The number of neutrons present in the more abundant isotope of boron is 'x'. Amorphous boron upon heating with air…

Strategy: Find x (neutrons in the most abundant boron isotope) and y (oxidation state of boron in its oxide with air), then sum them.

Step 1: Most Abundant Isotope of Boron Boron has two naturally occurring isotopes: $\ce{^{10}B}$ (~20%) and $\ce{^{11}B}$ (~80%). The more abundant isotope is $\ce{^{11}B}$.

• Atomic number = 5, Mass number = 11
• Neutrons = 11 − 5 = 6 So $x = 6$.

Step 2: Product of Boron + Air (Oxidation State) $\ce{4B + 3O2 \xrightarrow{\Delta} 2B2O3}$ In $\ce{B2O3}$: Let ox. state of B = y. Then $2y + 3(-2) = 0 \Rightarrow y = +3$. So $y = +3$.

Step 3: Sum $x + y = 6 + 3 = \boxed{9}$

$\boxed{\text{Answer: (B) 9}}$