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Structure of Atom — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryStructure of Atom5 Marks
Question
  1. How many electrons will present in sub-shell having spin quantum number value of $-\frac{1}{2}$ for n = 4?
  2. Which of the following transition will have minimum wavelength and why?
$\mathrm{n}_4 \rightarrow \mathrm{n}_1, \mathrm{n}_4 \rightarrow \mathrm{n}_2, \mathrm{n}_2 \rightarrow \mathrm{n}_1$.
  1. Give the number of radial nodes for 3s and 2p orbitals.
✓

Answer

  1. Number of orbitals in 4th shell = $n^2 = 4^2 = 16$; since each orbital has only one electron with $\text{m}_\text{s}=-\frac{1}{2},$ there will be total 32 electrons and 16 electrons will be with $\text{m}_\text{s}=+\frac{1}{2}.$
Hence, there will be 16 electrons with $\text{m}_\text{s}=-\frac{1}{2},$
  1. $n_4$ to $n_1$ transition is of maximum energy.
$\text{E}_4-\text{E}_1=\frac{1212}{16}-\Big(-\frac{1212}{1}\Big)$
$=-82+1312=1230\text{kJ}/ \text{ mol}^{-1}$
$\text{n}_4\rightarrow\text{n}_2$
$\Rightarrow\text{E}_4-\text{E}_2$
$=\frac{1312}{16}-\Big(-\frac{1312}{4}\Big)=\frac{-1312+5248}{16}-=\frac{3936}{16}$
$=246\text{kJ/ mol}^{-1}$
$\text{n}_2\rightarrow\text{n}_1$
$\Rightarrow\text{E}_2-\text{E}_1$
$=\frac{-1312}{4}-\Big(\frac{-1312}{1}\Big)=\frac{-1312+5248}{4}=\frac{3936}{4}$
$=984\text{kJ/ mol}^{-1}$
Since $\text{E}=\frac{\text{hc}}{\lambda},$ i.e. energy and wavelength are inversely proportionalto each other,
$\text{n}_4\xrightarrow{\ \ \ \ \ \ }\text{n}_1$ will have minimum wavelength.
$\because\text{E}=\frac{\text{hc}}{\lambda}\Rightarrow\text{E}\propto\frac{1}{\lambda}$
  1. Number of radial nodes = (n - l - 1)
    For 3s orbital, n = 3, l = 0
    When 'n' is principal quantum number, ‘l’ is azimuthal quantum number
    [l = 0 for s-orbital and l = 1 for p-orbital]
    Hence number of radial nodes = (3 - 0 - 1) = 2
    For 2p orbital, n = 2, l = 1
$\therefore$ Numbers of radial nodes = (2 - 1 - 1) = 0

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