Electronegativity Order and Oxidation State of Oxygen

To determine the correctness of the given statements, let us analyze them individually:

Analysis of Statement I: Electronegativity increases across a period from left to right in the periodic table due to an increase in nuclear charge and a decrease in atomic radius.

Nitrogen ($\text{N}$), Oxygen ($\text{O}$), and Fluorine ($\text{F}$) belong to Period 2 of the periodic table. Their electronegativity values on the Pauling scale are approximately:

• $\text{F} = 4.0$
• $\text{O} = 3.5$
• $\text{N} = 3.0$

Thus, the correct trend of electronegativity is: $\text{F} > \text{O} > \text{N}$

Therefore, Statement I is true.

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Analysis of Statement II:

1. In oxygen difluoride ($\text{OF}_2$): Since fluorine is more electronegative than oxygen, each fluorine atom is assigned an oxidation state of $-1$. Let $x$ be the oxidation state of oxygen: $x + 2(-1) = 0 \implies x = +2$ Hence, the oxidation state of oxygen in $\text{OF}_2$ is $+2$.

2. In sodium oxide ($\text{Na}_2\text{O}$): Sodium ($\text{Na}$) is an alkali metal and always exhibits an oxidation state of $+1$ in its compounds. Let $y$ be the oxidation state of oxygen: $2(+1) + y = 0 \implies y = -2$ Hence, the oxidation state of oxygen in $\text{Na}_2\text{O}$ is $-2$.

Therefore, Statement II is true.

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Conclusion: Since both Statement I and Statement II are true, the correct answer is option A.