Question

Which of the following options is the correct order of atomic radii?
A. \[Na < Be < B\]
B. \[{F^ - } < {O^{2 - }} < {N^{3 - }}\]
C. \[Na < Li < K\]
D. \[F{e^{3 + }} < F{e^{2 + }} < F{e^{4 + }}\]

Answer 1

Hint: Atomic radii increases down the group due to increase in number of shells and decreases along the period due to increases in nuclear charge. When the atoms have charge on then we'll check atomic radius by force of nucleus towards the electron or we can remember it by, nucleus repel the positive charge and attract the negative charge on the atom.

Complete step by step answer:
As we know atomic radius decreases in a period as we move left to right because, across a period, effective nuclear charge increases because electron shielding is constant. A higher effective nuclear charge causes greater attractions to the electrons and pulls the electrons closer to the nucleus which results in a smaller atomic radius and the number of energy levels increases as we move down a group as the number of electrons is increasing. Each subsequent energy level is further from the nucleus than the last. Therefore, the atomic radius increases as the group and energy levels increase.
Our 1st option is \[Na < Be < B\]. Be and B are of the same period and Be>B as B is on the right. So, this option is wrong. Our next option is \[{F^ - } < {O^{2 - }} < {N^{3 - }}\]. As we can see, these three are isoelectronic. The ionic radii of isoelectronic species decreases with increase in atomic number because the magnitude of nuclear charge increases with increase in atomic number. So the sequence will be \[{F^ - } < {O^{2 - }} < {N^{3 - }}\]. So, this option is correct. Option 3rd is \[Na < Li < K\]. As we know, these three are of the same group. So, the sequence will be \[Na < Li < K\]. So, this sequence is wrong. Option D is \[F{e^{3 + }} < F{e^{2 + }} < F{e^{4 + }}\]. As we can see in these three species no. The electron is decreasing with the same nucleus. So, the ion having less no. of electrons will be smallest and the order will be \[F{e^{3 + }} < F{e^{2 + }} < F{e^{4 + }}\] . So, this option is also wrong.
So, the correct option is B. \[{F^ - } < {O^{2 - }} < {N^{3 - }}\].

Note:
We can measure the ionic radius by measuring the distance between the nucleus and the electron in the outermost shell in which the last electron is present.