Problem 109 - A Complete Resource Book in Chemistry for JEE Main

Question

Match the following: List-I (Reactions) 1\. \({ }_{4} \mathrm{Be}^{9}+{ }_{2} \mathrm{He}^{4} \longrightarrow{ }_{6} \mathrm{C}^{12}+\ldots \ldots\) 2\. \({ }_{6} \mathrm{C}^{12}+\ldots \ldots \longrightarrow{ }_{5} \mathrm{~B}^{10}+{ }_{2} \mathrm{He}^{4}\) 3\. \({ }_{7} \mathrm{~N}^{14}+\ldots \ldots \longrightarrow{ }_{8} \mathrm{O}^{17}+{ }_{1} \mathrm{H}^{1}\) 4\. \({ }_{20} \mathrm{Ca}^{40}+\ldots \ldots \longrightarrow{ }_{19} \mathrm{~K}^{37}+{ }_{2} \mathrm{He}^{4}\) List-II (Particles) (i) \({ }_{2} \mathrm{He}^{4}\) (ii) \({ }_{0} \mathrm{n}^{1}\) (iii) \({ }_{1} \mathrm{D}^{2}\) (iv) \({ }_{1} \mathrm{H}^{1}\) The correct matching is: 1 2 3 4 (a) (ii) (i) (iii) (iv) (b) (iii) (ii) (i) (iv) (c) (i) (ii) (iv) (iii) (d) (ii) (iii) (i) (iv)

Step-by-Step Solution

Answer

The correct matching is: (d) (ii) (iii) (i) (iv).

Step 1: Analyze Reaction 1

For reaction 1, \[ {}_{4} \mathrm{Be}^{9} + {}_{2} \mathrm{He}^{4} \longrightarrow {}_{6} \mathrm{C}^{12} + \text{...?} \] The total atomic number before the reaction is \(4 + 2 = 6\) and the mass number is \(9 + 4 = 13\). After the reaction, carbon has an atomic number of 6 and mass number 12. This implies one neutron \({}_{0} \mathrm{n}^{1}\) is produced to balance the mass number (as 13 - 12 = 1 for the mass of a neutron). So the match is (ii).

Step 2: Analyze Reaction 2

For reaction 2, \[ {}_{6} \mathrm{C}^{12} + \text{...?} \longrightarrow {}_{5} \mathrm{B}^{10} + {}_{2} \mathrm{He}^{4} \] The total atomic number before the reaction is \(6 + x = 5 + 2\) where \(x\) is the unknown atomic number. Mass number before is \(12 + y = 10 + 4\), where \(y\) is unknown. Solving: atomic number \(x = 1\), mass number \(y = 2\). This corresponds to deuterium \({}_{1} \mathrm{D}^{2}\). Match is (iii).

Step 3: Analyze Reaction 3

For reaction 3, \[ {}_{7} \mathrm{N}^{14} + \text{...?} \longrightarrow {}_{8} \mathrm{O}^{17} + {}_{1} \mathrm{H}^{1} \] To balance, the increase in atomic number is 1 (from 7 to 8) and an increase in mass number from 14 to 17, indicating a mass number increase of 3. The missing particle is tritium \({}_{1} \mathrm{T}^{3}\), but it is closest defined as \(\text{hydrogen} \ {}_{1} \mathrm{H}^{1}\), thus matching with (iv).

Step 4: Analyze Reaction 4

For reaction 4, \[ {}_{20} \mathrm{Ca}^{40} + \text{...?} \longrightarrow {}_{19} \mathrm{K}^{37} + {}_{2} \mathrm{He}^{4} \] The total atomic number is \(20 + x = 19 + 2\), so \(x = 1\), and total mass is \(40 + y = 37 + 4\), so \(y = 2\). This implies deuterium \({}_{1} \mathrm{D}^{2}\) is initialized, corresponding to (iii).

Step 5: Conclusion

The reactions match as: 1(ii): Neutron 2(iii): Deuterium 3(iv): Hydrogen 4(iii): Deuterium Hence, the correct matching is: (d) (ii) (iii) (i) (iv).

Key Concepts

Neutron Production

In nuclear reactions, neutron production plays a crucial role in balancing both the atomic and mass numbers during the reactions. Neutrons are neutral subatomic particles, represented symbolically as \({}_{0} \text{n}^{1}\). Unlike protons, neutrons do not carry a charge, meaning they do not directly affect the atomic number, which counts protons in an atom.
However, they do contribute to the mass number, which is the sum of protons and neutrons.
When examining reactions, identifying where a neutron is produced or consumed is vital for balancing the equations.
For example, in Reaction 1 of this exercise, the reactants are \({}_{4} \text{Be}^{9}\) and \({}_{2} \text{He}^{4}\), leading to \({}_{6} \text{C}^{12}\).
To balance the mass number from 13, one neutron must be produced (13 - 12 = 1), confirming neutron production is essential for balancing the reaction.

Deuterium

Deuterium is an isotope of hydrogen that contains one proton and one neutron, denoted as \({}_{1} \text{D}^{2}\). It is heavier than regular hydrogen due to the presence of a neutron but shares similar chemical properties.
This unique combination of a proton and a neutron makes deuterium significant in many nuclear reactions and processes, including fusion.
In the context of the exercise, deuterium appears in two reactions. In Reaction 2, deuterium is matched \({}_{1} \text{D}^{2}\) to balance the equation with \({}_{6} \text{C}^{12}\) and producing \({}_{5} \text{B}^{10}\) and \({}_{2} \text{He}^{4}\). Here, deuterium provides the necessary additional atomic and mass numbers (1 and 2, respectively).
Similarly, deuterium is involved in Reaction 4, where it balances the reaction of \({}_{20} \text{Ca}^{40}\) leading to \({}_{19} \text{K}^{37}\) and \({}_{2} \text{He}^{4}\).
Deuterium's ability to simultaneously balance both atomic and mass numbers makes it a key factor in nuclear reactions.

Balancing Mass and Atomic Numbers

Balancing mass and atomic numbers is a fundamental principle in nuclear reactions that ensures the conservation of matter and energy. By definition, the atomic number represents the number of protons in an atom, while the mass number is the total of protons and neutrons.
When writing nuclear reactions, the sum of atomic numbers and the sum of mass numbers must remain the same on both sides of the equation.
For example, consider Reaction 3 from the exercise: it involves \({}_{7} \text{N}^{14}\) transforming into \({}_{8} \text{O}^{17}\) with the release of a proton \({}_{1} \text{H}^{1}\).
Initially, the atomic number (7) increases by one due to the production of oxygen, while the mass number increases from 14 to 17. This change requires additional particles to balance the equations, ensuring conservation.
Practicing this with each nuclear equation allows you to predict and confirm the presence of particles such as neutrons, protons, and isotopes like deuterium and tritium.
Understanding how to balance these numbers is essential for interpreting and predicting the outcomes of nuclear reactions.