The element with Z = 120 (not yet discovered) will be a/an

The electronic configuration of any element is obtained by filling the orbitals in the increasing order of their $$n+\ell$$ values, popularly called the Aufbau principle. First we write the order of filling up to the region of interest:

$$1s \,<\, 2s \,<\, 2p \,<\, 3s \,<\, 3p \,<\, 4s \,<\, 3d \,<\, 4p \,<\, 5s \,<\, 4d \,<\, 5p \,<\, 6s \,<\, 4f \,<\, 5d \,<\, 6p \,<\, 7s \,<\, 5f \,<\, 6d \,<\, 7p \,<\, 8s\ldots$$

We already know the electronic configuration of the heaviest experimentally confirmed noble-gas element, oganesson $$\bigl(Z = 118\bigr)$$. It ends with a completely filled $$7p^{6}$$ subshell, that is,

$$\text{Og: } [\text{Rn}]\, 5f^{14}\, 6d^{10}\, 7s^{2}\, 7p^{6}$$

Because $$Z = 118$$ exhausts the $$7p$$ subshell, the very next electrons must enter the subshell that comes after $$7p$$ in the above series, namely $$8s$$.

Hence, for $$Z = 119$$ we add one electron to obtain

$$Z = 119: [\text{Og}]\, 8s^{1}$$

and for $$Z = 120$$ we add the second electron of that $$8s$$ subshell:

$$Z = 120: [\text{Og}]\, 8s^{2}$$

An element whose outermost electronic configuration is $$ns^{1}$$ belongs to Group 1 (the alkali metals), whereas an element with $$ns^{2}$$ belongs to Group 2 (the alkaline-earth metals). We have just shown that the outermost configuration of the element with $$Z = 120$$ is $$8s^{2}$$.

Therefore, the element with atomic number $$120$$ will exhibit the characteristic properties of the Group 2 family, that is, it will be an alkaline earth metal.

Hence, the correct answer is Option 2.