The hybridisation of Cu in `(NH_(4))_(2)[CuCl_(4)] and Cs_(2)[CuCl_(4)]` is

Options

A. `dsp^(2)` in both
B. `dsp^(2) and sp^(3)` respectively
C. `sp^(3) and dsp^(2)` respectively
D. `sp^(3)` in both

Answer

To determine the hybridization of copper in the complexes \((NH_4)_2[CuCl_4]\) and \(Cs_2[CuCl_4]\), we will follow these steps:

### Step 1: Determine the oxidation state of copper in both complexes.
- In \((NH_4)_2[CuCl_4]\):
- Each Cl has an oxidation state of -1, and there are 4 Cl atoms, contributing a total of -4.
- Each ammonium ion \((NH_4^+)\) has a +1 charge, and there are 2 ammonium ions, contributing a total of +2.
- Let the oxidation state of Cu be \(x\). The equation becomes:
\[ x + (-4) + 2 = 0 \implies x - 4 + 2 = 0 \implies x = +2 \]

- In \(Cs_2[CuCl_4]\):
- Each Cl has an oxidation state of -1, contributing a total of -4.
- Each cesium ion \((Cs^+)\) has a +1 charge, and there are 2 cesium ions, contributing a total of +2.
- Similarly, we have:
\[ x + (-4) + 2 = 0 \implies x = +2 \]

### Step 2: Determine the electronic configuration of copper in the +2 oxidation state.
- The atomic number of copper (Cu) is 29, and its ground state electronic configuration is:
\[ [Ar] 3d^{10} 4s^1 \]
- In the +2 oxidation state, copper loses 2 electrons (1 from 4s and 1 from 3d):
\[ [Ar] 3d^9 \]

### Step 3: Analyze the ligand field and hybridization.
- The ligands in both complexes are Cl\(^-\), which is a weak field ligand. This means that it does not cause pairing of the d-electrons.
- The 3d\(^9\) configuration means there are 9 electrons in the 3d orbitals, and since Cl is a weak field ligand, no pairing occurs.

### Step 4: Determine the hybridization.
- In the case of \((NH_4)_2[CuCl_4]\) and \(Cs_2[CuCl_4]\), the copper ion is surrounded by 4 chloride ions.
- The hybridization can be determined by the number of orbitals involved in bonding:
- The 3d orbitals (3d) do not participate in hybridization due to the weak field nature of Cl.
- Instead, the 4s and 4p orbitals will hybridize to form 4 equivalent sp\(^3\) hybrid orbitals to accommodate the 4 Cl ligands.

### Conclusion:
- Therefore, the hybridization of copper in both \((NH_4)_2[CuCl_4]\) and \(Cs_2[CuCl_4]\) is **sp\(^3\)**.

### Final Answer:
The hybridization of Cu in both \((NH_4)_2[CuCl_4]\) and \(Cs_2[CuCl_4]\) is **sp\(^3\)**.

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