The correct trend in the first ionization enthalpies of the elements in the 3rd period of periodic table is :
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- $ Al$Si$S$AlShow Solution
Verified By CollegeduniaThe Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
Ionization enthalpy generally increases across a period. However, exceptions occur due to extra stability of half-filled subshells and penetration effects of s-electrons vs p-electrons.
Step 2: Key Formula or Approach:
1. Across 3rd Period: $Na<Mg>Al<Si<P>S<Cl<Ar$.
2. Exception 1: Mg (\(3s^2\)) $>$ Al (\(3p^1\)) due to fully filled s-subshell and higher penetration.
3. Exception 2: P (\(3p^3\)) $>$ S (\(3p^4\)) due to extra stability of half-filled p-subshell in Phosphorus.
Step 3: Detailed Explanation:
Arranging the given elements based on these rules: - Al (13) is lowest because it's at the start of the p-block and has lower effective nuclear charge than Si. - Si (14) follows Al. - S (16) is lower than P (15) because P has a stable \(3p^3\) half-filled configuration. - Cl (17) is the highest among these as it is furthest to the right. Thus: \(Al<Si<S<P<Cl\).
Step 4: Final Answer:
The correct trend is $Al"Was this answer helpful?00Top JEE Main Chemistry Questions
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