Question

Atomic radii of Fluorine and Neon in Angstrom unit are respectively given by

• A. $1.60,1.60$
• B. $0.75,1.60$
• C. $1.60,0.72$
• D. $0.72,0.72$

Answer 1

The correct option is B

$0.75,1.60$

The correct option is B

The explanation for the correct option:

1. Atomic radius, also known as atomic radii, is the total distance between an atom's nucleus and the outermost orbital of its electron
2. An element's atomic radius is entirely dependent on its electronic configuration
3. We can check the electron's orbital occupancy and the element's radius measurement condition using the electronic configuration.
4. The atomic number of Fluorine is $9$ and Neon is$10$. The electronic configuration of Fluorine is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^5$ and for Neon is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^6$
5. From the electronic configuration of Fluorine there are five electrons in the 2p orbital. The p orbital needs six electrons to form the stable configuration.
6. As a result, Fluorine has a smaller size than the following inert gas, Neon, due to its strong nuclear charge.
7. Also, Van der Waal's radius is used to measure the radius of Neon, whereas the covalent radius is used to measure the radius of Fluorine. It is known that the Van der Waal radius is greater than the covalent radius.

Final answer: Atomic radii of Fluorine and Neon in angstrom unit are respectively given by $0.75,1.60$