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

(C) The spin quantum number $(m_s)$ can only take values of $+1/2$ or $-1/2$.
In option $(C)$,the value of $m_s$ is given as $0$,which is not a valid value for the spin quantum number.
Therefore,the set $(n=3, l=2, m_l=1, m_s=0)$ is not possible.

(B) The ground state electronic configuration of neon $(Z = 10)$ is $1s^2 2s^2 2p^6$.
When an electron from the $2p$ orbital absorbs energy,it jumps to the $3s$ orbital.
Therefore,the excited state configuration of neon is $1s^2 2s^2 2p^5 3s^1$.

(C) The magnetic quantum number $(m)$ describes the spatial orientation of a particular orbital.
It is called the magnetic quantum number because the effect of different orientations of orbitals was first observed in the presence of a magnetic field.
The magnetic quantum number $(m)$ determines the number of orbitals and their orientation within a subshell.
Consequently,its value depends on the orbital angular momentum quantum number $l$.
For a given $l$,$m$ ranges from $-l$ to $+l$,including zero.

(C) The electronic configuration of $K$ $(Z=19)$ is $[Ar]4s^1$.
For $Cr$ $(Z=24)$,the stable configuration is $[Ar]3d^54s^1$ due to the extra stability of half-filled $d$-orbitals.
For $Cu$ $(Z=29)$,the stable configuration is $[Ar]3d^{10}4s^1$ due to the extra stability of fully-filled $d$-orbitals.
Therefore,the correct configuration is $K = [Ar]4s^1, Cr = [Ar]3d^54s^1, Cu = [Ar]3d^{10}4s^1$.

(A) The number of orbitals in a sub-shell is given by the formula $(2l + 1)$.
For different sub-shells,the azimuthal quantum number $l$ values are:
$s: l = 0$
$p: l = 1$
$d: l = 2$
$f: l = 3$
$g: l = 4$
$h: l = 5$
For the $h$ sub-shell,$l = 5$.
Therefore,the number of orbitals $= 2(5) + 1 = 11$.

(A) The given electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$,which corresponds to the element Chromium ($Cr$,atomic number $Z = 24$).
This configuration is the stable ground state of the Chromium atom,where the $3d$ subshell is half-filled,providing extra stability due to symmetry and exchange energy.

(D) For a given principal quantum number $n$,the azimuthal quantum number $l$ can have values from $0$ to $n-1$.
For $n = 4$,$l$ can be $0, 1, 2, 3$.
Option $A$ is incorrect because the spin quantum number $s$ can only be $+\frac{1}{2}$ or $-\frac{1}{2}$,not $0$.
Options $B$ and $C$ are incorrect because $l$ cannot be equal to $n$ ($l$ must be less than $n$).
Option $D$ is correct because $n = 4$,$l = 3$ (which is $< 4$),$m = -2$ (which is within the range $-l$ to $+l$),and $s = +\frac{1}{2}$ (which is a valid spin value).

(B) The principal quantum number is given as $n = 4$.
For a given $n$,the azimuthal quantum number $l$ can take values from $0$ to $n-1$.
When $l = 3$,it corresponds to the $f$-subshell.
The number of orbitals in a subshell is given by the formula $(2l + 1)$.
Substituting $l = 3$ into the formula: $2(3) + 1 = 6 + 1 = 7$.
Therefore,there are $7$ orbitals in the $4f$ subshell.

(D) $2p$ subshell contains a maximum of $6$ electrons.
According to Hund's rule and the Pauli exclusion principle,these $6$ electrons are distributed in $3$ orbitals $(2p_x, 2p_y, 2p_z)$.
Each orbital can hold a maximum of $2$ electrons with opposite spins ($+1/2$ and $-1/2$).
Therefore,there are $3$ electrons with spin $s = +1/2$ and $3$ electrons with spin $s = -1/2$.

(A) For a $4f$ orbital,the principal quantum number $n = 4$.
The azimuthal quantum number $l$ for an $f$ orbital is $3$ (since $l = 0, 1, 2, 3$ correspond to $s, p, d, f$ orbitals respectively).
The magnetic quantum number $m$ can take values from $-l$ to $+l$,i.e.,$m = \{-3, -2, -1, 0, +1, +2, +3\}$.
The spin quantum number $s$ can be either $+\frac{1}{2}$ or $-\frac{1}{2}$.
Comparing these with the given options,option $A$ $(n = 4, l = 3, m = +1, s = +\frac{1}{2})$ satisfies all conditions.

(B) The electronic configuration of $Cr$ $(Z = 24)$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^5, 4s^1$.
For $l = 1$ (which corresponds to $p$-orbitals),the electrons are in $2p^6$ and $3p^6$. Total electrons $= 6 + 6 = 12$.
For $l = 2$ (which corresponds to $d$-orbitals),the electrons are in $3d^5$. Total electrons $= 5$.
Therefore,the numbers of electrons with $l = 1$ and $l = 2$ are $12$ and $5$ respectively.

(C) The atomic number of potassium $(K)$ is $19$.
The electronic configuration is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^1$.
The valence electron is in the $4s$ orbital.
For the $4s^1$ electron:
Principal quantum number $(n)$ = $4$.
Azimuthal quantum number $(l)$ for $s$-orbital = $0$.
Magnetic quantum number $(m_l)$ = $0$.
Spin quantum number $(m_s)$ = $+\frac{1}{2}$.

(C) The ground state term symbol is determined by the electronic configuration,which is governed by the $Aufbau$ principle,$Hund's$ rule,and the $Pauli$ exclusion principle. However,among the given options,the $Aufbau$ principle is the fundamental rule for building up the electronic configuration in the ground state.

(D) The atomic number of the element is $24$ (Chromium).
According to the Aufbau principle,the expected configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^4 4s^2$.
However,half-filled $d$-orbitals are more stable due to exchange energy and symmetry.
Therefore,one electron from the $4s$ orbital shifts to the $3d$ orbital to make it half-filled.
The correct electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$.

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

(C) According to the Pauli exclusion principle,no two electrons in an atom can have the same set of all four quantum numbers $(n, l, m_l, m_s)$.
Therefore,the maximum number of quantum numbers that two electrons can have in common is $3$.

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

(D) According to the Aufbau principle,electrons fill orbitals in order of increasing energy.
Specifically,the $2p$ subshell must be completely filled before electrons enter the $3s$ subshell.
In the configuration $1s^2 2s^2 2p^2 3s^1$,the $2p$ subshell is only partially filled $(2p^2)$ while an electron has already entered the $3s$ orbital,which violates the Aufbau principle.
Therefore,this configuration is not possible.

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

(D) The maximum number of electrons in a subshell is given by the formula $2(2l + 1)$.
For an $f$-subshell,the azimuthal quantum number $l = 3$.
Substituting the value of $l$ in the formula: $2(2 \times 3 + 1) = 2(6 + 1) = 2(7) = 14$.
Therefore,the maximum number of electrons that can be accommodated in the $f$-subshell is $14$.

(B) According to the $Pauli$ exclusion principle,an orbital can accommodate a maximum of $2$ electrons with opposite spins.

(C) For an $f$ subshell,the azimuthal quantum number $l = 3$.
The number of orbitals in a subshell is given by $(2l + 1)$.
For $l = 3$,the number of orbitals $= 2(3) + 1 = 7$.
Each orbital can accommodate a maximum of $2$ electrons.
Therefore,the maximum number of electrons in $5f$ orbitals $= 7 \times 2 = 14$.

(D)

$Quantum \ Number$	$Possible \ Values$
Principal quantum number $(n)$	$1, 2, 3, 4, \ldots$
Azimuthal quantum number $(l)$	$0, 1, 2, 3, \ldots, (n-1)$
Magnetic quantum number $(m_l)$	$-l, \ldots, 0, \ldots, +l$
Spin quantum number $(m_s)$	$+1/2, -1/2$

For the given subshell where $l = 2$,the magnetic quantum number $m_l$ can take values from $-2$ to $+2$,which are $-2, -1, 0, +1, +2$.
This corresponds to $5$ orbitals in the $3d$ subshell.
Since each orbital can hold a maximum of $2$ electrons,the total number of electrons is $5 \times 2 = 10$.

(C) The total number of electrons in the given configuration is $2+2+5+1 = 10$.
An element with $10$ electrons in its neutral state is Neon $(Ne)$.
The ground state electronic configuration of Neon is $1s^2 2s^2 2p^6$.
In the configuration $1s^2 2s^2 2p^5 3s^1$,one electron from the $2p$ orbital has been promoted to the $3s$ orbital.
Therefore,this represents the excited state of the Neon atom.

(B) The electronic configuration of a sodium atom $(Z = 11)$ is $1s^2 2s^2 2p^6 3s^1$.
For $m = 0$,the electrons must be in $s$-orbitals or $p_z$-orbitals.
In $1s^2$,there are $2$ electrons with $m = 0$.
In $2s^2$,there are $2$ electrons with $m = 0$.
In $2p^6$,the orbitals are $2p_x, 2p_y, 2p_z$. The $2p_z$ orbital has $m = 0$ and contains $2$ electrons.
In $3s^1$,there is $1$ electron with $m = 0$.
Total number of electrons with $m = 0$ is $2 + 2 + 2 + 1 = 7$.

(D) According to the Pauli exclusion principle,any single orbital can accommodate a maximum of $2$ electrons with opposite spins.
Since $d_{z^2}$ is a single orbital,it can hold a maximum of $2$ electrons.

(C) The electronic configuration is given as $1s^2 2s^2 2p^3$.
According to Hund's rule of maximum multiplicity,the $2p$ subshell has three orbitals $(2p_x, 2p_y, 2p_z)$ which are singly occupied by three electrons.
Therefore,there are $3$ unpaired electrons in the $2p$ subshell.

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

(B) The electronic configuration is given as $1s^2, 2s^2, 2p^4$.
According to Hund's rule of maximum multiplicity,the $2p$ subshell has $3$ orbitals $(2p_x, 2p_y, 2p_z)$.
Filling $4$ electrons in these orbitals: the first $3$ electrons occupy each orbital singly,and the $4th$ electron pairs up in the first orbital.
This results in $1$ fully filled orbital and $2$ half-filled orbitals.
Therefore,there are $2$ unpaired electrons.

(B) The maximum number of electrons in any subshell is given by the formula $2(2l + 1)$.
For a $d$ subshell,the azimuthal quantum number $l = 2$.
Substituting the value of $l$ into the formula: $2(2 \times 2 + 1) = 2(5) = 10$.
Alternatively,a $d$ subshell consists of $5$ orbitals,and each orbital can accommodate a maximum of $2$ electrons according to the Pauli Exclusion Principle.
Therefore,the total number of electrons is $5 \times 2 = 10$.

(C) For a given sub-shell,the azimuthal quantum number is $l$.
Each sub-shell consists of $(2l + 1)$ orbitals.
According to the Pauli exclusion principle,each orbital can hold a maximum of $2$ electrons.
Therefore,the maximum number of electrons in a sub-shell is $2 \times (2l + 1)$.

(C) The atomic number of Beryllium $(Be)$ is $4$.
Its electronic configuration in the ground state is $1s^2, 2s^2$.
Since both the $1s$ and $2s$ orbitals are completely filled with paired electrons,there are no unpaired electrons in the ground state of the Beryllium atom.
Therefore,the number of unpaired electrons is $0$.

(B) The atomic number of $Ni$ is $28$. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.
When $Ni$ forms a $Ni^{2+}$ cation,it loses two electrons from the $4s$ orbital.
Thus,the electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
In the $3d^8$ configuration,the electrons are filled in the five $3d$ orbitals according to Hund's rule:
- The first five electrons occupy the five orbitals singly.
- The next three electrons pair up in the first three orbitals.
This leaves two orbitals with single electrons.
Therefore,there are $2$ unpaired electrons in the $Ni^{2+}$ cation.

(C) The $d$-subshell consists of $5$ degenerate orbitals $(d_{xy}, d_{yz}, d_{zx}, d_{x^2-y^2}, d_{z^2})$.
According to Hund's rule of maximum multiplicity,electrons fill these orbitals singly before pairing begins.
Therefore,the maximum number of unpaired electrons that can be accommodated in the $d$-subshell is $5$.

(C) The electronic configuration of $Cl^{-}$ is $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6}$.
Since the $3p$ subshell is completely filled $(3p^{6})$,the electron density distribution is spherically symmetric.
Other options like $Na$ $([Ne] 3s^{1})$,$C$ $(1s^{2} 2s^{2} 2p^{2})$,and $Fe$ $([Ar] 3d^{6} 4s^{2})$ do not have completely filled or half-filled $p$ or $d$ subshells that result in spherical symmetry.

(B) The atomic number of the element is $14$,which corresponds to Silicon $(Si)$.
The electronic configuration of $Si$ $(Z=14)$ is $1s^2 2s^2 2p^6 3s^2 3p^2$.
According to Hund's rule of maximum multiplicity,the two electrons in the $3p$ subshell occupy separate orbitals with parallel spins.
Therefore,there are $2$ unpaired electrons in the $3p$ subshell.

(A) The electronic configuration of the atom based on the given shell distribution $(K=2, L=8, M=6)$ is $1s^2 2s^2 2p^6 3s^2 3p^4$.
Identifying the $s$-orbitals: $1s^2$,$2s^2$,and $3s^2$.
The number of $s$-electrons is $2 + 2 + 2 = 6$.

(D) The ground state electronic configuration of carbon $(Z=6)$ is $1s^2, 2s^2, 2p_x^1, 2p_y^1$.
In the excited state,one electron from the $2s$ orbital is promoted to the empty $2p_z$ orbital.
The excited state electronic configuration becomes $1s^2, 2s^1, 2p_x^1, 2p_y^1, 2p_z^1$.
Therefore,the number of unpaired electrons in the excited state is $4$.

(B) The maximum number of electrons in a shell is given by the formula $2n^2$,where $n$ is the principal quantum number.
For the $N$ shell,the principal quantum number $n = 4$.
Therefore,the maximum number of electrons $= 2 \times (4)^2 = 2 \times 16 = 32$.

(D) The electronic configuration of $_{26}Fe$ is $[Ar] \, 3d^6 \, 4s^2$.
For $Fe^{2+}$,two electrons are removed from the $4s$ orbital,so the configuration is $[Ar] \, 3d^6$.
The number of $d$-electrons in $Fe^{2+}$ is $6$.
Now,let us check the options:
$A$: $Ne$ $(Z=10)$ is $1s^2 \, 2s^2 \, 2p^6$. Number of $p$-electrons $= 6$.
$B$: $Mg$ $(Z=12)$ is $1s^2 \, 2s^2 \, 2p^6 \, 3s^2$. Number of $s$-electrons $= 1s^2 + 2s^2 + 3s^2 = 6$.
$C$: $Fe$ $(Z=26)$ is $[Ar] \, 3d^6 \, 4s^2$. Number of $d$-electrons $= 6$.
$D$: $Cl^{-}$ $(Z=17+1=18)$ is $1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6$. Number of $p$-electrons $= 2p^6 + 3p^6 = 12$.
Therefore,the number of $d$-electrons in $Fe^{2+}$ is not equal to the number of $p$-electrons in $Cl^{-}$.

(C) The atomic number of bromine $(Br)$ is $35$.
The electronic configuration of $Br$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^{10}, 4s^2, 4p^5$.
The $p$-orbitals present are $2p$,$3p$,and $4p$.
The number of electrons in these orbitals are $2p^6$,$3p^6$,and $4p^5$.
Total number of electrons in $p$-orbitals $= 6 + 6 + 5 = 17$.

(B) The maximum number of electrons that can be accommodated in a subshell is given by the formula $2(2l + 1)$.
For a $d$-subshell,the azimuthal quantum number $l = 2$.
Substituting this value into the formula: $2(2 \times 2 + 1) = 2(4 + 1) = 2(5) = 10$.
Alternatively,there are $5$ orbitals in a $d$-subshell,and each orbital can hold a maximum of $2$ electrons according to the Pauli Exclusion Principle.
Therefore,the total number of electrons is $5 \times 2 = 10$.

(A) The atomic number of $Fe$ is $26$. The electronic configuration of $Fe$ is $[Ar] \ 3d^6 4s^2$.
To form $Fe^{3+}$,we remove $3$ electrons (two from $4s$ and one from $3d$):
$Fe^{3+} = [Ar] \ 3d^5 4s^0$.
The $3d$ subshell has $5$ orbitals,each containing one unpaired electron according to Hund's rule.
Therefore,the total number of unpaired electrons is $5$.

(B) The atomic number of nitrogen $(N)$ is $7$.
The electronic configuration of nitrogen is $1s^2, 2s^2, 2p^3$.
According to Hund's rule of maximum multiplicity,electrons fill the degenerate $p$-orbitals singly before pairing begins.
Thus,the $2p$ subshell contains $2p_x^1, 2p_y^1, 2p_z^1$.
Therefore,nitrogen has $3$ unpaired electrons.

(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 $2p$: $n=2, l=1$,so $(n+l) = 2+1 = 3$.
For $3p$: $n=3, l=1$,so $(n+l) = 3+1 = 4$.
For $2s$: $n=2, l=0$,so $(n+l) = 2+0 = 2$.
For $4d$: $n=4, l=2$,so $(n+l) = 4+2 = 6$.
Since the $2s$ orbital has the lowest $(n+l)$ value,it has the least energy.

(D) According to Pauli's exclusion principle,no two electrons in the same atom can have the same set of all four quantum numbers.

(B) The correct statement is $(b)$.
The second principal energy level $(n=2)$ consists of $2s$ and $2p$ subshells.
The $2s$ subshell has $1$ orbital and the $2p$ subshell has $3$ orbitals,totaling $4$ orbitals.
Each orbital can hold $2$ electrons,so the maximum capacity is $4 \times 2 = 8$ electrons.
Option $(a)$ is incorrect as there are theoretically infinite energy levels.
Option $(c)$ is incorrect because the $M$ shell $(n=3)$ can hold $2n^2 = 2(3)^2 = 18$ electrons.
Option $(d)$ is incorrect because according to the $(n+l)$ rule,$4s$ $(4+0=4)$ has lower energy than $3d$ $(3+2=5)$.

(C) According to the Aufbau principle,electrons fill orbitals in the order of increasing energy,which is determined by the $(n + l)$ rule.
For $5s$: $n = 5, l = 0$,so $(n + l) = 5 + 0 = 5$.
For $4d$: $n = 4, l = 2$,so $(n + l) = 4 + 2 = 6$.
For $5p$: $n = 5, l = 1$,so $(n + l) = 5 + 1 = 6$.
Since $5s$ has the lowest $(n + l)$ value,it has the lowest energy and will be filled first. The increasing order of energy is $5s < 4d < 5p$.

(D) The energy of an electron in a hydrogen-like atom depends on the principal quantum number $n$. For multi-electron atoms,it depends on the $(n + l)$ value.
For $2p_x$,$2p_y$,and $2p_z$ orbitals,the principal quantum number $n = 2$ and the azimuthal quantum number $l = 1$.
Since they have the same $(n + l)$ value $(2 + 1 = 3)$,they possess the same energy.
Such orbitals are known as degenerate orbitals.