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

(C) The magnetic quantum number $m$ can take values ranging from $-l$ to $+l$,including zero.
Therefore,the total number of allowed values for $m$ is calculated as $(l - (-l) + 1) = 2l + 1$.

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

(A) The shape of an orbital is determined by the azimuthal quantum number $(l)$.
For $s$-orbitals,$l = 0$,which corresponds to a spherical shape.
For $p$-orbitals,$l = 1$ (dumbbell shape).
For $d$-orbitals,$l = 2$ (double dumbbell shape).
For $f$-orbitals,$l = 3$ (complex shape).
Therefore,the $4s$ sub-shell is spherical.

(A) The atomic number of oxygen $(O)$ is $8$.
Its electronic configuration is $1s^2 2s^2 2p^4$.
Hund's rule of maximum multiplicity states that electron pairing in degenerate orbitals (orbitals of the same subshell) does not occur until each orbital is singly occupied.
For the $2p$ subshell containing $4$ electrons,the electrons will occupy the $2p_x, 2p_y,$ and $2p_z$ orbitals singly first,and then the fourth electron will pair up in one of the orbitals.
Thus,the configuration is $1s^2 2s^2 2p_x^2 2p_y^1 2p_z^1$ (or any equivalent arrangement like $2p_x^1 2p_y^2 2p_z^1$).
Therefore,option $A$ is correct.

(B) For a given azimuthal quantum number $l$,the magnetic quantum number $m$ can take values from $-l$ to $+l$,including zero.
The total number of values is given by the formula $(2l + 1)$.
Given $l = 2$,the possible values for $m$ are $-2, -1, 0, +1, +2$.
Thus,the total number of values is $2(2) + 1 = 5$.

(C) The $d$-orbitals have a double dumb-bell shape,except for the $d_{z^2}$ orbital which has a donut-shaped electron cloud around the center.
Specifically,the $d_{xy}$,$d_{yz}$,and $d_{xz}$ orbitals consist of four lobes oriented between the respective axes,forming a double dumb-bell shape.

(B) To determine the energy of a sub-shell,we use the $(n + l)$ rule.
For $3p$: $n = 3, l = 1$,so $(n + l) = 3 + 1 = 4$.
For $3d$: $n = 3, l = 2$,so $(n + l) = 3 + 2 = 5$.
For $4s$: $n = 4, l = 0$,so $(n + l) = 4 + 0 = 4$.
For $3s$: $n = 3, l = 0$,so $(n + l) = 3 + 0 = 3$.
Comparing the $(n + l)$ values,$3d$ has the highest value of $5$,therefore it has the highest energy.

(C) The $(n + l)$ rule states that orbitals are filled in increasing order of $(n + l)$ values.
For $4s$,$(n + l) = 4 + 0 = 4$.
For $3d$,$(n + l) = 3 + 2 = 5$.
According to the rule,$4s$ should be filled before $3d$.
In option $(C)$,the configuration $3d^5, 4s^1$ represents Chromium $(Z=24)$.
While this configuration is stable due to the half-filled $d$-subshell,it technically violates the Aufbau principle (which is based on the $(n+l)$ rule) because an electron is placed in $3d$ before $4s$ is completely filled with $2$ electrons.

(B) The electronic configuration of $K$ $(Z = 19)$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^1$.
The outermost electron is in the $4s$ orbital.
For the $4s$ orbital,the principal quantum number $n = 4$.
For an $s$-orbital,the azimuthal quantum number $l = 0$.
Since $l = 0$,the magnetic quantum number $m = 0$.
The spin quantum number $s$ for the single electron in the $4s$ orbital is $+ \frac{1}{2}$.
Thus,the set of quantum numbers is $n = 4, l = 0, m = 0, s = + \frac{1}{2}$.

(C) The number of orbitals in a sub-shell is given by $(2l + 1)$.
For a $p$ sub-shell,the azimuthal quantum number $l = 1$.
Therefore,the number of orbitals = $(2 \times 1 + 1) = 3$.
These orbitals are designated as $p_x, p_y,$ and $p_z$.

(C) The number of orbitals in a sub-shell is determined by the azimuthal quantum number,$\ell$.
For a $d$ sub-shell,the value of $\ell$ is $2$.
The number of orbitals in a sub-shell is given by the formula $(2\ell + 1)$.
Substituting $\ell = 2$ into the formula: $2(2) + 1 = 5$.
Thus,there are $5$ orbitals in the $d$ sub-shell.

(B) For a given azimuthal quantum number $l$,the number of orbitals is given by $(2l + 1)$.
For $l = 2$,the number of orbitals is $2(2) + 1 = 5$.
Each orbital can accommodate a maximum of $2$ electrons.
Therefore,the total number of electrons in the $d$-subshell $(l = 2)$ is $5 \times 2 = 10$ electrons.

(B) Pauli's exclusion principle states that 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 hold a maximum of two electrons,and they must have opposite spins.

(C) The azimuthal quantum number $(l)$ determines the shape of the orbital.
For different subshells,the values are:
$s$-orbital: $l = 0$
$p$-orbital: $l = 1$
$d$-orbital: $l = 2$
$f$-orbital: $l = 3$
Therefore,for $d$ electrons,the azimuthal quantum number is $2$.

(C) For $p$-orbital,the azimuthal quantum number $(l)$ is $1$.
The magnetic quantum number $(m_l)$ ranges from $-l$ to $+l$,including zero.
Therefore,for $l = 1$,the possible values of $m_l$ are $-1, 0, +1$.

(D) For a given principal quantum number $n$,the number of orbitals in a shell is given by the formula $n^2$.
For $n = 3$,the number of orbitals $= 3^2 = 9$.
These $9$ orbitals consist of one $3s$ orbital,three $3p$ orbitals,and five $3d$ orbitals $(1 + 3 + 5 = 9)$.

(D) The atomic number of $Ne$ is $10$,and its electronic configuration is $1s^2 \ 2s^2 \ 2p^6$.
$F^{-}$ has $9 + 1 = 10$ electrons: $1s^2 \ 2s^2 \ 2p^6$.
$Na^{+}$ has $11 - 1 = 10$ electrons: $1s^2 \ 2s^2 \ 2p^6$.
$Mg^{2+}$ has $12 - 2 = 10$ electrons: $1s^2 \ 2s^2 \ 2p^6$.
$Cl^{-}$ has $17 + 1 = 18$ electrons: $1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6$.
Therefore,$Cl^{-}$ does not have the configuration of $Ne$.

(C) The electronic configuration of $K^{+}$ $(Z=19)$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6$ (total $18$ electrons).
The electronic configuration of $Ca^{2+}$ $(Z=20)$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6$ (total $18$ electrons).
Since both $K^{+}$ and $Ca^{2+}$ have $18$ electrons,they are isoelectronic and possess the same electronic configuration.

(B) The azimuthal quantum number $l$ determines the shape of the orbital.
For $l = 0$,the orbital is an $s$-orbital,which has a spherical shape.
Therefore,the correct option is $(B)$.

(D) The electronic configuration of sodium ($Na$,atomic number $11$) 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 $n = 3$,the azimuthal quantum number $l = 0$,and the magnetic quantum number $m_l = 0$.
Thus,the magnetic quantum number for the valence electron is $0$.

(C) The atomic number of nitrogen $(N)$ is $7$.
According to the Aufbau principle,the electrons fill the orbitals in increasing order of energy: $1s, 2s, 2p$.
The electronic configuration is $1s^2 2s^2 2p^3$.
According to Hund's rule of maximum multiplicity,for degenerate orbitals (like $2p_x, 2p_y, 2p_z$),electrons occupy each orbital singly before pairing begins.
Therefore,the configuration is $1s^2, 2s^2, 2p_x^1, 2p_y^1, 2p_z^1$.

(D) In a multi-electron atom,the energy of an orbital depends on both the principal quantum number $(n)$ and the azimuthal quantum number $(l)$.
Orbitals with the same $n$ and $l$ values are degenerate (have the same energy) in the absence of external fields.
Comparing the given sets:
$1. n=1, l=0$ $(1s)$
$2. n=2, l=0$ $(2s)$
$3. n=2, l=1$ $(2p)$
$4. n=3, l=2$ $(3d)$
$5. n=3, l=2$ $(3d)$
Since $4$ and $5$ have the same $n=3$ and $l=2$,they represent the same subshell $(3d)$ and thus have the same energy.

(C) The atomic number of the element is $17$,which corresponds to Chlorine $(Cl)$.
According to the Aufbau principle,the electrons are filled in the order of increasing energy: $1s, 2s, 2p, 3s, 3p, 4s, \dots$
The electronic configuration for $17$ electrons is $1s^2, 2s^2, 2p^6, 3s^2, 3p^5$.
Therefore,the correct option is $(C)$.

(B) The $s$-orbital has a spherical shape due to its non-directional nature.

(A) According to the $(n+l)$ rule,the energy of an orbital is determined by the sum of its principal quantum number $(n)$ and azimuthal quantum number $(l)$.
For $4p$,$n=4$ and $l=1$,so $(n+l) = 4+1 = 5$.
For $4d$,$n=4$ and $l=2$,so $(n+l) = 4+2 = 6$.
For $4f$,$n=4$ and $l=3$,so $(n+l) = 4+3 = 7$.
Since the $4p$ orbital has the lowest $(n+l)$ value among the given options,the next electron will enter the $4p$-orbital.

(C) The atomic number of $Fe$ is $26$.
The electronic configuration of $Fe$ is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6$.
Grouping the electrons by principal quantum number $(n)$:
$n=1$: $2$ electrons $(1s^2)$
$n=2$: $8$ electrons $(2s^2 2p^6)$
$n=3$: $14$ electrons $(3s^2 3p^6 3d^6)$
$n=4$: $2$ electrons $(4s^2)$
Thus,the arrangement is $2, 8, 14, 2$.

(C) The atomic number of Copper $(Cu)$ is $29$.
The ground state electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$,which is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1$.
To form the $Cu^{2+}$ ion,two electrons are removed from the neutral $Cu$ atom.
First,one electron is removed from the $4s$ orbital,and then one electron is removed from the $3d$ orbital.
Therefore,the electronic configuration of $Cu^{2+}$ is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^9$.

(A) The atomic number of iron $(Fe)$ is $26$.
The ground state electronic configuration of $Fe$ is $[Ar] 3d^6 4s^2$ or $1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^6 4s^2$.
When $Fe$ forms the $Fe^{2+}$ ion,it loses two electrons from the outermost shell,which is the $4s$ orbital.
Therefore,the electronic configuration of $Fe^{2+}$ becomes $[Ar] 3d^6$ or $1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^6$.

(C) The number of electrons in a shell is given by the formula $2n^2$.
For the $3^{rd}$ quantum shell,$n = 3$.
Number of electrons = $2 \times (3)^2 = 2 \times 9 = 18$.
Therefore,the correct option is $(C)$.

(C) The given orbital diagram shows the electronic configuration as $1s^2 2s^2 2p^5$.
Total number of electrons = $2 + 2 + 5 = 9$.
The element with atomic number $9$ is Fluorine $(F)$.

(C) The magnetic quantum number $(m_l)$ depends on the azimuthal quantum number $(l)$.
For a given value of $l$,the possible values of $m_l$ range from $-l$ to $+l$,including zero.
Given $l = 3$,the possible values of $m_l$ are $-3, -2, -1, 0, +1, +2, +3$,which can be written as $0, \pm 1, \pm 2, \pm 3$.
Therefore,the correct option is $(C)$.

(D) The $37^{th}$ electron will enter the $5s$ sub-level.
According to the Aufbau principle,electrons fill orbitals in the order of increasing energy.
For the $4p$ orbital,$n=4$ and $l=1$,so $(n+l) = 4+1 = 5$.
For the $5s$ orbital,$n=5$ and $l=0$,so $(n+l) = 5+0 = 5$.
When $(n+l)$ values are equal,the orbital with the lower value of $n$ is filled first. Since $4p$ has $n=4$ and $5s$ has $n=5$,the $4p$ orbital is filled before the $5s$ orbital.
Therefore,after the $4p^6$ configuration is completed in Krypton,the next electron $(37^{th})$ enters the $5s$ orbital.

(D) The magnetic quantum number $m$ depends on the azimuthal quantum number $l$ such that $m$ ranges from $-l$ to $+l$.
For an $s$-orbital,$l = 0$,which implies $m = 0$.
Since the given magnetic quantum number is $m = - 1$,it is impossible for an electron to be in an $s$-orbital.

(B) The azimuthal quantum number $(l)$ determines the subshell to which the electron belongs.
It also defines the shape of the orbitals and is associated with the angular momentum of the electron.

(C) The principal quantum number $n = 2$ and azimuthal quantum number $l = 1$ corresponds to the $2p$ subshell.
For a given $l$,the number of orbitals is given by $(2l + 1)$.
For $l = 1$,the number of orbitals is $2(1) + 1 = 3$.
These $3$ orbitals are $p_x, p_y,$ and $p_z$.
Each orbital can accommodate a maximum of $2$ electrons.
Therefore,the total number of electrons = $3 \times 2 = 6$.

(A) The atomic number of Carbon $(C)$ is $6$.
Following the Aufbau principle,the electrons are filled in the orbitals in increasing order of energy: $1s, 2s, 2p$.
Thus,the electronic configuration is $1s^2 2s^2 2p^2$.

(A) The shape of an orbital is determined by the azimuthal quantum number $(l)$. For both $2p$ and $3p$ orbitals,the value of $l$ is $1$,which corresponds to a dumb-bell shape. Therefore,they have the same shape. They differ in size,energy,and the value of the principal quantum number $(n)$.

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

(C) The shape of the $p$-orbital is dumb-bell shaped.
This is due to the angular distribution of the electron density,which has two lobes separated by a nodal plane where the probability of finding an electron is zero.

(A) The atomic number of $Mn$ is $25$.
The ground state electronic configuration of neutral $Mn$ is $[Ar] \, 3d^5 4s^2$.
When $Mn$ forms a $Mn^{2+}$ ion,it loses two electrons from the outermost $4s$ orbital.
Therefore,the electronic configuration of $Mn^{2+}$ is $[Ar] \, 3d^5 4s^0$.

(B) The principal quantum number $(n)$ determines the main energy level or shell in which the electron resides.
It provides information about the average distance of the electron from the nucleus and the size of the orbital.
As the value of $n$ increases,the distance of the electron from the nucleus increases.

(C) The azimuthal quantum number $l$ determines the shape of the orbital.
For $l = 0$,the orbital is $s$ (spherically symmetrical).
For $l = 1$,the orbital is $p$ (dumb-bell shape).
Therefore,the correct option is $C$.

(B) The principal quantum number is $n = 3$.
The azimuthal quantum number is $l = 1$,which corresponds to the $p$ subshell.
The number of orbitals in a subshell is given by the formula $(2l + 1)$.
For $l = 1$,the number of orbitals $= 2(1) + 1 = 3$.
Each orbital can accommodate a maximum of $2$ electrons.
Therefore,the total number of electrons $= 3 \times 2 = 6$ electrons.

(D) The maximum number of electrons in a subshell is given by the formula $2(2l + 1)$.
For the azimuthal quantum number $l = 3$,the subshell is the $f$-subshell.
Substituting $l = 3$ into the formula: $2(2 \times 3 + 1) = 2(6 + 1) = 2(7) = 14$.
Therefore,the maximum number of electrons is $14$.

(A) The atomic number of copper $(Cu)$ is $29$.
Electronic configuration of $Cu$ is $[Ar] 3d^{10} 4s^1$ or $2, 8, 18, 1$.
When $Cu$ loses one electron to form $Cu^{+}$,the configuration becomes $2, 8, 18$.
In $Cu^{+}$,the outermost shell is the third shell $(n=3)$,which contains $18$ electrons.
Thus,$Cu^{+}$ is the ion with $18$ electrons in its outermost shell.

(B) According to the Aufbau principle,electrons fill orbitals in the order of increasing energy,which is determined by the $(n + l)$ rule.
For the given orbitals:
$4s: n=4, l=0 \implies n+l = 4$
$3d: n=3, l=2 \implies n+l = 5$
$4p: n=4, l=1 \implies n+l = 5$
$5s: n=5, l=0 \implies n+l = 5$
$4d: n=4, l=2 \implies n+l = 6$
Comparing the $(n+l)$ values and the $n$ values for orbitals with the same $(n+l)$,the correct order is $4s < 3d < 4p < 5s < 4d$.

(B) The azimuthal quantum number,denoted by $l$,determines the shape of the orbital.
For $l = 0$,the orbital is designated as $s$.
For $l = 1$,the orbital is designated as $p$.
For $l = 2$,the orbital is designated as $d$.
For $l = 3$,the orbital is designated as $f$.
Thus,the letters $s, p, d, f$ are used to represent the values of the azimuthal quantum number $l$.

(D) For a $4d$ electron:
$n = 4$ (principal quantum number).
For a $d$ orbital,the azimuthal quantum number $l = 2$.
The magnetic quantum number $m_l$ can take values from $-l$ to $+l$,i.e.,$-2, -1, 0, +1, +2$.
The spin quantum number $m_s$ can be $+\frac{1}{2}$ or $-\frac{1}{2}$.
Comparing this with the given options,option $D$ $(n=4, l=2, m_l=1, m_s=-\frac{1}{2})$ is a valid set of quantum numbers for a $4d$ electron.