Consider the following ionization enthalpies of two elements 'A' and 'B'.

Element	1st (kJ/mol)	2nd (kJ/mol)	3rd (kJ/mol)
A	899	1757	14847
B	737	1450	7731

Which of the following statements is correct?

First we write down the given data clearly in symbolic form.

For element $$A$$: $$IE_1(A)=899\;{\rm kJ\,mol^{-1}},\; IE_2(A)=1757\;{\rm kJ\,mol^{-1}},\; IE_3(A)=14847\;{\rm kJ\,mol^{-1}}.$$

For element $$B$$: $$IE_1(B)=737\;{\rm kJ\,mol^{-1}},\; IE_2(B)=1450\;{\rm kJ\,mol^{-1}},\; IE_3(B)=7731\;{\rm kJ\,mol^{-1}}.$$

To locate the position of the elements in the periodic table we examine where a very large jump in ionization enthalpy occurs. The principle we shall use is:

“When electrons are removed successively, a sudden, exceptionally large increase in ionization enthalpy indicates that the next electron is being taken from a new, inner shell which has a noble-gas configuration. The number of electrons removed before that jump equals the number of valence electrons, and hence equals the group number for the s-block.”

So we calculate the successive differences.

For $$A$$:

$$\Delta_1(A)=IE_2(A)-IE_1(A)=1757-899=858\;{\rm kJ\,mol^{-1}}$$

$$\Delta_2(A)=IE_3(A)-IE_2(A)=14847-1757=13090\;{\rm kJ\,mol^{-1}}$$

For $$B$$:

$$\Delta_1(B)=IE_2(B)-IE_1(B)=1450-737=713\;{\rm kJ\,mol^{-1}}$$

$$\Delta_2(B)=IE_3(B)-IE_2(B)=7731-1450=6281\;{\rm kJ\,mol^{-1}}$$

We observe that in each case $$\Delta_2\gg\Delta_1$$, that is, the third ionization enthalpy is enormously larger than the second, while the first two are of comparable, moderate magnitude. Thus only the first two electrons can be removed relatively easily, meaning that each element possesses two valence electrons. Hence both elements belong to the alkaline-earth family, i.e. Group-2.

Now we decide which one lies lower in the group. The rule we use is:

“Ionization enthalpy decreases down a group because atomic radius increases and the outer electrons are held less tightly.”

Comparing the first ionization enthalpies we find

$$IE_1(A)=899\;{\rm kJ\,mol^{-1}} \quad\text{and}\quad IE_1(B)=737\;{\rm kJ\,mol^{-1}}.$$

Because $$737<899$$, element $$B$$ loses its first electron more easily than element $$A$$. Therefore $$B$$ is situated below $$A$$ in the same group.

The only option that states “both in Group-2” and “$$B$$ comes below $$A$$” is Option D.

Hence, the correct answer is Option D.