1. Identify the metal from ionisation enthalpies

The given ionisation enthalpies are:

• First ionisation enthalpy $= 496\,\text{kJ mol}^{-1}$
• Second ionisation enthalpy $= 4560\,\text{kJ mol}^{-1}$

There is a very large jump from the first to the second ionisation enthalpy. This means that after losing one electron, the metal attains a stable noble gas configuration.

So the metal is a Group 1 metal and forms a monovalent ion: $M \to M^+ + e^-$

Hence, its hydroxide will be: $MOH$

2. Reaction with $HCl$

The reaction is: $MOH + HCl \to MCl + H_2O$

From the equation,

• $1$ mole of $MOH$ reacts with $1$ mole of $HCl$

So for $1$ mole of metal hydroxide, required $HCl = 1$ mole.

3. Reaction with $H_2SO_4$

Since $H_2SO_4$ is dibasic, the balanced reaction is: $2MOH + H_2SO_4 \to M_2SO_4 + 2H_2O$

From the equation,

• $2$ moles of $MOH$ react with $1$ mole of $H_2SO_4$

Therefore, for $1$ mole of $MOH$, required $H_2SO_4$ is: $\frac{1}{2}\text{ mole} = 0.5\text{ mole}$

4. Final answer

Thus, the required moles of $HCl$ and $H_2SO_4$, respectively, are: $1 \text{ and } 0.5$

So the correct option is B.