Atomic Structure MCQs with Answers – Part 3 (NEET Chemistry)

Explanation: Graph P represents the spatial probability density associated with the hydrogen \(1s\) wave function. The \(1s\) spatial probability density is greatest at the nucleus and decreases with increasing radial distance. Graph Q represents a different quantity: the radial probability-distribution function. For a spherically symmetric orbital, the probability in a thin spherical shell depends on both the local probability density and the volume of that shell. The relevant radial factor is \[ P(r)=4\pi r^2|\Psi_{1s}(r)|^2 \] The factor \(4\pi r^2\) accounts for the increasing surface area of spherical shells as their radius increases. At \(r=0\), this factor is zero even though the \(1s\) spatial probability density is finite and maximal. Therefore, the radial probability-distribution function starts from zero at the nucleus. At a small positive radius, the increasing shell-area factor initially causes the radial probability distribution to rise. At sufficiently large distances, the decreasing wave-function magnitude dominates, and the radial distribution falls towards zero. The two functions may consequently have their maxima at different radial distances. This does not represent a contradiction because the graphs describe different probability-related quantities. The value of the probability density at a point must not be confused with the probability of finding the electron within an entire spherical shell. Both graphs can therefore be correct representations derived from the same \(1s\) orbital.

Explanation: Statement I is correct because the wave function vanishes at the radial node. The probability density is proportional to \(|\Psi|^2\), which is zero wherever the wave function is zero. A radial node is therefore a surface of zero probability density in the ideal orbital description. Statement II is incorrect because probability density cannot be negative. Although the real \(2s\) wave function changes sign across its radial node, squaring its magnitude removes that sign. A negative value of the wave function in the outer region can still correspond to a positive probability density. Statement III is correct because the radial node is not an impenetrable physical barrier. The \(2s\) wave function is nonzero over regions on both sides of the node. Consequently, electron probability can occur in the outer region beyond the node. Statement IV is correct because a boundary chosen to enclose \(90\%\) of the probability leaves \(10\%\) of the total probability outside it. Such a boundary is a representation of a selected probability region, not a sharply defined limit of possible electron positions. The radial node and the chosen boundary surface therefore have different meanings. The node is associated with a zero of the wave function, whereas the outer boundary is associated with a selected cumulative probability. Confusing these surfaces would incorrectly treat the orbital as a collection of disconnected, rigid regions. The correct interpretation retains the zero probability density at the node while allowing nonzero probability in other regions beyond it. Thus, Statements I, III and IV are correct.

Explanation: Statement I is correct because electron probability density is determined by the squared magnitude of the wave function. For the two descriptions, \[ |\Psi_Q|^2=|-\Psi_P|^2=|\Psi_P|^2 \] Thus, reversing the overall sign does not change the probability density at any position. Statement II is correct because a radial node is a finite radial distance at which the wave function becomes zero. If \(\Psi_P\) vanishes at a particular radius, then \(-\Psi_P\) also vanishes at precisely that radius. Consequently, both descriptions have the same radial node. Statement III is incorrect because the sign of a real wave function does not determine the magnitude of electron probability. A negative wave-function amplitude may have a greater squared magnitude than a positive amplitude elsewhere. Comparing probabilities therefore requires comparing \(|\Psi|^2\), not the signs of the wave-function values. Statement IV is correct because changing the overall sign of a normalised real orbital wave function does not change its physical probability distribution. The positions of its nodes and the shape of its probability-density distribution remain unchanged. An overall sign change must also be distinguished from arbitrarily changing the sign in only one region of a wave function. The latter modification need not describe the same permitted atomic state. The question illustrates the distinction between the mathematical sign convention of a wave function and the physically meaningful probability density. Therefore, Statements I, II and IV are correct.

Explanation: \(\textbf{Distinguish local density from shell probability:}\) Spatial probability density describes probability per unit volume. The probability of finding an electron in a small region depends on the probability density and the volume of that region. For a thin spherical shell, \[ \Delta V\approx4\pi r^2\Delta r \] When the density is approximately constant across the shell thickness, \[ \Delta P\approx\rho(r)\,4\pi r^2\Delta r \] \(\textbf{Calculate the relative inner-shell probability:}\) The inner shell is centred at \(r_0\), where \[ \rho(r_0)=\frac{\rho_0}{2} \] Therefore, \[ \Delta P_{\mathrm{inner}}\approx\frac{\rho_0}{2}(4\pi r_0^2\Delta r) \] \(\textbf{Calculate the relative outer-shell probability:}\) The outer shell is centred at \(2r_0\), where \[ \rho(2r_0)=\frac{\rho_0}{8} \] Its approximate volume is \[ \Delta V_{\mathrm{outer}}\approx4\pi(2r_0)^2\Delta r \] \[ \Delta V_{\mathrm{outer}}\approx16\pi r_0^2\Delta r \] Hence, \[ \Delta P_{\mathrm{outer}}\approx\frac{\rho_0}{8}(16\pi r_0^2\Delta r) \] \(\textbf{Compare the two probabilities:}\) \[ \frac{\Delta P_{\mathrm{outer}}}{\Delta P_{\mathrm{inner}}} =\frac{(\rho_0/8)(16\pi r_0^2\Delta r)}{(\rho_0/2)(4\pi r_0^2\Delta r)} \] \[ \frac{\Delta P_{\mathrm{outer}}}{\Delta P_{\mathrm{inner}}} =\frac{2}{2} \] \[ \frac{\Delta P_{\mathrm{outer}}}{\Delta P_{\mathrm{inner}}}=1 \] \(\textbf{Physical interpretation:}\) The spatial probability density is four times smaller at the outer radius. However, the outer spherical shell has four times the volume of the inner shell when their radial thicknesses are equal and sufficiently small. These two factors cancel. Therefore, a smaller local probability density need not imply a smaller probability in an entire spherical shell. This is the reason that the radial probability-distribution function contains the geometrical factor \(4\pi r^2\). The most probable radial region need not coincide with the point of maximum spatial probability density. \(\textbf{Final answer:}\) \[ \boxed{\frac{\Delta P_{\mathrm{outer}}}{\Delta P_{\mathrm{inner}}}=1} \]

Explanation: \(\textbf{Identify the radial-node counts:}\) The number of radial nodes in an atomic orbital is \[ N_{\mathrm{radial}}=n-l-1 \] For the \(3s\) orbital, \(n=3\) and \(l=0\). Therefore, \[ N_{\mathrm{radial}}(3s)=3-0-1=2 \] For the \(3p\) orbital, \(l=1\): \[ N_{\mathrm{radial}}(3p)=3-1-1=1 \] For the \(3d\) orbital, \(l=2\): \[ N_{\mathrm{radial}}(3d)=3-2-1=0 \] \(\textbf{Compare hydrogen-like orbital energies:}\) In the elementary one-electron hydrogen-like description, orbital energy depends on the principal quantum number and nuclear charge. The three orbitals share \(n=3\). They consequently possess equal energies in that description. This equality is called degeneracy. \(\textbf{Account for the multi-electron environment:}\) In a multi-electron atom, electrons interact with one another as well as with the nucleus. Shielding and penetration affect the effective nuclear attraction experienced by electrons in different subshells. Within the same principal shell, \(s\) orbitals generally penetrate towards the nucleus more strongly than \(p\) orbitals. The \(p\) orbitals generally penetrate more strongly than the corresponding \(d\) orbitals. Consequently, the usual subshell energy ordering within this shell is \[ E_{3s}\lt E_{3p}\lt E_{3d} \] \(\textbf{Interpret the combined comparison:}\) The radial-node counts depend on the orbital quantum numbers and do not become reversed when other electrons are present. However, the energy degeneracy characteristic of the simple one-electron system is generally removed in the multi-electron atom. Orbitals can therefore possess different radial structures while remaining degenerate in a hydrogen-like species. Conversely, equal principal quantum numbers do not guarantee equal orbital energies in a multi-electron atom. The correct option must satisfy both the nodal relationships and the distinction between the two energy models.

Explanation: A normalised wave function represents a quantum state for which the total probability of finding the electron somewhere in space is one. The associated probability density is obtained from the squared magnitude of the wave function. For \(\Psi_P=-\Psi\), the probability density becomes \[ |\Psi_P|^2=|-\Psi|^2=|\Psi|^2 \] Thus, the overall sign reversal leaves the probability density unchanged. The normalisation is also preserved because the squared magnitude has not changed. The positions of all nodes remain the same because multiplying a wave function by \(-1\) does not change where it is zero. For the second construction, \(\Psi_Q=2\Psi\), the squared magnitude is \[ |\Psi_Q|^2=|2\Psi|^2=4|\Psi|^2 \] The resulting density is four times the original density at every point before normalisation. Its total integral over space is consequently four rather than one. It cannot therefore be used directly as a properly normalised probability density. Dividing \(\Psi_Q\) by its normalisation factor of two restores the original wave function. This does not create a new orbital with different nodes or a different spatial probability distribution. The question distinguishes changes in a mathematical amplitude from changes in the physical electronic state. An overall sign choice is physically immaterial for a single normalised orbital, while an arbitrary change of magnitude requires appropriate normalisation.

Explanation: The hydrogen \(2s\) orbital has one radial node at a finite distance from the nucleus. At this radial distance, the real wave function passes through zero and changes sign. Therefore, \[ \Psi(r_0)=0 \] The probability density is determined by the squared magnitude of the wave function: \[ \rho(r)=|\Psi(r)|^2 \] Consequently, the probability density becomes zero at the same radial node. However, it cannot become negative after the node because a squared magnitude is nonnegative. The radial probability-distribution function includes an additional geometrical factor: \[ P(r)=4\pi r^2|\Psi(r)|^2 \] At the radial node, the squared wave function is zero, so the radial probability distribution is also zero. At the nucleus, the \(2s\) wave function and its spatial probability density are nonzero. Nevertheless, the radial probability distribution is zero at \(r=0\) because the factor \(r^2\) vanishes. The zero of \(P(r)\) at the nucleus therefore has a different origin from its zero at the finite radial node. A radial probability-distribution graph must not be confused with a graph of spatial probability density against radial distance. The correct interpretation requires identifying both the wave-function node and the effect of spherical-shell geometry.

Explanation: \(\textbf{Determine the two spatial probability densities:}\) For a real orbital wave function, the probability density is proportional to the square of the wave-function amplitude. Therefore, \[ \rho=|\Psi|^2 \] In region P, \[ \Psi_P=+2u \] Hence, \[ \rho_P=|2u|^2=4u^2 \] In region Q, \[ \Psi_Q=-u \] Therefore, \[ \rho_Q=|-u|^2=u^2 \] The negative sign in region Q does not imply negative electron probability. \(\textbf{Include the unequal spatial volumes:}\) When probability density is approximately constant within a sufficiently small region, \[ P_{\mathrm{region}}\approx\rho V_{\mathrm{region}} \] For region P, \[ P_P\approx(4u^2)V \] For region Q, \[ P_Q\approx(u^2)(6V) \] \(\textbf{Calculate the probability ratio:}\) \[ \frac{P_P}{P_Q}=\frac{4u^2V}{6u^2V} \] \[ \frac{P_P}{P_Q}=\frac46 \] \[ \frac{P_P}{P_Q}=\frac23 \] \(\textbf{Interpret the result:}\) Region P has four times the probability density of region Q. Nevertheless, region Q occupies six times the volume of region P. The greater volume more than compensates for its smaller local density. Consequently, the total probability associated with Q is greater. The radial node separates regions in which the real wave function has opposite signs, but these signs do not determine the relative probabilities. A comparison of finite-region probabilities requires information about both the squared wave-function magnitude and the region volumes. \(\textbf{Final answer:}\) \[ \boxed{P_P:P_Q=2:3} \]

Explanation: Orbital P is spherically symmetric, which identifies it as an \(s\) orbital. Its real radial wave function has one finite radial node. The \(2s\) orbital has exactly one radial node, so P is identified as \(2s\). Orbital Q has two principal lobes separated by a nodal plane through the nucleus, characteristic of a \(p\) orbital. A \(2p\) orbital has one angular node and no radial node. Because P and Q belong to the same principal shell, Q is consequently identified as \(2p\). The two orbitals possess different spatial probability distributions and different types of nodes. However, in an isolated hydrogen atom, orbital energy depends on the principal quantum number in the elementary one-electron treatment. Both orbitals have \(n=2\), so they have equal energies within that treatment. Their different spatial shapes do not remove this hydrogen-like energy degeneracy. In a multi-electron atom, shielding and penetration influence subshell energies. An \(s\) orbital generally penetrates more strongly towards the nucleus than a \(p\) orbital in the same principal shell. It therefore experiences a greater effective nuclear attraction in the usual elementary comparison. Consequently, the \(2s\) orbital is generally lower in energy than the \(2p\) orbitals of a multi-electron atom. The correct option must distinguish orbital shape, nodal structure and the different energy rules for one-electron and multi-electron systems.

Explanation: The given description corresponds to the qualitative radial behaviour of a hydrogen \(2s\) orbital. The wave function changes sign at a finite radial node. At the node, the wave function is exactly zero. The electron probability density is determined by the squared magnitude of the wave function. Therefore, the probability density is zero at the node. Student Q correctly recognises both the zero at the node and the nonnegative character of probability density. Student P is incorrect because a negative real wave function does not produce negative probability density. Squaring a negative real amplitude gives a positive value wherever that amplitude is nonzero. Student R is also incorrect because a radial node is not an impenetrable boundary. The \(2s\) wave function remains nonzero over a region beyond its radial node. The probability of finding the electron in a suitable finite region beyond the node can therefore be nonzero. The fact that the wave function approaches zero at large distances does not imply that it vanishes throughout the outer region. The node identifies a particular spherical surface of zero probability density, not the outermost extent of the orbital. Only Student Q gives a physically consistent interpretation.

Explanation: Statement I is correct because the spatial probability density of the hydrogen \(1s\) orbital is greatest at the nucleus. Its value decreases as radial distance increases. Statement II is incorrect because spatial probability density and radial probability distribution are different functions. The radial probability-distribution function accounts for the volume associated with a spherical shell. The relevant relation is \[ P(r)=4\pi r^2\rho(r) \] Statement III is correct because the factor \(r^2\) becomes zero at the nucleus. Therefore, \(P(0)=0\), even though the \(1s\) spatial probability density at the nucleus is finite and maximal. Statement IV is correct because probability in a region depends on integrating probability density over that region's volume. A spherical shell at exactly zero radius has no finite volume. It cannot contain a finite probability simply because the local spatial probability density is nonzero. A small spherical region surrounding the nucleus can, however, contain a nonzero probability because it occupies finite volume. This distinction resolves the apparent difference between the maxima of the two graphs. The maximum of a spatial probability-density graph must not automatically be identified with the most probable radial distance. Thus, Statements I, III and IV are correct.

Explanation: The \(2s\) and \(2p\) orbitals belong to the same principal shell but differ in orbital angular momentum quantum number. The \(2s\) orbital is spherically symmetric and has one radial node. A \(2p\) orbital has a different spatial shape, with an angular nodal plane and no radial node. Their electron probability-density distributions are therefore different. In the elementary hydrogen-like model, orbital energy depends on the principal quantum number and nuclear charge. Both orbitals have \(n=2\), so they possess equal energies in hydrogen. The equality of their energies does not imply equality of their wave functions or probability distributions. In an ordinary multi-electron atom, electron-electron interactions modify the orbital-energy relationships. Shielding and penetration affect the effective nuclear attraction experienced by electrons occupying different subshells. The \(2s\) orbital generally penetrates more strongly than the \(2p\) orbitals. Consequently, the subshells usually differ in energy, with \(2s\) lying below \(2p\) in the elementary ordering. The radial node of \(2s\) does not, by itself, determine its energy relative to \(2p\). Likewise, the angular node of \(2p\) must not be mistaken for a radial node. The correct option recognises energy degeneracy in the hydrogen-like system while preserving the distinct orbital shapes and the effect of multi-electron interactions.

Explanation: \(\textbf{Use the orbital node relations:}\) For an atomic orbital with quantum numbers \(n\) and \(l\), the number of radial nodes is \[ N_{\mathrm{radial}}=n-l-1 \] The number of angular nodes is \[ N_{\mathrm{angular}}=l \] Therefore, the total number of nodes is \[ N_{\mathrm{total}}=n-1 \] \(\textbf{Identify orbital P:}\) P has zero angular nodes, so \[ l_P=0 \] This identifies an \(s\) orbital. It has two radial nodes: \[ n_P-0-1=2 \] Therefore, \[ n_P=3 \] Thus, P is \(3s\). \(\textbf{Identify orbital Q:}\) Q has one angular node: \[ l_Q=1 \] It is therefore a \(p\) orbital. Its one radial node gives \[ n_Q-1-1=1 \] \[ n_Q=3 \] Thus, Q is \(3p\). \(\textbf{Identify orbital R:}\) R has two angular nodes: \[ l_R=2 \] This identifies a \(d\) orbital. Its zero radial nodes give \[ n_R-2-1=0 \] \[ n_R=3 \] Thus, R is \(3d\). \(\textbf{Check the common-shell condition:}\) All three identifications have \(n=3\), as required. Each orbital also has two total nodes, although these are distributed differently between radial and angular types. Therefore, equal total node count does not imply identical orbital shape or nodal structure. The correct identification is P–\(3s\), Q–\(3p\), R–\(3d\).

Explanation: A radial node is a spherical surface at which the orbital wave function is zero. The probability density is consequently zero at the exact radial distance of the node. For a real wave function, this follows from the relation \(\rho(r)=|\Psi(r)|^2\). However, a spherical shell of finite thickness includes a range of radial distances. The stated shell extends both slightly inside and slightly outside the nodal surface. The wave function need not vanish throughout that entire interval. For a node at which the real wave function crosses zero, nonzero amplitudes occur on either side. Their squared magnitudes produce nonnegative, generally nonzero probability densities. Therefore, the shell can contain a nonzero probability even though the density is zero on its central nodal surface. The exact node must be distinguished from a surrounding region of finite volume. The sign change of the wave function does not cause probabilities on opposite sides of the node to cancel. Probabilities are determined from squared magnitudes, not by adding positive and negative wave-function amplitudes as probabilities. A radial node also does not constitute an impenetrable spherical barrier. The student's conclusion incorrectly extends a zero at one radial surface to an entire finite spatial region. Thus, option C provides the correct interpretation.

Explanation: Representation P concerns the real wave function itself. A wave function is a mathematical amplitude associated with the quantum state. Its sign can change across a radial node without generating negative probability. The overall sign of a real orbital wave function is not, by itself, an observable electron probability. Therefore, P matches 2. Representation Q is the spatial probability density. It is proportional to the squared magnitude of the wave function and is nonnegative. Probability in a finite spatial region is obtained by considering this density over the region's volume. Therefore, Q matches 3. Representation R includes the geometrical factor \(4\pi r^2\). This factor accounts for the area of spherical shells when probability is expressed as a function of radial distance. The function is consequently associated with probability in a thin spherical shell of specified radial thickness. Therefore, R matches 1. Representation S describes a conventional boundary-surface representation of an orbital. A surface chosen to enclose approximately \(90\%\) of the total probability does not exclude the remaining probability outside it. Therefore, S matches 4. The four representations serve different purposes and must not be treated as interchangeable. In particular, the signed wave function, nonnegative spatial probability density, radial probability distribution and chosen orbital boundary surface describe related but distinct features of the same quantum state.

Explanation: Statement I is correct because the angular part of an \(s\) orbital is spherically symmetric. Its spatial probability distribution does not depend on direction at a fixed radial distance. Statement II is correct because the number of radial nodes is given by \(n-l-1\). For an \(s\) orbital, \(l=0\). Therefore, an \(ns\) orbital has \(n-1\) radial nodes. Accordingly, \(1s\) has no radial node, \(2s\) has one and \(3s\) has two. Statement III is incorrect because the nucleus is not automatically a radial node of an \(s\) orbital. The real \(1s\) wave function is nonzero at the nucleus and has no radial node. The radial nodes of \(2s\) and \(3s\) are finite-distance zero surfaces, not compulsory sign changes at the nuclear position. Statement IV is correct because the elementary hydrogen energy expression is \[ E_n=-\frac{13.6}{n^2}\,\mathrm{eV} \] As \(n\) increases, the negative energy approaches zero from below. Thus, the \(1s\) state is more strongly bound than \(2s\), while \(2s\) is more strongly bound than \(3s\). The increasing number of radial nodes and the increasing electronic energy are associated with different permitted states, but the node count must not be mistaken for a separate formula that directly gives the energy. All three orbitals remain spherically symmetric despite possessing different radial structures. Therefore, Statements I, II and IV are correct.

Explanation: Graph X is identified as the spatial probability density of the hydrogen \(1s\) orbital. For this orbital, the spatial density is greatest at the nucleus and decreases as radial distance increases. Graph Y corresponds to the radial probability-distribution function. For the spherically symmetric \(1s\) orbital, this function is \[ P(r)=4\pi r^2\rho(r) \] At the nucleus, the geometrical factor \(r^2\) is zero. Consequently, the radial probability distribution begins at zero even though the spatial density is finite and maximal. At a finite radius, the competition between increasing spherical-shell area and decreasing spatial density produces a maximum in the radial probability distribution. Graph Z represents cumulative probability within a sphere of radius \(r\). As the sphere expands, it encloses more of the electron's total probability. The cumulative probability therefore increases from zero towards one. It does not decrease at large radial distances because increasing the enclosed region cannot remove probability already contained within it. At \(r=R\), the cumulative probability is \(0.90\), meaning that \(90\%\) of the total probability lies within the sphere of radius \(R\). The remaining probability lies outside that sphere. Thus, \[ P(\text{outside }R)=1-0.90=0.10 \] The three graphs represent related but different aspects of one orbital. Spatial probability density concerns local probability per unit volume, radial probability distribution accounts for spherical-shell geometry, and cumulative probability concerns the total probability enclosed within a chosen radius. The radius \(R\) does not define an impenetrable boundary because the electron can still be found outside it. The correct interpretation is therefore X as spatial density, Y as radial distribution and Z as cumulative probability, with \(10\%\) probability outside \(R\).

Explanation: \(\textbf{Calculate the photon energy for transition P:}\) The energy emitted during a transition is the difference between the initial and final electronic energies. For transition P, \[ E_{\gamma,P}=E_{3s}-E_{2p} \] Substituting the tabulated values, \[ E_{\gamma,P}=-5.50-(-10.0) \] \[ E_{\gamma,P}=4.50\,\mathrm{eV} \] \(\textbf{Calculate the photon energy for transition Q:}\) For transition Q, \[ E_{\gamma,Q}=E_{3p}-E_{2s} \] Therefore, \[ E_{\gamma,Q}=-5.00-(-12.0) \] \[ E_{\gamma,Q}=7.00\,\mathrm{eV} \] \(\textbf{Compare the emitted frequencies:}\) Planck's relation is \[ E_\gamma=h\nu \] Therefore, \[ \frac{\nu_Q}{\nu_P}=\frac{E_{\gamma,Q}}{E_{\gamma,P}} \] \[ \frac{\nu_Q}{\nu_P}=\frac{7.00}{4.50} \] \[ \frac{\nu_Q}{\nu_P}=\frac{14}{9} \] \(\textbf{Change the atomic system to hydrogen:}\) In the elementary hydrogen-like model, electronic energy depends on the principal quantum number rather than separately on the orbital type. Consequently, \[ E_{3s}=E_{3p}=E_3 \] and \[ E_{2s}=E_{2p}=E_2 \] Both transitions therefore involve the same energy difference: \[ E_{\gamma,P}=E_{\gamma,Q}=E_3-E_2 \] Hence, \[ \frac{\nu_Q}{\nu_P}=1 \] \(\textbf{Interpret the contrast:}\) The multi-electron energy data distinguish the \(s\) and \(p\) subshells within a principal shell. In hydrogen, the corresponding subshells are degenerate in the elementary one-electron treatment. The orbital labels alone do not determine the same frequency relationship in both kinds of atom. The difference arises from the different electronic interactions represented in the two atomic models. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{\nu_Q}{\nu_P}\right)_{\mathrm{multi-electron}}=\frac{14}{9},\qquad\left(\frac{\nu_Q}{\nu_P}\right)_{\mathrm{hydrogen}}=1} \]

Explanation: Statement I is correct because \(\mathrm{He^+}\) contains one electron attracted to a nucleus having charge \(+2e\). The electronic problem has the hydrogen-like form, with a single electron experiencing the attractive Coulomb potential of the nucleus. Within the stated non-relativistic model, its electronic wave functions and energies can be obtained exactly. Statement II is correct because neutral helium contains two electrons. Each electron experiences attraction towards the nucleus and repulsion from the other electron. The electron-electron interaction makes the exact mathematical treatment of the system more complicated. Statement III is incorrect because identical nuclear charge does not imply identical electronic structure when the numbers of electrons differ. Neutral helium has electron-electron repulsion, whereas \(\mathrm{He^+}\) does not. Their electronic states cannot therefore be treated as identical merely because both contain a helium nucleus. Statement IV is correct because approximate quantum mechanical methods are used to study multi-electron atoms. These methods permit useful descriptions of electronic orbitals and their energies. The important distinction is between a one-electron system and a system containing interacting electrons. It is not simply a distinction between nuclei of low and high atomic numbers. The success of the Schrödinger equation for hydrogen-like species provides a foundation for understanding more complicated atoms without implying that every multi-electron problem has an equally simple exact solution. Thus, Statements I, II and IV are correct.

Explanation: The Assertion is true because classical mechanics is highly successful in describing the motion of many ordinary macroscopic objects. Its usual trajectory-based treatment assigns definite positions and momenta to moving particles. However, microscopic electrons exhibit behaviour that cannot be described completely using such classical trajectories. The Reason is also true because electron diffraction provides evidence of matter-wave behaviour. De Broglie's hypothesis associates a wavelength with electron momentum. Heisenberg's uncertainty principle places a fundamental restriction on simultaneous position and corresponding momentum precision. These features are not included in an exclusively classical particle-trajectory description. The quantum mechanical model incorporates wave behaviour and gives a probability-based description of electronic states. This does not mean that classical mechanics becomes useless for macroscopic objects. For ordinary macroscopic motion, quantum effects are often negligible at the relevant observational scales. The important conclusion is that the appropriate physical description depends on the system and the significance of its quantum behaviour. Matter waves and uncertainty are direct reasons that a completely specified classical orbit is inadequate for an atomic electron. Therefore, the Reason correctly explains the Assertion.