In the following,which species does not conta | vedclass

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(D) To determine the hybridization,we calculate the steric number $(SN)$ using the formula: $SN = \frac{1}{2} (V + M - C + A)$,where $V$ is the number

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$CO_2$

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(D) To determine the hybridization,we calculate the steric number $(SN)$ using the formula: $SN = \frac{1}{2} (V + M - C + A)$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.$1$. For $SO_4^{2-}$: $SN = \frac{1}{2} (6 + 0 - 0 + 2) = 4$,which corresponds to $sp^3$ hybridization.$2$. For $CH_4$: $SN = \frac{1}{2} (4 + 4 - 0 + 0) = 4$,which corresponds to $sp^3$ hybridization.$3$. For $H_2O$: $SN = \frac{1}{2} (6 + 2 - 0 + 0) = 4$,which corresponds to $sp^3$ hybridization.$4$. For $CO_2$: $SN = \frac{1}{2} (4 + 0 - 0 + 0) = 2$,which corresponds to $sp$ hybridization.Therefore,$CO_2$ does not contain $sp^3$ hybridization.

Column $I$ (Molecule/ion)	Column $II$ (Hybridisation)
$A$. $H_3O^{+}$	$P$. $sp$
$B$. $NH_2^{-}$	$Q$. $sp^2$
$C$. $NO_3^{-}$	$R$. $sp^3$
$D$. $ClF_3$	$S$. $sp^3d$

In the following,which species does not contain $sp^3$ hybridization?

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The correct order of electronegativity of carbon in various hybridization states:

Which of the following compounds has the maximum $C-C$ single bond length?

The hybridisation of the central atom and the shape of $[IO_2F_5]^{2-}$ ion,respectively are

Identify the correct match:

Column $I$ (Molecule/ion) Column $II$ (Hybridisation)

$A$. $H_3O^{+}$ $P$. $sp$

$B$. $NH_2^{-}$ $Q$. $sp^2$

$C$. $NO_3^{-}$ $R$. $sp^3$

$D$. $ClF_3$ $S$. $sp^3d$

Assertion : Among the carbon allotropes,diamond is an insulator,whereas,graphite is a good conductor of electricity.
Reason : Hybridization of carbon in diamond and graphite are $sp^3$ and $sp^2$,respectively.

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