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

Heisenberg uncertainty principle

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Hund's rule of maximum multiplicity

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Pauli's exclusion principle

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Bohr's postulate of stationary orbits

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Solution

The correct option is C Pauli's exclusion principle
According to Pauli's exclusion principle, no two electrons can have all the four quantum number same. They should have different spins to exist. Since only two spin quantum numbers are possible, the electronic configuration would violate this rule.

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Similar questions

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

Q. If the nitrogen atom had electronic configuration

{1 s}^{7}

. This configuration has 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 but

{1 s}^{7}

is not observed because it violates

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:

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If carbon has electronic configuration 1s6, it would have lower energy than that of the normal ground state configuration 1s2 2s2 2p2, because the electrons would be closer to the nucleus. Yet 1s6 is not observed, since it violates

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