Question

The atomic radii of Li, F, Na and Si follow the order:
A. $Si>Li>Na>F$
B. $Li>F>Si>Na$
C. $Na>Si>F>Li$
D. $Na>Li>Si>F$

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.

Complete step by step answer:
In this question we have asked the sequence of atomic radius of $Li, F, Na, Si$.
As we know $Li$ and $Na$ are of the same group. $Li$ and $F$ are of the same period and $Na$ and $Si$ are of the same period. 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. As $Li$ and $Na$ are of the same group and $Li$ is above $Na$ so $Li$<$Na$. $Li$ and $F$ are of the same period and $Li$ is present left to $F$. So, $Li>F$. Same goes for $Na$ and $Si$. $Na>Si$, as $Na$ is left to $Si$. Now there is no relation between $Li$ and $Si$ and which have a bigger radius. Now we'll look at our option. The correct option is either $Na>Si>Li>F$ or $Na>Li>Si>F$. We have only one in the options and the correct option is D. $Na>Li>Si>F$.

Therefore, the correct answer is option D.

Note:
If we have both $Na>Si>Li>F$ and $Na>Li>Si>F$ as options then which one will we choose. So, the only way to solve such a question is you have to remember that $Li>Si$ or we can say $Si$ is of the 14 group while $Li$ is of the 1st group. So, $Li$ is larger.