Atomic-radii of fluorine and neon in angstrom units are, respectively, given by: A- 0.72, 1.60 B- 1.60, 1.60 C- 0.72, 0.72 D- 1.60, 0.72
Hint: We do not need the exact value of atomic radii of fluorine and neon we can just identify the relation between their radii. Identify the exact definition of atomic radii and how it is calculated.
Complete answer: We know that fluorine and neon are the elements of the periodic table belonging to the same period. Atomic radius the distance between the center of the nucleus of the atom to the boundary of the valence shell. But this calculation is very difficult so we follow different methods for calculation of atomic radii of atoms belonging to different categories. Ionic radius is the radius of the element in its ionic form. Ionic radius is less than the atomic radius for an element forming cation and ionic radius is more than the atomic radius for an element forming anion. Covalent radius is the radius of the element in its covalent compound. Metallic radius is the radius of the element in the metallic bond. Metals form metallic bonds. Fluorine molecule is a covalent compound. In the covalent compound, there will be a sharing of electrons so the distance between two nuclei will decrease. Inert gases mostly do not form any compounds so we cannot use any of the above radii for inert gases. Inert gases only have Vander Waal attraction forces. These attractions are very weak so 2 atoms cannot come close. Vander Waal radius is half the distance between two nuclei. So the covalent radius will be less than the Vander Waal radius. Radius of fluorine is 0.72 and the radius of neon is 1.60 in angstroms.
So, the correct answer is “Option A”.
Note: In neon atoms due to its fulfilled configuration there will be electronic repulsions so it will have more atomic radius than fluorine. Calculation of the atomic radii method also increases the radius of neon and decreases the radius of fluorine.