Which gas is a oxidizing agent as well as a reducing agent?
I'm guessing your teacher is looking for sulfur dioxide as the answer, but I don't see how or why you're supposed to be able to arrive to this answer logically. Either you'd need to read about it specifically, or maybe you're supposed to stare at a table of standard reduction potentials and notice that $\ce{SO2}$ appears as both a reagent and as a product with comparable voltages:

$$
\begin{array}{}
\ce{SO2(aq) + 4 H+(aq) + 4 e- &amp;&lt;=&gt;&amp; S(s) + 2 H2O(l)} &amp;  E^\circ = \pu{+0.50 V}\\
\ce{SO2(aq) + 2 H2O(l) &amp;&lt;=&gt;&amp; SO4^2-(aq) + 4 H+(aq) + 2 e-} &amp;  E^\circ = \pu{-0.17 V}
\end{array}
$$

(The bottom equation is the inverse of what you would find in a reduction potential table)

Extrapolating from these aqueous potentials, it appears $\ce{SO2}$ is a modest reductant and a weak oxidiser, but both reactions can be accessed in reasonable conditions. 

I think you're technically correct (as Ivan mentions), but what makes hydrogen gas a "less correct" answer is that it only rarely acts an oxidiser (in the presence of very strong reducing agents, such as alkali metals), whereas it can act as a reductant much more commonly. Compare these equations with the ones above:

$$
\begin{array}{}
\ce{H2(g) + 2 e- &amp;&lt;=&gt;&amp; 2 H^-(aq, extrapolated)} &amp;  E^\circ = \pu{-2.3 V}\\
\ce{H2(g) &amp;&lt;=&gt;&amp; 2 H+(aq) + 2 e-} &amp;  E^\circ = \pu{+0.00 V}
\end{array}
$$

(Reference for the hydride potential)

Note the much larger gap in potential between the two reactions. In this sense, the reducing nature of $\ce{H2}$ vastly overwhelms its oxidising nature, so most people will think of it as a reducing gas.