The number of neutrons present in the more ab | vedclass

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Question

$	ext { More abundant isotope }=\mathrm{B}^{11}$

${[	ext { Number of neutrons }=6]}$

$\mathrm{x}=6$

$\mathrm{~B}+\mathrm{O}_2 ightarrow \mathrm

Vedclass · Accepted Answer

$9$

Vedclass · Answer

$	ext { More abundant isotope }=\mathrm{B}^{11}$

${[	ext { Number of neutrons }=6]}$

$\mathrm{x}=6$

$\mathrm{~B}+\mathrm{O}_2 ightarrow \mathrm{B}_2 \mathrm{O}_3$

Oxidation state of $\mathrm{B}$ in $\mathrm{B}_2 \mathrm{O}_3=+3$

So, $\mathrm{y}=3$

Hence $x+y=9$

The number of neutrons present in the more abundant isotope of boron is ' $\mathrm{x}$ '. Amorphous boron upon heating with air forms a product, in which the oxidation state of boron is ' $y$ '. The value of $x+y$ is ...