Q.71. Number of paramagnetic complexes among the following is [MnBr₄]²⁻, [NiCl₄]²⁻, [Ni(CN)₄]²⁻, [Ni

Q: What is the correct answer for JEE Main Chemistry Q71?

The correct answer is 6. To determine if a complex is paramagnetic, we need to find the number of unpaired electrons. This depends on the central metal ion's oxidation state and d-electron configuration, and whether the ligands are strong field (SFL) or weak field (WFL). 1. **[MnBr₄]²⁻**: Mn is in +2 oxidation state ($d^5$). Br⁻ is a WFL. High spin complex. 5 unpaired electrons. Paramagnetic. 2. **[NiCl₄]²⁻**: Ni is in

Q: Q.71. Number of paramagnetic complexes among the following is [MnBr₄]²⁻, [NiCl₄]²⁻, [Ni(CN)₄]²⁻, [Ni(CO)₄], [CoF₆]³⁻, [Fe(CN)₆]⁴⁻, [Mn(CN)₆]³⁻, [Ti(CN)₆]³⁻, [Cu(H₂O)₆]²⁺, [Co(C₂O₄)₃]³⁻

The correct answer is 6. To determine if a complex is paramagnetic, we need to find the number of unpaired electrons. This depends on the central metal ion's oxidation state and d-electron configuration, and whether the ligands are strong field (SFL) or weak field (WFL). 1. **[MnBr₄]²⁻**: Mn is in +2 oxidation state ($d^5$). Br⁻ is a WFL. High spin complex. 5 unpaired electrons. Paramagnetic. 2. **[NiCl₄]²⁻**: Ni is in

Question

Q.71. Number of paramagnetic complexes among the following is [MnBr₄]²⁻, [NiCl₄]²⁻, [Ni(CN)₄]²⁻, [Ni(CO)₄], [CoF₆]³⁻, [Fe(CN)₆]⁴⁻, [Mn(CN)₆]³⁻, [Ti(CN)₆]³⁻, [Cu(H₂O)₆]²⁺, [Co(C₂O₄)₃]³⁻

FirstInTest · Accepted Answer

To determine if a complex is paramagnetic, we need to find the number of unpaired electrons. This depends on the central metal ion's oxidation state and d-electron configuration, and whether the ligands are strong field (SFL) or weak field (WFL).
1.  **[MnBr₄]²⁻**: Mn is in +2 oxidation state ($d^5$). Br⁻ is a WFL. High spin complex. 5 unpaired electrons. Paramagnetic.
2.  **[NiCl₄]²⁻**: Ni is in +2 oxidation state ($d^8$). Cl⁻ is a WFL. High spin complex. 2 unpaired electrons. Paramagnetic.
3.  **[Ni(CN)₄]²⁻**: Ni is in +2 oxidation state ($d^8$). CN⁻ is a SFL. Square planar complex. 0 unpaired electrons. Diamagnetic.
4.  **[Ni(CO)₄]**: Ni is in 0 oxidation state ($d^{10}$). CO is a SFL. Tetrahedral complex. 0 unpaired electrons. Diamagnetic.
5.  **[CoF₆]³⁻**: Co is in +3 oxidation state ($d^6$). F⁻ is a WFL. High spin complex. 4 unpaired electrons. Paramagnetic.
6.  **[Fe(CN)₆]⁴⁻**: Fe is in +2 oxidation state ($d^6$). CN⁻ is a SFL. Low spin complex. 0 unpaired electrons. Diamagnetic.
7.  **[Mn(CN)₆]³⁻**: Mn is in +3 oxidation state ($d^4$). CN⁻ is a SFL. Low spin complex. 2 unpaired electrons. Paramagnetic.
8.  **[Ti(CN)₆]³⁻**: Ti is in +3 oxidation state ($d^1$). CN⁻ is a SFL. 1 unpaired electron. Paramagnetic.
9.  **[Cu(H₂O)₆]²⁺**: Cu is in +2 oxidation state ($d^9$). H₂O is a WFL. 1 unpaired electron. Paramagnetic.
10. **[Co(C₂O₄)₃]³⁻**: Co is in +3 oxidation state ($d^6$). C₂O₄²⁻ (oxalate) is a SFL. Low spin complex. 0 unpaired electrons. Diamagnetic.
Paramagnetic complexes are: [MnBr₄]²⁻, [NiCl₄]²⁻, [CoF₆]³⁻, [Mn(CN)₆]³⁻, [Ti(CN)₆]³⁻, [Cu(H₂O)₆]²⁺. Total = 6.

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