A metal on combustion in excess air forms X. X upon hydrolysis with water yields $$H_2O_2$$ and $$O_2$$ along with another product. The metal is:

First, we recall the general behaviour of alkali-metal oxides. The possible oxides obtained on burning an alkali metal $$M$$ in excess air are:

$$M_2O$$   (normal oxide),   $$M_2O_2$$   (peroxide),   $$MO_2$$   (superoxide).

The nature of the oxide depends on the size of the metal cation. As we move down the group from Li to Cs, the ionic radius increases, and the larger cations can stabilise the larger $$O_2^-$$ superoxide ion more easily. Hence:

$$\text{Li} \longrightarrow \text{oxide}\;(Li_2O),$$
$$\text{Na} \longrightarrow \text{peroxide}\;(Na_2O_2),$$
$$\text{K},\;\text{Rb},\;\text{Cs} \longrightarrow \text{superoxide}\;(KO_2,\;RbO_2,\;CsO_2).$$

Now, the question states that the oxide $$X$$ on hydrolysis with water gives $$H_2O_2$$ and $$O_2$$ along with another product. Let us write the general hydrolysis reaction for a superoxide:

$$2\,MO_2 + 2\,H_2O \;\longrightarrow\; 2\,MOH + H_2O_2 + O_2.$$

Here we clearly see both $$H_2O_2$$ and $$O_2$$ appearing simultaneously, exactly as described. Thus, $$X$$ must be a superoxide $$MO_2$$.

Among the given metals, the one that most readily forms a superoxide on combustion in excess air is rubidium (Rb). Lithium (Li) forms only the normal oxide, sodium (Na) forms mainly the peroxide, magnesium (Mg) is not an alkali metal at all, while rubidium, being large, stabilises the superoxide $$RbO_2$$.

Therefore the metal whose combustion product fits the data is rubidium.

Hence, the correct answer is Option D.