Q14 / 90 · JEE Main 2024 (05 Apr Shift 1) · Chemistry

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
3
Correct answer
B
9
C
4
D
6

Solution

More abundant isotope $=\mathrm{B}^{11}$ [Number of neutrons $=6$ ] $\begin{aligned} & \mathrm{x}=6 \\ & \mathrm{~B}+\mathrm{O}_2 \rightarrow \mathrm{B}_2 \mathrm{O}_3 \end{aligned}$ Oxidation state of $\mathrm{B}$ in $\mathrm{B}_2 \mathrm{O}_3=+3$ So, $y=3$ Hence $x+y=9$

Hence, the correct answer is option: 'B'.

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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

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