To identify isostructural pairs, we need to determine the geometry of each species. Isostructural species have the same geometry and hybridization.

• ${\mathrm{NF}}_3$: Nitrogen (N) is the central atom. It has 5 valence electrons. Three F atoms form single bonds. N has one lone pair. Steric number = 3 (bond pairs) + 1 (lone pair) = 4. Hybridization is ${\mathrm{sp}}^3$. Geometry is trigonal pyramidal.
• ${\mathrm{NO}}_3^{-}$: Nitrogen (N) is the central atom. It has 5 valence electrons. Three O atoms are bonded. One O is double-bonded, and two are single-bonded (resonance structures exist). The negative charge is delocalized. Steric number = 3 (bond pairs, considering resonance) + 0 (lone pairs) = 3. Hybridization is ${\mathrm{sp}}^2$. Geometry is trigonal planar.
• ${\mathrm{BF}}_3$: Boron (B) is the central atom. It has 3 valence electrons. Three F atoms form single bonds. B has no lone pairs. Steric number = 3 (bond pairs) + 0 (lone pairs) = 3. Hybridization is ${\mathrm{sp}}^2$. Geometry is trigonal planar.
• ${\mathrm{H}}_3{\mathrm{O}}^{+}$: Oxygen (O) is the central atom. It has 6 valence electrons. Three H atoms form single bonds. The positive charge means O has lost one electron, so it effectively has 5 valence electrons for bonding. O has one lone pair. Steric number = 3 (bond pairs) + 1 (lone pair) = 4. Hybridization is ${\mathrm{sp}}^3$. Geometry is trigonal pyramidal.
• ${\mathrm{NH}}_3$: Nitrogen (N) is the central atom. It has 5 valence electrons. Three H atoms form single bonds. N has one lone pair. Steric number = 3 (bond pairs) + 1 (lone pair) = 4. Hybridization is ${\mathrm{sp}}^3$. Geometry is trigonal pyramidal.

Based on the geometries:

• Trigonal pyramidal: ${\mathrm{NF}}_3$, ${\mathrm{H}}_3{\mathrm{O}}^{+}$, ${\mathrm{NH}}_3$
• Trigonal planar: ${\mathrm{NO}}_3^{-}$, ${\mathrm{BF}}_3$

Therefore, the isostructural pairs are $\left[{\mathrm{NF}}_3,{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$ (both trigonal pyramidal) and $\left[{\mathrm{NO}}_3^{-},{\mathrm{BF}}_3\right]$ (both trigonal planar).

Option Analysis:

• A) $\left[{\mathrm{NF}}_3,{\mathrm{NO}}_3{}^{-}\right]$ and $\left[{\mathrm{BF}}_3,{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$: Incorrect. ${\mathrm{NF}}_3$ is trigonal pyramidal, ${\mathrm{NO}}_3{}^{-}$ is trigonal planar. ${\mathrm{BF}}_3$ is trigonal planar, ${\mathrm{H}}_3{\mathrm{O}}^{+}$ is trigonal pyramidal.
• B) $\left[{\mathrm{NF}}_3,{\mathrm{BF}}_3\right]$ and $\left[{\mathrm{NO}}_3^{-},{\mathrm{BF}}_3\right]$: Incorrect. ${\mathrm{NF}}_3$ is trigonal pyramidal, ${\mathrm{BF}}_3$ is trigonal planar.
• C) $\left[{\mathrm{NF}}_3,{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$ and $\left[{\mathrm{NO}}_3^{-},{\mathrm{BF}}_3\right]$: Correct. Both pairs have the same geometry.
• D) $\left[{\mathrm{NF}}_3,{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$ and $\left[{\mathrm{NH}}_3,{\mathrm{BF}}_3\right]$: Incorrect. While $\left[{\mathrm{NF}}_3,{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$ is a correct pair, ${\mathrm{NH}}_3$ is trigonal pyramidal and ${\mathrm{BF}}_3$ is trigonal planar, so $\left[{\mathrm{NH}}_3,{\mathrm{BF}}_3\right]$ is not an isostructural pair.

Correct Answer: (C)