Question

Which of the following is correct order of size of the given species?

• A. $I>I^{-}>I^{+}$
• B. $I^{+}>I^{-}>I$
• C. $I>I^{+}>I^{-}$
• D. $I^{-}>I>I^{+}$

Answer 1

The correct option is D $I^{-}>I>I^{+}$

For any neutral species $a$, the order of radius of corresponding anion neutral species and the anion is,

$a^{-}>a>a^{+}$

Explanation: For any anionic species the no of electrons is greater than the no of protons. Hence the effective nuclear attractive force towards the valence electrons will be less than the neutral atom. Hence size will be maximum.

For any cationic species, the no of electrons is lesser than the no of protons. Hence the effective nuclear attractive force towards the valence electrons will be greater than the neutral atom. So the atomic size or radius will be minimum.

So the correct sequence of radius or size will be,

$I^{-}>I>I^{+}$. Option D