In compound X, all the bond angles are exactly $109°{28}^{\text{'}};\mathrm{X}$ is

It is given to find out the compound X, all the bond angles are exactly $109°{28}^{\text{'}};\mathrm{X}$. 
Explanation

In the compound $\mathrm{X}$, all the bond angles are exactly $109°{28}^{\text{'}}\mathrm{X}$ is carbon tetrachloride.
${\mathrm{CCl}}_4$ has ${\mathrm{sp}}^3$ - hybridization with 4 bond pairs and 0 lone pairs of electrons giving regular tetrahedron geometry as all the four $\mathrm{C}-\mathrm{Cl}$ bonds are equivalent. In others, the geometry is a little distorted in spite of ${\mathrm{sp}}^3$ hybridization due to different atoms on the vertices of a tetrahedron.

So, option C) is correct.