The correct order regarding the electronegativity of hybrid orbitals of carbon is

Question

The correct order regarding the electronegativity of hybrid orbitals of carbon is (A) s p < s p2 < s p3 (B) s p >s p2 < s p3 (C) s p>s p2>s p3 (D) s p < s p2 > s p3

Tardigrade Admin · Accepted Answer

More then s-character more closer towards the nucleus more will be the electronegativity

Q. The correct order regarding the electronegativity of hybrid orbitals of carbon is

$s p < s p^{2} < s p^{3}$

$s p > s p^{2} < s p^{3}$

$s p > s p^{2} > s p^{3}$

$s p < s p^{2} > s p^{3}$

More then s-character more closer towards the nucleus more will be the electronegativity

Q. The correct order regarding the electronegativity of hybrid orbitals of carbon is

sp<sp2<sp3

sp>sp2<sp3

sp>sp2>sp3

sp<sp2>sp3

More then s-character more closer towards the nucleus more will be the electronegativity

$s p < s p^{2} < s p^{3}$

$s p > s p^{2} < s p^{3}$

$s p > s p^{2} > s p^{3}$

$s p < s p^{2} > s p^{3}$