p-Block elementsHard
Question
The electrons identified by quantum numbers n and l :-
(a) n = 4 , l = 1 (b) n = 4, l = 0
(c) n = 3, l = 2 (d) n = 3, l = 1
Can be placed in order of increasing energy as
(a) n = 4 , l = 1 (b) n = 4, l = 0
(c) n = 3, l = 2 (d) n = 3, l = 1
Can be placed in order of increasing energy as
Options
A.(a) < (c) < (b) < (d)
B.(c) < (d) < (b) < (a)
C.(d) < (b) < (c) < (a)
D.(b) < (d) < (a) < (c)
Solution
Energy ∞ (n + l) a {from (n + l) rule}. If (n + l) is same than energy will be higher for higher value of ′n′.
(a) n + l = 4 + 1 = 5
(b) 4 + 0 = 4
(c) 3 + 2 = 5
(d) 3 + 1 = 4
In (a) & (c) n + l = 5, n is more in (a) So (a) > (c) and in (b) & (d) n + l = 4, n is more in (b) so - (b) > (d) so order of energy (a) > (c) > (b) > (d)
(a) n + l = 4 + 1 = 5
(b) 4 + 0 = 4
(c) 3 + 2 = 5
(d) 3 + 1 = 4
In (a) & (c) n + l = 5, n is more in (a) So (a) > (c) and in (b) & (d) n + l = 4, n is more in (b) so - (b) > (d) so order of energy (a) > (c) > (b) > (d)
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