Statement (I): The first ionization energy of Pb is greater than that of Sn. Statement (II): The first ionization energy of Ge is greater than that of Si.
In light of the above statements, choose the correct answer from the options given below:
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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 is lower than that of Sn because Pb is lower in the periodic table and has a higher atomic size. - Statement (II) is true: Ge has a higher ionisation energy than Si because it is in the same group but higher in the periodic table, so its electrons are closer to the nucleus. Final Answer: Statement I is false but Statement II is true.
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Approach Solution -2
Step 1: Analyze Statement (I).
Statement (I) states that the first ionization energy of Pb (Lead) is greater than that of Sn (Tin). However, this is incorrect. Although Pb and Sn are in the same group (Group 14), lead (Pb) is lower in the periodic table compared to tin (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 is less than that of Sn, making Statement (I) false.
Step 2: Analyze Statement (II).
Statement (II) states that the first ionization energy of Ge (Germanium) is greater than that of Si (Silicon). This is true. Ge is in Period 4 and Si is in Period 3, so 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 is greater than that of Si, making Statement (II) true.
Step 3: Conclusion.
Statement (I) is false, and Statement (II) is true.
Final Answer:
\[
\boxed{\text{Statement I is false but Statement II is true.}}
\]
