Respuesta :
Answer: Option (B) is the correct answer.
Explanation:
Both oxygen and fluorine are period 2 elements and when we move across a period then there occurs a decrease in atomic size of the atoms. Hence, the atomic radius of a neutral fluorine atom is smaller than a neutral oxygen atom.
Moreover, atomic number of fluorine is 9 and it has higher nuclear charge due to which it will cause more attraction of electrons. As a result, size of a fluorine ion will be smaller.
On the other hand, size of oxygen atom is larger and has small nuclear charge due to which attraction of electrons by its nucleus will not be strong enough. Hence, the size of [tex]O^{2-}[/tex] will be larger.
Thus, we can conclude that the statement [tex]F^{-}[/tex] has a larger nuclear charge than [tex]O^{2-}[/tex] has, is correct for the fact that the [tex]F^{-}[/tex] ion is smaller than the [tex]O^{2-}[/tex] ion.
F⁻ has a larger nuclear charge than O²⁻ has.
• Both oxygen and fluorine are period 2 elements.
• As one moves from left to right in a period, one can witness an increase in the effective nuclear charge.
• The effective nuclear charge of fluorine is more than oxygen.
• Effective nuclear charge is inversely proportional to atomic size. Thus, if effective nuclear charge increases the atomic size decreases and vice versa.
• The F⁻ has a larger nuclear charge in comparison to O²⁻, thus, the size of F⁻ ion will be smaller in comparison to O²⁻ ion.
Thus, the correct answer is option B, that is, F⁻ has a larger nuclear charge than O²⁻.
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