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

(B) $1.$ For $S_1$ (spherically symmetric,$\ell = 0$):
$\text{Radial node} = n - \ell - 1 = 1$ $\Rightarrow n - 0 - 1 = 1$ $\Rightarrow n = 2.$
Thus,state $S_1$ is $2s$.
For $S_2$,energy $E_{S_2} = E_H(\text{ground}) = -13.6 \ eV$.
$E_{S_2} = \frac{-13.6 \times Z^2}{n^2} = -13.6 \ eV$ $\Rightarrow \frac{3^2}{n^2} = 1$ $\Rightarrow n = 3.$
Given $S_2$ has one radial node: $n - \ell - 1 = 1$ $\Rightarrow 3 - \ell - 1 = 1$ $\Rightarrow \ell = 1$.
$2.$ Energy of $S_1$ $(n=2, Z=3)$: $E_{S_1} = \frac{-13.6 \times 3^2}{2^2} = -13.6 \times 2.25 \ eV$.
In units of $E_H(\text{ground}) = -13.6 \ eV$,the energy is $2.25$.
$3.$ For $S_2$,$\ell = 1$ (as calculated above).

(C) For a shell with principal quantum number $n$,the total number of orbitals is $n^2$.
For $n=3$,the total number of orbitals is $3^2 = 9$.
Each orbital can hold a maximum of $2$ electrons,one with $m_s = +1/2$ and one with $m_s = -1/2$.
Therefore,for $n=3$,there are $9$ orbitals,and each can accommodate exactly one electron with $m_s = -1/2$.
Thus,the maximum number of electrons with $n=3$ and $m_s = -1/2$ is $9$.

(A) For a hydrogen-like species,the energy is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \ eV$.
For $He^+$,$Z = 2$. Given $E = -3.4 \ eV$,we have $-3.4 = -13.6 \times \frac{2^2}{n^2}$,which simplifies to $n^2 = 16$,so $n = 4$.
Given azimuthal quantum number $\ell = 2$,the state is $4d$.
Number of angular nodes $= \ell = 2$.
Number of radial nodes $= n - \ell - 1 = 4 - 2 - 1 = 1$.
Since $He^+$ is a single-electron species,the nuclear charge experienced is the full nuclear charge $Z = 2e$,so statement $(4)$ is incorrect.
Thus,statements $(1)$ and $(3)$ are true.

(C) For $n=4$,the possible subshells are $4s, 4p, 4d, 4f$.
Condition $|m_{\ell}|=1$ implies $m_{\ell} = +1$ or $m_{\ell} = -1$.
- In $4p$ subshell $(\ell=1)$: $m_{\ell} = -1, 0, +1$. The orbitals with $|m_{\ell}|=1$ are $m_{\ell} = -1$ and $m_{\ell} = +1$ ($2$ orbitals).
- In $4d$ subshell $(\ell=2)$: $m_{\ell} = -2, -1, 0, +1, +2$. The orbitals with $|m_{\ell}|=1$ are $m_{\ell} = -1$ and $m_{\ell} = +1$ ($2$ orbitals).
- In $4f$ subshell $(\ell=3)$: $m_{\ell} = -3, -2, -1, 0, +1, +2, +3$. The orbitals with $|m_{\ell}|=1$ are $m_{\ell} = -1$ and $m_{\ell} = +1$ ($2$ orbitals).
Total number of orbitals with $|m_{\ell}|=1$ is $2+2+2 = 6$. Since each orbital can hold one electron with $m_s=-1/2$,the total number of electrons is $6$.

(C) For a multi-electron species like $H^-$,the electronic configuration is $1s^2$.
Ground state: $1s^2$.
First excited state: One electron is promoted to the next available orbital,$1s^1 2s^1$.
Second excited state: One electron is promoted to the $2p$ subshell,$1s^1 2p^1$.
The $2p$ subshell consists of three degenerate orbitals $(2p_x, 2p_y, 2p_z)$.
Since the $1s$ electron is fixed,the degeneracy is determined by the number of available $2p$ orbitals,which is $3$.

(D) Degenerate orbitals are orbitals that have the same energy level.
In a hydrogen-like atom,$2p_x$ and $2p_y$ orbitals possess the same energy,making them degenerate.
$A$ spectral line is observed only when an electron transitions between orbitals of different energy levels (i.e.,$\Delta E \neq 0$).
Since $2p_x$ and $2p_y$ have the same energy,the transition $2p_x \rightarrow 2p_y$ involves no change in energy $(\Delta E = 0)$,and thus no spectral line is observed.
Therefore,Statement-$I$ is false and Statement-$II$ is true.

(C) Statement $I$ is true: For a given principal quantum number $n$,the total number of orbitals in a shell is calculated as $n^2$.
Statement $II$ is true: The magnetic quantum number $m_l$ determines the spatial orientation of orbitals and ranges from $-l$ to $+l$ including zero,which corresponds to the number of orbitals in a subshell.

(A) In a hydrogen atom,the energy of an orbital depends only on the principal quantum number $(n)$.
For a given $n$,all orbitals are degenerate (have the same energy).
The given orbitals have the following $n$ values:
$A$. $4s$ $(n=4)$
$B$. $3p_x$ $(n=3)$
$C$. $3d_{x^2-y^2}$ $(n=3)$
$D$. $3d_{z^2}$ $(n=3)$
$E$. $4p_z$ $(n=4)$
Since the energy increases with $n$,the orbitals with $n=3$ $(3p_x, 3d_{x^2-y^2}, 3d_{z^2})$ have the lowest energy.
Therefore,the correct option is $B, C$ and $D$ only.

(D) The energy of an orbital in a multielectron atom depends on the principal quantum number $(n)$ and the azimuthal quantum number $(\ell)$.
$A: n=1, \ell=0, m_{\ell}=0 \rightarrow 1s$
$B: n=2, \ell=0, m_{\ell}=0 \rightarrow 2s$
$C: n=2, \ell=1, m_{\ell}=1 \rightarrow 2p$
$D: n=3, \ell=2, m_{\ell}=1 \rightarrow 3d$
$E: n=3, \ell=2, m_{\ell}=0 \rightarrow 3d$
In the absence of external electric and magnetic fields,all orbitals belonging to the same subshell (same $n$ and $\ell$) are degenerate,meaning they have the same energy.
Since both $D$ and $E$ correspond to the $3d$ subshell,they have the same energy.

(C) For a single-electron species like the hydrogen atom,the energy of an orbital depends only on the principal quantum number $(n)$.
Therefore,orbitals with the same $n$ value have the same energy,regardless of the azimuthal quantum number $(l)$.
Thus,$2s = 2p$,$3s = 3p = 3d$,and $4s = 4p = 4d = 4f$.
Evaluating the given statements:
$A$. $1s < 2p < 3d < 4s$ is incorrect because $3d = 4s$ is not true; actually $3d < 4s$ is false as they have different $n$ values,but $3d$ and $4s$ are not equal. Wait,for $H$,$E_n \propto 1/n^2$,so $E_{3d} < E_{4s}$. Thus $A$ is correct.
$B$. $1s < 2s = 2p < 3s = 3p$ is correct.
$C$. $1s < 2s < 2p < 3s < 3p$ is incorrect because $2s = 2p$ and $3s = 3p$.
$D$. $1s < 2s < 4s < 3d$ is incorrect because $3d < 4s$.
Since $C$ and $D$ are not correct,the answer is $C$ and $D$ only.

(D) The azimuthal (subsidiary) quantum number $\ell$ determines the shape of the orbital.
$(B)$ The provided image represents the boundary surface diagram of the $2p_x$ orbital.
$(C)$ The $+$ and $-$ signs in the wave function of the $2p_x$ orbital represent the phase of the wave function,not the electrical charge.
$(D)$ The wave function of the $2p_x$ orbital is zero in the $yz$ plane,which acts as the nodal plane for this orbital.
Therefore,statements $(A)$ and $(B)$ are correct.

(D) The orbital angular momentum is given by the formula $\sqrt{\ell(\ell+1)} \frac{h}{2 \pi}$.
For the $2s$ orbital,the azimuthal quantum number $\ell = 0$.
Therefore,the orbital angular momentum $= \sqrt{0(0+1)} \frac{h}{2 \pi} = 0$.
For the $2p$ orbital,the azimuthal quantum number $\ell = 1$.
Therefore,the orbital angular momentum $= \sqrt{1(1+1)} \frac{h}{2 \pi} = \sqrt{2} \frac{h}{2 \pi}$.

(C) The electronic configuration of $Cr$ $(Z=24)$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^5, 4s^1$.
Azimuthal quantum number $l=1$ corresponds to $p$-orbitals. The electrons in $p$-orbitals are in $2p^6$ and $3p^6$,which gives $6+6 = 12$ electrons.
Azimuthal quantum number $l=2$ corresponds to $d$-orbitals. The electrons in $d$-orbitals are in $3d^5$,which gives $5$ electrons.
Therefore,the number of electrons with $l=1$ is $12$ and with $l=2$ is $5$.

(B) The extra stability of half-filled subshells is attributed to the following factors:
$(I)$ Symmetrical distribution of electrons: Electrons are distributed symmetrically in the orbitals,which leads to lower energy.
$(II)$ Larger exchange energy: The number of possible exchanges for electrons with parallel spins is maximum in half-filled and fully-filled configurations,resulting in higher exchange energy and greater stability.
$(III)$ Smaller coulombic repulsion: Due to the symmetrical arrangement,the inter-electronic repulsion is minimized.
$(IV)$ Smaller shielding of electrons by one another: The effective nuclear charge experienced by the electrons is optimized in these configurations.
Therefore,statements $(A), (B), (D),$ and $(E)$ are correct.

(B) The element with atomic number $9$ is Fluorine $(F)$,with electronic configuration $1s^2 2s^2 2p^5$.
$(A)$ In $1s^2 2s^2 2p^5$,there are $9$ electrons. Filling them: $1s$ $(2)$,$2s$ $(2)$,$2p$ $(5)$. Total up-spins $(m_s = +\frac{1}{2})$ are $1+1+3 = 5$ and down-spins $(m_s = -\frac{1}{2})$ are $1+1+2 = 4$. Statement $A$ is correct.
$(B)$ The $2p^5$ configuration means $p_x^2 p_y^2 p_z^1$ or similar. The unpaired electron can be in any of the three $p$ orbitals,so $B$ is not necessarily true. Statement $B$ is incorrect.
$(C)$ The last electron enters the $2p$ subshell where $n=2$ and $l=1$. Statement $C$ is correct.
$(D)$ Angular nodes = $l$. For $1s$ $(l=0)$,$2s$ $(l=0)$,$2p$ $(l=1)$. Total angular nodes = $0+0+1+1+1 = 3$. Statement $D$ is incorrect.
Therefore,only $A$ and $C$ are correct.

(B) The energy of an electron in an atom is determined by the $(n+l)$ rule.
For electron $X$: $n=3, l=2$. Therefore,$(n+l) = 3+2 = 5$.
For electron $Y$: $n=3, l=0$. Therefore,$(n+l) = 3+0 = 3$.
Since the $(n+l)$ value for $X$ $(5)$ is greater than the $(n+l)$ value for $Y$ $(3)$,the energy of $X$ is greater than the energy of $Y$.

(C) Hund's rule of maximum multiplicity states that for a given electron configuration,the term with the maximum multiplicity has the lowest energy. This means that electrons should occupy degenerate orbitals singly as far as possible before pairing occurs.
$(A)$ $2s^2 2p^2$: Electrons are singly occupied in $2p$ orbitals. This follows Hund's rule.
$(B)$ $2s^2 2p^3$: The $2s$ orbital has two electrons with parallel spins,which violates the Pauli exclusion principle. The $2p$ orbital follows Hund's rule.
$(C)$ $2s^2 2p^2$: Electrons are paired in one $2p$ orbital instead of being singly occupied. This violates Hund's rule.
$(D)$ $2s^2 2p^3$: This is the correct ground state configuration following both Pauli's exclusion principle and Hund's rule.
Therefore,only Hund's rule is violated in option $(C)$.

(D) Statement $(a)$ is incorrect: When an electron moves from a higher energy level $(n=4)$ to a lower energy level $(n=2)$,it emits energy,it does not gain it.
Statement $(b)$ is incorrect: Energy of an orbital increases with the principal quantum number $(n)$. Thus,a $3p$ orbital has higher energy than a $2p$ orbital.
Statement $(c)$ is correct: Any $s$-orbital (or subshell) can hold a maximum of $2$ electrons.
Statement $(d)$ is incorrect: The maximum number of electrons in an orbit is given by $2n^2$. For the $3^{rd}$ orbit $(n=3)$,the maximum number of electrons is $2(3)^2 = 2(9) = 18$ electrons.

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

(C) The condition given is $(n+\ell) \leq 3$.
For $3p$,the principal quantum number $n = 3$ and the azimuthal quantum number $\ell = 1$.
Therefore,$(n+\ell) = 3 + 1 = 4$.
Since $4 > 3$,the $3p$ subshell is not possible under the given condition.

(A) $1$. Angular nodes are equal to the azimuthal quantum number $(l)$. For $d$-orbitals,$l=2$,so they have $2$ angular nodes. Thus,$A$ matches $P$ $(4d_{x^2-y^2})$.
$2$. The $d_{z^2}$ orbital has zero nodal planes (it has two conical nodal surfaces). Thus,$B$ matches $Q$ $(3d_{z^2})$.
$3$. Radial nodes are calculated by the formula $(n - l - 1)$. For $3s$,$n=3, l=0$,so radial nodes $= 3-0-1 = 2$. Thus,$C$ matches $S$ $(3s)$.
$4$. Angular nodes for $f$-orbitals are $l=3$. Thus,$D$ matches $R$ $(4f)$.
Therefore,the correct match is $A-P, B-Q, C-S, D-R$.

(B) In a hydrogen atom ($H$-atom),the energy of an orbital depends only on the principal quantum number $(n)$. Therefore,$3d$ $(n=3)$ has lower energy than $4s$ $(n=4)$.
However,the Assertion states the energy of $3d$ is smaller than $4s$,which is true for the $H$-atom.
The Reason provided describes the $(n+\ell)$ rule,which applies only to multi-electron species,not to the $H$-atom.
For the $H$-atom,the energy order is $1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f$.
Since the Reason explains the energy order for multi-electron species and not the $H$-atom,it is not the correct explanation for the Assertion.
Thus,both statements are true,but the Reason is not the correct explanation for the Assertion.

(A) The formula for orbital angular momentum is $\sqrt{\ell(\ell+1)} \frac{h}{2 \pi}$.
For a $3s$ subshell,the azimuthal quantum number $\ell = 0$. Thus,the orbital angular momentum $= \sqrt{0(0+1)} \frac{h}{2 \pi} = 0$.
For a $3p$ subshell,the azimuthal quantum number $\ell = 1$. Thus,the orbital angular momentum $= \sqrt{1(1+1)} \frac{h}{2 \pi} = \sqrt{2} \frac{h}{2 \pi} = \frac{h}{\sqrt{2} \pi}$.

(A) For an orbital with principal quantum number $n$ and azimuthal quantum number $l$:
$1.$ Number of angular nodes (nodal planes) $= l$
$2.$ Number of radial nodes $= n - l - 1$
$3.$ Total number of nodes $= (n - l - 1) + l = n - 1$
Given: Number of nodal planes $= 2$,so $l = 2$ (which corresponds to a $d$-orbital).
Given: Total nodes $= 4$,so $n - 1 = 4$,which implies $n = 5$.
Therefore,the orbital is $5d$.

(B) The electronic configuration of Sodium $(Z=11)$ is $1s^2, 2s^2, 2p^6, 3s^1$.
The valence shell electron is in the $3s$ orbital.
For the $3s$ orbital,the principal quantum number $n = 3$.
For an $s$-orbital,the azimuthal quantum number $\ell = 0$.
Since $\ell = 0$,the magnetic quantum number $m = 0$.
The spin quantum number $s$ for the first electron in an orbital is typically taken as $+\frac{1}{2}$.

(A) The electronic configuration of the atom is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$.
We need to find the number of electrons with magnetic quantum number $m=0$ and spin quantum number $s=+\frac{1}{2}$.
- $1s^2$: One electron has $m=0, s=+\frac{1}{2}$.
- $2s^2$: One electron has $m=0, s=+\frac{1}{2}$.
- $2p^6$: The $2p_z$ orbital has $m=0$. It contains $2$ electrons,one of which has $s=+\frac{1}{2}$.
- $3s^2$: One electron has $m=0, s=+\frac{1}{2}$.
- $3p^6$: The $3p_z$ orbital has $m=0$. It contains $2$ electrons,one of which has $s=+\frac{1}{2}$.
- $4s^2$: One electron has $m=0, s=+\frac{1}{2}$.
Total number of electrons with $m=0$ and $s=+\frac{1}{2}$ is $1+1+1+1+1+1 = 6$.

(B) The $3d_{x^2-y^2}$ orbital has lobes along the $x$ and $y$ axes. The electron density is concentrated along these axes,not in the $xy$ plane. Thus,the statement is correct.
$(b)$ The $3d_{z^2}$ orbital has lobes along the $z$-axis and a ring of electron density in the $xy$ plane. Thus,the statement that electron density in the $xy$ plane is zero is incorrect.
$(c)$ The number of radial nodes is given by $(n - l - 1)$. For $2s$,$n=2, l=0$,so nodes $= 2 - 0 - 1 = 1$. Thus,the statement is correct.
$(d)$ For $2p_z$ orbital,the nodal plane is the $xy$ plane (where $z=0$). The $yz$ plane is not a nodal plane for $2p_z$. Thus,the statement is incorrect.
Therefore,statements $(b)$ and $(d)$ are incorrect.

(C) The energy of an orbital is determined by the $(n + \ell)$ rule.
For orbitals with the same $(n + \ell)$ value,the orbital with the lower $n$ value has lower energy.
Calculating $(n + \ell)$ values:
$3s: 3 + 0 = 3$
$3p: 3 + 1 = 4$
$4s: 4 + 0 = 4$
$3d: 3 + 2 = 5$
$4p: 4 + 1 = 5$
$5s: 5 + 0 = 5$
$4d: 4 + 2 = 6$
$5p: 5 + 1 = 6$
Comparing these,the correct increasing order is $3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p$.

(A) The electronic configuration of ${}_{24}Cr$ is $[Ar] 3d^5 4s^1$ and that of ${}_{29}Cu$ is $[Ar] 3d^{10} 4s^1$.
These configurations are considered abnormal because they deviate from the expected filling order of the Aufbau principle.
The stability of these configurations arises because half-filled $(d^5)$ and fully-filled $(d^{10})$ subshells possess extra stability due to symmetry and exchange energy.
While they are indeed $d-$block elements,the primary reason for the specific electronic arrangement is the stability of the filled/half-filled subshells.

(C) The electronic configuration of the elements provided in the options are as follows:
$Cu (Z=29): 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^{10}$
$Cr (Z=24): 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5$
$Zn (Z=30): 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10}$
$K (Z=19): 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1$
Among the given options,only $Zn$ $(Z=30)$ has a $4s^2$ configuration in its outermost shell.

(C) The electronic configuration of Argon ($Ar$,$Z=18$) is $1s^2 2s^2 2p^6 3s^2 3p^6$.
For a $p$-orbital,the magnetic quantum number $m$ takes values $-1, 0, +1$.
In the $2p^6$ subshell,there are $3$ orbitals $(2p_x, 2p_y, 2p_z)$. Each orbital contains $2$ electrons.
The orbital with $m = +1$ contains $2$ electrons.
Similarly,in the $3p^6$ subshell,there are $3$ orbitals. The orbital with $m = +1$ contains $2$ electrons.
Total number of electrons with $m = +1$ is $2 + 2 = 4$.

(D) The $p$-orbitals $(p_x, p_y, p_z)$ are oriented along the $x, y,$ and $z$ axes respectively.
Among the $d$-orbitals,$d_{x^2-y^2}$ and $d_{z^2}$ are oriented along the axes.
Specifically,$d_{x^2-y^2}$ has lobes along the $x$ and $y$ axes,and $d_{z^2}$ has lobes along the $z$ axis.
In contrast,$d_{xy}, d_{yz},$ and $d_{zx}$ orbitals have lobes located between the axes.
Therefore,$p_z$ and $d_{x^2-y^2}$ are both located along the axes.

(C) The $4^{th}$ shell $(n = 4)$ contains subshells corresponding to $\ell = 0, 1, 2, 3$.
For $\ell = 2$,the subshell is $4d$.
The number of orbitals in a subshell is given by the formula $(2\ell + 1)$.
Substituting $\ell = 2$,we get $2(2) + 1 = 5$ orbitals.

(D) In the provided image,the third orbital (the first box of the $p$-subshell) contains two electrons with the same spin (both pointing upwards).
According to Pauli's exclusion principle,no two electrons in an atom can have the same set of four quantum numbers.
This implies that in a single orbital,electrons must have opposite spins.
Since the two electrons in the same orbital have the same spin,Pauli's exclusion principle is violated.

(D) The energy of an orbital is determined by the $(n+l)$ rule. Higher the sum of $(n+l)$ value of an orbital,the higher is its energy.
If two orbitals have the same value of $(n+l)$,then the orbital with the higher value of $n$ will have more energy.
Let us calculate $(n+l)$ for each:
$(A)$ $2p$: $n=2, l=1 \implies n+l = 3$
$(B)$ $3s$: $n=3, l=0 \implies n+l = 3$
$(C)$ $3d$: $n=3, l=2 \implies n+l = 5$
$(D)$ $4p$: $n=4, l=1 \implies n+l = 5$
Comparing $3d$ and $4p$,both have $(n+l) = 5$. Since $4p$ has a higher principal quantum number $(n=4)$ compared to $3d$ $(n=3)$,$4p$ has the highest energy.

(A) The energy of an orbital is determined by the $(n+l)$ rule,where $n$ is the principal quantum number and $l$ is the azimuthal quantum number. The orbital with the lowest $(n+l)$ value has the lowest energy. If $(n+l)$ values are equal,the orbital with the lower $n$ value has lower energy.
Calculating $(n+l)$ for each:
$2p: n=2, l=1 \implies n+l = 3$
$3s: n=3, l=0 \implies n+l = 3$
$3d: n=3, l=2 \implies n+l = 5$
$4p: n=4, l=1 \implies n+l = 5$
Comparing $2p$ and $3s$,both have $(n+l) = 3$. Since $2p$ has a lower $n$ value $(n=2)$ compared to $3s$ $(n=3)$,$2p$ has the lowest energy.

(A) The atomic number of copper $(Cu)$ is $29$.
The ground state electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$.
In the $3d$ subshell,all $10$ electrons are paired.
In the $4s$ subshell,there is $1$ electron,which is unpaired.
Therefore,the number of unpaired electrons in $Cu$ is $1$.

(A) The element located in period-$4$ and group-$12$ is Zinc $(Zn)$.
$Zn$ has an atomic number of $30$.
The electronic configuration of $Zn$ in the ground state is $[Ar] \ 3d^{10} \ 4s^2$.
In this configuration,all $3d$ and $4s$ orbitals are completely filled,meaning there are $0$ unpaired electrons.
In the excited state (e.g.,$Zn^{2+}$ ion),the configuration is $[Ar] \ 3d^{10}$,which also contains $0$ unpaired electrons.
Therefore,the total number of unpaired electrons is $0$.

(C) The atomic number of $Ti$ is $22$. The electronic configuration of $Ti$ is $[Ar] 3d^2 4s^2$.
In the $+2$ oxidation state,$Ti^{2+}$ loses two electrons from the $4s$ orbital.
Therefore,the electronic configuration of $Ti^{2+}$ is $[Ar] 3d^2$.
Thus,there are $2$ electrons present in the $3d$ orbital.

(D) The electronic configuration of $Cr (Z=24)$ is $[Ar] 3d^5 4s^1$.
This configuration is adopted because half-filled $d$-orbitals $(d^5)$ provide extra stability.
In the $3d$ subshell,there are $5$ electrons,each occupying one of the $5$ $d$-orbitals singly.
Therefore,the number of unpaired electrons in the $d$-orbitals is $5$.

(C) The atomic number of Chromium $(Cr)$ is $24$.
Its expected electronic configuration is $[Ar] 3d^4 4s^2$.
However,due to the extra stability of half-filled $d$-orbitals,the actual electronic configuration is $[Ar] 3d^5 4s^1$.
In this configuration,there are $5$ unpaired electrons in the $3d$ subshell and $1$ unpaired electron in the $4s$ subshell.
Therefore,the total number of unpaired electrons is $5 + 1 = 6$.

(B) The number of radial nodes in an orbital is given by the formula: $\text{Radial nodes} = n - l - 1$.
For the $2s$ orbital,the principal quantum number $n = 2$ and the azimuthal quantum number $l = 0$.
Substituting these values into the formula: $\text{Radial nodes} = 2 - 0 - 1 = 1$.
Therefore,the $2s$ orbital has $1$ radial node.

(D) The $d$-orbitals are five in number: $d_{xy}$,$d_{yz}$,$d_{xz}$,$d_{x^2-y^2}$,and $d_{z^2}$.
Among these,the first four ($d_{xy}$,$d_{yz}$,$d_{xz}$,and $d_{x^2-y^2}$) have a double-dumbbell shape.
The $d_{z^2}$ orbital has a unique shape consisting of a dumbbell along the $z$-axis with a doughnut-shaped ring of electron density in the $xy$-plane.
Therefore,$d_{z^2}$ has a different shape compared to the others.

(C) The value of $(n+l)$ for the $3d$ orbital is calculated as follows:
For $3d$,$n=3$ and $l=2$,so $(n+l) = 3+2 = 5$.
Now,let us calculate the $(n+l)$ values for the given options:
$A) 4s$: $n=4, l=0 \implies (n+l) = 4+0 = 4$
$B) 3s$: $n=3, l=0 \implies (n+l) = 3+0 = 3$
$C) 4p$: $n=4, l=1 \implies (n+l) = 4+1 = 5$
$D) 2p$: $n=2, l=1 \implies (n+l) = 2+1 = 3$
Comparing these values,the $4p$ orbital has the same $(n+l)$ value of $5$ as the $3d$ orbital.

(C) The principal quantum number $n$ represents the shell number,and the azimuthal quantum number $l$ represents the subshell type.
For $l=0$,the subshell is $s$.
For $l=1$,the subshell is $p$.
For $l=2$,the subshell is $d$.
For $l=3$,the subshell is $f$.
Given $n=3$ and $l=2$,the orbital is $3d$.

(D) The $N$ shell corresponds to the principal quantum number $n = 4$.
For a given $n$,the possible values of the azimuthal quantum number $l$ range from $0$ to $n-1$.
Thus,for $n = 4$,$l = 0, 1, 2, 3$ (corresponding to $4s, 4p, 4d, 4f$ subshells).
The number of orbitals in a subshell is given by $2l + 1$.
For $l = 0$ $(4s)$: $2(0) + 1 = 1$ orbital.
For $l = 1$ $(4p)$: $2(1) + 1 = 3$ orbitals.
For $l = 2$ $(4d)$: $2(2) + 1 = 5$ orbitals.
For $l = 3$ $(4f)$: $2(3) + 1 = 7$ orbitals.
Total number of orbitals in $N$ shell $= 1 + 3 + 5 + 7 = 16$.
Alternatively,the total number of orbitals in a shell is given by $n^2 = 4^2 = 16$.

(D) The orbital designation is determined by the principal quantum number $n$ and the azimuthal quantum number $l$.
For $l=0$,the orbital is $s$.
For $l=1$,the orbital is $p$.
For $l=2$,the orbital is $d$.
For $l=3$,the orbital is $f$.
Given $n=4$ and $l=3$,the orbital is $4f$.

(C) In a hydrogen atom,the energy of an orbital depends only on the principal quantum number $(n)$.
Since $2s$ and $2p$ both have $n = 2$,they have the same energy.
Orbitals having the same energy are called degenerate orbitals.
Therefore,$2s$ and $2p$ are degenerate orbitals.

(B) According to the Pauli Exclusion Principle,any single orbital can accommodate a maximum of $2$ electrons with opposite spins.
Since the question specifies a single $p$-orbital defined by the quantum numbers $n=5$ and $m=1$,it can hold a maximum of $2$ electrons.

(B) The $M$-shell corresponds to the principal quantum number $n = 3$.
For $n = 3$,the possible values of the azimuthal quantum number $l$ are $0, 1, 2$,which correspond to the $3s, 3p,$ and $3d$ subshells.
The number of orbitals in a subshell is given by $2l + 1$.
For $l = 0$ $(3s)$: $2(0) + 1 = 1$ orbital.
For $l = 1$ $(3p)$: $2(1) + 1 = 3$ orbitals.
For $l = 2$ $(3d)$: $2(2) + 1 = 5$ orbitals.
Total number of orbitals $= 1 + 3 + 5 = 9$.
Each orbital can accommodate a maximum of $2$ electrons.
Therefore,the maximum number of electrons in the $M$-shell $= 9 \times 2 = 18$.