Mix Examples-Structure of atom Questions in English

(B) node is a region where the probability of finding an electron is zero,which corresponds to the point where the wave function $\Psi$ crosses the axis (i.e.,$\Psi = 0$).
In graph $B$,the wave function $\Psi$ crosses the axis once,indicating one node.
Graph $A$ has no nodes.
Graph $C$ has two nodes.
Graph $D$ has no nodes (it represents a probability density or a different orbital function).

(C) From the energy level diagram,the energy of the transition from $E_3$ to $E_1$ is the sum of the energies of the transitions from $E_3$ to $E_2$ and $E_2$ to $E_1$.
$E_{3-1} = E_{3-2} + E_{2-1}$
Since $E = h\nu = \frac{hc}{\lambda}$,we have $h\nu_3 = h\nu_2 + h\nu_1$,which simplifies to $\nu_3 = \nu_2 + \nu_1$.

(D) quantum of energy is emitted when an electron transitions from a higher energy level to a lower energy level.
In all the given transitions ($n = 4 \rightarrow n = 2$,$n = 3 \rightarrow n = 1$,and $n = 4 \rightarrow n = 1$),the electron moves from a higher principal quantum number to a lower one,resulting in the emission of energy in the form of a photon (quantum).
Therefore,all the given transitions involve the emission of one quantum of energy.

(C) For $He^+$ $(Z=2)$:
$(A)$ Ground state energy $(n=1)$: $E_n = -13.6 \times (Z^2/n^2) \text{ eV} = -13.6 \times (4/1) = -54.4 \text{ eV}$. Thus,$A-iv$.
$(B)$ Potential energy of $H$ atom $(Z=1)$ in $I$ orbit $(n=1)$: $PE = 2 \times E_n = 2 \times (-13.6 \text{ eV}) = -27.2 \text{ eV}$. Thus,$B-ii$.
$(C)$ Kinetic energy of $II$ excited state $(n=3)$ of $He^+$: $KE = |E_n| = 13.6 \times (Z^2/n^2) = 13.6 \times (4/9) \approx 6.04 \text{ eV}$. Thus,$C-i$.
$(D)$ Ionization potential of $He^+$: Energy required to remove electron from $n=1$ to $\infty$,which is $54.4 \text{ eV}$. Converting to Joules: $54.4 \times 1.602 \times 10^{-19} \text{ J} \approx 8.72 \times 10^{-18} \text{ J}$. Thus,$D-iii$.
Therefore,the correct match is $A-iv, B-ii, C-i, D-iii$.

(C) The kinetic energy $(K.E.)$ of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula:
$K.E. = \frac{13.6 \times Z^2}{n^2} \text{ eV}$
Where $Z$ is the atomic number and $n$ is the principal quantum number.
From this relation,we can see that $K.E. \propto \frac{1}{n^2}$.
As $n$ increases,$K.E.$ decreases rapidly.
Graph $(C)$ shows the variation of $K.E.$ with $n$,where $K.E.$ is inversely proportional to $n^2$,which matches the curve shown in option $(C)$.

(D) : The electronic configuration of $Cr$ $(Z=24)$ is $[Ar] \ 3d^5 4s^1$ due to the extra stability of half-filled $d$-orbitals. This is correct.
$B$: The magnetic quantum number $(m_l)$ ranges from $-l$ to $+l$,including zero,so it can have negative values. This is correct.
$C$: For $Ag$ $(Z=47)$,the configuration is $[Kr] \ 4d^{10} 5s^1$. The $4d^{10}$ subshell has $5$ pairs of electrons (total $10$ electrons,$5$ spin up,$5$ spin down). The $5s^1$ has $1$ electron. In the filled inner shells ($[Kr]$ has $36$ electrons,$18$ up,$18$ down),we have $18$ up and $18$ down. Adding the $4d^{10}$ ($5$ up,$5$ down) and $5s^1$ ($1$ up),we get $18+5+1 = 24$ electrons of one spin and $18+5 = 23$ of the opposite spin. This is correct.
Therefore,all statements are correct.

(C) For hydrogen-like atoms (single-electron species),the energy of orbitals depends only on the principal quantum number $n$.
Therefore,the energy of $2s$ orbital is equal to the energy of $2p$ orbital $(E_{2s} = E_{2p})$.
Thus,the statement that the energy of $2s$ orbital is less than $2p$ orbital is incorrect.

(D) Statement $(I)$ is $CORRECT$ according to Hund's Rule of Maximum Multiplicity,which states that for a given electron configuration,the term with the maximum multiplicity has the lowest energy.
Statement $(II)$ is $CORRECT$ because the Pauli Exclusion Principle only states that no two electrons in an atom can have the same set of four quantum numbers. If two electrons are in different orbitals within the same shell,they can have the same spin.
Statement $(III)$ is $CORRECT$ because $He$ has a valence configuration of $1s^2$,while other noble gases like $Ne, Ar, Kr, Xe$ have $ns^2 np^6$ configurations.
Therefore,all three statements are correct.

(D) Let us analyze each statement:
$1$. For a given subshell,the principal quantum number $n$ is always greater than the azimuthal quantum number $l$ $(n > l)$. Thus,the azimuthal quantum number cannot have the highest numerical value.
$2$. Generally,as electronegativity increases,the effective nuclear charge increases,which makes it harder to remove an electron,thus increasing the ionisation energy. Therefore,the statement that it decreases is incorrect.
$3$. If the intermolecular axis is the $x$-axis,the $p_x$ orbital can only form a sigma $(\sigma)$ bond due to head-on overlap. $A$ $d_{xy}$ orbital has lobes in the $xy$ plane; it cannot form a $\pi$ bond with $p_x$ along the $x$-axis because the symmetry does not allow for lateral overlap.
$4$. Since all statements are incorrect,the correct option is $D$.

(B) The time period $T$ of an electron in the $n^{th}$ orbit is given by the formula $T \propto \frac{n^3}{Z^2}$.
For the $3^{rd}$ orbit of $H^{-}$ atom $(n_1 = 3, Z_1 = 1)$:
$T_1 \propto \frac{3^3}{1^2} = 27$.
For the $2^{nd}$ orbit of $He^{+}$ ion $(n_2 = 2, Z_2 = 2)$:
$T_2 \propto \frac{2^3}{2^2} = \frac{8}{4} = 2$.
The ratio of time periods is $\frac{T_1}{T_2} = \frac{27}{2}$.

(D) For an electron in a hydrogen-like atom,the relationship between potential energy $(PE)$ and kinetic energy $(KE)$ is given by $PE = -2 \ KE$.
Initially,$KE_1 = y$,so $PE_1 = -2 \ y$.
Finally,$KE_2 = y/4$,so $PE_2 = -2 \ (y/4) = -y/2$.
Change in potential energy $\Delta PE = PE_2 - PE_1 = -y/2 - (-2 \ y) = -y/2 + 2 \ y = \frac{3 \ y}{2}$.

(B) Statement $(a)$ is incorrect because a photon is a massless,neutral particle of light.
Statement $(b)$ is correct; the nucleus is extremely small $(10^{-15} \ m)$ compared to the atom $(10^{-10} \ m)$.
Statement $(c)$ is incorrect because diffraction demonstrates the wave nature of light,while the photoelectric effect demonstrates the particle nature.
Statement $(d)$ is correct; the dimensions of Planck's constant $(h)$ are $[ML^2T^{-1}]$,which are the same as the dimensions of angular momentum $(L = mvr = [M][LT^{-1}][L] = [ML^2T^{-1}])$.
Therefore,the correct statements are $(b)$ and $(d)$.

(B) The radius of the orbit $r$ is given by $r = 0.529 \times n^2 \ \mathring{A}$ for a hydrogen atom.
Given diameter $= 16.92 \ \mathring{A}$,so radius $r = 8.46 \ \mathring{A}$.
Equating the two: $0.529 \times n^2 = 8.46$.
$n^2 = \frac{8.46}{0.529} \approx 16$.
Thus,$n = 4$.
The maximum number of electrons in an orbit is given by the formula $2n^2$.
Number of electrons $= 2 \times (4)^2 = 2 \times 16 = 32$.

(A) For the $2s$ orbital,the radial probability density function $R^2(r)$ (or $4\pi r^2 \psi^2$) represents the probability of finding the electron at a distance $r$ from the nucleus.
For an $s$-orbital,the probability density is maximum at the nucleus $(r=0)$.
As $r$ increases,the probability density decreases,reaches a node (where probability is zero) at a certain distance,and then increases to a smaller secondary peak before decreasing again.
Therefore,the graph starts at a maximum value at $r=0$,touches the $r$-axis at the node,and then shows a smaller peak,which matches the curve in option $A$.

(D) The Aufbau principle states that electrons fill lower energy orbitals first.
According to Hund's rule,pairing of electrons in degenerate orbitals (like $2p$) only occurs after each orbital is singly occupied.
Option $A$ shows a correctly filled $2s$ and $2p$ orbital for Nitrogen $(1s^2 2s^2 2p^3)$.
Option $B$ violates the Aufbau principle because the $2s$ orbital is not fully filled before electrons enter the $2p$ orbital.
Option $C$ follows both rules for Carbon $(1s^2 2s^2 2p^2)$.
Option $D$ violates Hund's rule because electrons are paired in the first $2p$ orbital before the second $2p$ orbital is singly occupied.

(D) Both the assertion and the reason are false.
$2p_x$ and $2p_y$ orbitals are degenerate,meaning they have the same energy.
Since there is no energy difference between these orbitals,no electronic transition occurs.
Consequently,no energy is released,and no spectral line is observed.

(B) The correct matches are:
$(1)$ $\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$ is the Heisenberg Uncertainty Principle $(B)$.
$(2)$ $mvr = \frac{nh}{2\pi}$ represents the quantization of angular momentum $(D)$.
$(3)$ $\lambda = \frac{h}{\sqrt{2m(KE)}}$ is the De Broglie equation relating wavelength to kinetic energy $(A)$.
$(4)$ $\nu = 3.29 \times 10^{15} \left( \frac{1}{n_i^2} - \frac{1}{n_f^2} \right)$ is the Rydberg frequency equation for the $H$-spectrum $(C)$.
Therefore,the correct sequence is $(1-B, 2-D, 3-A, 4-C)$.

(C) The correct matches are:
$(1)$ Dalton proposed the $(C)$ Atomic theory.
$(2)$ Maxwell proposed the $(D)$ Electromagnetic radiation theory.
$(3)$ Huygens proposed the $(B)$ Wave theory.
Therefore,the correct sequence is $(1-C, 2-D, 3-B)$.

(C) The correct matching is: $(1-C, 2-E, 3-A, 4-D)$.
$(1)$ Hund's Rule states that in degenerate orbitals,electrons prefer to remain unpaired with parallel spins as much as possible.
$(2)$ Aufbau Principle states that in the ground state of atoms,electrons fill orbitals in the order of increasing energy.
$(3)$ Pauli Exclusion Principle states that no two electrons in an atom can have the same set of all four quantum numbers.
$(4)$ Heisenberg's Uncertainty Principle states that it is impossible to determine simultaneously the exact position and exact momentum of a microscopic particle like an electron.

(D) The correct matching is: $(1-D, 2-D, 3-B, 3-C, 4-A)$.
$(1)$ Photon exhibits wave-particle duality,thus it is associated with wavelength and momentum.
$(2)$ Electron also exhibits wave-particle duality,thus it is associated with wavelength and momentum.
$(3)$ $\psi^2$ represents the probability density of finding an electron and is always a positive value.
$(4)$ The principal quantum number $(n)$ for the $N$ shell is $4$.

(D) Statement $I$ is true: Rutherford's model could not explain the stability of the atom or the origin of the line spectrum of hydrogen.
Statement $II$ is true: Bohr's model assumes that electrons move in well-defined circular orbits with fixed radii and velocities,which directly contradicts Heisenberg's uncertainty principle,which states that it is impossible to determine both the position and momentum of a subatomic particle simultaneously with absolute precision.
Therefore,both statements are true.

(C) The maximum number of emission lines produced when an electron drops from an excited state $n$ to the ground state is calculated using the formula:
$\text{Number of lines} = \frac{n(n-1)}{2}$
Given $n = 5$,substituting the value into the formula:
$\text{Number of lines} = \frac{5(5-1)}{2} = \frac{5 \times 4}{2} = \frac{20}{2} = 10 \text{ lines.}$

(C) The kinetic energy $K$ of the emitted electron is given by the de Broglie wavelength formula: $\lambda = \frac{h}{\sqrt{2mK}}$.
Rearranging for $K$: $K = \frac{h^2}{2m\lambda^2}$.
Substituting the given values: $K = \frac{(6.626 \times 10^{-34})^2}{2 \times 9.1 \times 10^{-31} \times (3.3 \times 10^{-10})^2} \approx 2.215 \times 10^{-18} \ J$.
The minimum energy required to escape from the ground state of $H$ atom is the ionization energy: $E_{req} = 13.6 \ eV = 13.6 \times 1.602 \times 10^{-19} \ J \approx 2.179 \times 10^{-18} \ J$.
The total energy absorbed is $E_{abs} = E_{req} + K$.
The ratio is $\frac{E_{abs}}{E_{req}} = 1 + \frac{K}{E_{req}} = 1 + \frac{2.215 \times 10^{-18}}{2.179 \times 10^{-18}} \approx 1 + 1.016 = 2.016$.
Rounding to the nearest integer,the value is $2$.

(A) - Atoms consist of three fundamental particles: electrons,protons,and neutrons. Statement $A$ is incorrect.
- The mass of the electron is $9.10939 \times 10^{-31} \ kg$. Statement $B$ is correct.
- All isotopes of a given element have the same number of electrons and protons,thus they show the same chemical properties. Statement $C$ is correct.
- Protons and neutrons present in the nucleus are collectively called nucleons. Statement $D$ is incorrect.
- Dalton's atomic theory regarded the atom as the ultimate particle of matter. Statement $E$ is correct.
Therefore,the correct statements are $B, C,$ and $E$.

(B, D, D) $1$. For $2s$ orbital $(n=2, l=0)$,the number of radial nodes is $(n-l-1) = (2-0-1) = 1$. The graph $(P)$ represents the radial wave function of $2s$ orbital,which has one radial node. Thus,$(II)(ii)(P)$ is correct.
$2$. For $He^{+}$ ion,$1s$ orbital is $\psi \propto (\frac{Z}{a_0})^{3/2} e^{-(Zr/a_0)}$. Option $(D)$ is incorrect because $1s$ orbital has no angular dependence (no $\theta$ function),and $(R)$ is incorrect because the probability density for $1s$ is maximum at the nucleus,but the combination $(I)$(iii)$(R)$ is fundamentally wrong due to the angular function $(iii)$.
$3$. For the hydrogen atom,$1s$ orbital is $\psi \propto (\frac{Z}{a_0})^{3/2} e^{-(Zr/a_0)}$. The energy required for $n=2 \to n=4$ is $E_4 - E_2 = -13.6 Z^2(\frac{1}{16} - \frac{1}{4}) = 13.6 Z^2(\frac{3}{16})$. The energy for $n=2 \to n=6$ is $E_6 - E_2 = -13.6 Z^2(\frac{1}{36} - \frac{1}{4}) = 13.6 Z^2(\frac{8}{36}) = 13.6 Z^2(\frac{2}{9})$. The ratio is $\frac{3/16}{2/9} = \frac{27}{32}$. Thus,$(I)(i)(S)$ is correct.

(A) . Heisenberg's uncertainty principle states that it is impossible to determine simultaneously the exact position and momentum of an electron,thus ruling out definite paths.
$B$. Electrons in orbitals have negative potential energy due to attraction by the nucleus. An electron at infinite distance has zero energy. Since negative values are lower than zero,the statement is correct.
$C$. The energy of an electron is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \ eV$. For $n=1$,the energy is most negative,indicating the most stable state.
$D$. The velocity of an electron in Bohr's model is given by $V_n = 2.19 \times 10^6 \times \frac{Z}{n} \ m/s$. As $n$ increases,the velocity $V_n$ decreases,not increases.

(D) The wave function for the $1s$ orbital of a hydrogen atom is given by $\Psi_{1s} = \frac{1}{\sqrt{\pi a_0^3}} e^{-r/a_0}$.
$A.$ The probability density $\Psi^2$ is proportional to $e^{-2r/a_0}$. At $r = 0$ (nucleus),$\Psi^2$ is maximum. This statement is correct.
$B.$ The electron has a non-zero probability of being found at any distance $r$ from the nucleus,including $2a_0$. This statement is correct.
$C.$ The $1s$ orbital has no angular dependence,making it spherically symmetrical. This statement is correct.
$D.$ The total energy of an electron in a hydrogen atom is given by $E_n = -\frac{13.6 \ eV}{n^2}$. This energy is constant for a given orbital and does not depend on the distance $r$ from the nucleus. The energy is maximum (zero) at $r = \infty$. Therefore,this statement is incorrect.

(B) $1$. Pauli's exclusion principle states that 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 with opposite spins.
$2$. Hund's rule of maximum multiplicity states that electron pairing in degenerate orbitals (like $p, d, f$) does not occur until each orbital is singly occupied.
$3$. In option $B$,the first orbital contains two electrons with the same spin (violates Pauli's principle),and the second orbital also contains two electrons with the same spin (violates Pauli's principle). Furthermore,pairing occurs before all orbitals are singly occupied (violates Hund's rule).
$4$. Therefore,option $B$ violates both principles.

(D) The energy of stationary states in hydrogen-like atoms is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \ eV$,which is inversely proportional to $n^2$. Thus,option $A$ is correct.
The radius of the $n^{th}$ orbit is given by $r_n = 0.529 \times \frac{n^2}{Z} \ \mathring{A}$. For $H$ $(Z=1, n=1)$,$r_1 = 0.529 \ \mathring{A}$. For $He^{+}$ $(Z=2, n=1)$,$r_1 = 0.529 \times \frac{1^2}{2} = 0.2645 \ \mathring{A}$. Since $0.2645$ is half of $0.529$,option $B$ is correct.
The angular quantum number $(l)$ determines the shape of the orbital. Thus,option $C$ is correct.
The total number of nodes in an orbital is given by $n - 1$. For the $3s$ orbital,$n = 3$,so the total number of nodes is $3 - 1 = 2$. Option $D$ states the number of nodes is $3$,which is incorrect.

(A) Statement $I$ is incorrect because the electron spin quantum number describes the orientation of the spin of the electron,not the nucleus.
Statement $II$ is correct because orbitals with the same $n$ and $l$ values (degenerate orbitals) have the same energy in the absence of an external magnetic field.
Statement $III$ is incorrect because the energy of a photon $(E = h\nu = \frac{hc}{\lambda})$ is inversely proportional to the wavelength $(\lambda)$ and directly proportional to the wave number $(\bar{\nu} = \frac{1}{\lambda})$.
Statement $IV$ is correct because the Lyman series corresponds to electronic transitions to the $n=1$ energy level,which emits radiation in the ultra-violet region.
Therefore,statements $II$ and $IV$ are correct.

(A) . Nodes $(III)$ $|\psi|^2$ is zero,as $|\psi|^2$ represents the region where the probability of finding an electron is zero.
$B$. Subsidiary Quantum Number $(I)$ determines the three-dimensional shape of the orbital. It is also called the azimuthal quantum number.
$C$. White Light $(V)$ produces a continuous spectrum because it contains all wavelengths of visible light,appearing as a continuous pattern.
$D$. Heisenberg's Uncertainty Principle $(II)$ is significant only for the motion of microscopic objects,as it states that it is impossible to determine simultaneously the exact position and exact momentum of a subatomic particle.
Therefore,the correct matching is $A-III, B-I, C-V, D-II$.

(B) Statement $A$ is incorrect because the notation ${}_{12}^{24}Mg$ represents $12$ protons (atomic number $Z=12$) and $12$ neutrons $(A-Z = 24-12=12)$.
Statement $C$ is incorrect because energy $E$ is inversely proportional to wavelength $\lambda$ $(E = hc/\lambda)$. Therefore, $E_1/E_2 = \lambda_2/\lambda_1 = 300/900 = 1/3$, which means $E_1 : E_2 = 1 : 3$.
Statement $B$ is correct: $\lambda = c/\nu = (3 \times 10^8 \ m/s) / (4.5 \times 10^{15} \ s^{-1}) = 6.67 \times 10^{-8} \ m \approx 6.7 \times 10^{-8} \ m$.
Statement $D$ is correct: $n = E\lambda / hc = (1 \ J \times 2000 \times 10^{-12} \ m) / (6.626 \times 10^{-34} \ J \cdot s \times 3 \times 10^8 \ m/s) \approx 1.006 \times 10^{16}$.

(B) is correct: Heisenberg uncertainty principle applies to microscopic particles like electrons.
$B$ is correct: The size of an orbital increases with an increase in the principal quantum number $(n)$. Since $n=3 > n=2$,the size of $3p_x$ is greater than $2p_x$.
$C$ is incorrect: For multi-electron atoms,the energy of an orbital depends on both $n$ and $l$. For hydrogen-like species,energy depends on $Z^2$. Since $H$ $(Z=1)$ and $Li$ $(Z=3)$ have different atomic numbers,their $2s$ orbitals have different energies.
$D$ is correct: The electronic configuration of $Cr$ $(Z=24)$ is an exception to the Aufbau principle,resulting in $[Ar] 3d^5 4s^1$ due to the extra stability of half-filled $d$-orbitals.