The increasing order of O-N-O bond angle in t | vedclass

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(D) The bond angles for the given species are as follows:$1.$ $NO_2^+$: It has $sp$ hybridization and a linear geometry, resulting in a bond angle of

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None of these

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(D) The bond angles for the given species are as follows:$1.$ $NO_2^+$: It has $sp$ hybridization and a linear geometry, resulting in a bond angle of $180^\circ$.$2.$ $NO_2$: It has $sp^2$ hybridization with one unpaired electron, resulting in a bond angle of approximately $134^\circ$.$3.$ $NO_2^-$: It has $sp^2$ hybridization with one lone pair, resulting in a bond angle of approximately $115^\circ$.Comparing these values, the increasing order of the bond angle is $NO_2^- Since this specific order is not listed in options $A$, $B$, or $C$, the correct answer is $D$.

List-$I$ (Molecule)	List-$II$ (Shape)
$A$. $BrF_5$	$I$. $T$-shape
$B$. $H_2O$	$II$. See-saw
$C$. $ClF_3$	$III$. Bent
$D$. $SF_4$	$IV$. Square pyramidal

Molecule	$P-X$ (axial) bond length / $P-X$ (equatorial) bond length
$PF_5$	$a / b$
$PF_4CH_3$	$c / d$
$PF_3(CH_3)_2$	$e / f$
$PCl_5$	$g / h$

The increasing order of $O-N-O$ bond angle in the species $NO_2$, $NO_2^+$ and $NO_2^-$ is

Similar Questions

The shape of a molecule of $NH_3$,in which the central atom contains a lone pair of electrons,is

Match List-$I$ with List-$II$:

List-$I$ (Molecule) List-$II$ (Shape)

$A$. $BrF_5$ $I$. $T$-shape

$B$. $H_2O$ $II$. See-saw

$C$. $ClF_3$ $III$. Bent

$D$. $SF_4$ $IV$. Square pyramidal

In which of the following molecules/ions are all the bond angles not equal?

Consider the following information $(X = F$ or $Cl)$:

Molecule $P-X$ (axial) bond length / $P-X$ (equatorial) bond length

$PF_5$ $a / b$

$PF_4CH_3$ $c / d$

$PF_3(CH_3)_2$ $e / f$

$PCl_5$ $g / h$

According to the given information,choose the incorrect order of bond lengths.

The geometry of the ammonia molecule can be best described as:

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