Use the pattern in successive ionization energies

• A very large jump from the $1^{st}$ to the $2^{nd}$ ionization energy means that after removing the first electron, the atom has reached a stable noble-gas configuration.
• This is characteristic of an element with one valence electron, i.e. an alkali metal.

Identify X

For $X$: $IE_1 = 495\ \text{kJ/mol}, \qquad IE_2 = 4563\ \text{kJ/mol}$

There is a huge jump: $4563 - 495 \text{ is very large}$

So, $X$ must have one valence electron  Group 1 element.

Among the options, the Group 1 candidate is Na.

Also, the known first ionization energy of sodium is approximately: $IE_1(\text{Na}) \approx 496\ \text{kJ/mol}$

This matches very well.

Hence, $X = \text{Na}$

Identify Y

For $Y$: $IE_1 = 731\ \text{kJ/mol}, \qquad IE_2 = 1450\ \text{kJ/mol}$

The increase is moderate, not enormous. This suggests that removing the second electron is still from the valence shell, so the element likely has two valence electrons.

That corresponds to a Group 2 element.

Among the options, the Group 2 candidate is Mg.

Also, known values are approximately: $IE_1(\text{Mg}) \approx 738\ \text{kJ/mol}, \qquad IE_2(\text{Mg}) \approx 1451\ \text{kJ/mol}$

This matches closely.

Hence, $Y = \text{Mg}$

Check options

• A: $X=F, Y=Mg$  incorrect, because $X$ shows alkali-metal behavior, not fluorine.
• B: $X=Mg, Y=F$  incorrect.
• C: $X=Na, Y=Mg$  correct.
• D: $X=Mg, Y=Na$  incorrect.

Final answer

$\boxed{X = \text{Na}, \quad Y = \text{Mg}}$

So the correct option is: $\boxed{\text{C}}$