The correct answer is Option C:

$$ \mathrm{Br}^{-}<\mathrm{F}^{-}<\mathrm{H}_2 \mathrm{O}<\mathrm{NH}_3 $$.

Here's why:

The spectrochemical series is a list of ligands arranged in order of increasing field strength. This means that ligands higher on the series cause a larger splitting of the d-orbitals in a transition metal complex, resulting in a larger crystal field stabilization energy (CFSE).

Here's a breakdown of the factors influencing the spectrochemical series:

• Charge Density: Ligands with a higher charge density (more negative charge concentrated in a smaller space) will interact more strongly with the metal ion's d-orbitals. For example, $$ \mathrm{F}^{-} $$ has a higher charge density than $$ \mathrm{Br}^{-} $$, making it a stronger field ligand.
• Electronegativity: More electronegative atoms within a ligand draw electron density away from the metal center, weakening the metal-ligand bond and reducing field strength.
• π-Backbonding: Ligands that can participate in π-backbonding (donation of electrons from the metal d-orbitals to empty ligand orbitals) can strengthen the metal-ligand bond and increase field strength. For example, $$ \mathrm{CN}^{-} $$ is a strong field ligand due to its ability to engage in π-backbonding with transition metals.

Let's analyze why the other options are incorrect:

• Option A: This order is incorrect because $$ \mathrm{NH}_3 $$ is a stronger field ligand than halides, which is not reflected in this arrangement.
• Option B: This order is incorrect because $$ \mathrm{CN}^{-} $$ is a much stronger field ligand than $$ \mathrm{NH}_3 $$.
• Option D: This order is incorrect because $$ \mathrm{CN}^{-} $$ is a much stronger field ligand than $$ \mathrm{Br}^{-} $$.

Therefore, the correct order in increasing field strength is: $$ \mathrm{Br}^{-}<\mathrm{F}^{-}<\mathrm{H}_2 \mathrm{O}<\mathrm{NH}_3 $$.