Question

Arrange \[S,\,Al,\,Ar,\,Mg,\,P\] according to decreasing order of atomic radii.
(i) \[Mg > Al > P > S > Ar\]
(ii) $Ar > S > P > Al > Mg$
(iii) $Mg > S > Al > P > Ar$
(iv) $Ar > P > S > Al > Mg$$$

Answer 1

Hint:All the elements given in the question belong to the second period. As they all belong to the same period the atomic number gradually increases across the period. Correspondingly there is an increase in the number of protons and electrons. Check where the extra electrons and protons are added and find the trend in the atomic radii.

Complete answer:
The atomic number of $S$is $16$ and its electronic configuration is $\left[ {Ne} \right]3{s^2}3{p^4}$.
Hence $S$ is an element of Second Period, Group$16$.
The atomic number of $Al$is $13$ and its electronic configuration is $\left[ {Ne} \right]3{s^2}3{p^1}$.
Hence $Al$ is an element of Second Period, Group$13$.
The atomic number of $Ar$is $18$ and its electronic configuration is $\left[ {Ne} \right]3{s^2}3{p^6}$.
Hence $Ar$ is an element of Second Period, Group$18$.
The atomic number of $Mg$is $12$ and its electronic configuration is $\left[ {Ne} \right]3{s^2}$.
Hence $Mg$ is an element of Second Period, Group$2$.
The atomic number of $P$is $15$ and its electronic configuration is $\left[ {Ne} \right]3{s^2}3{p^3}$.
Hence $P$ is an element of Second Period, Group$15$.
Therefore we can see the given elements belong to different groups but to the same period.
As we move across a period the atomic number increases which results in corresponding increase in the number of protons and electrons. As the number of protons increases inside the nucleus the effective nuclear charge increases. But since the electrons are added to the same outer shell the increase in effective nuclear charge is much more pronounced. So with the increase in atomic number due to greater effective nuclear charge, the nucleus pulls the electrons towards itself to a much more extent resulting in a corresponding decrease in atomic radii.
Since the atomic number of $Ar > S > P > Al > Mg$.
Therefore the decreasing order of atomic radii is \[Mg > Al > P > S > Ar\].
Hence the correct answer is (i) \[Mg > Al > P > S > Ar\].

Note:Try to learn the periodic table for these kinds of questions as you need to know the atomic number of the given elements correctly. If the atomic number goes wrong you would interpret wrongly and mark the wrong option.