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Correct order of -I effect is
A.$-CN>-NC$
B. $-F>-C{{F}_{3}}$
C.$-OH_{2}^{+}>-OH$
D. $N{{O}_{2}}>-NH_{3}^{+}$

Answer
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Hint: The inductive effect is known as polarisation induced when the different groups are attached to the C-C bond after replacing the hydrogen atom.
When the groups attached are more electronegative and withdraw the sigma bond electrons towards them, it is called negative I effect(-I) examples are- $N{{H}_{4}},CN,CHO\,etc$
When the groups attached are electron donating and they donate the electrons to the sigma bond, then are called positive I effect, examples are $C{{H}_{3}}\,,C{{H}_{2}}\,,C{{H}_{3}},etc$.

Complete step by step answer:
The negative inductive effect will be better shown by a group who is more electronegative and attracts the electron pair towards itself.
In option A, the CN has more -I effect than NC because the carbon draws electrons towards itself due to the presence of resonance and also, it is given in the electronegative series.
In the option B, it is also correct that fluoride anion shows more electronegative effect than carbon tetrafluoride because the dipole moment of $C{{F}_{3}}$ is also zero which makes it less viable to draw electron towards itself.
In the option C, the negative I effect of $O{{H}_{2}}$ should not be greater than OH because of the electronegativity difference between the two atoms.
In option D and according to the series, it is wrong that $N{{O}_{2}}$ has a greater negative effect of Induction rather it is just the opposite of that.
So, the correct answer is “Option A and B”.

Note: The series in decreasing order of -I effect which can help in understanding
$N{{H}_{3}},N{{O}_{2}},CN,S{{O}_{3}}H,CHO,CO,COOH,COCl,COOR,CON{{H}_{2}},F,Cl,Br,I,OR,OH,N{{H}_{2}},{{C}_{6}}{{H}_{5}},H$