Question

The correct order of atomic size of $C,{\text{ }}N,{\text{ }}P,{\text{ }}S$ follows the order -
a) $N{\text{ }} < {\text{ }}C{\text{ }} < {\text{ }}S{\text{ }} < {\text{ }}P$
b) $N{\text{ }} < {\text{ }}C{\text{ }} < {\text{ }}P{\text{ }} < {\text{ }}S$
c) $C{\text{ }} < {\text{ }}N{\text{ }} < {\text{ }}S{\text{ }} < {\text{ }}P$
d) $C{\text{ }} < {\text{ }}N{\text{ }} < {\text{ }}P{\text{ }} < {\text{ }}S$

Answer 1

In a multi-electron atom the valence electron experiences attraction by the nucleus and repulsion by the electrons of inner-shells. This combined effect of attraction and repulsion force acting on the valence-electron, makes the electron experience less attraction from the nucleus. This is called Shielding Effect.

Complete step by step solution:
Shielding is when electrons in the inner electron shells of an atom can shield the outer electrons from the pull of the nucleus. The nucleus can pull the outer electrons in tighter when the attraction is strong and less tight when the attraction is weakened. So, the more shielding occurs.
Shielding effect increases down a group because the nuclear core is farther removed from the valence electrons. Because complete electron shells shield the nuclear charge very effectively. To consider the atomic radius, period by period. Across the Period, from left to right, the atomic radius progressively decreases. The nitrogen atom is larger than the oxygen, which is larger than the fluorine atom.
So the correct order of atomic size of $C,{\text{ }}N,{\text{ }}P,{\text{ }}S$ follows the order
$N{\text{ }} < {\text{ }}C{\text{ }} < {\text{ }}S{\text{ }} < {\text{ }}P$.
On going from $C{\text{ }}to{\text{ }}N$ the nuclear charge increases. But the additional electron is added in the same energy level. Hence, due to shielding effect, the effective nuclear charge increases. The attraction of the nucleus for the valence electrons increases. Hence, the atomic size decreases. The atomic size of $C$ is larger than that of $N.$ For the same reason, the atomic size of $P$ is larger than that of $S$. The atomic size of $S$ is larger than that of $C$ as the electron is added to a new energy level.

**So the option (a) is correct.

Note:**
The size of $S$ and $P$ orbital are very less compared to the $d{\text{ }} \,and\, {\text{ }}f$ orbital because the electrons of $d{\text{ }} and {\text{ }}f$ are spread in large area and hence there is less effective shielding. Structures of the orbital are also responsible for that.