Question

The decreasing order of -I effect of the following hybridisation of an atom is:
I. $sp$ II. $s{p}^2$ III. $s{p}^3$

• A. I > II > III
• B. III > II > I
• C. I > III > II
• D. II > III > I

Answer 1

The correct option is A I > II > III

More the % s-character means more electronegativity of the attached carbon, so more will be -I. So in between $sp,s{p}^{2}ands{p}^3$, sp is the most electronegative because of more $\%$ s-character So, the negative inductive order is $sp\left(50\%s\right)>s{p}^2\left(33\%s\right)>s{p}^3\left(25\%s\right).$