The correct order of ligands arranged in increasing field strength.

<p>The correct answer is <strong>Option C</strong>: <p>$\mathrm{Br}^{-}<\mathrm{F}^{-}<\mathrm{H}_2 \mathrm{O}<\mathrm{NH}_3$.</p> </p> <p>Here's why:</p> <p>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).</p> <p>Here's a breakdown of the factors influencing the spectrochemical series:</p> <ul> <li><b>Charge Density</b>: 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.</li> <li><b>Electronegativity</b>: More electronegative atoms within a ligand draw electron density away from the metal center, weakening the metal-ligand bond and reducing field strength. </li> <li><b>π-Backbonding</b>: 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.</li> </ul> <p>Let's analyze why the other options are incorrect:</p> <ul> <li><b>Option A</b>: This order is incorrect because $\mathrm{NH}_3$ is a stronger field ligand than halides, which is not reflected in this arrangement. <li><b>Option B</b>: This order is incorrect because $\mathrm{CN}^{-}$ is a much stronger field ligand than $\mathrm{NH}_3$. <li><b>Option D</b>: This order is incorrect because $\mathrm{CN}^{-}$ is a much stronger field ligand than $\mathrm{Br}^{-}$.</li> </ul> <p>Therefore, the correct order in increasing field strength is: $\mathrm{Br}^{-}<\mathrm{F}^{-}<\mathrm{H}_2 \mathrm{O}<\mathrm{NH}_3$.</p> <p></p>