Taking Bohr's radius as r0 = 53 pm, the ground state radius of Li++ ion, on the basis of Bohr's model, will be about
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 18 pm
In the Bohr model the orbit radius for a one-electron ion of atomic number Z is
\( \displaystyle r_n = \frac{r_0 n^{2}}{Z} \)
Li++ is lithium (Z = 3) with only one electron left, so the Bohr model applies. For the ground state, n = 1:
\( \displaystyle r_1 = \frac{53\,\mathrm{pm} \times 1^{2}}{3} \)
\( \displaystyle r_1 \approx 17.7\,\mathrm{pm} \approx 18\,\mathrm{pm} \)
Note: This question is from the NCERT Exemplar. NTA sets numerical questions from the Exemplar too, so do not ignore it while practising numericals.
NCERT Exemplar:
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