Question

For (i) I⁻, (ii) Cl⁻, (iii) Br⁻, the increasing order of
nucleophilicity would be
(a) Cl⁻ < Br⁻ < I⁻ (b) I⁻ < Cl⁻ < Br⁻
(c) Br⁻ < Cl⁻ < I⁻ (d) I⁻ < Br⁻ < Cl⁻

Answer 1

Answer:

d) I⁻ < Br⁻ < Cl⁻ order of nucleophilicity

Answer 2

For a polar protic solvent, it's Cl⁻ < Br⁻ < I⁻(Option A) , while for a polar aprotic solvent, its I⁻ < Br⁻ < Cl⁻(Option D)

Explanation:

• The nucleophilicity of the ion depends on its charge to mass ratio.

• Iodide is a much bigger ion which same charge that of a much smaller chloride ion.

• So, iodide is much less potent nucleophile than bromide and lesser than chloride.

• This is the normal scenario in the polar aprotic solvents.

• But this senario changes in the polar protic solvents.

• As the protic solvent can donate hydrogen ions, it can form hydrogen bonds with the ions.

• Iodine being of smaller charge dense, has lesser protic molecules around them, so the shell is smaller.

• The chlorine being more charge dense, has more thick solvent shell around it.

• This makes the effective size of chlorine ion bigger than iodide ion, thereby making it less charge dense.

• So the nucleophilicity of chloride ion becomes lesser than bromide and lesser than iodide.

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