Question

Incorrect order of ionic radius is;
(A) $ La{}^{ + 3} > Gd{}^{ + 3} > Eu{}^{ + 3} > Lu{}^{ + 3} $
(B) $ V{}^{ + 2} > V{}^{ + 3} > V{}^{ + 4} > V{}^{ + 5} $
(C) $ K{}^ + > Sc{}^{ + 3} > V{}^{ + 5} > Mn{}^{ + 7} $
(D) $ In{}^ + > Sn{}^{ + 2} > Sb{}^{ + 3} $

Answer 1

Hint :Ionic size is measured in terms of crystal lattice, which requires a solid form for the compound. We should know about shielding effect, effective nuclear charge, where is the given element placed in the periodic table and its trend.

Complete Step By Step Answer:
The trend of ionic size is, the charge on the nucleus increases and the outermost shell remains the same that means the effective nuclear charge will increase resulting in a decrease in ionic radius. Down the group the shell keeps on adding resulting in an increase in the ionic radius.
In option B, the vanadium has $ 2{}^ + \& 3{}^ + $ oxidation. In $ V{}^{ + 3} $ the effective nuclear charge is more due to the less number of electrons as compared to $ V{}^{ + 2} $ . Hence indicating that it is in the correct order of ionic radius. This shows that $ V{}^{ + 2} $ is greater than $ V{}^{ + 3} $ . Keeping all the points in mind let us see all the options,
Option A talks about f-block elements, this block element has a property called lanthanoid contraction that is when it moves across the period from $ La \to Lu $ the ionic as well as atomic size constantly decreases. So according to the order given in the periodic table, $ Gd{}^{ + 3} $ comes after $ Eu{}^{ + 3} $ indicating that $ Eu{}^{ + 3} $ is greater than $ Gd{}^{ + 3} $ hence option A has the incorrect order of ionic radius.
The correct option is; Option A.

Note :
The greater the charge on the element the more will be its ionic radius. In the f-series because of the not so perfect shielding of f-orbital and more of diffused nature of f-orbital outermost electron will experience more effective nuclear charge and the size will decrease down the group due to more effective nuclear charge.