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Structure of Atom — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryStructure of Atom3 Marks
Question
The first shell may contain up to 2 electrons, the second shell up to 8, the third shell up to 18, and the fourth shell up to 32. Explain this arrangement in terms of quantum numbers.
✓

Answer

For the fust shell $\text{n}=1,\text{l}=0,\text{m}_\text{l}=0,\text{m}_\text{s}=+\frac{1}{2}$ and $-\frac{1}{2}.$ It can have 2 electrons both with opposite spins.
  1. For $\text{n}=2,$
$\text{l}=0,\text{m}_\text{l}=0,\text{m}_\text{s}=+\frac{1}{2},-\frac{1}{2},$

$\text{l}=1,\text{m}_\text{l}=1,0+1,\text{m}_\text{s}=+\frac{1}{2},-\frac{1}{2}.$

Therefore, a total of 2 + 6 = 8 electrons are present.
  1. For $\text{n}=3,$ when
$\text{l}=0,\text{m}_\text{l}=0,\text{m}_\text{s}=+\frac{1}{2},-\frac{1}{2}$

$\text{l}=1,\text{m}_\text{l}=-1,0+1,\text{m}_\text{s}=\frac{1}{2},-\frac{1}{2}$

$\text{l}=2,\text{m}_\text{l}=-2,-1,0,+1,+2,,\text{m}_\text{s}$

$=+\frac{1}{2},-\frac{1}{2}$

Therefore, a total of 2 + 6 + 10 = 18 electlons aie present.
  1. For $\text{n}=4,$ when
$\text{l}=0,\text{m}_\text{l}=0,\text{m}_\text{s}=+\frac{1}{2},-\frac{1}{2}$

$\text{l}=1,\text{m}_\text{l}=-1,0+1,\text{m}_\text{s}=+\frac{1}{2},-\frac{1}{2}$

$\text{l}=3,\text{m}_\text{l}=-2,-1,0+1,+2,\text{m}_\text{s}=+\frac{1}{2},-\frac{1}{2}$

$\text{l}=3,\text{m}_\text{l}=-3,-2,-1,0,+1,+2,+3,\text{m}_\text{s}$

$=+\frac{1}{2},-\frac{1}{2}$

Therefore, a total of 2 + 6 + 10 + 14 = 32 electrons are present.

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