JEE Mains · Chemistry · STD 11 -p-Block elements - I
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 \(9\)
Answer & Solution
Correct Answer
(D) \(9\)
Step-by-step Solution
Detailed explanation
\(\text { More abundant isotope }=\mathrm{B}^{11}\) \({[\text { Number of neutrons }=6]}\) \(\mathrm{x}=6\) \(\mathrm{~B}+\mathrm{O}_2 \rightarrow \mathrm{B}_2 \mathrm{O}_3\) Oxidation state of \(\mathrm{B}\) in \(\mathrm{B}_2 \mathrm{O}_3=+3\) So, \(\mathrm{y}=3\) Hence…
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