Help with a kaplan logic

[Labrat07]

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From the passage info, I thought the lower the s percent the higher ionization and I conclude high electronegativity. Why 's the answer not sp3 but sp (answer is d). Please help

 

[amy_k]

They are saying in the passage that the higher the percent s character, the lower the energy state (stability) of the orbital and thus the higher the ionization energy for an electron in that orbital. This is because the s orbital is closer to the nucleus than p, so the orbitals ranked by increasing distance from the nucleus would be s < sp3 < sp2 < sp < p, and the ionization energy decreases with increasing distance to the nucleus.

Increasing s character increases ionization energy. The example used in the passage is p (with 0% s character) versus sp3 (with 25% s character). The sp3 has higher ionization energy than p. The sp orbital in answer D has 50% s character so will have the highest ionization energy for an electron in that orbital.

 

[Labrat07]

I see what you're saying now. I misunderstood the passage completely. Thank you for clarifying it.

 

[amy_k]

No problem, glad it helped!