Question

The atomic number (Z) of an element is 24. In its ground state how many electrons are present in the ‘N’ shell?
(a) 4
(b) 2
(c) 1
(d) 3

Answer 1

Hint: Chromium is a d-block element with the atomic number of 24. It is present in the sixth group and fourth period of the periodic table. The ‘N’ shell is actually the valence shell.
Complete step by step answer:
Chromium is the element with atomic number 24. It is the sixth element in the fourth periods, and shows +2, +3 and +6 oxidation states. It is one of the d-block elements. The first 18 electrons will be placed in exactly the same manner as is placed in Argon since its atomic number is 18.
Rules for filling the electrons:
Rule 1 –Aufbau Principle- Lowest energy orbitals are filled first. This can be determined by using the formula n+l where n is the principal quantum number and l is the orbital angular momentum quantum number. If the n+l value is the same then look at their n value to determine their order of filling. Therefore, the filling pattern is 1s, 2s, 2p, 3s, 3p, 4s, 3d, etc. The orbitals within a subshell are degenerate which implies that the entire subshell of a particular orbital type is filled before moving to the next subshell of higher energy.
Rule 2 – Pauli’s Exclusion Principle – “Only two electrons are permitted per orbital and they must be of opposite spin. If one electron within an orbital possesses a clockwise spin, then the second electron within that orbital will possess a counter clockwise spin”.
Rule 3- Hund's Rule of maximum multiplicity – “All orbitals within a subshell are of equal energy. Electrons are repulsive to one another and only pair after all of the orbitals have been singly filled”.
Keeping all these rules in mind, the electronic configuration up to 18 electrons will be equal to that of the Argon.
$\left[ Ar \right] ={ 1s }^{ 2 }{ 2s }^{ 2 }{ 2p }^{ 6 }{ 3s }^{ 2 }{ 3p }^{ 6 }$
Now 6 more electrons remain. After the filling of the 3p orbital, these 6 electrons will end up in 3d and 4s orbitals. According to the n+l rule, the lowest energy orbital among them is 4s. Therefore the 2 electrons should enter the 4s orbital. But only one electron goes into the 4s electron. This is because half-filled and fully-filled orbitals are highly stable. If only one electron enters into the 4s orbital, then the remaining 5 electrons will enter into the five 3d orbitals and hence the 3d orbitals will be half-filled which will be exceptionally stable.
Therefore the electronic configuration will be: $\left[ Ar \right] { 3d }^{ 5 }{ 4s }^{ 1 }$
The ‘N’ shell will be the valence shell, which is equal to 4. In the fourth shell only one electron is present.
Hence, the correct answer is (c) 1.
Note: If according to the n+l rule, the order of filling for two orbitals comes out to be the same example, both 4f and 5d orbitals have the same n+l value, then their n value has to be taken into account. The orbital with lower n value will be filled first.