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(A) The ionization energy depends on the penetration power of the orbitals. Electrons in orbitals closer to the nucleus are more strongly attracted by

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$s$-orbital

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(A) The ionization energy depends on the penetration power of the orbitals. Electrons in orbitals closer to the nucleus are more strongly attracted by the nucleus. The order of penetration power of orbitals is $s > p > d > f$. Since the $s$-orbital is closest to the nucleus,it has the highest penetration power,making it the most difficult to remove an electron from. Therefore,the ionization energy is highest for the $s$-orbital.

If other factors remain the same,the ionization energy required to remove an electron from which of the following orbitals will be the highest?

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