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

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$A-4, B-1, C-2, D-3$

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(A) To find the hybridisation,we use the formula: $	ext{Steric number} = \frac{1}{2} [V + M - C + A]$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.$A. NO_3^-$$\frac{1}{2}(5 + 0 - 0 + 1) = 3 ightarrow sp^2$$B. NH_2^-$$\frac{1}{2}(5 + 2 - 0 + 1) = 4 ightarrow sp^3$$C. SCN^-$$\frac{1}{2}(4 + 0 - 0 + 2) = 3$ (Note: For $SCN^-$,central $C$ has $2$ sigma bonds,$sp$ hybridisation)$D. ICl_2^-$$\frac{1}{2}(7 + 2 - 0 + 1) = 5 ightarrow sp^3d$Correct matching is $A-4, B-1, C-2, D-3$.

Match the following ions with the corresponding hybridisation of their central atoms.
$A. NO_3^-$ $1. sp^3$
$B. NH_2^-$ $2. sp$
$C. SCN^-$ $3. sp^3d$
$D. ICl_2^-$ $4. sp^2$

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$A. NO_3^-$	$1. sp^3$
$B. NH_2^-$	$2. sp$
$C. SCN^-$	$3. sp^3d$
$D. ICl_2^-$	$4. sp^2$

Match the following ions with the corresponding hybridisation of their central atoms.$A. NO_3^-$$1. sp^3$$B. NH_2^-$$2. sp$$C. SCN^-$$3. sp^3d$$D. ICl_2^-$$4. sp^2$