Question

In the absence of Aufbau rule, what block would K(19), Sc(21) and Ca(20) belong to:
(A) s-block
(B) p-block
(C) d-block
(D) f-block

Answer 1

Hint: Aufbau rule states that the arrangement of electrons in ground state electronic configuration occurs in a way that electrons first occupy those orbitals that have less energy. The number of electrons present in the ground state of an element is equal to its atomic number.


Complete step by step answer:
Let’s know something about the Aufbau rule first.
- Aufbau rule states that the arrangement of electrons in ground state electronic configuration occurs in a way that electrons first occupy those orbitals that have less energy.
- It is an approximation that energy of orbitals can be shown by the (n+l) rule where n is the principal quantum number and l is the azimuthal quantum number of the orbital. Higher the (n+l) number, higher will be its energy. e.g. Electrons will occupy 4s orbital first than 3d because the (n+l) number for 4s is (4+0)=4 and for 3d orbital it is (3+2)=5.
- So, we do not use Aufbau’s rule in arrangement of electrons in orbitals, then (n+l) rule will not be followed as well as it is an Aufbau’s approximation, so electrons will occupy the orbitals in the order of principal quantum number only. So, let’s try and find the new electronic configurations of given elements as Aufbau’s rule is not followed.

i) K(19):
Here, the atomic number of Potassium is given 19. So, we can get that the number of electrons present in the ground state will be equal to the number of protons present which is 19. So, electrons will arrange in orbitals in an increasing order of principal quantum number only. So, new electronic configuration of K will be:
$$K:1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^1$$
Here, we can see that the last electron is in the d-orbital of the compound. So, this compound can be of d-block because d-block compounds are those compounds who have their last electron filled in any d-orbitals.

ii) Sc(21):
We know that Sc will have 21 electrons in its ground state electronic configuration. So, if we arrange all these electrons as Aufbau’s rule not being followed, we will get its new electronic configuration as
$$Sc:1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^3$$
Here, we can see that the last electron is filled in the d-orbital. So, it could be a d-block element.

iii) Ca(20):
We know that total electrons in the ground state of Ca will be 20. So, if we arrange them in a way that Aufbaus rule not being followed, it will get its new electronic configuration as follows:
$$Ca:1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^2$$
We can see that the last electron is in d-orbital, so this compound will be called a d-block element.

So, from the above discussion, we can say that if Aufbau rule is not being followed, then K, Sc and Ca will be d-block elements.

So, the correct answer is option (C) d-block.

Note:
Do not get confused between Hund's rule and Aufbau rule, Hund rule states that electrons will be arranged in the orbitals having the same energy in a way that the resulting spin multiplicity will be highest.