The geometry of a complex species can be understood from the knowledge of type of hybridisation of orbitals of central atom. The hybridisation of orbitals of central atom in `[B(OH_(4))]^(-)` and the geometry of the complex are respectively.

Options

A. `sp^(3)`, tetrahedral
B. `sp^(3)`, square planar
C. `sp^(3), d^(2)` octahedral
D. `dsp^(2)` square plannar.

Explanation

To determine the hybridization and geometry of the complex ion \([B(OH_4)]^{-}\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Central Atom**:
The central atom in the complex ion \([B(OH_4)]^{-}\) is boron (B).

**Hint**: Always start by identifying the central atom in the complex.

2. **Determine the Oxidation State of Boron**:
In the complex, there are four hydroxide ions (OH\(^-\)). The overall charge of the complex is -1.
Let the oxidation state of boron be \(x\):
\[ x + 4(-1) = -1 \implies x - 4 = -1 \implies x = +3 \]
Therefore, the oxidation state of boron in \([B(OH_4)]^{-}\) is +3.

**Hint**: Use the charge balance to find the oxidation state of the central atom.

3. **Write the Electronic Configuration of Boron**:
The electronic configuration of boron (atomic number 5) is:
\[ 1s^2 \, 2s^2 \, 2p^1 \]

**Hint**: Knowing the atomic number helps in writing the electronic configuration.

4. **Identify the Available Orbitals for Hybridization**:
In the +3 oxidation state, boron will have the following configuration:
\[ 1s^2 \, 2s^2 \, 2p^0 \quad (\text{since one } 2p \text{ electron is removed}) \]
Boron will have 4 orbitals available for hybridization (1s and 3p orbitals).

**Hint**: Determine which orbitals are involved in bonding based on the oxidation state.

5. **Determine the Hybridization**:
Since boron is forming four bonds with the four hydroxide ions, it undergoes \(sp^3\) hybridization. This involves the mixing of one s orbital and three p orbitals to form four equivalent \(sp^3\) hybrid orbitals.

**Hint**: The number of bonds formed indicates the type of hybridization.

6. **Determine the Geometry**:
The geometry associated with \(sp^3\) hybridization is tetrahedral. Therefore, the geometry of the complex \([B(OH_4)]^{-}\) is tetrahedral.

**Hint**: The hybridization type directly correlates with the geometry of the molecule.

### Final Answer:
The hybridization of the central atom in \([B(OH_4)]^{-}\) is \(sp^3\) and the geometry of the complex is tetrahedral.