Question

Ionic radius is the effective distance from the nucleus of an ion up to which it has its influence on its electron cloud. A cation is always much smaller than the corresponding atom. Further more the number of electrons removed, smaller will be the size of the resulting positive ion. For example $r_{Fe}>r_{F{e}^{2+}}>r_{F{e}^{3+}}$
A cation formed by the loss of electrons may result in the complete disappearance of the outer shell and since, the remaining inner shells do not extend so far in space, the cation is much smaller than the metal atom. For example
$\underset{(2,8,1)}{\mathrm{Na}}\to \underset{(2,8)}{N{a}^{+}}$
In case of isoelectronic ions, the greater the nuclear charge, the greater is the attraction for electrons and smaller is ionic radius
The size of the species $Pb,P{b}^{2+}$ and $P{b}^{4+}$ decreases as

• A. $P{b}^{4+}>P{b}^{2+}>Pb$
• B. $Pb>P{b}^{2+}>P{b}^{4+}$
• C. $P{b}^{4+}>Pb>P{b}^{2+}$
• D. $P{b}^{2+}>P{b}^{4+}>Pb$

Answer 1

Answer: (B)

Higher is the positive charge, smaller is the size of the atom