Question

Ionic radius of:

• A. $T{i}^{4+}<M{n}^{7+}$
• B. ${}^{35}C{l}^{-}{<}^{37}C{l}^{-}$
• C. $K^{+}>C{l}^{-}$
• D. $P^{3+}>P^{5+}$

Answer 1

The correct option is D $P^{3+}>P^{5+}$

(a) Atomic radius of $T{i}^{4+}>M{n}^{7+}$, because as we move from left to right of the periodic table, atomic radius decreases and also having greater charge will further decrease the atomic radius of $M{n}^{7+}$.

(b) ${}^{35}Cl$ and ${}^{37}Cl$ are the isotopes, which differ by the number of neutrons only, thus will have similar atomic radius.

(c) The atomic radius of $K^{+}$ ion is less than that of $C{l}^{-}$ ion.

(d) Larger the charge on the ion, smaller will be the size of ion for the same element. For $P^{3+}$, there are 15 protons for 12 electrons while for $P^{5+}$, there are 15 protons for 10 electrons. Thus in $P^{5+}$ the valence electrons are more strongly attracted by the nucleus than those of $P^{3+}$. As a result, there is a larger contraction in the size of $P^{5+}$ than that of $P^{3+}$. Hence $P^{3+}$ is bigger than $P^{5+}$.