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

(D) For an electron,the magnetic quantum number $m$ is constrained by the azimuthal quantum number $l$ such that $|m| \leq l$.
Given $m = 2$,the possible values for $l$ must satisfy $l \geq 2$.
Since the principal quantum number is $n = 4$,the possible values for $l$ are $0, 1, 2, 3$.
Combining these constraints,$l$ can only be $2$ or $3$.
Therefore,the statement that $l$ can be $0, 1, 2, 3$ is incorrect because $l$ cannot be $0$ or $1$ when $m = 2$.

(A) For any principal quantum number $n$,the azimuthal quantum number $l$ can have values from $0$ to $n-1$.
For $n = 1$,the only possible value for $l$ is $0$.
Therefore,the set $n = 1, l = 1$ is not possible because $l$ cannot be equal to or greater than $n$.
Thus,option $A$ is the correct answer.

(A) The atomic number of $Fe$ is $26$. The ground state electronic configuration of $Fe$ is $[Ar] 3d^6 4s^2$.
To form $Fe^{3+}$,we remove $3$ electrons: $2$ from the $4s$ orbital and $1$ from the $3d$ orbital.
Thus,the configuration becomes $[Ar] 3d^5$,which is $1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^5$.

(D) The atomic number of $Na$ is $11$.
The electronic configuration of $Na$ is $1s^2 2s^2 2p^6 3s^1$.
The last electron enters the $3s$ orbital.
For an $s$-orbital,the azimuthal quantum number $(l)$ is $0$.
Therefore,the correct option is $(D)$.

(C) The total number of nodes is given by $n-1$.
The number of spherical (radial) nodes is given by $n-l-1$.
The number of non-spherical (angular) nodes is given by $l$.
For a $3p$ orbital,the principal quantum number $n = 3$ and the azimuthal quantum number $l = 1$.
Therefore,the number of spherical nodes $= n-l-1 = 3-1-1 = 1$.
The number of non-spherical nodes $= l = 1$.
Thus,a $3p$ orbital has one spherical and one non-spherical node.

(B) For the $4p$ sub-shell,the principal quantum number $n = 4$ and the azimuthal quantum number $l = 1$ (since $p$-orbital corresponds to $l = 1$).
All electrons in this sub-shell must have $n = 4$ and $l = 1$.
The magnetic quantum number $m$ can take values $-1, 0, +1$,and the spin quantum number $s$ can be $\pm \frac{1}{2}$.
Among the given options,$l = 1$ is the only quantum number that is common to all electrons in the $4p$ sub-shell.

(D) The atomic number of the element is $27$. The electronic configuration is determined using the Aufbau principle,which states that electrons fill orbitals in order of increasing energy.
For $Z = 27$,the configuration is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^7$.
In terms of orbital notation,the $3d$ subshell has $5$ orbitals,where $7$ electrons are filled as $(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow)(\uparrow)(\uparrow)$,and the $4s$ orbital is filled as $(\uparrow\downarrow)$.
Thus,option $D$ is correct.

(A) For a given principal quantum number $n$,the azimuthal quantum number $l$ can take values from $0$ to $n-1$.
For $n = 3$,$l$ can be $0, 1, 2$.
For each value of $l$,the magnetic quantum number $m$ can take values from $-l$ to $+l$ including zero.
If $l = 0$,$m = 0$.
If $l = 1$,$m = -1, 0, +1$.
If $l = 2$,$m = -2, -1, 0, +1, +2$.
Therefore,option $A$ correctly represents these values.

(C) The principal quantum number $(n)$ determines the size and energy of the shell.
The azimuthal quantum number $(l)$ determines the shape of the orbital.
The magnetic quantum number $(m_l)$ describes the spatial orientation of the orbital in space.
The spin quantum number $(m_s)$ describes the spin of the electron.

(D) Degenerate orbitals are those that have the same energy level.
In a hydrogen-like atom or in the absence of an external field,orbitals belonging to the same subshell (same $n$ and $l$ values) are degenerate.
For the set $2p_x, 2p_y, 2p_z$,all three orbitals have the same principal quantum number $(n = 2)$ and the same azimuthal quantum number $(l = 1)$,meaning they belong to the same $2p$ subshell.
Therefore,they have the same energy and are degenerate.

(A) The given quantum numbers $n = 3$,$l = 0$,$m = 0$,$s = -1/2$ correspond to an electron in the $3s$ orbital.
For $Mg$ $(Z = 12)$,the electronic configuration is $1s^2 2s^2 2p^6 3s^2$.
The two electrons in the $3s$ orbital have quantum numbers $(3, 0, 0, +1/2)$ and $(3, 0, 0, -1/2)$.
Thus,the set belongs to $Mg$.

(C) For a given principal quantum number $n$,the number of sub-shells is equal to $n$.
Given $n = 3$,the number of sub-shells is $3$ (which are $3s, 3p, 3d$).
The total number of orbitals in a shell is given by the formula $n^2$.
For $n = 3$,the number of orbitals $= 3^2 = 9$.
Therefore,the number of sub-shells is $3$ and the number of orbitals is $9$.

(D) The atomic number of $Cu$ is $29$. The electronic configuration of neutral $Cu$ is $[Ar] \ 3d^{10} 4s^1$.
To form $Cu^{2+}$,two electrons are removed,first from the $4s$ orbital and then from the $3d$ orbital.
Thus,$Cu^{2+} = [Ar] \ 3d^9 4s^0$ or simply $[Ar] \ 3d^9$.

(A) The atomic number of Titanium $(Ti)$ is $Z = 22$.
According to the Aufbau principle,electrons are filled in orbitals in increasing order of energy: $1s, 2s, 2p, 3s, 3p, 4s, 3d$.
Filling $22$ electrons: $1s^2, 2s^2, 2p^6, 3s^2, 3p^6$ accounts for $18$ electrons.
The remaining $4$ electrons are filled as $4s^2$ and $3d^2$.
Thus,the configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^2$.

(D) The atomic number of iron $(Fe)$ is $26$.
Following the Aufbau principle,the electrons are filled in the order of increasing energy: $1s, 2s, 2p, 3s, 3p, 4s, 3d$.
Filling $26$ electrons: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6$.
Therefore,the correct configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6$.

(D) The energy of an orbital is determined by the $(n + l)$ rule.
For option $A$: $n = 4, l = 0$,so $(n + l) = 4 + 0 = 4$.
For option $B$: $n = 3, l = 0$,so $(n + l) = 3 + 0 = 3$.
For option $C$: $n = 3, l = 1$,so $(n + l) = 3 + 1 = 4$.
For option $D$: $n = 3, l = 2$,so $(n + l) = 3 + 2 = 5$.
Comparing the $(n + l)$ values,option $D$ has the highest value of $5$,which corresponds to the $3d$ orbital. Therefore,it has the highest energy.

(B) The $l$ (Azimuthal) quantum number determines the shape of the orbital or subshell.
$n$ (Principal) determines the size and energy of the orbital.
$m_l$ (Magnetic) determines the orientation of the orbital in space.
$m_s$ (Spin) determines the spin of the electron.

(C) The number of orbitals in a given energy level $n$ is given by the formula $n^2$.
For the $n = 2$ energy level,the number of orbitals is $2^2 = 4$.
These orbitals are one $2s$ orbital and three $2p$ orbitals $(2p_x, 2p_y, 2p_z)$.

(B) The ground state electronic configuration follows the Aufbau principle,where electrons fill orbitals in increasing order of energy.
Option $B$ represents the configuration of Rubidium ($Rb$,$Z=37$),which is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 5s^1$.
This configuration correctly follows the order of energy levels $(1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s)$.
Other options represent either excited states or incorrect filling sequences.

(B) The azimuthal quantum number $(l)$ can take integer values ranging from $0$ to $(n - 1)$.
Given $n = 3$,the possible values for $l$ are $0, 1, \text{and } (3 - 1) = 2$.
Therefore,the possible values of $l$ are $0, 1, 2$.

(B) For a given principal quantum number $n$,the azimuthal quantum number $l$ can take values from $0$ to $n-1$. For $n = 5$,$l$ can be $0, 1, 2, 3, 4$.
The magnetic quantum number $m$ ranges from $-l$ to $+l$.
If $m = +3$,then $l$ must be at least $3$ (i.e.,$l = 3$ or $l = 4$).
Option $A$ $(l = 4)$ is possible because $m = +3$ is allowed for $l = 4$.
Option $C$ $(l = 3)$ is possible because $m = +3$ is allowed for $l = 3$.
Option $B$ suggests $l = 0, 1, 3$. While $l = 3$ is valid for $m = +3$,$l = 0$ and $l = 1$ are not valid for $m = +3$ because for $l=0, m=0$ and for $l=1, m=-1, 0, +1$.
Therefore,the statement in option $B$ is incorrect.

(D) The given electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{1}$.
Counting the total number of electrons: $2 + 2 + 6 + 1 = 11$ electrons.
For $Al^{3+}$ (atomic number $13$),the number of electrons is $13 - 3 = 10$. Configuration: $1s^{2}2s^{2}2p^{6}$.
For $Ne$ (atomic number $10$),the ground state configuration is $1s^{2}2s^{2}2p^{6}$. An excited state would involve promotion of an electron to a higher orbital,such as $1s^{2}2s^{2}2p^{5}3s^{1}$.
For $Mg^{+}$ (atomic number $12$),the number of electrons is $12 - 1 = 11$. The ground state configuration is $1s^{2}2s^{2}2p^{6}3s^{1}$.
Since the configuration $1s^{2}2s^{2}2p^{6}3s^{1}$ represents the ground state of $Mg^{+}$,and none of the options correctly identify this as a ground state species,the correct choice is $D$.

(B) The electronic configuration of $P$ $(Z=15)$ is $1s^2 2s^2 2p^6 3s^2 3p^3$.
The valence electrons are in $3s$ and $3p$ orbitals.
For the $3s$ orbital,the electrons are $A$ and $B$. Since they are in the same orbital $(n=3, l=0, m=0)$,they share three quantum numbers $(n, l, m)$ and differ only in spin $(s)$. Given $s_B = +\frac{1}{2}$,then $s_A = -\frac{1}{2}$. Thus,$A$ and $B$ share three quantum numbers.
For the $3p$ orbital,the electrons are $X, Y, Z$. They are in different sub-orbitals $(m = -1, 0, +1)$,so they have different $m$ values. Thus,they do not share three quantum numbers.
Therefore,the group of electrons sharing three identical quantum numbers is $AB$.

(A) The atomic number of Scandium $(Sc)$ is $21$.
According to the Aufbau principle,electrons are filled in orbitals in increasing order of their energy $(1s < 2s < 2p < 3s < 3p < 4s < 3d)$.
Filling $21$ electrons: $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^1$.
Thus,the correct configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1$.

(B) For a given value of $n + l = 6$,the possible values for $n$ and $l$ are determined by the condition $l < n$.
For $n = 6, l = 0$ ($6s$ subshell).
For $n = 5, l = 1$ ($5p$ subshell).
For $n = 4, l = 2$ ($4d$ subshell).
For $n = 3, l = 3$ ($3f$ subshell).
Since $l$ must be less than $n$,the possible subshells are $6s, 5p, 4d, 3f$. Thus,the total number of subshells is $4$.

(D) The principal quantum number $n = 4$ indicates the shell is $4$.
For the azimuthal quantum number $l$,the values correspond to orbitals: $l = 0$ $(s)$,$l = 1$ $(p)$,$l = 2$ $(d)$,and $l = 3$ $(f)$.
Since $l = 3$,the electron is in an $f$ orbital.
Combining $n = 4$ and $l = 3$,the electron is in the $4f$ orbital.

(B) For any given principal quantum number $n$,the azimuthal quantum number $l$ can have values ranging from $0$ to $n-1$.
In option $B$,$n = 1$,so $l$ can only be $0$.
Since the value of $l$ is given as $1$,which is not less than $n$,this set of quantum numbers is not allowed.

(D) The energy of an orbital is determined by the $(n+l)$ rule.
Calculating $(n+l)$ values for each set:
$A: 3+2=5$
$B: 5+0=5$
$C: 4+1=5$
$D: 4+2=6$
Since orbital $D$ has the highest $(n+l)$ value of $6$,it has the highest energy.

(D) The magnetic quantum number $m_l$ ranges from $-l$ to $+l$ for a given azimuthal quantum number $l$.
For $m_l = -3$,the minimum value of $l$ must be $3$ (since $l \geq |m_l|$).
The principal quantum number $n$ must be greater than $l$ $(n > l)$.
Therefore,the minimum value for $n$ is $l + 1 = 3 + 1 = 4$.
Thus,the correct option is $(D)$.

(D) To completely describe an electron in an atom,we need $4$ quantum numbers: the principal quantum number $(n)$,the azimuthal quantum number $(l)$,the magnetic quantum number $(m_l)$,and the spin quantum number $(m_s)$.
These $4$ quantum numbers define the energy,size,shape,orientation,and spin of the electron,respectively.
According to Pauli's exclusion principle,no two electrons in an atom can have the same set of all $4$ quantum numbers.

(B) The total number of electrons in the given configuration is $2 + 2 + 1 + 1 + 1 = 7$.
An element with $7$ electrons is Nitrogen $(N)$.
The electronic configuration of Nitrogen $(Z = 7)$ is $1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1$.

(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 $+l$,including zero.
Therefore,for $l = 1$,the possible values for $m$ are $-1, 0, +1$.
In option $(C)$,$m = 2$,which is not possible because $m$ cannot be greater than $l$ (i.e.,$|m| \leq l$).

(A) For a given principal quantum number $n$,the possible values of the azimuthal quantum number $l$ range from $0$ to $n - 1$.
For the $f$-orbital,the value of $l$ is $3$.
In the $n = 3$ shell,the maximum value of $l$ is $n - 1 = 3 - 1 = 2$.
Since $l = 3$ is not possible for $n = 3$,the $3f$ orbital does not exist.

(B) For any principal quantum number $n$,the azimuthal quantum number $l$ can have values from $0$ to $n-1$.
For $n = 1$,the only possible value for $l$ is $0$.
Therefore,the set $n = 1, l = 1$ is not possible because $l$ cannot be equal to $n$.
Thus,option $B$ is the incorrect set of quantum numbers.

(A) The atomic number of iron $(Fe)$ is $26$.
The electronic configuration of neutral $Fe$ is $[Ar] \, 3d^6 4s^2$.
To form the ferric ion $(Fe^{3+})$,three electrons are removed: two from the $4s$ orbital and one from the $3d$ orbital.
Thus,the electronic configuration of $Fe^{3+}$ is $[Ar] \, 3d^5$.

(C) The maximum number of electrons that can be accommodated in a shell with principal quantum number $n$ is given by the formula $2n^2$.
For the highest principal quantum number $n = 4$,the maximum number of electrons is $2 \times (4)^2 = 2 \times 16 = 32$.

(B) The given electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$.
This configuration corresponds to the element Chromium $(Cr)$,which has an atomic number of $24$.
According to the Aufbau principle,the expected configuration is $3d^4 4s^2$,but due to the extra stability of half-filled $d$-orbitals,one electron from the $4s$ orbital shifts to the $3d$ orbital.
Therefore,$3d^5 4s^1$ is the stable ground state configuration of $Cr$.

(A) For a set of quantum numbers to be valid,the following rules must be satisfied:
$1$. $n$ is a positive integer $(1, 2, 3, \dots)$.
$2$. $l$ 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 $-1/2$.
Checking the options:
$(A)$ $n=3, l=2, m=2, s=+1/2$: Here $l < n$ $(2 < 3)$ and $|m| \le l$ $(2 \le 2)$. This set is valid.
$(B)$ $n=3, l=4$: Invalid because $l$ must be less than $n$.
$(C)$ $n=4, l=0, m=2$: Invalid because $|m|$ must be $\le l$ $(2 \not\le 0)$.
$(D)$ $n=4, l=4$: Invalid because $l$ must be less than $n$.

(A) For a valid set of quantum numbers,the following rules must be satisfied:
$1$. $n$ (principal quantum number) must be a positive integer $(1, 2, 3, ...)$.
$2$. $l$ (azimuthal quantum number) can have values from $0$ to $n-1$.
$3$. $m$ (magnetic quantum number) can have values from $-l$ to $+l$ including $0$.
In option $A$,$n = 1$. The allowed value for $l$ is only $0$ $(n-1 = 1-1 = 0)$. Since $l = 2$ is given,this set is invalid because $l$ cannot be greater than or equal to $n$.

(C) The electronic configuration is determined by the number of electrons.
$Cl^{-}$ has $17 + 1 = 18$ electrons.
$Ar$ has $18$ electrons.
Since both have $18$ electrons,they have the same electronic configuration $(1s^2 2s^2 2p^6 3s^2 3p^6)$.

(C) For a given value of $l$,the magnetic quantum number $m$ can range from $-l$ to $+l$ including $0$.
In option $(C)$,$n = 3$ and $l = 0$ (which corresponds to the $3s$ orbital).
For $l = 0$,the only possible value for $m$ is $0$.
Therefore,$m = -1$ is not possible when $l = 0$.
Thus,the set of quantum numbers in option $(C)$ is invalid.

(C) The atomic number of chromium $(Cr)$ is $24$.
The electronic configuration of $Cr$ is $[Ar] 3d^5 4s^1$.
The $19^{th}$ electron is the first electron to enter the $4s$ orbital.
For the $4s^1$ electron: $n=4$,$l=0$ (for $s$-orbital),$m=0$,and $s=+1/2$.
Therefore,the correct set of quantum numbers is $n=4, l=0, m=0, s=1/2$.

(C) The magnetic quantum number $m_l$ can take values from $-l$ to $+l$ including zero.
Given the azimuthal quantum number $l = 3$,the possible values for $m_l$ are calculated as follows:
$m_l = -3, -2, -1, 0, +1, +2, +3$.
Therefore,the correct option is $(C)$.

(C) The principal quantum number $n = 2$ indicates the second shell.
The azimuthal quantum number $l = 1$ corresponds to the $p$ subshell.
Therefore,the combination $n = 2, l = 1$ represents the $2p$ orbital.

(D) The atomic number of sodium $(Na)$ is $11$.
The electronic configuration of $Na$ is $1s^2 2s^2 2p^6 3s^1$.
The valence electron is in the $3s$ orbital.
For an $s$-orbital,the azimuthal quantum number $(l)$ is $0$.
The magnetic quantum number $(m_l)$ ranges from $-l$ to $+l$,so for $l = 0$,$m_l = 0$.

(C) The azimuthal quantum number $(l)$ determines the shape of the orbital and the angular momentum of the electron in a given orbital. The magnitude of angular momentum is given by the formula $\sqrt{l(l+1)} \frac{h}{2\pi}$.

(B) The energy of an electronic configuration is determined by the sum of the energies of all the electrons present in the orbitals.
For a given principal quantum number $n=3$,the energy of the orbitals follows the order $3s < 3p < 3d$.
To find the configuration with the maximum energy,we compare the total number of electrons in the higher energy orbitals ($3p$ and $3d$).
Option $A$: $2$ electrons in $3s$,$2$ in $3p$,$0$ in $3d$. Total electrons in $3p+3d = 2$.
Option $B$: $2$ electrons in $3s$,$3$ in $3p$,$2$ in $3d$. Total electrons in $3p+3d = 5$.
Option $C$: $2$ electrons in $3s$,$3$ in $3p$,$1$ in $3d$. Total electrons in $3p+3d = 4$.
Option $D$: $2$ electrons in $3s$,$3$ in $3p$,$1$ in $3d$. Total electrons in $3p+3d = 4$.
Comparing the total number of electrons in the higher energy orbitals,Option $B$ has the highest number of electrons in the $3p$ and $3d$ orbitals,thus representing the state with the maximum energy.

(D) For a $d$-orbital,the azimuthal quantum number $l = 2$.
The number of magnetic quantum numbers $(m_l)$ is given by the formula $(2l + 1)$.
Substituting $l = 2$ into the formula: $m_l = (2 \times 2 + 1) = 5$.
Thus,there are $5$ possible values for the magnetic quantum number for a $d$-orbital.

(A) The atomic number of chlorine $(Cl)$ is $17$.
Its electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^5$.
Thus,the outer electronic structure is $3s^2 3p^5$.