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Question

(A) To determine the hybridisation,we use the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons,$M$ is the number

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$A-R, B-R, C-Q, D-S$

Vedclass · Answer

(A) To determine the hybridisation,we use the formula: $H = \frac{1}{2}(V + M - C + A)$,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.$1$. $H_3O^{+}$: $H = \frac{1}{2}(6 + 3 - 1 + 0) = 4$ ($sp^3$ hybridisation).$2$. $NH_2^{-}$: $H = \frac{1}{2}(5 + 2 - 0 + 1) = 4$ ($sp^3$ hybridisation).$3$. $NO_3^{-}$: $H = \frac{1}{2}(5 + 0 - 0 + 1) = 3$ ($sp^2$ hybridisation).$4$. $ClF_3$: $H = \frac{1}{2}(7 + 3 - 0 + 0) = 5$ ($sp^3d$ hybridisation).Thus,the correct match is $A-R, B-R, C-Q, D-S$.

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

Identify the correct match:

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

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

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

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

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

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