The correct order of atomic radii of the elements $\mathrm{O}, \mathrm{N}, \mathrm{S}$ and $\mathrm{P}$ is

The correct order of atomic radii of the elements $\mathrm{O}, \mathrm{N}, \mathrm{S}$ and $\mathrm{P}$ is
  1. $\mathrm{N} < \mathrm{P} < \mathrm{S} < \mathrm{O}$
  2. $\mathrm{N} < \mathrm{O} < \mathrm{P} < \mathrm{S}$
  3. $\mathrm{O} < \mathrm{N} < \mathrm{P} < \mathrm{S}$
  4. $O < N < S < P$

Solution

The atomic size decreases on moving from left to right in a period as the number of valence electrons in the same shell increases due to which the effective nuclear charge increases. Thus, size of $\mathrm{O}$ is less than $\mathrm{N}$ and also size of $\mathrm{S}$ is less than $\mathrm{P}$. Now, on moving down in a group atomic size increases because each time one new shell is added and electrons in the outermost shell move away from the nucleus. Thus, size of $\mathrm{O} < \mathrm{S}$ and $\mathrm{N} < \mathrm{P}$. Thus, the overall order is $\mathrm{O} < \mathrm{N} < \mathrm{S} < \mathrm{P}$.

Asked in: AP EAMCET 2022 (08 Jul Shift 2)

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