Atomic Structure MCQs with Answers – Part 6 (NEET Chemistry)

Explanation: The labels \(p_x,p_y,p_z\) describe spatial orientations relative to a chosen set of Cartesian axes. They do not denote three different principal shells. All three belong to the \(p\) subshell of the specified principal shell. The magnetic quantum number is associated with orbital angular-momentum projection relative to a chosen quantisation axis. Changing the coordinate axes used to describe a system is not the same as physically moving an electron into a different energy level. In the stipulated experiment, the atom itself has not been perturbed. Only the reference axes used in its mathematical description have changed. The physical spatial probability distribution consequently remains the same. Its mathematical representation may use different coordinate components or directional labels. A change in representation does not by itself require absorption or emission of a photon. It also does not require the electron's spin projection to change. The distinction is particularly important when interpreting the familiar real \(p\)-orbital drawings. The three orbital orientations are a convenient basis for describing spatial states, rather than a rule that every Cartesian letter must possess one fixed numerical \(m_l\) assignment independent of representation. A genuine electronic transition requires a physical change between electronic states, not merely relabelling the same state after rotating the laboratory coordinate system. The investigator has therefore confused coordinate description with electronic dynamics.

Explanation: \(\textbf{Identify P from its angular shape:}\) A two-lobed orbital separated by a nodal plane is characteristic of a real \(p\) orbital. Because P belongs to the second principal shell, its subshell identity is \(2p\). For \(2p\), the angular momentum quantum number is \(l=1\). Therefore, it has one angular node. Its radial-node count is \[ N_{\mathrm{radial},P}=2-1-1=0 \] \(\textbf{Identify Q from its lobe directions:}\) The real \(d_{xy}\) orbital has four principal lobes between the \(x\)- and \(y\)-axes. Since Q belongs to the third principal shell, it is \(3d_{xy}\). Its angular momentum quantum number is \(l=2\). Therefore, it has two angular nodes. Its radial-node count is \[ N_{\mathrm{radial},Q}=3-2-1=0 \] \(\textbf{Identify R from its spherical symmetry:}\) A spherically symmetric orbital belongs to an \(s\) subshell. Since R belongs to the second principal shell, it is \(2s\). Its angular momentum quantum number is zero. Thus, it has no angular nodes. Its radial-node count is \[ N_{\mathrm{radial},R}=2-0-1=1 \] \(\textbf{Compare the three different nodal patterns:}\) P has a planar angular node but no finite radial node. Q has two angular nodes and no finite radial node. R has one spherical radial node but no angular node. All three orbitals possess a total of \(n-1\) nodes, giving respectively one, two and one. However, identical total nodal counts do not imply identical spatial shapes. The nature and geometry of the nodes depend on both \(n\) and \(l\). The comparison combines information from lobe direction, spherical symmetry, shell identity and the radial-node relation. \(\textbf{Final answer:}\) \[ \boxed{P=2p:(1,0),\qquad Q=3d_{xy}:(2,0),\qquad R=2s:(0,1)} \]

Explanation: \(\textbf{Determine the separation of two successive Balmer lines:}\) For a hydrogen transition terminating at \(n=2\), the emission wavenumber is \[ \bar{\nu}_n=R_H\left(\frac14-\frac1{n^2}\right) \] For successive upper levels \(n\) and \(n+1\), subtraction eliminates the common lower-level term: \[ \Delta\bar{\nu}=R_H\left(\frac1{n^2}-\frac1{(n+1)^2}\right) \] \(\textbf{Evaluate pair P:}\) For upper levels five and six, \[ \Delta\bar{\nu}_P=1.097\times10^7\left(\frac1{25}-\frac1{36}\right) \] \[ \Delta\bar{\nu}_P=1.097\times10^7\left(\frac{11}{900}\right) \] \[ \Delta\bar{\nu}_P\approx1.341\times10^5\,\mathrm{m^{-1}} \] This separation exceeds the instrument's resolution threshold. Therefore, pair P is resolved. \(\textbf{Evaluate pair Q:}\) For upper levels six and seven, \[ \Delta\bar{\nu}_Q=1.097\times10^7\left(\frac1{36}-\frac1{49}\right) \] \[ \Delta\bar{\nu}_Q=1.097\times10^7\left(\frac{13}{1764}\right) \] \[ \Delta\bar{\nu}_Q\approx8.085\times10^4\,\mathrm{m^{-1}} \] This separation is smaller than the specified threshold. Therefore, pair Q is not resolved. \(\textbf{Connect the observation with spectral convergence:}\) Higher hydrogen energy levels are progressively closer together. Successive lines belonging to a fixed spectral series consequently become more closely spaced in wavenumber as the upper principal quantum number increases. The inability to resolve pair Q does not mean that its two transitions have identical photon energies. It means only that their separation falls below the instrument's stated resolving capability. The result distinguishes the physical existence of discrete spectral lines from their experimental resolution. \(\textbf{Final answer:}\) \[ \boxed{\Delta\bar{\nu}_P\approx1.341\times10^5\,\mathrm{m^{-1}},\qquad\Delta\bar{\nu}_Q\approx8.085\times10^4\,\mathrm{m^{-1}}} \]

Explanation: \(\textbf{Identify the initial energy level:}\) For hydrogen, \[ E_n=-\frac{13.6}{n^2}\,\mathrm{eV} \] The initial energy is \[ E_2=-\frac{13.6}{4} \] \[ E_2=-3.400\,\mathrm{eV} \] \(\textbf{Test the first photon for bound-state excitation:}\) The energy of the \(n=4\) state is \[ E_4=-\frac{13.6}{16} \] \[ E_4=-0.850\,\mathrm{eV} \] Therefore, \[ \Delta E_{2\rightarrow4}=E_4-E_2 \] \[ \Delta E_{2\rightarrow4}=-0.850-(-3.400) \] \[ \Delta E_{2\rightarrow4}=2.550\,\mathrm{eV} \] The first photon exactly matches this excitation energy. The atom can consequently reach \(n=4\). \(\textbf{Recalculate the binding energy before the second photon arrives:}\) The electron's ionisation threshold from \(n=4\) is \[ E_{\mathrm{ion},4}=0-E_4 \] \[ E_{\mathrm{ion},4}=0.850\,\mathrm{eV} \] The second photon has energy \(2.550\,\mathrm{eV}\), exceeding this new threshold. \(\textbf{Determine the photoelectron kinetic energy:}\) \[ K=E_{\gamma,2}-E_{\mathrm{ion},4} \] \[ K=2.550-0.850 \] \[ K=1.700\,\mathrm{eV} \] \(\textbf{Check the combined energy transfer:}\) The two photons together supply \[ E_{\mathrm{absorbed,total}}=2(2.550)=5.100\,\mathrm{eV} \] The initial electron was bound by \(3.400\,\mathrm{eV}\). The remaining energy is \[ 5.100-3.400=1.700\,\mathrm{eV} \] This agrees with the sequential calculation. \(\textbf{Interpret the changed electronic condition:}\) The first photon changes the occupied energy level and therefore the binding energy relevant to the second photon. The same photon energy can produce excitation in one stage and ionisation in another. It is incorrect to apply the original \(n=2\) ionisation threshold after the electron has already moved to \(n=4\). The result depends on the stipulated order and timing of the photon interactions. \(\textbf{Final answer:}\) \[ \boxed{2\rightarrow4\rightarrow\infty,\qquad K=1.700\,\mathrm{eV}} \]

Explanation: \(\textbf{Recover the principal quantum number:}\) Bohr's angular-momentum condition is \[ L=n\hbar \] The measured magnitude is \(2\hbar\). Therefore, \[ n=2 \] \(\textbf{Derive the acceleration dependence:}\) For circular motion, centripetal acceleration is \[ a=\frac{v^2}{r} \] The Bohr speed dependence is \[ v\propto\frac Zn \] The radius dependence is \[ r\propto\frac{n^2}{Z} \] Combining these relationships, \[ a\propto\frac{Z^2/n^2}{n^2/Z} \] \[ a\propto\frac{Z^3}{n^4} \] \(\textbf{Apply the measured acceleration ratio:}\) For hydrogen's first orbit, \[ Z_{\mathrm H}=1,\qquad n_{\mathrm H}=1 \] Thus, \[ \frac{a_{\mathrm{ion}}}{a_{\mathrm{H},1}}=\frac{Z^3/n^4}{1} \] Substituting the measured ratio and \(n=2\), \[ \frac12=\frac{Z^3}{2^4} \] \[ \frac12=\frac{Z^3}{16} \] \[ Z^3=8 \] \[ Z=2 \] The one-electron ion is \(\mathrm{He^+}\). \(\textbf{Determine its binding-energy magnitude:}\) \[ |E_n|=13.6\frac{Z^2}{n^2}\,\mathrm{eV} \] Substituting, \[ |E_2|=13.6\frac{2^2}{2^2} \] \[ |E_2|=13.6\,\mathrm{eV} \] \(\textbf{Check the independent physical constraints:}\) The inferred state satisfies the measured angular momentum because its principal quantum number is two. Its acceleration ratio is \[ \frac{a_{\mathrm{He^+},2}}{a_{\mathrm{H},1}}=\frac{2^3}{2^4}=\frac12 \] Both observations are consistent with the same species and orbit. The inverse inference requires combining quantisation with the force-related acceleration dependence. Acceleration cannot be inferred from orbital radius alone without also accounting for speed. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{He^+},\qquad n=2,\qquad |E_2|=13.6\,\mathrm{eV}} \]

Explanation: Potassium has atomic number 19. Its neutral ground-state configuration is \([Ar]\,4s^1\). Scandium has atomic number 21 and the neutral configuration \([Ar]\,3d^1\,4s^2\). The first two electrons removed during formation of \(\mathrm{Sc^{2+}}\) are the two \(4s\) electrons. Therefore, \(\mathrm{Sc^{2+}}\) has the configuration \([Ar]\,3d^1\). It contains \(21-2=19\) electrons, the same number as neutral potassium. However, equal electron counts do not force the occupied outer subshells to be identical. The nuclear charges and the electronic environments of the two species differ. The familiar \(4s\)-before-\(3d\) rule is a neutral-atom filling guide, not an instruction to preserve a \(4s\) electron throughout every ionisation sequence. Removing another electron from \(\mathrm{Sc^{2+}}\) removes its remaining \(3d\) electron. The resulting \(\mathrm{Sc^{3+}}\) ion has the closed-shell configuration \([Ar]\). The argon core is not disturbed during these first three specified removals. The comparison demonstrates why electronic configurations of ions must be derived from their parent configurations and electron-removal sequence rather than from electron count alone. It also distinguishes the concept of isoelectronic species from that of identical outer-subshell configurations.

Explanation: \(\textbf{Account for copper's single electron loss:}\) Neutral copper has atomic number 29. Its observed configuration is \[ [Ar]\,3d^{10}\,4s^1 \] Removing the single \(4s\) electron gives \[ \mathrm{Cu^+}:[Ar]\,3d^{10} \] The ion has 28 electrons. \(\textbf{Account for zinc's two electron losses:}\) Neutral zinc has atomic number 30. Its configuration is \[ [Ar]\,3d^{10}\,4s^2 \] Removing both \(4s\) electrons produces \[ \mathrm{Zn^{2+}}:[Ar]\,3d^{10} \] This ion also has 28 electrons. \(\textbf{Track the different outer subshells of gallium:}\) Neutral gallium has atomic number 31. Its configuration is \[ [Ar]\,3d^{10}\,4s^2\,4p^1 \] Under the specified electron-removal order, the \(4p\) electron is removed before the two \(4s\) electrons. Therefore, formation of \(\mathrm{Ga^{3+}}\) leaves \[ \mathrm{Ga^{3+}}:[Ar]\,3d^{10} \] Its electron count is \[ N_e=31-3=28 \] \(\textbf{Compare the final electron occupancies:}\) All three ions have the same electron count and occupied-subshell notation. The outer \(3d\) subshell contains ten electrons distributed as five opposite-spin pairs. Consequently, none has an unpaired \(3d\) electron in the specified ground-state arrangement. \(\textbf{Interpret their isoelectronic relationship:}\) The neutral atoms differ in both nuclear charge and outer electron configuration. Their indicated ionisation processes nevertheless produce the same total electron population and subshell occupancy. Isoelectronic configurations do not imply identical nuclear charges or identical ionic sizes. The essential reasoning is to account for the exceptional copper ground state and remove the appropriate outer electrons from each different parent atom. \(\textbf{Final answer:}\) \[ \boxed{N_e(\mathrm{Cu^+})=N_e(\mathrm{Zn^{2+}})=N_e(\mathrm{Ga^{3+}})=28} \]

Explanation: \(\textbf{Examine the initial phosphorus occupancy:}\) Phosphorus has the outer configuration \(3s^2\,3p^3\). The three \(3p\) electrons occupy separate degenerate orbitals with parallel spins: \[ 3p^3:\quad[\uparrow]\ [\uparrow]\ [\uparrow] \] This is a half-filled \(p\) subshell. The distribution has the special stability associated with half-filled subshells in the usual NCERT treatment. \(\textbf{Examine the initial sulfur occupancy:}\) Sulfur has the outer configuration \(3s^2\,3p^4\). Its \(3p\) arrangement is \[ 3p^4:\quad[\uparrow\downarrow]\ [\uparrow]\ [\uparrow] \] One \(p\) orbital contains an opposite-spin electron pair. The paired-electron repulsion contributes to the relative ease of removing an electron from sulfur. \(\textbf{Relate the occupancies to the observed ionisation energies:}\) Across a period, increasing nuclear charge generally tends to increase first ionisation energy. However, this trend is not perfectly monotonic. The half-filled \(3p^3\) occupancy of phosphorus is comparatively stable. Removal from sulfur's paired \(3p\) occupancy can be easier despite sulfur's greater nuclear charge. This explains the stated anomaly without assigning the two elements to different principal shells. \(\textbf{Determine the resulting ionic occupancies:}\) Removal from phosphorus gives \[ \mathrm{P^+}:[Ne]\,3s^2\,3p^2 \] Hund's rule gives two unpaired \(3p\) electrons. Removal of one electron from sulfur's paired \(3p\) orbital gives \[ \mathrm{S^+}:[Ne]\,3s^2\,3p^3 \] The resulting half-filled \(3p^3\) subshell has three unpaired electrons. \(\textbf{Interpret the contrasting changes:}\) Phosphorus loses one of three initially unpaired electrons. Sulfur loses an electron from a pair and thereby increases its number of unpaired electrons. Thus, the ionisation-energy anomaly and the ionic unpaired-electron counts are both explained by the initial orbital occupancy. The conclusion depends on Hund's rule, electron pairing and relative subshell stability rather than nuclear charge alone. \(\textbf{Final answer:}\) \[ \boxed{N_u(\mathrm{P^+})=2,\qquad N_u(\mathrm{S^+})=3} \]

Explanation: \(\textbf{Examine the ground-state distribution:}\) Configuration P has a completely filled \(2s\) orbital and three electrons in the \(2p\) subshell. Its \(2p^3\) electrons occupy three different degenerate spatial orbitals singly with parallel spins. Thus, P has three unpaired electrons. \(\textbf{Examine the promoted configuration:}\) Configuration Q has one electron remaining in \(2s\). That electron is unpaired. The \(2p^4\) occupancy contains one paired orbital and two singly occupied orbitals. Consequently, its \(2p\) subshell contributes two unpaired electrons. The combined unpaired-electron count for Q is therefore three. \(\textbf{Distinguish electron count from energy:}\) Both configurations contain seven electrons. Both may be represented by Pauli-permitted orbital occupancies. However, Q requires promotion of an electron from \(2s\) to a higher-energy \(2p\) subshell in the usual multi-electron ordering. It is consequently an excited configuration relative to the stated ground state P. The identical numbers of unpaired electrons do not mean the two configurations have identical energies. Nor does the Pauli exclusion principle require a singly occupied \(2s\) orbital to be invalid. A spatial orbital is allowed to contain zero, one or two electrons. Hund's rule determines the usual spin arrangement within the degenerate \(2p\) subshell, but it does not independently decide the energetic cost of the \(2s\rightarrow2p\) promotion. This example shows why the magnetic behaviour implied by an unpaired-electron count cannot always identify the electronic configuration uniquely. The ground-state distinction requires the occupancy pattern and orbital-energy ordering in addition to spin counting.

Explanation: \(\textbf{Establish the common ionic electron count:}\) Oxygen has atomic number eight and gains two electrons to form \(\mathrm{O^{2-}}\). Fluorine has atomic number nine and gains one electron to form \(\mathrm{F^-}\). Sodium has atomic number eleven and loses one electron to form \(\mathrm{Na^+}\). Magnesium has atomic number twelve and loses two electrons to form \(\mathrm{Mg^{2+}}\). In each case, the resulting ion contains ten electrons. \(\textbf{Identify the common configuration:}\) A ten-electron ground-state configuration is \[ 1s^2\,2s^2\,2p^6 \] This is the neon configuration. Its \(2p\) subshell contains three opposite-spin electron pairs. Therefore, all four resulting ions have zero unpaired electrons. \(\textbf{Inspect neutral oxygen and fluorine:}\) Neutral oxygen has \(2p^4\), giving two unpaired electrons under Hund's rule. Adding two electrons completes the \(2p^6\) subshell. Neutral fluorine has \(2p^5\), giving one unpaired electron. Adding one electron likewise completes \(2p^6\). \(\textbf{Inspect neutral sodium and magnesium:}\) Neutral sodium has \([Ne]\,3s^1\). Its single \(3s\) electron is unpaired. Removing that electron produces \([Ne]\). Neutral magnesium has \([Ne]\,3s^2\), with its two \(3s\) electrons paired. It therefore has no unpaired electrons before ionisation. Removing those two electrons also produces \([Ne]\). \(\textbf{Interpret the different routes to the same configuration:}\) The four ions are isoelectronic despite being produced by both electron gain and electron loss. Their neutral parents have different numbers of unpaired electrons. The common final configuration does not imply identical ionic charges or nuclear attractions. The comparison requires electron counting, appropriate gain-or-loss decisions and Hund-rule occupancy rather than merely recognising the neon symbol.

Explanation: \(\textbf{Relate the ionic occupancies to the neutral configurations:}\) The neutral configurations are stipulated to be \[ [Ar]\,3d^n\,4s^2 \] Formation of each \(2+\) ion removes its two \(4s\) electrons. Therefore, the resulting ionic \(3d\) occupation remains \(3d^n\). \(\textbf{Use the Hund-rule occupancy trend:}\) A \(d\) subshell contains five spatial orbitals. For \(d^1\) through \(d^5\), electrons occupy separate orbitals singly before pairing. For \(d^6\) through \(d^{10}\), additional electrons pair with those already occupying the five orbitals. The corresponding unpaired-electron counts for \(d^6,d^7,d^8\) are four, three and two. \(\textbf{Determine the consecutive electron occupancies:}\) The observed pattern decreases from four to three to two. The consecutive configurations \[ 3d^6,\quad3d^7,\quad3d^8 \] satisfy this pattern. The alternative sequence \(3d^4,3d^3,3d^2\) would also show counts four, three and two, but it runs in decreasing, rather than increasing, \(d\)-electron occupancy. The stem specifies increasing \(n\) across consecutive elements. Therefore, the \(d^6,d^7,d^8\) sequence is required. \(\textbf{Identify the neutral atoms:}\) Iron has the ground-state configuration \[ \mathrm{Fe}:[Ar]\,3d^6\,4s^2 \] Cobalt has \[ \mathrm{Co}:[Ar]\,3d^7\,4s^2 \] Nickel has \[ \mathrm{Ni}:[Ar]\,3d^8\,4s^2 \] Their \(2+\) ions accordingly have four, three and two unpaired electrons. \(\textbf{Interpret the reverse inference:}\) An unpaired-electron count does not always identify a unique \(d\)-subshell occupancy by itself. For example, four unpaired electrons can occur in either \(d^4\) or \(d^6\). The consecutive-element condition and the specified direction of increasing electron occupancy remove this ambiguity. The solution therefore requires combining Hund's rule with electronic-configuration progression rather than matching one spin count to one element.

Explanation: \(\textbf{Examine the neutral zinc atom:}\) Zinc has the configuration \([Ar]\,3d^{10}\,4s^2\). The \(3d^{10}\) subshell is completely filled. Its five spatial orbitals contain five opposite-spin electron pairs. The \(4s^2\) electrons are also paired. Thus, neutral zinc has no unpaired electrons. \(\textbf{Apply the first zinc ionisation:}\) Removal of one \(4s\) electron gives \[ \mathrm{Zn^+}:[Ar]\,3d^{10}\,4s^1 \] The remaining \(4s\) electron is unpaired. The number of unpaired electrons therefore becomes one. \(\textbf{Apply the second zinc ionisation:}\) Removal of the final \(4s\) electron gives \[ \mathrm{Zn^{2+}}:[Ar]\,3d^{10} \] All occupied orbitals are paired. The unpaired-electron count returns to zero. \(\textbf{Analyse the copper ionisation independently:}\) Neutral copper has \([Ar]\,3d^{10}\,4s^1\). Its filled \(3d\) subshell contributes no unpaired electrons. The single \(4s\) electron contributes one. Removal of this electron produces \[ \mathrm{Cu^+}:[Ar]\,3d^{10} \] The resulting ion has no unpaired electrons. \(\textbf{Interpret the contrasting sequences:}\) Zinc first acquires an unpaired electron count of one through removal of an electron from a pair. Its second ionisation removes the remaining unpaired \(4s\) electron. Copper begins with a singly occupied \(4s\) orbital and loses that electron directly. Both routes lead to a filled \(3d^{10}\) arrangement, but the changes in unpaired-electron count are different. The comparison shows that an electron-removal process need not always decrease the number of unpaired electrons. It also distinguishes complete subshell occupancy from the spin state of a separate outer orbital. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{Zn}:0\rightarrow1\rightarrow0;\qquad\mathrm{Cu}:1\rightarrow0} \]

Explanation: The given ground-state configuration is \([Ne]\,3s^2\,3p^2\). In the ordinary ground state, the \(3s\) orbital contains an opposite-spin pair. The two \(3p\) electrons occupy different degenerate orbitals singly with parallel spins. Arrangement P instead transfers one electron from \(3s\) into \(3p\). Its outer occupancy becomes \(3s^1\,3p^3\). The \(3s\) orbital is singly occupied, which is permitted by Pauli's principle. The three \(3p\) electrons occupy separate orbitals singly with parallel spins, satisfying Hund's rule within that subshell. The rearrangement therefore does not violate an occupancy prohibition. However, it requires excitation from the lower-energy \(3s\) subshell to \(3p\). It is not the ordinary ground-state arrangement. Arrangement Q leaves the subshell electron counts unchanged. It places both \(3p\) electrons into the same spatial orbital with opposite spins. That pair satisfies the Pauli exclusion principle. Nevertheless, the other two degenerate \(3p\) orbitals remain empty. For the normal maximum-multiplicity arrangement, the electrons should occupy separate degenerate orbitals before pairing. Therefore, Q does not satisfy Hund's rule for the ground-state distribution. The two arrangements demonstrate distinct ways in which a proposed occupancy can differ from the normal ground state. A Pauli-permitted configuration may be energetically excited, or it may depart from Hund's maximum-multiplicity arrangement. Neither distinction should automatically be described as a violation of the Pauli exclusion principle.

Explanation: \(\textbf{Examine neutral gallium:}\) Neutral gallium has atomic number 31 and ground-state electronic configuration \[ \mathrm{Ga}:[Ar]\,3d^{10}\,4s^2\,4p^1 \] The \(3d^{10}\) subshell is completely filled, so its electrons are paired. The \(4s^2\) electrons are also paired. The single \(4p\) electron remains unpaired. Therefore, neutral gallium has one unpaired electron. The highest occupied principal shell is \(n=4\). \(\textbf{Examine the formation of }\mathrm{Ga^+}\textbf{:}\) The first electron is removed from the \(4p\) subshell. The resulting configuration is \[ \mathrm{Ga^+}:[Ar]\,3d^{10}\,4s^2 \] All occupied orbitals contain paired electrons. Hence, the number of unpaired electrons becomes zero. However, electrons still occupy the \(4s\) orbital. Therefore, the highest occupied principal shell remains \(n=4\). \(\textbf{Examine the formation of }\mathrm{Ga^{2+}}\textbf{:}\) The second electron is removed from the \(4s\) subshell. The resulting configuration is \[ \mathrm{Ga^{2+}}:[Ar]\,3d^{10}\,4s^1 \] The remaining \(4s\) electron is unpaired. Therefore, the number of unpaired electrons becomes one. Since the \(4s\) orbital is still occupied, the highest occupied principal shell remains \(n=4\). \(\textbf{Examine the formation of }\mathrm{Ga^{3+}}\textbf{:}\) The third electron is removed from the remaining \(4s\) orbital. The configuration becomes \[ \mathrm{Ga^{3+}}:[Ar]\,3d^{10} \] The \(3d^{10}\) subshell is completely filled. Thus, no unpaired electrons remain. The fourth principal shell is now unoccupied, while the third principal shell contains occupied orbitals. Therefore, the highest occupied principal shell changes from \(n=4\) to \(n=3\) upon formation of \(\mathrm{Ga^{3+}}\). \(\textbf{Determine the complete unpaired-electron sequence:}\) The unpaired-electron counts for neutral Ga, \(\mathrm{Ga^+}\), \(\mathrm{Ga^{2+}}\) and \(\mathrm{Ga^{3+}}\) are \[ N_u:1\rightarrow0\rightarrow1\rightarrow0 \] \(\textbf{Interpret the successive changes:}\) Removing an electron does not necessarily decrease the number of unpaired electrons. The first removal eliminates the unpaired \(4p\) electron. The second removal breaks the \(4s^2\) pair, leaving one unpaired electron. The third removal eliminates that remaining \(4s\) electron. Similarly, the highest occupied principal shell is determined by whether any electrons remain in that shell, not by the number of unpaired electrons. Consequently, the highest occupied shell changes only after the last \(n=4\) electron has been removed. \(\textbf{Final conclusion:}\) The unpaired-electron sequence is \(1\rightarrow0\rightarrow1\rightarrow0\). The highest occupied principal shell changes from \(n=4\) to \(n=3\) upon formation of \(\mathrm{Ga^{3+}}\), when the final \(4s\) electron is removed.

Explanation: \(\textbf{Interpret the neutral unpaired-electron count:}\) The filled \(3d^{10}\) subshell contributes no unpaired electrons. The \(4s^2\) electrons are also paired. Therefore, the two unpaired electrons must occur within the \(4p\) subshell. \(\textbf{Recognise the ambiguity in the neutral occupancy:}\) A \(p\) subshell contains three degenerate spatial orbitals. The \(p^2\) arrangement contains two singly occupied orbitals. The \(p^4\) arrangement also contains two singly occupied orbitals, because the fourth electron forms a pair. Thus, two unpaired \(p\) electrons do not uniquely determine whether the neutral occupancy is \(p^2\) or \(p^4\). \(\textbf{Apply the observed ionisation result:}\) The question states that one \(4p\) electron is removed. If the neutral atom were \(4p^2\), the ion would have \(4p^1\), contributing only one unpaired electron. That contradicts the observed ionic count of three. If the neutral atom were \(4p^4\), removal of the electron from the paired orbital would leave \(4p^3\). The three remaining electrons would occupy separate \(p\) orbitals singly with parallel spins. The resulting ion would therefore have three unpaired electrons, as observed. \(\textbf{Identify the element:}\) The neutral configuration must be \[ [Ar]\,3d^{10}\,4s^2\,4p^4 \] The total electron count is \[ N_e=18+10+2+4 \] \[ N_e=34 \] Atomic number 34 corresponds to selenium. The singly positive ion has the outer occupancy \[ 4s^2\,4p^3 \] \(\textbf{Interpret the reverse inference:}\) The neutral unpaired-electron count alone is insufficient because both \(p^2\) and \(p^4\) satisfy it. The changed spin count after a specified electron removal distinguishes the alternatives. The half-filled \(p^3\) occupancy explains the three unpaired electrons in the resulting ion. This is an example of using two related electronic observations to infer a configuration rather than identifying an element from one magnetic property alone. \(\textbf{Final answer:}\) \[ \boxed{Z=34,\qquad N_{u,\mathrm{neutral}}=2,\qquad N_{u,\mathrm{ion}}=3} \]

Explanation: \(\textbf{Establish the energy reference and the two arrangements:}\) Take the one-electron energy of orbital P as zero. Orbitals Q and R each have energy \(\Delta\). Arrangement I occupies P and Q singly. The two electrons are unpaired and have parallel spins. Its stipulated stabilisation lowers the energy by \(0.20\,\mathrm{eV}\). Therefore, \[ E_I=0+\Delta-0.20 \] \[ E_I=\Delta-0.20\,\mathrm{eV} \] Arrangement II places both electrons in P with opposite spins. Its stipulated pairing penalty is \(1.20\,\mathrm{eV}\). Hence, \[ E_{II}=1.20\,\mathrm{eV} \] \(\textbf{Compare the arrangements before the splitting changes:}\) Initially, \[ \Delta=0.80\,\mathrm{eV} \] Therefore, \[ E_{I,\mathrm{initial}}=0.80-0.20 \] \[ E_{I,\mathrm{initial}}=0.60\,\mathrm{eV} \] Since \[ 0.60\lt 1.20 \] Arrangement I is lower in energy. It contains two unpaired electrons. \(\textbf{Apply the larger orbital-energy separation:}\) After the change, \[ \Delta=1.60\,\mathrm{eV} \] Thus, \[ E_{I,\mathrm{final}}=1.60-0.20 \] \[ E_{I,\mathrm{final}}=1.40\,\mathrm{eV} \] The energy of Arrangement II remains \[ E_{II}=1.20\,\mathrm{eV} \] Since \[ 1.20\lt 1.40 \] Arrangement II is now energetically preferred. Its two electrons are paired in P, so there are no unpaired electrons. \(\textbf{Interpret the changed-condition result:}\) Both arrangements satisfy the Pauli exclusion principle. The maximum-multiplicity arrangement is preferred initially because its energy benefit outweighs the specified orbital-energy separation. However, the larger splitting makes occupation of the higher-energy orbital less favourable. Hund's rule is most directly stated for electrons filling degenerate orbitals and must not be treated as an absolute prohibition on pairing whenever any vacant orbital exists at a different energy. The problem explicitly supplies the energetic contributions needed to determine the preferred arrangement in each condition. \(\textbf{Final answer:}\) \[ E_{I,\mathrm{initial}}=0.60\,\mathrm{eV}\lt E_{II}=1.20\,\mathrm{eV}\lt E_{I,\mathrm{final}}=1.40\,\mathrm{eV} \]

Explanation: The configuration of X corresponds to neutral chromium. Its observed ground state is \([Ar]\,3d^5\,4s^1\). The \(3d^5\) subshell is half-filled, with one electron in each of five \(d\) orbitals under Hund's rule. These five \(d\) electrons are unpaired. The additional \(4s^1\) electron is also unpaired in the neutral atom. Removal of that \(4s\) electron produces \(\mathrm{Cr^+}\) with configuration \([Ar]\,3d^5\). The resulting ion retains five unpaired electrons. The configuration of Y corresponds to neutral copper. Its observed ground state is \([Ar]\,3d^{10}\,4s^1\). The \(3d^{10}\) subshell is completely filled, with five opposite-spin electron pairs. Therefore, its \(3d\) subshell contributes no unpaired electrons. The neutral copper atom has one unpaired electron in \(4s\). Removal of that electron produces \(\mathrm{Cu^+}\) with configuration \([Ar]\,3d^{10}\). The resulting copper ion has no unpaired electrons. Both ionic products retain their respective half-filled or completely filled \(d\) occupancies. However, these two stable occupancy patterns have very different spin arrangements. The statement that both neutral atoms initially have one \(4s\) electron is insufficient to predict equal final unpaired-electron counts. The occupied \(3d\) subshell must also be examined. A half-filled \(d^5\) subshell contains five unpaired electrons, whereas a completely filled \(d^{10}\) subshell contains none. Thus, the relevant distinction is not simply whether \(4s\) is occupied, but how the electrons are distributed among the five \(d\) orbitals after ionisation.

Explanation: \(\textbf{Evaluate the Aufbau ordering parameters:}\) The standard guide compares subshells using the sum \[ n+l \] The orbital angular momentum quantum numbers for \(s,p,d,f\) are respectively \(0,1,2,3\). Therefore, \[ 4f:\quad n+l=4+3=7 \] \[ 5d:\quad n+l=5+2=7 \] \[ 6p:\quad n+l=6+1=7 \] \[ 7s:\quad n+l=7+0=7 \] \(\textbf{Resolve the equal-sum condition:}\) When \(n+l\) is equal, the standard Aufbau guide places the subshell with the smaller \(n\) first. The prescribed order is consequently \[ 4f\rightarrow5d\rightarrow6p\rightarrow7s \] \(\textbf{Calculate the capacity of the first subshell:}\) The number of spatial orbitals in a subshell is \[ N_{\mathrm{orbitals}}=2l+1 \] For \(4f\), \(l=3\), so \[ N_{4f}=2(3)+1=7 \] Each orbital holds at most two electrons with opposite spins. Therefore, \[ C_{4f}=2(7)=14 \] Fourteen of the supplied electrons fill \(4f\). \(\textbf{Allocate the remaining electrons:}\) The number left is \[ 24-14=10 \] The \(5d\) subshell has five spatial orbitals and a capacity of ten electrons. Thus, the remaining ten electrons completely fill \(5d\). No electron is left to occupy \(6p\) or \(7s\). \(\textbf{Interpret the guide correctly:}\) The exercise explicitly restricts the available subshells and invokes the standard ordering rule. It does not claim that an actual neutral atom must possess this arrangement without considering other subshells or observed configuration exceptions. The tie-breaking rule and electron capacities are separate requirements, and both must be applied. \(\textbf{Final answer:}\) \[ \boxed{4f^{14}\,5d^{10};\qquad N_{6p}=0} \]

Explanation: Vanadium has 23 electrons. After its eighteen-electron argon core, five electrons remain. Its ordinary neutral ground-state configuration is \([Ar]\,3d^3\,4s^2\). Arrangement P places two opposite-spin electrons in \(4s\) and three parallel-spin electrons singly in three different \(3d\) orbitals. It satisfies the usual Aufbau occupancy, the Pauli exclusion principle and Hund's rule. Therefore, P correctly represents the stipulated ground state. Arrangement Q also contains \(4s^2\,3d^3\), but two \(3d\) electrons occupy the same spatial orbital while another \(3d\) orbital is singly occupied. The paired electrons have opposite spins, so Pauli's principle is satisfied. However, two of the five degenerate \(3d\) orbitals remain empty. The usual Hund-rule arrangement would occupy three different \(3d\) orbitals singly before pairing. Thus, Q is Pauli-permitted but Hund-unfavoured. Arrangement R places two upward-spin electrons in the same \(4s\) spatial orbital. These electrons would have identical spatial quantum numbers and identical spin projections. That violates the Pauli exclusion principle regardless of how the \(3d\) electrons are arranged. Arrangement S contains \(4s^1\,3d^4\). Its five electrons are distributed among permitted orbital states, and the four \(3d\) electrons occupy different orbitals singly. Thus, S is not Pauli-forbidden. Nevertheless, relative to vanadium's stated ordinary ground state, it involves redistribution of an electron from \(4s\) to \(3d\). It is a promoted configuration rather than the normal ground-state occupancy of neutral vanadium. The four arrangements illustrate why electron count alone cannot establish a valid ground-state orbital diagram. Pauli exclusion, Hund's rule and the energetically preferred subshell occupancy must be checked independently.

Explanation: \(\textbf{Establish the electron-pair counting condition:}\) A doubly occupied spatial orbital contains two electrons with opposite spins. If a subshell contains \(N\) electrons, \(p\) opposite-spin pairs and \(u\) singly occupied orbitals, then \[ N=2p+u \] Thus, \[ u=N-2p \] The number of unpaired electrons must have the same odd-or-even character as the total subshell electron count. \(\textbf{Examine the seven-electron }d\textbf{ subshell:}\) A \(d\) subshell contains five spatial orbitals. For seven electrons, at least two orbitals must contain pairs. The usual Hund-rule arrangement has two pairs and three singly occupied orbitals. Therefore, \(d^7\) ordinarily has three unpaired electrons. Proposal P claims two unpaired electrons. However, seven electrons cannot be divided into opposite-spin pairs plus exactly two single electrons, because the total would have to be even. P is impossible under the stipulated occupancy model. Proposal Q claims one unpaired electron. This is possible with three paired orbitals, one singly occupied orbital and one empty orbital. The total is \[ 3(2)+1=7 \] It obeys Pauli's restriction but pairs more electrons than Hund's usual maximum-multiplicity arrangement. \(\textbf{Examine the four-electron }p\textbf{ subshell:}\) A \(p\) subshell has three spatial orbitals. The usual \(p^4\) arrangement contains one pair and two singly occupied orbitals. Therefore, it has two unpaired electrons. Proposal R, with zero unpaired electrons, is possible by placing two opposite-spin pairs in two orbitals and leaving the third empty. It is Pauli-permitted but violates the normal Hund-rule filling preference. Proposal S claims three unpaired electrons for a four-electron subshell. Three singly occupied orbitals account for three electrons, leaving one electron that must form a pair. This reduces the number of singly occupied orbitals to two. Thus, exactly three unpaired electrons are impossible for \(p^4\). \(\textbf{Interpret the distinction:}\) A Pauli-forbidden occupancy cannot exist because it violates the permitted electron states. A Hund-unfavoured occupancy can obey Pauli exclusion while failing to minimise energy through the ordinary maximum-multiplicity arrangement. Q and R belong to the latter category. The question therefore requires checking both occupancy feasibility and ground-state preference rather than equating all non-Hund arrangements with Pauli violations.