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(B) The given order is $sp  sp^2$ $(3

Vedclass · Accepted Answer

$(iii)$,$(iv)$

Vedclass · Answer

(B) The given order is $sp $(i)$ Electronegativity: It is directly proportional to $s$-character. Since $sp$ $(50\% \ s)$ $> sp^2$ $(33.3\% \ s)$ $> sp^3$ $(25\% \ s)$,the order is $sp > sp^2 > sp^3$. $(ii)$ Bond angle: $sp$ $(180^{\circ})$ $> sp^2$ $(120^{\circ})$ $> sp^3$ $(109.5^{\circ})$. $(iii)$ Size: Higher $s$-character leads to more compact orbitals,hence smaller size. Thus,$sp $(iv)$ Energy level: $s$-orbitals are lower in energy than $p$-orbitals. Higher $p$-character leads to higher energy. Thus,$sp Therefore,$(iii)$ and $(iv)$ follow the given order.

List-$I$	List-$II$
$(1)$ $SF_4$	$(A)$ $sp^3d^2$
$(2)$ $IF_5$	$(B)$ $d^2sp^3$
$(3)$ $NO_2^+$	$(C)$ $sp^3d$
$(4)$ $NH_4^+$	$(D)$ $sp^3$
	$(E)$ $sp$

Choose the correct code of characteristics for the given order of hybrid orbitals of the same atom,$sp < sp^2 < sp^3$:
$(i)$ Electronegativity
$(ii)$ Bond angle between same hybrid orbitals
$(iii)$ Size
$(iv)$ Energy level

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Match the species in List-$I$ with their correct hybrid orbitals in List-$II$.

List-$I$ List-$II$

$(1)$ $SF_4$ $(A)$ $sp^3d^2$

$(2)$ $IF_5$ $(B)$ $d^2sp^3$

$(3)$ $NO_2^+$ $(C)$ $sp^3d$

$(4)$ $NH_4^+$ $(D)$ $sp^3$

$(E)$ $sp$

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Choose the correct code of characteristics for the given order of hybrid orbitals of the same atom,$sp < sp^2 < sp^3$: $(i)$ Electronegativity $(ii)$ Bond angle between same hybrid orbitals $(iii)$ Size $(iv)$ Energy level