoms and has no lone pairs. The hybridization is $sp^2$,and the geometry is trigonal planar with bond angles of $120^{\circ}$.$4$. $PH_3$: The central atom $P$ has $5$ valence electrons. It forms $3$ bonds with $H$ atoms and has $1$ lone pair. The geometry is pyramidal with bond angles close to $93^{\circ}$.Therefore,$BCl_3$ is the correct species.

List-$I$	List-$II$
$(a)$ $SF_6$	$(i)$ $sp^3d^2$
$(b)$ $PCl_5$	(ii) $sp^3d$
$(c)$ $XeF_4$	(iii) $sp^3d^3$
$(d)$ $IF_7$	(iv) $sp^3d^2$

$Column\, I$	$Column\, II$
$(A)\, XeF_6$	$(i)\, \text{Distorted octahedral}$
$(B)\, XeO_3$	$(ii)\, \text{Square planar}$
$(C)\, XeOF_4$	$(iii)\, \text{Pyramidal}$
$(D)\, XeF_4$	$(iv)\, \text{Square pyramidal}$

	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	$0$

The species having bond angles of $120^{\circ}$ is

Similar Questions

Which of the following statements is correct regarding the molecular formula and the hybridization of Xenon in its compound?

Hybridization is due to the overlapping of

Match the following:
List-$I$ List-$II$
$(a)$ $SF_6$ $(i)$ $sp^3d^2$
$(b)$ $PCl_5$ (ii) $sp^3d$
$(c)$ $XeF_4$ (iii) $sp^3d^3$
$(d)$ $IF_7$ (iv) $sp^3d^2$

The correct answer is:

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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