JEE Main 2023 · 15 April · morning · Single correct · +4 / −1
Which of the following expressions is correct in case of a CsCl unit cell (edge length 'a')?
- (a) r_(Cs⁺)+r_(Cl⁻)=√3/2 a
- (b) r_(Cs⁺)+r_(Cl⁻)=a/√2
- (c) r_(Cs⁺)+r_(Cl⁻)=a
- (d) r_(Cs⁺)+r_(Cl⁻)=a/2
Answer and solution
Answer (a) r_(Cs⁺)+r_(Cl⁻)=√3/2 a
Solution
Chemistry ma’am · 4 steps
In CsCl the ions touch along the body diagonal, so half of √3a gives the sum of the radii.
- Picture the cell. CsCl has a body-centred arrangement: Cl⁻ ions sit at the 8 corners and one Cs⁺ ion sits at the body centre. Each ion has 8 nearest neighbours of the opposite charge.
- Where ions touch. The central Cs⁺ touches the corner Cl⁻ ions along the body diagonal, not along an edge or a face diagonal. The body diagonal of a cube of edge a has length √3a.
- Half the diagonal. Going from one corner to the body centre covers half the body diagonal. That distance is one Cl⁻ radius plus one Cs⁺ radius.
r_(Cs⁺) + r_(Cl⁻) = √3a/2
- So. The correct expression is r_(Cs⁺) + r_(Cl⁻) = √3/2a, which is option (a).
The idea to keep. First find the line along which the cation and anion actually touch. Body-centred CsCl touches along the body diagonal; rock-salt NaCl touches along the edge, giving r₊ + r₋ = a/2.
Answer from the NTA final answer key. Solution by PrepWiser.