Question

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Which of the following correctly compares the radii of F and F−?
A) F has a smaller radius than F− because an additional electron causes greater repulsion in F−.
B) F has a larger radius than F− because an additional electron causes greater repulsion in F.
C) F has a smaller radius than F− because F− has an additional energy shell.
D) F has a larger radius than F− because F has an additional energy shell.

Answer 1

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F has a smaller radius than F− because an additional electron causes greater repulsion in F−.bcz. Thus F- will be larger than Ne (and larger than Na+). The general trends will continue to hold. From top to bottom of the periodic table ions will increase in radii. However, now left to right the radius is more of a function of the number of electrons...

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Answer 2

The statement, F has a smaller radius than $F^{-}$ because an additional electron causes greater repulsion in $F^{-}$.

Explanation:

• When a neutral atom tends to gain an electron then it acquires a negative charge because of which it forms an anion. Generally, non-metals tend to gain electrons in order to gain stability.

• Atomic number of fluorine is 9 and its electronic configuration is 2, 7. Hence, when it gains an electron then neutral atom of fluorine forms $F^{-}$ ion.

• Shells in $F^{-}$ will remain the same but due to addition of an electron there will be electron-electron repulsion. As a result, radius of $F^{-}$ will be greater than the radius of F.

Learn more about atomic radius:

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Learn more about anion:

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