It simply depends on the shape of the marginal external damages. If the total damages are linear in output, then the marginal damages are constant, and the gradient of the marginal social cost is the same as that of the marginal private cost.
If the total damages are quadratic (or some other form with increasing marginal damages), the marginal social cost has a steeper slope than the marginal private cost.
In a simple example: if the marginal private costs are $MPC(Q)=70+2Q$, where $Q$ is quantity produced and total damages are $D(Q)=50Q$, then marginal external damages are $MED=50$ and the marginal social costs of production would be:
If in contrast the damage function would be $D(Q)=50Q+2Q^2$, then marginal external damages would be $MED=50+4Q$ and the marginal social costs of production:
As you can see the latter has a steeper slope than the original MPC, whereas the former does not.