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

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$(i v) < (i i) < (i i i) < (i)$

$(i i) < (i v) < (i) < (i i i)$

$(i) < (i i i) < (i i) < (i v)$

$(i i i) < (i) < (i v) < (i i)$

answer is A.

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Detailed Solution

(n +l) rule higher the value of (n + l), the higher is the energy. When (n + l) value is the same see value of n

$∴ IV<II<III<I$

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