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 planar

Explanation

To determine the hybridization of the central atom in the complex ion \([B(OH)_4]^-\) and its geometry, we can follow these steps:

### Step 1: Identify the central atom and its valence electrons
The central atom in the complex is boron (B). Boron has an atomic number of 5, which means it has 3 valence electrons (the electronic configuration is 1s² 2s² 2p¹).

**Hint:** Remember that the number of valence electrons can be determined from the atomic number and the electron configuration.

### Step 2: Determine the number of bonds
In the complex \([B(OH)_4]^-\), boron is bonded to four hydroxyl groups (OH). Each hydroxyl group contributes one bond.

**Hint:** Count the number of bonds formed by the central atom with surrounding atoms or groups.

### Step 3: Calculate the steric number
The steric number (SN) can be calculated using the formula:
\[ \text{Steric Number} = \frac{1}{2} \left( V + M - C + A \right) \]
Where:
- \(V\) = number of valence electrons of the central atom (3 for boron)
- \(M\) = number of monovalent bonds (4 from the four OH groups)
- \(C\) = charge of cation (0, since there is no cation here)
- \(A\) = charge of anion (1, since the overall charge is -1)

Substituting the values:
\[ \text{Steric Number} = \frac{1}{2} \left( 3 + 4 - 0 + 1 \right) = \frac{1}{2} \times 8 = 4 \]

**Hint:** The steric number helps determine the hybridization and geometry of the molecule.

### Step 4: Determine the hybridization
A steric number of 4 corresponds to \(sp^3\) hybridization.

**Hint:** Recall that different steric numbers correspond to different types of hybridization (e.g., SN=2 → sp, SN=3 → sp², SN=4 → sp³).

### Step 5: Determine the geometry
For \(sp^3\) hybridization, the geometry is tetrahedral. This is because there are four bond pairs and no lone pairs around the central atom.

**Hint:** The geometry can often be inferred from the hybridization type.

### Final Answer
The hybridization of the central atom in \([B(OH)_4]^-\) is \(sp^3\) and the geometry of the complex is tetrahedral.

**Summary:**
- Hybridization: \(sp^3\)
- Geometry: Tetrahedral