Question

The increasing order of atomic radii of the following group $13$ element is:
$A.\,\,Al\, < \,Ga\, < \,In\, < \,Tl$
$B.\,\,Ga\, < \,Al\, < \,In\, < \,Tl$
$C.\,\,Al\, < \,In\, < \,Ga\, < \,Tl$
$D.\,\,Al\, < \,Ga\, < \,Tl\, < \,In$

Answer 1

In periodic table, the metals are present on the left side and non-metals are present on the right side. The metallic character depends upon the ability to lose the electrons. As we move down to the group, the ability to lose the electron increases and as we move across the period, the ability to lose the electron decreases.

Complete step-by-step answer: As we move across the period from the left to right, the atomic size decreases. Since the electron is added in the same shell due to which the atomic size decreases and there is an increase in the valence electrons. Therefore, the tendency to lose electrons from the outermost shell gets tough from the left to right.
The metallic character is depending upon the losing of valence electrons. Due to this reason, the ability of an element to lose valence electrons from top to bottom increases. Whereas, in the case of across the period it decreases due to the decrease in the ability to lose valence electrons.
As we move down to the group the atomic size increases. Since the electron gets added in a new shell due to which the atomic size gets increased. Even the tendency to lose electrons from the outermost valence shell also gets increased as we move down to the group. So, the metallic character gets increased as we move down to the group.
As we move down to the group, the exception is observed. When the electrons enter in the d orbital poor shielding of the nuclear charge by $3d$ electrons $Ga$ has $10d$ electrons which can’t effectively screen the nuclear charge. The electrons in the outermost shell of $Ga$ experience more force of attraction towards the nucleus than $Al$ which decreases atomic radius. Therefore $Ga$ , despite having a higher atomic size, has lower atomic radii than $Al$ .

Therefore, the correct sequence is $Ga < Al < In < Tl$ . Hence the correct option is option (B).

Note: As we move down to the group, there is an increase in atomic size and less energy is required to remove the outermost electron. Whereas, due to the decrease in the atomic size across the period from left to right, the outermost electrons experience the greater nuclear charge and becomes difficult to remove the outermost electrons. As we move in the group the metallic character increases and in the period the metallic character decreases.