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

Question

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

Step 1: Analyze Reaction 1

The reaction given is: \( _{4} \mathrm{Be}^{9} + _{2} \mathrm{He}^{4} \rightarrow _{6} \mathrm{C}^{12} + \ldots \) This equation must be balanced for mass and atomic number. On the left side, the atomic numbers add up to 6 and the mass numbers add up to 13. The right side has \( _{6} \mathrm{C}^{12} \) with atomic number 6 and mass number 12. So, we need a particle with zero atomic number and mass number 1 to balance it: \( _{0} \mathrm{n}^{1} \). Thus, the missing particle in Reaction 1 is \( \mathrm{n}^{1} \) which matches to (ii).

Step 2: Analyze Reaction 2

The reaction given is: \( _{6} \mathrm{C}^{12} + \ldots \rightarrow _{5} \mathrm{B}^{10} + _{2} \mathrm{He}^{4} \)On the left side, the atomic number is 6 and the mass number is 12. On the right side, \( _{5} \mathrm{B}^{10} + _{2} \mathrm{He}^{4} \) has a total atomic number of 7 and a mass number of 14. To balance this, the missing particle must have an atomic number of 1 and a mass number of 2, which corresponds to \( _{1} \mathrm{D}^{2} \). Hence, Reaction 2 matches (iii).

Step 3: Analyze Reaction 3

The reaction given is: \( _{7} \mathrm{N}^{14} + \ldots \rightarrow _{8} \mathrm{O}^{17} + _{1} \mathrm{H}^{1} \) The left side has \( _{7} \mathrm{N}^{14} \). The right side, we have \( _{8} \mathrm{O}^{17} + _{1} \mathrm{H}^{1} \), total atomic number of 9 and mass number of 18. The missing particle on the left side needs to have an atomic number of 2 and a mass number of 4 to balance. Therefore, \( _{2} \mathrm{He}^{4} \) is the particles that matches Reaction 3, which corresponds to (i).

Step 4: Analyze Reaction 4

The reaction given is: \( _{20} \mathrm{Ca}^{40} + \ldots \rightarrow _{19} \mathrm{K}^{37} + _{2} \mathrm{He}^{4} \)The left side has \( _{20} \mathrm{Ca}^{40} \), and the right side \( _{19} \mathrm{K}^{37} + _{2} \mathrm{He}^{4} \) has a total atomic number of 21 and mass number of 41. The missing particle needs to have an atomic number of 1 and a mass number of 1, which is a proton \( _{1} \mathrm{H}^{1} \). Thus, Reaction 4 matches (iv).

Step 5: Determine the Matching

After analyzing each reaction, we have identified the matches:- Reaction 1 matches (ii) \( \mathrm{n}^{1} \)- Reaction 2 matches (iii) \( _{1} \mathrm{D}^{2} \)- Reaction 3 matches (i) \( _{2} \mathrm{He}^{4} \)- Reaction 4 matches (iv) \( _{1} \mathrm{H}^{1} \)Thus, the correct matching is (c): 1(ii), 2(iii), 3(i), 4(iv).

Key Concepts

Particle Reactions

In nuclear chemistry, particle reactions involve changes at the core of an atom, unlike chemical reactions that involve electrons or chemical bonds. These reactions occur in the atom's nucleus and often result in the transformation of elements. Understanding particle reactions is essential for comprehending processes such as nuclear fission and fusion. In our exercise, each nuclear reaction involves a change induced by a particle like a neutron, a deuteron (heavy hydrogen), or an alpha particle (helium nucleus).
- In Reaction 1, the neutron ( ^{1}) completes the nuclear reaction by balancing the mass and atomic numbers. - Reaction 2 uses a deuteron ( _{1} ext{D}^{2}) to achieve balance, which is a useful particle in nuclear reactions for adding both a proton and a neutron. - An alpha particle ( _{2} ext{He}^{4}) is involved in Reaction 3, demonstrating typical behavior as it commonly participates due to its stability. - Finally, a proton ( _{1} ext{H}^{1}) is involved in Reaction 4, pointing to the simplest form of hydrogen, crucial in many nuclear reactions. Understanding these particles and their interactions allows for predictions and descriptions of new nuclear reactions.

Mass Number Balance

Balancing a nuclear equation requires careful attention to both mass numbers, indicated as the superscript in nuclear symbols. In a nuclear reaction, mass balance ensures the total mass number on the reactant side equals the total mass number on the product side. This is crucial for predicting the resultant particles.
- For Reaction 1, the total mass on the left (9 from Beryllium and 4 from Helium) equals 13, needing an atomic arrangement on the right that also totals 13. - In Reaction 2, the initial setup of mass 12 balances against a total output mass of 14, implying the missing particle would have a mass number 2. - Reaction 3 poses a similar challenge, with the resultant mass increased to 18, hence demanding a balancing mass on the left including the particle involved. - Reaction 4 calls for precision, with changed nuclear configurations maximizing just one unit of mass evident in the contrast between inputs and outputs. The essence of mass number balance reflects the conservation of nucleons during nuclear transformations.

Atomic Number Balance

The atomic number, shown as the subscript in nuclear notations, represents the number of protons and defines the element. When balancing a nuclear reaction, the atomic numbers must be equal on both sides to maintain the law of conservation of charge. - In Reaction 1, the atomic number changes from a sum of 6 to 6, necessitating a null change with a neutron ( ^{1}). - Reaction 2 transitions from an atomic number of 6 (from _{6} ext{C}^{12}) to 7 (combination of _{5} ext{B}^{10} and _{2} ext{He}^{4}), needing an accessory particle with atomic number 1. - Reaction 3 illustrates a jump in the atomic number from 7 to 9 by the presence of _{8} ext{O}^{17} and _{1} ext{H}^{1}. - While Reaction 4 highlights increment bracketing, 20 transitioning to 21 due to the interaction quantum. These atomic transformations illustrate the symmetry in charge and the dynamic interactions within nuclear systems.

Nuclear Equations

Nuclear equations are a symbolic representation of nuclear reactions, balancing mass and atomic numbers to predict the products and reactants.
These equations are fundamental in understanding the transition of one element to another due to changes in the nucleus. - Reaction 1: The equation _{4} ext{Be}^{9} + _{2} ext{He}^{4} ightarrow _{6} ext{C}^{12} + ^{1}, demonstrates conservation of both mass and atomic numbers through the completion of neutron interaction. - Reaction 2: _{6} ext{C}^{12} + _{1} ext{D}^{2} ightarrow _{5} ext{B}^{10} + _{2} ext{He}^{4} maintains equilibrium via a deuteron's involvement. - Reaction 3: Highlights are given by _{7} ext{N}^{14} + _{2} ext{He}^{4} ightarrow _{8} ext{O}^{17} + _{1} ext{H}^{1}, where an alpha particle is integral. - Reaction 4: _{20} ext{Ca}^{40} + _{1} ext{H}^{1} ightarrow _{19} ext{K}^{37} + _{2} ext{He}^{4}, exhibits a proton's pivotal balancing role. By writing nuclear equations this way, we can predict and verify the outcomes of nuclear interactions.