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
<p>The most abundant isotope of Boron is ${ }_5 \mathrm{~B}^{11}$.</p>
<p>No. of neutrons in it $=x=6$</p>
<p>$$2 \mathrm{B}(\mathrm{s})+\mathrm{N}_2(\mathrm{g}) \xrightarrow{\Delta} 2 \mathrm{BN}(\mathrm{s})$$</p>
<p>Oxidation state of boron in $\mathrm{B N=y=+3}$</p>
<p>So, $x+y=6+3$</p>
<p>$=9$</p>
About this question
Subject: Chemistry · Chapter: p-Block Elements · Topic: Group 13: Boron Family
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