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Question: What is electronic configuration of elements of \({\text{IIIrd}}\) group? A. \(1{s^2},\;2{s^2}2p{s...

What is electronic configuration of elements of IIIrd{\text{IIIrd}} group?
A. 1s2,  2s22ps31{s^2},\;2{s^2}2p{s^3}
B. 1s2,  2s22p6,  3s23p11{s^2},\;2{s^2}2{p^6},\;3{s^2}3{p^1}
C. 1s2,  2s22p61{s^2},\;2{s^2}2{p^6}
D. 1s2,  2s22p6,3s11{s^2},\;2{s^2}2{p^6},3{s^1}

Explanation

Solution

Electronic configuration of an element represents how the electrons are distributed among the orbits, shells and subshells of an atom. It is a very important aspect for an element to determine its physical, chemical, electrical and magnetic behaviour. The electronic configuration of an element obeys three rules which are Aufbau’s principle, Hund’s rule and Pauli exclusion principle.

Complete answer: The three rules which determine the electronic configuration of an element are discussed below:
Aufbau’s Principle:
It states that electrons in an atom are always filled in the increasing order of the atomic orbitals i.e., the orbital with lower energy will be filled first than the orbital which has comparatively higher energy. The order in which the energy of the orbitals increases can easily be determined by the help of (n+l)\left( {n + l} \right) rule, according to which lower the value of (n+l)\left( {n + l} \right), lower will be its orbital energy.
Hund’s Rule:
It states that every orbital in a subshell must be singly occupied before the pairing of electrons so as to minimize the repulsion of electrons present within the same orbital. All singly occupied electrons in the subshell must be of the same spin.
Pauli exclusion principle:
It states that in a single atom, no two electrons cannot have the same set of quantum numbers and as only two electrons can occupy a single orbital, they must be present in the opposite spin.
The IIIrd{\text{IIIrd}} group of the modern periodic table consist of Boron, Aluminium, Gallium, indium and thallium. The electronic configuration of each element according to the given rules are as follows:
Boron (B):
Atomic number =5 = 5
Electronic configuration =1s2,  2s22p1 = 1{s^2},\;2{s^2}2{p^1}
Aluminium (Al):
Atomic number =13 = 13
Electronic configuration =1s2,  2s22p6,  3s23p1 = 1{s^2},\;2{s^2}2{p^6},\;3{s^2}3{p^1}
Gallium (Ga):
Atomic number =31 = 31
Electronic configuration =[Ar]4s24p1 = [Ar]4{s^2}4{p^1}
Indium (In):
Atomic number =49 = 49
Electronic configuration =[Kr]5s25p1 = [Kr]5{s^2}5{p^1}
Thallium (Tl):
Atomic number =81 = 81
Electronic configuration =[Xe]6s26p1 = [Xe]6{s^2}6{p^1}
Hence, among given options the electronic configuration for an element of third group is 1s2,  2s22p6,  3s23p11{s^2},\;2{s^2}2{p^6},\;3{s^2}3{p^1}.
Thus, option (B) is the correct answer.

Note:
It is important to note that in (n+l)\left( {n + l} \right) rule, n denotes the principal quantum number which describes the energy of an electron and the most probable distance between the nucleus and the electron whereas l denotes azimuthal quantum number which determines the shape and orbital angular momentum.