To solve the problem of matching the complex ions with their geometries, we will analyze each complex ion one by one:
### Step 1: Analyze `[Ag(NH₃)₂]⁺`
1. **Determine the oxidation state of Ag**:
- Ammonia (NH₃) is a neutral ligand, so the oxidation state of Ag in `[Ag(NH₃)₂]⁺` is +1.
2. **Electronic configuration of Ag**:
- Silver (Ag) has an atomic number of 47, and its electronic configuration is [Kr] 4d¹⁰ 5s². In the +1 oxidation state, it loses one electron, resulting in [Kr] 4d¹⁰.
3. **Hybridization**:
- With two ligands (NH₃), the hybridization is sp, leading to a linear geometry.
4. **Conclusion**: The geometry is correctly stated as linear.
### Step 2: Analyze `[MnCl₄]²⁺`
1. **Determine the oxidation state of Mn**:
- Chlorine (Cl) has a -1 charge. Therefore, for `[MnCl₄]²⁺`, we have \( x + 4(-1) = +2 \) → \( x = +2 \).
2. **Electronic configuration of Mn**:
- Manganese (Mn) has an atomic number of 25, with the configuration [Ar] 3d⁵ 4s². In the +2 state, it becomes [Ar] 3d⁵.
3. **Hybridization**:
- With four ligands (Cl), the hybridization is sp³, which corresponds to a tetrahedral geometry.
4. **Conclusion**: The geometry is correctly stated as tetrahedral.
### Step 3: Analyze `[Cu(NH₃)₄]²⁺`
1. **Determine the oxidation state of Cu**:
- Again, NH₃ is neutral, so for `[Cu(NH₃)₄]²⁺`, we have \( x + 0 = +2 \) → \( x = +2 \).
2. **Electronic configuration of Cu**:
- Copper (Cu) has an atomic number of 29, with the configuration [Ar] 3d¹⁰ 4s¹. In the +2 state, it becomes [Ar] 3d⁹.
3. **Hybridization**:
- With four ligands (NH₃), and since NH₃ is a strong field ligand, the hybridization is dsp², leading to a square planar geometry.
4. **Conclusion**: The geometry is correctly stated as square planar.
### Step 4: Analyze `[Ni(CN)₄]²⁻`
1. **Determine the oxidation state of Ni**:
- Cyanide (CN) has a -1 charge. Therefore, for `[Ni(CN)₄]²⁻`, we have \( x + 4(-1) = -2 \) → \( x = +2 \).
2. **Electronic configuration of Ni**:
- Nickel (Ni) has an atomic number of 28, with the configuration [Ar] 3d⁸ 4s². In the +2 state, it becomes [Ar] 3d⁸.
3. **Hybridization**:
- With four ligands (CN), which are also strong field ligands, the hybridization is dsp², leading to a square planar geometry.
4. **Conclusion**: The geometry is correctly stated as square planar.
### Final Conclusion
After analyzing all four complex ions, we find that all the geometries provided are correct:
1. `[Ag(NH₃)₂]⁺` - Linear
2. `[MnCl₄]²⁺` - Tetrahedral
3. `[Cu(NH₃)₄]²⁺` - Square planar
4. `[Ni(CN)₄]²⁻` - Square planar
Thus, the correct answer is that all the matches are correct.
