The position of an ant, denoted as $ \mathrm{S} $ in meters, moving in the $ \mathrm{Y} $-$ \mathrm{Z} $ plane is given by $ S = 2t^2 \hat{j} + 5 \hat{k} $, where $ t $ is in seconds. To determine the magnitude and direction of the ant's velocity at $ t = 1 $ second, we need to differentiate the position function with respect to time.

The velocity $ \overrightarrow{\mathrm{v}} $ is given by:

$$\overrightarrow{\mathrm{v}} = \frac{d\mathrm{S}}{dt} = \frac{d}{dt} (2t^2 \hat{j} + 5 \hat{k})$$

On differentiating, we get:

$$\overrightarrow{\mathrm{v}} = 4t \hat{j}$$

At $ t = 1 $ second:

$$\overrightarrow{\mathrm{v}} = 4 \cdot 1 \hat{j} = 4 \hat{j}$$

Therefore, the magnitude of the velocity is $ 4 \mathrm{~m/s} $ and it is directed along the $ y $-axis.