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NCERT Exemplar · Q29

Q.Assertion (A): Among halogens fluorine is the best oxidant.
Reason (R): Fluorine is the most electronegative atom.

(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) Both A and R are false.
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Both statements are individually true — fluorine IS the best oxidant among halogens,

and it IS the most electronegative atom — but electronegativity is not the reason

for its oxidising supremacy. That is set by the thermodynamics of

12F2+e−→F−(aq)\frac{1}{2}\text{F}_2 + e^- \to \text{F}^-(aq): a weak F–F bond plus the enormous

hydration enthalpy of the tiny F⁻ ion. The correct option is (ii).

Evaluating the Assertion

Oxidising power in solution is measured by the standard reduction potential:

  • F2/F−\text{F}_2/\text{F}^-: +2.87+2.87 V — the highest of all the halogens (Cl2\text{Cl}_2: +1.36+1.36 V, Br2\text{Br}_2: +1.09+1.09 V, I2\text{I}_2: +0.54+0.54 V).

Fluorine therefore has by far the strongest tendency to be reduced: Assertion (A) is true.

Evaluating the Reason

Fluorine tops the Pauling electronegativity scale (≈ 4.0) — the most electronegative

element there is. Reason (R) is true.

Is R the correct explanation of A?

Electronegativity describes an atom's pull on a shared pair inside a bond — a useful

qualitative signal, but not what fixes the reduction potential. The potential of

12X2+e−→X−(aq)\frac{1}{2}X_2 + e^- \to X^-(aq) is set by three thermodynamic terms:

  1. Bond dissociation enthalpy of X₂ (energy to free the atom),
  2. Electron gain enthalpy of X (energy released on adding the electron),
  3. Hydration enthalpy of X⁻ (energy released on solvating the ion).

For fluorine the decisive contributions are the unusually weak F–F bond (lone-pair

repulsion between the small atoms) and the exceptionally large hydration enthalpy of …

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