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

(A) The electronic configuration of Nitrogen $(N)$ is $1s^2 2s^2 2p^3$.
In the valence shell $(n=2)$,the $p$-subshell has $3$ electrons.
Since the $p$-subshell can hold a maximum of $6$ electrons,$3$ electrons represent a half-filled configuration $(p^3)$.

(A) For a set of quantum numbers to be valid,the following rules must be satisfied:
$1$. $n$ is a positive integer $(1, 2, 3, ...)$.
$2$. $l$ can have values from $0$ to $n-1$.
$3$. $m$ can have values from $-l$ to $+l$ (including $0$).
$4$. $s$ can be $+1/2$ or $-1/2$.
Evaluating the options:
$(A)$ $n=3, l=2, m=-3$: Since $l=2$,$m$ can only range from $-2$ to $+2$. Thus,$m=-3$ is invalid.
$(B)$ $n=4, l=3, m=3, s=+1/2$: Here $l < n$ $(3 < 4)$ and $m$ is within the range $-3$ to $+3$. This set is valid.
$(C)$ $n=2, l=1, m=0, s=-1/2$: Here $l < n$ $(1 < 2)$ and $m$ is within the range $-1$ to $+1$. This set is also valid.
$(D)$ $n=4, l=3, m=2, s=+1/2$: Here $l < n$ $(3 < 4)$ and $m$ is within the range $-3$ to $+3$. This set is also valid.
Note: In many textbook contexts,this question is presented as a multiple-choice question where only one is invalid or specific criteria are applied. Given the standard rules,options $B, C,$ and $D$ are physically possible. However,if the question implies selecting the most standard set or identifying an error,$A$ is the only one that violates the rules.

(D) For a hydrogen atom,the energy of an orbital depends only on the principal quantum number $(n)$.
Since all the orbitals mentioned $(3s, 3p_x, 3d_{xy}, 3d_{yz})$ have the same principal quantum number $(n = 3)$,they are degenerate (have the same energy).
Therefore,transitions between these orbitals do not involve any change in energy,meaning no absorption or emission of energy occurs.
Thus,all the given transitions are possible without energy change.

(B) The atomic number of potassium $(K)$ is $19$.
Its electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1$.
The last electron enters the $4s$ orbital.
For the $4s$ orbital,the principal quantum number $n = 4$.
Since it is an $s$-orbital,the azimuthal quantum number $l = 0$.
For $l = 0$,the magnetic quantum number $m = 0$.
Therefore,the values are $n = 4, l = 0, m = 0$.

(C) According to the Aufbau principle,electrons fill orbitals in order of increasing energy. The energy order is determined by the $(n + l)$ rule. For $3d$,$(n + l) = 3 + 2 = 5$. After $3d$ is filled,the next orbitals to be filled are $4p$ $(n + l = 4 + 1 = 5)$ and $5s$ $(n + l = 5 + 0 = 5)$. However,comparing $3d$ $(n=3, l=2)$ and $4p$ $(n=4, l=1)$,the $4p$ orbital has a higher energy than $3d$. Thus,after the $3d$ subshell is completely filled,the next electron enters the $4p$ orbital.

(C) The electronic configuration of an element with atomic number $17$ $(Cl)$ is $1s^2 2s^2 2p^6 3s^2 3p^5$.
The valence shell is the $n=3$ shell.
The orbitals in the valence shell are $3s$ and $3p$.
The $3s$ orbital contains $2$ electrons (one pair).
The $3p$ subshell has $3$ orbitals $(3p_x, 3p_y, 3p_z)$.
With $5$ electrons in the $3p$ subshell,the distribution is $(3p_x)^2 (3p_y)^2 (3p_z)^1$.
Thus,there are two pairs in the $3p$ subshell and one pair in the $3s$ orbital.
Total number of orbitals containing electron pairs = $1$ $(3s)$ + $2$ $(3p_x, 3p_y)$ = $3$.

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

(B) The electronic configuration of an element with atomic number $Z = 104$ (Rutherfordium) is $[Rn] \ 5f^{14} \ 6d^2 \ 7s^2$.
We need to identify orbitals where $(n + l) = 8$:
For $5f$: $n = 5, l = 3 \implies n + l = 5 + 3 = 8$.
For $6d$: $n = 6, l = 2 \implies n + l = 6 + 2 = 8$.
For $7p$: $n = 7, l = 1 \implies n + l = 7 + 1 = 8$.
For $8s$: $n = 8, l = 0 \implies n + l = 8 + 0 = 8$.
In the configuration $[Rn] \ 5f^{14} \ 6d^2 \ 7s^2$:
- The $5f$ orbital has $14$ electrons.
- The $6d$ orbital has $2$ electrons.
- The $7p$ and $8s$ orbitals are empty.
Total electrons with $(n + l) = 8$ are $14 + 2 = 16$.

(C) For $n = 3$ and $l = 2$,the orbital is $3d$.
The $l = 2$ value corresponds to the $d$-subshell,which has $(2l + 1) = 2(2) + 1 = 5$ orbitals.
Each orbital can accommodate a maximum of $2$ electrons.
Therefore,the total number of electrons = $5 \times 2 = 10$ electrons.

(B) The electronic configuration of $Ca^{2+}$ is $(2, 8, 8)$.
In this configuration,the outermost shell $(n=3)$ contains $8$ electrons and the penultimate shell $(n=2)$ also contains $8$ electrons.
Thus,$Ca^{2+}$ has an equal number of electrons in the outermost and penultimate shells.

(C) The ground state electronic configuration of a neon atom $(Z = 10)$ is $1s^{2}\, 2s^{2}\, 2p^{6}$.
When one electron from the $2p$ orbital is promoted to the $3s$ orbital,it results in the first excited state of neon: $1s^{2}\, 2s^{2}\, 2p^{5}\, 3s^{1}$.

(C) To determine if a species is paramagnetic,we check for the presence of unpaired electrons.
$1$. $Be^-$ ($4 + 1 = 5$ electrons): $1s^2 2s^2 2p^1$ (Unpaired electron present,paramagnetic).
$2$. $Ne^{2+}$ ($10 - 2 = 8$ electrons): $1s^2 2s^2 2p^4$ (Unpaired electrons present,paramagnetic).
$3$. $Cl^-$ ($17 + 1 = 18$ electrons): $1s^2 2s^2 2p^6 3s^2 3p^6$ (All electrons are paired,diamagnetic).
$4$. $As^+$ ($33 - 1 = 32$ electrons): The electronic configuration ends in $4p^2$,which has unpaired electrons (paramagnetic).
Therefore,$Cl^-$ is not paramagnetic.

(C) The atomic number of $Fe$ is $26$.
The electronic configuration of neutral $Fe$ atom is $[Ar] 3d^6 4s^2$.
When $Fe$ forms $Fe^{2+}$ ion,it loses two electrons from the $4s$ orbital.
Therefore,the electronic configuration of $Fe^{2+}$ is $[Ar] 3d^6$.
Thus,the number of $d$-electrons in $Fe^{2+}$ is $6$.

(C) 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 to make it half-filled $(3d^5)$.
The correct electronic configuration is $[Ar] \, 3d^5 \, 4s^1$.

(C) The electronic configuration of the $M^{2+}$ ion is $[Ar] \, 3d^8$.
This means the ion has lost $2$ electrons from its neutral state.
The neutral atom $M$ would have the configuration $[Ar] \, 3d^8 \, 4s^2$.
Adding the electrons: $18$ (from Argon) $+ 8 + 2 = 28$.
Therefore,the atomic number of the element is $28$.

(C) Given the configuration: $(n - 1)s^2 (n - 1)p^6 (n - 1)d^x ns^2$.
Substituting $n = 4$ and $x = 5$,we get the configuration of the valence and inner shells as $3s^2 3p^6 3d^5 4s^2$.
The total number of electrons is the sum of electrons in all shells: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^2$.
Total electrons $= 2 + 2 + 6 + 2 + 6 + 5 + 2 = 25$.
Since the atom is neutral,the number of protons equals the number of electrons,which is $25$.

(D) The atomic number of copper $(Cu)$ is $29$.
According to the Aufbau principle,the expected configuration is $[Ar] \, 3d^9 \, 4s^2$.
However,a fully filled $d$-subshell is more stable due to symmetry and exchange energy.
Therefore,one electron from the $4s$ orbital jumps to the $3d$ orbital to make it $3d^{10}$.
The correct electronic configuration is $[Ar] \, 3d^{10} \, 4s^1$.

(B) The atomic number of iron $(Fe)$ is $26$.
The electronic configuration of $Fe$ is $[Ar] 3d^6 4s^2$.
When $Fe$ forms a ferrous ion $(Fe^{2+})$,it loses two electrons from the $4s$ orbital.
The electronic configuration of $Fe^{2+}$ becomes $[Ar] 3d^6$.
In the $3d$ subshell,there are $5$ orbitals. According to Hund's rule,the $6$ electrons are filled as: one electron in each of the $5$ orbitals and the $6^{th}$ electron pairs up in the first orbital.
Thus,there are $4$ unpaired electrons in the $Fe^{2+}$ ion.

(C) The atomic number of $Fe$ is $26$.
The electronic configuration of $Fe$ is $[Ar] 3d^6 4s^2$.
When $Fe$ forms an $Fe^{2+}$ ion,it loses two electrons from the $4s$ orbital.
The electronic configuration of $Fe^{2+}$ becomes $[Ar] 3d^6 4s^0$.
Therefore,the number of $d$-electrons present in $Fe^{2+}$ is $6$.

(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 valence shell configuration is $3s^2 3p^5$.
In the $3p$ subshell,there are $3$ orbitals $(3p_x, 3p_y, 3p_z)$. According to Hund's rule,the $5$ electrons are filled as follows: $3p_x^2, 3p_y^2, 3p_z^1$.
The unpaired electron is in the $3p_z$ orbital.
For this electron:
Principal quantum number $(n)$ = $3$
Azimuthal quantum number $(l)$ for $p$-orbital = $1$
Magnetic quantum number $(m)$ can be $-1, 0, +1$. By convention,the unpaired electron is often assigned $m = +1$.
Thus,the set of quantum numbers is $n=3, l=1, m=1$.

(B) According to the Aufbau principle,electrons fill orbitals in order of increasing energy. The $2s$ orbital has lower energy than the $2p$ orbitals.
Therefore,the $2s$ orbital must be completely filled (containing $2$ electrons) before any electrons enter the $2p$ orbitals.
In option $(B)$,the $2s$ orbital contains only $1$ electron while the $2p$ orbitals are already being filled,which violates the Aufbau principle.

(D) The $1s$ orbital is the ground state of the hydrogen atom,which has the lowest energy.
An electron in the $1s$ orbital can absorb a photon to transition to a higher energy level (excited state).
However,since there is no energy level lower than $1s$,an electron in the $1s$ orbital cannot emit a photon to transition to a lower state.

(C) The correct electronic configuration of $Cu$ $(Z=29)$ is $[Ar] 4s^1 3d^{10}$ due to the extra stability of the fully filled $d$-subshell.
Option $C$ shows the configuration as $4s^2 3d^9$,which is incorrect for the ground state of $Cu$.

(A) The magnetic quantum number $m$ can take values from $-l$ to $+l$ including zero.
For option $A$,$n = 3$ and $l = 2$.
The possible values for $m$ are $-2, -1, 0, +1, +2$.
Since $m = -3$ is outside this range,this set is inconsistent with quantum mechanical theory.

(B) The orbital angular momentum of an electron is given by the formula: $\text{Angular momentum} = \sqrt{l(l + 1)} \frac{h}{2\pi}$.
For an $s$ orbital,the azimuthal quantum number $l = 0$.
Substituting $l = 0$ into the formula: $\text{Angular momentum} = \sqrt{0(0 + 1)} \frac{h}{2\pi} = 0$.
Therefore,the orbital angular momentum of an electron in an $s$ orbital is zero.

(D) The quantum numbers $n = 4, l = 1$ indicate that the last electron enters the $4p$ orbital.
For $l = 1$,the possible values of $m$ are $-1, 0, +1$.
The electron is in the $4p$ orbital with $m = +1$ and $s = -1/2$,which corresponds to the last electron of the $p$-block element.
The electronic configuration of the atom is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6$.
Summing the electrons: $2 + 2 + 6 + 2 + 6 + 2 + 10 + 6 = 36$.
Therefore,the atomic number is $36$,which corresponds to Krypton $(Kr)$.

(A) The atomic number of nitrogen is $Z = 7$. The 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.
Therefore,the three electrons in the $2p$ orbitals must occupy separate orbitals with parallel spins.
This corresponds to the configuration $1s^2, 2s^2, 2p_x^1, 2p_y^1, 2p_z^1$,which is represented by option $A$.

(D) The spin quantum number $(m_s)$ describes the intrinsic angular momentum of an electron.
Historically,it was visualized as the electron spinning on its axis,where $+1/2$ corresponds to clockwise rotation and $-1/2$ corresponds to anticlockwise rotation.
Alternatively,it represents the orientation of the electron's magnetic moment,where $+1/2$ indicates the magnetic moment is pointing up and $-1/2$ indicates it is pointing down.
Therefore,both statements $A$ and $C$ are accepted interpretations.

(B) The atomic number of Cesium $(Cs)$ is $55$. The electronic configuration of $Cs$ is: $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^{10}, 4s^2, 4p^6, 4d^{10}, 5s^2, 5p^6, 6s^1$.
For the cesium ion $(Cs^+)$,one electron is removed from the $6s$ orbital: $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^{10}, 4s^2, 4p^6, 4d^{10}, 5s^2, 5p^6$.
Total number of electrons in $s$-orbitals $(1s, 2s, 3s, 4s, 5s)$: $2 + 2 + 2 + 2 + 2 = 10$.
Total number of electrons in $p$-orbitals $(2p, 3p, 4p, 5p)$: $6 + 6 + 6 + 6 = 24$.
Total number of electrons in $d$-orbitals $(3d, 4d)$: $10 + 10 = 20$.
Thus,the counts are $10, 24, 20$.

(B) For $n = 3$ and $l = 1$,the subshell is $3p$.
According to the Pauli exclusion principle,any single orbital can accommodate a maximum of $2$ electrons with opposite spins.
Therefore,a single $3p$ orbital can hold $2$ electrons.

(C) The $d-$orbitals are classified into two sets based on their orientation relative to the coordinate axes:
$1$. $e_g$ set: These orbitals have electron density along the axes. These include $d_{z^2}$ and $d_{x^2 - y^2}$.
$2$. $t_{2g}$ set: These orbitals have electron density between the axes. These include $d_{xy}, d_{yz},$ and $d_{xz}$.
Therefore,the pair of $d-$orbitals that have electron density along the axes is $d_{z^2}$ and $d_{x^2 - y^2}$.

(B) According to the Pauli Exclusion Principle,no two electrons in an atom can have the same set of all four quantum numbers.
For two electrons in the same orbital,the principal quantum number $(n)$,azimuthal quantum number $(l)$,and magnetic quantum number $(m_l)$ are identical.
Therefore,they must differ in their spin quantum number $(m_s)$,which can be either $+1/2$ or $-1/2$.

(C) The electronic configuration of Titanium $(Z = 22)$ is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^2$.
According to the $(n+l)$ rule,the energy of an orbital increases as the value of $(n+l)$ increases.
For $3s$: $n+l = 3+0 = 3$.
For $3p$: $n+l = 3+1 = 4$.
For $4s$: $n+l = 4+0 = 4$.
For $3d$: $n+l = 3+2 = 5$.
Since $3p$ and $4s$ have the same $(n+l)$ value,the one with the lower $n$ value has lower energy. Thus,$3p < 4s$.
The correct order of increasing energy is $3s < 3p < 4s < 3d$.

(D) The electronic configuration of $Fe^{2+}$ is $[Ar] 3d^{6} 4s^{0}$. Thus,the number of $d-$ electrons is $6$.
$A) \ Fe \ (Z = 26): [Ar] 3d^{6} 4s^{2}$. Number of $d-$ electrons is $6$.
$B) \ Ne \ (Z = 10): 1s^{2} 2s^{2} 2p^{6}$. Number of $p-$ electrons is $6$.
$C) \ Mg \ (Z = 12): 1s^{2} 2s^{2} 2p^{6} 3s^{2}$. Total number of $s-$ electrons is $2 + 2 + 2 = 6$.
$D) \ Cl \ (Z = 17): 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{5}$. Total number of $p-$ electrons is $6 + 5 = 11$.
Since $11 \neq 6$,the number of $p-$ electrons in $Cl$ is not equal to the number of $d-$ electrons in $Fe^{2+}$.

(C) The formula for the angular momentum of an electron is given by $\sqrt{l(l+1)} \ \hbar$.
For a $d$ orbital,the azimuthal quantum number $l = 2$.
Substituting the value of $l$ into the formula:
Angular momentum $= \sqrt{2(2+1)} \ \hbar = \sqrt{2 \times 3} \ \hbar = \sqrt{6} \ \hbar$.

(A) The quantum numbers $n = 3$,$l = 1$,and $m_l = 0$ uniquely define a single orbital.
$n = 3$ indicates the third shell.
$l = 1$ indicates a $p$-subshell.
$m_l = 0$ specifies a particular orbital within the $p$-subshell (typically $3p_z$).
Since all three quantum numbers are fixed,they correspond to exactly $1$ orbital.

(B) The set of quantum numbers $n = 3, l = 1$ and $m = -1$ defines a specific orbital,which is the $3p_x$ or $3p_y$ orbital (depending on the convention for $m$).
According to the Pauli Exclusion Principle,any single orbital can hold a maximum of $2$ electrons with opposite spins.
Therefore,the maximum number of electrons associated with this specific set of quantum numbers is $2$.

(A) $n$ represents the main energy level and $l$ represents the subshell.
If $n = 4$ and $l = 3,$ the subshell is $4f.$
In an $f$ subshell,there are $2l + 1 = 2(3) + 1 = 7$ orbitals.
Since each orbital can accommodate a maximum of $2$ electrons,the maximum number of electrons in the $4f$ subshell is $7 \times 2 = 14$.

(C) The electronic configuration of rubidium atom $(Z = 37)$ is $[Kr] 5s^1$.
The valence electron is in the $5s$ orbital.
For the $5s$ orbital,the principal quantum number $(n)$ is $5$.
For an $s$-orbital,the azimuthal quantum number $(l)$ is $0$.
Since $l = 0$,the magnetic quantum number $(m_l)$ is $0$.
The spin quantum number $(m_s)$ can be either $+1/2$ or $-1/2$.
Thus,the set of quantum numbers is $(n = 5, l = 0, m_l = 0, m_s = +1/2)$.
Therefore,the correct option is $C$.

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

(B) The total number of atomic orbitals in a given energy level (shell) with principal quantum number $n$ is given by the formula $n^{2}$.
For the fourth energy level,$n = 4$.
Therefore,the number of atomic orbitals $= 4^{2} = 16$.

(A) According to the $(n + l)$ rule,electrons fill orbitals in order of increasing $(n + l)$ values.
For $n = 6$:
$ns$ corresponds to $6s$ $(n+l = 6+0 = 6)$
$(n-2)f$ corresponds to $4f$ $(n+l = 4+3 = 7)$
$(n-1)d$ corresponds to $5d$ $(n+l = 5+2 = 7)$
$np$ corresponds to $6p$ $(n+l = 6+1 = 7)$
Comparing the $(n+l)$ values and the $n$ values for orbitals with the same $(n+l)$,the filling order is $6s$ $\rightarrow 4f$ $\rightarrow 5d$ $\rightarrow 6p$.

(D) For a given azimuthal quantum number $l$,the number of orbitals in a subshell is given by $(2l + 1)$.
Since each orbital can accommodate a maximum of $2$ electrons,the total number of electrons in a subshell is $2 \times (2l + 1) = 4l + 2$.

(B) For an electron in an atom,the quantum numbers must satisfy the following rules:
$1$. $n$ 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 be $+1/2$ or $-1/2$.
In option $B$,$n = 3$ and $l = 2$. The allowed values for $m$ are $-2, -1, 0, +1, +2$. Since $m = -3$ is outside this range,this arrangement is not permissible.

(B) The rules for quantum numbers are: $n > 0$,$0 \le l < n$,$-l \le m \le +l$,and $s = \pm 1/2$.
$(i)$ $(3, 2, 1, +1/2)$ is possible $(n=3, l=2, m=1)$.
$(ii)$ $(2, 2, 1, +1/2)$ is impossible because $l$ must be less than $n$ $(l < n)$.
$(iii)$ $(4, 3, -2, -1/2)$ is possible $(n=4, l=3, m=-2)$.
$(iv)$ $(1, 0, -1, -1/2)$ is impossible because $m$ must be between $-l$ and $+l$ (here $l=0$,so $m$ must be $0$).
$(v)$ $(3, 2, 3, +1/2)$ is impossible because $m$ must be between $-l$ and $+l$ (here $l=2$,so $m$ can only be $-2, -1, 0, 1, 2$).
Thus,sets $(ii), (iv)$ and $(v)$ are not possible.

(D) The magnetic quantum number is denoted by $m_l$.
It provides information about the spatial orientation of the orbital with respect to the standard set of coordinate axes.
Explanation for incorrect options:
- The azimuthal quantum number $(l)$ determines the shape of the orbital.
- The spin quantum number $(m_s)$ denotes the orientation of the spin of the electron.
- The principal quantum number $(n)$ determines the size and,to a large extent,the energy of the orbital.
Therefore,options $(A)$,$(B)$,and $(C)$ are incorrect.
Hence,option $(D)$ is correct.

(C) According to the $(n+l)$ rule,the energy of an orbital increases as the sum of the principal quantum number $(n)$ and the azimuthal quantum number $(l)$ increases.
For option $A$: $n+l = 3+0 = 3$
For option $B$: $n+l = 3+1 = 4$
For option $C$: $n+l = 3+2 = 5$
For option $D$: $n+l = 4+0 = 4$
Since option $C$ has the highest $(n+l)$ value of $5$,it represents the highest energy state.