Question

Which amongst the following atoms has the smallest size?
(A) $Na$
(B) $Al$
(C) $N{a^ + }$
(D) $A{l^{3 + }}$

Answer 1

The size here refers to the atomic size. The atomic size of an element is defined as the distance of the center of the nucleus to the outermost electron of the orbital. The atomic size of the element is changed when the electron is added or removed from the isolated gaseous or neutral atom.

Complete answer:

Now, we have the basic knowledge of the atomic size of elements. First, we will discuss some factors which affect the atomic size of the element. One of the basic and important factors which affect the atomic radius is the effective nuclear charge. The effective nuclear charge of an element or atom depends on the total number of electrons present in the outermost electron or the total number of the electron that atom can hold in its outermost orbital.
Now, we are back to our question. We have understood that the atomic size of an atom or element depends on the number of protons and electrons present in an atom. According to Fajan’s rule, it was observed that the size of the cation is smaller than anions. As we know that cation is formed when electrons are removed from an atom.
So, we can conclude that $N{a^ + }$ and $A{l^{3 + }}$ are smaller in size as compared to $Na$ and $Al$. We know that the greater the charge on the cation, the smaller will be its size. So when we compare $N{a^ + }$ and $A{l^{3 + }}$ the charge over aluminium is greater than sodium. Hence, the smallest size will be $A{l^{3 + }}$ .

Therefore, the correct option is (D).

Note: In general, the atomic size of atoms follows the order $anion > neutral > cation$. But it is valid only when the number of electrons in atoms is different. For isoelectronic species having the same number of electrons, we use effective nuclear charge.