Which of the following complexes are diamagnetic ?
`{:([Pt(NH_3)_4]^(2+),[Co(SCN)_4]^(2-),[Cu(en)_2]^(2+),[HgI_4]^(2-)),("square planar","tetrahedral","square planar","tetrahedral"),((i),(ii),(iii),"iv"):}`

Options

A. (i) and (ii)
B. (ii) and (iii)
C. (i) and (iv)
D. (iii) and (iv)

Explanation

To determine which of the given complexes are diamagnetic, we need to analyze the electron configurations and the presence of unpaired electrons in each complex. Let's go through each complex step by step.

### Step 1: Analyze the first complex \([Pt(NH_3)_4]^{2+}\)

1. **Oxidation State**: Platinum (Pt) is in the +2 oxidation state.
2. **Electron Configuration**: The electron configuration of Pt is \([Xe] 4f^{14} 5d^8\).
3. **Hybridization**: The complex has a square planar geometry, which corresponds to \(dsp^2\) hybridization.
4. **Electron Pairing**: In a square planar complex, the 5d electrons will pair up. The 5d^8 configuration means there are no unpaired electrons after pairing.
5. **Conclusion**: Since there are no unpaired electrons, \([Pt(NH_3)_4]^{2+}\) is **diamagnetic**.

### Step 2: Analyze the second complex \([Co(SCN)_4]^{2-}\)

1. **Oxidation State**: Cobalt (Co) is in the +2 oxidation state.
2. **Electron Configuration**: The electron configuration of Co is \([Ar] 3d^7\).
3. **Hybridization**: The complex has a tetrahedral geometry, which corresponds to \(sp^3\) hybridization.
4. **Electron Pairing**: In tetrahedral complexes, there is less splitting of the d-orbitals, and thus pairing does not occur. The 3d^7 configuration will have three unpaired electrons.
5. **Conclusion**: Since there are unpaired electrons, \([Co(SCN)_4]^{2-}\) is **paramagnetic**.

### Step 3: Analyze the third complex \([Cu(en)_2]^{2+}\)

1. **Oxidation State**: Copper (Cu) is in the +2 oxidation state.
2. **Electron Configuration**: The electron configuration of Cu is \([Ar] 3d^{10} 4s^1\), but in +2 state, it becomes \([Ar] 3d^9\).
3. **Hybridization**: The complex has a square planar geometry, which corresponds to \(dsp^2\) hybridization.
4. **Electron Pairing**: The 3d^9 configuration will have one unpaired electron.
5. **Conclusion**: Since there is one unpaired electron, \([Cu(en)_2]^{2+}\) is **paramagnetic**.

### Step 4: Analyze the fourth complex \([HgI_4]^{2-}\)

1. **Oxidation State**: Mercury (Hg) is in the +2 oxidation state.
2. **Electron Configuration**: The electron configuration of Hg is \([Xe] 4f^{14} 5d^{10} 6s^2\), and in +2 state, it becomes \([Xe] 4f^{14} 5d^{10}\).
3. **Hybridization**: The complex has a tetrahedral geometry, which corresponds to \(sp^3\) hybridization.
4. **Electron Pairing**: The 5d^10 configuration is fully filled, meaning there are no unpaired electrons.
5. **Conclusion**: Since there are no unpaired electrons, \([HgI_4]^{2-}\) is **diamagnetic**.

### Final Conclusion

The complexes that are diamagnetic are:
- \([Pt(NH_3)_4]^{2+}\)
- \([HgI_4]^{2-}\)

Thus, the answer is that the diamagnetic complexes are **1 and 4**.