Atomic Structure Result Question 6-1
6. The quantum number of four electrons are given below:
I. $n=4, l=2, m_l=-2, m_s=-\dfrac{1}{2}$
II. $n=3, l=2, m_l=1, m_s=+\dfrac{1}{2}$
III. $n=4, l=1, m_l=0, m_s=+\dfrac{1}{2}$
IV. $n=3, l=1, m_l=1, m_s=-\dfrac{1}{2}$
The correct order of their increasing energies will be
(2019 Main, 8 April I)
(a) IV $<$ III $<$ II $<$ I
(b) I $<$ II $<$ III $<$ IV
(c) IV $<$ II $<$ III $<$ I
(d) I $<$ III $<$ II $<$ IV
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Answer:
Correct Answer: 6. ( c )
Solution:
- Smaller the value of $(n+l)$, smaller the energy. If two or more sub-orbits have same values of $(n+l)$, sub-orbits with lower values of $n$ has lower energy. The $(n+l)$ values of the given options are as follows :
I. $n=4, l=2 ; n+l=6$
II. $n=3, l=2 ; n+l=5$
III. $n=4, l=1, n+l=5$
IV. $n=3, l=1, n+l=4$
Among II and III, $n=3$ has lower value of energy. Thus, the correct order of their increasing energies will be
$\text { IV }<\text { II }<\text { III }<\text { I }$
BETA
