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 _________ .

Solution

Chemistry ma’am · 3 steps

Two small facts to recall: B-11 is the common isotope, and boron burns in air to B₂O₃ at +3.

1. Find x. Natural boron is about 80% ¹¹B and 20% ¹⁰B, so ¹¹B is the more abundant isotope. Boron has atomic number 5, so neutrons = 11 - 5 = 6. That gives x = 6.
2. Find y. Amorphous boron heated in air gives B₂O₃. It also gives some BN, because boron reacts with the nitrogen in air. In B₂O₃, 2y + 3(-2) = 0, so y = +3. Boron is also +3 in BN, so the answer is the same either way.

   4B + 3O₂ →(Δ) 2B₂O₃

3. So. x + y = 6 + 3 = 9, which is option (a).

The idea to keep. Neutrons = mass number minus atomic number, and you must use the isotope the question names. Boron shows +3 in its common compounds: B₂O₃, BN, BF₃, borates.

Answer from the NTA final answer key. Solution by PrepWiser.