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

(B) The correct answer is $(B)$.
According to Pauli's exclusion principle,no two electrons in an atom can have the same set of all four quantum numbers $(n, l, m_l, m_s)$.
Since an orbital is defined by the first three quantum numbers $(n, l, m_l)$,the only way to distinguish two electrons in the same orbital is by their spin quantum number $(m_s)$,which can only have values of $+1/2$ or $-1/2$.
Therefore,an orbital can accommodate a maximum of two electrons with opposite spins.

(A) According to the Aufbau principle,orbitals are filled in the order of increasing energy,meaning the lowest energy level is occupied first before moving to higher energy levels.
Therefore,the statement follows the Aufbau principle.

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

(B) The Aufbau principle states that electrons fill lower energy orbitals before higher energy ones.
Exceptions to this rule occur due to the extra stability provided by half-filled or fully-filled $d$-orbitals.
For $Cr$ $(Z=24)$,the configuration is $[Ar] 3d^5 4s^1$ instead of $[Ar] 3d^4 4s^2$.
For $Cu$ $(Z=29)$,the configuration is $[Ar] 3d^{10} 4s^1$ instead of $[Ar] 3d^9 4s^2$.
Similarly,$Ag$ $(Z=47)$ shows an exceptional configuration of $[Kr] 4d^{10} 5s^1$.
Therefore,the pair $Cu$ and $Ag$ exhibits deviations from the standard Aufbau filling order.

(C) The Aufbau principle states that in the ground state of an atom,the orbitals are filled in the increasing order of their energies,starting from the lowest energy orbital.

(A) According to the Aufbau principle,electrons are filled in the orbitals in the increasing order of their energy levels $(1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s)$.
In option $(A)$,the configuration $1s^2 2s^2 2p^6$ follows the correct order of energy filling.
In option $(B)$,$3p$ is filled before $3s$,which violates the principle.
In option $(C)$,the $2s$ and $2p$ orbitals are skipped.
In option $(D)$,the $2p$ orbital is skipped.
Therefore,the correct configuration is $1s^2 2s^2 2p^6$.

(D) The electronic configuration of $Cr$ $(Z=24)$ is $[Ar] \ 3d^5 \ 4s^1$.
According to Hund's rule,the $3d$ subshell has $5$ unpaired electrons and the $4s$ subshell has $1$ unpaired electron.
Total number of unpaired electrons = $5 + 1 = 6$.
Thus,$Cr$ contains $6$ unpaired electrons.

(B) The $(n + l)$ rule,also known as the $n+l$ rule or the Madelung rule,is a part of the $Aufbau$ principle.
According to the $Aufbau$ principle,electrons fill atomic orbitals in the order of increasing energy,starting from the orbital with the lowest $(n + l)$ value.
If two orbitals have the same $(n + l)$ value,the electron enters the orbital with the lower $n$ value.

(B) The $Aufbau$ principle,also called the $Aufbau$ rule,states that in the ground state of an atom or ion,electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels.

(B) According to the $(n+l)$ rule,the energy of an orbital increases with the value of $(n+l)$.
For $5p$: $n=5, l=1$,so $(n+l) = 6$.
For $6s$: $n=6, l=0$,so $(n+l) = 6$.
For $4f$: $n=4, l=3$,so $(n+l) = 7$.
For $5d$: $n=5, l=2$,so $(n+l) = 7$.
If $(n+l)$ values are equal,the orbital with the lower $n$ value has lower energy.
Comparing $5p$ and $6s$: both have $(n+l) = 6$,but $5p$ has a lower $n$ value,so $5p < 6s$.
Comparing $4f$ and $5d$: both have $(n+l) = 7$,but $4f$ has a lower $n$ value,so $4f < 5d$.
Thus,the correct order is $5p < 6s < 4f < 5d$.

(D) The energy of an orbital is determined by the $(n + l)$ rule,where $n$ is the principal quantum number and $l$ is the azimuthal quantum number.
For $3d$: $n=3, l=2$,so $(n+l) = 3+2 = 5$.
For $5p$: $n=5, l=1$,so $(n+l) = 5+1 = 6$.
For $4s$: $n=4, l=0$,so $(n+l) = 4+0 = 4$.
For $6d$: $n=6, l=2$,so $(n+l) = 6+2 = 8$.
Since the orbital with the highest $(n+l)$ value has the maximum energy,$6d$ has the maximum energy.

(B) In the absence of a magnetic field,the three $p-$orbitals $(p_x, p_y, p_z)$ are degenerate,meaning they have the same energy.
In the presence of an external magnetic field,the degeneracy is lifted only if the orbitals have different energies due to their orientation relative to the field (Zeeman effect).
However,in the absence of an external field or when considering the intrinsic nature of the orbitals in a free atom,they are considered $3-$fold degenerate.
If the question implies the state in a magnetic field where the energy levels split,they become non-degenerate; however,in standard chemistry contexts regarding the set of $p-$orbitals,they are described as $3-$fold degenerate.

(B) The orbital angular momentum of an electron is given by the formula: $\text{Angular Momentum} = \sqrt{l(l + 1)} \frac{h}{2\pi}$.
For a $d$-electron,the azimuthal quantum number $l = 2$.
Substituting the value of $l$ into the formula:
$\text{Angular Momentum} = \sqrt{2(2 + 1)} \frac{h}{2\pi} = \sqrt{2(3)} \frac{h}{2\pi} = \sqrt{6} \frac{h}{2\pi}$.
Thus,the correct option is $B$.

(A) The number of radial nodes in an orbital is given by the formula $(n - l - 1)$.
For a $2s$ orbital,the principal quantum number $n = 2$ and the azimuthal quantum number $l = 0$.
Therefore,the number of radial nodes $= 2 - 0 - 1 = 1$.
Thus,the correct option is $A$.

(D) The orbital angular momentum is given by the formula $\sqrt{l(l + 1)} \times \frac{h}{2\pi }$.
For an $s$-orbital,the azimuthal quantum number $l = 0$.
Substituting $l = 0$ into the formula:
$\text{Angular momentum} = \sqrt{0(0 + 1)} \times \frac{h}{2\pi } = 0 \times \frac{h}{2\pi } = 0$.

(D) The azimuthal quantum number $l = 3$ corresponds to the $f$-subshell.
The number of orbitals in a subshell is given by $(2l + 1)$.
For $l = 3$,the number of orbitals is $2(3) + 1 = 7$.
Since each orbital can hold a maximum of $2$ electrons,the total number of electrons is $7 \times 2 = 14$.

(B) According to the Aufbau principle,electrons fill orbitals in increasing order of their $(n+l)$ values. If two orbitals have the same $(n+l)$ value,the one with the lower $n$ value is filled first.
For option $(B)$,the order is $1s (n+l=1) < 2p (n+l=3) < 4s (n+l=4) < 3d (n+l=5)$. This sequence follows the increasing order of energy levels correctly.

(A) Deuterium is an isotope of hydrogen with atomic number $Z = 1$.
Since the atomic number represents the number of protons (and electrons in a neutral atom),deuterium has $1$ electron.
The electronic configuration of a $1$-electron system is $1s^1$.

(A) The number of nodal planes in an orbital is given by the azimuthal quantum number,$l$.
For a $p$ orbital,the value of $l$ is $1$.
Therefore,the number of nodal planes in a $p_x$ orbital is $l = 1$.
The nodal plane for a $p_x$ orbital is the $yz$-plane.

(C) The electronic configuration of $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$.
$1$. For $Fe$: $3d^6 4s^2$. The $3d$ subshell has $6$ electrons,which means $4$ unpaired electrons.
$2$. For $Fe^{2+}$: $[Ar] 3d^6$. The $3d$ subshell has $6$ electrons,which means $4$ unpaired electrons.
$3$. For $Fe^{3+}$: $[Ar] 3d^5$. The $3d$ subshell has $5$ electrons,which means $5$ unpaired electrons.
$4$. For $Fe^{4+}$: $[Ar] 3d^4$. The $3d$ subshell has $4$ electrons,which means $4$ unpaired electrons.
Thus,$Fe^{3+}$ has the maximum number of unpaired electrons $(5)$.

(B) Non-metals readily form diatomic molecules by sharing electrons to complete their octets.
Element $M$ has the configuration $1s^2, 2s^2 2p^5$,meaning it has $7$ electrons in its valence shell.
It requires one more electron to complete its octet.
Therefore,two atoms of $M$ share one electron each to form a diatomic molecule $(M_2)$,similar to the formation of $F_2$ from fluorine atoms.

(C) The atomic number of element $X$ is $7.$
Its electronic configuration is $1s^2 2s^2 2p^3,$ which means it has $5$ valence electrons.
According to Lewis dot structure,the $5$ valence electrons are represented by placing dots around the symbol $X.$
Two electrons are paired,and three are unpaired,represented as $:\,\mathop {\mathop X\limits^. }\limits_. .$.

(A) completely filled $d^{10}$ subshell consists of five $d$ orbitals,each containing two electrons with opposite spins. Due to the uniform distribution of electron density in all directions,the resulting electron cloud is spherically symmetrical.

(B) For multi-electron atoms,the energy of an orbital depends on both the principal quantum number $(n)$ and the azimuthal quantum number $(l)$.
According to the $(n+l)$ rule,the energy of an orbital increases as the value of $(n+l)$ increases.
For $2s$ orbital: $n=2, l=0$,so $(n+l) = 2+0 = 2$.
For $2p$ orbital: $n=2, l=1$,so $(n+l) = 2+1 = 3$.
Since the $(n+l)$ value for $2p$ is greater than that for $2s$,the $2p$ orbital has higher energy than the $2s$ orbital.

(C) The number of angular nodes (nodal planes) for an orbital is given by the value of the azimuthal quantum number,$l$.
For a $d$ orbital,the value of $l$ is $2$.
Therefore,a $d$ orbital has $2$ nodal planes.

(B) The principal quantum number $n = 2$ indicates that the element belongs to the $2^{nd}$ period.
The azimuthal quantum number $l = 1$ indicates a $p$-orbital.
For $l = 1$,the possible values of $m$ are $+1, 0, -1$. The given values $m = 1$ and $s = -1/2$ correspond to the last electron in the $2p$ subshell.
The configuration $2p^6$ represents a fully filled subshell,which corresponds to the noble gas Neon $(Ne)$.
Elements with a $p^6$ configuration belong to the $18^{th}$ group,which is also known as Group $0$.

(C) The electronic configuration is $1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^5, 4s^1$.
Since the last electron enters the $3d$ subshell,the element belongs to the $d$-block.
Therefore,the correct option is $(c)$.

(B) noble gas has a stable,fully filled valence shell configuration,typically represented as $ns^2 np^6$ for elements in the $p$-block.
Option $B$ $(1s^2 2s^2 2p^6 3s^2 3p^6)$ corresponds to Argon $(Ar)$,which is a noble gas with a complete octet in its outermost shell $(n=3)$.

(B) The modern periodic law states that the physical and chemical properties of elements are periodic functions of their atomic numbers. Since the atomic number is equal to the number of protons,which determines the arrangement of electrons in an atom,these properties are essentially periodic functions of their electronic configuration.

(B) The total number of electrons in the given configuration $1s^2 2s^2 2p^6 3s^2$ is $2 + 2 + 6 + 2 = 12$.
Since the atom is neutral,the atomic number is equal to the number of electrons,which is $12$.
The element with atomic number $12$ is Magnesium $(Mg)$.

(C) The electronic configuration $[Ne] \ 3s^2$ corresponds to Magnesium $(Mg)$.
$Mg$ has two valence electrons in the $3s$-orbital.
It can easily lose these two electrons to achieve a stable noble gas configuration,forming a dipositive ion $(Mg^{2+})$.
Therefore,it has the maximum tendency to form dipositive ions among the given options.

(C) The electronic configuration of the element is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1$.
Since the last electron enters the $3d$ subshell,the element belongs to the $d$-block of the periodic table.
Therefore,the correct option is $(C)$.

(D) The number of unpaired electrons for each element in option $D$ is as follows:
$N$ $(Z=7)$: $1s^2 2s^2 2p^3$ has $3$ unpaired electrons.
$P$ $(Z=15)$: $1s^2 2s^2 2p^6 3s^2 3p^3$ has $3$ unpaired electrons.
$V$ $(Z=23)$: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3$ has $3$ unpaired electrons.
Since all three elements in option $D$ have $3$ unpaired electrons,this is the correct set.

(A) According to the $(n+l)$ rule,the energy of an orbital increases as the value of $(n+l)$ increases.
For a given $(n+l)$ value,the orbital with the lower $n$ value has lower energy.
Following the Aufbau principle,the sequence of energy levels is $ns < (n-2)f < (n-1)d < np$ is incorrect; the standard filling order is $ns < (n-2)f < (n-1)d < np$ depending on the specific $n$ value.
However,based on the $(n+l)$ rule,the general increasing order of energy is $ns < (n-2)f < (n-1)d < np$ is not universally true for all $n$. Given the options,the standard order of increasing energy is $ns < (n-2)f < (n-1)d < np$ is often represented as $ns < (n-2)f < (n-1)d < np$.

(D) The atomic number of calcium $(Ca)$ is $20$.
Its electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$.
The valence shell is the $4^{th}$ shell $(n=4)$,which contains $2$ electrons.
Therefore,the correct option is $(D)$.

(D) The correct answer is $(D)$.
$He$ (Helium) has an atomic number of $2$,meaning its electronic configuration is $1s^2$.
It has only $2$ electrons in its outermost shell (duplet) and does not possess an octet ($8$ electrons) like other noble gases.

(B) The atomic number of oxygen $(O)$ is $8$.
The electronic configuration of oxygen is $1s^2 2s^2 2p^4$.
In the $2p$-subshell,there are $4$ electrons.
According to Hund's rule of maximum multiplicity,the electrons are filled as follows:
$2p_x^2, 2p_y^1, 2p_z^1$.
Thus,there are $2$ unpaired electrons in the $p$-subshell.

(B) Argon is a noble gas with atomic number $18$.
The electronic configuration of argon is $1s^2 2s^2 2p^6 3s^2 3p^6$.
The outermost shell configuration is $3s^2 3p^6$,which corresponds to the general noble gas configuration $ns^2 np^6$ where $n=3$.

(D) Argon $(Ar)$ has an atomic number of $18$.
The electronic configuration of argon is $1s^2 2s^2 2p^6 3s^2 3p^6$.
The distribution of electrons in shells is $K=2, L=8, M=8$.
Therefore,the last orbit (valence shell) of argon contains $8$ electrons.

(B) The atomic number of neon $(Ne)$ is $10$.
Following the Aufbau principle,the electrons are filled in the orbitals in the order of increasing energy: $1s, 2s, 2p$.
Thus,the electronic configuration is $1s^2 2s^2 2p^6$.

(C) Noble gases have a stable valence shell configuration,typically $ns^2 np^6$ (except for Helium,which is $1s^2$).
Option $C$ represents the configuration $1s^2, 2s^2 2p^6$,which corresponds to the element Neon ($Ne$,atomic number $10$).
This configuration has a completely filled valence shell,making it a noble gas.

(D) The correct answer is $(D)$.
$A$: Mass defect is directly related to binding energy by the equation $BE = \Delta m \times c^2$.
$B$: Hideki Yukawa proposed the existence of mesons in $1935$.
$C$: The radius of a nucleus is typically in the range of $10^{-15} \ m$ to $10^{-14} \ m$,which is $10^{-13} \ cm$ to $10^{-12} \ cm$.
$D$: The magnetic quantum number $(m_l)$ determines the orientation of the orbital in space,whereas the azimuthal quantum number $(l)$ is a measure of the orbital angular momentum of the electron.

(B) The atomic number of iron $(Fe)$ is $26$.
The electronic configuration of $Fe$ is $[Ar] 3d^6 4s^2$.
When $Fe$ forms a ferrous ion $(Fe^{2+})$,it loses two electrons from the $4s$ orbital.
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: $4$ orbitals contain $1$ electron each,and $1$ orbital contains $2$ electrons.
Therefore,the number of unpaired electrons in $Fe^{2+}$ is $4$.

(B) The atomic number of $Fe$ is $26$. The electronic configuration of $Fe$ is $[Ar] 3d^6 4s^2$.
For $Fe^{+3}$ ion,three electrons are removed (two from $4s$ and one from $3d$),resulting in the configuration $[Ar] 3d^5$.
In the $3d^5$ subshell,there are $5$ orbitals,each containing one electron according to Hund's rule.
Therefore,the number of unpaired electrons in $Fe^{+3}$ is $5$.

(B) The atomic number of cobalt $(Co)$ is $27$.
Following the Aufbau principle,the electrons are filled in the order of increasing energy levels: $1s, 2s, 2p, 3s, 3p, 4s, 3d$.
The distribution of $27$ electrons is: $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^7$.
Thus,the correct electronic configuration is $1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^7, 4s^2$.

(B) The atomic number of cobalt $(Co)$ is $27$.
The electronic configuration of $Co$ is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^7 4s^2$.
In the $3d$ subshell,there are $7$ electrons. According to Hund's rule,these electrons fill the five $d$-orbitals as follows: $\uparrow\downarrow, \uparrow\downarrow, \uparrow, \uparrow, \uparrow$.
This results in $3$ unpaired electrons in the $3d$ subshell.
Therefore,the correct option is $B$.

(B) The total number of electrons in the given configuration is $2 + 2 + 6 + 2 + 6 + 6 = 24$.
$Fe$ has an atomic number of $26$.
$Fe^{2+}$ is formed by the loss of two electrons from the $4s$ orbital of $Fe$ $(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6)$.
Thus,the electronic configuration of $Fe^{2+}$ is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^6$.

(C) The ground state electronic configuration of a chromium $(Cr)$ atom $(Z=24)$ is $[Ar] 3d^5 4s^1$.
$(A)$ It has $5$ electrons in $3d$ and $1$ electron in $4s$ orbitals,which is correct.
$(B)$ The valence electrons are in the $3d$ $(n=3)$ and $4s$ $(n=4)$ orbitals,which is correct.
$(C)$ It has $5$ electrons in the $3d$ orbital,not $6$. Thus,this statement is incorrect.
$(D)$ For $4s$ electrons,$l=0$ ($s$-orbital),and for $3d$ electrons,$l=2$ ($d$-orbital),which is correct.
Therefore,the incorrect statement is $(C)$.

(B) . The expected electronic configuration of $Cu$ $(Z = 29)$ is $[Ar] 3d^{9} 4s^{2}$,but the actual configuration is $[Ar] 3d^{10} 4s^{1}$.
This occurs because fully filled $d$-orbitals are more stable than partially filled $d$-orbitals.
Therefore,one electron from the $4s$ orbital migrates to the $3d$ orbital to achieve a more stable,fully filled $d^{10}$ configuration.

(D) $P(0)$: Phosphorus is a $p-$block element (Group $15$). Its valence shell is the $3rd$ shell $(3s^2 3p^3)$,and it does not have electrons in the $3d-$subshell.
$Fe(III)$: The electronic configuration is $[Ar] 3d^5$. It has valence electrons in the $3d-$subshell.
$Mn(II)$: The electronic configuration is $[Ar] 3d^5$. It has valence electrons in the $3d-$subshell.
$Cr(I)$: The electronic configuration is $[Ar] 3d^5$. It has valence electrons in the $3d-$subshell.
Therefore,$P(0)$ is the correct answer.