Question

In which of the following pairs, the first atom or ion is not larger than the second?
(A) $F{{e}^ {2+}}, F{{e}^ {3+}}$
(B) $O, S$
(C) $N, O$
(D) $C{{l}^ {-}}, Cl$

Answer 1

To solve this question refer to the atomic number of the given element so that we can find the electronic configuration of these elements to find their size.

Complete answer:
We have been provided with few pairs of elements,
we need to find in which of these pairs the first element, ion or atom is smaller than the second,
so, for that:
the first pair that we have is: $F{{e}^ {2+}}, F{{e}^ {3+}}$,
so, these are the ions of Fe having positive charge +2 and +3 respectively,
so, the atomic number of Fe is 26,
so, the electronic configuration of $F{{e}^ {2+}} =\left [ Ar \right]3{{d}^ {6}}$
and the electronic configuration of $F{{e}^ {3+}} =\left [ Ar \right]3{{d}^ {5}}$,
so, as in this pair, $F{{e}^ {2+}}$ has a greater number of electrons, so its size would be more than that of $F{{e}^ {3+}}$,
the next pair that we have is: $O, S$,
so, these elements are oxygen and sulphur with atomic number 8 and 16 respectively,
so, the electronic configuration of $O=1{{s}^{2}}2{{s}^{2}}2{{p}^ {4}}$
and the electronic configuration of $S=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^ {4}}$,
so, as in this pair, Sulphur has a greater number of electrons, so its size would be more than that of oxygen,
the next pair that we have is: $N, O$,
so, these elements are nitrogen and oxygen with atomic number 7 and 8 respectively,
so, the electronic configuration of $N=1{{s}^{2}}2{{s}^{2}}2{{p}^ {3}}$
and the electronic configuration of $O=1{{s}^{2}}2{{s}^{2}}2{{p}^ {4}}$,
so, as in this pair, nitrogen has a smaller number of electrons than oxygen but when we talk about the overall size of electron in ${{O}_ {2}}$ is smaller than that of ${{N}_ {2}}$,
the last pair that we have is: $C{{l}^ {-}}, Cl$,
so, in this pair we have an ion of chlorine with a negative charge of -1 and the other is chlorine,
the atomic number of chlorine is 17,
so, the electronic configuration of ${Cl^-} =\left [ Ne \right]3{{s}^{2}}3{{p}^ {6}}$
and the electronic configuration of $Cl=\left [ Ne \right]3{{s}^{2}}3{{p}^ {5}}$,
so, as in this pair, $C{{l}^ {-}}$ has a greater number of electrons, so its size would be more than that of chlorine.

So, from this we can say that the correct option is (B).

Note:
In the case of nitrogen and oxygen, even though oxygen has a higher number of electrons its size is smaller than that of nitrogen. The reason is that in oxygen due to high atomic number the positive charge is more and attraction towards the nucleus is also greater, which results in the smaller size of oxygen.