Question

Determine the maximum number of the electron that has been having each of the following designations: $\mathbf{2}\mathit{f}\mathbf{,}\mathbf{2}{\mathit{d}}_{\mathbf{x}\mathbf{y}}\mathbf{,}\mathbf{3}\mathit{p}\mathbf{,}\mathbf{5}{\mathit{d}}_{\mathbf{y}\mathbf{z}}\mathbf{,}\mathbf{4}\mathit{p}$.

Short answer

Answer:

Quantum Numbers may be defined as a set of four identification numbers which are needed to have full information about various electrons present in an atom.

Step-by-step solution

Principal Quantum Number (n) :

It describes the average distance of the electron from the nucleus

Value of$n$	1	2	3	4	5	6
Designation of Shell	K	L	M	N	O	P

Azimuthal or Angular Quantum Number (l):

It gives information about various subshells in a particular principal shell.

$$\begin{gathered} l\mathit{=}\mathit{0}\mathit{,}s\mathit{-}subshell \\ l\mathit{=}\mathit{1}\mathit{,}p\mathit{-}subshell \\ l\mathit{=}\mathit{2}\mathit{,}d\mathit{-}subshell \\ l\mathit{=}\mathit{3}\mathit{,}f\mathit{-}subshell \end{gathered}$$

Magnetic Quantum Number:

It is designated by $M_{\mathit{i}}$and it gives information about orbitals associated with different$l$ values and for $M_{\mathit{i}}\mathit{,}l$value varies$\mathit{=}\mathit{-}l\mathit{}to\mathit{}\mathit{+}l$

Spin Quantum Number (Ms):

It tells about electron orientation or spin which is clockwise, upspin or anticlockwise downspin.

As per Pauli’s exclusion principle, only 2 electrons with opposite spin can be filled in an orbital like

Case I:

$\mathit{2}f\mathit{=}$it has no electron because for $n\mathit{=}\mathit{2}$there is no existence of $f\mathit{-}$orbital because $f\mathit{-}$orbital has $l\mathit{=}\mathit{3}$so the value of $n$should be 4 to have $f\mathit{-}$orbital.

Case II:

For $\mathit{2}{d}_{\mathit{x}\mathit{y}}$: It has no electron because $d\mathit{-}$orbital are not present for $n\mathit{=}\mathit{2}$.

Case III:

For $\mathit{3}p\mathit{=}$it has maximum $\mathit{6}{e}^{\mathit{-}}$because $p\mathit{-}$subshell has $\mathit{3}p\mathit{-}$orbitals and accommodates $\mathit{2}{e}^{\mathit{-}}$in each.

Case IV:

For $\mathit{5}{d}_{\mathit{y}\mathit{z}}\mathit{=}$it has a maximum of two electrons because an orbital can accommodate only two electrons with opposite spin.

Case V:

For $\mathit{4}p\mathit{-}$: it has maximum 6 electrons because $p\mathit{-}$the subshell has $\mathit{3}p\mathit{-}$orbitals $\left(p_x,p_y,p_z\right)$. Therefore each orbital can accommodate two electrons in each.