Which element has an outer electron configuration of ${\mathrm{s}}^2{\mathrm{p}}^4$?

The chemical element selenium has the symbol Se and atomic number 34.

 The electronic configuration of this element can be given as,

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^4$

So Se has an outer electron configuration of ${\mathrm{s}}^2{\mathrm{p}}^4$.