Question

The atomic number of an element $M$ is $26$. How many electrons are present in the $M - shell$ of the element in its ${M^{3 + }}$ state?
A. $11$
B. $15$
C. $14$
D. $13$

Answer 1

The atomic number $26$ in the periodic table is for the element $Fe$ known as Iron, and $K,{\text{ }}L,M,N$are the shells in which the electron are filled for which the value is as follows:
$K = 1,{\text{ }}L = 2,{\text{ }}M = 3,{\text{ }}N = 4$and so on.

Complete answer:
For this question one must know what are the rules for filling the electrons to obtain electronic configurations;
i. Aufbau Principle: The electron which is added will always occupy the orbital with the lowest energy first.
ii. Pauli Exclusion Principle: Each orbital can hold a maximum of two electrons of opposite spins.
iii. Hund’s Rule of Multiplicity: When filling a sub-shell, each orbital must be occupied singly (keeping electron spins the same) before they are occupied in pairs.
And the electrons are filled in the order as follows: $1s < 2s < 2p < 3s < 3p < 4s < 3d$ and further. The $1,{\text{ }}2$ and $3$ in the configuration are the $K,{\text{ }}L,{\text{ }}M$ shells and the $s,{\text{ }}p,{\text{ }}d$ are the sub-shells where the $s$ sub-shell can have $2$ electrons, $p$ sub-shell can have $6$ electrons and $d$ sub-shell can hold a maximum of $10$ electrons.
Now for the element $26$ the electronic configuration will be as follows:
Electronic configuration is $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^6}$
Now as we are asked for the ${M^{3 + }}$ state which means we have to remove three electrons and the first $2$ electrons are removed from the $4s$ then from the 3d this is because as we know that the $4s$ has lower energy than $3d$ but still $4s$ is the outermost electron as its experiences repulsion, so the ${M^{3 + }}$ electronic configuration will be
${M^{3 + }}$ electronic configuration will be $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^5}$
Now having a look at the configuration the electrons present for$M{\text{ }} = {\text{ }}3$ , the number of electrons present will be $2 + 6 + 5 = 13\;$ electrons.

Hence, the correct option is D.

Note: The energy for $4s$ is less than $3d$ because the overall $\left( {n + l} \right)$ value is less for $4s$ than the $3d$, now the value of $l$ means the value of sub-shells; $s,{\text{ }}p,{\text{ }}d$and $f$are $0,{\text{ }}1,{\text{ }}2,{\text{ }}3$ respectively and for $4s$ the $\left( {n + l} \right)$ value comes out to be $4 + 0{\text{ }} = {\text{ }}4$ and for $3d$ is $3 + 2{\text{ }} = {\text{ }}5.$