Question

Maximum possible electron(s) in Mn, for which $n + l + m = 5$ is/are:
(A) 1
(B) 2
(C) 3
(D) 10

Answer 1

We know that, manganese is the twenty fifth element of the periodic table and hence its electronic configuration would be $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^5}$ which is nothing but a systematic representation of the size, shape and orientation of the number of orbitals. We also know that the electron filling in each atomic orbital takes place by filling the complete orbitals in a subshell with one electron each and then filling those orbitals with another electron of opposite spin to stabilise it.

Complete Step by Step answer
We know that, there are 4 quantum numbers that represent the state of an atom, and the electrons that constitute them. These quantum numbers are systematically represented in the form of electronic configuration and are arranged into the periodic table accordingly.
There are four quantum numbers and they are:
Principle quantum numbers: This quantum number describes the size of the orbital of an atom and it is represented by $n$ in $ns,np,nd$ . The value of $n$ can be 1,2,3,4 and so on.
Azimuthal quantum numbers: This quantum number represents the shape of the orbital and it is represented by $l$ . The values of $l$ ranges from $0$ to $\left( {n - 1} \right)$ . In an electronic configuration it is the representation of the orbitals $s,p,d,f$ where $s = 0,p = 1,d = 2,f = 3$ and so on.
Magnetic quantum number: It represents the orientation of the orbitals and is represented by ${m_l}$ . The values of magnetic quantum numbers range from $- l$ to $+ l$ .
Spin Quantum number: It represents direction of electron spin and is represented by ${m_s}$ .The values of spin quantum numbers are $+ \dfrac{1}{2}$ for upward spin and $- \dfrac{1}{2}$ for downward spin.
We now know that manganese has the electron configuration $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^5}$ . Now we are supposed to calculate the maximum possible electrons in the $Mn$ for $n + l + m = 5$ .
From the electronic configuration we can infer that the values of quantum numbers will be,
$n = 1$ to $n = 3$ , $l = 0$ to $l = \left( {n - 1} \right) = 3 - 1 = 2$ , and ${m_l} = - 2, - 1,0, + 1, + 2$ . We can also say that the value of $n + l + m = 5$ occurs only at orbitals $3p$ and $3d$ .
For $3p$ , we have $n = 3,l = 1$ so the corresponding values of ${m_l}$ will be $- 1,0, + 1$ out of which ${m_l} = + 1$ is the desirable value here. There will be only 2 electrons in the $3p$ orbital with orientation ${m_l} = + 1$ .
For $3d$ , we have $n = 3,l = 2$ and the corresponding values of ${m_l}$ will be $- 2, - 1,0, + 1, + 2$ out of which ${m_l} = 0$ is the only desirable value here. But we know that the $3d$ orbital which has the capacity of 10 electrons is only half filled in $Mn$ from the configuration [only $3{d^5}$ ]. Thus, there will be only 1 electron in the $3d$ orbital with orientation ${m_l} = 0$ .
Thus from $3p$ and $3d$ we will have a maximum number of 3 electrons, i.e. option C.

Note
We know electrons are negatively charged and hence repel each other; thus, an atomic orbital is completely filled before the electrons pair up to repel each other thus affecting the ionic stability of the atom.