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

(A) The angular momentum of an electron in an orbital is given by the formula: $\text{Angular momentum} = \sqrt{l(l+1)} \frac{h}{2\pi}$.
If the angular momentum is $0$,then $\sqrt{l(l+1)} \frac{h}{2\pi} = 0$,which implies $l(l+1) = 0$.
This means the azimuthal quantum number $l$ must be equal to $0$.
The $s$-orbital corresponds to $l = 0$.
Therefore,the $3s$ orbital is the correct answer.

(D) For an orbital,the principal quantum number $n$ represents the shell,and the azimuthal quantum number $l$ represents the subshell.
Given $n=4$ and $l=1$.
Since $l=1$ corresponds to the $p$ subshell,the orbital is $4p$.
Comparing with the options:
$4s \rightarrow n=4, l=0$
$3d \rightarrow n=3, l=2$
$4d \rightarrow n=4, l=2$
$4p \rightarrow n=4, l=1$
Therefore,the correct option is $D$.

(D) The symbol $n$ represents the principal quantum number (shell). The symbol $l$ represents the azimuthal quantum number (subshell),and $m_l$ represents the magnetic quantum number (orientation of the orbital).
Given $n=3$,$l=1$,and $m_l=-1$.
For a given set of quantum numbers $(n, l, m_l)$,only one specific orbital is defined.
Here,$n=3$ and $l=1$ corresponds to the $3p$ subshell,and $m_l=-1$ specifies one of the three $p$-orbitals (e.g.,$p_x$ or $p_y$).
Thus,only $1$ orbital is possible for this specific set of quantum numbers.
Therefore,option $(D)$ is the correct answer.

(C) The electronic configuration of xenon $(Xe)$ with atomic number $54$ is: $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^{10}, 4s^2, 4p^6, 4d^{10}, 5s^2, 5p^6$.
Counting the orbitals:
$1s$ ($1$ orbital),$2s$ ($1$ orbital),$2p$ ($3$ orbitals),$3s$ ($1$ orbital),$3p$ ($3$ orbitals),$3d$ ($5$ orbitals),$4s$ ($1$ orbital),$4p$ ($3$ orbitals),$4d$ ($5$ orbitals),$5s$ ($1$ orbital),$5p$ ($3$ orbitals).
Total number of orbitals = $1 + 1 + 3 + 1 + 3 + 5 + 1 + 3 + 5 + 1 + 3 = 27$.

(C) For $n=4$,total orbitals = $n^2 = 16$. Each orbital can have one electron with $m_s=+\frac{1}{2}$. Thus,$16$ electrons are possible. (Correct)
b) For $n=5$,subshells are $s, p, d, f, g$ $(l=0, 1, 2, 3, 4)$,so there are $5$ subshells. (Incorrect)
c) For $n=2$,$l$ can be $0, 1$. If $l=1$,$m_l$ can be $-1, 0, +1$. Thus,$n=2, l=1, m_l=0, m_s=-\frac{1}{2}$ is valid. (Correct)
d) Radial nodes = $n-l-1$. For $3s$,$n=3, l=0$. Nodes = $3-0-1 = 2$. (Correct)
Therefore,statements $a, c,$ and $d$ are correct.

(C) For a $3d$-orbital,the principal quantum number $n=3$ and the azimuthal quantum number $l=2$.
For $l=2$,the magnetic quantum number $m$ can take any integer value from $-l$ to $+l$,which means $m \in \{-2, -1, 0, +1, +2\}$.
The spin quantum number $s$ can be either $+\frac{1}{2}$ or $-\frac{1}{2}$.
Comparing these conditions with the given options,option $C$ $(n=3, l=2, m=-2, s=+\frac{1}{2})$ satisfies all the criteria.

(D) The electronic configuration of an element with atomic number $(Z) = 25$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^5$.
For $(n+l) = 3$:
- $2p$ orbital $(n=2, l=1)$: $(n+l) = 2+1 = 3$. Number of electrons = $6$.
- $3s$ orbital $(n=3, l=0)$: $(n+l) = 3+0 = 3$. Number of electrons = $2$.
So,$x = 6 + 2 = 8$.
For $(n+l) = 4$:
- $3p$ orbital $(n=3, l=1)$: $(n+l) = 3+1 = 4$. Number of electrons = $6$.
- $4s$ orbital $(n=4, l=0)$: $(n+l) = 4+0 = 4$. Number of electrons = $2$.
So,$y = 6 + 2 = 8$.
The value of $(x+y) = 8 + 8 = 16$.

(C) According to Hund's rule of maximum multiplicity,pairing of electrons in the orbitals of a particular subshell does not take place until each orbital of that subshell is singly occupied.
In option $(C)$,the $p$-subshell has one orbital paired while the second orbital has only one electron and the third is empty. This violates Hund's rule because the electron should have occupied the third orbital singly before pairing occurred in the first orbital.

(B) The total number of electron exchanges is given by the formula $\frac{n(n-1)}{2}$,where $n$ is the number of electrons with the same spin in the same subshell.
For a $d^5$ configuration,all $5$ electrons have the same spin (parallel spin) in $5$ different orbitals.
Substituting $n = 5$ into the formula:
$\text{Total exchanges} = \frac{5(5-1)}{2} = \frac{5 \times 4}{2} = \frac{20}{2} = 10$.

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

(C) For $n=1$,the orbital is $1s$ ($1$ orbital).
For $n=2$,the orbitals are $2s$ and $2p$ ($1+3=4$ orbitals).
For $n=3$,the orbitals are $3s$,$3p$,and $3d$ ($1+3+5=9$ orbitals).
The total number of orbitals for $n$ up to $3$ is $1 + 4 + 9 = 14$.
Each orbital can accommodate exactly one electron with a spin quantum number $m_s = -\frac{1}{2}$.
Therefore,the total number of electrons with $m_s = -\frac{1}{2}$ is $14$.
Hence,the correct option is $C$.

(B) For a given principal quantum number $n=4$,the possible subshells are $4s$,$4p$,$4d$,and $4f$.
Each subshell contains at least one orbital with $m_l=0$ (specifically the $ns$,$np_z$,$nd_{z^2}$,and $nf_{z^3}$ orbitals).
There are $4$ such orbitals in the $n=4$ shell that have $m_l=0$.
Since each orbital can hold only one electron with a specific spin $m_s=\frac{1}{2}$,the total number of electrons is $4 \times 1 = 4$.

(A) The ratio of $s$-electrons to $p$-electrons is determined by the electronic configuration of each species.
For $K^{+}$ $(Z=19)$: $1s^2 2s^2 2p^6 3s^2 3p^6$. Total $s$-electrons = $2+2+2 = 6$. Total $p$-electrons = $6+6 = 12$. Ratio = $6:12 = 1:2$.
For $Cr^{3+}$ $(Z=24)$: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^3$. Total $s$-electrons = $2+2+2 = 6$. Total $p$-electrons = $6+6 = 12$. Ratio = $6:12 = 1:2$.
Since both $K^{+}$ and $Cr^{3+}$ have the same ratio of $1:2$,the correct option is $A$.

(B) For $P$ $(Z=15)$: $1s^2 2s^2 2p^6 3s^2 3p^3$. Total $s$-electrons = $2+2+2 = 6$. Total $p$-electrons = $6+3 = 9$. Ratio = $6/9 = 2/3$.
For $Ca$ $(Z=20)$: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$. Total $s$-electrons = $2+2+2+2 = 8$. Total $p$-electrons = $6+6 = 12$. Ratio = $8/12 = 2/3$.
Thus,the pair is $P$ and $Ca$.

(B) Given,$E = \left( \frac{-R_{H}}{n^2} \right) = \left( \frac{-R_{H}}{16} \right)$.
Comparing the denominators,we get $n^2 = 16$,which implies $n = 4$.
For a hydrogen atom,the energy depends only on the principal quantum number $n$.
The degeneracy of a level with principal quantum number $n$ is given by the formula $n^2$.
For $n = 4$,the degeneracy is $4^2 = 16$.
This corresponds to the $16$ orbitals present in the $n = 4$ shell: $4s$ ($1$ orbital),$4p$ ($3$ orbitals),$4d$ ($5$ orbitals),and $4f$ ($7$ orbitals),totaling $1 + 3 + 5 + 7 = 16$.

(C) The wave function $\psi$ itself does not have any direct physical meaning. It is a mathematical function that describes the state of an electron in an atom. The physical significance of $\psi$ is that the square of its magnitude,$|\psi|^2$,represents the probability density of finding an electron at a particular point in space. Therefore,the statement that the total information about an electron in an atom is stored in its $\psi$ is incorrect because $\psi$ is just a mathematical amplitude,whereas the physical probability is given by $|\psi|^2$.

(A) The electronic configuration of $Sr$ $(Z=38)$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^{10}, 4p^6, 5s^2$.
Electrons with $l=0$ (s-orbitals) are: $1s^2, 2s^2, 3s^2, 4s^2, 5s^2$. Total electrons $x = 2+2+2+2+2 = 10$.
Electrons with $l=2$ (d-orbitals) are: $3d^{10}$. Total electrons $y = 10$.
Therefore,$(x-y) = 10 - 10 = 0$.

(C) The number of radial nodes in an orbital is given by the formula: $\text{Radial nodes} = n - l - 1$.
Angular nodes are given by: $\text{Angular nodes} = l$.
The total number of nodes is the sum of radial and angular nodes: $\text{Total nodes} = (n - l - 1) + l = n - 1$.

(C) The rules for quantum numbers are:
$1$. The principal quantum number '$n$' can be any positive integer $(1, 2, 3, \dots)$.
$2$. The azimuthal quantum number '$l$' can have values from $0$ to $(n-1)$.
$3$. The magnetic quantum number '$m$' can have values from $-l$ to $+l$ including $0$.
$4$. The spin quantum number '$s$' can be $\pm\frac{1}{2}$.
In option $C$,$n=3$,so the maximum value of '$l$' is $(n-1) = 2$.
Since the given value is $l=3$,which is not possible for $n=3$,this set is impossible.

(C) The atomic number $Z = 24$ corresponds to Chromium $(Cr)$.
The electronic configuration of $Cr$ is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$.
For $(n+\ell) = 3$: The orbitals are $2p$ $(n=2, \ell=1)$ and $3s$ $(n=3, \ell=0)$. Total electrons $= 6 + 2 = 8$.
For $(n+\ell) = 4$: The orbitals are $3p$ $(n=3, \ell=1)$ and $4s$ $(n=4, \ell=0)$. Total electrons $= 6 + 1 = 7$.
For $(n+\ell) = 5$: The orbital is $3d$ $(n=3, \ell=2)$. Total electrons $= 5$.

(D) The orbital angular momentum $(L)$ of an electron is given by the formula: $L = \sqrt{l(l+1)} \hbar$,where $l$ is the azimuthal quantum number and $\hbar = \frac{h}{2\pi}$.
For a $d$ orbital,the value of the azimuthal quantum number $l$ is $2$.
Substituting $l = 2$ into the formula:
$L = \sqrt{2(2+1)} \hbar$
$L = \sqrt{2(3)} \hbar$
$L = \sqrt{6} \hbar$
Therefore,the orbital angular momentum of an electron in a $d$ orbital is $\sqrt{6} \hbar$.

(D) For a given energy level $n$,the total number of orbitals is given by $n^2$.
For $n=4$,the total number of orbitals $= 4^2 = 16$.
Each orbital can hold a maximum of $2$ electrons with opposite spins,i.e.,$+\frac{1}{2}$ and $-\frac{1}{2}$.
Since there are $16$ orbitals,the total number of electrons is $16 \times 2 = 32$.
Out of these $32$ electrons,exactly half will have a spin value of $+\frac{1}{2}$.
Therefore,the number of electrons with spin $+\frac{1}{2} = \frac{32}{2} = 16$.
Thus,the values are $16$ and $16$.

(D) The orbital angular momentum of an electron is given by the formula: $\sqrt{l(l+1)} \frac{h}{2 \pi}$.
For a $d$-orbital,the azimuthal quantum number $(l)$ is $2$.
Substituting the value of $l$ into the formula:
Orbital angular momentum $= \sqrt{2(2+1)} \frac{h}{2 \pi} = \sqrt{2(3)} \frac{h}{2 \pi} = \sqrt{6} \frac{h}{2 \pi}$.
Thus,the correct option is $D$.

(B) The number of angular nodes is given by the azimuthal quantum number $l$. Here,angular nodes $= 1$,so $l = 1$,which corresponds to a $p$-subshell.
The number of radial nodes is given by the formula $n - l - 1$,where $n$ is the principal quantum number.
Given radial nodes $= 4$ and $l = 1$,we have $n - 1 - 1 = 4$,which gives $n = 6$.
Therefore,the orbital is $6p$ (e.g.,$6p_{z}$).

(B) The number of angular nodes is given by the azimuthal quantum number,$l$. Given that the orbital has one angular node,$l = 1$,which corresponds to a $p$-orbital.
The number of maxima in the radial probability distribution curve is given by the formula $n - l$.
Given that the number of maxima is $3$,we have $n - l = 3$.
Substituting $l = 1$ into the equation: $n - 1 = 3$,which gives $n = 4$.
Therefore,the orbital is $4p$.

(B) The number of radial nodes is calculated using the formula: $\text{Radial nodes} = n - l - 1$.
For the $3s$ orbital,$n = 3$ and $l = 0$. Thus,$\text{Radial nodes} = 3 - 0 - 1 = 2$.
For the $2p$ orbital,$n = 2$ and $l = 1$. Thus,$\text{Radial nodes} = 2 - 1 - 1 = 0$.
Therefore,the number of radial nodes in $3s$ and $2p$ orbitals are $2$ and $0$ respectively.

(A) The energy of an orbital is determined by the $(n+l)$ rule. According to this rule,the orbital with a higher $(n+l)$ value has higher energy.
If the $(n+l)$ values are the same,the orbital with the higher value of $n$ has higher energy.
Calculating $(n+l)$ values for each:
$A: n=5, l=2 \implies (n+l) = 5+2 = 7$
$B: n=5, l=0 \implies (n+l) = 5+0 = 5$
$C: n=4, l=3 \implies (n+l) = 4+3 = 7$
$D: n=4, l=1 \implies (n+l) = 4+1 = 5$
Comparing the values:
For $A$ and $C$,$(n+l) = 7$. Since $A$ has a higher $n$ $(5 > 4)$,$A > C$.
For $B$ and $D$,$(n+l) = 5$. Since $B$ has a higher $n$ $(5 > 4)$,$B > D$.
Therefore,the decreasing order of energy is $A > C > B > D$.

(A) The subshell is defined by the azimuthal quantum number $l$. For $l=2$,the subshell is the $d$-subshell.
The number of orbitals in a subshell is given by $(2l+1)$.
For $l=2$,the number of orbitals $= 2(2)+1 = 5$.
Since each orbital can hold a maximum of $2$ electrons,the maximum number of electrons in the $d$-subshell is $5 \times 2 = 10$.

(C) The rules for quantum numbers are:
$1$. $n = 1, 2, 3, \dots$
$2$. $l = 0$ to $(n-1)$
$3$. $m_l = -l$ to $+l$
$4$. $m_s = \pm \frac{1}{2}$
Check each option:
$(A)$ For $n=1, l=0$,the only possible value for $m_l$ is $0$. Thus,$m_l=+1$ is incorrect.
$(B)$ For $n=1$,$l$ can only be $0$. Thus,$l=1$ is incorrect.
$(C)$ For $n=3$,$l$ can be $0, 1, 2$. If $l=1$,$m_l$ can be $-1, 0, +1$. Since $m_l=0$ and $m_s=+\frac{1}{2}$ are valid,this set is possible.
$(D)$ For $n=3$,$l$ can only be $0, 1, 2$. Thus,$l=3$ is incorrect.
Therefore,option $(C)$ is correct.

(C) The maximum number of electrons in a subshell is given by the formula $2(2l + 1)$.
For a subshell with $n=4$ and $l=3$,the subshell is $4f$.
Substituting $l=3$ into the formula:
$= 2(2 \times 3 + 1) = 2(6 + 1) = 2 \times 7 = 14$ electrons.

(C) $1$. In $d_{x^2-y^2}$ orbital,electrons are present along the $x$ and $y$ axes,so the electron density in the $XY$ plane is not zero.
$2$. According to the $(n+l)$ rule,the energy increases as the sum of $(n+l)$ increases.
For $3p$,$(n+l) = 3+1 = 4$.
For $2p$,$(n+l) = 2+1 = 3$.
Thus,$3p$ has higher energy than $2p$.
$3$. The number of angular nodes is equal to the azimuthal quantum number $l$.
For $p$-orbitals,$l = 1$,so $3p_z$ has one angular node.
$4$. The number of radial nodes is given by $(n-l-1)$.
For $4f$,$n = 4$ and $l = 3$,so radial nodes $= 4-3-1 = 0$.

(B) The formula for calculating the number of radial nodes is given by: $\text{Radial nodes} = n - l - 1$.
For the $3p$ orbital,the principal quantum number $n = 3$ and the azimuthal quantum number $l = 1$.
Substituting these values into the formula: $\text{Radial nodes} = 3 - 1 - 1 = 1$.

(A) For $4d$ orbital,the principal quantum number $n = 4$ and the azimuthal quantum number $l = 2$.
Number of angular nodes $= l = 2$.
Number of radial nodes $= n - l - 1 = 4 - 2 - 1 = 1$.
Therefore,the number of angular and radial nodes are $2$ and $1$ respectively.

(A) The $Spin$ quantum number $(s)$ explains the fine structure of spectral lines,such as the doublets observed in the spectra of $H$ and alkali metals,and the doublets and triplets observed in the spectra of alkaline earth metals,due to the electron spin angular momentum.

(C) For the $N$ shell,the principal quantum number $n = 4$.
Number of sub-levels (sub-shells) is equal to $n$,so there are $4$ sub-levels.
Number of orbitals is given by $n^2 = 4^2 = 16$.
Number of electrons is given by $2n^2 = 2 \times 4^2 = 32$.
Thus,the values are $4, 16, 32$.

(C) In a hydrogen atom,which is a single-electron system,the energy of orbitals depends only on the principal quantum number $(n)$. Therefore,$2s$ and $2p$ orbitals have the same energy.
In multi-electron atoms like $He$,the energy of orbitals depends on both the principal quantum number $(n)$ and the azimuthal quantum number $(l)$. Due to the shielding effect,the $2s$ orbital has lower energy than the $2p$ orbital.
Thus,statement $I$ is correct,and statement $II$ is incorrect.

(C) The number of elements in a period is equal to the number of electrons that can be filled in the orbitals being filled for that period.
For the $6^{th}$ period,the orbitals being filled are $6s$,$4f$,$5d$,and $6p$.
The maximum number of electrons that can be accommodated in these orbitals is:
$6s: 2$ electrons
$4f: 14$ electrons
$5d: 10$ electrons
$6p: 6$ electrons
Total number of elements = $2 + 14 + 10 + 6 = 32$.
Thus,the $6^{th}$ period contains $32$ elements,ranging from atomic number $(Z) = 55$ to $(Z) = 86$.

(B) According to the Aufbau principle,electrons are filled in orbitals in increasing order of their energy,which is determined by the $(n+l)$ rule.
For $Ti$ $(Z=22)$,the electronic configuration is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^2$.
The order of filling of orbitals is $1s, 2s, 2p, 3s, 3p, 4s, 3d$.

(A) The given electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1$.
Summing the electrons: $2+2+6+2+6+10+1 = 29$.
The element with atomic number $29$ is Copper $(Cu)$.
This configuration represents the stable ground state of $Cu$ where the $3d$ subshell is completely filled.

(A) The extra stability of half-filled and fully filled electronic configurations can be explained in terms of symmetry and exchange energy.
All the orbitals of the same subshell that are either completely filled or half-filled have a more symmetrical distribution of electrons.
Consequently,the shielding of electrons is relatively small,and the electrons are more strongly attracted by the nucleus.
Furthermore,the number of possible exchanges is maximum in these configurations,which leads to higher exchange energy and,therefore,greater stability of the atom.
Thus,both $(A)$ and $(R)$ are correct,and $(R)$ is the correct explanation of $(A)$.

(B) The electronic configuration of an element with atomic number $Z$ in $+2$ oxidation state is equivalent to an atom with $Z-2$ electrons.
Given $Z-2 = 24$,the atomic number $Z = 26$. The element is $Fe$.
The ground state electronic configuration of $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$.
The configuration of $Fe^{2+}$ is $[Ar] 3d^6$.
In the $3d^6$ configuration,the electrons are distributed in $5$ orbitals of $d$-subshell according to Hund's rule: $d_{xy}^2, d_{yz}^1, d_{zx}^1, d_{x^2-y^2}^1, d_{z^2}^1$.
Thus,there are $4$ unpaired electrons.

(D) The order of increasing energy is determined by the $(n+l)$ rule.
If two orbitals have the same value of $(n+l)$,the orbital with the lower value of $n$ has lower energy.
$(i)$ For $n=4, l=1$,$(n+l) = 4+1 = 5$.
(ii) For $n=4, l=0$,$(n+l) = 4+0 = 4$.
(iii) For $n=3, l=2$,$(n+l) = 3+2 = 5$.
(iv) For $n=3, l=1$,$(n+l) = 3+1 = 4$.
Comparing the values: (iv) and (ii) have $(n+l) = 4$. Since (iv) has lower $n$,$(iv) < (ii)$.
$(i)$ and (iii) have $(n+l) = 5$. Since (iii) has lower $n$,$(iii) < (i)$.
Thus,the increasing order of energy is $(iv) < (ii) < (iii) < (i)$.

(C) The electronic configurations of the elements are as follows:
For $X$ $(Z=19)$: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1$. The $M$-shell $(n=3)$ contains $2+6 = 8$ electrons.
For $Y$ $(Z=21)$: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^1 4s^2$. The $M$-shell $(n=3)$ contains $2+6+1 = 9$ electrons.
For $Z$ $(Z=25)$: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^2$. The $M$-shell $(n=3)$ contains $2+6+5 = 13$ electrons.
Comparing the number of electrons in the $M$-shell: $Z (13) > Y (9) > X (8)$.
Thus,the correct order is $Z > Y > X$.

(A) The electronic configurations of the given elements are as follows:
$K (Z=19): 1s^2, 2s^2 2p^6, 3s^2 3p^6, 4s^1$. The $M$ shell $(n=3)$ has $2+6 = 8$ electrons.
$Mn (Z=25): 1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^5, 4s^2$. The $M$ shell $(n=3)$ has $2+6+5 = 13$ electrons.
$Ni (Z=28): 1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^8, 4s^2$. The $M$ shell $(n=3)$ has $2+6+8 = 16$ electrons.
$Sc (Z=21): 1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^1, 4s^2$. The $M$ shell $(n=3)$ has $2+6+1 = 9$ electrons.
Comparing the number of electrons in the $M$ shell: $K (8) < Sc (9) < Mn (13) < Ni (16)$.
Thus,$K$ has the least number of electrons in its $M$ shell.