JEE MAIN - Chemistry (2005 - No. 56)
In a multi-electron atom, which of the following orbitals described by the three quantum members will have the same energy in the absence of magnetic and electric fields?
(A) n = 1, l = 0, m = 0
(B) n = 2, l = 0, m = 0
(C) n = 2, l = 1, m = 1
(D) n = 3, l = 2, m = 1
(E) n = 3, l = 2, m = 0
(A) n = 1, l = 0, m = 0
(B) n = 2, l = 0, m = 0
(C) n = 2, l = 1, m = 1
(D) n = 3, l = 2, m = 1
(E) n = 3, l = 2, m = 0
(D) and (E)
(C) and (D)
(B) and (C)
(A) and (B)
Explanation
As here is no electric and magnetic field so ignore m to calculate the energy of orbital.
As here atom is multi-electron so ( n + l ) rule is applicable here. This rule says those orbitals which have more value of ( n + l ) will have more energy.
In (A), n + l = 1 + 0 = 1
In (B), n + l = 2 + 0 = 2
In (C), n + l = 2 + 1 = 3
In (D), n + l = 3 + 2 = 5
In (E), n + l = 3 + 2 = 5
So (D) and (E) will be of equal energy.
As here atom is multi-electron so ( n + l ) rule is applicable here. This rule says those orbitals which have more value of ( n + l ) will have more energy.
In (A), n + l = 1 + 0 = 1
In (B), n + l = 2 + 0 = 2
In (C), n + l = 2 + 1 = 3
In (D), n + l = 3 + 2 = 5
In (E), n + l = 3 + 2 = 5
So (D) and (E) will be of equal energy.
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