Question

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 ...

• A. 4
• B. 6
• C. 3
• D. 9

Answer 1

The Correct Option is D

The more abundant isotope of boron is ¹¹B, which contains 6 neutrons (x = 6). When boron reacts with oxygen:

B + O₂ → B₂O₃

In B₂O₃, boron has an oxidation state of +3 (y = 3). Thus, x + y = 6 + 3 = 9.

Answer 2

Step 1: Identify the more abundant isotope of boron
Boron has two stable isotopes:

• \( ^{10}\text{B} \) with 19.9% abundance
• \( ^{11}\text{B} \) with 80.1% abundance

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

Step 2: Calculate the number of neutrons in \( ^{11}\text{B} \)
Number of neutrons \( = \text{mass number} - \text{atomic number} \).
For boron, atomic number = 5.
Hence, for \( ^{11}\text{B} \): \[ x = 11 - 5 = 6 \] So, the more abundant isotope of boron has **6 neutrons**.

Step 3: Write the reaction of amorphous boron with air (oxygen)
When amorphous boron is heated in air, it forms boron trioxide (\( \text{B}_2\text{O}_3 \)): \[ 4\text{B (s)} + 3\text{O}_2(g) \rightarrow 2\text{B}_2\text{O}_3(s) \]
Step 4: Determine the oxidation state of boron in B₂O₃
Let the oxidation state of boron be \( y \). Since oxygen has oxidation state –2: \[ 2y + 3(-2) = 0 \Rightarrow 2y - 6 = 0 \Rightarrow y = +3. \] Thus, boron has an oxidation state of **+3** in the oxide.

Step 5: Calculate the required value
\[ x + y = 6 + 3 = 9. \]

Final answer

9