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

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$(iii)$,$(iv)$

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(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.

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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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