Find $x$: neutrons in the more abundant isotope of boron

Boron has two common isotopes:

• ${}^{10}\text{B}$
• ${}^{11}\text{B}$

The more abundant isotope is ${}^{11}\text{B}$.

Atomic number of boron $= 5$.

Hence number of neutrons: $x = 11 - 5 = 6$

Find $y$: oxidation state of boron in the product formed on heating amorphous boron with air

On heating amorphous boron in air, it forms boron oxide: $4B + 3O_2 \rightarrow 2B_2O_3$

In $B_2O_3$, let oxidation state of boron be $y$.

Oxygen has oxidation state $-2$.

So, $2y + 3(-2) = 0$ $2y - 6 = 0$ $2y = 6$ $y = +3$

Compute $x+y$

$x+y = 6+3 = 9$

Match with options

Option A: 9 is correct.