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

(B) For the orbital $3d_{z^2}$:
$1$. The principal quantum number $n$ is given by the coefficient,so $n = 3$.
$2$. For a $d$-orbital,the azimuthal quantum number $\ell = 2$.
$3$. The magnetic quantum number $m$ for $d_{z^2}$ is $0$.
$4$. The spin quantum number $s$ can be either $+\frac{1}{2}$ or $-\frac{1}{2}$ for any electron.
Thus,the correct values are $n = 3, \ell = 2, m = 0, s = +\frac{1}{2} \text{ or } -\frac{1}{2}$.

(A) The electronic configuration of argon $(Z = 18)$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6$.
For $p$-orbitals,the azimuthal quantum number $l = 1$,so the magnetic quantum number $m_{\ell}$ can take values $-1, 0, +1$.
In the $2p^6$ subshell,there are $2$ electrons in the $2p_x$ orbital (where $m_{\ell} = 1$ or $-1$ depending on convention).
In the $3p^6$ subshell,there are also $2$ electrons in the $3p_x$ orbital (where $m_{\ell} = 1$ or $-1$ depending on convention).
Thus,the total number of electrons with $m_{\ell} = 1$ is $2 + 2 = 4$.

(B) The electronic configuration of $Br$ (atomic number $35$) is $[Ar]\, 3d^{10}\,4s^2\,4p^5$.
$Br^{-}$ is formed by adding one electron to $Br$,so $Br^{-}$ is $[Ar]\, 3d^{10}\,4s^2\,4p^6$.
$Br^{+2}$ is formed by removing two electrons from the neutral $Br$ atom. Electrons are removed from the outermost shell ($4p$ orbital first).
$Br^{+2}$ configuration is $[Ar]\, 3d^{10}\,4s^2\,4p^3$.
The element with atomic number $33$ is Arsenic $(As)$,which has the configuration $[Ar]\, 3d^{10}\,4s^2\,4p^3$.
Therefore,the configuration of $Br^{+2}$ is identical to $As$.

(D) The electronic configuration of Barium $(Z = 56)$ is $[Xe] \, 6s^2$.
According to the Aufbau principle,the order of filling of orbitals after $6s$ is determined by the $(n+l)$ rule.
The sequence of energy levels is $6s < 4f < 5d < 6p < 7s$.
Therefore,after the $6s$ orbital is filled,the next orbitals to be filled are $4f$,$5d$,and $6p$.

(B) The number of elements in a period is given by the number of electrons that can be filled in the orbitals corresponding to that period.
For a given period $n$,the orbitals filled are $ns$,$(n-2)f$,$(n-1)d$,and $np$.
The total number of electrons is $2 + 2(2l+1) + 2(2l+1) + 6 = \frac{(n+2)^2}{2}$.
Alternatively,for a period $n$,the maximum azimuthal quantum number $l$ is related to the number of orbitals.
Given the formula $\frac{(n+2)^2}{2}$,we equate this to the capacity of orbitals.
For the $n^{th}$ period,the maximum value of the azimuthal quantum number $l$ is $\frac{n}{2}$ if $n$ is even,or $\frac{n+1}{2}$ if $n$ is odd.
However,checking the options against the standard derivation for this specific theoretical model,the correct expression is $\frac{n}{2}$.

(A) If an orbital can accommodate $3$ electrons,then the capacity of subshells changes as follows:
$s$-subshell ($1$ orbital) = $3$ electrons.
$p$-subshell ($3$ orbitals) = $9$ electrons.
$d$-subshell ($5$ orbitals) = $15$ electrons.
$f$-subshell ($7$ orbitals) = $21$ electrons.
Filling the electrons for atomic number $50$ in the order of energy:
$1s^3, 2s^3, 2p^9, 3s^3, 3p^9, 3d^{15}, 4s^3, 4p^5$.
Total electrons = $3 + 3 + 9 + 3 + 9 + 15 + 3 + 5 = 50$.
The valence shell is the $4^{th}$ shell,so it belongs to the $4^{th}$ period.
The last electron enters the $p$-subshell,so it belongs to the $p$-block.

(D) In the $d$-orbital set,the orbitals are classified based on their orientation relative to the axes.
$1$. The $d_{xy}, d_{yz},$ and $d_{zx}$ orbitals have their lobes directed between the axes.
$2$. The $d_{x^2-y^2}$ and $d_{z^2}$ orbitals have their lobes directed along the axes.
Therefore,the two orbitals located along the axes are $d_{x^2-y^2}$ and $d_{z^2}$.

(B) The energy of an electron in a multi-electron atom is determined by the $(n + l)$ rule.
According to this rule,the orbital with the lower $(n + l)$ value has lower energy.
If the $(n + l)$ values are equal,the orbital with the lower $n$ value has lower energy.
Calculating $(n + l)$ for each option:
$A: n = 3, l = 2 \implies n + l = 3 + 2 = 5$
$B: n = 4, l = 0 \implies n + l = 4 + 0 = 4$
$C: n = 4, l = 1 \implies n + l = 4 + 1 = 5$
$D: n = 5, l = 0 \implies n + l = 5 + 0 = 5$
Comparing the values,option $B$ has the lowest $(n + l)$ value of $4$.
Therefore,the electron in option $B$ is associated with the least energy.

(D) According to the $(n + \ell)$ rule:
$1$. The orbital with the lower $(n + \ell)$ value has lower energy.
$2$. If two orbitals have the same $(n + \ell)$ value,the one with the lower $n$ value has lower energy.
Statement $D$ is incorrect because if $(n + \ell)$ is the same,the orbital with the higher $n$ value has higher energy,not lower energy.
Statement $B$ is also technically incorrect in a multi-electron system as energy depends on $Z_{eff}$,but $D$ is a direct violation of the Aufbau principle rule.

(B) The electronic configuration of Niobium $(Z = 41)$ is: $[Kr] 4d^4 5s^1$ or $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^{10}, 4s^2, 4p^6, 4d^4, 5s^1$.
For $m = -1$,we consider the orbitals where the magnetic quantum number $m$ can be $-1$:
$2p$ subshell $(2p_y)$: $2$ electrons
$3p$ subshell $(3p_y)$: $2$ electrons
$3d$ subshell ($3d_{xy}$ or $3d_{yz}$): $2$ electrons
$4p$ subshell $(4p_y)$: $2$ electrons
$4d$ subshell ($4d_{xy}$ or $4d_{yz}$): $0$ to $1$ electron depending on the specific arrangement of the $4$ electrons in $4d$.
Summing these,we get $2 + 2 + 2 + 2 + (0 \text{ to } 1) = 8$ or $9$ electrons.

(B) To determine the paramagnetism,we look at the number of unpaired electrons in the ground state of each atom:
$1. \text{Oxygen } (O): 1s^2 2s^2 2p^4$. The $2p$ subshell has $2$ unpaired electrons.
$2. \text{Nitrogen } (N): 1s^2 2s^2 2p^3$. The $2p$ subshell has $3$ unpaired electrons.
$3. \text{Aluminum } (Al): [Ne] 3s^2 3p^1$. The $3p$ subshell has $1$ unpaired electron.
$4. \text{Calcium } (Ca): [Ar] 4s^2$. All electrons are paired,so it is diamagnetic ($0$ unpaired electrons).
Comparing the number of unpaired electrons: $N (3) > O (2) > Al (1) > Ca (0)$.
Therefore,the correct order of decreasing paramagnetism is $N > O > Al > Ca$.

(C) The electronic configuration of an element with three $2p$ electrons is $1s^2 \, 2s^2 \, 2p^3$.
Summing the electrons: $2 + 2 + 3 = 7$.
The element with atomic number $7$ is Nitrogen $(N)$.

(D) An excited state occurs when an electron is promoted to a higher energy orbital than it would occupy in the ground state configuration.
In option $D$,the configuration $1s^2 2s^2 2p^5 3s^1$ represents an excited state of the neon atom $(1s^2 2s^2 2p^6)$.
Here,one electron from the $2p$ orbital has been promoted to the $3s$ orbital,which is a higher energy level.

(C) The correct answer is $(c)$.
$Cr$ $(Z=24)$: The electronic configuration is $[Ar] 3d^5 4s^1$. Each of the five $d$-orbitals $(d_{xy}, d_{yz}, d_{zx}, d_{x^2-y^2}, d_{z^2})$ contains $1$ electron.
$Fe^{3+}$ $(Z=26)$: The electronic configuration is $[Ar] 3d^5$. Each of the five $d$-orbitals contains $1$ electron.
$Cu^{+}$ $(Z=29)$: The electronic configuration is $[Ar] 3d^{10}$. Each of the five $d$-orbitals contains $2$ electrons.
Therefore,the $d_{xy}$ orbital in $Cu^{+}$ contains $2$ electrons,which is the maximum among the given options.

(C) An atom is in an excited state when an electron jumps from a lower energy orbital to a higher energy orbital,violating the Aufbau principle.
In the configuration $1s^2, 2s^2 2p^2, 3s^1$,the ground state for an atom with $7$ electrons would be $1s^2, 2s^2 2p^3$.
Since one electron has been promoted from the $2p$ orbital to the $3s$ orbital,this represents an excited state.

(C) The electronic configuration of $Ca$ $(Z = 20)$ according to the Aufbau principle is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$.
According to the Aufbau principle,the $4s$ orbital is filled before the $3d$ orbital due to its lower energy.
If the Aufbau principle were not followed,the $3d$ orbital would be filled before the $4s$ orbital.
In that case,the electronic configuration of $Ca$ would be $1s^2 2s^2 2p^6 3s^2 3p^6 3d^2$.
Since the last electron enters the $3d$ orbital,the element would be placed in the $d-$block.

(A) The maximum number of unpaired electrons in the $4p$ subshell is $3$,which corresponds to the $4p^3$ configuration.
The electronic configuration for this element is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^{10}, 4p^3$.
Summing the electrons: $2 + 2 + 6 + 2 + 6 + 2 + 10 + 3 = 33$.
Therefore,the atomic number is $33$.

(D) The phosphide ion is $P^{3-}$. The atomic number of phosphorus is $15$. So,the number of electrons in $P^{3-}$ is $15 + 3 = 18$.
Comparing this with the options:
$N^{3-} = 7 + 3 = 10$ electrons.
$F^- = 9 + 1 = 10$ electrons.
$Na^+ = 11 - 1 = 10$ electrons.
$Cl^- = 17 + 1 = 18$ electrons.
Thus,the phosphide ion $(P^{3-})$ has the same number of electrons as the chloride ion $(Cl^-)$.

(B) The electronic configuration of Chlorine ($Cl$,atomic number $17$) is $[Ne] 3s^2 3p^5$.
Thus,the element with the outer electronic configuration $3s^2 3p^5$ is $Cl$.

(A) The atomic number of nickel $(Ni)$ is $28$. The electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^8$.
For $m = 0$:
- $1s^2$: $2$ electrons $(l=0, m=0)$
- $2s^2$: $2$ electrons $(l=0, m=0)$
- $2p^6$: $2$ electrons ($m=0$ for $2p_z$)
- $3s^2$: $2$ electrons $(l=0, m=0)$
- $3p^6$: $2$ electrons ($m=0$ for $3p_z$)
- $4s^2$: $2$ electrons $(l=0, m=0)$
- $3d^8$: In $d$-orbitals,$m$ values are $-2, -1, 0, +1, +2$. For $d^8$,filling $8$ electrons: $d_{xy}^2, d_{yz}^2, d_{zx}^2, d_{x^2-y^2}^1, d_{z^2}^1$ (or similar distribution). The $d_{z^2}$ orbital corresponds to $m=0$. With $8$ electrons,$d_{z^2}$ will have $2$ electrons.
Total electrons with $m=0$ is $2+2+2+2+2+2+2 = 14$.

(A) The electronic configuration of element $A$ is $1s^2, 2s^2, 2p^6$.
This configuration corresponds to the neon $(Ne)$ atom,which is a noble gas.
Noble gases have a completely filled valence shell,making them chemically inert and stable in their monoatomic form.
Therefore,the stable form of $A$ is represented as $A$.

(D) The $d$ orbitals are defined by their spatial orientation relative to the coordinate axes.
$d_{xy}$ orbital has lobes oriented between the $X$ and $Y$ axes.
$d_{yz}$ orbital has lobes oriented between the $Y$ and $Z$ axes.
$d_{xz}$ orbital has lobes oriented between the $X$ and $Z$ axes.
$d_{x^2-y^2}$ orbital has lobes oriented along the $X$ and $Y$ axes.
$d_{z^2}$ orbital has lobes oriented along the $Z$ axis.
Therefore,the $d$ orbital with orientation along the $X$ and $Y$ axes is $d_{x^2-y^2}$.

(C) The correct answer is $(C)$.
For an orbital to have zero probability of finding an electron in the $XY$ plane,the $XY$ plane must be a nodal plane for that orbital.
$1$. For $3p_z$,the nodal plane is the $XY$ plane (since the electron density is along the $z$-axis).
$2$. For $4d_{xz}$,the nodal planes are the $XY$ plane and the $YZ$ plane (since the lobes lie in the $XZ$ plane).
$3$. Therefore,in the pair $(4d_{xz}, 3p_z)$,both orbitals have the $XY$ plane as a nodal plane,meaning the probability of finding an electron in the $XY$ plane is zero for both.

(A) Spin multiplicity is calculated using the formula $2S + 1$,where $S$ is the total spin of the electrons in the valence shell.
For $2p^3$,there are $3$ unpaired electrons,so $S = 3 \times (1/2) = 3/2$. Multiplicity $= 2(3/2) + 1 = 4$.
For $2p^4$,there are $2$ unpaired electrons,so $S = 2 \times (1/2) = 1$. Multiplicity $= 2(1) + 1 = 3$.
For $2p^5$,there is $1$ unpaired electron,so $S = 1/2$. Multiplicity $= 2(1/2) + 1 = 2$.
For $2p^6$,there are $0$ unpaired electrons,so $S = 0$. Multiplicity $= 2(0) + 1 = 1$.
Thus,the configuration $1s^2, 2s^2, 2p^3$ has the maximum spin multiplicity.

(A) The ground state of an element follows the Aufbau principle,Pauli's exclusion principle,and Hund's rule of maximum multiplicity.
According to Hund's rule,electrons fill degenerate orbitals (like $p$-orbitals) singly first with parallel spins before pairing begins.
Option $A$ shows the first two orbitals (likely $1s$ and $2s$) fully paired,and the next three orbitals (likely $2p$) each containing one electron with parallel spins (all upward),which correctly represents the ground state configuration (e.g.,for Nitrogen,$1s^2 2s^2 2p^3$).

(B) The correct matching is as follows:
$(a)$ $4s$: $s$-orbitals are spherically symmetric and non-directional,so $(a) - (q)$.
$(b)$ $4p$: The angular momentum of an orbital is given by $\sqrt{l(l+1)} \frac{h}{2\pi}$. For $p$-orbital $(l=1)$,angular momentum $= \sqrt{1(1+1)} \frac{h}{2\pi} = \sqrt{2} \frac{h}{2\pi}$. However,in the context of this specific matching problem,$d$-orbitals $(l=2)$ have angular momentum $\sqrt{2(2+1)} \frac{h}{2\pi} = \sqrt{6} \frac{h}{2\pi}$. Re-evaluating the options based on standard textbook problems: $1s$ is the ground state $(n=1, l=0)$,which has zero radial nodes $(n-l-1 = 1-0-1 = 0)$. Thus,$(c) - (s)$.
$(d)$ $3d$: $l=2$. Angular momentum $= \sqrt{l(l+1)} \frac{h}{2\pi} = \sqrt{6} \frac{h}{2\pi}$.
Given the options provided,the most accurate match is $(a) - (q)$,$(b) - (r)$,$(c) - (s)$,$(d) - (p)$ is not listed,but checking $(b) - (r)$ where $l=1$ is often simplified in some contexts,or $(d) - (r)$ where $l=2$.
Correct mapping: $(a) - (q)$ [$s$-orbital is non-directional],$(b) - (r)$ [Angular momentum related],$(c) - (s)$ [Radial node $n-l-1 = 1-0-1 = 0$],$(d) - (p)$ [Circular orbit concept].
Therefore,the correct option is $B$.

(A) For the wave function $\Psi_{n\ell m}$,the quantum numbers are $n = 3$,$\ell = 1$,and $m = 0$.
Number of radial nodes $= n - \ell - 1 = 3 - 1 - 1 = 1$.
Number of angular nodes $= \ell = 1$.
Therefore,$\Psi_{310}$ has $1$ radial node and $1$ angular node.

(C) The number of orientations of a subshell is determined by the magnetic quantum number,denoted by $m_l$.
For a given azimuthal quantum number $l$,the possible values of $m_l$ range from $-l$ to $+l$,including zero.
The total number of such values is given by the formula $(2l + 1)$.
Each value of $m_l$ corresponds to a specific orientation of an orbital in space.

(C) According to the $(n+l)$ rule,the energy of an orbital is determined by the sum of the principal quantum number $(n)$ and the azimuthal quantum number $(l)$.
For $4f$: $n=4, l=3$,so $n+l = 4+3 = 7$.
For $5d$: $n=5, l=2$,so $n+l = 5+2 = 7$.
For $3s$: $n=3, l=0$,so $n+l = 3+0 = 3$.
For $6p$: $n=6, l=1$,so $n+l = 6+1 = 7$.
Since $3s$ has a different $(n+l)$ value compared to the others,it is the odd one.

(D) For a set of quantum numbers to be allowed,they must satisfy the following rules:
$1$. $n$ (principal quantum number) must be a positive integer $(1, 2, 3, \dots)$.
$2$. $l$ (azimuthal quantum number) can have values from $0$ to $n-1$.
$3$. $m$ (magnetic quantum number) can have values from $-l$ to $+l$ including $0$.
$4$. $s$ (spin quantum number) can only be $+1/2$ or $-1/2$.
Evaluating the options:
- Option $A$: $n=1, l=1$ is not allowed because $l$ must be less than $n$ $(l < n)$.
- Option $B$: $s=1$ is not allowed because $s$ must be $\pm 1/2$.
- Option $C$: $n=3, l=0, m=-1$ is not allowed because if $l=0$,then $m$ must be $0$.
- Option $D$: $n=3, l=1, m=1, s=1/2$ satisfies all conditions ($l < n$,$-l \le m \le l$,and $s = \pm 1/2$).
Therefore,the correct set is $D$.

(C) The atomic number of $Ca$ is $20$.
According to the $Aufbau$ principle,the electronic configuration is $1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2$.
If the $Aufbau$ rule were not followed and electrons filled orbitals in the order of increasing principal quantum number $(n)$,the configuration would be $1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^2$.
Since the last electron enters the $3d$ orbital,$Ca$ would belong to the $d-$ block.

(C) The electronic configuration of ${}_{24}Cr$ is $[Ar] 3d^5 4s^1$.
This configuration is more stable than $[Ar] 3d^4 4s^2$ due to two main reasons:
$1$. Higher exchange energy: The $3d^5$ configuration allows for a greater number of possible exchanges between electrons with parallel spins,which releases more energy and increases stability.
$2$. Symmetrical distribution of electron: The half-filled $d$-subshell $(d^5)$ provides a symmetrical distribution of electron density,which leads to lower electron-electron repulsion and higher stability.

(C) The $(n + l)$ rule (also known as the Aufbau principle) states that electrons fill orbitals in order of increasing $(n + l)$ values.
For $La$ $(Z=57)$,the configuration is $[Xe] 5d^1 6s^2$. According to the $(n + l)$ rule,the $4f$ orbital $(n=4, l=3, n+l=7)$ should be filled before the $5d$ orbital $(n=5, l=2, n+l=7)$. However,the electron enters the $5d$ orbital instead of the $4f$ orbital.
Similarly,for $Ac$ $(Z=89)$,the configuration is $[Rn] 6d^1 7s^2$. The electron enters the $6d$ orbital instead of the $5f$ orbital.
Thus,both $La$ and $Ac$ are exceptions to the $(n + l)$ rule.

(C) The $4$ quantum numbers are used to describe an electron in an atom:
$1$. Principal quantum number $(n)$ determines the size and energy of the orbital.
$2$. Azimuthal quantum number $(l)$ determines the shape of the orbital.
$3$. Magnetic quantum number $(m_l)$ determines the orientation of the orbital in space.
$4$. Spin quantum number $(m_s)$ describes the spin of the electron.
Therefore,the orientation of an atomic orbital is given by the magnetic quantum number.

(A) The number of orbitals in a subshell is given by $(2l + 1)$.
Since each orbital can hold a maximum of $2$ electrons,
The maximum number of electrons in a subshell $= 2 \times (2l + 1) = 4l + 2$.

(C) The wave function is given as $\Psi_{n,l,m_l}$.
Comparing $\Psi_{310}$ with $\Psi_{n,l,m_l}$,we get $n = 3$,$l = 1$,and $m_l = 0$.
For $n = 3$ and $l = 1$,the orbital is $3p$.
The magnetic quantum number $m_l = 0$ for a $p$-orbital corresponds to the $p_z$ orbital.
Therefore,$\Psi_{310}$ represents the $3p_z$ orbital.

(A) For a $g$ subshell,the azimuthal quantum number $l = 4$.
The number of orbitals in a subshell is given by the formula $(2l + 1)$.
Substituting $l = 4$ into the formula:
Number of orbitals $= 2(4) + 1 = 8 + 1 = 9$.

(B) For the $3d^5$ configuration,the number of unpaired electrons $(n)$ is $5$.
The formula for spin multiplicity is $2S + 1$,where $S = \sum s$.
Since each of the $5$ electrons has a spin of $+1/2$,the total spin $S = 5 \times (1/2) = 5/2$.
Spin multiplicity $= 2(5/2) + 1 = 5 + 1 = 6$.

(C) 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 $-1/2$.
In option $C$,$n = 4$ and $l = 3$. The possible values for $m$ range from $-3$ to $+3$. Since the given value is $m = 4$,which is greater than $l$,this set is invalid.

(D) The rules for quantum numbers are as follows:
$1$. $n$ is the principal quantum number $(n = 1, 2, 3, ...)$.
$2$. $l$ is the azimuthal quantum number ($l = 0$ to $n-1$).
$3$. $m$ is the magnetic quantum number ($m = -l$ to $+l$ including $0$).
$4$. $s$ is the spin quantum number ($s = +1/2$ or $-1/2$).
Evaluating the options:
- Option $A$: $n=4, l=0, m=0, s=-1/2$. This is valid ($l < n$ and $m$ is within range).
- Option $B$: $n=3, l=2, m=-2, s=+1/2$. This is valid ($l < n$ and $m$ is within range).
- Option $C$: $n=5, l=4, m=0, s=-1/2$. This is valid ($l < n$ and $m$ is within range).
- Option $D$: $n=4, l=3, m=-4, s=-1/2$. Here,$m$ must be in the range $-3$ to $+3$. Since $m = -4$ is outside this range,this set is incorrect.

(C) To define an orbital,we need the principal quantum number $(n)$,the azimuthal quantum number $(l)$,and the magnetic quantum number $(m)$.
$n$ determines the size and energy of the orbital.
$l$ determines the shape of the orbital.
$m$ determines the orientation of the orbital in space.
The spin quantum number $(s)$ describes the spin of the electron within the orbital,not the orbital itself.

(D) The state of an electron in an atom is fully described by four quantum numbers: principal quantum number $(n)$,azimuthal quantum number $(l)$,magnetic quantum number $(m_l)$,and spin quantum number ($s$ or $m_s$).
For a $3d_{z^2}$ orbital,the values are $n=3$ and $l=2$.
The magnetic quantum number $(m_l)$ for $d$-orbitals ranges from $-2$ to $+2$.
However,when specifying the state of an electron in a specific orbital,we define the orbital using $n, l,$ and $m_l$. The spin quantum number $(s)$ is also required to completely define the state of the electron.
Since the question asks for the quantum numbers that describe the state of an electron in a specific orbital,$n, l,$ and $m$ are the spatial quantum numbers defining the orbital itself. Given the options,$n, l,$ and $m$ are the primary set defining the orbital state.

(C) For a given principal quantum number $n$,the possible values of the azimuthal quantum number $l$ range from $0$ to $n-1$.
For $n = 5$,$l$ can be $0, 1, 2, 3, 4$.
The magnetic quantum number $m_l$ depends on $l$ and ranges from $-l$ to $+l$.
We need to find the orbitals where $m_l = +2$ is possible:
- If $l = 0$ ($s$-orbital): $m_l = 0$ (not possible)
- If $l = 1$ ($p$-orbital): $m_l = -1, 0, +1$ (not possible)
- If $l = 2$ ($d$-orbital): $m_l = -2, -1, 0, +1, +2$ (possible)
- If $l = 3$ ($f$-orbital): $m_l = -3, -2, -1, 0, +1, +2, +3$ (possible)
- If $l = 4$ ($g$-orbital): $m_l = -4, -3, -2, -1, 0, +1, +2, +3, +4$ (possible)
Thus,for $n = 5$,the orbitals with $m_l = +2$ are $5d, 5f,$ and $5g$.
Each orbital can hold a maximum of $2$ electrons.
Total electrons = $3 \text{ orbitals} \times 2 \text{ electrons/orbital} = 6 \text{ electrons}$.

(C) The value of $(n + l)$ for an orbital determines its energy according to the Aufbau principle.
For $6s$: $n = 6, l = 0$,so $(n + l) = 6 + 0 = 6$.
For $3d$: $n = 3, l = 2$,so $(n + l) = 3 + 2 = 5$.
For $4f$: $n = 4, l = 3$,so $(n + l) = 4 + 3 = 7$.
For $5p$: $n = 5, l = 1$,so $(n + l) = 5 + 1 = 6$.
Comparing the values,$4f$ has the maximum value of $(n + l) = 7$.

(C) The angular momentum of an electron is given by the formula: $L = \sqrt{l(l+1)} \frac{h}{2\pi}$,where $l$ is the azimuthal quantum number.
For an $s$ orbital,the value of $l$ is $0$.
Therefore,the angular momentum of an electron in a $2s$ orbital is $\sqrt{0(0+1)} \frac{h}{2\pi} = 0$.
Among the given options,only $s$ orbitals have $l = 0$.
Thus,the angular momentum of an electron in a $4s$ orbital is also $\sqrt{0(0+1)} \frac{h}{2\pi} = 0$.
Hence,the angular momentum of a $2s$ electron is equal to that of a $4s$ electron.

(B) The angular momentum $(L)$ of an electron in an orbital is calculated using the formula: $L = \sqrt{l(l+1)} \frac{h}{2\pi}$.
For a $d$ orbital,the azimuthal quantum number $(l)$ is $2$.
Substituting $l = 2$ into the formula:
$L = \sqrt{2(2+1)} \frac{h}{2\pi} = \sqrt{2 \times 3} \frac{h}{2\pi} = \sqrt{6} \frac{h}{2\pi}$.
Since $\sqrt{6} = \sqrt{2 \times 3} = \sqrt{2} \times \sqrt{3}$,we can write this as $\sqrt{2} \times \sqrt{3} \frac{h}{2\pi} = \frac{\sqrt{3}h}{\sqrt{2}\pi}$.
However,looking at the standard representation $\sqrt{6} \frac{h}{2\pi}$,if we simplify $\frac{\sqrt{6}}{2} = \frac{\sqrt{6}}{\sqrt{4}} = \sqrt{1.5}$,the value is $\sqrt{1.5} \frac{h}{\pi}$.
Given the options provided in the prompt,there seems to be a discrepancy in the standard notation. Re-evaluating the provided options,option $B$ is the closest mathematical equivalent to $\sqrt{6} \frac{h}{2\pi}$ if written as $\frac{\sqrt{3}h}{\sqrt{2}\pi}$.

(B) The wave function $\psi(r, \theta, \phi)$ for a hydrogen-like atom can be separated into a radial part $R(r)$ and an angular part $Y(\theta, \phi)$.
The radial part $R(r)$ depends on the principal quantum number $(n)$ and the azimuthal quantum number $(l)$.
Therefore,the correct option is $B$.

(B) The spin magnetic moment $(\mu_s)$ of an electron is given by the formula $\mu_s = \sqrt{4s(s + 1)} \text{ BM}$,where $s$ is the spin quantum number.
For a single electron,$s = 1/2$.
Substituting this value,$\mu_s = \sqrt{4(1/2)(1/2 + 1)} = \sqrt{2(3/2)} = \sqrt{3} \approx 1.732 \text{ BM}$.
This is also expressed as $\mu = \sqrt{n(n + 2)} \text{ BM}$,where $n$ is the number of unpaired electrons.

(D) In an isolated atom,all five $d$ orbitals ($d_{xy}$,$d_{yz}$,$d_{zx}$,$d_{x^2-y^2}$,and $d_{z^2}$) are degenerate.
However,their shapes are not identical.
The $d_{xy}$,$d_{yz}$,and $d_{zx}$ orbitals have a cloverleaf shape (four lobes between the axes).
The $d_{x^2-y^2}$ orbital also has a cloverleaf shape but with lobes along the axes.
The $d_{z^2}$ orbital has a unique shape with two lobes along the $z$-axis and a ring of electron density in the $xy$-plane.
Therefore,the shapes of $d_{x^2-y^2}$ and $d_{z^2}$ are different.