JEE MAIN - Chemistry (2024 - 5th April Morning Shift - No. 13)
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 _________ .
9
6
4
3
Explanation
The most abundant isotope of Boron is $${ }_5 \mathrm{~B}^{11}$$.
No. of neutrons in it $$=x=6$$
$$2 \mathrm{B}(\mathrm{s})+\mathrm{N}_2(\mathrm{g}) \xrightarrow{\Delta} 2 \mathrm{BN}(\mathrm{s})$$
Oxidation state of boron in $$\mathrm{B N=y=+3}$$
So, $$x+y=6+3$$
$$=9$$
Comments (0)
