Question

The correct order of radii is:
(a)- $N$ <$Be$ <$B$
(b)- ${{F}^{-}}$ <${{O}^{2-}}$ <${{N}^{3-}}$
(c)- $Na$ <$Li$ <$K$
(d)- $F{{e}^{3+}}$ <$F{{e}^{2+}}$ <$F{{e}^{4+}}$

Answer 1

Hint: In isoelectronic species, as the nuclear charge increases, the nucleus pulls the electrons due to which the radii decrease. As we move down the group the atomic radii increases and along the period it decreases.

Complete step by step answer:
As we move down the group the atomic radii increases and along the period it decreases.
Isoelectronic species or ions are the ions of different elements that differ in the magnitude of the nuclear charge but have the same number of electrons.
(a)- $N$ <$Be$ <$B$
They all belong to the same period. As we move along the period the atomic size decreases, because of the nuclear charge increases. The atomic number of nitrogen is 7, the atomic number of beryllium is 4, and the atomic number of boron is 5.
The order of radii will be: $Be$ <$B$ <$N$
(b)- ${{F}^{-}}$ <${{O}^{2-}}$ <${{N}^{3-}}$
These all are isoelectronic species. They all have 10 electrons. As the negative charge increases, the atomic size increases. The order of radii will be: ${{F}^{-}}$ <${{O}^{2-}}$ <${{N}^{3-}}$
(c)- $Na$ <$Li$ <$K$
They belong to the same group and as we move down the group the atomic radii increases due to the increase in the number of shells. The order of radii will be: $Li$ <$Na$ <$K$.
(d)- $F{{e}^{3+}}$ <$F{{e}^{2+}}$ <$F{{e}^{4+}}$
As the positive charge increases on the atom, the atomic radius decreases because the number of electrons decreases. The order of radii will be: $F{{e}^{4+}}$ <$F{{e}^{3+}}$ <$F{{e}^{2+}}$.

So, the correct answer is an option (b)- ${{F}^{-}}$ <${{O}^{2-}}$ <${{N}^{3-}}$


Note: The atomic radius of noble gas in a period is the largest because it has a complete octet. It must be noted that as the positive charge on the cation increases the size decreases and as the negative charge on the anion increases the size increases.