MCQEasyJEE 2025Trends in Atomic Radius & Ionisation Energy

JEE Chemistry 2025 Question with Solution

Given below are two statements:

Statement (I): The first ionization energy of Pb\text{Pb} is greater than that of Sn\text{Sn}. Statement (II): The first ionization energy of Ge\text{Ge} is greater than that of Si\text{Si}.

\text{In light of the above statements, choose the correct answer from the options given below:}

  • A

    Statement I is true but Statement II is false

  • B

    Both Statement I and Statement II are false

  • C

    Statement I is false but Statement II is true

  • D

    Both Statement I and Statement II are true

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

  • Statement (I): The first ionization energy of Pb\text{Pb} is greater than that of Sn\text{Sn}.
  • Statement (II): The first ionization energy of Ge\text{Ge} is greater than that of Si\text{Si}.

Find: Which statement combination is correct.

Ionisation energy generally decreases down a group due to increasing atomic size and distance of electrons from the nucleus.

  • Statement (I) is false: The ionisation energy of Pb\text{Pb} is lower than that of Sn\text{Sn} because Pb\text{Pb} is lower in the periodic table and has a higher atomic size.
  • Statement (II) is true: Ge\text{Ge} has a higher ionisation energy than Si\text{Si} because it is in the same group but higher in the periodic table, so its electrons are closer to the nucleus.

Therefore, the correct option is C.

Stepwise Analysis

Given:

  • Compare the first ionization energies of Pb\text{Pb} and Sn\text{Sn}.
  • Compare the first ionization energies of Ge\text{Ge} and Si\text{Si}.

Find: Determine the truth values of Statement (I) and Statement (II).

Step 1: Analyze Statement (I). Statement (I) states that the first ionization energy of Pb\text{Pb} (Lead) is greater than that of Sn\text{Sn} (Tin). However, this is incorrect. Although Pb\text{Pb} and Sn\text{Sn} are in the same group (Group 14), lead (Pb\text{Pb}) is lower in the periodic table compared to tin (Sn\text{Sn}). As we move down a group, the atomic radius increases, and the outermost electrons are farther from the nucleus, leading to a lower ionization energy. Therefore, the first ionization energy of Pb\text{Pb} is less than that of Sn\text{Sn}, making Statement (I) false.

Step 2: Analyze Statement (II). Statement (II) states that the first ionization energy of Ge\text{Ge} (Germanium) is greater than that of Si\text{Si} (Silicon). This is true. Ge\text{Ge} is in Period 4 and Si\text{Si} is in Period 3, so Si\text{Si} has a smaller atomic radius and the outermost electrons are held more tightly by the nucleus, leading to a higher ionization energy. Therefore, the first ionization energy of Ge\text{Ge} is greater than that of Si\text{Si}, making Statement (II) true.

Step 3: Conclusion. Statement (I) is false, and Statement (II) is true.

Statement I is false but Statement II is true.\boxed{\text{Statement I is false but Statement II is true.}}

Therefore, the correct option is C.

Common mistakes

  • Assuming ionization energy always increases down a group is incorrect because the general trend is a decrease due to larger atomic size and greater distance of the valence electron from the nucleus. Compare positions in the periodic table before deciding.

  • Confusing atomic size with effective nuclear attraction leads to wrong conclusions. A lower element in the same group usually has a larger radius, so its outer electron is removed more easily. Use periodic trends, not just atomic number.

  • Reading the statement pair carelessly and selecting the option matching only one statement can cause error. Evaluate Statement (I) and Statement (II) separately, then match the exact combination in the options.

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