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In an atom the order of increasing energy of electrons with quantum numbers
(i) $n=4, l=1$
(ii) $n=4, l=0$
(iii) $n=3, l=2$ and (iv) $n=3, l=1$ is
Options:
(i) $n=4, l=1$
(ii) $n=4, l=0$
(iii) $n=3, l=2$ and (iv) $n=3, l=1$ is
Solution:
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Verified Answer
The correct answer is:
iv $ < \mathrm{ii} < \mathrm{iii} < \mathrm{i}$
The order of increase of energy can be calculated from $(n+I)$ rule. If two orbitals have same value of $(n+I)$, the orbital with lower value of $n$ will be filled first.
(i) For $n=4, I=1,(n+I)=4+1=5$
(ii) For $n=4, l=0,(n+l)=4+0=4$
(iii) For $n=3, l=2,(n+l)=3+2=5$
(iv) For $n=3, I=1,(n+I)=3+1=4$
Therefore correct order is
(iv) $ < $ (ii) $ < $ (iii) $ < $ (i).
(i) For $n=4, I=1,(n+I)=4+1=5$
(ii) For $n=4, l=0,(n+l)=4+0=4$
(iii) For $n=3, l=2,(n+l)=3+2=5$
(iv) For $n=3, I=1,(n+I)=3+1=4$
Therefore correct order is
(iv) $ < $ (ii) $ < $ (iii) $ < $ (i).
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