The radius of the 2ⁿᵈ orbit of Li²⁺ is x. The expected radius of the 3ʳᵈ orbit of Be³⁺ is

Solution

Chemistry ma’am · 3 steps

Bohr radius goes as n²/Z, so set up a ratio and let the constant 0.529 Å cancel.

1. Bohr radius formula. For a one-electron species, the radius of the nth orbit is rₙ = 0.529 × n²/Z Å. A bigger n makes the orbit larger. A bigger nuclear charge Z pulls it in.

   r_(Be³⁺)/r_(Li²⁺) = (n₂²/Z₂)/(n₁²/Z₁) = (3²/4)/(2²/3)

2. Put in values. For Li²⁺, Z = 3 and n = 2, so n²/Z = 4/3. For Be³⁺, Z = 4 and n = 3, so n²/Z = 9/4. The ratio is 9/4 × 3/4 = 27/16.
3. So. The radius of the 3rd orbit of Be³⁺ is 27/16x, which is option (d).

The idea to keep. For hydrogen-like species, r ∝ n²/Z. Comparing two species means taking the ratio of n²/Z for each, and the constant 0.529 Å drops out.

Answer from the NTA final answer key. Solution by PrepWiser.