The correct order regarding the electro negativity of hybrid orbital's of carbon is

sp<sp²<sp³

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sp<sp²> sp³

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sp>sp²>sp³

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sp>sp²<sp³

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Solution

The correct option is C
sp>sp²>sp³

With the increase in % of s - character in hybrid orbitals the electro negativity of the hybrid orbital's increase.

sp - 50% s character, ${s p}^{2}$ - 33.3 % s character

${s p}^{3}$ - 25% s character.

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C_{2} H_{2} - -- \to C_{2} H_{4} - -- \to C_{2} H_{6}

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