The correct regarding `CuCl_(5)^(-3)` compound is

Options

A. Hybridisation is `sp^(3)d`
B. Axial bond length is larger than equitorial bond length
C. Equatorial bond length is longer than axial bond length
D. Both (1) & (3)

Answer

To solve the question regarding the compound \( \text{CuCl}_5^{-3} \), we will follow these steps:

### Step 1: Determine the Oxidation State of Copper
Let the oxidation state of copper be \( X \). The charge on each chlorine atom is \(-1\), and since there are 5 chlorine atoms, the total contribution from chlorine is \(-5\). The overall charge of the complex is \(-3\).

Setting up the equation:
\[ X + (-5) = -3 \]
Solving for \( X \):
\[ X - 5 = -3 \\ X = -3 + 5 \\ X = +2 \]
So, the oxidation state of copper in \( \text{CuCl}_5^{-3} \) is \( +2 \).

### Step 2: Determine the Electron Configuration of Copper
The ground state electron configuration of copper (Cu) is:
\[ \text{[Ar]} \, 3d^{10} \, 4s^1 \]
In the \( +2 \) oxidation state, copper loses two electrons: one from the \( 4s \) orbital and one from the \( 3d \) orbital. Therefore, the electron configuration becomes:
\[ \text{[Ar]} \, 3d^9 \]

### Step 3: Determine the Hybridization
To determine the hybridization, we need to consider the ligands and the electron configuration. Chlorine is a weak field ligand, which means it does not cause pairing of electrons in the \( 3d \) orbitals.

- The \( 3d \) subshell has 9 electrons, which can be arranged as follows:
- 5 unpaired electrons will occupy the 5 \( 3d \) orbitals.
- The remaining 4 electrons will pair up in the \( 3d \) orbitals.

Since there are 5 chlorine ligands, we need to hybridize one \( s \), three \( p \), and one \( d \) orbital:
\[ \text{Hybridization} = sp^3d \]

### Step 4: Determine the Geometry
The geometry associated with \( sp^3d \) hybridization is trigonal bipyramidal. In this geometry:
- There are 5 positions: 3 equatorial and 2 axial.
- The axial bond lengths are typically shorter than the equatorial bond lengths due to the arrangement of the ligands.

### Step 5: Analyze the Bond Lengths
In a trigonal bipyramidal geometry:
- The axial bond lengths are shorter than the equatorial bond lengths. Therefore, the statement that "the axial bond length is larger than the equatorial bond length" is incorrect.

### Conclusion
1. The hybridization of \( \text{CuCl}_5^{-3} \) is \( sp^3d \) (Correct).
2. The geometry is trigonal bipyramidal (Correct).
3. The equatorial bond length is longer than the axial bond length (Correct).

Thus, the correct statements about \( \text{CuCl}_5^{-3} \) are:
- The hybridization is \( sp^3d \).
- The equatorial bond length is longer than the axial bond length.

### Final Answer
The correct options regarding \( \text{CuCl}_5^{-3} \) are statements 1 and 3.

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