Borax `[Na_(2)B_(4)O_(7).10H_(2)O]` when heated on platinum loop it gives a dark transparent glass like bead. The hot bead is dipped in the salt till it reacts with transition metal oxide. It produces characteristic bead of meta borate.
`{:(" Colour of the bead"," lon"),("(a) Blue green or light blue",Cu^(+2)),("(b) Yellow",Fe^(+2)or Fe^(+3)),("(c) Green",Cr^(+3)),("(d) Violet",Mn^(+2)),("(e) Dark blue",Co^(+2)),("(f) Brown",Ni^(+2)):}`
The hybridisation of B in Borax is

Options

A. `sp^(2)`
B. `sp^(3)`
C. Both (1) & (2)
D. sp

Answer

To determine the hybridization of boron in borax \([Na_2B_4O_7 \cdot 10H_2O]\), we will analyze the structure of borax and calculate the hybridization for the two types of boron atoms present in the molecule.

### Step 1: Understand the Structure of Borax
Borax consists of boron (B), oxygen (O), sodium (Na), and water molecules. The boron atoms in borax are bonded to oxygen atoms, forming a network structure. There are two types of boron atoms in borax:
- Boron labeled as 1 (B1)
- Boron labeled as 2 (B2)

### Step 2: Determine the Valence Electrons
Each boron atom has 3 valence electrons.

### Step 3: Identify the Monovalent Atoms Attached
In borax, each boron atom is attached to hydroxyl groups (OH) and oxygen atoms. The number of monovalent atoms (which are typically hydrogen atoms in hydroxyl groups) attached to each boron atom needs to be counted.

### Step 4: Calculate the Steric Number for B1
For boron labeled as 1:
- Valence electrons = 3
- Monovalent atoms = 3 (from 3 OH groups)
- Cationic charge = 0
- Anionic charge = 0

Using the formula for hybrid number (steric number):
\[ \text{Steric Number} = \frac{1}{2} \left( \text{Valence electrons} + \text{Monovalent atoms} - \text{Cationic charge} + \text{Anionic charge} \right) \]
\[ \text{Steric Number for B1} = \frac{1}{2} (3 + 3 - 0 + 0) = \frac{1}{2} \times 6 = 3 \]

### Step 5: Determine Hybridization for B1
A steric number of 3 corresponds to \(sp^2\) hybridization.

### Step 6: Calculate the Steric Number for B2
For boron labeled as 2:
- Valence electrons = 3
- Monovalent atoms = 4 (3 from OH groups and 1 from the bridging oxygen)
- Cationic charge = 0
- Anionic charge = 1 (due to the negative charge on the borate ion)

Using the same formula:
\[ \text{Steric Number for B2} = \frac{1}{2} (3 + 4 - 0 + 1) = \frac{1}{2} \times 8 = 4 \]

### Step 7: Determine Hybridization for B2
A steric number of 4 corresponds to \(sp^3\) hybridization.

### Conclusion
In borax, we have:
- Boron labeled as 1 (B1) is \(sp^2\) hybridized.
- Boron labeled as 2 (B2) is \(sp^3\) hybridized.

Thus, the hybridization of boron in borax consists of both \(sp^2\) and \(sp^3\) hybridizations.

### Final Answer
The hybridization of B in borax is both \(sp^2\) and \(sp^3\).

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