Chemistry · p-Block (I) (Grp. 13, 14)

JEE Main 2024 — 5 April, Shift 1 — Question 76

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:

Option A:
$4$
Option B:
$6$
Option C:
$3$
Option D:
$9$
Correct

Answer: D

Step-by-step solution

The more abundant isotope of boron is $\mathrm{^{11}B}$ . Number of neutrons, $x = 11 - 5 = 6$

On heating amorphous boron in air, we have

\mathrm{4B + 3O_2 \rightarrow 2B_2O_3}

Let oxidation state of boron = $y$

2y + 3(-2) = 0 \Rightarrow y = +3

Thus, $x+y = 6+3 = 9$ and the correct answer is Option D.

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 5 April, Shift 1
Subject: Chemistry
Chapter: p-Block (I) (Grp. 13, 14)
Topic: Introduction & Properties of Group 13: Boron Family

Given below are two statements, one is labelled as Statement I and the other is labelled as Statement II. Statement I : In group 13, the…

Answer: D

Step-by-step solution

Practise p-Block (I) (Grp. 13, 14)

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