Question

: The electronic configuration of the element which is just above the element with atomic number $43$ in the same periodic group is:
A.$1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^5},4{s^2}$
B.$1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^{10}},4{s^2}4{p^5}$
C.$1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^6},4{s^1}$
D.$1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^{10}},4{s^1}4{p^6}$

Answer 1

Electronic configuration is the arrangement of electrons in different orbits and in its orbitals. Orbits i.e. $1,2,3...$ and orbitals i.e. $s - ,p - ,d -$. The element which is just above the element with atomic number $43$ will be element having atomic number $25$.

Complete step by step answer:
Let us see how to calculate the atomic number with the given element.
The method is as follows:
The given atomic number of elements is $43$. Now we have to calculate the element just above this element in the same periodic group. So to calculate this subtract $18$ from the given element atomic number. So the atomic number of the required element will be $43 - 18 = 25$.
Orbits: These are the spaces in the atom where electrons revolve. The orbits are as $1,2,3,...$
Orbitals: The space where electrons are likely to be found. The orbitals are as $s,p,d$ and $f$ orbitals.
For $s -$orbitals maximum number of electrons it can have is two.
For $p -$orbitals maximum number of electrons it can have is six
For $d -$ orbitals maximum number of electrons it can have is ten.
For $f -$ orbitals maximum number of electrons it can have is fourteen.
Now the electronic configuration will be as: The number of electrons is $25$. Two electrons will be in $1s$ so it will become $1{s^2}$. Then the next two electrons will go into $2s$ and now the configuration will be as: $1{s^2},2{s^2}$. Now we are left with $21$ electrons. The next six electrons will go into $2p$orbitals. Now the configuration is as: $1{s^2},2{s^2}2{p^6}$. Now the left electrons are $15$. Similarly next eight electrons will go into $3s$ and $3p$ orbitals. Now the configuration is as: $1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}$. Now the total electrons left are seven. Now the next two electrons will go into $4s$. And configuration will be as: $1{s^2},2{s^2}2{p^6},3{s^2}3{p^6},4{s^2}$. And finally the last five remaining electrons will go into $3d$ orbitals. And finally the electronic configuration the element will be as $1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^5},4{s^2}$.

Hence option A is correct.

Note:
For a given orbital maximum number of electrons it can hold is determined as $2(2l + 1)$ where $l$ is azimuthal quantum number. For $s$ the value of $l$ is zero, for $p$ the value of $l$ is one and so on. So the maximum number of electrons in $s$ orbitals is two.