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If the given four electronic configurations.
(i) $n=4, l=1$
(ii) $n=4, l=0$
(iii) $n=3, l=2$
(iv) $n=3, l=1$
are arranged in order of increasing energy, then the order will be
Options:
(i) $n=4, l=1$
(ii) $n=4, l=0$
(iii) $n=3, l=2$
(iv) $n=3, l=1$
are arranged in order of increasing energy, then the order will be
Solution:
2918 Upvotes
Verified Answer
The correct answer is:
(iv) $ < $ (ii) $ < $ (iii) $ < $ (i)
According to $(n+l)$ rule;
(i) More be the sum of $(n+l)$ value, more be the energy.
(ii) For same value of sum, more be the value of $n$, more be the energy.
For
(i) $n+l \rightarrow 4+1=5$
(ii) $n+l \rightarrow 4+0=4$
(iii) $n+l \rightarrow 3+2=5$
(iv) $n+l \rightarrow 3+1=4$
Hence, order of energy will be
(iv) $ < $ (ii) $ < $ (iii) $ < $ (i)
(i) More be the sum of $(n+l)$ value, more be the energy.
(ii) For same value of sum, more be the value of $n$, more be the energy.
For
(i) $n+l \rightarrow 4+1=5$
(ii) $n+l \rightarrow 4+0=4$
(iii) $n+l \rightarrow 3+2=5$
(iv) $n+l \rightarrow 3+1=4$
Hence, order of energy will be
(iv) $ < $ (ii) $ < $ (iii) $ < $ (i)
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