The quantum numbers of four electrons are given below : I. n=4 ; I=2 ; m_1=-2 ; s=- 1 2 II. n=3 ; I=2 ; m_1=1 ; s=+ 1 2 III. n=4 ; I=1 ; m_1=0 ; s=+ 1 2 IV. n=3 ; I=1 ; m_1=-1 ; s=+ 1 2 The correct decreasing order of energy of these electrons is
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)
Related: Chemistry — Structure of Atom · All PYQ Banks