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

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

(C) The atomic number of the element is $15$.
The electronic configuration of the element (Phosphorus,$P$) is $1s^2, 2s^2, 2p^6, 3s^2, 3p^3$.
In terms of shells,the configuration is $2, 8, 5$.
The outermost orbit is the $3^{rd}$ shell,which contains $5$ electrons.
Therefore,the correct option is $(C)$.

(C) The electronic configuration of an element with $4$ electrons in the $2p$ orbital is $1s^2 2s^2 2p^4$.
The total number of electrons is $2 + 2 + 4 = 8$.
Since the number of electrons equals the atomic number $(Z)$ for a neutral atom,$Z = 8$.
The element with atomic number $8$ is Oxygen $(O)$.

(B) The given valence shell electronic configuration is $4s^2 4p^6$. This indicates that the $4th$ shell is completely filled,which corresponds to a noble gas configuration.
The complete electronic configuration is $[Ar]^{18} 3d^{10} 4s^2 4p^6$.
Total number of electrons $= 18 + 10 + 2 + 6 = 36$.
Since the number of electrons in a neutral atom equals the number of protons,the atomic number $Z = 36$.

(C) The electronic configuration of an element with atomic number $17$ $(Cl)$ is: $1s^2 2s^2 2p^6 3s^2 3p^5$.
Its valence shell is the $n=3$ shell,which has the configuration $3s^2 3p^5$.
Representing the orbitals:
$3s^2$: One orbital containing a pair of electrons.
$3p^5$: Three orbitals,where two orbitals contain electron pairs and one orbital contains a single electron.
Total number of orbitals containing electron pairs in the valence shell $= 1 (\text{from } 3s) + 2 (\text{from } 3p) = 3$.

(A) The atomic number of the element is $Z = 35$,which corresponds to Bromine $(Br)$.
The electronic configuration of Bromine is $1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{10} \, 4s^2 \, 4p^5$ or $[Ar] \, 3d^{10} \, 4s^2 \, 4p^5$.
The outermost shell is the $n = 4$ shell.
The number of electrons in the $4s$ and $4p$ orbitals is $2 + 5 = 7$.
Therefore,the number of electrons in the outermost shell is $7$.

(C) The electronic configuration of an element with atomic number $Z = 35$ is:
$1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^5$
The $p-$ orbitals present are $2p$,$3p$,and $4p$.
The number of electrons in these orbitals are:
$2p^6 = 6$ electrons
$3p^6 = 6$ electrons
$4p^5 = 5$ electrons
Total number of electrons in all $p-$ orbitals $= 6 + 6 + 5 = 17$.

(D) The electronic configuration of neutral calcium ($Ca$,atomic number $20$) is $2, 8, 8, 2$.
When it forms the $Ca^{2+}$ ion,it loses $2$ electrons from the outermost shell.
The electronic configuration of $Ca^{2+}$ becomes $2, 8, 8$.
In $Ca^{2+}$,the outermost shell (third shell) has $8$ electrons,and the penultimate shell (second shell) also has $8$ electrons.
Therefore,$Ca^{2+}$ is the species that has the same number of electrons in its outermost and penultimate shells.

(B) Hund's Rule of Maximum Multiplicity states that for a given electron configuration,the term with the greatest multiplicity has the lowest energy. In simpler terms,every orbital in a subshell is singly occupied with one $e^-$ before any one orbital is doubly occupied,and all $e^-$ in singly occupied orbitals have the same spin.

(A) The $1s$ orbital is the lowest energy state (ground state) for an electron in an atom.
Since the electron is already at the minimum possible energy level,it cannot lose energy to transition to a lower state.
Therefore,any electronic transition starting from the $1s$ orbital must involve the absorption of energy to move the electron to a higher energy orbital.

(C) The $d_{xy}$ orbital has four lobes that are oriented in the $xy-$ plane,specifically between the $x$ and $y$ axes.
The lobes bisect the angle between the axes,meaning they are located at an angle of $45^{\circ}$ from both the $x$ and $y$ axes.
Therefore,the probability of finding an electron is maximum along these directions.

(D) The spin quantum number is not derived from the Schrodinger wave equation.
The principal,azimuthal,and magnetic quantum numbers arise as a natural consequence of solving the Schrodinger equation for the hydrogen atom.
The spin quantum number was introduced later to explain the magnetic properties and the fine structure of spectral lines,taking values of $s = +1/2$ or $-1/2$.

(B) An atomic orbital is a mathematical function that describes the wave-like behavior of an electron in an atom. It represents the region in space around the nucleus where the probability of finding an electron is maximum.

(C) The atomic number of $Be$ is $4$.
The electronic configuration of $Be$ is $1s^2 2s^2$.
The $4^{th}$ electron enters the $2s$ orbital.
For the $2s$ orbital,the principal quantum number $n = 2$.
For an $s$ orbital,the azimuthal quantum number $l = 0$.
Since $l = 0$,the magnetic quantum number $m = 0$.
Assuming the first electron in the $2s$ orbital has $s = +1/2$,the second electron (the $4^{th}$ electron of $Be$) will have $s = -1/2$ according to the Pauli exclusion principle.

(A) The $Principal \ quantum \ number$ $(n)$ determines the main energy level or shell of an electron.
It specifies the size and the energy of the orbital in which the electron is located.
Therefore,it provides information about both the location (shell) and the energy of the electron.

(B) The shape of an orbital is determined by the azimuthal quantum number,denoted by '$l$'.
$n$ represents the principal quantum number (size and energy).
$m$ represents the magnetic quantum number (orientation).
$s$ represents the spin quantum number (spin of the electron).

(D) The Pauli Exclusion Principle states that in an atom or molecule,no two electrons can have the same set of all four quantum numbers $(n, l, m_l, m_s)$.
This principle implies that an orbital can hold a maximum of two electrons,and these two electrons must have opposite spins.

(C) - Hund's rule of maximum multiplicity states that for a given electron configuration,the term with maximum multiplicity has the lowest energy. Therefore,pairing of electrons in the orbitals of a subshell (orbitals of equal energy) starts only when each orbital is singly filled.

(C) noble gas has a stable,fully filled valence shell configuration,typically represented as $ns^2 np^6$ (except for Helium,which is $1s^2$).
Option $(C)$ $1s^2, 2s^2 2p^6$ corresponds to Neon $(Ne)$,which is a noble gas with a complete octet in its second shell.

(A) The principal quantum number $(n)$ determines the size of the orbital.
The azimuthal quantum number $(l)$ determines the shape of the orbital.
The magnetic quantum number $(m)$ determines the orientation of the orbital in space.
Therefore,they are respectively related to size,shape,and orientation.

(A) The atomic number of rubidium $(Rb)$ is $37$.
The electronic configuration of $Rb$ is: $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^{10}, 4s^2, 4p^6, 5s^1$.
The valence electron is in the $5s$ orbital.
For the $5s^1$ electron:
Principal quantum number $(n) = 5$.
Azimuthal quantum number $(l)$ for $s$-orbital $= 0$.
Magnetic quantum number $(m_l) = 0$.
Spin quantum number $(m_s) = +\frac{1}{2}$ (or $-\frac{1}{2}$).
Thus,the set of quantum numbers is $(5, 0, 0, +\frac{1}{2})$.

(A) The atomic number of chromium $(Cr)$ is $24$.
According to the Aufbau principle,the expected configuration is $[Ar] \, 3d^4 \, 4s^2$.
However,a $3d$ subshell that is half-filled $(3d^5)$ is more stable due to exchange energy and symmetry.
Therefore,one electron from the $4s$ orbital jumps into the $3d$ orbital to achieve the stable configuration $[Ar] \, 3d^5 \, 4s^1$.

(C) In the $2p$ orbital,the coefficient $2$ represents the principal quantum number $(n = 2)$.
The letter $p$ represents the azimuthal quantum number $(l)$,where $l = 0$ for $s$,$l = 1$ for $p$,$l = 2$ for $d$,and $l = 3$ for $f$.
Therefore,for $2p$,$n = 2$ and $l = 1$.

(C) Electronic configuration of $H^{-}$ is $1s^2$. It has $2$ electrons in the extra-nuclear space.

(A) The given quantum numbers $n = 2, l = 0, m = 0, s = +\frac{1}{2}$ correspond to the $2s^1$ orbital.
This indicates that the outermost electron is in the $2s$ subshell.
The electronic configuration of the element is $1s^2, 2s^1$.
Since the total number of electrons is $3$,the atomic number $Z = 3$,which corresponds to Lithium $(Li)$.

(A) The principal quantum number,denoted by $n$,determines the main energy level or shell of an electron.
It primarily provides information about the size of the orbital and the average distance of the electron from the nucleus.
Therefore,the correct option is $A$.

(D) The electronic configuration of the element is $1s^2, 2s^2 2p^6, 3s^2 3p^2$.
The valence shell is the $3^{rd}$ shell $(n=3)$.
The number of electrons in the $3^{rd}$ shell is $2 (3s) + 2 (3p) = 4$.
Therefore,the number of valence electrons is $4$.

(C) The magnetic quantum number $(m_l)$ describes the spatial orientation of the orbitals in space relative to the chosen axis.

(C) The correct answer is $(C)$.
For any given value of the azimuthal quantum number $l$,the magnetic quantum number $m$ must range from $-l$ to $+l$ (including zero).
In option $(C)$,$l = 2$,so the possible values for $m$ are $-2, -1, 0, +1, +2$.
Since $m = -3$ is outside this range,this set of quantum numbers is impossible.

(D) For a given principal quantum number $n$,the azimuthal quantum number $l$ can have values ranging from $0$ to $n - 1$.
Given $n = 3$,the possible values for $l$ are $0, 1, 2$.
Therefore,the value $l = 3$ is incorrect.

(C) The atomic number of the element is $29$,which corresponds to Copper $(Cu)$.
The electronic configuration of $Cu$ $(Z=29)$ is $[Ar] \ 3d^{10} 4s^1$.
In the $3d^{10}$ subshell,all $10$ electrons are paired.
Therefore,the total number of unpaired electrons in the $d$-orbitals is $0$.

(C) The $s$ orbital is spherical in shape.
The $p$ orbitals are dumb-bell shaped.
Therefore,the shape of the $2p$ orbital is dumb-bell.

(NONE) For a given principal quantum number $n = 2$,the possible values of the azimuthal quantum number $l$ are $0$ and $1$.
For $l = 0$ ($s$-orbital),the magnetic quantum number $m_l$ can only be $0$.
For $l = 1$ ($p$-orbital),the magnetic quantum number $m_l$ can be $-1, 0, +1$.
Thus,for $n = 2$,the possible values for the magnetic quantum number are $0, -1, 0, +1$,which gives a total of $4$ values.

(C) The atomic number of chromium $(Cr)$ is $24$. The expected electronic configuration is $[Ar] \, 3d^4 \, 4s^2$. However,due to the extra stability associated with half-filled $d$-orbitals,one electron from the $4s$ orbital shifts to the $3d$ orbital. Thus,the actual ground state configuration is $[Ar] \, 3d^5 \, 4s^1$. This corresponds to $5$ unpaired electrons in the $3d$ subshell and $1$ unpaired electron in the $4s$ subshell,which is represented by option $(C)$.

(C) The atomic number of carbon is $6$.
Its electronic configuration is $1s^2 2s^2 2p^2$ or $2, 4$.
Carbon has $2$ electrons in the $K$-shell and $4$ electrons in the $L$-shell,which is the outermost shell.
Therefore,the number of valence electrons is $4$.

(B) The azimuthal quantum number $l$ determines the shape of the orbital.
For $l = 0$,the orbital is $s$ (spherical).
For $l = 1$,the orbital is $p$ (dumb-bell shape).
For $l = 2$,the orbital is $d$ (double dumb-bell shape).
Therefore,for a dumb-bell shaped orbital,the value of $l$ is $1$.

(D) $Cr$ $(Z=24)$ has the electronic configuration $[Ar] \, 4s^1 3d^5$.
This is because half-filled $d$-orbitals $(d^5)$ provide extra stability due to symmetry and exchange energy.
Therefore,one electron from the $4s$ orbital shifts to the $3d$ orbital to achieve this stable half-filled configuration.

(D) The atomic number of calcium $(Ca)$ is $20$. The electronic configuration of neutral $Ca$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2$.
To form the calcium ion $(Ca^{2+})$,the atom loses two electrons from its outermost shell,which is the $4s$ orbital.
Therefore,the electronic configuration of $Ca^{2+}$ becomes $1s^2, 2s^2, 2p^6, 3s^2, 3p^6$.

(D) The $K$ shell corresponds to the principal quantum number $n = 1$.
For $n = 1$,the only possible value for the azimuthal quantum number $l$ is $0$ ($s$-orbital).
For $l = 0$,the only possible value for the magnetic quantum number $m_l$ is $0$.
According to the Pauli exclusion principle,no two electrons in an atom can have the same set of all four quantum numbers.
Therefore,the two electrons in the $1s$ orbital must differ in their spin quantum number ($m_s = +1/2$ and $m_s = -1/2$).

(A) completely filled $d^{10}$ subshell consists of five $d$-orbitals,each containing two electrons with opposite spins.
Because all five $d$-orbitals $(d_{xy}, d_{yz}, d_{zx}, d_{x^2-y^2}, d_{z^2})$ are fully occupied,the total electron density distribution becomes uniform in all directions.
This uniform distribution results in a spherically symmetrical electron cloud.
Therefore,the correct option is $A$.

(C) The magnetic quantum number $m_l$ ranges from $-l$ to $+l$ for a given azimuthal quantum number $l$.
Given $m_l = -3$,the minimum value of $l$ must be $3$ (since $l \geq |m_l|$).
For a given $l$,the principal quantum number $n$ must be at least $l + 1$.
Therefore,$n \geq 3 + 1$,which means $n \geq 4$.
The minimum possible value for the principal quantum number is $4$.

(D) The maximum number of electrons in an energy level with principal quantum number $n$ is given by the formula $2n^2$.
Since each orbital can accommodate a maximum of $2$ electrons,the total number of orbitals in that energy level is calculated by dividing the total number of electrons by $2$.
Therefore,the number of orbitals $= \frac{2n^2}{2} = n^2$.

(D) The number of orbitals in a shell with principal quantum number $n$ is given by the formula $n^2$.
For the fourth principal quantum number,$n = 4$.
Therefore,the number of orbitals $= n^2 = 4^2 = 16$.

(C) For any principal quantum number $n$,the azimuthal quantum number $l$ can take values from $0$ to $n - 1$.
For $n = 3$,the possible values for $l$ are $0, 1, 2$.
In option $C$,$l = 3$,which is not possible for $n = 3$ because $l$ must be less than $n$.
Therefore,the set of quantum numbers in option $C$ is invalid.

(B) The atomic number of sodium $(Na)$ is $11$.
The electronic configuration is $1s^2, 2s^2, 2p^6, 3s^1$.
The valence electron is in the $3s$ orbital.
For the $3s$ orbital,the principal quantum number $(n) = 3$.
Since it is an $s$-orbital,the azimuthal quantum number $(l) = 0$.
Consequently,the magnetic quantum number $(m) = 0$.
For the single electron in the $s$-orbital,the spin quantum number $(s) = +\frac{1}{2}$.
Thus,the set of quantum numbers is $(n = 3, l = 0, m = 0, s = +\frac{1}{2})$.

(B) Hund's rule states that pairing of electrons in the orbitals of a subshell (orbitals of equal energy) starts only when each orbital is singly filled. Since nitrogen has an atomic number of $7$,its electronic configuration is $1s^2 2s^2 2p^3$. According to Hund's rule,the three electrons in the $2p$ subshell occupy separate orbitals,resulting in three unpaired electrons.

(C) The region between the $1s$ and $2s$ orbitals is known as a radial node or nodal surface.
At this surface,the probability of finding an electron is zero.

(D) For an $s$ orbital,the azimuthal quantum number $l = 0$.
Since the magnetic quantum number $m_l$ ranges from $-l$ to $+l$,for $l = 0$,the only possible value is $m_l = 0$.

(C) The maximum number of electrons in a shell is given by the formula $2n^2$,where $n$ is the principal quantum number.
For the $K$ shell,$n = 1$.
For the $L$ shell,$n = 2$.
For the $M$ shell,$n = 3$.
Therefore,the maximum number of electrons in the $M$ shell $= 2 \times (3)^2 = 2 \times 9 = 18$.