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Chemistry · Classification of Elements and Periodicity in Properties

हिन्दी

The first ionization enthalpy of boron (\( Z = 5 \)) is slightly less than that of beryllium (\( Z = 4 \)). Which formula/expression correctly reflects the electronic configuration difference responsible for this anomaly during ionization?

  1. \( \text{B}: 1s^2 2s^2 2p^1 \rightarrow e^{-} \text{ removed from } 2s \text{ vs. Be}: 1s^2 2p^2 \rightarrow e^{-} \text{ removed from } 2p \)
  2. \( \text{B}: 1s^2 2s^2 2p^1 \rightarrow e^{-} \text{ removed from } 2p \text{ vs. Be}: 1s^2 2s^2 \rightarrow e^{-} \text{ removed from } 2s \)
  3. \( \text{B}: 1s^2 2p^3 \rightarrow e^{-} \text{ removed from } 2p \text{ vs. Be}: 1s^2 2s^2 \rightarrow e^{-} \text{ removed from } 2s \)
  4. \( \text{B}: 1s^2 2s^1 2p^2 \rightarrow e^{-} \text{ removed from } 2p \text{ vs. Be}: 1s^2 2s^2 \rightarrow e^{-} \text{ removed from } 2p \)

Correct answer · Option 2

\( \text{B}: 1s^2 2s^2 2p^1 \rightarrow e^{-} \text{ removed from } 2p \text{ vs. Be}: 1s^2 2s^2 \rightarrow e^{-} \text{ removed from } 2s \)

Explanation

In Be, the electron is removed from a 2s orbital which has higher penetration than a 2p electron removed from B, making Be's ionization enthalpy unexpectedly higher.
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