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

(C) Given the configuration: $(n - 1)s^2 (n - 1)p^6 (n - 1)d^x ns^2$.
For $n = 4$ and $x = 5$,the configuration becomes $3s^2 3p^6 3d^5 4s^2$.
This corresponds to the element Manganese $(Mn)$,which has an atomic number of $25$.
The total number of electrons is calculated by summing the electrons in all shells:
$1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^2$.
Total electrons $= 2 + 2 + 6 + 2 + 6 + 5 + 2 = 25$.
Since the number of protons equals the number of electrons in a neutral atom,the number of protons is $25$.

(D) The electronic configuration of $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$.
For $Fe^{2+}$,the configuration is $[Ar] 3d^6$. Thus,the number of $d-$ electrons is $6$.
$(A)$ $Ne$ $(Z=10)$: $1s^2 2s^2 2p^6$. Number of $p-$ electrons $= 6$.
$(B)$ $Mg$ $(Z=12)$: $1s^2 2s^2 2p^6 3s^2$. Number of $s-$ electrons $= 2+2+2 = 6$.
$(C)$ $Fe$ $(Z=26)$: $[Ar] 3d^6 4s^2$. Number of $d-$ electrons $= 6$.
$(D)$ $Cl^{-}$ $(Z=17)$: $1s^2 2s^2 2p^6 3s^2 3p^6$. Number of $p-$ electrons $= 6+6 = 12$.
Since $6 \neq 12$,the number of $d-$ electrons in $Fe^{2+}$ is not equal to the number of $p-$ electrons in $Cl^{-}$.

(D) The atomic number of phosphorus $(P)$ is $15$. The phosphide ion $(P^{3-})$ is formed by gaining $3$ electrons,so it has $15 + 3 = 18$ electrons.
Its electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6$,which is the same as that of the noble gas Argon $(Ar)$.
Among the given options,the chloride ion $(Cl^-)$ has an atomic number of $17$ and gains $1$ electron to have $17 + 1 = 18$ electrons.
Thus,the electronic configuration of $P^{3-}$ ($18$ electrons) is similar to that of $Cl^-$ ($18$ electrons).

(A) For a given shell (principal quantum number $n$),the energy of orbitals increases with the increase in the azimuthal quantum number $(l)$.
The values of $l$ for $s, p, d, f$ orbitals are $0, 1, 2, 3$ respectively.
Therefore,the order of energy is $s < p < d < f$.

(A) The $p_x$ orbital is oriented along the $x$-axis.
Since the $yz$ plane is perpendicular to the $x$-axis,the electron density for the $p_x$ orbital in the $yz$ plane is zero.
Therefore,the correct option is $A$.

(C) The five $d$-orbitals are $d_{xy}, d_{yz}, d_{xz}, d_{x^2-y^2},$ and $d_{z^2}$.
The first four orbitals $(d_{xy}, d_{yz}, d_{xz}, d_{x^2-y^2})$ have a similar shape,which is 'double-dumbbell'.
The fifth orbital,$d_{z^2}$,has a different shape,which is 'doughnut-shaped' or 'dumbbell with a ring'.
Therefore,the correct statement is that the first four orbitals have similar shapes,while the fifth orbital has a different shape.

(B) For a $4d$ orbital,the principal quantum number $n = 4$ and the azimuthal quantum number $l = 2$ (since $s=0, p=1, d=2, f=3$).
Since $l$ must be $2$ for a $d$-orbital,the value $l = 1$ is incorrect for a $4d$ orbital.
Therefore,the set of quantum numbers containing $l = 1$ is not possible.

(B) The $2p$ subshell consists of three orbitals $(2p_x, 2p_y, 2p_z)$.
According to Hund's rule of maximum multiplicity,electrons fill these orbitals singly before pairing begins.
The maximum number of unpaired electrons in the $2p$ subshell is $3$,which occurs in the Nitrogen atom $(Z = 7)$.
The electronic configuration of Nitrogen is $1s^2 2s^2 2p^3$.
Since the atomic number $(Z)$ equals the number of protons,the number of protons is $7$.

(B) The magnetic quantum number $(m)$ can take values from $-l$ to $+l$,including zero.
The total number of values for a given $l$ is $(2l + 1)$.
Let the total number of magnetic quantum numbers be $m_{total} = 2l + 1$.
Rearranging the formula to solve for $l$:
$m_{total} - 1 = 2l$
$l = \frac{(m_{total} - 1)}{2}$.
Thus,the correct relation is $l = \frac{(m - 1)}{2}$.

(A) For orbital $A$: $n = 3, l = 2$. The orbital is $3d$. The energy is determined by $(n + l)$ rule. $(n + l) = 3 + 2 = 5$.
For orbital $B$: $n = 5, l = 0$. The orbital is $5s$. The energy is determined by $(n + l)$ rule. $(n + l) = 5 + 0 = 5$.
Since both orbitals have the same $(n + l)$ value of $5$,the orbital with the lower $n$ value has lower energy.
Therefore,$3d$ $(A)$ has lower energy than $5s$ $(B)$,or energy of $B$ is greater than $A$.

(D) The principal quantum number $n = 4$ indicates the shell is the $4^{th}$ shell.
The azimuthal quantum number $l = 3$ corresponds to the $f$ subshell.
Therefore,the electron is present in the $4f$ orbital.

(D) The shape of an orbital is determined by the azimuthal quantum number $(l)$.
For $s$-orbitals,$l = 0$,which corresponds to a spherical shape.
Spherical shapes are non-directional because the probability density of finding an electron is the same in all directions from the nucleus.
Therefore,the $3s$ orbital is non-directional.

(A) For a hydrogen atom,the energy of an orbital depends only on the principal quantum number $(n)$.
Since all the given orbitals ($3s$,$3p$,and $3d$) have the same principal quantum number $(n = 3)$,they possess the same energy in the hydrogen atom.
Therefore,the correct statement is that $3s$,$3p$,and $3d$ orbitals have the same energy.

(A) The magnetic quantum number $m_l$ depends on the azimuthal quantum number $l$,where $m_l$ ranges from $-l$ to $+l$.
For an $s$-orbital,$l = 0$,so $m_l$ can only be $0$.
Since the given value is $m_l = -1$,it is impossible for an electron to be in an $s$-orbital because $m_l$ cannot be $-1$ when $l = 0$.
Therefore,the correct option is $A$.

(A) To find the element with the maximum number of unpaired electrons,we write the electronic configuration for each option in the ground state:
$A) \ Z = 15 \ (P): 1s^2 2s^2 2p^6 3s^2 3p^3$. The $3p$ subshell has $3$ unpaired electrons.
$B) \ Z = 10 \ (Ne): 1s^2 2s^2 2p^6$. All electrons are paired ($0$ unpaired electrons).
$C) \ Z = 12 \ (Mg): 1s^2 2s^2 2p^6 3s^2$. All electrons are paired ($0$ unpaired electrons).
$D) \ Z = 8 \ (O): 1s^2 2s^2 2p^4$. The $2p$ subshell has $2$ unpaired electrons.
Comparing the number of unpaired electrons,$Z = 15$ has the maximum number ($3$ unpaired electrons). Therefore,the correct option is $A$.

(C) For a $3p$ orbital,the principal quantum number $n = 3$ and the azimuthal quantum number $l = 1$.
Number of radial nodes (spherical nodes) = $n - l - 1 = 3 - 1 - 1 = 1$.
Number of angular nodes (planar nodes) = $l = 1$.
Therefore,the $3p$ orbital has one radial node and one angular node.

(B) To determine the subshell of the $39^{th}$ electron,we write the electronic configuration of the element with atomic number $Z = 39$ (Yttrium).
The order of filling orbitals according to the Aufbau principle is: $1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, \dots$
The configuration is: $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^{10}, 4p^6, 5s^2, 4d^1$.
The total number of electrons is $2+2+6+2+6+2+10+6+2+1 = 39$.
The $39^{th}$ electron enters the $4d$ subshell.

(B) The wave function is given as $\Psi_{nlm}$.
Comparing $\Psi_{310}$ with $\Psi_{nlm}$,we get:
$n = 3$ (Principal quantum number),
$\ell = 1$ (Azimuthal quantum number,which corresponds to the $p$ orbital),
$m = 0$ (Magnetic quantum number,which corresponds to the $p_z$ orbital).
Therefore,the orbital is $3p_z$.

(B) The given electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1$.
This configuration corresponds to the element Potassium ($K$,atomic number $19$).
Since the electrons are filled in the orbitals according to the Aufbau principle,Hund's rule,and Pauli's exclusion principle,this represents the stable ground state of the atom.

(B) The orbital angular momentum of an electron is given by the formula: $L = \sqrt{l(l+1)} \frac{h}{2\pi}$.
For an $s$-orbital,the azimuthal quantum number $(l)$ is $0$.
Substituting $l = 0$ into the formula: $L = \sqrt{0(0+1)} \frac{h}{2\pi} = 0$.
Therefore,the orbital angular momentum of an electron in a $2s$-orbital is $0$.

(C) For $n = 2$ and $l = 1$,the orbital is $2p$.
$2p$ subshell consists of three orbitals: $2p_x, 2p_y, 2p_z$.
Each orbital can accommodate a maximum of $2$ electrons.
Therefore,the total number of electrons in the $2p$ subshell is $3 \times 2 = 6$ electrons.

(B) According to the $(n+l)$ rule,the energy of an orbital increases as the value of $(n+l)$ increases. If two orbitals have the same $(n+l)$ value,the one with the lower $n$ value has lower energy.
Calculating $(n+l)$ for each:
$(1) \, n=4, l=1 \implies n+l = 5$
$(2) \, n=4, l=0 \implies n+l = 4$
$(3) \, n=3, l=2 \implies n+l = 5$
$(4) \, n=3, l=1 \implies n+l = 4$
Comparing the values:
For $(n+l) = 4$: $(4)$ has $n=3$ and $(2)$ has $n=4$. Thus,$(4) < (2)$.
For $(n+l) = 5$: $(3)$ has $n=3$ and $(1)$ has $n=4$. Thus,$(3) < (1)$.
Combining these,the increasing order of energy is $(4) < (2) < (3) < (1)$.

(B) The element with atomic number $15$ is Phosphorus $(P)$.
The electronic configuration of Phosphorus is $1s^2 2s^2 2p^6 3s^2 3p^3$ or $2, 8, 5$.
Therefore,the number of electrons in the outermost shell (valence shell) is $5$.

(B) The atomic number of sodium is $Z = 11$.
The electronic configuration is $1s^2 2s^2 2p^6 3s^1$.
The valence electron is in the $3s$ orbital.
For the $3s$ orbital:
Principal quantum number $(n) = 3$.
Azimuthal quantum number $(l) = 0$ (for $s$-orbital).
Magnetic quantum number $(m) = 0$.
Spin quantum number $(s) = +1/2$ or $-1/2$.
Thus,the correct set is $n = 3, l = 0, m = 0, s = +1/2$.

(C) The number of angular nodal planes in an orbital is given by the azimuthal quantum number,$l$.
For an $s$-orbital,$l = 0$ (zero nodal planes).
For a $p$-orbital,$l = 1$ (one nodal plane).
For a $d$-orbital,$l = 2$ (two nodal planes).
For an $f$-orbital,$l = 3$ (three nodal planes).
Therefore,the $d$-orbital has two angular nodal planes.

(A) is the statement of the Pauli Exclusion Principle,which states that no two electrons in an atom can have the same set of all four quantum numbers $(n, l, m_l, m_s)$.
$R$ is also a true statement because,according to the Pauli Exclusion Principle,if two electrons occupy the same orbital (same $n, l, m_l$),they must have opposite spins ($m_s = +1/2$ and $-1/2$).
Since the restriction on the spins is the direct reason why the four quantum numbers cannot be identical for two electrons in the same orbital,$R$ is the correct explanation of $A$.

(B) The electronic configuration of the element is $1s^2 2s^2 2p^3$.
Total number of electrons = $2 + 2 + 3 = 7$.
Since the atom is neutral,the atomic number $(Z)$ = $7$.
The element with atomic number $7$ is Nitrogen $(N)$.
Atomic weight = $2 \times Z = 2 \times 7 = 14$,which is the atomic mass of Nitrogen.

(C) According to the Aufbau principle,the energy of an orbital is determined by the $(n + l)$ rule.
For $4s$: $n = 4, l = 0$,so $(n + l) = 4 + 0 = 4$.
For $3d$: $n = 3, l = 2$,so $(n + l) = 3 + 2 = 5$.
For $4p$: $n = 4, l = 1$,so $(n + l) = 4 + 1 = 5$.
Since $4s$ has the lowest $(n + l)$ value,it has the lowest energy.
Between $3d$ and $4p$,both have $(n + l) = 5$,but $3d$ has a lower $n$ value $(3 < 4)$,so $3d$ has lower energy than $4p$.
Thus,the correct order is $4s < 3d < 4p$.

(B) The energy of an orbital is determined by the $(n + l)$ rule.
For option $A$: $(n + l) = 3 + 1 = 4$.
For option $B$: $(n + l) = 7 + 0 = 7$.
For option $C$: $(n + l) = 3 + 2 = 5$.
For option $D$: $(n + l) = 3 + 0 = 3$.
Comparing the $(n + l)$ values,$7 > 5 > 4 > 3$.
Therefore,the orbital with the highest energy is represented by the set of quantum numbers in option $B$.

(A) For a given shell (principal quantum number $n$),the energy of orbitals depends on the azimuthal quantum number $(l)$.
As the value of $l$ increases,the energy of the orbital increases.
The values of $l$ for $s, p, d, f$ orbitals are $0, 1, 2, 3$ respectively.
Therefore,the order of energy is $s < p < d < f$.

(B) The electronic configuration of $Cr$ $(Z = 24)$ is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5$.
For azimuthal quantum number $l = 1$,the electrons are in $p$-orbitals ($2p^6$ and $3p^6$). Total electrons = $6 + 6 = 12$.
For azimuthal quantum number $l = 2$,the electrons are in $d$-orbitals $(3d^5)$. Total electrons = $5$.
Therefore,the number of electrons for $l = 1$ and $l = 2$ are $12$ and $5$ respectively.

(C) The total number of electrons in an atom is the sum of electrons in all its shells.
Given:
$K$-shell = $2$ electrons
$L$-shell = $8$ electrons
$M$-shell = $6$ electrons
Total electrons = $2 + 8 + 6 = 16$ electrons.
Therefore,the correct option is $C$.

(A) The atomic number of $Cr$ (Chromium) is $24$.
According to the Aufbau principle,the expected configuration is $[Ar] 4s^2 3d^4$.
However,half-filled and fully-filled $d$-orbitals are more stable due to exchange energy and symmetry.
Therefore,one electron from the $4s$ orbital jumps to the $3d$ orbital to make the $d$-subshell half-filled $(3d^5)$.
The correct electronic configuration is $[Ar] 4s^1 3d^5$,which is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5$.

(B) According to the $(n + l)$ rule,the energy of an orbital is determined by the sum of the principal quantum number $(n)$ and the azimuthal quantum number $(l)$.
For $(A): n + l = 3 + 2 = 5$.
For $(B): n + l = 4 + 2 = 6$.
For $(C): n + l = 4 + 1 = 5$.
For $(D): n + l = 5 + 0 = 5$.
Since the orbital in option $(B)$ has the highest $(n + l)$ value of $6$,it possesses the highest energy.

(A) The principal quantum number '$n$' determines the main energy level or shell of an electron in an atom.
It primarily dictates the effective size of the orbital and the average distance of the electron from the nucleus.
Therefore,'$n$' is associated with the effective size of the orbitals.

(C) The stability of an electronic configuration is determined by the symmetry and exchange energy of the orbitals.
According to Hund's rule,half-filled and fully-filled orbitals are more stable due to their symmetrical distribution and higher exchange energy.
The configuration $3d^5 4s^1$ represents a half-filled $d$-subshell $(d^5)$,which provides extra stability compared to the other given configurations.

(C) For the $4f$ orbital:
$1$. The principal quantum number $n$ is given by the coefficient,so $n = 4$.
$2$. For an $f$ subshell,the azimuthal quantum number $l$ is $3$.
$3$. The magnetic quantum number $m$ can range from $-l$ to $+l$,i.e.,$-3, -2, -1, 0, +1, +2, +3$. Thus,$m = +1$ is a valid value.
$4$. The spin quantum number $s$ can be either $+\frac{1}{2}$ or $-\frac{1}{2}$.
Comparing these with the options,option $C$ satisfies all conditions: $n = 4, l = 3, m = +1, s = +\frac{1}{2}$.

(C) To determine the electronic configuration,we count the total number of electrons in each species:
$1$. $F^{+}$: Atomic number $9$,so $9 - 1 = 8$ electrons.
$Ne$: Atomic number $10$,so $10$ electrons.
$2$. $Li^{+}$: Atomic number $3$,so $3 - 1 = 2$ electrons.
$He^{-}$: Atomic number $2$,so $2 + 1 = 3$ electrons.
$3$. $Cl^{-}$: Atomic number $17$,so $17 + 1 = 18$ electrons.
$Ar$: Atomic number $18$,so $18$ electrons.
$4$. $Na$: Atomic number $11$,so $11$ electrons.
$K$: Atomic number $19$,so $19$ electrons.
Since $Cl^{-}$ and $Ar$ both have $18$ electrons,they have the same electronic configuration.

(A) The $Aufbau$ principle states that in the ground state of the atoms,the orbitals are filled in the order of their increasing energies. Therefore,electrons enter the orbital with the lowest energy first.

(C) The $d_{xy}$ orbital has four lobes that lie in the $xy$-plane.
These lobes are oriented between the $x$ and $y$ axes.
The maximum electron density for the $d_{xy}$ orbital is located at an angle of $45^{\circ}$ with respect to both the $x$ and $y$ axes.

(B) The electronic configuration of nitrogen $(Z = 7)$ is $1s^2 2s^2 2p^3$.
According to Hund's rule of maximum multiplicity,pairing of electrons in the orbitals belonging to the same subshell $(p, d, f)$ does not take place until each orbital is singly occupied.
Therefore,the three electrons in the $2p$ subshell occupy separate orbitals,resulting in three unpaired electrons.

(A) The atomic number of $Cr$ is $24$. Its electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5$.
For $l = 1$ (p-orbitals),we consider the $2p$ and $3p$ subshells. The number of electrons is $6 + 6 = 12$.
For $l = 2$ (d-orbitals),we consider the $3d$ subshell. The number of electrons is $5$.
Therefore,the number of electrons with $l = 1$ and $l = 2$ are $12$ and $5$ respectively.

(B) The electronic configuration is $1s^2 2s^2 2p^3$.
For $n = 2$ and $l = 1$,the electrons are in the $2p$ subshell.
The $2p$ subshell has $3$ orbitals $(p_x, p_y, p_z)$ containing $3$ electrons.
According to Hund's rule,these $3$ electrons will occupy separate orbitals with parallel spins.
Thus,all $3$ electrons in the $2p$ subshell can have a spin quantum number $s = +1/2$.

(B) The ground state electronic configuration of Neon ($Ne$,atomic number $10$) is $1s^2, 2s^2 2p^6$.
When an electron from the $2p$ orbital is promoted to the $3s$ orbital,the configuration becomes $1s^2, 2s^2 2p^5, 3s^1$.
This represents the excited state of the $Ne$ atom.

(D) The nitride ion is represented as $N^{3-}$.
The atomic number of nitrogen $(N)$ is $7$,and its electronic configuration is $1s^2, 2s^2, 2p^3$.
When it gains $3$ electrons to form the nitride ion $(N^{3-})$,its electronic configuration becomes $1s^2, 2s^2, 2p^6$.
The valence shell is the $n=2$ shell,which contains $2 + 6 = 8$ electrons.
Therefore,$1$ mole of $N^{3-}$ ions contains $8$ moles of valence electrons.

(C) In the given configuration,one orbital contains three electrons (two with parallel spins and one with opposite spin),which is physically impossible as an orbital can hold a maximum of two electrons with opposite spins.
According to $Pauli's$ exclusion principle,no two electrons in an atom can have the same set of four quantum numbers. This implies that an orbital can hold a maximum of two electrons,and they must have opposite spins.
Since the configuration shows an orbital with three electrons,it violates $Pauli's$ exclusion principle.

(A) The electronic configuration follows the order of shells: $K=2, L=8, M=18, N=32$.
Given that the $M$ shell has $13$ electrons and the $N$ shell has $1$ electron,the configuration is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^5, 4s^1$.
This corresponds to the element $Cr$ (Chromium).
In the $3d$ subshell,there are $5$ unpaired electrons,and in the $4s$ subshell,there is $1$ unpaired electron.
Therefore,the total number of unpaired electrons is $5 + 1 = 6$.

(A) For a subshell,the value of $(n + l)$ is given as $7$.
Possible combinations of $(n, l)$ such that $n > l$ are:
$1$. If $n = 7, l = 0$ (subshell $7s$)
$2$. If $n = 6, l = 1$ (subshell $6p$)
$3$. If $n = 5, l = 2$ (subshell $5d$)
$4$. If $n = 4, l = 3$ (subshell $4f$)
Thus,there are $4$ possible subshells.

(A) According to the $(n+l)$ rule,the energy of an orbital increases as the value of $(n+l)$ increases.
If $(n+l)$ values are equal,the orbital with the lower $n$ value has lower energy.
Calculating $(n+l)$ for each:
$(i) n=4, l=1 \implies n+l = 5$
$(ii) n=4, l=0 \implies n+l = 4$
$(iii) n=3, l=2 \implies n+l = 5$
$(iv) n=3, l=1 \implies n+l = 4$
Comparing the values:
For $(iv)$ and $(ii)$,$(n+l) = 4$. Since $n(iv) < n(ii)$,energy $(iv) < (ii)$.
For $(i)$ and $(iii)$,$(n+l) = 5$. Since $n(iii) < n(i)$,energy $(iii) < (i)$.
Combining these,the increasing order of energy is $(iv) < (ii) < (iii) < (i)$.

(C) For an azimuthal quantum number $l = 3$,the subshell is an $f$-subshell.
The number of orbitals in a subshell is given by the formula $(2l + 1)$.
Substituting $l = 3$:
Number of orbitals $= 2(3) + 1 = 7$.
Thus,for $l = 3$,there are $7$ orbitals of the $f$-type.