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

(B) Statement $A$ is based on the Pauli Exclusion Principle,which states that no two electrons in an atom can have the same set of four quantum numbers. This implies that an orbital can hold a maximum of two electrons with opposite spins.
Reason $R$ is also a true statement because electrons have spin angular momentum,which generates a magnetic moment. Electrons with opposite spins produce magnetic fields that are oriented in opposite directions.
However,the reason $R$ is not the direct explanation for why an orbital can only hold two electrons; the Pauli Exclusion Principle is the fundamental reason. Therefore,both statements are true,but $R$ is not the correct explanation of $A$.

(C) - The $Aufbau$ principle states that electrons fill orbitals in the order of increasing energy.

(A) The electronic configuration of the given elements in their ground state is as follows:
$1.$ For atomic number $8$ $(O)$: $1s^2 2s^2 2p^4$. Number of unpaired $3p$ electrons = $0$.
$2.$ For atomic number $10$ $(Ne)$: $1s^2 2s^2 2p^6$. Number of unpaired $3p$ electrons = $0$.
$3.$ For atomic number $12$ $(Mg)$: $1s^2 2s^2 2p^6 3s^2$. Number of unpaired $3p$ electrons = $0$.
$4.$ For atomic number $15$ $(P)$: $1s^2 2s^2 2p^6 3s^2 3p^3$. The $3p$ subshell has $3$ orbitals,each containing $1$ unpaired electron according to Hund's rule.
Thus,the element with atomic number $15$ has the maximum number of unpaired $3p$ electrons ($3$ unpaired electrons).

(C) For any given orbital,the maximum number of electrons it can hold is $2$,according to the Pauli Exclusion Principle.
Since the question specifies a single orbital (defined by $n=6$,$l=1$ for $P$,and $m=0$),it can accommodate a maximum of $2$ electrons.

(B) The ground state of an atom is the state with the lowest possible energy configuration,following the Aufbau principle,Pauli exclusion principle,and Hund's rule.
Comparing the given configurations:
$(A)$ $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1$ corresponds to Scandium ($Sc$,$Z=21$).
$(B)$ $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^0$ corresponds to Calcium ($Ca$,$Z=20$).
$(C)$ $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^1$ is an excited state of Scandium.
$(D)$ $1s^2 2s^2 2p^6 3s^2 3p^6 4s^0 3d^2$ is an excited state of Scandium.
Since all options represent valid electronic configurations,the question typically refers to the standard filling order. Option $B$ represents the stable ground state of Calcium.

(B) The atomic number of Nitrogen $(N)$ is $7$. The electronic configuration of $N$ is $1s^2 2s^2 2p^3$.
For the $N^{+2}$ ion,two electrons are removed from the $2p$ orbital.
The electronic configuration of $N^{+2}$ becomes $1s^2 2s^2 2p^1$.
In the $2p^1$ orbital,there is $1$ unpaired electron.
Therefore,the number of unpaired electrons is $1$.

(C) The electronic configuration of an element with the outermost orbital $4s^2$ implies that the $3d$ subshell is completely filled.
Following the Aufbau principle,the complete electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2$.
Summing the electrons: $2 + 2 + 6 + 2 + 6 + 10 + 2 = 30$.
Therefore,the atomic number is $30$,which corresponds to Zinc $(Zn)$.

(C) The total number of orbitals in a shell with principal quantum number $n$ is given by the formula $n^2$.
For $n = 4$,the total number of orbitals is $4^2 = 16$.

(A) The atomic number of fluorine $(F)$ is $9$.
The electronic configuration of fluorine is $1s^2 2s^2 2p^5$.
In the $2p$ subshell,there are $5$ electrons.
According to Hund's rule,the electrons are filled as: $2p_x^2, 2p_y^2, 2p_z^1$.
The paired electrons have spins of $+1/2$ and $-1/2$,which cancel each other out.
There is only one unpaired electron in the $2p_z$ orbital.
The total spin $(S)$ is equal to $n \times (1/2)$,where $n$ is the number of unpaired electrons.
Here,$n = 1$,so $S = 1 \times (1/2) = 0.5$.

(C) The atomic number of chlorine $(Cl)$ is $17$.
The electronic configuration of $Cl$ is $1s^2 2s^2 2p^6 3s^2 3p^5$.
The unpaired electron is present in the $3p$ orbital.
For the $3p$ orbital,the principal quantum number $n = 3$ and the azimuthal quantum number $l = 1$.
The magnetic quantum number $m$ can be $-1, 0, +1$,and the spin quantum number $s$ can be $\pm 1/2$.
Comparing this with the given options,option $C$ $(n = 3, l = 1, m = 1, s = 1/2)$ represents a valid set of quantum numbers for an electron in the $3p$ subshell.

(B) The electronic configuration of $_{15}P$ is $[Ne] 3s^2 3p^3$.
In the $3s$ orbital,there are two electrons ($A$ and $B$) with opposite spins.
For electron $A$,$n=3, l=0, m=0, s=-1/2$.
For electron $B$,$n=3, l=0, m=0, s=+1/2$.
Since $A$ and $B$ share the same principal $(n)$,azimuthal $(l)$,and magnetic $(m)$ quantum numbers,they have three identical quantum numbers.

(C) Hund's rule of maximum multiplicity states that for a given electron configuration,the term with the maximum multiplicity has the lowest energy.
In the case of nitrogen $(Z = 7)$,the electronic configuration is $1s^2 2s^2 2p^3$.
According to Hund's rule,electrons fill degenerate orbitals (like $2p_x, 2p_y, 2p_z$) singly first before pairing occurs.
Therefore,$1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1$ is the correct ground state configuration,while $1s^2 2s^2 2p_x^2 2p_y^1 2p_z^0$ represents an excited state.

(D) In a hydrogen atom,the energy of an orbital depends only on the principal quantum number $(n)$.
Since all the given orbitals ($3s$,$3p$,and $3d$) have the same principal quantum number $(n = 3)$,they are degenerate,meaning they possess the same energy.
Therefore,the correct statement is that all $3s$,$3p$,and $3d$ orbitals have the same energy.

(A) The energy of an orbital in a multi-electron atom depends on the effective nuclear charge $(Z_{eff})$.
As the atomic number $(Z)$ increases,the attraction between the nucleus and the electrons increases,which lowers the energy of the orbital.
For the $2s$-orbital,the energy decreases as the nuclear charge increases.
The atomic numbers are $H (Z=1)$,$Li (Z=3)$,$Na (Z=11)$,and $K (Z=19)$.
Since $Z_{eff}$ increases from $H$ to $K$,the energy of the $2s$-orbital decreases in the order: $E_{2s(H)} > E_{2s(Li)} > E_{2s(Na)} > E_{2s(K)}$.

(A) Statement $1$ is true because,in a hydrogen-like atom,the energy of an orbital depends only on the principal quantum number $(n)$,given by the formula $E_n = -R_H \times (Z^2 / n^2)$.
Statement $2$ is also true because the principal quantum number $(n)$ defines the shell and determines the average distance of the electron from the nucleus.
Since the energy level of the electron is directly related to its distance from the nucleus (as defined by the shell $n$),Statement $2$ provides the physical basis for Statement $1$.

(C) For any electron,the spin quantum number $(s)$ can only have two possible values: $+1/2$ or $-1/2$.
It cannot be $0$.
Therefore,the set $n = 3, l = 2, m = 1, s = 0$ is not possible.

(B) The atomic number of $Na$ is $11$.
The electronic configuration of $Na$ is $1s^2 2s^2 2p^6 3s^1$.
The valence electron is the one in the $3s$ orbital.
For the $3s$ orbital:
Principal quantum number $(n)$ = $3$.
Azimuthal quantum number $(l)$ for $s$-orbital = $0$.
Magnetic quantum number $(m)$ = $0$.
Spin quantum number $(s)$ = $+1/2$ or $-1/2$.
Thus,the correct set is $(n = 3, l = 0, m = 0, s = 1/2)$.

(C) The $1s$ orbital can accommodate a maximum of $2$ electrons according to the Pauli exclusion principle,which states that no two electrons in an atom can have the same set of all four quantum numbers.
Since an $s$-orbital has only one subshell $(m_l = 0)$ and can hold electrons with opposite spins ($m_s = +1/2$ and $-1/2$),it cannot hold $7$ electrons.
Therefore,the configuration $1s^7$ violates the Pauli exclusion principle.

(A) The energy of an electron in a hydrogen-like atom is given by the formula $E_n = -\frac{13.6 \ Z^2}{n^2} \ eV$,which depends on the principal quantum number $n$.
Thus,statement $A$ is correct.
The principal quantum number $n$ defines the shell and determines the average distance of the electron from the nucleus.
Thus,statement $R$ is correct and it explains why the energy depends on $n$ (as energy is a function of distance from the nucleus).
Therefore,the correct option is $A$.

(B) The $l = 2$ orbital corresponds to the $d$-subshell.
$1$. For $_{26}Fe^{2+}$: Electronic configuration is $[Ar] 3d^6$. In $3d^6$,there is $1$ paired electron set (containing $2$ electrons) and $4$ unpaired electrons. Number of paired electrons = $2$.
$2$. For $_{27}Co^{2+}$: Electronic configuration is $[Ar] 3d^7$. In $3d^7$,there are $2$ paired electron sets (containing $4$ electrons) and $3$ unpaired electrons. Number of paired electrons = $4$.
$3$. For $_{28}Ni^{2+}$: Electronic configuration is $[Ar] 3d^8$. In $3d^8$,there are $3$ paired electron sets (containing $6$ electrons) and $2$ unpaired electrons. Number of paired electrons = $6$.
Total number of paired electrons = $2 + 4 + 6 = 12$.

(C) The given electronic configuration shows orbitals with paired electrons while an adjacent orbital is empty.
According to $Hund's$ rule of maximum multiplicity,electrons fill degenerate orbitals singly first before pairing occurs.
Since the electrons are paired in the first two orbitals while the third orbital in the set is empty,this violates $Hund's$ rule.

(B) For a $d$-orbital,the azimuthal quantum number $l = 2$.
The magnetic quantum number $m_l$ ranges from $-l$ to $+l$ including zero.
Therefore,for $l = 2$,the values of $m_l$ are $-2, -1, 0, +1, +2$.

(A) According to the $(n+l)$ rule,the energy of an orbital increases as the value of $(n+l)$ increases. If two orbitals have the same $(n+l)$ value,the one with the lower $n$ value has lower energy.
Calculating $(n+l)$ for each:
$(1) n = 4, l = 1 \implies n+l = 5 \text{ (4p orbital)}$
$(2) n = 4, l = 0 \implies n+l = 4 \text{ (4s orbital)}$
$(3) n = 3, l = 1 \implies n+l = 4 \text{ (3p orbital)}$
$(4) n = 3, l = 2 \implies n+l = 5 \text{ (3d orbital)}$
Comparing the values:
For $(3)$ and $(2)$,both have $(n+l) = 4$. Since $(3)$ has $n=3$ and $(2)$ has $n=4$,$(3) < (2)$.
For $(4)$ and $(1)$,both have $(n+l) = 5$. Since $(4)$ has $n=3$ and $(1)$ has $n=4$,$(4) < (1)$.
Combining these,the increasing order of energy is $(3) < (2) < (4) < (1)$.

(D) For a given principal quantum number $n$,the total number of orbitals is given by $n^2$.
For $n = 3$,the total number of orbitals is $3^2 = 9$.
Each orbital can hold a maximum of two electrons,one with $m_s = +1/2$ and one with $m_s = -1/2$.
Therefore,the number of electrons with $m_s = -1/2$ is equal to the total number of orbitals,which is $9$.

(A) The magnetic quantum number $m$ ranges from $-l$ to $+l$. Given $m = 0, \pm 1$,the number of values is $3$,which corresponds to $l = 1$ (the $p$-orbital). The $p$-orbital starts from the second principal energy level,i.e.,$n = 2$. However,in the context of standard multiple-choice questions regarding the $p$-subshell,$n=2$ is the minimum value.

(B) The correct option is $B$.
The electronic configuration of $Cr$ $(Z=24)$ is: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$.
The $M$ shell corresponds to the principal quantum number $n=3$.
Summing the electrons in the $3s$,$3p$,and $3d$ orbitals: $2 (3s) + 6 (3p) + 5 (3d) = 13$ electrons.
Thus,$Cr$ satisfies the given condition.

(A) The atomic number of nitrogen $(N)$ is $7$. Its electronic configuration is $1s^2 2s^2 2p^3$.
According to Hund's rule of maximum multiplicity,for a given electron configuration,the term with the maximum multiplicity has the lowest energy. This means that electrons in the same subshell should occupy separate orbitals with parallel spins as far as possible.
In the $2p$ subshell,there are $3$ electrons. Therefore,they should occupy the three $2p$ orbitals singly with parallel spins.
Option $A$ (image $203-$a248) correctly shows $1s^2$ (paired),$2s^2$ (paired),and $2p^3$ (three unpaired electrons with parallel spins).

(A) For the $4s$ orbital,$(n + l) = 4 + 0 = 4$.
For the $3d$ orbital,$(n + l) = 3 + 2 = 5$.
According to the $(n + l)$ rule,the orbital with the lower $(n + l)$ value has lower energy and is filled first.
Since $4 < 5$,the $19^{th}$ electron enters the $4s$ orbital.
Therefore,both Statement $A$ and Reason $R$ are true,and $R$ is the correct explanation of $A$.

(D) The set of quantum numbers $n = 3, l = 2, m = +2, s = +1/2$ uniquely identifies a single specific orbital and a specific spin state of an electron within that orbital.
$n = 3$ represents the third shell.
$l = 2$ represents the $d$-subshell ($3d$ orbital).
$m = +2$ specifies one of the five $d$-orbitals.
$s = +1/2$ specifies the spin of the electron.
According to the Pauli Exclusion Principle,each unique set of four quantum numbers corresponds to exactly one electron.
Therefore,only $1$ electron can have this specific set of quantum numbers.

(D) According to the $Pauli$ exclusion principle,an orbital can accommodate a maximum of two electrons,and these two electrons must have opposite spins. Therefore,the correct option is $D$.

(B) The atomic number of Phosphorus $(P)$ is $15$.
The electronic configuration of $P$ is $1s^2 2s^2 2p^6 3s^2 3p^3$.
For $l = 1$,the orbitals are $2p$ and $3p$.
In $2p^6$,all three orbitals $(m = -1, 0, +1)$ are fully filled,so there is one pair in $m = 0$.
In $3p^3$,according to Hund's rule,each orbital has one electron,so there is no pair in $m = 0$.
Thus,the total number of electron pairs with $l = 1$ and $m = 0$ is $1$ (from the $2p$ subshell).

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

(A) Pauli's exclusion principle states that no two electrons in an atom can have the same set of four quantum numbers,meaning an orbital can hold a maximum of two electrons with opposite spins. Hund's rule states that electron pairing in degenerate orbitals (like $p$-orbitals) does not occur until each orbital is singly occupied.
In option $(A)$,the $2p_x$ orbital contains two electrons with parallel spins,which violates the Pauli exclusion principle. Furthermore,since the $2p_y$ orbital is empty while $2p_x$ is paired,it also violates Hund's rule regarding the order of filling.
Therefore,option $(A)$ violates both principles.

(D) For a given azimuthal quantum number $l$,the number of orbitals is given by $(2l + 1)$.
For $l = 3$,the number of orbitals is $(2 \times 3 + 1) = 7$.
Each orbital can accommodate a maximum of $2$ electrons.
Therefore,the maximum number of electrons = $7 \times 2 = 14$.

(C) For a hydrogen atom (a single-electron system),the energy of an orbital depends only on the principal quantum number $n$.
Since all the subshells $(3s, 3p, 3d)$ have the same principal quantum number $n = 3$,they are degenerate,meaning they have the same energy.
Therefore,the energy required to remove an electron from these subshells to $n = \infty$ is the same.
Thus,$E_1 = E_2 = E_3$.

(B) The $Pauli$ exclusion principle states that no two electrons in an atom can have the same set of all four quantum numbers. As a consequence,an orbital can accommodate a maximum of two electrons,and these must have opposite spins.

(D) For a given principal quantum number $n$,the azimuthal quantum number $l$ can take values from $0$ to $n-1$. For $n = 5$,$l$ can be $0, 1, 2, 3, 4$.
The magnetic quantum number $m$ can take values from $-l$ to $+l$.
If $m = 2$,then $l$ must be at least $2$ (i.e.,$l = 2, 3, 4$).
Option $D$ is incorrect because it includes $l = 0$ and $l = 1$,for which $m = 2$ is not possible.

(D) The atomic number of $Ne$ is $10$. Its electronic configuration is $1s^2 2s^2 2p^6$.
$F^{-}$ has $9 + 1 = 10$ electrons. Configuration: $1s^2 2s^2 2p^6$.
$Na^{+}$ has $11 - 1 = 10$ electrons. Configuration: $1s^2 2s^2 2p^6$.
$Mg^{2+}$ has $12 - 2 = 10$ electrons. Configuration: $1s^2 2s^2 2p^6$.
$Cl^{-}$ has $17 + 1 = 18$ electrons. Configuration: $1s^2 2s^2 2p^6 3s^2 3p^6$.
Therefore,$Cl^{-}$ does not have the same electronic configuration as $Ne$.

(B) The Aufbau principle states that electrons fill orbitals in order of increasing energy.
Option $A$ represents a valid configuration for $N$ or $O$ (depending on total electrons).
Option $B$ violates the Aufbau principle because the $2s$ orbital is only half-filled while the $2p$ orbitals are already being filled,which is energetically unfavorable compared to filling the $2s$ orbital completely first.
Option $C$ represents the ground state of Nitrogen $(1s^2 2s^2 2p^3)$.
Option $D$ represents the ground state of Fluorine $(1s^2 2s^2 2p^5)$.
Therefore,option $B$ is the correct answer.

(A) In a multi-electron atom,the energy of an orbital depends on both the principal quantum number $(n)$ and the azimuthal quantum number $(l)$.
Orbitals with the same $n$ and $l$ values are degenerate (have the same energy) in the absence of an external field.
Analyzing the given orbitals:
$(a) n=1, l=0$ $(1s)$
$(b) n=3, l=0$ $(3s)$
$(c) n=2, l=1$ $(2p)$
$(d) n=3, l=2$ $(3d)$
$(e) n=3, l=2$ $(3d)$
Comparing the sets,orbitals $(d)$ and $(e)$ both have $n=3$ and $l=2$,meaning they are both $3d$ orbitals.
Therefore,they have the same energy.

(A) The spin angular momentum $(L_s)$ of an electron is calculated using the formula:
$L_s = \sqrt{S(S + 1)} \frac{h}{2\pi}$
where $S$ is the spin quantum number and $h$ is Planck's constant.
Thus,the correct option is $A$.

(D) The electronic configuration of the atom based on the given shell distribution $(K=2, L=8, M=7)$ is as follows:
$K$ shell $(n=1)$: $1s^2$ (contains $2$ electrons)
$L$ shell $(n=2)$: $2s^2 2p^6$ (contains $8$ electrons)
$M$ shell $(n=3)$: $3s^2 3p^5$ (contains $7$ electrons)
Total electrons in $p$-orbitals = (electrons in $2p$) + (electrons in $3p$)
Total = $6 + 5 = 11$ electrons.
Therefore,the correct option is $D$.

(C) The orbital angular momentum is given by the formula $\sqrt{l(l + 1)} \cdot \frac{h}{2\pi}$.
For an $s$-orbital,the azimuthal quantum number $l = 0$.
Substituting $l = 0$ into the formula: $\sqrt{0(0 + 1)} \cdot \frac{h}{2\pi} = \sqrt{0} \cdot \frac{h}{2\pi} = 0$.
Therefore,the orbital angular momentum for an $s$-electron is $0$.

(C) The principal quantum number $n = 5$ indicates the $5^{th}$ shell.
The azimuthal quantum number $l = 2$ corresponds to the $d$-subshell.
The magnetic quantum number $m = 0$ for a $d$-subshell specifically corresponds to the $d_{z^2}$ orbital.
Therefore,the orbital is $5d_{z^2}$.

(D) The atomic number of Chromium $(Cr)$ is $24$.
According to the Aufbau principle,the expected configuration is $[Ar] 3d^4 4s^2$.
However,due to the extra stability of half-filled $d$-orbitals,one electron from the $4s$ orbital shifts to the $3d$ orbital.
Thus,the actual electronic configuration is $[Ar] 3d^5 4s^1$.
In terms of shells,this corresponds to $2, 8, 13, 1$ $(K=2, L=8, M=13, N=1)$.

(C) The maximum number of electrons in a shell is given by the formula $2n^2$.
For the shell with principal quantum number $n = 4$:
Maximum electrons $= 2 \times (4)^2 = 2 \times 16 = 32$.

(B) The atomic number of chlorine $(Cl)$ is $17$.
The electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^5$.
The unpaired electron is in the $3p$ orbital.
For $3p$ orbital:
Principal quantum number $(n)$ = $3$.
Azimuthal quantum number $(l)$ = $1$.
Magnetic quantum number $(m)$ can be $-1, 0, +1$.
Spin quantum number $(s)$ = $+1/2$ or $-1/2$.
Among the given options,$n = 3, l = 1, m = 0, s = +1/2$ is a valid set of quantum numbers for one of the $p$-orbitals.

(C) The electronic configuration of $Fe$ $(Z = 26)$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^6$.
The $20^{th}$ electron is the second electron entering the $4s$ orbital.
For the $4s$ orbital,the principal quantum number $n = 4$,azimuthal quantum number $l = 0$,magnetic quantum number $m_l = 0$,and spin quantum number $m_s = +\frac{1}{2}$ (assuming the first electron is $+\frac{1}{2}$).

(A) The $p$-orbital with $n = 6$ is the $6p$ subshell.
For a given $p$-orbital,the magnetic quantum number $m$ can take values $-1, 0, +1$.
The orbital corresponding to $m = 0$ is the $6p_z$ orbital.
According to the Pauli Exclusion Principle,any single orbital can accommodate a maximum of $2$ electrons.