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The atomic radii of fluorine and neon in angstrom unit are respectively given by:A.0.75, 1.60B.1.60, 1.60C.0.72, 0.72D.1.60, 0.72

Last updated date: 13th Aug 2024
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Answer
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Hint:The atomic radii of an element is completely depending upon the electronic configuration of the element. As we know the electronic configuration of the given element, we can check the orbital occupancy of the electron as well as the radius measurement condition of the given element by which the answer can be given.

Complete step by step answer:
The atomic number of fluorine is 9 and neon is 10.
The electronic configuration of fluorine (Group-17 element: Halogen) is
The electronic configuration of neon (Group-18 element: Inert gas) is
Thus, we can clearly see fluorine has 5 electrons in 2p orbital. To form an ideal configuration the p orbital should be filled with 6 electrons. In fluorine one electron is less.
Hence, the nuclear charge is high in case of fluorine that leads to the smaller size of fluorine than that of the next inert gas neon.
According to the given options, only option A shows the smaller atomic radius of fluorine than the neon.
The atomic radius of fluorine and neon is 0.75 angstrom and 1.60 angstrom respectively.

So, option A. 0.75, 1.60 is the correct answer.

Note:
In other words, the atomic radius of fluorine is measured by its covalent radius and in case neon the atomic radius is measured by the van der Waals radius. As, the van der Waals radius is always greater than covalent radius, so the fluorine has a smaller radius than neon. Among the four option, only option-A satisfy the condition of $\text{radiu}{{\text{s}}_{\text{vanderWaals}}}$>$\text{radiu}{{\text{s}}_{\text{covalent}}}$. So, using the covalent and van der Waals radius concept this question can be solved.