For the reaction K + Br₂ → KBr + Br, which follows the harpoon mechanism, the reactive cross section is closest to

(Use \(\rm\frac{e^2}{4 \pi \varepsilon_0}\) = 2.3 × 10⁻²⁸ J m, Ionization energy of K = 422.5 kJ mol⁻¹, electron affinity of Br₂ = 250 kJ mol⁻¹ and N_A = 6 × 10²³ mol⁻¹)

Concept:

Harpoon mechanism:-

One electron from the metal atom jumps to the nonmetal atom when two reactant molecules—one metal and one nonmetal—approach each other at a distance R (the radius of the cross-section where the collision occurs). The process is known as the Harpoon mechanism, and that electron is known as the Harpoon. Following that, non-metals become negatively charged ions, whereas metals become positively charged ions.

     K + Br2 → KBr + Br

When K approaches Br the valence electron(Harpoon) of K moves into Br.

This mechanism involves the following interactions;

• Ionization Energy (\(K\rightarrow K^{+} + e^{-}\) )
• Electron affinity (\(Br + e^{-}\rightarrow Br^{-}\) )
• Coulombic interaction (\(-\frac{e^2}{4\pi \epsilon _0 R}\) , R= separation between K⁺  and Br^-)

So the reaction will be energetically favorable when 

\(I.E-E.A-\frac{e^2}{4\pi\epsilon _0 R}=0\)

Example:

  K + Br2 → KBr + Br

\(K\rightarrow K^+ \Rightarrow I.E\)

\(Br_2\rightarrow 2Br^-\Rightarrow E.A\)

According to the theory,

\(I.E.-E.A.=\frac{e^2}{4\pi \epsilon _0 R}\) , (where R= radius of cross-section in which reaction occurs,  r₁ and r₂ = radius of each reactant molecule

 \(R> r_1+r_2\) )

Explanation:-

\(\Rightarrow R= \frac{e^2}{4\pi \epsilon _0} \times\frac{1}{I.E-E.A.}\)

=\(2.3\times 10^{-28}Jm\: \times\frac{1}{(422.5-250){KJ.mol^{-1}}}\)

=\(2.3\times 10^{-28}Jm\: \times\frac{1\times6.023\times 10^{23}}{(422.5-250) \times10^{3}J}\)

= \(8.03\times10^{-10}\)

Area of cross-section

= \(\pi R^2\)

= \(\pi (8.03\times10^{-10})^{2}\)

= 2 × 10−18 m2

Conclusion:-

Hence the correct answer is 2 × 10−18 m2.