Question

The electronic configurations of four elements A, B, C, D are given below:
(A) 1s22s22p63s1
(B) 1s22s22p63s23p1
(C) 1s22s22p63s2
(D) 1s22s22p63s23p2

• A. \( 1s^2 2s^2 2p^6 3s^1 \)
• B. \( 1s^2 2s^2 2p^6 3s^2 3p^1 \)
• C. \( 1s^2 2s^2 2p^6 3s^2 \)
• D. \( 1s^2 2s^2 2p^6 3s^2 3p^2 \) \textbf{The correct order of first ionization enthalpy of these elements is:}

Hint:

For elements in the same period, ionization enthalpy increases from left to right, as nuclear attraction strengthens.

Answer 1

The Correct Option is D

Step 1: Understanding Ionization Enthalpy Ionization enthalpy is the energy required to remove the outermost electron from an atom. It increases across a period and decreases down a group. Step 2: Analyzing the Configurations - Element A has one electron in the 3s orbital, making it easiest to remove (\( \text{low ionization energy} \)). - Element B has a 3p^1 electron, which is relatively easy to ionize but requires more energy than A. - Element C has a filled 3s^2 configuration, requiring higher ionization energy. - Element D has a 3p^2 electron, which is harder to remove due to increased nuclear attraction. Step 3: Ordering the Ionization Enthalpies Since D has the most nuclear attraction, followed by C, B, and A, the correct order is: \[ D>C>B>A \]