Question

From the given set of species, point out the species from each set having least atomic radius :

(a) $O^{2-},F^{-},N{a}^{+}$

(b) $Ni,Cu,Zn$

(c) $Li,Be,Mg$

(d) $He,L{i}^{+},H^{-}$

• A. $O^{-2},Cu,Li,H^{-}$
• B. $N{a}^{+},Ni,Be,L{i}^{+}$
• C. $F^{-},Zn,Mg,He$
• D. $N{a}^{+},Cu,Be,He$

Answer 1

The correct option is B $N{a}^{+},Ni,Be,L{i}^{+}$

(a) The species $O^{2-},F^{-},N{a}^{+}$ are isoelectronic species. Each has 10 electrons.

The effective nuclear charge for $N{a}^{+}$ is maximum. Hence, $N{a}^{+}$ has a least atomic radius.

(b) $Ni,Cu,Zn$ belong to 3d transition series. The atomic radii of $Ni,Cu$ and $Zn$ are 125, 128 and 137 pm respectively. Thus, $Ni$ has a lowest atomic radius.

(c) Among $Li,Be,Mg$ the element with lowest atomic radius is $Be$.
On moving down the group, the atomic radius increases and on moving from left to right in a period, the atomic radius decreases.

(d) The species $He,L{i}^{+},H^{-}$ are isoelectronic species. Each has 2 electrons.

The effective nuclear charge for $L{i}^{+}$ is maximum. Hence, $L{i}^{+}$ has a least atomic radius.

Therefore, the correct option is B.