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Correct decreasing order of energy of the orbitals having following set of quantum numbersA) n = 3, l = 0, m = 0B) n = 4, l = 0, m = 0C) n = 3, l = 1, m = 0D) n = 3, l = 2, m = 1(a) A > C > B > D(b) D > B > C > A(c) A > B > C > D(d) D > C > B > A |
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Answer» (b) D > B > C > A Following the principle of (n+1) rule, the correct order of energy is D > B > C > The subshells with the lowest (n+1) value has the lowest energy. When two or more subshells have the same (n+1) value, then the subshell with the lowest value of n have lowest energy |
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