Question

Which of the following orders are correct with respect to their size?

(1) $A{l}^{3+}<M{g}^{2+}<N{a}^{+}<F^{-}$

(2) $T{e}^{2-}>I^{-}>C{s}^{+}>B{a}^{2+}$

(3) $F{e}^{3+}<F{e}^{2+}<F{e}^{4+}$

(4) $Mg>Al>Si>P$

• A. $\left(1\right),\left(2\right)$ & $\left(4\right)$
• B. $\left(2\right),\left(3\right)$ & $\left(4\right)$
• C. $\left(1\right),\left(3\right)$ & $\left(4\right)$
• D. $\left(1\right),\left(2\right),\left(3\right)$ & $\left(4\right)$
• E. $\left(2\right)$ & $\left(4\right)$

Answer 1

The correct option is A $\left(1\right),\left(2\right)$ & $\left(4\right)$

(1) $M{g}^{2+},N{a}^{+},A{l}^{3+}$ and $F^{-}$ are isoelectronic and thus follow the order ${}_{13}A{l}^{3+}{<}_{12}M{g}^{2+}{<}_{11}N{a}^{+}{<}_9{F}^{-}$.

(2) Atomic number : $Te=52;I=53;Cs=55;Ba=56.$ All the ions given are isoelectronic and have the xenon electron configuration.

Ionic radius$\propto \frac{1}{nuclearcharge}$
Hence, the given order is correct.

(3) More the positive charge on the atom, less is the size.
The correct order would be

$F{e}^{+2}<F{e}^{+3}<F{e}^{+4}$

Option 3 is incorrect.

(4) Across the period, the size of atoms decreases. $Mg=160$ pm; $Al=143$ pm; $Si=118$; $P=110$ pm.

Option A is the correct answer.