The preference of three unpaired electrons in the ground state of phosphorous can be explained by:

Pauli's exclusion principle

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Aufbau principle

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Heisenberg's uncertainty principle

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Hund's rule

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Solution

The correct option is D Hund's rule
According to Hund's rule of maximum multiplicity, electron pairing does not occur until each degenerate orbital has one electron each.
In Phosphorus, atomic number = 15
So. electronic configuration = $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p_{x}^{1} 3 p_{y}^{1} 3 p_{z}^{1}$

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Similar questions

Q. The preference of three unpaired electrons in the ground state of phosphorous can be explained by:

If carbon atom has electronic configuration 1 $s^{6}$ , it would have energy lower than that of the normal ground state configuration 1 $s^{2}$ 2 $s^{2}$ 2 $p^{2}$ . Yet 1 $s^{6}$ is not observed because it violates

Q. If the nitrogen atom had an electronic configuration of

1 s^{7}

it would have lower energy than that of the normal ground state configuration

1 s^{2}, 2 s^{2}, 2 p^{3}

because the electrons would be closer to the nucleus.

1 s^{7}

is not observed because it violates:

Q. If carbon has electronic configuration

1 s^{6}

, it would have lower energy than that of the normal ground state configuration

1 s^{2} 2 s^{2} 2 p^{2}

, because the electrons would be closer to the nucleus. Yet

1 s^{6}

is not observed, since it violates

In the ground state, carbon atom has how many unpaired electron(s)?

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Hund's Rule

Standard XII Chemistry

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The preference of three unpaired electrons in the ground state of phosphorous can be explained by:

The correct option is D Hund's ruleAccording to Hund's rule of maximum multiplicity, electron pairing does not occur until each degenerate orbital has one electron each. In Phosphorus, atomic number = 15 So. electronic configuration = 1s2 2s2 2p6 3s2 3p1x 3p1y 3p1z