Q. If "l" can have total (n+2) values,then find which one will have the highest ionisation energy?
["l" is azimuthal quantum number.]
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When l can have n+2 values, its possible values would be 0 to n+1, then the corresponding orbitals would have been as follows:
n l orbital designation
----- -------- ---------------------
1 0, 1, 2 1s, 1p, 1d
2 0, 1, 2, 3 2s, 2p, 2d, 2f
and so on.
The electronic configuration of the given elements, would then be as follows (Note the energy ordering of orbitals, based on (n+l) rule):
`He(2) rarr 1s^2`
`Xe(54) rarr 1s^2,1p^6,2s^2,1d^10, 2p^6,3s^2,2d^10,3p^6, 4s^2,2f^8.`
`Ne(10) rarr 1s^2,1p^6,2s^2`
`O(8) rarr 1s^2,1p^6`
Therefore, having its outermost 'octet' filled, oxygen would have the highest ionization energy. Correct answer is option D).
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