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Q. If "l" can have total (n+2) values,then find which one will have the highest...

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user8235304 | Student, Grade 11 | Valedictorian

Posted July 25, 2013 at 5:22 PM via web

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Q. If "l" can have total (n+2) values,then find which one will have the highest ionisation energy?

A) He

B) Xe

C) Ne

D) O

["l" is azimuthal quantum number.]

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llltkl | College Teacher | Valedictorian

Posted July 25, 2013 at 6:27 PM (Answer #1)

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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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