For the reaction, $2\mathrm{H}\left(g\right)\to {\mathrm{H}}_2\left(g\right)$ what are the signs of $\mathrm{\Delta }{H}^{\circ }$ and $\mathrm{\Delta }{S}^{\circ }?$

The given reaction is,

$2\mathrm{H}\left(g\right)\to {\mathrm{H}}_2\left(g\right)$

In this reaction, number of moles reduces and hence entropy decreases. Therefore it will be negative.

$\mathrm{\Delta }{S}^{\circ }<0$

Now, for the spontaneous process, the Gibbs energy is negative and hence,

$\mathrm{\Delta }{H}^{\circ }<0$