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 .
- 9
- 6
- 4
- 3
Answer
(1) 9
Solution
Boron: 10B ( 19\%) and 11B ( 81\%) ⇒ more abundant 115B
⇒ x = 11 - 5 = 6 neutrons
4B + 3O2 2B2O3 (with N2 of air: BN) ⇒ B is +3 in both ⇒ y = 3
∴ x + y = 6 + 3 = 9
Answer: (1)