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

(N/A) The stability of completely filled and half-filled subshells is attributed to the following factors:
$1$. Symmetrical distribution of electrons: The completely filled or half-filled subshells have a symmetrical distribution of electrons,which leads to greater stability. In these configurations,the shielding of electrons from one another is relatively small,and the electrons are more strongly attracted by the nucleus.
$2$. Exchange energy: The stabilizing effect arises whenever two or more electrons with the same spin are present in degenerate orbitals. They can exchange their positions,and the energy released due to this exchange is called exchange energy. The number of possible exchanges is maximum for half-filled and completely filled configurations,leading to maximum stability.
For example,in a $d^{5}$ electron configuration,all five electrons have the same spin. This allows for the maximum number of exchanges,resulting in the maximum release of energy,which makes the $d^{5}$ configuration highly stable.
In summary,the extra stability of half-filled and completely filled subshells (like $3d^{5}$ and $3d^{10}$) is due to:
$i$. Relatively small shielding.
$ii$. Smaller coulombic repulsion energy.
$iii$. Larger exchange energy.

(A) The energy of an orbital is determined by the $(n + l)$ rule. If $(n + l)$ values are equal,the orbital with the lower $n$ value has lower energy.
$(1)$ $n=4, l=2 \rightarrow n+l = 6$ $(4d)$
$(2)$ $n=3, l=2 \rightarrow n+l = 5$ $(3d)$
$(3)$ $n=4, l=1 \rightarrow n+l = 5$ $(4p)$
$(4)$ $n=3, l=2 \rightarrow n+l = 5$ $(3d)$
$(5)$ $n=3, l=1 \rightarrow n+l = 4$ $(3p)$
$(6)$ $n=4, l=1 \rightarrow n+l = 5$ $(4p)$
Comparing $(n+l)$ values:
$(5)$ $3p$ $(n+l=4)$
$(2)$ and $(4)$ $3d$ $(n+l=5)$
$(3)$ and $(6)$ $4p$ $(n+l=5)$
$(1)$ $4d$ $(n+l=6)$
Since $(2)$ and $(4)$ have the same $(n+l)$ and $n$,they have the same energy. Similarly,$(3)$ and $(6)$ have the same energy.
Increasing order of energy: $(5) < (2) = (4) < (3) = (6) < (1)$.

(N/A) $P$ $(Z=15)$: Electronic configuration is $[Ne] \, 3s^2 \, 3p^3$. The $3p$ subshell has $3$ electrons in separate orbitals,so there are $3$ unpaired electrons.
$(b)$ $Si$ $(Z=14)$: Electronic configuration is $[Ne] \, 3s^2 \, 3p^2$. The $3p$ subshell has $2$ electrons in separate orbitals,so there are $2$ unpaired electrons.
$(c)$ $Cr$ $(Z=24)$: Electronic configuration is $[Ar] \, 3d^5 \, 4s^1$. All $5$ electrons in $3d$ and $1$ electron in $4s$ are unpaired,totaling $6$ unpaired electrons.
$(d)$ $Fe$ $(Z=26)$: Electronic configuration is $[Ar] \, 3d^6 \, 4s^2$. In $3d^6$,one orbital is paired and four are unpaired,totaling $4$ unpaired electrons.
$(e)$ $Kr$ $(Z=36)$: Electronic configuration is $[Ar] \, 3d^{10} \, 4s^2 \, 4p^6$. All orbitals are completely filled,so there are $0$ unpaired electrons.

(N/A) The electronic configuration of oxygen (atomic number $8$) is $1s^2 2s^2 2p^4$.
Following Hund's rule and the Pauli exclusion principle,the orbital diagram is:
$1s$: [↑↓]
$2s$: [↑↓]
$2p$: [↑↓] [↑] [↑]
From the orbital diagram,it is observed that there are two unpaired electrons in the $2p$ orbitals.

(B) The electronic configuration of neutral nickel $(_{28}Ni)$ is $1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{6}, 3d^{8}, 4s^{2}$.
When forming a cation,electrons are removed from the orbital with the highest principal quantum number $(n)$ first.
Since the $4s$ orbital has a higher principal quantum number $(n=4)$ compared to the $3d$ orbital $(n=3)$,the two electrons are lost from the $4s$ orbital.
Therefore,the configuration of $Ni^{2+}$ is $1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{6}, 3d^{8}, 4s^{0}$.

(C) Degenerate orbitals are those that have the same energy level.
For a given atom,orbitals with the same principal quantum number $(n)$ and same azimuthal quantum number $(l)$ are degenerate.
In the given set,$3d_{xy}, 3d_{z^2}, 3d_{yz}$ all have $n=3$ and $l=2$,so they are degenerate.
Similarly,$4d_{xy}, 4d_{yz}, 4d_{z^2}$ all have $n=4$ and $l=2$,so they are also degenerate.
Therefore,both sets are degenerate.

(A) For $3p$ orbital,the principal quantum number $n = 3$ and the azimuthal quantum number $l = 1$.
The number of angular nodes is given by $l = 1$.
The number of radial nodes is given by $n - l - 1 = 3 - 1 - 1 = 1$.
Therefore,the $3p$ orbital has $1$ angular node and $1$ radial node.

(A) $(I)$ $(a)$ $1s < 2s < 2p < 3s$ (Energy values: $1s(1+0=1), 2s(2+0=2), 2p(2+1=3), 3s(3+0=3)$)
$(b)$ $3s < 3p < 4s < 4d$ (Energy values: $3s(3+0=3), 3p(3+1=4), 4s(4+0=4), 4d(4+2=6)$)
$(c)$ $4d < 5p < 6s < 4f < 5d$ (Energy values: $4d(4+2=6), 5p(5+1=6), 6s(6+0=6), 4f(4+3=7), 5d(5+2=7)$)
$(d)$ $7s < 5f < 6d < 7p$ (Energy values: $7s(7+0=7), 5f(5+3=8), 6d(6+2=8), 7p(7+1=8)$)
$(II)$ $(a)$ $5s$ (Energy values: $4d(6), 4f(7), 5s(5), 5p(6)$)
$(b)$ $6p$ (Energy values: $5p(6), 5d(7), 5f(8), 6s(6), 6p(7)$)

(A) The electronic configuration of $Cu$ $(Z=29)$ is $[Ar] 3d^{10} 4s^{1}$.
This is because completely filled and half-filled orbitals possess extra stability due to symmetry and exchange energy.
In $3d^{10} 4s^{1}$,the $d$-subshell is completely filled,which provides extra stability to the atom compared to the $3d^{9} 4s^{2}$ configuration.

(N/A)

$1$. An orbit is a defined circular path around the nucleus in which the electrons revolve.	$1$. An orbital is the three-dimensional space around the nucleus within which the probability of finding an electron is maximum (up to $90 \%$).
$2$. All orbits are circular and disc-like.	$2$. Different orbitals have different shapes.
$3$. The concept of an orbit is not in accordance with the wave character of electrons and the uncertainty principle.	$3$. The concept of an orbital is in accordance with the wave character of electrons and the uncertainty principle.
$4$. The maximum number of electrons in any orbit is given by $2n^{2}$,where $n$ is the number of the orbit.	$4$. The maximum number of electrons in any orbital is two.

(N/A) In a hydrogen atom,the energy of an electron depends only on the principal quantum number $(n)$.
In multielectron atoms,the presence of other electrons causes shielding and repulsion,which makes the energy of an electron dependent on both the principal quantum number $(n)$ and the azimuthal quantum number $(l)$.
According to the $(n+l)$ rule,for a given principal quantum number $(n)$,the energy of orbitals increases as the value of $l$ increases.
Therefore,in multielectron atoms,the orbitals of the same principal quantum number $(n)$ have different energies,following the order $E_s < E_p < E_d < E_f$.

(N/A) The energy of an electron in an orbit is determined by the atomic number $(Z)$ and the principal quantum number $(n)$.
$1$. As the atomic number $(Z)$ increases,the energy becomes more negative (lower),indicating stronger attraction between the nucleus and the electron.
$2$. As the principal quantum number $(n)$ increases,the energy becomes less negative (higher),indicating the electron is further from the nucleus.

(N/A) The energy of hydrogen and hydrogen-like species depends only on the principal quantum number $n$.
It does not depend on the azimuthal quantum number $l$ or the magnetic quantum number $m_{l}$.

(N/A) For a multi-electron atom,the energy of an orbital depends on both the principal quantum number $(n)$ and the azimuthal quantum number $(l)$. The energy increases with an increase in the value of $(n + l)$. If two orbitals have the same $(n + l)$ value,the orbital with the lower value of $n$ has lower energy.

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

(N/A) Atomic orbitals are identified by their quantum numbers: the principal quantum number $(n)$,the azimuthal quantum number $(l)$,and the magnetic orbital quantum number $(m_{l})$.

(N/A) $(n)$ Principal quantum number: Represents the energy of the electron in an atom and the average distance of the electron from the nucleus.
$(l)$ Azimuthal quantum number: Represents the shape of the orbital.
$(m_l)$ Magnetic quantum number: Determines the spatial orientation of the orbital.

(N/A) nodal surface (or node) is a region in an orbital where the probability density of finding an electron is zero.
For an orbital with principal quantum number $n$ and azimuthal quantum number $l$,the total number of nodes is given by $(n - 1)$.
These are divided into radial nodes $(n - l - 1)$ and angular nodes $(l)$.

(N/A) The $s$ orbitals start from $n=1$,and the first $s$ orbital is $1s$.
The $p$ orbitals start from $n=2$,and the first $p$ orbital is $2p$.
The $d$ orbitals start from $n=3$,and the first $d$ orbital is $3d$.
The $f$ orbitals start from $n=4$,and the first $f$ orbital is $4f$.

(C) The orbitals $1p, 1d, 2d, 1f, 2f, 3f$ are not possible.
For any orbital,the principal quantum number $n$ must be greater than the azimuthal quantum number $l$.
For $p$ orbital,$l=1$,so $n$ must be $\ge 2$. Thus,$1p$ is not possible.
For $d$ orbital,$l=2$,so $n$ must be $\ge 3$. Thus,$1d$ and $2d$ are not possible.
For $f$ orbital,$l=3$,so $n$ must be $\ge 4$. Thus,$1f, 2f,$ and $3f$ are not possible.

(D) The differences between $2s$ and $2p$ orbitals are as follows:
$1$. Shape: The $2s$ orbital is spherical,whereas the $2p$ orbital is dumbbell-shaped.
$2$. Energy: For a given principal quantum number $n$,the energy of the $s$-orbital is lower than that of the $p$-orbital. Thus,$E_{2s} < E_{2p}$.
$3$. Quantum Numbers: For $2s$,the azimuthal quantum number $l = 0$ and magnetic quantum number $m_l = 0$. For $2p$,$l = 1$ and $m_l = -1, 0, +1$.
$4$. Nodes: The $2s$ orbital has $n - l - 1 = 2 - 0 - 1 = 1$ radial node,while the $2p$ orbital has $n - l - 1 = 2 - 1 - 1 = 0$ radial nodes.

(N/A) The difference lies in their orientation in space and their magnetic quantum number $(m_l)$ values. They are oriented along the $x, y,$ and $z$ axes respectively. Their $m_l$ values are distinct and can be any value from the set ${+1, 0, -1}$ depending on the convention used.

(D) The orbitals $2p_x, 2p_y$ and $2p_z$ belong to the same principal energy level $(n=2)$ and subshell $(l=1)$.
They possess the same shape (dumbbell shape),same size,and same energy in the absence of an external magnetic field (degenerate orbitals).

(A) The energy of an orbital in a multi-electron atom is determined by the $(n + l)$ rule.
For $p$ orbitals,the azimuthal quantum number $l = 1$.
Energy increases as the principal quantum number $n$ increases.
Thus,the order of energy is $2p < 3p < 4p < 5p$.

(N/A) The number of radial nodes is given by the formula $(n - l - 1)$.
For $p$-orbitals,the azimuthal quantum number $l = 1$.
For $2p$ $(n=2)$: $2 - 1 - 1 = 0$.
For $3p$ $(n=3)$: $3 - 1 - 1 = 1$.
For $4p$ $(n=4)$: $4 - 1 - 1 = 2$.
For $5p$ $(n=5)$: $5 - 1 - 1 = 3$.
Thus,the number of radial nodes are $0, 1, 2,$ and $3$ respectively.

(A) The number of orbitals in a subshell is given by the formula $(2l + 1)$,where $l$ is the azimuthal quantum number.
This represents the total number of values of the magnetic quantum number $(m_l)$ for a given value of $l$.

(A) The number of orbitals in both $2p$ and $3p$ subshells is the same,which is $3$ $(p_x, p_y, p_z)$.
The number of orbitals in a subshell is given by the formula $(2l + 1)$.
Since the number of orbitals depends only on the value of the azimuthal quantum number $(l)$,and for any $p$-subshell $l = 1$,both $2p$ and $3p$ contain $3$ orbitals.

(N/A) The number of orbitals in a subshell depends on the azimuthal quantum number $(l)$ and is equal to the number of possible values of the magnetic quantum number $(m_{l})$,which is given by $(2l + 1)$.
It does not depend on the principal quantum number $(n)$ or the spin quantum number $(s)$.

(A) The nodal plane is the plane in which the probability of finding an electron is zero.
For $p_x$ orbital,the nodal plane is the $yz$-plane.
For $p_y$ orbital,the nodal plane is the $zx$-plane.
For $p_z$ orbital,the nodal plane is the $xy$-plane.
Therefore,the nodal planes are $yz, zx$,and $xy$ respectively.

(B) The $d_{xy}$ orbital has two nodal planes.
These are the planes where the probability of finding the electron is zero.
For the $d_{xy}$ orbital,the electron density is concentrated between the $x$ and $y$ axes.
The nodal planes are the $xz$-plane and the $yz$-plane.

(N/A) The number of angular nodes is determined by the azimuthal quantum number,$l$.
For $s$ orbital: $l = 0$,therefore $0$ angular node.
For $p$ orbital: $l = 1$,therefore $1$ angular node.
For $d$ orbital: $l = 2$,therefore $2$ angular nodes.
For $f$ orbital: $l = 3$,therefore $3$ angular nodes.

(A) In a hydrogen atom,the energy of an orbital depends only on the principal quantum number $(n)$. Therefore,$2s$ and $2p$ orbitals have the same energy (degenerate).
In multi-electron atoms,the energy of an orbital depends on both $n$ and $l$ (azimuthal quantum number). Due to shielding effects,the $2s$ orbital has lower energy than the $2p$ orbital.
Specifically,the $2s$ orbital of hydrogen has higher energy than the $2s$ orbital of other atoms because the effective nuclear charge $(Z_{eff})$ in multi-electron atoms is higher,which stabilizes the $2s$ orbital.

(B) The energy of an orbital in a multi-electron atom depends on the effective nuclear charge $(Z_{eff})$.
As the atomic number $(Z)$ increases,the attraction between the nucleus and the electron increases,which lowers the energy of the orbital.
For the $2s$ orbital,the energy decreases as the nuclear charge increases.
Therefore,the order of energy is $E_{2s}(H) > E_{2s}(Li) > E_{2s}(Na) > E_{2s}(K)$.

(B) The energy of the $3d$ orbital is higher than that of the $4s$ orbital.
This is determined by the $(n+l)$ rule:
For $4s$: $(n+l) = (4+0) = 4$.
For $3d$: $(n+l) = (3+2) = 5$.
Since the $(n+l)$ value for $3d$ is greater than that for $4s$,the $3d$ orbital has higher energy.

(A) The $2p$ orbital has lower energy. According to the $(n + l)$ rule,if the value of $(n + l)$ is the same for two orbitals,the orbital with the lower value of $n$ (principal quantum number) has lower energy. For $2p$,$n = 2$,and for $3s$,$n = 3$. Since $2 < 3$,the $2p$ orbital has lower energy.

(C) In a hydrogen atom,there is only one electron. According to quantum mechanics,the energy of orbitals in a hydrogen atom depends only on the principal quantum number $(n)$. Therefore,all orbitals with the same $n$ value have the same energy. Consequently,all orbitals such as $1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f, \dots$ are present in the hydrogen atom,although they are unoccupied in the ground state.

(N/A) In the ground state of a hydrogen atom,the electron resides in the $1s$ orbital,which is the state of minimum energy and maximum stability.
In the excited state of a hydrogen atom,the electron occupies any orbital other than $1s$,such as $2s, 2p, 3s, 3p, \dots$ orbitals.
These excited states are unstable and the electron tends to return to the ground state by emitting energy.

The subshell is determined by the value of $n$ and $l$. The magnetic quantum number $m_l$ ranges from $-l$ to $+l$,and the number of orbitals is given by $(2l + 1)$.

Quantum Numbers	Subshell,$m_l$ values,and Number of Orbitals
$(i) \ n = 3, l = 2$	Subshell: $3d$,$m_l: -2, -1, 0, +1, +2$,Number of orbitals: $5$
$(ii) \ n = 4, l = 3$	Subshell: $4f$,$m_l: -3, -2, -1, 0, +1, +2, +3$,Number of orbitals: $7$
$(iii) \ n = 2, l = 0$	Subshell: $2s$,$m_l: 0$,Number of orbitals: $1$
$(iv) \ n = 5, l = 1$	Subshell: $5p$,$m_l: -1, 0, +1$,Number of orbitals: $3$

(N/A) $(i)$ $4f$ experiences greater attraction because for both $4f$ and $5d$,$(n+l) = 7$,but $4f$ has a lower principal quantum number $n$.
$(ii)$ $5s$ experiences greater attraction because for $4d$,$(n+l) = 6$,while for $5s$,$(n+l) = 5$. Lower $(n+l)$ value indicates stronger attraction.
$(iii)$ $7s$ experiences greater attraction because for $7s$,$(n+l) = 7$,while for $5f$,$(n+l) = 8$.
$(iv)$ $5p$ experiences greater attraction because both are $p$-orbitals,but $5p$ has a lower principal quantum number $n$ compared to $7p$.
$(v)$ $3f$ does not exist,so $3d$ is the only valid orbital.

(A) The atomic number of $He$ is $2$.
Its electronic configuration is $1s^2$.
According to the Pauli exclusion principle and Hund's rule,the two electrons in the $1s$ orbital must have opposite spins.
Therefore,the correct representation is $(i)$ $\boxed{\uparrow\downarrow}$ in the $1s$ orbital.

(B) The total number of electrons in the given electronic configuration $1s^2, 2s^2, 2p^2$ is $2 + 2 + 2 = 6$.
Since the number of electrons is $6$,the atomic number of the element is $6$.
The element with atomic number $6$ is Carbon $(C)$.

(A) The electronic configuration of $Ca$ $(Z = 20)$ is $[Ar]^{18} 4s^2$.
The two outermost electrons are in the $4s$ orbital.
For the $4s$ orbital,the principal quantum number $n = 4$ and the azimuthal quantum number $l = 0$.
Since $l = 0$,the magnetic quantum number $m_l = 0$.
The spin quantum numbers for the two electrons are $m_s = +1/2$ and $m_s = -1/2$.
Thus,the four quantum numbers are $(4, 0, 0, +1/2)$ and $(4, 0, 0, -1/2)$.

(B) The atomic number $Z = 16$ corresponds to Sulfur $(S)$.
The electronic configuration is: $1s^2 2s^2 2p^6 3s^2 3p^4$.
Total $p$ electrons: $2p^6 + 3p^4 = 10$ electrons.
Orbital distribution:
$1s$: $2$ electrons (fully filled)
$2s$: $2$ electrons (fully filled)
$2p$: $6$ electrons (three $2p$ orbitals fully filled)
$3s$: $2$ electrons (fully filled)
$3p$: $4$ electrons ($3p_x^2, 3p_y^1, 3p_z^1$ - one orbital fully filled,two half-filled).
Total fully filled orbitals: $1s, 2s, 2p_x, 2p_y, 2p_z, 3s, 3p_x = 7$.
Total half-filled orbitals: $3p_y, 3p_z = 2$.

(A) The given electronic configuration is $1s^2, 2s^2, 2p^6$,which corresponds to a total of $10$ electrons.
Since the ion has a charge of $-2$,it means it has gained $2$ electrons.
Subtracting these $2$ electrons gives the neutral atom configuration: $1s^2, 2s^2, 2p^4$,which corresponds to $8$ electrons.
The element with atomic number $8$ is Oxygen $(O)$.
Therefore,the ion is $O^{2-}$.

(C) The electronic configuration of an atom with atomic number $Z = 25$ is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^2$.
The $s$ orbitals present are $1s, 2s, 3s,$ and $4s$.
Each of these $s$ orbitals is completely filled with $2$ electrons.
Total number of $s$ electrons $= 2 (1s) + 2 (2s) + 2 (3s) + 2 (4s) = 8$ electrons.

(A) For a given principal quantum number $n$,the total number of orbitals is given by $n^2$.
For $n = 3$,the number of orbitals = $3^2 = 9$.
The total number of electrons is given by $2n^2$.
For $n = 3$,the number of electrons = $2 \times (3^2) = 2 \times 9 = 18$.

(A) For the configuration $2p^4$: The total electrons are $1s^2 2s^2 2p^4$,so $Z = 8$,which corresponds to Oxygen $(O)$.
For the configuration $3s^2$: The total electrons are $1s^2 2s^2 2p^6 3s^2$,so $Z = 12$,which corresponds to Magnesium $(Mg)$.
For the configuration $2p^3$: The total electrons are $1s^2 2s^2 2p^3$,so $Z = 7$,which corresponds to Nitrogen $(N)$.
Therefore,the atoms are $O, Mg, N$.

(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 the same,the orbital with the lower $n$ value has lower energy.
$(i)$ $n=2, l=0 \implies n+l = 2+0 = 2$ ($2s$ orbital)
$(ii)$ $n=2, l=1 \implies n+l = 2+1 = 3$ ($2p$ orbital)
$(iii)$ $n=3, l=2 \implies n+l = 3+2 = 5$ ($3d$ orbital)
$(iv)$ $n=3, l=1 \implies n+l = 3+1 = 4$ ($3p$ orbital)
Comparing the $(n+l)$ values: $2 < 3 < 4 < 5$.
Therefore,the increasing order of energy is: $(i) < (ii) < (iv) < (iii)$.

(A) The number of radial nodes in an orbital is given by the formula: $\text{Radial nodes} = n - l - 1$.
For $4s$ orbital: $n = 4$,$l = 0$. Therefore,$\text{Radial nodes} = 4 - 0 - 1 = 3$.
For $3s$ orbital: $n = 3$,$l = 0$. Therefore,$\text{Radial nodes} = 3 - 0 - 1 = 2$.
Thus,the number of radial nodes for $4s$ and $3s$ are $3$ and $2$ respectively.

(A) The distance of an orbital from the nucleus is determined by the principal quantum number $(n)$.
As the value of $n$ increases,the average distance of the electron from the nucleus increases.
For $2p$,$n = 2$.
For $3p$,$n = 3$.
Since $2 < 3$,the $2p$ orbital is closer to the nucleus than the $3p$ orbital.