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

(B) The $M$ shell corresponds to the $n=3$ principal energy level.
For an element to have $13$ electrons in the $M$ shell,the electronic configuration must include $3s^2, 3p^6, 3d^5$.
Considering the Aufbau principle and stability,the configuration is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^5, 4s^1$.
The total number of electrons is $2+2+6+2+6+5+1 = 24$.
The element with atomic number $24$ is chromium $(Cr)$.

(B) The atomic number of $Cu$ is $29$.
According to the Aufbau principle,the expected configuration is $[Ar] 3d^{9} 4s^{2}$.
However,fully filled $d$-orbitals are more stable than partially filled ones.
Therefore,one electron from the $4s$ orbital shifts to the $3d$ orbital to make it $3d^{10}$.
The correct electronic configuration is $[Ar] 3d^{10} 4s^{1}$.

(C) According to the $Pauli$ $Exclusion$ $Principle$,an orbital can accommodate a maximum of two electrons,and these electrons must have opposite spins.
The representation $\uparrow \uparrow$ shows two electrons with parallel spins in the same orbital,which violates this principle.

(A) For an electron in the $5d$ orbital,the principal quantum number $n = 5$ and the azimuthal quantum number $l = 2$ (since $s=0, p=1, d=2, f=3$).
The magnetic quantum number $m_l$ can take any integer value from $-l$ to $+l$,which is $-2, -1, 0, 1, 2$.
Both $(5, 2, 1)$ and $(5, 2, -1)$ are valid sets of quantum numbers for a $5d$ electron.
However,in multiple-choice questions of this type,if multiple options are mathematically valid,we select the one provided in the set. Since both $A$ and $D$ are valid,but $D$ is often cited in standard examples,we identify the valid range.

(B) The rules for quantum numbers are as follows:
$1$. Principal quantum number $n$ can be any positive integer $(1, 2, 3, ...)$.
$2$. Azimuthal quantum number $\ell$ ranges from $0$ to $n-1$.
$3$. Magnetic quantum number $m$ ranges from $-\ell$ to $+\ell$.
$4$. Spin quantum number $s$ can be $+\frac{1}{2}$ or $-\frac{1}{2}$.
Evaluating the options:
$A$: $(3, 0, -1, +\frac{1}{2})$ is incorrect because $m$ cannot be $-1$ when $\ell=0$.
$B$: $(4, 3, -2, -\frac{1}{2})$ is correct because for $n=4$,$\ell=3$ is allowed,and for $\ell=3$,$m=-2$ is allowed.
$C$: $(3, 1, -2, -\frac{1}{2})$ is incorrect because $m$ cannot be $-2$ when $\ell=1$.
$D$: $(4, 2, -3, +\frac{1}{2})$ is incorrect because $m$ cannot be $-3$ when $\ell=2$.

(A) The orbital angular momentum is given by the formula: $\text{Orbital angular momentum} = \sqrt{\ell(\ell+1)} \frac{h}{2 \pi}$.
For a $4f$ orbital,the azimuthal quantum number $\ell = 3$. Therefore,the orbital angular momentum is $\sqrt{3(3+1)} \frac{h}{2 \pi} = \sqrt{12} \frac{h}{2 \pi} = 2 \sqrt{3} \frac{h}{2 \pi}$.
For a $4s$ orbital,the azimuthal quantum number $\ell = 0$. Therefore,the orbital angular momentum is $\sqrt{0(0+1)} \frac{h}{2 \pi} = 0$.
The difference between the two is $2 \sqrt{3} \frac{h}{2 \pi} - 0 = 2 \sqrt{3} \frac{h}{2 \pi}$.

(D) Key point: Species with one or more unpaired electrons exhibit a non-zero magnetic moment.
Electronic configurations:
$Na^{+} (Z=11) = 1s^{2} 2s^{2} 2p^{6}$ (All electrons paired)
$Mg (Z=12) = 1s^{2} 2s^{2} 2p^{6} 3s^{2}$ (All electrons paired)
$F^{-} (Z=9) = 1s^{2} 2s^{2} 2p^{6}$ (All electrons paired)
$Ar^{+} (Z=18) = 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{5}$ (One unpaired electron in $3p$ orbital)
Since $Ar^{+}$ has an unpaired electron,it possesses a non-zero magnetic moment. Therefore,$(d)$ is the correct option.

(D) According to the Pauli Exclusion Principle,no two electrons in an atom can have the same set of all four quantum numbers $(n, l, m, s)$.
Therefore,all four quantum numbers are necessary to uniquely identify and describe the state of an electron in an atom.
$n$ (Principal quantum number) defines the shell and energy.
$l$ (Azimuthal quantum number) defines the subshell and orbital shape.
$m$ (Magnetic quantum number) defines the spatial orientation.
$s$ (Spin quantum number) defines the electron spin direction.

(A) The $yz$ plane is defined by the equation $x = 0$.
An orbital will have zero probability of finding an electron in the $yz$ plane if it has a nodal plane coinciding with the $yz$ plane.
The $p_x$ orbital has its electron density distributed along the $x$-axis,and its nodal plane is the $yz$ plane.
Therefore,the probability of finding an electron in the $yz$ plane for the $p_x$ orbital is zero.

(D) The quantum numbers $n=3$ and $\ell=2$ correspond to the $3d$ subshell.
Since the last electron is filled in the $3d$ subshell,we look for the maximum number of electrons possible when the $3d$ subshell is completely filled.
The electronic configuration for a completely filled $3d$ subshell is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10}$.
Counting the total number of electrons: $2+2+6+2+6+2+10 = 30$.
This corresponds to the element Zinc $(Zn)$ with atomic number $30$.

(C) For a given value of azimuthal quantum number $l$,the magnetic quantum number $m$ can have values ranging from $-l$ to $+l$ (including zero).
If $l=0$,then $m$ must be $0$.
In option $(c)$,$l=0$ but $m=-1$,which violates the rule $-l \leq m \leq l$.
Therefore,the arrangement $n=2, l=0, m=-1$ is impossible and absurd.

(C) For a given value of $n$,the azimuthal quantum number $l$ can have values from $0$ to $n-1$.
For a given value of $l$,the magnetic quantum number $m$ can have values from $-l$ to $+l$ including $0$.
In option $C$,$n=2$ and $l=0$.
For $l=0$,the only possible value for $m$ is $0$.
Therefore,$m=-1$ is not possible for $l=0$.

(B) The atomic number of $Ni$ is $28$.
Its electronic configuration is: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^8$.
In the $3d$ subshell,there are $8$ electrons.
According to Hund's rule,these $8$ electrons are filled in the $5$ $d$-orbitals as follows:
First,$5$ electrons are filled singly,and then the remaining $3$ electrons are paired.
This results in $3$ paired orbitals and $2$ singly occupied (unpaired) orbitals.
Therefore,the number of unpaired electrons in $Ni$ is $2$.

(A) According to the $(n+l)$ rule:
$1$. The orbital with a lower $(n+l)$ value has lower energy.
$2$. If the $(n+l)$ values are the same,the orbital with the lower $n$ value has lower energy.
Calculating $(n+l)$ values:
$(i)$ $n+l = 4+1 = 5$
$(ii)$ $n+l = 4+0 = 4$
$(iii)$ $n+l = 3+2 = 5$
$(iv)$ $n+l = 3+1 = 4$
Comparing the values:
For $(iv)$ and $(ii)$,both have $(n+l) = 4$. Since $(iv)$ has $n=3$ and $(ii)$ has $n=4$,$(iv) < (ii)$.
For $(iii)$ and $(i)$,both have $(n+l) = 5$. Since $(iii)$ has $n=3$ and $(i)$ has $n=4$,$(iii) < (i)$.
Combining these,the order of increasing energy is $(iv) < (ii) < (iii) < (i)$.

(B) The electronic configuration of $Cr$ $(Z=24)$ is $[Ar]_{18} 4s^1 3d^5$.
Electrons $1$ to $18$ are filled in the $Ar$ core.
The $19^{\text{th}}$ electron is the first electron to enter the $4s$ orbital.
For the $4s$ orbital: principal quantum number $n=4$,azimuthal quantum number $l=0$,magnetic quantum number $m_l=0$,and spin quantum number $m_s=+\frac{1}{2}$.
Thus,the set of quantum numbers is $(4, 0, 0, +\frac{1}{2})$.

(D) For $n=5$:
For each subshell $l$,the number of orbitals with $m_l=-1$ is $1$ (if $l \ge 1$).
Subshells present in $n=5$ are $l=0, 1, 2, 3, 4$.
$m_l=-1$ is possible for $l=1, 2, 3, 4$.
Each orbital can hold $2$ electrons.
Total electrons $= 4 \text{ orbitals} \times 2 \text{ electrons/orbital} = 8$ electrons.
$(B)$ For $n=3, l=2, m_l=-1, m_s=+\frac{1}{2}$:
This set of quantum numbers defines a specific orbital and a specific spin.
Only $1$ electron can have these exact quantum numbers.

(C) node is a region where the probability of finding an electron is zero. For a radial wave function,this corresponds to the point where the wave function $\psi(r)$ crosses the x-axis,i.e.,$\psi(r) = 0$.
In the radial wave function plot for a $2s$ orbital,the curve crosses the x-axis at a specific distance from the nucleus. Based on standard representations of this graph,point $C$ typically marks the intersection where $\psi(r) = 0$,representing the spherical node.

(D) The energy of atomic orbitals in multi-electron atoms is determined by the $(n+l)$ rule.
$1$. According to the $(n+l)$ rule,the orbital with the lower $(n+l)$ value has lower energy.
$2$. If the $(n+l)$ values are equal,the orbital with the lower value of $n$ has lower energy.
Calculating $(n+l)$ for each orbital:
$A: n=3, l=2 \implies n+l = 3+2 = 5$
$B: n=4, l=0 \implies n+l = 4+0 = 4$
$C: n=6, l=1 \implies n+l = 6+1 = 7$
$D: n=5, l=1 \implies n+l = 5+1 = 6$
$E: n=2, l=1 \implies n+l = 2+1 = 3$
Comparing the $(n+l)$ values: $3 (E) < 4 (B) < 5 (A) < 6 (D) < 7 (C)$.
Therefore,the order of increasing energy is $E < B < A < D < C$.

(D) The number of radial nodes is given by the formula $(n-l-1)$ and the number of angular nodes (nodal planes) is given by $l$.
$A$. For $2s$ orbital: $n=2, l=0$. Radial nodes $= 2-0-1 = 1$. Nodal planes $= l = 0$. This matches $IV$.
$B$. For $3s$ orbital: $n=3, l=0$. Radial nodes $= 3-0-1 = 2$. Nodal planes $= l = 0$. This matches $III$.
$C$. For $3p$ orbital: $n=3, l=1$. Radial nodes $= 3-1-1 = 1$. Nodal planes $= l = 1$. This matches $II$.
$D$. For $4d$ orbital: $n=4, l=2$. Radial nodes $= 4-2-1 = 1$. Nodal planes $= l = 2$. This matches $I$.
Therefore,the correct matching is $A-IV, B-III, C-II, D-I$.

(D) The electronic configuration of $Cr$ $(Z=24)$ is $[Ar] 3d^5 4s^1$.
The first $18$ electrons occupy the Argon core $(1s^2 2s^2 2p^6 3s^2 3p^6)$.
According to the Aufbau principle,the $19^{th}$ electron enters the $4s$ orbital.
For the $4s$ orbital,the principal quantum number $n=4$,the azimuthal quantum number $l=0$,the magnetic quantum number $m=0$,and the spin quantum number $s=+\frac{1}{2}$.

(B) is true: According to the $(n+l)$ rule,if two orbitals have the same $(n+l)$ value,the one with the lower $n$ value has lower energy.
$B$ is false: In a multi-electron atom,the energy of an orbital depends on the effective nuclear charge $(Z_{eff})$,not just the atomic number.
$C$ is true: The size of an orbital increases as the principal quantum number $n$ increases. Since $n=2$ for $2p_x$ and $n=3$ for $3p_x$,$2p_x$ is smaller.
$D$ is false: The number of radial nodes is given by the formula $(n-l-1)$.
For $5f$: $5-3-1 = 1$ radial node.
For $6s$: $6-0-1 = 5$ radial nodes.
For $4d$: $4-2-1 = 1$ radial node.
For $5p$: $5-1-1 = 3$ radial nodes.
For $5d$: $5-2-1 = 2$ radial nodes.
Since $5d$ has $2$ radial nodes,statement $D$ is incorrect.
Therefore,only statements $A$ and $C$ are true.

(A) The orbital is represented as $nl$,where $n$ is the principal quantum number and $l$ is the azimuthal quantum number.
The values of $l$ correspond to orbitals: $l=0$ $(s)$,$l=1$ $(p)$,$l=2$ $(d)$,$l=3$ $(f)$.
$(A)$ $n=2, l=1$ corresponds to $2p$.
$(B)$ $n=4, l=0$ corresponds to $4s$.
$(C)$ $n=5, l=3$ corresponds to $5f$.
$(D)$ $n=3, l=2$ corresponds to $3d$.
Therefore,the correct matching is $A-II, B-III, C-IV, D-I$.