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The hybridization of atomic orbitals of nitrogen in NO2+ , NO2- and NH4+ are
- a)sp2, sp3 and sp2 respectively
- b)sp, sp2 and sp3 respectively
- c)sp2, sp and sp3 respectively
- d)sp2, sp3 and sp respectively
Correct answer is option 'B'. Can you explain this answer?
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The hybridization of atomic orbitals of nitrogen in NO2+ , NO2- and ...
Hybridization of Nitrogen in NO2, NO2- and NH4
NO2:
- Nitrogen in NO2 has one lone pair of electrons and two sigma bonds with oxygen atoms.
- The electronic configuration of nitrogen in its ground state is 1s2 2s2 2p3.
- To form two sigma bonds with oxygen, nitrogen must hybridize three of its orbitals.
- Therefore, the hybridization of nitrogen in NO2 is sp2.
NO2-:
- NO2- is a negatively charged ion, which means it has an additional electron.
- This extra electron occupies one of the nitrogen's hybridized orbitals, leading to the formation of a new sigma bond with the oxygen atom.
- Therefore, the hybridization of nitrogen in NO2- is sp3.
NH4:
- Nitrogen in NH4 has four sigma bonds with four hydrogen atoms.
- The electronic configuration of nitrogen in its ground state is 1s2 2s2 2p3.
- To form four sigma bonds with hydrogen, nitrogen must hybridize all its orbitals.
- Therefore, the hybridization of nitrogen in NH4 is sp3.
In summary, the hybridization of nitrogen in NO2, NO2- and NH4 are sp2, sp3 and sp3 respectively.
NO2:
- Nitrogen in NO2 has one lone pair of electrons and two sigma bonds with oxygen atoms.
- The electronic configuration of nitrogen in its ground state is 1s2 2s2 2p3.
- To form two sigma bonds with oxygen, nitrogen must hybridize three of its orbitals.
- Therefore, the hybridization of nitrogen in NO2 is sp2.
NO2-:
- NO2- is a negatively charged ion, which means it has an additional electron.
- This extra electron occupies one of the nitrogen's hybridized orbitals, leading to the formation of a new sigma bond with the oxygen atom.
- Therefore, the hybridization of nitrogen in NO2- is sp3.
NH4:
- Nitrogen in NH4 has four sigma bonds with four hydrogen atoms.
- The electronic configuration of nitrogen in its ground state is 1s2 2s2 2p3.
- To form four sigma bonds with hydrogen, nitrogen must hybridize all its orbitals.
- Therefore, the hybridization of nitrogen in NH4 is sp3.
In summary, the hybridization of nitrogen in NO2, NO2- and NH4 are sp2, sp3 and sp3 respectively.
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The hybridization of atomic orbitals of nitrogen in NO2+ , NO2- and ...
NO2+ = 1/2 [5 + 0 + 0 - 1] = 2 sp;
NO2- = 1/2 [5 + 0 + 1 - 0] = 3 sp2;
NO4+ = 1/2 [5 + 4 + 0 - 1] = 4 sp3;
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The hybridization of atomic orbitals of nitrogen in NO2+ , NO2- and NH4+ area)sp2, sp3 and sp2 respectivelyb)sp, sp2 and sp3 respectivelyc)sp2, sp and sp3 respectivelyd)sp2, sp3 and sp respectivelyCorrect answer is option 'B'. Can you explain this answer?
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