A s-block element (M) reacts with oxygen to form an oxide of the formula MO$$_2$$. The oxide is pale yellow in colour and paramagnetic. The element (M) is:

We are told that an s-block element M reacts with oxygen to form an oxide of formula MO$$_2$$, which is pale yellow in colour and paramagnetic.

Among s-block elements, when reacted with excess oxygen: lithium forms Li$$_2$$O (normal oxide), sodium forms Na$$_2$$O$$_2$$ (peroxide), potassium forms KO$$_2$$ (superoxide), rubidium and caesium also form superoxides. Calcium, barium, and strontium can form peroxides CaO$$_2$$, BaO$$_2$$, etc., and magnesium forms MgO.

The formula MO$$_2$$ with M being an s-block element points to either a peroxide (like Na$$_2$$O$$_2$$, CaO$$_2$$) or a superoxide (KO$$_2$$, RbO$$_2$$). A superoxide KO$$_2$$ has the formula MO$$_2$$ where M is K (potassium, atomic mass ~39), and potassium is indeed an s-block element. Superoxides contain the O$$_2^-$$ ion, which has an unpaired electron, making it paramagnetic. KO$$_2$$ is also known to be pale yellow in colour.

In contrast, peroxides like Na$$_2$$O$$_2$$ have a different formula (M$$_2$$O$$_2$$, not MO$$_2$$) and CaO$$_2$$ is a peroxide that is white and diamagnetic. Superoxides contain O$$_2^-$$ with one unpaired electron, explaining the paramagnetism. The pale yellow colour is characteristic of KO$$_2$$.

Therefore, element M is potassium (K), which is option 4.