The number of neutrons present in the more abundant isotope of boron is '$$x$$'. Amorphous boron upon heating with air forms a product, in which the oxidation state of boron is '$$y$$'. The value of $$x + y$$ is ______

Find x (neutrons in more abundant isotope of boron).

Boron has two isotopes: $$^{10}B$$ (19.9%) and $$^{11}B$$ (80.1%). The more abundant isotope is $$^{11}B$$.

Number of neutrons in $$^{11}B = 11 - 5 = 6$$. So $$x = 6$$.

Find y (oxidation state of boron in the product with air).

When amorphous boron is heated with air, it forms boron trioxide ($$B_2O_3$$):

$$ 4B + 3O_2 \rightarrow 2B_2O_3 $$

In $$B_2O_3$$, the oxidation state of boron is $$+3$$. So $$y = +3$$.

Therefore, $$x + y = 6 + 3 = 9$$.

The correct answer is Option (2): 9.