The number of complexes from the following with no electrons in the t$_{2g}$ orbital is _______.
TiCl$_4$, [MnO$_4$]$^{-}$, [FeO$_4$]$^{2-}$, [FeCl$_4$]$^{-}$, [CoCl$_4$]$^{2-}$
TiCl$_4$, [MnO$_4$]$^{-}$, [FeO$_4$]$^{2-}$, [FeCl$_4$]$^{-}$, [CoCl$_4$]$^{2-}$
- 3
- 1
- 4
- 2
The Correct Option is A
Approach Solution - 1
To determine the number of complexes with no electrons in the \(t_2\) orbital, analyze each complex and its oxidation state, electronic configuration, and crystal field splitting.
Step 1: Analyze each complex
1.\(\text{TiCl}_4\):
- Oxidation state of Ti: \(+4 \, (\text{Ti}^{4+})\).
- Electronic configuration of \(\text{Ti}^{4+}\): \(3d^0\).
- No electrons in \(t_2\) orbitals.
2.\([\text{MnO}_4]^-\):
- Oxidation state of Mn: \(+7 \, (\text{Mn}^{7+})\).
- Electronic configuration of \(\text{Mn}^{7+}\): \(3d^0\).
- No electrons in \(t_2\) orbitals.
3.\([\text{FeO}_4]^{2-}\):
- Oxidation state of Fe: \(+6 \, (\text{Fe}^{6+})\).
- Electronic configuration of \(\text{Fe}^{6+}\): \(3d^0\).
- No electrons in \(t_2\) orbitals.
4.\([\text{FeCl}_4]^-\):
- Oxidation state of Fe: \(+2 \, (\text{Fe}^{2+})\).
- Electronic configuration of \(\text{Fe}^{2+}\): \(3d^6\).
- \(t_2\) orbitals are populated with electrons (\(t_2^3e^3\)).
5.\([\text{CoCl}_4]^{2-}\):
- Oxidation state of Co: \(+2 \, (\text{Co}^{2+})\).
- Electronic configuration of \(\text{Co}^{2+}\): \(3d^7\).
- \(t_2\) orbitals are populated with electrons (\(t_2^4e^3\)).
Step 2: Count complexes with no \(t_2\) electrons
- \(\text{TiCl}_4\), \([\text{MnO}_4]^-\), and \([\text{FeO}_4]^{2-}\) have no electrons in \(t_2\) orbitals.
- \([\text{FeCl}_4]^-\) and \([\text{CoCl}_4]^{2-}\) have electrons in \(t_2\) orbitals.
Conclusion:
The number of complexes with no electrons in the \(t_2\) orbital is:
\[3 \, (\text{TiCl}_4, \, [\text{MnO}_4]^-, \, [\text{FeO}_4]^{2-}).\]
Final Answer: (1).
Approach Solution -2
TiCl₄, [MnO₄]⁻, [FeO₄]²⁻, [FeCl₄]⁻, [CoCl₄]²⁻
Step 2: Determine oxidation states and electronic configurations.
1. TiCl₄:
Oxidation state of Ti = +4 → Ti⁴⁺
Atomic number of Ti = 22 → Ti: [Ar] 3d² 4s² → Ti⁴⁺: [Ar] (no 3d electrons)
Hence, t₂g⁰ configuration.
2. [MnO₄]⁻:
Let oxidation state of Mn = x.
x + 4(–2) = –1 → x = +7
Mn⁷⁺: [Ar] 3d⁰
Hence, t₂g⁰ configuration.
3. [FeO₄]²⁻:
Let oxidation state of Fe = x.
x + 4(–2) = –2 → x = +6
Fe⁶⁺: [Ar] 3d²
In a strong field (as in ferrate ion), both 3d electrons are paired and elevated due to strong oxidation, effectively approximated as t₂g⁰ configuration.
4. [FeCl₄]⁻:
Fe oxidation state = +3
Fe³⁺: [Ar] 3d⁵ (high-spin complex with weak field ligands like Cl⁻)
t₂g³ e_g² → not t₂g⁰.
5. [CoCl₄]²⁻:
Co oxidation state = +2
Co²⁺: [Ar] 3d⁷ (weak field, high-spin complex)
t₂g⁵ e_g² → not t₂g⁰.
Step 3: Identify complexes with t₂g⁰ configuration.
TiCl₄ → t₂g⁰
[MnO₄]⁻ → t₂g⁰
[FeO₄]²⁻ → t₂g⁰
Step 4: Final Answer.
Number of complexes with no electrons in t₂g orbital = 3
Final Answer: 3
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