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The electronic configuration of oxygen is written as 1s2 2s2 2pXx2 2pXy1 2pXz1 and not as 1s2 2s2 2pXx2, 2pXy2 2pXz0, Explain.

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Question

The electronic configuration of oxygen is written as 1s2 2s2 \[\ce{2p^2_{{x}}}\] \[\ce{2p^1_{{y}}}\] \[\ce{2p^1_{{z}}}\] and not as 1s2 2s2 \[\ce{2p^2_{{x}}}\], \[\ce{2p^2_{{y}}}\] \[\ce{2p^0_{{z}}}\], Explain.

Answer in Brief
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Solution

  1. According to Hund’s rule of maximum multiplicity “Pairing of electrons in the orbitals belonging to the same subshell does not occur unless each orbital belonging to that subshell has got one electron each.”
  2. Oxygen has 8 electrons. The first two electrons will pair up in the 1s orbital, the next two electrons will pair up in the 2s orbital and this leaves 4 electrons, which must be placed in the 2p orbitals.
  3. Each of the three degenerate p-orbitals must get one electron of parallel spin before anyone of them receives the second electron of opposite spin. Therefore, two p orbitals have one electron each and one p-orbital will have two electrons.
    Thus, the electronic configuration of oxygen is written as 1s2 2s2 \[\ce{2p^2_{{x}}}\] \[\ce{2p^1_{{y}}}\] \[\ce{2p^1_{{z}}}\] and not as 1s2 2s2 \[\ce{2p^2_{{x}}}\], \[\ce{2p^2_{{y}}}\] \[\ce{2p^0_{{z}}}\]
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Chapter 4: Structure of Atom - Exercises [Page 54]

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Balbharati Chemistry [English] Standard 11 Maharashtra State Board
Chapter 4 Structure of Atom
Exercises | Q 6. (P) | Page 54

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