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

(A) The magnetic quantum number $(m_l)$ describes the orientation of an orbital in $3D$ space relative to the chosen coordinate system.

(A) The principal quantum number $n=4$ indicates the $4th$ shell.
The azimuthal quantum number $\ell=3$ corresponds to the $f$-subshell.
Therefore,the orbital is $4f$.

(B) For a given azimuthal quantum number $\ell$,the magnetic quantum number $m_\ell$ can take values ranging from $-\ell$ to $+\ell$,including zero.
Therefore,the total number of possible values for the magnetic quantum number is $(2 \ell + 1)$.

(A) The azimuthal quantum number $\ell=2$ corresponds to the $d$-subshell.
Each orbital can hold a maximum of $2$ electrons.
The number of orbitals in a subshell is given by $(2\ell+1)$.
For $\ell=2$,the number of orbitals $= 2(2)+1 = 5$.
Therefore,the maximum number of electrons $= 5 \times 2 = 10$ electrons.

(A) The statement "No two electrons in an atom can have the identical set of four quantum numbers" is the definition of $Pauli's \ exclusion \ principle$.
According to this principle,each orbital can hold a maximum of two electrons,and these two electrons must have opposite spins.

(C) To find the number of unpaired electrons,we write the electronic configuration for each element:
$1$. Fluorine ($F$,$Z=9$): $1s^2 2s^2 2p^5$. The $2p$ subshell has $5$ electrons,resulting in $1$ unpaired electron.
$2$. Sodium ($Na$,$Z=11$): $1s^2 2s^2 2p^6 3s^1$. The $3s$ subshell has $1$ unpaired electron.
$3$. Nitrogen ($N$,$Z=7$): $1s^2 2s^2 2p^3$. According to Hund's rule,the $2p$ subshell has $3$ electrons in separate orbitals,resulting in $3$ unpaired electrons.
$4$. Oxygen ($O$,$Z=8$): $1s^2 2s^2 2p^4$. The $2p$ subshell has $4$ electrons,resulting in $2$ unpaired electrons.
Comparing the values,Nitrogen has the maximum number of unpaired electrons $(3)$.

(B) The statement describes $Hund's \ rule \ of \ maximum \ multiplicity$.
According to this rule,for a given electron configuration,the term with maximum multiplicity has the lowest energy.
In simpler terms,pairing of electrons in the orbitals of a given subshell $(p, d, f)$ does not take place until each orbital is singly occupied.

(A) According to Pauli's exclusion principle,no two electrons in an atom can have the same set of four quantum numbers $(n, l, m_l, m_s)$.
This implies that an orbital can hold a maximum of two electrons,and they must have opposite spins.

(C) Chromium $(Cr)$ has an atomic number of $24$.
Its expected electronic configuration is $[Ar] 3d^4 4s^2$.
However,due to the extra stability of half-filled $d$-orbitals,the observed 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 $4s$ orbital.
Therefore,the total number of unpaired electrons is $5 + 1 = 6$.

(D) The given electronic configuration is $[Ar] 3d^{10} 4s^{2}$.
Summing the electrons: $18 (Ar) + 10 (3d) + 2 (4s) = 30$.
The atomic number $Z = 30$ corresponds to the element Zinc $(Zn)$.

(D) The electronic configuration of $Cu (Z=29)$ is $[Ar] 3d^{10} 4s^{1}$.
This occurs because a fully filled $3d$ subshell provides extra stability to the atom.
Therefore,$Cu$ has only one electron in the $4s$ orbital.

(C) The electronic configuration of the $M^{2+}$ ion is $[Ar] 3d^8$.
This means the neutral atom $M$ has lost $2$ electrons to form the $M^{2+}$ ion.
To find the configuration of the neutral atom $M$,we add the $2$ electrons back to the $3d^8$ configuration.
The configuration of the neutral atom $M$ is $[Ar] 3d^8 4s^2$.
The total number of electrons in the neutral atom is $18$ (from $[Ar]$) $+ 8 + 2 = 28$.
Therefore,the atomic number of the element is $28$.

(B) The atomic number of Chromium $(Cr)$ is $24$.
According to the Aufbau principle,the expected configuration is $[Ar] 3d^4 4s^2$.
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,resulting in the stable configuration $[Ar] 3d^5 4s^1$.
Thus,the correct option is $B$.

(D) The elements with an unpaired electron in their $s$-subshell are those that have an $ns^1$ configuration in their ground state.
These include the alkali metals $(H, Li, Na, K)$ and elements where an electron is promoted to the $s$-subshell to achieve stability,such as $Cr$ $([Ar] 3d^5 4s^1)$ and $Cu$ $([Ar] 3d^{10} 4s^1)$.
Thus,the elements are $H$ $(Z=1)$,$Li$ $(Z=3)$,$Na$ $(Z=11)$,$K$ $(Z=19)$,$Cr$ $(Z=24)$,and $Cu$ $(Z=29)$.
Total number of such elements is $6$.

(C) The electronic configuration of a chlorine atom $(Z=17)$ is $1s^2 2s^2 2p^6 3s^2 3p^5$.
The unpaired electron is present in the $3p$ subshell.
For the $3p$ orbital: Principal quantum number $(n)$ = $3$,Azimuthal quantum number $(l)$ = $1$.
The $3p^5$ configuration fills the orbitals as follows: $m_l = -1$ ($2$ electrons),$m_l = 0$ ($2$ electrons),$m_l = +1$ ($1$ electron).
Thus,for the unpaired electron: $n=3$,$l=1$,$m_l=1$,and $m_s = \pm \frac{1}{2}$.

(B) The atomic number of copper $(Cu)$ is $29$.
The ground state electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$.
To form the $Cu^{2+}$ ion,two electrons are removed,one from the $4s$ orbital and one from the $3d$ orbital.
Therefore,the electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9 4s^0$.

(A) The shell '$N$' corresponds to the principal quantum number $n = 4$.
The total number of orbitals in a shell is given by the formula $n^2$.
Substituting $n = 4$ into the formula,we get $4^2 = 16$.
Therefore,the number of orbitals associated with the '$N$' shell is $16$.

(B) For the $3p$ orbital,the principal quantum number $n = 3$ and the azimuthal quantum number $l = 1$.
The number of angular nodes is given by $l = 1$.
The number of radial nodes is given by the formula $(n - l - 1)$.
Substituting the values: $3 - 1 - 1 = 1$.
Therefore,the number of angular and radial nodes in the $3p$ orbital are $1$ and $1$ respectively.

(D) The rules for quantum numbers are as follows:
$1$. The principal quantum number $n$ can be any positive integer.
$2$. The azimuthal quantum number $l$ can have values from $0$ to $n-1$.
$3$. The magnetic quantum number $m_l$ can have values from $-l$ to $+l$.
$4$. The spin quantum number $s$ can only be $+1/2$ or $-1/2$.
Evaluating the options:
- In option $D$,for $n = 3$,the maximum value of $l$ is $2$. If $l = 2$,the possible values for $m_l$ are $-2, -1, 0, +1, +2$.
- Therefore,$m_l = -3$ is not permissible for $l = 2$.
- Additionally,the spin quantum number $s$ must be $\pm 1/2$. The original options provided $s = \pm 2$,which is physically impossible.

(B) For electron $P$,the quantum numbers are $n = 3$ and $l = 2$.
Thus,the value of $(n + l) = 3 + 2 = 5$.
For electron $Q$,the quantum numbers are $n = 3$ and $l = 0$.
Thus,the value of $(n + l) = 3 + 0 = 3$.
According to the $(n + l)$ rule (Aufbau principle),the orbital with a higher $(n + l)$ value has higher energy.
Therefore,$P$ has greater energy than $Q$.

(A) For any orbital,the principal quantum number $n$ and the azimuthal quantum number $l$ must satisfy the condition $l < n$.
For the $3f$ orbital,$n = 3$ and for an $f$-subshell,$l = 3$.
Since $l$ cannot be equal to or greater than $n$ ($l < n$ is required),the $3f$ orbital is impossible.
In the other options:
$2p$: $n=2, l=1$ ($1 < 2$ is valid).
$4d$: $n=4, l=2$ ($2 < 4$ is valid).
$2s$: $n=2, l=0$ ($0 < 2$ is valid).

(C) The electronic configuration of sodium $(Z=11)$ is $1s^2, 2s^2, 2p^6, 3s^1$.
For the outermost electron,which is in the $3s$ orbital:
Principal quantum number $(n) = 3$.
Azimuthal quantum number $(l)$ for $s$-orbital $= 0$.
Magnetic quantum number $(m_l) = 0$.
Spin quantum number $(m_s) = +\frac{1}{2}$ (or $-\frac{1}{2}$).
Thus,the set of quantum numbers is $(3, 0, 0, \frac{1}{2})$.

(C) According to the $(n+l)$ rule,the energy of an orbital increases as the value of $(n+l)$ increases.
For option $A$: $n+l = 4+0 = 4$.
For option $B$: $n+l = 3+1 = 4$.
For option $C$: $n+l = 3+2 = 5$.
For option $D$: $n+l = 3+0 = 3$.
Since the value of $(n+l)$ is maximum for option $C$,it represents the highest energy level.

(C) The electronic configuration of the $Cu$ $(Z=29)$ atom is $[Ar] 3d^{10} 4s^{1}$.
Since the outermost shell is $4s$,the outermost electron resides 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 $(m_s)$ = $+\frac{1}{2}$.
Thus,the set of quantum numbers is $(4, 0, 0, +\frac{1}{2})$.

(C) The electronic configuration of potassium $(Z=19)$ is $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{1}$.
The outermost electron is in the $4s$ orbital.
For the $4s$ orbital,the principal quantum number $(n) = 4$.
The azimuthal quantum number $(l)$ for an $s$-orbital is $0$.
The magnetic quantum number $(m_l)$ is $0$.
The spin quantum number $(m_s)$ is $+\frac{1}{2}$ (or $-\frac{1}{2}$).
Thus,the correct set of quantum numbers is $4, 0, 0, \frac{1}{2}$.

(B) Given,azimuthal quantum number $(l) = 2$.
The number of orbitals is given by the formula $(2l + 1)$.
Substituting the value of $l$:
Number of orbitals $= (2 \times 2 + 1) = 4 + 1 = 5$.

(C) The distance of an orbital from the nucleus depends on the principal quantum number $(n)$.
For a given $n$,the penetration power of orbitals follows the order $s > p > d > f$.
Among the given options,$4s$ has the lowest principal quantum number $(n=4)$ and the highest penetration power,making it the closest to the nucleus.

(C) For multi-electron atoms,the energy of the orbitals increases with an increase in their $(n+l)$ value.
According to the $(n+l)$ rule:
For $5p$: $n=5, l=1$,$(n+l) = 6$
For $6s$: $n=6, l=0$,$(n+l) = 6$
For $4f$: $n=4, l=3$,$(n+l) = 7$
For $5d$: $n=5, l=2$,$(n+l) = 7$
When $(n+l)$ values are equal,the orbital with the lower $n$ value has lower energy.
Comparing $5p$ and $6s$: $5p$ has lower $n$,so $5p < 6s$.
Comparing $4f$ and $5d$: $4f$ has lower $n$,so $4f < 5d$.
Thus,the correct increasing order of energy is: $5p < 6s < 4f < 5d$.

(C) The electronic configuration of $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$. Thus,$Fe$ has $6$ electrons in the $d$ orbital.
Electronic configurations of the given elements:
$Mg$ $(Z=12)$: $1s^2 2s^2 2p^6 3s^2$. Total '$s$' electrons = $2+2+2 = 6$.
$Cl$ $(Z=17)$: $1s^2 2s^2 2p^6 3s^2 3p^5$. Total '$p$' electrons = $6+5 = 11$.
$Ne$ $(Z=10)$: $1s^2 2s^2 2p^6$. Total '$p$' electrons = $6$.
Comparing these values,the number of $d$ electrons in $Fe$ $(6)$ is equal to the total '$s$' electrons of $Mg$ $(6)$ and the total '$p$' electrons of $Ne$ $(6)$.
Therefore,statements $(i)$ and $(iii)$ are correct.

(C) The atomic number of $Fe$ is $26$. The ground state electronic configuration of a neutral $Fe$ atom is $[Ar] 3d^6 4s^2$.
To form the $Fe^{3+}$ ion,we must remove $3$ electrons from the neutral atom.
Electrons are removed from the outermost shell first,which is the $4s$ orbital,followed by the $3d$ orbital.
Removing $2$ electrons from $4s$ and $1$ electron from $3d$ results in the configuration: $[Ar] 3d^5 4s^0$.

(A) . Chromium $(Z=24)$: Electronic configuration $(EC) = [Ar] 3d^5 4s^1$. Number of unpaired electrons $= 6$.
$B$. Iron $(Z=26)$: $EC = [Ar] 3d^6 4s^2$. Number of unpaired electrons $= 4$.
$C$. Manganese $(Z=25)$: $EC = [Ar] 3d^5 4s^2$. Number of unpaired electrons $= 5$.
$D$. Vanadium $(Z=23)$: $EC = [Ar] 3d^3 4s^2$. Number of unpaired electrons $= 3$.
Therefore,chromium $(Cr)$ has the maximum number of unpaired electrons.

(B) The atom has a valence electronic configuration of $3d^6$ in its $+3$ oxidation state. This corresponds to the $Co^{3+}$ ion $(Z=27)$.
To find the ground state configuration of the neutral atom $(Co)$,we add back the $3$ electrons removed during oxidation.
The neutral $Co$ atom has the configuration $[Ar] 3d^7 4s^2$.
In the $3d^7$ subshell,there are $5$ orbitals. According to Hund's rule,the $7$ electrons are filled as follows: $5$ electrons occupy the orbitals singly,and $2$ electrons pair up.
This leaves $5 - 2 = 3$ unpaired electrons in the $3d$ subshell.
Therefore,the number of unpaired electrons in the ground state of the atom is $3$.

(C) For any electron,the value of the azimuthal quantum number $l$ must range from $0$ to $n-1$.
In option $C$,$n=3$,which implies that $l$ can only take values $0, 1, \text{or } 2$.
Since $l=3$ is given,this set of quantum numbers is impossible because $l$ cannot be equal to or greater than $n$.
Therefore,the correct option is $C$.

(A) The magnetic quantum number $m_l$ ranges from $-l$ to $+l$. An orbital can contain a maximum of $2$ electrons (with opposite spins).
$1$. For $3d$ orbital $(l=2)$: $m_l$ values are $-2, -1, 0, +1, +2$. The value $m_l = -2$ is possible. Max electrons = $2$.
$2$. For $6d$ orbital $(l=2)$: $m_l$ values are $-2, -1, 0, +1, +2$. The value $m_l = -2$ is possible. Max electrons = $2$.
$3$. For $5s$ orbital $(l=0)$: $m_l$ value is $0$. The value $m_l = -2$ is not possible. Max electrons = $0$.
$4$. For $4f$ orbital $(l=3)$: $m_l$ values are $-3, -2, -1, 0, +1, +2, +3$. The value $m_l = -2$ is possible. Max electrons = $2$.
Total electrons = $2 + 2 + 0 + 2 = 6$.

(A) For $Z=24$ (Chromium): The electronic configuration is $[Ar] 3d^5 4s^1$.
Electrons with $m_l=0$:
$1s^2$ ($2$ electrons),$2s^2$ ($2$ electrons),$2p^6$ ($2$ electrons in $2p_z$),$3s^2$ ($2$ electrons),$3p^6$ ($2$ electrons in $3p_z$),$3d^5$ ($1$ electron in $3d_{z^2}$),$4s^1$ ($1$ electron).
Total = $2+2+2+2+2+1+1 = 12$.
For $Z=29$ (Copper): The electronic configuration is $[Ar] 3d^{10} 4s^1$.
Electrons with $m_l=0$:
$1s^2$ $(2)$,$2s^2$ $(2)$,$2p^6$ $(2)$,$3s^2$ $(2)$,$3p^6$ $(2)$,$3d^{10}$ ($2$ electrons in $3d_{z^2}$),$4s^1$ $(1)$.
Total = $2+2+2+2+2+2+1 = 13$.
Thus,the values are $12$ and $13$.

(A) The atomic number of strontium $(Sr)$ is $38$. The electronic configuration is $[Kr] 5s^2$.
The valence electrons are present in the $5s$ orbital.
For the $5s$ orbital:
Principal quantum number $(n) = 5$.
Azimuthal quantum number $(l) = 0$ (for $s$-orbital).
Magnetic orbital quantum number $(m_l) = 0$.
Electron spin quantum number $(m_s) = +\frac{1}{2}$ or $-\frac{1}{2}$.
Thus,the correct set of quantum numbers is $n=5, l=0, m_l=0, m_s=+\frac{1}{2}$.

(B) The region where the probability density function,$\psi^2(r)$ reduces to zero is called a nodal surface or simply a node.
For an $ns$ orbital,the number of radial nodes is given by $(n-1)$.
The provided graph shows one point where the probability density touches the $r$-axis (one radial node).
For $1s$ orbital: $(n-1) = (1-1) = 0$ nodes.
For $2s$ orbital: $(n-1) = (2-1) = 1$ node.
For $3s$ orbital: $(n-1) = (3-1) = 2$ nodes.
Since the graph has one node,it represents the $2s-$orbital.

(B) The number of radial nodes is given by the formula: $\text{Radial nodes} = n - l - 1$.
The number of angular nodes is given by the value of the azimuthal quantum number: $\text{Angular nodes} = l$.
For a $4d$ orbital,the principal quantum number $n = 4$ and the azimuthal quantum number $l = 2$.
Therefore,$\text{Angular nodes} = 2$.
$\text{Radial nodes} = 4 - 2 - 1 = 1$.
The sum of angular and radial nodes is $2 + 1 = 3$.

(D) An orbital in an atom is defined by the first three quantum numbers: the principal quantum number $(n)$,the azimuthal quantum number $(l)$,and the magnetic orbital quantum number $(m_l)$.
These three quantum numbers specify the size,shape,and orientation of the orbital.
The fourth quantum number,the spin quantum number $(m_s)$,is used to describe the spin of the electron within the orbital,not the orbital itself.

(C) The maximum number of electrons that can be accommodated in an orbit with principal quantum number $n$ is given by the formula $2n^2$.
For $n=6$,the number of electrons $= 2(6)^2 = 2 \times 36 = 72$.

(A) The number of angular nodes for an orbital is given by the azimuthal quantum number $l$.
For the third shell $(n=3)$,the possible subshells are $3s$,$3p$,and $3d$.
For $3s$ orbital,$l=0$,so the number of angular nodes is $0$.
For $3p$ orbital,$l=1$,so the number of angular nodes is $1$.
For $3d$ orbital,$l=2$,so the number of angular nodes is $2$.
Therefore,the total number of angular nodes for all orbitals in the third shell is $0 + 1 + 2 = 3$.

(B) For a $4f$ orbital,the principal quantum number $n=4$ and the azimuthal quantum number $\ell=3$.
For a given value of $\ell$,the magnetic quantum number $m_l$ can take values from $-\ell$ to $+\ell$,i.e.,$-3, -2, -1, 0, +1, +2, +3$.
Thus,$m_l=+1$ is a valid value.
The spin quantum number $m_s$ can be either $+\frac{1}{2}$ or $-\frac{1}{2}$.
Therefore,the set $n=4, \ell=3, m_l=+1, m_s=+1/2$ is correct.

(A) For any orbital,the number of angular nodes is equal to the azimuthal quantum number $(l)$.
For a $4f$-orbital,the principal quantum number $n = 4$ and the azimuthal quantum number $l = 3$.
Angular nodes $= l = 3$.
The number of radial nodes is given by the formula: $\text{Radial nodes} = n - l - 1$.
Substituting the values: $\text{Radial nodes} = 4 - 3 - 1 = 0$.
Therefore,the number of radial nodes is $0$ and the number of angular nodes is $3$.

(B) The azimuthal quantum number $(l)$ determines the shape of the orbital.
It is a quantum number for an atomic orbital that determines its orbital angular momentum and describes its shape.
The azimuthal quantum number is the second of a set of quantum numbers representing an electron's unique quantum state.
For example,$l = 0$ represents a spherical orbital ($s$-orbital),and $l = 1$ represents a dumbbell-shaped orbital ($p$-orbital).
Hence,option $B$ is the correct answer.

(D) According to the $Pauli$ exclusion principle,any single orbital can hold a maximum of $2$ electrons with opposite spins.
The quantum numbers $n=4$ and $l=3$ specify a $4f$ subshell,which contains $7$ distinct orbitals.
However,the question asks for the maximum number of electrons in an orbital (singular),not the entire subshell.
Therefore,regardless of the values of $n$ and $l$,any single orbital can accommodate a maximum of $2$ electrons.

(A) The total number of electrons in a shell with principal quantum number $n$ is given by $2n^2$.
For $n=4$,the total number of electrons is $2 \times 4^2 = 32$.
Since each orbital can hold a maximum of $2$ electrons with opposite spins ($+\frac{1}{2}$ and $-\frac{1}{2}$),exactly half of the total electrons will have a spin quantum number of $+\frac{1}{2}$.
Therefore,the number of electrons with spin $+\frac{1}{2}$ is $\frac{32}{2} = 16$.

(B) For an electron in a $3p$-orbital,the principal quantum number $n = 3$.
For a $p$-orbital,the azimuthal quantum number $l = 1$.
The magnetic quantum number $m$ can take values from $-l$ to $+l$,which means $m = -1, 0, +1$.
The spin quantum number $s$ can be $+1/2$ or $-1/2$.
Comparing these with the given options,in option $(b)$,the value of $m$ is given as $-2$,which is not possible for $l = 1$.
Therefore,the set $(3, 1, -2, -1/2)$ is incorrect.

(A) The maximum number of electrons in a shell with principal quantum number $n$ is given by $2n^2$.
For $n=4$,the total number of electrons is $2(4)^2 = 32$. The number of electrons with $m_s = +\frac{1}{2}$ is $32/2 = 16$.
For $n=3$,the total number of electrons is $2(3)^2 = 18$. The number of electrons with $m_s = -\frac{1}{2}$ is $18/2 = 9$.
The sum of these electrons is $16 + 9 = 25$.

(A) The azimuthal quantum number $(l)$ determines the subshell type. For a given subshell,the number of orbitals is given by $(2l + 1)$.
Each orbital can hold a maximum of $2$ electrons.
Therefore,the maximum number of electrons in a subshell is $2 \times (2l + 1)$.
Given $n=3$ and $l=2$,this corresponds to the $3d$ subshell.
Substituting $l=2$ into the formula: $2 \times (2(2) + 1) = 2 \times (4 + 1) = 2 \times 5 = 10$ electrons.
Thus,the $3d$ subshell can accommodate a maximum of $10$ electrons.

(B) The number of angular nodes is given by the azimuthal quantum number $l$. For $d$-orbitals,$l = 2$,so they have $2$ angular nodes.
The number of radial nodes is calculated using the formula: $\text{Radial nodes} = n - l - 1$.
For the $3d$ orbital: $n = 3$ and $l = 2$.
$\text{Radial nodes} = 3 - 2 - 1 = 0$.
Thus,the $3d$ orbital has $0$ radial nodes and $2$ angular nodes.