The quantum numbers of four electrons are given below : I. n=4 ; I=2 ; m₁=-2 ; s=- 1 2 II. n=3 ; I=2 ; m₁=1 ; s=+ 1 2 III. n=4 ; I=1 ; m₁=0 ; s=+ 1 2 IV. n=3 ; I=1 ; m₁=-1 ; s=+ 1 2 The correct decreasing order of energy of these electrons

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The quantum numbers of four electrons are given below : I. n=4 ; I=2 ; m₁=-2 ; s=- 1 2 II. n=3 ; I=2 ; m₁=1 ; s=+ 1 2 III. n=4 ; I=1 ; m₁=0 ; s=+ 1 2 IV. n=3 ; I=1 ; m₁=-1 ; s=+ 1 2 The correct decreasing order of energy of these electrons is

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Correct answer: B. I III II IV

(I) n=4, I=2, m_I=-2, s=- 1 2 ; represents 4 d(n+I=6) (II) n=3, I=2, m_l=1, s=+ 1 2 ; represents 3 d(n+I=5) (III) n=4, I=1, m_I=0, s=+ 1 2 ; represents 4 p(n+I=5) (IV) n=3, I=1, m_l=-1, s=+ 1 2 ; represents 3 p(n+I=4) Order of energy depends on the (n+I) , greater is the (n+I) value greater is the energy, if (n+I) is same, then it depends on n ; if ' n ' is more, energy is more. Step-1: According to ( n + I ) Energy =( I ) ( II )=( III ) ( IV ) Step-2 : If n , then energy increases Energy =( I ) ( III ) ( II ) ( IV

The quantum numbers of four electrons are given below : I. n=4 ; I=2 ; m₁=-2 ; s=- 1 2 II. n=3 ; I=2 ; m₁=1 ; s=+ 1 2 III. n=4 ; I=1 ; m₁=0 ; s=+ 1 2 IV. n=3 ; I=1 ; m₁=-1 ; s=+ 1 2 The correct decreasing order of energy of these electrons is

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