In the given electrolysis we have moles of Mg produced is equal to moles of Cl2 produced
Moles of Mg = (given mass / molecular mass) = (6.5/24) = 0.27
therefore moles of Cl2 produced = 0.27
at NTP condition volume of 1 mole of gas = 22.4 lit
hence, volume of 0.27 moles of Cl2 = 0.27 × 22.4 = 6.048 which is nearly equal to 5.99

So, here correct answer is 5.99

$$Hope it's help... $$

To calculate the volume of Cl2 gas produced during the electrolysis of MgCl2, we need to determine the number of moles of Mg produced and use the stoichiometry of the reaction to find the number of moles of Cl2 produced.

**1. Calculate the number of moles of Mg produced:**
- Given: Mass of Mg = 6.5 g
- The molar mass of Mg = 24.31 g/mol
- Number of moles of Mg = Mass of Mg / Molar mass of Mg = 6.5 g / 24.31 g/mol ≈ 0.2675 mol

**2. Use the stoichiometry of the reaction to find the number of moles of Cl2 produced:**
The balanced equation for the electrolysis of MgCl2 is:
2MgCl2 → 2Mg + Cl2

From the equation, we see that 2 moles of MgCl2 produce 1 mole of Cl2.
Therefore, the number of moles of Cl2 produced is also 0.2675 mol.

**3. Calculate the volume of Cl2 gas at NTP:**
At NTP (Normal Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 L.
Therefore, the volume of Cl2 gas produced is:
Volume of Cl2 = Number of moles of Cl2 × Molar volume at NTP
= 0.2675 mol × 22.4 L/mol
≈ 5.99 L

Hence, the volume of Cl2 gas produced during the electrolysis of MgCl2 at NTP is approximately 5.99 L.

Therefore, the correct option is A. 5.99 L.