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Question
The electrons identified by quantum numbers n and l:
(P) n = 5, l = 1 (Q) n = 3, l = 0
(R) n = 4, l = 2 (S) n = 2, l = 1
can be placed in decreasing order of energy as :
The electrons identified by quantum numbers n and l:
(P) n = 5, l = 1 (Q) n = 3, l = 0
(R) n = 4, l = 2 (S) n = 2, l = 1
can be placed in decreasing order of energy as :
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Solution
The correct option is A P>R>Q>S
The n+l values of P, Q, R, and S respectively are 6, 3, 6, 3.
According to Bohr Bury's n+l rule , greater the value of n+l greate the energy. If two subshells have the same n+l value , then one with greater value of n has the greater energy.
The n+l values of P, Q, R, and S respectively are 6, 3, 6, 3.
According to Bohr Bury's n+l rule , greater the value of n+l greate the energy. If two subshells have the same n+l value , then one with greater value of n has the greater energy.
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