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

oxygen is acting as an oxidizing agent in all of the following reactions excepta. \\( 2 c(s) + o_2(g) \\rightarrow 2 co(g) \\)b. \\( s(s) + o_2(g) \\rightarrow so_2(g) \\)c. \\( 2 f_2(g) + o_2(g) \\rightarrow 2 of_2(g) \\)d. \\( 2 na(s) + o_2(g) \\rightarrow na_2o_2(s) \\)e. \\( 2 mg(s) + o_2(g) \\rightarrow 2 mgo(s) \\)

oxygen is acting as an oxidizing agent in all of the following reactions excepta. \\( 2 c(s) + o_2(g) \\rightarrow 2 co(g) \\)b. \\( s(s) + o_2(g) \\rightarrow so_2(g) \\)c. \\( 2 f_2(g) + o_2(g) \\rightarrow 2 of_2(g) \\)d. \\( 2 na(s) + o_2(g) \\rightarrow na_2o_2(s) \\)e. \\( 2 mg(s) + o_2(g) \\rightarrow 2 mgo(s) \\)

Answer

Brief Explanations:

To determine in which reaction oxygen is not acting as an oxidizing agent, we analyze the oxidation states of oxygen in each reaction:

  • Oxidizing agent: A substance that gets reduced (oxidation state decreases).
  • Reduction: Gain of electrons (oxidation state decreases).
  • Oxidation: Loss of electrons (oxidation state increases).

Reaction A: ( \boldsymbol{2\ C(s) + O_2(g) \rightarrow 2\ CO(g)} )

  • Oxidation state of ( O ) in ( O_2 ): ( 0 ).
  • Oxidation state of ( O ) in ( CO ): ( -2 ).
  • ( O ) gains electrons (oxidation state decreases from ( 0 ) to ( -2 )) → ( O_2 ) is reduced (oxidizing agent).

Reaction B: ( \boldsymbol{S(s) + O_2(g) \rightarrow SO_2(g)} )

  • Oxidation state of ( O ) in ( O_2 ): ( 0 ).
  • Oxidation state of ( O ) in ( SO_2 ): ( -2 ).
  • ( O ) gains electrons (oxidation state decreases from ( 0 ) to ( -2 )) → ( O_2 ) is reduced (oxidizing agent).

Reaction C: ( \boldsymbol{2\ F_2(g) + O_2(g) \rightarrow 2\ OF_2(g)} )

  • Oxidation state of ( O ) in ( O_2 ): ( 0 ).
  • Oxidation state of ( O ) in ( OF_2 ): Fluorine is more electronegative than oxygen, so ( F ) has ( -1 ). Let oxidation state of ( O ) be ( x ). Then ( x + 2(-1) = 0 ) → ( x = +2 ).
  • ( O ) loses electrons (oxidation state increases from ( 0 ) to ( +2 )) → ( O_2 ) is oxidized (not an oxidizing agent).

Reaction D: ( \boldsymbol{2\ Na(s) + O_2(g) \rightarrow Na_2O_2(s)} )

  • Oxidation state of ( O ) in ( O_2 ): ( 0 ).
  • Oxidation state of ( O ) in ( Na_2O_2 ): ( Na ) is ( +1 ). Let oxidation state of ( O ) be ( x ). Then ( 2(+1) + 2x = 0 ) → ( x = -1 ).
  • ( O ) gains electrons (oxidation state decreases from ( 0 ) to ( -1 )) → ( O_2 ) is reduced (oxidizing agent).

Reaction E: ( \boldsymbol{2\ Mg(s) + O_2(g) \rightarrow 2\ MgO(s)} )

  • Oxidation state of ( O ) in ( O_2 ): ( 0 ).
  • Oxidation state of ( O ) in ( MgO ): ( -2 ).
  • ( O ) gains electrons (oxidation state decreases from ( 0 ) to ( -2 )) → ( O_2 ) is reduced (oxidizing agent).

In reaction C, oxygen’s oxidation state increases (it is oxidized), so ( O_2 ) is not acting as an oxidizing agent here.

Answer:

C. ( 2\ F_2(g) + O_2(g) \rightarrow 2\ OF_2(g) )