
In case of nitrogen, if ${{{M}}_{{1}}}$ represents spin multiplicity if Hund’s rule is followed and ${{{M}}_{{2}}}$ represents spin multiplicity if only Hund’s Rule is violated then the value of $\dfrac{{{{{M}}_{{1}}}}}{{{{{M}}_{{2}}}}}$ will be:
Answer
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Hint: Pairing of electrons in the orbitals belonging to the same subshell (p, d or f) does not take place until each orbital belonging to that subshell has got one electron each and spin multiplicity is given as ${{n = 2s + 1}}$.
Complete step by step answer:
According to Hund’s rule pairing of electrons occurs only after all the available degenerate orbitals are filled with one electron each.
Atomic number of Nitrogen is 7 and the ground state electronic configuration is ${{1}}{{{s}}^{{2}}}{{2}}{{{s}}^{{2}}}{{2}}{{{p}}_{{x}}}^{{1}}{{2}}{{{p}}_{{y}}}^{{1}}{{2}}{{{p}}_{{z}}}^{{1}}$
We have spin multiplicity= ${{2s + 1}}$ where s is the total spin
In Nitrogen the number of unpaired electrons is 3
Thus we have ${{3 \times }}\dfrac{{{1}}}{{{2}}}{{ = }}\dfrac{{{3}}}{{{2}}}$
On substituting n=${{2 \times }}\dfrac{{{3}}}{{{2}}}{{ + 1 = 4}}$
Thus spin multiplicity of Nitrogen, ${{{M}}_{{1}}}{{ = 4}}$
If Hund’s rule is violated then
${{s = }}\dfrac{{{1}}}{{{2}}}{{ - }}\dfrac{{{1}}}{{{2}}}{{ + }}\dfrac{{{1}}}{{{2}}}$
${{n = 2}}\left( {\dfrac{{{1}}}{{{2}}}{{ + 1}}} \right){{ = 3}}$
${{{M}}_{{2}}}{{ = 3}}$
Thus the ratio $\dfrac{{{{{M}}_{{1}}}}}{{{{{M}}_{{2}}}}}{{ = }}\dfrac{{{4}}}{{{3}}}$.
Additional information:
Spin multiplicity=${{2s + 1}}$ where s is the total spin= ${{ \pm }}\dfrac{{{1}}}{{{2}}}{{ \times }}$ number of unpaired electrons
Spin multiplicity = ${{n + 1}}$ where n is the number of unpaired electron
First formula is used when the spin is specified and the second formula is used when spin is not specified. The spin multiplicity is generally given as 1, 2, 3, 4, 5, 6.
Note: The maximum possible orientation of spin angular momentum in a given orbital is said to be spin multiplicity and it is generally used to make the prediction of spin multiplicity value. The states having 1, 2, 3, 4, 5 multiplicity are called singlet, doublets, triplets, quartets and quintets.
Complete step by step answer:
According to Hund’s rule pairing of electrons occurs only after all the available degenerate orbitals are filled with one electron each.
Atomic number of Nitrogen is 7 and the ground state electronic configuration is ${{1}}{{{s}}^{{2}}}{{2}}{{{s}}^{{2}}}{{2}}{{{p}}_{{x}}}^{{1}}{{2}}{{{p}}_{{y}}}^{{1}}{{2}}{{{p}}_{{z}}}^{{1}}$
We have spin multiplicity= ${{2s + 1}}$ where s is the total spin
In Nitrogen the number of unpaired electrons is 3
Thus we have ${{3 \times }}\dfrac{{{1}}}{{{2}}}{{ = }}\dfrac{{{3}}}{{{2}}}$
On substituting n=${{2 \times }}\dfrac{{{3}}}{{{2}}}{{ + 1 = 4}}$
Thus spin multiplicity of Nitrogen, ${{{M}}_{{1}}}{{ = 4}}$
If Hund’s rule is violated then
${{s = }}\dfrac{{{1}}}{{{2}}}{{ - }}\dfrac{{{1}}}{{{2}}}{{ + }}\dfrac{{{1}}}{{{2}}}$
${{n = 2}}\left( {\dfrac{{{1}}}{{{2}}}{{ + 1}}} \right){{ = 3}}$
${{{M}}_{{2}}}{{ = 3}}$
Thus the ratio $\dfrac{{{{{M}}_{{1}}}}}{{{{{M}}_{{2}}}}}{{ = }}\dfrac{{{4}}}{{{3}}}$.
Additional information:
Spin multiplicity=${{2s + 1}}$ where s is the total spin= ${{ \pm }}\dfrac{{{1}}}{{{2}}}{{ \times }}$ number of unpaired electrons
Spin multiplicity = ${{n + 1}}$ where n is the number of unpaired electron
First formula is used when the spin is specified and the second formula is used when spin is not specified. The spin multiplicity is generally given as 1, 2, 3, 4, 5, 6.
Note: The maximum possible orientation of spin angular momentum in a given orbital is said to be spin multiplicity and it is generally used to make the prediction of spin multiplicity value. The states having 1, 2, 3, 4, 5 multiplicity are called singlet, doublets, triplets, quartets and quintets.
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