- Home
- Standard 11
- Chemistry
p-Block Elements - I
medium
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 ...
A
$4$
B
$6$
C
$3$
D
$9$
(JEE MAIN-2024)
Solution
$\text { More abundant isotope }=\mathrm{B}^{11}$
${[\text { Number of neutrons }=6]}$
$\mathrm{x}=6$
$\mathrm{~B}+\mathrm{O}_2 \rightarrow \mathrm{B}_2 \mathrm{O}_3$
Oxidation state of $\mathrm{B}$ in $\mathrm{B}_2 \mathrm{O}_3=+3$
So, $\mathrm{y}=3$
Hence $x+y=9$
Standard 11
Chemistry
