Let's analyze each option:

Option A: The bond angle in hydrides of Group 15 elements decreases down the group. This is due to the decrease in electronegativity of the central atom, which leads to a decrease in the s-character of the bonding orbitals and an increase in the p-character. Also, as the size of the central atom increases, the bond pair-bond pair repulsion decreases, leading to smaller bond angles. Therefore, the correct order should be ${\mathrm{NH}}_3>{\mathrm{PH}}_3>{\mathrm{AsH}}_3>{\mathrm{SbH}}_3$. The given order is incorrect.

Option B: In ${\mathrm{H}}_2\mathrm{O}$, the central atom is oxygen, and the bond angle is approximately 104.5°. In ${\mathrm{OF}}_2$, the central atom is oxygen, but fluorine is more electronegative than oxygen. This pulls the bonding electron pairs away from the central oxygen atom, reducing the bond pair-bond pair repulsion and thus decreasing the bond angle (approx. 103.2°). In ${\mathrm{Cl}}_2\mathrm{O}$, chlorine is less electronegative than oxygen. The bonding electron pairs are closer to the central oxygen atom, increasing bond pair-bond pair repulsion and thus increasing the bond angle (approx. 110.9°). Therefore, the correct order should be ${\mathrm{OF}}_2<{\mathrm{H}}_2\mathrm{O}<{\mathrm{Cl}}_2\mathrm{O}$. The given order is incorrect.

Option C: These are hydronium ion analogues. The central atom in these species (${\mathrm{O}}^{+},{\mathrm{S}}^{+},{\mathrm{Se}}^{+},{\mathrm{Te}}^{+}$) belongs to Group 16. As we move down the group, the electronegativity of the central atom decreases, and its size increases. Similar to Group 15 hydrides, the bond angle decreases down the group due to reduced s-character in bonding orbitals and decreased bond pair-bond pair repulsion. Therefore, the correct order for increasing bond angle is ${\mathrm{H}}_3{\mathrm{Te}}^{+}<{\mathrm{H}}_3{\mathrm{Se}}^{+}<{\mathrm{H}}_3{\mathrm{S}}^{+}<{\mathrm{H}}_3{\mathrm{O}}^{+}$. This order is correct.

Option D: In boron trihalides (${\mathrm{BF}}_3,{\mathrm{BCl}}_3,{\mathrm{BBr}}_3,{\mathrm{BI}}_3$), the central boron atom is ${\mathrm{sp}}^2$ hybridized, and all have a trigonal planar geometry with ideal bond angles of 120°. There is no significant change in bond angle due to the size of the halogen atoms because the geometry is determined by the ${\mathrm{sp}}^2$ hybridization and the absence of lone pairs on boron. The given order implies varying bond angles, which is incorrect.

Correct Answer: (C)