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AIEEE 2012 Solved Paper

Section: Chemistry

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Question : 1 of 89

Marks: +1, -0

The electrons identified by quantum numbers

n

I

n=4, l=1

n=4, l=0

n=3, l=2

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 $)$
$(a) < (c) < (b) < (d)$

Solution:

(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.

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