JEE MAIN - Chemistry (2012 - No. 22)
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 :
(c) < (d) < (b) < (a)
(d) < (b) < (c) < (a)
(b) < (d) < (a) < (c)
(a) < (c) < (b) < (d)
Explanation
(a) n = 4, $$l$$ = 1 (p-subshell), so 4p
(b) n = 4, $$l$$ = 0 (s-subshell), so 4s
(c) n = 3, $$l$$ = 2 (d-subshell), so 3d
(d) n = 3, $$l$$ = 1 (p-subshell), so 3p
Accroding to the Bohr ( n + $$l$$ ) rule,
Enery order of the subshell : 3p < 4s < 3d < 4p
Note: When two orbital have the same value of ( n + $$l$$ ) then the orbital which have lower value of n will be filled first.
(b) n = 4, $$l$$ = 0 (s-subshell), so 4s
(c) n = 3, $$l$$ = 2 (d-subshell), so 3d
(d) n = 3, $$l$$ = 1 (p-subshell), so 3p
Accroding to the Bohr ( n + $$l$$ ) rule,
Enery order of the subshell : 3p < 4s < 3d < 4p
Note: When two orbital have the same value of ( n + $$l$$ ) then the orbital which have lower value of n will be filled first.
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