In the reaction
`3Cl_(2)+6NaOHrarrNaClO_(3)+5NaCl+3H_(2)O`
the element which loses as well as gains electrons is

Options

A. `Na`
B. `O`
C. `Cl`
D. None of these

Explanation

To determine which element in the reaction `3Cl₂ + 6NaOH → NaClO₃ + 5NaCl + 3H₂O` both loses and gains electrons, we need to analyze the oxidation states of the elements involved, particularly chlorine (Cl).

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
- Reactants: \(3Cl_2\) and \(6NaOH\)
- Products: \(NaClO_3\), \(5NaCl\), and \(3H_2O\)

2. **Assign Oxidation States**:
- In \(Cl_2\), the oxidation state of chlorine is 0 (since it is in its elemental form).
- In \(NaClO_3\):
- Sodium (Na) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.
- Let the oxidation state of chlorine in \(NaClO_3\) be \(x\).
- The equation for the oxidation states is: \(1 + x + 3(-2) = 0\)
- This simplifies to: \(x - 5 = 0\) → \(x = +5\).
- In \(NaCl\):
- Sodium (Na) is +1, and chlorine (Cl) is -1.
- In \(H_2O\):
- Hydrogen (H) is +1, and oxygen (O) is -2.

3. **Determine Changes in Oxidation States**:
- Chlorine in \(Cl_2\) (0) is converted to chlorine in \(NaClO_3\) (+5) and in \(NaCl\) (-1).
- This indicates that chlorine is undergoing both oxidation and reduction:
- **Oxidation**: \(Cl^0 \to Cl^{+5}\) (loses electrons)
- **Reduction**: \(Cl^0 \to Cl^{-1}\) (gains electrons)

4. **Conclusion**:
- The element that loses electrons (oxidation) and gains electrons (reduction) is chlorine (Cl).

### Final Answer:
The element which loses as well as gains electrons in the reaction is **Chlorine (Cl)**.