Question

The ionic radii of ${Li}^{+}$ ${Be}^{2+}$ and $B^{3+}$ follow the order:

• A. ${Be}^{2+}>B^{3+}>{Li}^{\ast }$
• B. ${Li}^{+}>B^{3+}>{Be}^{2+}$
• C. $B^{3+}>B{e}^{2+}>L{i}^{+}$
• D. $L{i}^{+}>B{e}^{2+}>B^{3+}$

Answer 1

Answer: (D)

$L{i}^{+}>B{e}^{2+}>B^{3+}$

While the $L{i}^{+}$ will be the largest because the nuclear charge on it (+1) is least out of the three. $B{e}^{2+}$ will have the size in between the two cations. Therefore, the order of ionic radius is ​$B^{3+}<B{e}^{2+}<L{i}^{+}$ even though the three cations have equal no. of electrons

Hence option D is correct