Statement I: $\text{SnCl}_2$ does not act as a ligand, but $\text{SnCl}_3^-$ acts as a good ligand.
Statement II: On addition of $\text{Cl}^-$ to $\text{SnCl}_2$, the orbital containing the lone pair acquires less s-character.
Both statements are correct:
Statement I is true. $\text{SnCl}_2$ is not considered a ligand because it does not readily form coordinate bonds with metal centers. On the other hand, $\text{SnCl}_3^-$ behaves as a good ligand because it can donate its lone pair of electrons to coordinate with metal ions.
Statement II is also true. When a chloride ion ($\text{Cl}^-$) is added to $\text{SnCl}_2$, it forms $\text{SnCl}_3^-$ by donating its lone pair to the $\text{Sn}$ atom. In this process, the orbital containing the lone pair acquires less s-character. This is because the addition of a negatively charged electron-rich species ($\text{Cl}^-$) to $\text{SnCl}_2$ leads to the polarization of the orbital containing the lone pair, causing it to have less s-character.
Statement II is the correct explanation for Statement I. The change in s-character of the orbital upon addition of $\text{Cl}^-$ to $\text{SnCl}_2$ enables $\text{SnCl}_2$ to act as a ligand in the form of $\text{SnCl}_3^-$, which has a stable coordination complex. This explains why $\text{SnCl}_2$ does not act as a ligand on its own but $\text{SnCl}_3^-$ does.
Therefore, the correct answer is Option 1: If both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I. |
