oxygen is acting as an oxidizing agent in all of the following reactions except\na. $2\\ c(s) + o_2(g)…

oxygen is acting as an oxidizing agent in all of the following reactions except\na. $2\\ c(s) + o_2(g) \\rightarrow 2\\ co(g)$\nb. $s(s) + o_2(g) \\rightarrow so_2(g)$\nc. $2\\ f_2(g) + o_2(g) \\rightarrow 2\\ of_2(g)$\nd. $2\\ na(s) + o_2(g) \\rightarrow na_2o_2(s)$\ne. $2\\ mg(s) + o_2(g) \\rightarrow 2\\ mgo(s)$

oxygen is acting as an oxidizing agent in all of the following reactions except\na. $2\\ c(s) + o_2(g) \\rightarrow 2\\ co(g)$\nb. $s(s) + o_2(g) \\rightarrow so_2(g)$\nc. $2\\ f_2(g) + o_2(g) \\rightarrow 2\\ of_2(g)$\nd. $2\\ na(s) + o_2(g) \\rightarrow na_2o_2(s)$\ne. $2\\ mg(s) + o_2(g) \\rightarrow 2\\ mgo(s)$

Answer

Explanation:

Step1: Define oxidizing agent role

An oxidizing agent gains electrons, so its oxidation number decreases.

Step2: Find O oxidation number in reactants

In $\text{O}_2$, oxidation number of O is $0$.

Step3: Calculate O oxidation number in products for each reaction

Reaction A: $\text{CO}$

O oxidation number: $-2$ (decreases from $0$; $\text{O}_2$ is oxidizing agent)

Reaction B: $\text{SO}_2$

O oxidation number: $-2$ (decreases from $0$; $\text{O}_2$ is oxidizing agent)

Reaction C: $\text{OF}_2$

Let oxidation number of O be $x$. F has oxidation number $-1$. $$x + 2(-1) = 0 \implies x = +2$$ Oxidation number increases from $0$ to $+2$; $\text{O}_2$ is reducing agent.

Reaction D: $\text{Na}_2\text{O}_2$

O oxidation number: $-1$ (decreases from $0$; $\text{O}_2$ is oxidizing agent)

Reaction E: $\text{MgO}$

O oxidation number: $-2$ (decreases from $0$; $\text{O}_2$ is oxidizing agent)

Answer:

C. $2 \text{F}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2 \text{OF}_2(\text{g})$