Match List-I with List-II : List-I (Spec | vedclass

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(D) To determine the hybridisation,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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$(a)-(iii), (b)-(i), (c)-(v), (d)-(iv)$

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(D) To determine the hybridisation,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 of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.$(a)$ $SF_{4}$: $SN = \frac{1}{2} (6 + 4) = 5$. This corresponds to $sp^{3}d$ hybridisation $(iii)$.$(b)$ $IF_{5}$: $SN = \frac{1}{2} (7 + 5) = 6$. This corresponds to $sp^{3}d^{2}$ hybridisation $(i)$.$(c)$ $NO_{2}^{+}$: $SN = \frac{1}{2} (5 + 0 - 1) = 2$. This corresponds to $sp$ hybridisation $(v)$.$(d)$ $NH_{4}^{+}$: $SN = \frac{1}{2} (5 + 4 - 1) = 4$. This corresponds to $sp^{3}$ hybridisation $(iv)$.Thus,the correct match is $(a)-(iii), (b)-(i), (c)-(v), (d)-(iv)$.

List-$I$ (Species)	List-$II$ (Hybrid Orbitals)
$(a)$ $SF_{4}$	$(i)$ $sp^{3}d^{2}$
$(b)$ $IF_{5}$	$(ii)$ $d^{2}sp^{3}$
$(c)$ $NO_{2}^{+}$	$(iii)$ $sp^{3}d$
$(d)$ $NH_{4}^{+}$	$(iv)$ $sp^{3}$
	$(v)$ $sp$

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

List-$I$ (Species) List-$II$ (Hybrid Orbitals)

$(a)$ $SF_{4}$ $(i)$ $sp^{3}d^{2}$

$(b)$ $IF_{5}$ $(ii)$ $d^{2}sp^{3}$

$(c)$ $NO_{2}^{+}$ $(iii)$ $sp^{3}d$

$(d)$ $NH_{4}^{+}$ $(iv)$ $sp^{3}$

$(v)$ $sp$

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Match List-$I$ with List-$II$ : List-$I$ (Species) List-$II$ (Hybrid Orbitals) $(a)$ $SF_{4}$ $(i)$ $sp^{3}d^{2}$ $(b)$ $IF_{5}$ $(ii)$ $d^{2}sp^{3}$ $(c)$ $NO_{2}^{+}$ $(iii)$ $sp^{3}d$ $(d)$ $NH_{4}^{+}$ $(iv)$ $sp^{3}$ $(v)$ $sp$ Choose the correct answer from the options given below :