Question

The number of isoelectronic species among \(\ce{S^{2-}}, \ce{C^{4-}}, \ce{Mn^{2+}}, \ce{Co^{3+}}\) and \(\ce{Fe^{3+}}\) is ‘n’. If ‘n’ moles of AgCl is formed during the reaction of complex with formula \(\ce{CoCl2(en)2NH3\) with excess of AgNO\(_3\) solution, then the number of electrons present in the \(t_{2g}\) orbital of the complex is ________.}

Hint:

For coordination complexes: - Identify oxidation state of metal. - Check whether ligands are strong or weak field. - Use crystal field theory to distribute electrons in \(t_{2g}\) and \(e_g\).

Answer 1

Correct Answer: 6

Step 1: Check isoelectronic species. - \(\ce{S^{2-}}\): 18 e\(^{-}\). - \(\ce{C^{4-}}\): 10 e\(^{-}\). - \(\ce{Mn^{2+}}\): 25 - 2 = 23 e\(^{-}\). - \(\ce{Co^{3+}}\): 27 - 3 = 24 e\(^{-}\). - \(\ce{Fe^{3+}}\): 26 - 3 = 23 e\(^{-}\). Isoelectronic species: \(\ce{Mn^{2+}}\) and \(\ce{Fe^{3+}}\) (both 23 e\(^{-}\)). So, \(n = 2\).
Step 2: Complex formula. \(\ce{CoCl2(en)2NH3}\). - Coordination number = 6 (2 Cl\(^-\), 2 en, 1 NH\(_3\)). - Charge balance: en and NH\(_3\) are neutral, Cl\(^-\) are anionic. \[ \text{Oxidation state of Co} = +3 \]
Step 3: Reaction with AgNO\(_3\). 2 Cl\(^-\) ions are outside coordination sphere → precipitate with AgNO\(_3\). So, \(n = 2\) moles AgCl formed.
Step 4: Electronic configuration of Co\(^{3+}\). Co: [Ar] 3d\(^7\)4s\(^2\). Co\(^{3+}\): [Ar] 3d\(^6\). In octahedral field with strong ligands (en, NH\(_3\)): low-spin complex. So, configuration: \(t_{2g}^6 e_g^0\).
Step 5: Number of electrons in \(t_{2g}\). \[ t_{2g}^6 \quad \Rightarrow \quad 6 \,\text{electrons} \] Final Answer: \[ \text{Electrons in } t_{2g} = 6 \]