Question

Ionic radii of

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

Answer 1

The correct option is B

${}^{35}C{l}^{-}<{}^{37}C{l}^{-}$

Lets take a look at all the options.

(a)$T{i}^{4+}\Rightarrow 18electrons,22protons$ but will have same configuration as $Ar$.

$M{n}^{7+}\Rightarrow 10electrons,17protons$ but will have same electronic configuration as $Ne$. $Ar$ has one shell more than $Ne$.

$\therefore$ $T{i}^{4+}>M{n}^{7+}$

(b)${}^{35}C{l}^{-}<{}^{3}7C{l}^{-}$

Because of isotopic effect.

(c)$K^{+}\to 18electrons,19protons$
But has an electronic configuration of $Ar$

$C{l}^{-}\to 18electrons,17protons$

But also has an electronic configuration of $Ar$

We know that in $K^{+}$ $19protons$ are attracting $18electrons$ while in $C{l}^{-}$ $17protons$ are attracting $18electrons$. Therefore, ionic radii $\Rightarrow K^{+}<C{l}^{‑}$.

(d) $P^{3+}>P^{5+}$

$P^{3+}\to 12electrons,15protons$

$P^{5+}\to 10electrons,15protons$.

$P^{3+}>P^{5+}$

Because $15protons$ will be able to attract $10electrons$ more, rather than $15electrons$.