Quantum number, Electronic configuration and Shape of orbitals Questions in English

(A) To determine the energy order,we use the $(n+l)$ rule:
$A. n=4, l=1 \implies n+l = 4+1 = 5$ ($4p$ orbital)
$B. n=4, l=0 \implies n+l = 4+0 = 4$ ($4s$ orbital)
$C. n=3, l=2 \implies n+l = 3+2 = 5$ ($3d$ orbital)
$D. n=3, l=1 \implies n+l = 3+1 = 4$ ($3p$ orbital)
According to the $(n+l)$ rule,orbitals with lower $(n+l)$ values have lower energy.
If $(n+l)$ values are equal,the orbital with the lower $n$ value has lower energy.
Comparing the values:
$D (n+l=4, n=3) < B (n+l=4, n=4) < C (n+l=5, n=3) < A (n+l=5, n=4)$
Thus,the order of increasing energy is $D < B < C < A$.

(A) The electronic configuration of Rubidium $(Z=37)$ is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^1$.
The valence electron is present in the $5s$ orbital.
For the $5s$ orbital,the principal quantum number $n=5$.
Since it is an $s$-orbital,the azimuthal quantum number $l=0$.
For $l=0$,the magnetic quantum number $m=0$.
The spin quantum number $s$ can be $+1/2$ or $-1/2$. Thus,the set is $(5, 0, 0, +1/2)$.

(B) The electronic configuration of $Al$ $(Z = 13)$ is $1s^2 2s^2 2p^6 3s^2 3p^1$.
The last electron enters the $3p$ orbital.
For the $3p$ orbital,the quantum numbers are:
Principal quantum number $(n)$ = $3$.
Azimuthal quantum number $(l)$ = $1$.
Magnetic quantum number $(m)$ can be $-1, 0, +1$.
Spin quantum number $(s)$ can be $+\frac{1}{2}$ or $-\frac{1}{2}$.
Option $B$ represents the quantum numbers for an electron in the $3s$ orbital $(n=3, l=0, m=0, s=+\frac{1}{2})$,which is not the last electron of $Al$. Therefore,this set is not possible for the last electron.

(D) The atomic number of Palladium $(Pd)$ is $46$.
The electronic configuration of $Pd$ in its ground state is $[Kr] 4d^{10} 5s^0$.
The value $l = 2$ corresponds to the $d$-subshell.
In the $4d^{10}$ configuration,there are $10$ electrons in the $d$-subshell.
Therefore,the number of electrons with $l = 2$ is $10$.

(C) The given angular wave function is $\psi_{(\theta, \phi)} = \left( \frac{15}{4\pi} \right)^{1/2} \sin \theta \cos \theta \sin \phi$.
This corresponds to the $d_{yz}$ orbital.
In spherical coordinates,$y = r \sin \theta \sin \phi$ and $z = r \cos \theta$.
The product $yz$ is proportional to $\sin \theta \sin \phi \cos \theta$,which matches the angular part provided.
The nodal planes for a $d_{yz}$ orbital are the $XY$ plane $(z=0)$ and the $XZ$ plane $(y=0)$,where the probability of finding the electron is zero.
Therefore,the correct option is $C$.

(B) The electronic configuration of $Cu$ $(Z=29)$ is $1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6 \ 4s^1 \ 3d^{10}$.
For $s$-orbitals $(l=0)$,$m_l$ is always $0$. There are $2$ electrons in $1s$,$2$ in $2s$,$2$ in $3s$,and $1$ in $4s$.
For $p$-orbitals $(l=1)$,$m_l$ values are $-1, 0, +1$. Each $p$-subshell has $6$ electrons,so $2$ electrons have $m_l=0$ in each ($2p$ and $3p$).
For $d$-orbitals $(l=2)$,$m_l$ values are $-2, -1, 0, +1, +2$. In a filled $3d^{10}$ subshell,$2$ electrons have $m_l=0$.
Total electrons with $m_l = 0$ are: $2 (1s) + 2 (2s) + 2 (2p) + 2 (3s) + 2 (3p) + 1 (4s) + 2 (3d) = 13$.

(D) The number of radial nodes in an orbital is given by the formula $n - l - 1$.
For a $2s$ orbital $(n=2, l=0)$,radial nodes $= 2 - 0 - 1 = 1$. The plot should show $1$ node.
For a $2p$ orbital $(n=2, l=1)$,radial nodes $= 2 - 1 - 1 = 0$. The plot should show $0$ nodes.
For a $3s$ orbital $(n=3, l=0)$,radial nodes $= 3 - 0 - 1 = 2$. The plot should show $2$ nodes.
For a $3p$ orbital $(n=3, l=1)$,radial nodes $= 3 - 1 - 1 = 1$. The plot should show $1$ node.
Looking at the provided plots:
- Plot $A$ $(2s)$ shows $1$ node (Correct).
- Plot $B$ $(2p)$ shows $0$ nodes (Correct).
- Plot $C$ $(3s)$ shows $2$ nodes (Correct).
- Plot $D$ $(3p)$ shows $1$ node,but the label is $3p$. Wait,let's re-examine the plots. Plot $D$ shows $1$ node,which is correct for $3p$. However,looking closely at the options provided in standard chemistry problems of this type,the plot for $2p$ should have $0$ nodes,and the plot for $3p$ should have $1$ node. All plots appear to match their labels based on the radial node formula. Let's re-evaluate the question. If all are correct,there might be a subtle error in the provided image or the question expects identifying the one that is commonly mislabeled in textbooks. Actually,the $2p$ plot $(B)$ shows $0$ nodes,which is correct. The $3p$ plot $(D)$ shows $1$ node,which is correct. The $3s$ plot $(C)$ shows $2$ nodes,which is correct. The $2s$ plot $(A)$ shows $1$ node,which is correct. Given the options,if one must be incorrect,we re-verify the $3p$ plot. The $3p$ plot should have $1$ radial node. The image $D$ shows $1$ node. It seems all are correctly labeled. However,often in such questions,the $2p$ plot is incorrectly shown with a node. Since all are correct,the question might be flawed,but based on standard curriculum,all are correctly labeled.

(A) The electronic configuration of $Cr$ $(Z = 24)$ is $[Ar] 3d^5 4s^1$.
In this configuration,the electrons are present in the $1s, 2s, 2p, 3s, 3p, 3d,$ and $4s$ orbitals.
Option $A$ corresponds to $n = 4, l = 1$,which represents a $4p$ orbital.
Since there are no electrons in the $4p$ orbital for a chromium atom in its ground state,this set of quantum numbers is not associated with any electron in $Cr$.

(B) The $3^{rd}$ principal energy level $(n=3)$ consists of $3s$,$3p$,and $3d$ subshells. For this level to be completely filled,it must contain $2 + 6 + 10 = 18$ electrons.
In $Ar$ $(Z=18)$,the configuration is $1s^2 2s^2 2p^6 3s^2 3p^6$. Here,the $3d$ subshell is empty,so the $3^{rd}$ level is not completely filled.
In $Zn$ $(Z=30)$,the configuration is $[Ar] 3d^{10} 4s^2$. The $3^{rd}$ level $(3s^2 3p^6 3d^{10})$ contains $18$ electrons and is completely filled.
In $Cu$ $(Z=29)$,the configuration is $[Ar] 3d^{10} 4s^1$. The $3^{rd}$ level is completely filled.
In $Cu^{+2}$ $(Z=29)$,the configuration is $[Ar] 3d^9$. The $3^{rd}$ level is not completely filled.
In $Zn^{+2}$ $(Z=30)$,the configuration is $[Ar] 3d^{10}$. The $3^{rd}$ level is completely filled.
Comparing the options,both $Cu$ and $Zn$ have a completely filled $3^{rd}$ principal energy level.

(C) For any given orbital,the magnetic quantum number $m$ must satisfy the condition $-l \leqslant m \leqslant +l$.
In option $C$,$l=2$,so the possible values for $m$ are $-2, -1, 0, 1, 2$.
The value $m=-3$ is not possible for $l=2$.
Therefore,the arrangement $n=3, l=2, m=-3, s=1/2$ is impossible.

(D) The angular wave function for $p$-orbitals is given by:
$A(\theta, \phi) = \sqrt{\frac{3}{4\pi}} \cos \theta$ for $p_z$ orbital.
Given expression is $\frac{1}{2}\sqrt{\frac{3}{\pi}}\cos \theta = \sqrt{\frac{1}{4}} \cdot \sqrt{\frac{3}{\pi}} \cos \theta = \sqrt{\frac{3}{4\pi}} \cos \theta$.
Since the angular part depends only on $\theta$ and involves $\cos \theta$,it corresponds to the $p_z$ orbital,which is oriented along the $Z$-axis.
For $p_z$ orbital,the nodal plane is the $XY$ plane,where $\theta = 90^{\circ}$ and $\cos 90^{\circ} = 0$.

(C) The number of radial nodes is given by the formula: $n - \ell - 1 = 1$.
Given $n = 3$,we substitute the value: $3 - \ell - 1 = 1$.
Solving for $\ell$: $2 - \ell = 1$,which gives $\ell = 1$.
The orbital angular momentum is calculated using the formula: $\sqrt{\ell(\ell+1)} \frac{h}{2\pi}$.
Substituting $\ell = 1$: $\sqrt{1(1+1)} \frac{h}{2\pi} = \sqrt{2} \frac{h}{2\pi}$.

(A) $1$. For a $4d$ orbital,the number of spherical (radial) nodes is given by the formula $n - l - 1$. Here,$n = 4$ and $l = 2$. So,$4 - 2 - 1 = 1$. The statement saying it is $3$ is $FALSE$.
$2$. For $p_y$ orbital,the electron density is zero in the $xz$ plane,so it acts as an angular node. This is $TRUE$.
$3$. For $s$ orbitals,the probability density $\psi^2$ is maximum at the nucleus $(r = 0)$. This is $TRUE$.
$4$. The angular wave function for $p_z$ is proportional to $\cos \theta$,where $\theta$ is the angle with the $z$-axis. This is $TRUE$.

(A) According to Hund's rule of maximum multiplicity,the most stable configuration (lowest energy) is the one with the maximum number of unpaired electrons with parallel spins.
In the given options,the configuration with three electrons in three separate orbitals (as shown in option $A$) represents the half-filled state,which is more stable and has lower energy compared to the configurations where electrons are paired in the same orbital,as pairing electrons requires overcoming inter-electronic repulsion.

(B) The azimuthal quantum number $(l)$ determines the shape of the orbital and the subshell.
The values of $l$ correspond to subshells as follows:
$l = 0$ for $s$-subshell
$l = 1$ for $p$-subshell
$l = 2$ for $d$-subshell
$l = 3$ for $f$-subshell
$l = 4$ for $g$-subshell
Therefore,for the $g$-subshell,the value of the azimuthal quantum number is $4$.

(C) For hydrogen-like species (single electron systems),the energy of an electron depends only on the principal quantum number $n$.
The energy of an orbital is given by the formula:
$E_n = -13.6 \times \frac{Z^2}{n^2} \ eV$
In the given question,the sub-shells $3s$,$3p$,and $3d$ all belong to the same principal shell $(n=3)$.
Since they have the same value of $n$,they are degenerate,meaning they have the same energy.
Therefore,the energy required to remove an electron from these sub-shells is equal.
$E_1 = E_2 = E_3$
Thus,option $C$ is correct.

(D) For any orbital,the principal quantum number $n$ must be greater than the azimuthal quantum number $l$. Since $l = 4$ for $g$ orbitals,the lowest possible value for $n$ is $n = l + 1 = 4 + 1 = 5$.
The magnetic quantum number $m_l$ ranges from $-l$ to $+l$. For $l = 4$,$m_l$ can take values $\{-4, -3, -2, -1, 0, +1, +2, +3, +4\}$.
The number of orbitals in a subshell is given by $2l + 1$. For $l = 4$,there are $2(4) + 1 = 9$ orbitals.
Each orbital can hold a maximum of $2$ electrons,so the total number of electrons is $9 \times 2 = 18$.

(C) The energy of an orbital is determined by the $(n + l)$ rule.
For $(i): n = 4, l = 0 \implies n + l = 4 + 0 = 4 \ (4s \text{ orbital})$.
For $(ii): n = 3, l = 1 \implies n + l = 3 + 1 = 4 \ (3p \text{ orbital})$.
For $(iii): n = 3, l = 2 \implies n + l = 3 + 2 = 5 \ (3d \text{ orbital})$.
For $(iv): n = 3, l = 0 \implies n + l = 3 + 0 = 3 \ (3s \text{ orbital})$.
According to the $(n + l)$ rule,if $(n + l)$ values are different,the orbital with the higher $(n + l)$ value has higher energy.
If $(n + l)$ values are the same,the orbital with the higher $n$ value has higher energy.
Comparing the values: $(iii) \ (5) > (i) \ (4, n=4) > (ii) \ (4, n=3) > (iv) \ (3)$.
Thus,the decreasing order of energy is $(iii) > (i) > (ii) > (iv)$.

(B) $p$ orbital is dumbbell-shaped.
The electrons present in a $p$ orbital can have any one of three values of magnetic quantum numbers: $0, +1$,or $-1$.

(A) The $s$-orbital has a spherical shape,which means the probability of finding an electron is the same in all directions around the nucleus. Therefore,the $s$-orbital is considered non-directional. In contrast,$p$,$d$,and $f$ orbitals have specific directional orientations in space.

(A) The orbitals having the same energy but different in orientation are called degenerate orbitals.
For example,the $3d$-orbitals have $l = 2$,which gives $m = -2, -1, 0, +1, +2$.
This means there are five different orientations represented by $d_{xy}, d_{yz}, d_{zx}, d_{x^2-y^2}$,and $d_{z^2}$,all of which possess the same energy.

(C) The principal quantum number $n = 4$ indicates the shell is the $4^{th}$ shell.
The azimuthal quantum number $l = 3$ corresponds to the $f$ subshell ($l=0$ is $s$,$l=1$ is $p$,$l=2$ is $d$,$l=3$ is $f$).
Combining these,the electron occupies the $4f$ subshell.
Therefore,the correct option is $C$.

(B) The rules for quantum numbers are as follows:
$1$. $n$ (principal quantum number) can be any positive integer $(1, 2, 3, \dots)$.
$2$. $l$ (azimuthal quantum number) can range from $0$ to $n-1$.
$3$. $m$ (magnetic quantum number) can range from $-l$ to $+l$,including $0$.
$4$. $s$ (spin quantum number) can be $+1/2$ or $-1/2$.
Evaluating the options:
- Option $A$: $n=2, l=1, m=-1, s=-1/2$. Here $l < n$ and $|m| \le l$. This is possible.
- Option $B$: $n=3, l=2, m=-3, s=+1/2$. Here $l < n$,but $m$ must be between $-2$ and $+2$. Since $|m| = 3 > l$,this set is not possible.
- Option $C$: $n=2, l=0, m=0, s=+1/2$. Here $l < n$ and $|m| \le l$. This is possible.
- Option $D$: $n=3, l=2, m=-2, s=+1/2$. Here $l < n$ and $|m| \le l$. This is possible.
Therefore,the set in option $B$ is not possible.

(C) For any principal quantum number $n$,the azimuthal quantum number $l$ can have values from $0$ to $n-1$.
In option $A$,$n=2$,so $l$ can be $0$ or $1$. Since $l=0$,this is correct.
In option $B$,$n=4$,so $l$ can be $0, 1, 2, 3$. Since $l=3$,this is correct.
In option $C$,$n=2$,so $l$ can only be $0$ or $1$. The value $l=2$ is not possible for $n=2$.
Therefore,the set of quantum numbers in option $C$ is incorrect.

(A) For a valid set of quantum numbers,the following rules must be satisfied:
$1$. $n$ is a positive integer $(1, 2, 3, \dots)$.
$2$. $l$ can have values from $0$ to $n-1$.
$3$. $m$ can have values from $-l$ to $+l$ (including $0$).
$4$. $s$ can only be $+1/2$ or $-1/2$.
Evaluating the options:
Option $A$: $n=5, l=2, m=+2, s=-1/2$. Here $l < n$ $(2 < 5)$ and $|m| \le l$ $(2 \le 2)$. This is a valid set.
Option $B$: $n=5, l=5$. This is invalid because $l$ must be less than $n$.
Option $C$: $s=0$. This is invalid because $s$ must be $\pm 1/2$.
Option $D$: $m=+3, l=2$. This is invalid because $|m|$ cannot exceed $l$ $(3 > 2)$.

(A) The principal quantum number $n$ determines the number of subshells present in a given energy level.
For any principal quantum level $n$,the number of subshells is equal to $n$.
For example,if $n = 1$,there is $1$ subshell $(1s)$.
If $n = 2$,there are $2$ subshells ($2s$ and $2p$).
Therefore,the number of subshells in the $n^{th}$ quantum level is $n$.

(D) The number of orbitals in a given energy level with principal quantum number $n$ is given by the formula $n^2$.
For $n = 3,$ the number of orbitals is $3^2 = 9$.
These $9$ orbitals consist of $1$ orbital in the $3s$ subshell,$3$ orbitals in the $3p$ subshell,and $5$ orbitals in the $3d$ subshell $(1 + 3 + 5 = 9)$.

(C) For a given principal quantum number $n$,the number of subshells is equal to $n$. Therefore,for $n = 3$,the number of subshells is $3$ (which are $3s, 3p, 3d$).
The total number of orbitals in a shell is given by the formula $n^2$. Therefore,for $n = 3$,the number of orbitals is $3^2 = 9$.

(B) The azimuthal quantum number $l$ determines the subshell type:
$l = 0$ corresponds to $s$-orbital,
$l = 1$ corresponds to $p$-orbital,
$l = 2$ corresponds to $d$-orbital,
$l = 3$ corresponds to $f$-orbital.
Given $l = 2$,the orbital type is $d$.
The number of orbitals in a subshell is given by the formula $(2l + 1)$.
For $l = 2$,the number of orbitals $= 2(2) + 1 = 5$.
Thus,for $l = 2$,there are $5$ orbitals of type $d$.

(C) The $d_{xy}$ orbital has four lobes that lie in the $xy-$ plane.
These lobes are oriented between the $x$ and $y$ axes.
The maximum electron density (probability) for the $d_{xy}$ orbital is located at an angle of $45^o$ from both the $x$ and $y$ axes,which corresponds to the bisectors of the axes.

(D) The electron spin quantum number $(m_s)$ describes the intrinsic angular momentum of an electron.
While it is often visualized as the electron spinning on its axis,this is a simplified model.
In quantum mechanics,the values $+1/2$ and $-1/2$ represent two distinct quantum mechanical spin states that do not have a direct classical analogue.

(D) For a given value of $n$,the azimuthal quantum number $l$ can have values from $0$ to $(n-1)$.
The magnetic quantum number $m$ can have values from $-l$ to $+l$.
Given $m = -3$,the minimum value of $l$ must be $3$ (since $|m| \leq l$).
If $l = 3$,then the principal quantum number $n$ must be at least $n = l + 1 = 3 + 1 = 4$.
Therefore,the principal quantum number $n$ can be $4$ or higher. Among the given options,$4$ is the correct value.

(A) For a given azimuthal quantum number $l$,the magnetic quantum number $m$ ranges from $-l$ to $+l$.
For $s$ orbital,$l = 0$,so $m$ can only be $0$.
For $p$ orbital,$l = 1$,so $m$ can be $-1, 0, +1$.
For $d$ orbital,$l = 2$,so $m$ can be $-2, -1, 0, +1, +2$.
For $f$ orbital,$l = 3$,so $m$ can be $-3, -2, -1, 0, +1, +2, +3$.
Since the given magnetic quantum number is $m = -1$,it is impossible for the electron to be in an $s$ orbital because $m$ must be $0$ for $l = 0$.

(D) For a given orbital,the magnetic quantum number $m_l$ depends on the azimuthal quantum number $l$ such that $m_l$ ranges from $-l$ to $+l$.
Given $m_l = 2$,the possible values for $l$ must satisfy the condition $l \ge |m_l|$.
Therefore,$l$ must be at least $2$ (i.e.,$l = 2$ or $l = 3$ for $n = 4$).
Option $D$ states that $l$ can be $0, 1, 2, \text{ or } 3$,which is incorrect because $l$ cannot be $0$ or $1$ when $m_l = 2$.

(D) The rules for quantum numbers are as follows:
$1$. The principal quantum number $(n)$ can be any positive integer $(1, 2, 3, \dots)$.
$2$. For a given $n$,the azimuthal quantum number $(l)$ can have values from $0$ to $(n-1)$.
$3$. For a given $l$,the magnetic quantum number $(m)$ can have values from $-l$ to $+l$,including $0$.
Evaluating the options:
- Option $A$: For $n=1$,$l$ can be $0$. For $l=0$,$m$ can be $0$. This set is correct.
- Option $B$: For $n=2$,$l$ can be $0$ or $1$. If $l=0$,$m=0$. If $l=1$,$m=-1, 0, +1$. This set is correct.
- Option $C$: For $n=3$,$l$ can be $0, 1, 2$. If $l=0$,$m=0$. If $l=1$,$m=-1, 0, +1$. If $l=2$,$m=-2, -1, 0, +1, +2$. This set is correct.
Since all provided sets follow the rules of quantum mechanics,the correct answer is $D$.

(B) According to the Pauli Exclusion Principle,no two electrons in an atom can have the same set of all four quantum numbers $(n, l, m_l, m_s)$.
Option $A$ states that two electrons may have identical values of $n, l,$ and $m$. This is possible if they have different spin quantum numbers ($m_s = +1/2$ and $-1/2$).
Option $B$ states a $4d$ electron has $n = 4$ and $l = 3$. For a $d-$orbital,the azimuthal quantum number $l$ must be $2$. Therefore,$l = 3$ corresponds to an $f-$orbital,making this statement impossible.
Option $C$ and $D$ are possible according to Hund's Rule and the Pauli Exclusion Principle for electrons in different $p-$orbitals or the same orbital.

(B) The orbital angular momentum of an electron in the $2s$-orbital is zero.
For a given subshell with azimuthal quantum number $l$,the orbital angular momentum is given by the expression:
$L = \frac{h}{2\pi} \sqrt{l(l+1)}$
For $2s$ orbital,the azimuthal quantum number $l = 0$.
Substituting the value of $l$ in the formula:
$L = \frac{h}{2\pi} \sqrt{0(0+1)} = 0$

(B) The formula for orbital angular momentum is given by $\frac{h}{2 \pi} \sqrt{\ell(\ell+1)}$.
For an $s$ orbital,the azimuthal quantum number $\ell = 0$.
Substituting $\ell = 0$ into the formula: $\text{Angular momentum} = \frac{h}{2 \pi} \sqrt{0(0+1)} = 0$.

(A) The orbital angular momentum $(L)$ is given by the formula $L = \sqrt{l(l+1)} \hbar$.
For a $d$ electron,the azimuthal quantum number $(l)$ is $2$.
Substituting the value of $l$ into the formula:
$L = \sqrt{2(2+1)} \hbar = \sqrt{2 \times 3} \hbar = \sqrt{6} \hbar$.
Therefore,the orbital angular momentum for a $d$ electron is $\sqrt{6} \hbar$.

(D) To completely describe an electron in an atom,we require $4$ quantum numbers: the principal quantum number $(n)$,the azimuthal quantum number $(l)$,the magnetic quantum number $(m_l)$,and the spin quantum number $(m_s)$.
This is consistent with Pauli's exclusion principle,which states that no two electrons in an atom can have the same set of all $4$ quantum numbers.

(D) The principal quantum number $(n)$ determines the main energy level or shell of an electron in an atom.
It is primarily related to the size of the orbital and the average distance of the electron from the nucleus.
As the value of $n$ increases,the size and energy of the orbital increase.

(C) The magnetic quantum number $(m_l)$ is related to the orientation of orbitals in space.
It determines the preferred spatial orientation of orbitals.
It specifies the number of orbitals in a subshell.
It also explains the splitting of spectral lines in the presence of magnetic ($Zeeman$ effect) and electric ($Stark$ effect) fields.

(C) The principal quantum number,denoted by $n$,determines the main energy level or shell in which an electron resides.
It provides information about the size of the orbital and the average distance of the electron from the nucleus.
As the value of $n$ increases,the electron is,on average,further away from the nucleus.

(D) Atomic orbitals are defined as the regions of space around the nucleus of an atom where the probability of finding an electron is maximum.
Unlike the Bohr orbits which are circular paths,atomic orbitals represent the wave function of an electron in a three-dimensional space.

(D) According to the Pauli Exclusion Principle,any single orbital can accommodate a maximum of $2$ electrons,and these electrons must have opposite spins.
Therefore,any $p-$ orbital can accommodate a maximum of $2$ electrons with opposite spins.

(B) The $p$-orbital is dumb-bell shaped.
Electrons present in the $p$-orbital can have any one of three values of the magnetic quantum number: $0, +1$,or $-1$.

(D) For any subshell,the number of orbitals is given by $(2l + 1)$.
For an $f-$ subshell,the azimuthal quantum number $l = 3$.
Number of orbitals $= 2(3) + 1 = 7$.
Since each orbital can accommodate a maximum of $2$ electrons,the maximum number of electrons in the $f-$ subshell is $7 \times 2 = 14$.

(C) For any given principal quantum number $n$,the azimuthal quantum number $l$ can have values from $0$ to $n-1$.
For a given $l$,the magnetic quantum number $m$ can have values from $-l$ to $+l$ including zero.
In option $C$,$n=3$ and $l=2$. The allowed values for $m$ are $-2, -1, 0, +1, +2$.
Since $m=-3$ is outside the range of allowed values for $l=2$,this arrangement is impossible.

(D) For any orbital,the principal quantum number $n$ is given by the coefficient of the orbital symbol. For $2p$,$n = 2$.
The azimuthal quantum number $l$ represents the subshell type. For $s$-orbitals,$l = 0$; for $p$-orbitals,$l = 1$; for $d$-orbitals,$l = 2$; and for $f$-orbitals,$l = 3$.
Therefore,for $2p$-orbitals,$n = 2$ and $l = 1$.

(D) For a $4d$ electron,the principal quantum number $n = 4$.
For a $d$-orbital,the azimuthal quantum number $l = 2$.
The magnetic quantum number $m$ can have any integer value from $-l$ to $+l$,i.e.,$-2, -1, 0, +1, +2$.
The spin quantum number $s$ can be either $+1/2$ or $-1/2$.
Comparing this with the given options,option $D$ $(n=4, l=2, m=1, s=-1/2)$ satisfies all the conditions for a $4d$ electron.