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
(iv) < (ii) < (iii) < (i)
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B
(ii) < (iv) < (i) < (iii)
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C
(i) < (iii) < (ii) < (iv)
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D
(iii) < (i) < (iv) < (ii)
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Solution
The correct option is C
(i) < (iii) < (ii) < (iv)
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).