Question

The electrons, identified by quantum number n and l,
(i) n = 4, l = 1(ii) n = 4, l = 0(iii) n = 3, l = 2 and (iv) n = 3, l = 1
Can be placed in order of increasing energy, from lowest to highest as

• A. (i) < (iii) < (ii) < (iv)
• B. (ii) < (iv) < (i) < (iii)
• C. (iv) < (ii) < (iii) < (i)
• D. (iii) < (i) < (iv) < (ii)

Answer 1

The correct option is C

(iv) < (ii) < (iii) < (i)

First just calculate the (n + l) values of all.

(i) n + l = 4 + 1 = 5
(ii) n + l = 4 + 0 = 4
(iii) n + l = 3 + 2 = 5
(iv) n + l = 3 + 1 = 4

Now, we know, for same (n + l) values, the one having higher energy will be having higher principal quantum number (n).

Therefore,
(iv) < (ii) < (iii) < (i)