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(A) The atomic number of oxygen $(O)$ is $8$. Its electronic configuration is $1s^2 2s^2 2p^4$. Hund's rule of maximum multiplicity states that electr

Vedclass · Accepted Answer

$1s^2 2s^2 2p_x^2 2p_y^1 2p_z^1$

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(A) The atomic number of oxygen $(O)$ is $8$. Its electronic configuration is $1s^2 2s^2 2p^4$. Hund's rule of maximum multiplicity states that electron pairing in degenerate orbitals (orbitals of the same subshell) does not occur until each orbital is singly occupied. For the $2p$ subshell containing $4$ electrons,the electrons will occupy the $2p_x, 2p_y,$ and $2p_z$ orbitals singly first,and then the fourth electron will pair up in one of the orbitals. Thus,the configuration is $1s^2 2s^2 2p_x^2 2p_y^1 2p_z^1$ (or any equivalent arrangement like $2p_x^1 2p_y^2 2p_z^1$). Therefore,option $A$ is correct.

Which electronic configuration for oxygen is correct according to Hund's rule of maximum multiplicity?

Similar Questions

Nitrogen has the electronic configuration $1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1$ and not $1s^2 2s^2 2p_x^2 2p_y^1 2p_z^0$,which is determined by:

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Which of the following gases does not have an octet or eight electrons in the outer shell?

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