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Arrabge in decreasing order, the energy of 2s orbital in the following atoms H, Li, Na, K

A

`E_(2s(H))gtE_(2s(Li))gtE_(2s(Na))gtE_(2s(K))`

B

`E_(2s(H))gtE_(2s(Na))gtE_(2s(Li))gtE_(2s(K))`

C

`E_(2s(H))gtE_(2s(Na))=E_(2s(K))gtE_(2s(Li))`

D

`E_(2s(K))ltE_(2s(Na))ltE_(2s(Li))ltE_(2s(H))`

Text Solution

Verified by Experts

The correct Answer is:
A
For larger atoms, electron in 2s orbital is more tightly bound to the nucleus because of more positive charge in nucleus. So energy of electron is more negative. Also energy of electron `E_(n)=-13.6(Z^(2)/n^(2))`
Here .n. is constant so for large atom the energy will be more negative i.e. energy will decrease. So
`E_(2s(H))gtE_(2s(Li))gtE_(2s(Na))gtE_(2s(K))`.
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