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

(B) The distance of an electron from the nucleus depends on the principal quantum number $n$.
In the configuration $4s^2 3d^{10}$,the outermost electrons are in the $n=4$ shell.
In the configuration $4s^0 3d^{10}$,the outermost electrons are in the $n=3$ shell.
Since $n=3 < n=4$,the electrons in the $3d^{10}$ configuration are closer to the nucleus compared to those in the $4s^2 3d^{10}$ configuration.

(B) Both $3d_{xy}$ and $3d_{x^2-y^2}$ orbitals have the same shape,consisting of $4$ lobes.
However,they differ in their spatial orientation.
The lobes of the $3d_{xy}$ orbital lie in the $xy$-plane between the $x$ and $y$ axes.
The lobes of the $3d_{x^2-y^2}$ orbital lie along the $x$ and $y$ axes.
Thus,they are rotated by $45^{\circ}$ relative to each other.

(A) For a $5f$ orbital,the principal quantum number $n = 5$ and the azimuthal quantum number $l = 3$.
The number of radial nodes is calculated using the formula: $n - l - 1 = 5 - 3 - 1 = 1$.
The number of angular nodes is equal to the azimuthal quantum number $l$,which is $3$.
Therefore,the number of radial nodes is $1$ and the number of angular nodes is $3$.

(C) According to the Pauli Exclusion Principle,no two electrons in an atom can have the same set of all four quantum numbers.
For Helium ($He$,atomic number $Z = 2$),the electronic configuration is $1s^2$.
The two electrons in the $1s$ orbital have the same principal quantum number $(n=1)$,azimuthal quantum number $(l=0)$,and magnetic quantum number $(m_l=0)$.
Therefore,they must have different spin quantum numbers $(m_s)$ to satisfy the principle.
Thus,the spins are $+1/2$ and $-1/2$.

(C) Degenerate orbitals are those that have the same energy level (same principal quantum number $n$ and same subshell $l$).
$(i)$ The orbitals $3d_{xy}, 3d_{z^2}, 3d_{xz}$ all have $n=3$ and $l=2$,so they are degenerate.
$(ii)$ The orbitals $4d_{xy}, 4d_{yz}, 4d_{z^2}$ all have $n=4$ and $l=2$,so they are degenerate.
Therefore,both sets are degenerate.

(A) The maximum number of electrons in an orbital subshell is given by the formula $2(2l + 1)$.
For a $d$ orbital,the azimuthal quantum number $l = 2$. Therefore,the maximum number of electrons is $2(2 \times 2 + 1) = 2(5) = 10$.
For an $f$ orbital,the azimuthal quantum number $l = 3$. Therefore,the maximum number of electrons is $2(2 \times 3 + 1) = 2(7) = 14$.
Thus,the maximum number of electrons in $d$ and $f$ orbitals are $10$ and $14$,respectively.

(C) For the $3d$ orbital,the principal quantum number $n = 3$.
For a $d$ subshell,the azimuthal quantum number $l = 2$.
Therefore,the value of $(n + l) = 3 + 2 = 5$.

(N/A) For the $4f$ orbital:
$n = 4$ (Principal quantum number).
$l = 3$ (Azimuthal quantum number for $f$-orbital).
$m_{l}$ can take $7$ values: $-3, -2, -1, 0, +1, +2, +3$.
$m_{s}$ can take values of $+\frac{1}{2}$ and $-\frac{1}{2}$.

(C) For any subshell,the number of orbitals is given by the formula $(2l + 1)$.
For an $f$ subshell,the azimuthal quantum number $l = 3$.
Therefore,the number of orbitals $= 2(3) + 1 = 7$.

(B) The $f$-subshell first appears in the $4^{th}$ energy level.
According to the $(n+l)$ rule,the lowest energy $f$-subshell is $4f$.
For $4f$,the principal quantum number $n = 4$.
For an $f$-subshell,the azimuthal quantum number $l = 3$.

$(a) H^-$: Total electrons = $1 + 1 = 2$. Electronic configuration: $1s^2$.
$(b) Na^+$: Total electrons = $11 - 1 = 10$. Electronic configuration: $1s^2 2s^2 2p^6$.
$(c) O^{2-}$: Total electrons = $8 + 2 = 10$. Electronic configuration: $1s^2 2s^2 2p^6$.
$(d) F^-$: Total electrons = $9 + 1 = 10$. Electronic configuration: $1s^2 2s^2 2p^6$.

(N/A) For the outermost configuration $3s^1$,the complete electronic configuration is $1s^2 2s^2 2p^6 3s^1$. The total number of electrons is $2+2+6+1 = 11$. Thus,the atomic number $Z = 11$.
$(b)$ For the outermost configuration $2p^3$,the complete electronic configuration is $1s^2 2s^2 2p^3$. The total number of electrons is $2+2+3 = 7$. Thus,the atomic number $Z = 7$.
$(c)$ For the outermost configuration $3p^5$,the complete electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^5$. The total number of electrons is $2+2+6+2+5 = 17$. Thus,the atomic number $Z = 17$.

(N/A) The electronic configuration is $[He] \, 2s^1$. The atomic number $Z = 2 + 1 = 3$. Thus,the element is Lithium $(Li)$.
$(b)$ The electronic configuration is $[Ne] \, 3s^2 \, 3p^3$. The atomic number $Z = 10 + 2 + 3 = 15$. Thus,the element is Phosphorus $(P)$.
$(c)$ The electronic configuration is $[Ar] \, 4s^2 \, 3d^1$. The atomic number $Z = 18 + 2 + 1 = 21$. Thus,the element is Scandium $(Sc)$.

(C) For any orbital,the azimuthal quantum number $l$ determines its type. For $g$ orbitals,$l = 4$.
The principal quantum number $n$ must be greater than $l$. Specifically,for a given $l$,the minimum value of $n$ is $n = l + 1$.
Substituting $l = 4$:
$n = 4 + 1 = 5$.
Thus,the minimum value of $n$ for $g$ orbitals is $5$.

(A) For $3d$ orbitals,the principal quantum number $n = 3$.
For $d$ subshell,the azimuthal quantum number $l = 2$.
The magnetic quantum number $m_l$ can take values from $-l$ to $+l$,including zero.
Therefore,for $l = 2$,the possible values of $m_l$ are $-2, -1, 0, +1, +2$.

(B) $(i)$ In a neutral atom,the number of electrons is equal to the number of protons. Therefore,the number of protons = $29$.
$(ii)$ The atomic number $Z$ is $29$. The electronic configuration is written by filling orbitals in increasing order of energy. For $Z = 29$ (Copper),the configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1$ because the fully filled $3d$ subshell provides extra stability.

(N/A) For $n = 3$,the possible values of the azimuthal quantum number $l$ range from $0$ to $(n - 1)$.
Thus,$l = 0, 1, 2$.
For each value of $l$,the magnetic quantum number $m_l$ ranges from $-l$ to $+l$ including $0$.
If $l = 0$,$m_l = 0$.
If $l = 1$,$m_l = -1, 0, +1$.
If $l = 2$,$m_l = -2, -1, 0, +1, +2$.
Total number of $m_l$ values $= 1 + 3 + 5 = 9$.

(A) For a $3d$ orbital,the principal quantum number $n = 3$.
For a $d$ subshell,the azimuthal quantum number $l = 2$.
The magnetic quantum number $m_{l}$ ranges from $-l$ to $+l$,including zero.
Therefore,for $l = 2$,the possible values for $m_{l}$ are $-2, -1, 0, +1, +2$.

(B) For an orbital to be possible,the principal quantum number $n$ and the azimuthal quantum number $l$ must satisfy the condition $l < n$.
For $1p$: $n=1, l=1$. Since $l$ is not less than $n$,$1p$ is not possible.
For $2s$: $n=2, l=0$. Since $l < n$,$2s$ is possible.
For $2p$: $n=2, l=1$. Since $l < n$,$2p$ is possible.
For $3f$: $n=3, l=3$. Since $l$ is not less than $n$,$3f$ is not possible.
Therefore,$2s$ and $2p$ are the possible orbitals.

(N/A) The orbital notation is given by $nl$,where $n$ is the principal quantum number and $l$ is the azimuthal quantum number ($l=0$ is $s$,$l=1$ is $p$,$l=2$ is $d$,$l=3$ is $f$).

Quantum Numbers	Orbital
$(a) n = 1, l = 0$	$1s$
$(b) n = 3, l = 1$	$3p$
$(c) n = 4, l = 2$	$4d$
$(d) n = 4, l = 3$	$4f$

(A, C, E) The following sets of quantum numbers are not possible:
$(a) n = 0$ is not possible because the principal quantum number $n$ must be a positive integer $(n = 1, 2, 3, \dots)$.
$(c) n = 1, l = 1$ is not possible because for a given $n$,the azimuthal quantum number $l$ can only range from $0$ to $n-1$. Thus,if $n = 1$,$l$ must be $0$.
$(e) n = 3, l = 3$ is not possible because $l$ must always be less than $n$ $(l < n)$. Therefore,$l$ can only take values $0, 1, 2$ when $n = 3$.

(A) For a given principal quantum number $n$,the total number of electrons is given by $2n^2$.
For $n = 4$,total electrons $= 2(4)^2 = 2 \times 16 = 32$.
In any shell,half of the electrons have $m_s = +\frac{1}{2}$ and half have $m_s = -\frac{1}{2}$.
Therefore,the number of electrons with $m_s = -\frac{1}{2}$ is $32 / 2 = 16$.
$(b)$ For $n = 3$ and $l = 0$,the orbital is $3s$.
$A$ single orbital can hold a maximum of $2$ electrons.
Therefore,the number of electrons is $2$.

(D) The correct matching is:
$(1)$ Principal quantum number $(n)$ determines the size of the orbital $(D)$.
$(2)$ Azimuthal quantum number $(l)$ determines the shape of the orbital $(A)$.
$(3)$ Magnetic quantum number $(m_l)$ determines the orientation of the orbital in space $(C)$.
$(4)$ Spin quantum number $(m_s)$ determines the spin of the electron $(B)$.
Therefore,the correct sequence is $(1-D, 2-A, 3-C, 4-B)$.

(A) $(1)$ Principal quantum number $(n)$ determines the energy and size of the orbital.
$(2)$ Azimuthal quantum number $(l)$ determines the shape of the orbital.
$(3)$ Magnetic quantum number $(m_l)$ determines the spatial orientation of the orbital.
$(4)$ Spin quantum number $(m_s)$ determines the spin of the electron.
Therefore,the correct match is $1-B, 2-D, 3-A, 4-C$.

(B) An electron in an atom is identified by its four quantum numbers: the principal quantum number $(n)$,the azimuthal quantum number $(l)$,the magnetic quantum number $(m_l)$,and the spin quantum number $(m_s)$.

(B) The distribution of electrons into different orbitals of an atom is called its electronic configuration.

(B) In atomic structure,a shell is defined as a set of orbitals that share the same principal quantum number $(n)$.
Each shell represents a specific energy level for electrons orbiting the nucleus.
As the value of $n$ increases,the shell is further from the nucleus and has higher energy.

(B) The electronic configuration of an atom describes the distribution of electrons in various atomic orbitals. These orbitals are grouped into subshells denoted as $s, p, d,$ and $f$.

(C) The position of an element in the periodic table is determined by its $Electronic \ configuration$,which describes the distribution of electrons in various orbitals. The $Principal \ quantum \ number$ $(n)$ determines the period,while the $subshell$ in which the last electron enters determines the block and group.

(N/A) For $n = 6$,the subshells available are $6s, 6p, 5d, 4f$. The $5g$ and $6g$ subshells are not filled in the ground state of known elements because their energy is significantly higher than the $7s$ orbital.

$n=6$ subshells	$6s, 6p, 5d, 4f$
Number of orbitals	$1 (6s) + 3 (6p) + 5 (5d) + 7 (4f) = 16$ orbitals
Total elements	$16 \times 2 = 32$ elements

(A) For $Z = 117$ (Tennessine):
The electronic configuration is $[Rn] 5f^{14} 6d^{10} 7s^2 7p^5$.
For $Z = 120$ (Unbinilium):
The electronic configuration is $[Og] 8s^2$,where $[Og]$ represents the noble gas Oganesson with $Z = 118$.

(A) The $2s$ electrons are more strongly attracted to the nucleus than the $2p$ electrons. This is because the $2s$ orbital is closer to the nucleus and has a higher probability density near the nucleus compared to the $2p$ orbital,which experiences greater shielding from the inner shell electrons.

(N/A) In an atom,electrons are filled according to the $(n+l)$ rule in order of increasing energy.
For $3d$: $n=3, l=2$,so $(n+l) = 3+2 = 5$.
For $4s$: $n=4, l=0$,so $(n+l) = 4+0 = 4$.
Since the $(n+l)$ value for $4s$ is lower than $3d$,the $4s$ orbital has lower energy and is filled first.
During ionization,electrons are removed from the orbital with the highest principal quantum number $(n)$ first. Since $4s$ has $n=4$ and $3d$ has $n=3$,the $4s$ electrons are further from the nucleus and are removed first.

(D) For a given principal quantum number $n=4$,the possible values of azimuthal quantum number $\ell$ are $0, 1, 2, 3$.
These correspond to the subshells $4s, 4p, 4d, 4f$ respectively.
The magnetic quantum number $m$ ranges from $-\ell$ to $+\ell$.
For $4s$ $(\ell=0)$: $m=0$.
For $4p$ $(\ell=1)$: $m=-1, 0, +1$.
For $4d$ $(\ell=2)$: $m=-2, -1, 0, +1, +2$.
For $4f$ $(\ell=3)$: $m=-3, -2, -1, 0, +1, +2, +3$.
The value $m=-2$ is present in the $4d$ and $4f$ subshells.
Therefore,the number of subshells associated with $n=4$ and $m=-2$ is $2$.

(B) Given $l$ ranges from $0$ to $(n+1)$.
For $n=1$,$l=0, 1, 2$ (orbitals: $1s, 1p, 1d$).
For $n=2$,$l=0, 1, 2, 3$ (orbitals: $2s, 2p, 2d, 2f$).
Applying the $(n+l)$ rule for energy order: $1s (1+0=1) < 1p (1+1=2) < 2s (2+0=2) < 1d (1+2=3) < 2p (2+1=3) < 3s (3+0=3) < 2d (2+2=4)$.
Electronic configuration for atomic number $9$: $1s^2 1p^6 2s^1$.
Since it has $1$ electron in the outermost $s$-orbital $(2s^1)$,it behaves like an alkali metal.
Thus,$9$ is the first alkali metal.

(B) Probability density is defined as $|\psi|^2$. Since it is a square of the wave function,it can never be negative.
Nodes are regions where the probability density is zero.
For a $3p$ orbital,the radial node is present where the radial wave function $R(r)$ becomes zero,which is also where the probability density $|R(r)|^2$ becomes zero,as shown in the provided graph.
Therefore,the correct statement is that it can be zero for $3p$ orbital.

(D) According to the $(n + \ell)$ rule,the energy of orbitals increases in the order of their $(n + \ell)$ values. For the $6^{th}$ period,the orbitals filled are $6s$ $(n+\ell = 6+0=6)$,$4f$ $(n+\ell = 4+3=7)$,$5d$ $(n+\ell = 5+2=7)$,and $6p$ $(n+\ell = 6+1=7)$.
Thus,the correct order of filling is $6s, 4f, 5d, 6p$.

(B) For any orbital,the number of angular nodes is given by the azimuthal quantum number $l$.
For an $s$-orbital,$l = 0$,so the number of angular nodes is $0$.
The number of radial nodes is given by the formula $n - l - 1$.
For the $3s$ orbital,the principal quantum number $n = 3$ and $l = 0$.
Therefore,the number of radial nodes $= 3 - 0 - 1 = 2$.
Thus,the $3s$ orbital has $0$ angular nodes and $2$ radial nodes.

(C) For a $4p$ electron,the principal quantum number $n=4$ and the azimuthal quantum number $l=1$.
The magnetic quantum number $m$ can take values from $-l$ to $+l$,which means for $l=1$,$m$ can be $-1, 0, +1$.
The value $m=2$ is not possible for $l=1$ because $m$ must satisfy the condition $|m| \leq l$.
Therefore,the set $n=4, l=1, m=2, m_{s}=+\frac{1}{2}$ is not possible.

(A) The magnetic quantum number $m$ for an $s$-orbital is always $0$.
For $Na$ $(Z=11)$,the electronic configuration is $1s^{2} 2s^{2} 2p^{6} 3s^{1}$. The last electron enters the $3s$ orbital,which has $l=0$,therefore $m=0$.
For $O$ $(Z=8)$,the configuration is $1s^{2} 2s^{2} 2p^{4}$. The last electron enters a $2p$ orbital $(l=1)$,where $m$ can be $-1, 0, +1$.
For $Cl$ $(Z=17)$,the configuration is $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{5}$. The last electron enters a $3p$ orbital $(l=1)$,where $m$ can be $-1, 0, +1$.
For $N$ $(Z=7)$,the configuration is $1s^{2} 2s^{2} 2p^{3}$. The last electron enters a $2p$ orbital $(l=1)$,where $m$ can be $-1, 0, +1$.
Thus,only $Na$ has its last electron in an $s$-orbital where $m$ is strictly $0$.

(B) For a given orbital,the magnetic quantum number $m_{\ell}$ ranges from $-\ell$ to $+\ell$.
Given $m_{\ell} = -3$,the azimuthal quantum number $\ell$ must be at least $3$.
Since $n = 4$ and $\ell = 3$,this corresponds to a $4f$ orbital.
The formula for the number of radial nodes is $n - \ell - 1$.
Substituting the values: $4 - 3 - 1 = 0$.

(B) The number of angular nodes is given by the azimuthal quantum number $l$.
Since there are no angular nodes,$l = 0$,which corresponds to an '$s$' orbital.
The number of radial nodes is given by the formula $n - l - 1$.
Given that the number of radial nodes is $2$,we have $n - 0 - 1 = 2$.
Solving for $n$,we get $n = 3$.
Therefore,the orbital is $3s$.

(A) For a given principal quantum number $n = 5$,the possible values of azimuthal quantum number $\ell$ are $0, 1, 2, 3, 4$.
The magnetic quantum number $m_{l}$ ranges from $-\ell$ to $+\ell$ for each $\ell$.
We need to find the number of orbitals where $m_{l} = +2$:
$1$. For $\ell = 0$ $(5s)$,$m_{l} = 0$.
$2$. For $\ell = 1$ $(5p)$,$m_{l} = \{-1, 0, +1\}$.
$3$. For $\ell = 2$ $(5d)$,$m_{l} = \{-2, -1, 0, +1, +2\}$. Here,$m_{l} = +2$ is possible ($1$ orbital).
$4$. For $\ell = 3$ $(5f)$,$m_{l} = \{-3, -2, -1, 0, +1, +2, +3\}$. Here,$m_{l} = +2$ is possible ($1$ orbital).
$5$. For $\ell = 4$ $(5g)$,$m_{l} = \{-4, -3, -2, -1, 0, +1, +2, +3, +4\}$. Here,$m_{l} = +2$ is possible ($1$ orbital).
Total number of orbitals with $m_{l} = +2$ is $1 + 1 + 1 = 3$.

(B) The electronic configuration of $Fe$ $(Z = 26)$ is $[Ar] 3d^6 4s^2$.
In the $3d$ subshell,there are $6$ electrons. According to Hund's rule,the distribution is:

$\uparrow \downarrow$

$\uparrow$

$\uparrow$

$\uparrow$

$\uparrow$

Number of unpaired electrons $(n) = 4$.
The spin-only magnetic moment $(\mu)$ is given by the formula $\mu = \sqrt{n(n + 2)} \,BM$.
$\mu = \sqrt{4(4 + 2)} = \sqrt{24} \,BM$.
Using $\sqrt{24} \approx 4.899 \,BM$.
$\mu = 48.99 \times 10^{-1} \,BM$.
Rounding off to the nearest integer,we get $49$.

(C) The number of radial nodes for an orbital is given by the formula: $\text{Number of radial nodes} = n - \ell - 1$.
For the $3s$ orbital,the principal quantum number $n = 3$ and the azimuthal quantum number $\ell = 0$.
Substituting these values: $\text{Number of radial nodes} = 3 - 0 - 1 = 2$.
$A$ radial distribution function plot for an orbital with $2$ radial nodes must show $2$ points where the probability density touches the $r$-axis (excluding $r=0$ and $r=\infty$).
Looking at the provided plots:
- Plot $(A)$ has $0$ radial nodes (corresponds to $1s$).
- Plot $(B)$ has $1$ radial node (corresponds to $2s$).
- Plot $(C)$ has $0$ radial nodes.
- Plot $(D)$ has $2$ radial nodes.
Therefore,the correct plot for the $3s$ orbital is $(D)$.

(D) The number of radial nodes is given by the formula $n-l-1$ and the number of angular nodes is given by $l$.
Given that radial nodes = $2$ and angular nodes = $2$,we have $l = 2$.
Substituting $l = 2$ into the radial node formula: $n-2-1 = 2$,which gives $n = 5$.
Since $l = 2$ corresponds to the $d$ orbital,the orbital is $5d$.

(D) The electronic configuration of $Ge$ $(Z=32)$ is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^2$.
We need to identify completely filled orbitals where the magnetic quantum number $m_l = 0$.
- $1s$ orbital $(m_l=0)$: $1$ orbital (completely filled)
- $2s$ orbital $(m_l=0)$: $1$ orbital (completely filled)
- $2p$ subshell $(m_l=-1, 0, +1)$: $1$ orbital with $m_l=0$ (completely filled)
- $3s$ orbital $(m_l=0)$: $1$ orbital (completely filled)
- $3p$ subshell $(m_l=-1, 0, +1)$: $1$ orbital with $m_l=0$ (completely filled)
- $4s$ orbital $(m_l=0)$: $1$ orbital (completely filled)
- $3d$ subshell $(m_l=-2, -1, 0, +1, +2)$: $1$ orbital with $m_l=0$ (completely filled)
- $4p$ subshell: Not completely filled.
Total count of completely filled orbitals with $m_l=0$ is $1+1+1+1+1+1+1 = 7$.
Thus,$x = 7$.

(B) The atomic number of $Zn$ is $30$. The electronic configuration of $Zn$ is $[Ar] 3d^{10} 4s^{2}$.
For $Zn^{+}$ ion,one electron is removed from the $4s$ orbital,so the configuration becomes $[Ar] 3d^{10} 4s^{1}$.
The outermost electron is in the $4s$ subshell.
For an $s$-orbital,the azimuthal quantum number $l = 0$.
The magnetic quantum number $m$ ranges from $-l$ to $+l$,so for $l = 0$,$m = 0$.

(A) The atomic number of $Ga$ is $31$. The electronic configuration of $Ga$ is $[Ar] 3d^{10} 4s^{2} 4p^{1}$.
For the $Ga^{+}$ ion,one electron is removed from the outermost $4p$ orbital. Thus,the electronic configuration of $Ga^{+}$ is $[Ar] 3d^{10} 4s^{2}$.
The valence electrons are in the $4s$ subshell.
The azimuthal quantum number $(l)$ for an $s$-orbital is $0$.

(D) The electronic configuration of Vanadium $(Z=23)$ is: $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{3} 4s^{2}$.
The p-orbitals present are $2p$ and $3p$.
Number of electrons in $2p$ orbital = $6$.
Number of electrons in $3p$ orbital = $6$.
Total number of electrons in p-orbitals = $6 + 6 = 12$.