The number of unpaired d-electrons in $\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ is ________.

<p>First, let us determine the $d$-electron count of the cobalt center in $\bigl[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6\bigr]^{3+}$.</p> <h2>1. Oxidation state and $d$-electron count</h2> <p><p><strong>Neutral cobalt (Co)</strong> has an atomic number of 27 and an electronic configuration: </p> <p>$ \mathrm{Co} \; \bigl[\mathrm{Ar}\bigr]\, 3d^7\, 4s^2. $ </p></p> <p><p>In the <strong>+3 oxidation state</strong>, cobalt has lost a total of 3 electrons:</p></p> <p><p>First two electrons are removed from the $4s$ orbital.</p></p> <p><p>The third electron is removed from the $3d$ orbitals.</p></p> <p>Thus, $\mathrm{Co}^{3+}$ has</p> <p>$ 3d^{7-1} = 3d^6. $</p> <p>So in $\bigl[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6\bigr]^{3+}$, the cobalt center is $d^6$.</p> <h2>2. High-spin vs Low-spin for $\mathrm{Co}^{3+}$</h2> <p>Even though $\mathrm{H}_2\mathrm{O}$ is generally considered a <strong>weak field</strong> ligand, the $\mathrm{Co}^{3+}$ ion has a relatively high charge (+3). A higher charge on the metal center usually <strong>increases</strong> the ligand-field splitting ($\Delta_\mathrm{o}$) significantly compared to lower oxidation states of the same metal. </p> <p>In most typical octahedral complexes of $\mathrm{Co}^{3+}$, the splitting $\Delta_\mathrm{o}$ is large enough that the complex ends up being <strong>low spin</strong>. </p> <h3>Electronic configuration in an octahedral field</h3> <p>For a $d^6$ octahedral <strong>low-spin</strong> complex, all six electrons pair up in the lower-energy $\mathrm{t_{2g}}$ orbitals:</p> <p>$ \mathrm{t_{2g}^6}\, \mathrm{e_g^0} $</p> <p>That leaves <strong>0 unpaired</strong> electrons.</p> <h2>3. Final answer</h2> <p>Therefore, the number of unpaired $d$-electrons in </p> <p>$\bigl[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6\bigr]^{3+}$ is $ \boxed{0}. $</p> <p><strong>Correct Option: A (0)</strong></p>