Q14 / 160 · TS EAMCET 2014 · Chemistry

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

  • A
    $\mathrm{iii} < $ (i) $ < \mathrm{iv} < $ ii
  • B
    $\mathrm{ii} < \mathrm{iv} < \mathrm{i} < \mathrm{iii}$
  • C
    $\mathrm{i} < \mathrm{iii} < \mathrm{ii} < \mathrm{iv}$
  • Correct answer
    D
    iv $ < \mathrm{ii} < \mathrm{iii} < \mathrm{i}$

Solution

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).

Hence, the correct answer is option: 'D'.

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