Consider the following sets of quantum numbers for four different electrons in a multi-electron atom: I. n=6 ; l=0 ; m_l=0 ; s=+ 1 2 II. n=5 ; l=1 ; m_l=-1 ; s=- 1 2 III. n=5 ; l=2 ; m_l=+2 ; s=+ 1 2 IV. n=4 ; l=3 ; m_l=-3 ; s=- 1 2 Identify the correct increasing order of energy for these electrons.

Step-by-step solution

The energy of an electron in a multi-electron atom is determined by the (n+l) rule. Electrons with a lower value of (n+l) have lower energy. If two electrons have the same (n+l) value, the electron with the lower principal quantum number n has lower energy. Calculating the (n+l) values for the given electrons: For I: n=6, l=0 n+l = 6 For II: n=5, l=1 n+l = 6 For III: n=5, l=2 n+l = 7 For IV: n=4, l=3 n+l = 7 Comparing I and II, both have n+l = 6 . Since n=5 for II is less than n=6 for I, the energy of II is less th