301. In an independent-orbital description of an atom, two electrons initially occupy the same \(2p_z\) spatial orbital with opposite spin quantum numbers. One electron is then removed, leaving the other electron in the same spatial orbital. Consider two fixed spatial positions: P lies along a principal \(2p_z\) lobe, while Q lies in the orbital's nodal \(xy\)-plane.
Which option correctly describes the combined electron number density contributed by this orbital at P and Q before and after electron removal?
ⓐ. At P, the density is unchanged because the wave function remains the same; at Q, it becomes nonzero because the electrons are no longer spin-paired.
ⓑ. At P, the density becomes one-fourth of its original value; at Q, it remains zero.
ⓒ. At P, the density becomes half its original value; at Q, it remains zero, while the remaining electron is unpaired.
ⓓ. At P, the density becomes half its original value; at Q, it also becomes half of a previously nonzero density.
Correct Answer: At P, the density becomes half its original value; at Q, it remains zero, while the remaining electron is unpaired.
Explanation: Initially, the \(2p_z\) spatial orbital contains two electrons with opposite spin quantum numbers.
The two electrons occupy the same spatial orbital and consequently have the same one-electron spatial probability density.
Let the normalised spatial orbital wave function be \(\Psi\).
For one electron, the probability density is \(|\Psi|^2\).
For two electrons occupying that spatial orbital, the combined electron number density is
\[
\rho_{\mathrm{initial}}=2|\Psi|^2
\]
The point P lies along a principal lobe where the wave function is nonzero.
Therefore, the initial density contributed by the two electrons at P is twice the single-electron contribution.
After one electron is removed, the orbital remains occupied by only one electron.
Under the stated unchanged-orbital assumption, its density contribution becomes
\[
\rho_{\mathrm{final}}=|\Psi|^2
\]
Thus, the density at P becomes half its original value.
Point Q lies in the angular nodal plane of the \(2p_z\) orbital.
The wave function at Q is zero, so its probability density is also zero.
Multiplying the zero density by two does not make it nonzero.
Therefore, the density at Q remains zero both before and after removal.
The remaining electron is unpaired because the orbital no longer contains an electron of opposite spin.
Removing an electron changes the occupancy and combined electron number density, but it does not eliminate the angular node of the unchanged spatial orbital.
The question therefore distinguishes occupancy, intrinsic spin and the fixed nodal geometry of an atomic orbital.
302. At a fixed nonzero radial distance from the nucleus, the angular part of a real \(3d_{z^2}\) wave function is proportional to \(3\cos^2\theta-1\), where \(\theta\) is measured from the positive \(z\)-axis. Assume that the radial part is nonzero at the selected distance.
Which option correctly identifies its angular nodal condition and compares its spatial probability densities along the positive \(z\)-axis and positive \(x\)-axis at that distance?
ⓐ. \(\cos^2\theta=1/2\); density ratio \(2:1\)
ⓑ. \(\cos^2\theta=1/3\); density ratio \(2:1\)
ⓒ. \(\cos^2\theta=1/2\); density ratio \(4:1\)
ⓓ. \(\cos^2\theta=1/3\); density ratio \(4:1\)
Correct Answer: \(\cos^2\theta=1/3\); density ratio \(4:1\)
Explanation: \(\textbf{Identify the angular nodal condition:}\)
The supplied angular dependence is
\[
\Psi_{\mathrm{angular}}\propto3\cos^2\theta-1
\]
An angular node occurs where this angular factor becomes zero.
Therefore,
\[
3\cos^2\theta-1=0
\]
Rearranging,
\[
3\cos^2\theta=1
\]
\[
\cos^2\theta=\frac13
\]
These angular conditions describe the two conical nodal surfaces of the real \(d_{z^2}\) orbital.
\(\textbf{Evaluate the wave function along the }z\textbf{-axis:}\)
Along the positive \(z\)-axis,
\[
\theta=0^\circ
\]
Therefore,
\[
3\cos^2 0^\circ-1=3(1)-1
\]
\[
=2
\]
The angular wave-function factor has magnitude two in the chosen proportionality.
\(\textbf{Evaluate the wave function along the }x\textbf{-axis:}\)
Along the positive \(x\)-axis,
\[
\theta=90^\circ
\]
Hence,
\[
3\cos^2 90^\circ-1=3(0)-1
\]
\[
=-1
\]
The angular wave-function factor has magnitude one.
\(\textbf{Compare the probability densities:}\)
Spatial probability density is proportional to the squared magnitude of the complete wave function.
At the same radial distance, the common radial factor cancels in the ratio.
Therefore,
\[
\frac{\rho_z}{\rho_x}=\frac{|2|^2}{|-1|^2}
\]
\[
\frac{\rho_z}{\rho_x}=\frac41
\]
\(\textbf{Interpret the orbital geometry:}\)
The wave-function amplitudes along the selected axes have opposite signs.
Nevertheless, both corresponding probability densities are positive.
The angular nodal cones occur at intermediate directions rather than along the selected axes.
The \(d_{z^2}\) orbital therefore has a different nodal geometry from real \(d\) orbitals whose angular nodes are two planes.
All remain \(d\) orbitals with \(l=2\), but their angular probability distributions are different.
\(\textbf{Final answer:}\)
\[
\boxed{\cos^2\theta_{\mathrm{node}}=\frac13,\qquad\rho_z:\rho_x=4:1}
\]
303. A real hydrogen \(3p_z\) orbital is described using laboratory axes \(x,y,z\). The researcher then changes only the coordinate system: the new positive \(x'\)-axis is chosen to lie along the original positive \(z\)-axis. The nucleus, electron state and physical orbital probability distribution remain unchanged.
Consider the following statements.
Statement I: Relative to the new coordinate axes, the same orbital can be described as having its principal lobes along the \(x'\)-axis.
Statement II: Changing the coordinate axes without changing the physical state changes the electron's measured probability density at a fixed physical point.
Statement III: The orbital retains one radial node because its principal and angular momentum quantum numbers remain \(n=3\) and \(l=1\).
Statement IV: A passive change of coordinate axes does not by itself change the electronic energy of the isolated atom.
Which statements are correct?
ⓐ. I, II and III only
ⓑ. I, III and IV only
ⓒ. II, III and IV only
ⓓ. I, II and IV only
Correct Answer: I, III and IV only
Explanation: Statement I is correct because orbital labels such as \(p_x\), \(p_y\) and \(p_z\) describe orientations relative to a chosen coordinate system.
The physical \(3p_z\) orbital has its principal lobes along the original \(z\)-axis.
When the new \(x'\)-axis is defined to coincide with that direction, the orbital is aligned with the new \(x'\)-axis.
It may consequently be described using the corresponding primed-coordinate orbital orientation.
Statement II is incorrect because the change is only a relabelling of coordinates.
The physical electronic state and its spatial probability distribution are stipulated to remain unchanged.
Therefore, the probability density measured at a fixed physical point remains the same.
Changing the numerical coordinates assigned to that point does not change the underlying probability.
Statement III is correct because the number of radial nodes is determined by
\[
N_{\mathrm{radial}}=n-l-1
\]
For the \(3p\) orbital,
\[
N_{\mathrm{radial}}=3-1-1=1
\]
A coordinate transformation does not alter the principal quantum number or the radial nodal surface.
Statement IV is correct because an alteration of coordinate axes alone is not a physical change to the atom.
It does not supply or remove electronic energy.
The distinction is between rotating the physical electronic state relative to a fixed environment and changing the coordinate labels used to describe the same state.
In the present isolated system, the orbital has not physically changed.
The interpretation illustrates why Cartesian orbital labels are descriptions of orientation rather than independently measurable changes produced by merely renaming axes.
Thus, Statements I, III and IV are correct.
304. Two one-electron systems are compared. In system P, the electron occupies the hydrogen \(2s\) state. In system Q, the electron occupies the \(3p\) state of \(\mathrm{He^+}\). Both orbitals have exactly one radial node, but their principal quantum numbers, angular structures and nuclear charges differ.
Which option correctly compares their ionisation energies from the occupied states and their angular-node counts?
ⓐ. \(E_{\mathrm{ion},Q}/E_{\mathrm{ion},P}=4/9\); P has no angular node and Q has one
ⓑ. \(E_{\mathrm{ion},Q}/E_{\mathrm{ion},P}=16/9\); both have one angular node
ⓒ. \(E_{\mathrm{ion},Q}/E_{\mathrm{ion},P}=16/9\); P has no angular node and Q has one
ⓓ. \(E_{\mathrm{ion},Q}/E_{\mathrm{ion},P}=9/16\); P has one angular node and Q has none
Correct Answer: \(E_{\mathrm{ion},Q}/E_{\mathrm{ion},P}=16/9\); P has no angular node and Q has one
Explanation: \(\textbf{Identify the electronic states:}\)
System P contains a hydrogen electron in \(2s\).
Therefore,
\[
Z_P=1,\qquad n_P=2,\qquad l_P=0
\]
System Q contains a \(\mathrm{He^+}\) electron in \(3p\).
Therefore,
\[
Z_Q=2,\qquad n_Q=3,\qquad l_Q=1
\]
\(\textbf{Verify the common radial-node count:}\)
The number of radial nodes is
\[
N_{\mathrm{radial}}=n-l-1
\]
For P,
\[
N_{\mathrm{radial},P}=2-0-1=1
\]
For Q,
\[
N_{\mathrm{radial},Q}=3-1-1=1
\]
Thus, the stated common radial-node count is consistent.
\(\textbf{Determine P's occupied-state ionisation energy:}\)
For a hydrogen-like species,
\[
E_{\mathrm{ion},n}=13.6\frac{Z^2}{n^2}\,\mathrm{eV}
\]
For P,
\[
E_{\mathrm{ion},P}=13.6\frac{1^2}{2^2}
\]
\[
E_{\mathrm{ion},P}=3.40\,\mathrm{eV}
\]
\(\textbf{Determine Q's occupied-state ionisation energy:}\)
For Q,
\[
E_{\mathrm{ion},Q}=13.6\frac{2^2}{3^2}
\]
\[
E_{\mathrm{ion},Q}=\frac{54.4}{9}\,\mathrm{eV}
\]
\(\textbf{Calculate the energy ratio:}\)
\[
\frac{E_{\mathrm{ion},Q}}{E_{\mathrm{ion},P}}
=\frac{13.6(4/9)}{13.6(1/4)}
\]
\[
\frac{E_{\mathrm{ion},Q}}{E_{\mathrm{ion},P}}=\frac{16}{9}
\]
\(\textbf{Compare angular nodes:}\)
An \(s\) orbital has \(l=0\), so P has no angular node.
A \(p\) orbital has \(l=1\), so Q has one angular node.
Their equal radial-node counts therefore do not imply identical angular structures or equal ionisation energies.
The comparison requires treating radial nodes, nuclear charge and principal quantum number as separate pieces of information.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{E_{\mathrm{ion},Q}}{E_{\mathrm{ion},P}}=\frac{16}{9},\qquad N_{\mathrm{angular},P}=0,\qquad N_{\mathrm{angular},Q}=1}
\]
305. Two electrons in the same atom are described by the following quantum numbers:
Electron P: \((n,l,m_l,m_s)=(3,0,0,+\tfrac12)\).
Electron Q: \((n,l,m_l,m_s)=(3,1,0,+\tfrac12)\).
For orbital angular momentum, \(L=\sqrt{l(l+1)}\hbar\) and \(L_z=m_l\hbar\).
Which option correctly evaluates their orbital angular momenta, radial nodes and compatibility with the Pauli exclusion principle?
ⓐ. Both have zero orbital angular momentum magnitude and two radial nodes; their coexistence violates Pauli because their spin projections are equal.
ⓑ. P has \(L=0\) and two radial nodes; Q has \(L=\sqrt2\hbar\) and one radial node; both have \(L_z=0\), and their coexistence is Pauli-permitted.
ⓒ. P has \(L=0\) and one radial node; Q has \(L=\sqrt2\hbar\) and two radial nodes; their coexistence violates Pauli because both have \(m_l=0\).
ⓓ. P has \(L=\hbar\) and two radial nodes; Q has \(L=\sqrt2\hbar\) and one radial node; both must have opposite spin projections.
Correct Answer: P has \(L=0\) and two radial nodes; Q has \(L=\sqrt2\hbar\) and one radial node; both have \(L_z=0\), and their coexistence is Pauli-permitted.
Explanation: \(\textbf{Identify the spatial orbitals:}\)
Electron P has
\[
n=3,\qquad l=0
\]
It therefore occupies a \(3s\) spatial orbital.
Electron Q has
\[
n=3,\qquad l=1
\]
It occupies a \(3p\) spatial orbital.
The different values of \(l\) establish that the electrons belong to different spatial orbital states.
\(\textbf{Compare the orbital angular momentum magnitudes:}\)
The magnitude is
\[
L=\sqrt{l(l+1)}\hbar
\]
For P,
\[
L_P=\sqrt{0(1)}\hbar=0
\]
For Q,
\[
L_Q=\sqrt{1(2)}\hbar
\]
\[
L_Q=\sqrt2\hbar
\]
\(\textbf{Compare their axial projections:}\)
Both electrons have \(m_l=0\).
Therefore,
\[
L_{z,P}=0\hbar=0
\]
\[
L_{z,Q}=0\hbar=0
\]
The equal projections do not imply equal orbital angular momentum magnitudes.
\(\textbf{Determine the radial nodes:}\)
For P,
\[
N_{\mathrm{radial},P}=3-0-1=2
\]
For Q,
\[
N_{\mathrm{radial},Q}=3-1-1=1
\]
Their spatial orbital structures are consequently different.
\(\textbf{Apply the Pauli exclusion principle:}\)
Both electrons have \(m_s=+\tfrac12\), but their complete four-quantum-number sets are not identical.
They differ in the angular momentum quantum number \(l\).
Therefore, their simultaneous presence in the atom is compatible with the Pauli exclusion principle.
Equal spin projections are forbidden only when they cause the complete quantum-number sets of two electrons to become identical.
The result distinguishes spin projection, orbital angular momentum magnitude, axial orbital angular momentum and radial nodal structure without treating them as interchangeable properties.
\(\textbf{Final answer:}\)
\[
\boxed{L_P=0,\quad L_Q=\sqrt2\hbar,\quad L_{z,P}=L_{z,Q}=0,\quad N_{\mathrm{radial},P}=2,\quad N_{\mathrm{radial},Q}=1}
\]
306. Four subshells are considered while constructing the ground-state electronic configuration of an ordinary multi-electron atom using the standard Aufbau \((n+l)\) rule: \(3p\), \(4s\), \(3d\) and \(4p\).
Which sequence correctly represents their order of filling according to this rule?
ⓐ. \(3p\rightarrow3d\rightarrow4s\rightarrow4p\)
ⓑ. \(4s\rightarrow3p\rightarrow3d\rightarrow4p\)
ⓒ. \(3p\rightarrow4s\rightarrow3d\rightarrow4p\)
ⓓ. \(3p\rightarrow4s\rightarrow4p\rightarrow3d\)
Correct Answer: \(3p\rightarrow4s\rightarrow3d\rightarrow4p\)
Explanation: \(\textbf{Apply the two parts of the Aufbau ordering rule:}\)
In the standard Aufbau guide, a subshell with smaller \(n+l\) is filled before one with larger \(n+l\).
When two subshells have equal values of \(n+l\), the subshell with smaller \(n\) is placed first.
The relevant quantum-number values are:
\[
3p:\quad n+l=3+1=4
\]
\[
4s:\quad n+l=4+0=4
\]
\[
3d:\quad n+l=3+2=5
\]
\[
4p:\quad n+l=4+1=5
\]
\(\textbf{Resolve the first equality:}\)
The \(3p\) and \(4s\) subshells both have \(n+l=4\).
Because \(3p\) has the smaller principal quantum number, it precedes \(4s\).
\(\textbf{Resolve the second equality:}\)
The \(3d\) and \(4p\) subshells both have \(n+l=5\).
Since \(3d\) has \(n=3\), it precedes \(4p\), which has \(n=4\).
\(\textbf{Combine the two comparisons:}\)
Both subshells having \(n+l=4\) precede those having \(n+l=5\).
The resulting order is
\[
3p\rightarrow4s\rightarrow3d\rightarrow4p
\]
\(\textbf{Interpret the limitation of the rule:}\)
The Aufbau principle provides a useful guide for constructing ordinary ground-state electronic configurations.
It should not be interpreted as a universally exceptionless prediction of relative orbital energies in every neutral atom and ion.
In particular, the familiar filling of \(4s\) before \(3d\) does not mean that \(4s\) electrons must always be removed after \(3d\) electrons during ionisation.
The correct sequence for the stated filling-rule comparison is option C.
307. A neutral atom has the ground-state electronic configuration \([Ne]\,3s^2\,3p^2\). Three electrons are represented in the \(3p\) subshell of a proposed excited arrangement, while only one electron remains in \(3s\):
\[
[Ne]\,3s^1\,3p^3
\]
Assume that the three \(3p\) electrons occupy different orbitals with parallel spins.
Which conclusion correctly compares the two arrangements?
ⓐ. The proposed arrangement is impossible because a \(p\) subshell cannot contain three electrons.
ⓑ. The proposed arrangement violates the Pauli exclusion principle because all three \(3p\) electrons have parallel spins.
ⓒ. The proposed arrangement is the ground state because a half-filled \(3p\) subshell must always take priority over filling \(3s\).
ⓓ. The proposed arrangement is a valid excited-state arrangement with the same total electron count; it has more unpaired electrons than the stated ground state.
Correct Answer: The proposed arrangement is a valid excited-state arrangement with the same total electron count; it has more unpaired electrons than the stated ground state.
Explanation: \(\textbf{Check electron conservation:}\)
The neon core contains ten electrons.
In the ground-state arrangement, the additional electrons are
\[
3s^2\,3p^2
\]
Thus, the atom contains fourteen electrons.
In the proposed arrangement, the additional electrons are
\[
3s^1\,3p^3
\]
This also represents fourteen electrons.
The proposed change therefore transfers an electron between subshells without changing the identity or charge of the atom.
\(\textbf{Apply the Aufbau principle:}\)
For the ordinary ground-state construction, \(3s\) is lower in the filling order than \(3p\).
The stated ground configuration consequently fills \(3s\) before placing the remaining electrons in \(3p\).
The alternative arrangement leaves a vacancy in \(3s\) while placing an additional electron in \(3p\).
It represents an excitation rather than the ordinary ground-state filling pattern.
\(\textbf{Check the Pauli and Hund restrictions:}\)
A \(p\) subshell contains three spatial orbitals.
Three electrons can occupy these orbitals singly with parallel spins.
No two electrons then have identical sets of all four quantum numbers.
The proposed \(3p^3\) occupancy is therefore compatible with both Pauli's exclusion principle and Hund's rule.
\(\textbf{Compare the unpaired-electron counts:}\)
In the ground state, \(3s^2\) is paired, while \(3p^2\) has two unpaired electrons.
In the proposed excited arrangement, the \(3s^1\) electron is unpaired and all three \(3p^3\) electrons are unpaired.
Therefore, the unpaired-electron count increases from two to four.
A configuration can satisfy the Pauli and Hund restrictions without being the lowest-energy arrangement of the atom.
The correct classification is a Pauli-permitted excited state, not an impossible configuration or the ordinary ground state.
308. Three electrons occupy a set of three degenerate \(p\) orbitals. Four arrangements are proposed, where each pair of square brackets represents one spatial orbital and arrows denote electron spin projections.
| Arrangement | Orbital 1 | Orbital 2 | Orbital 3 |
|---|
| P | ↑↓ | ↑ | Empty |
| Q | ↑ | ↑ | ↑ |
| R | ↑↑ | ↑ | Empty |
| S | ↑ | ↓ | ↑ |
Which assessment is correct for the lowest-energy occupancy under the usual Hund and Pauli rules?
ⓐ. P is preferred because electron pairing in the first orbital minimises the number of occupied orbitals.
ⓑ. R is preferred because all three electrons have parallel spins.
ⓒ. S is preferred because singly occupied orbitals must contain both spin orientations.
ⓓ. Q is preferred; P and S do not satisfy the maximum-multiplicity arrangement, while R violates the Pauli exclusion principle.
Correct Answer: Q is preferred; P and S do not satisfy the maximum-multiplicity arrangement, while R violates the Pauli exclusion principle.
Explanation: The three \(p\) orbitals are specified to be degenerate, meaning they have equal energies under the stated conditions.
Hund's rule requires electrons to occupy degenerate orbitals singly with parallel spins before pairing occurs in the lowest-energy arrangement.
Arrangement Q places one electron in each orbital.
All three singly occupied orbitals have the same spin projection, so Q satisfies Hund's rule.
It also satisfies the Pauli exclusion principle because the electrons occupy different spatial orbitals.
Arrangement P pairs two electrons while leaving one degenerate orbital empty.
Although its electron count is correct and the pair has opposite spins, its arrangement does not follow Hund's rule for the lowest-energy distribution.
Arrangement R places two electrons with the same spin projection in the same spatial orbital.
Those electrons would have identical spatial quantum numbers and identical spin quantum numbers.
Consequently, R violates the Pauli exclusion principle.
Arrangement S places one electron in each orbital, so its spatial distribution satisfies the single-occupancy part of Hund's rule.
However, the spins are not parallel, and the arrangement does not have the maximum spin multiplicity.
Therefore, S is not the preferred lowest-energy arrangement under the stated rules.
The distinction is that Pauli's principle determines whether a proposed simultaneous occupancy is permitted, whereas Hund's rule helps select the preferred arrangement among permitted occupancies of degenerate orbitals.
Only Q satisfies both requirements for the stated lowest-energy arrangement.
309. An atom has the ground-state configuration \([Ne]\,3s^2\,3p^4\). The neutral atom loses one electron, and the resulting singly positive ion retains its electrons in the lowest available orbital arrangement. Assume that electron removal occurs from the outer \(3p\) subshell.
Which option correctly compares the number of unpaired electrons in the neutral atom and its ion and identifies the reason for the change?
ⓐ. Two in the neutral atom and three in the ion; removal of a paired \(3p\) electron leaves three singly occupied \(3p\) orbitals.
ⓑ. Two in the neutral atom and one in the ion; removing any electron must reduce the unpaired-electron count by one.
ⓒ. Four in the neutral atom and three in the ion; the four \(3p\) electrons initially occupy four separate orbitals.
ⓓ. Two in the neutral atom and zero in the ion; the three remaining \(3p\) electrons become paired to minimise spin.
Correct Answer: Two in the neutral atom and three in the ion; removal of a paired \(3p\) electron leaves three singly occupied \(3p\) orbitals.
Explanation: \(\textbf{Identify the neutral atom's outer-subshell occupancy:}\)
The neutral configuration ends in
\[
3s^2\,3p^4
\]
The \(3s^2\) subshell contains a paired set of electrons.
The \(3p\) subshell contains three spatial orbitals.
Following Hund's rule, four electrons occupy these orbitals with one pair and two singly occupied orbitals.
A representative arrangement is
\[
3p^4:\quad[\uparrow\downarrow]\ [\uparrow]\ [\uparrow]
\]
The neutral atom therefore has two unpaired electrons.
\(\textbf{Apply the specified ionisation:}\)
Removal of one outer \(3p\) electron changes the occupancy to
\[
3p^4\rightarrow3p^3
\]
The lowest-energy arrangement of three electrons in three degenerate \(p\) orbitals places one electron in each orbital with parallel spins:
\[
3p^3:\quad[\uparrow]\ [\uparrow]\ [\uparrow]
\]
The resulting ion therefore has three unpaired electrons.
\(\textbf{Interpret the counterintuitive change:}\)
Although the ion has one fewer electron, its unpaired-electron count is greater.
This occurs because the removed electron came from a paired orbital.
Its removal leaves three singly occupied \(p\) orbitals.
The half-filled \(p^3\) arrangement is compatible with Hund's maximum-multiplicity rule.
The total electron count and the unpaired-electron count are different quantities.
A decrease in the former does not necessarily require a decrease in the latter.
The correct comparison is two unpaired electrons before ionisation and three after ionisation.
\(\textbf{Final answer:}\)
\[
\boxed{N_{\mathrm{unpaired,neutral}}=2,\qquad N_{\mathrm{unpaired,ion}}=3}
\]
310. An element has atomic number 26. Its neutral ground-state configuration is represented as \([Ar]\,3d^6\,4s^2\). The element forms a \(2+\) ion by removal of its two outer \(4s\) electrons.
Consider the following statements.
Statement I: The ion has the electronic configuration \([Ar]\,3d^6\).
Statement II: Because \(4s\) fills before \(3d\) in the usual neutral-atom Aufbau sequence, the first two electrons removed during ionisation must necessarily come from \(3d\).
Statement III: The \(3d^6\) subshell has four unpaired electrons in its usual Hund-rule arrangement.
Statement IV: The difference between neutral-atom filling order and electron-removal order illustrates why the simple Aufbau sequence must not be interpreted as an immutable energy ordering for all atoms and ions.
Which statements are correct?
ⓐ. I, II and III only
ⓑ. II, III and IV only
ⓒ. I, III and IV only
ⓓ. I, II and IV only
Correct Answer: I, III and IV only
Explanation: Statement I is correct because the neutral atom contains six electrons in \(3d\) and two in \(4s\).
The specified formation of the \(2+\) ion removes both \(4s\) electrons.
The resulting configuration is therefore \([Ar]\,3d^6\).
Statement II is incorrect because orbital filling order in the neutral atom does not automatically determine electron-removal order during ionisation.
The fact that \(4s\) is commonly filled before \(3d\) is not a rule requiring \(3d\) electrons to be removed first.
For the stated first-transition-series atom, the two \(4s\) electrons are removed in forming the \(2+\) ion.
Statement III is correct because a \(d\) subshell contains five spatial orbitals.
Under Hund's rule, the first five electrons occupy different orbitals singly with parallel spins.
The sixth electron pairs with one of those electrons.
Consequently, the \(3d^6\) subshell contains one paired orbital and four singly occupied orbitals.
Statement IV is correct because relative orbital energies depend on the electronic environment.
The usual Aufbau order is a useful guide to constructing neutral-atom ground-state configurations.
It does not imply that the order of electron removal must be the exact reverse of that filling sequence.
The ion has a different electronic population and consequently a different orbital-energy environment.
This distinction is important when interpreting configurations of transition-metal ions.
The correct statements are I, III and IV.
311. Four hypothetical electronic arrangements have the same total number of electrons and the same noble-gas core \([Ne]\). The arrangements are:
| Arrangement | Configuration beyond \([Ne]\) |
|---|
| P | \(3s^2\,3p^3\) |
| Q | \(3s^1\,3p^4\) |
| R | \(3s^2\,3p^2\,4s^1\) |
| S | \(3s^2\,3p^5\) |
Which statement correctly identifies the arrangements having the same total electron count and distinguishes their ground-state or excited-state interpretation?
ⓐ. P, Q and R all contain fifteen electrons; P is the ordinary ground-state arrangement, while Q and R are excited arrangements. S contains seventeen electrons.
ⓑ. P and Q contain fifteen electrons, but R contains sixteen because \(4s\) belongs to the fourth shell.
ⓒ. All four contain fifteen electrons; their configurations differ only in the spatial orientation of orbitals.
ⓓ. P, Q and S contain fifteen electrons; R is forbidden because an electron cannot occupy \(4s\) while a \(3p\) orbital is not completely filled.
Correct Answer: P, Q and R all contain fifteen electrons; P is the ordinary ground-state arrangement, while Q and R are excited arrangements. S contains seventeen electrons.
Explanation: \(\textbf{Count electrons independently of shell labels:}\)
The neon core contains ten electrons.
Arrangement P contributes
\[
2+3=5
\]
additional electrons and therefore represents fifteen electrons in total.
Arrangement Q contributes
\[
1+4=5
\]
additional electrons, so it also contains fifteen electrons.
Arrangement R contributes
\[
2+2+1=5
\]
additional electrons and likewise contains fifteen electrons.
Arrangement S contributes
\[
2+5=7
\]
additional electrons, giving seventeen electrons altogether.
\(\textbf{Identify the ordinary ground-state arrangement:}\)
For a fifteen-electron neutral atom, the conventional Aufbau sequence fills \(3s\) before \(3p\), and places the remaining electrons in the \(3p\) subshell.
This produces
\[
[Ne]\,3s^2\,3p^3
\]
Therefore, P is the ordinary ground-state configuration.
\(\textbf{Interpret the alternative arrangements:}\)
In Q, one electron has been promoted from \(3s\) into \(3p\).
In R, an electron has been promoted into \(4s\), leaving the lower-energy \(3p\) subshell incompletely occupied.
Neither promotion changes the total electron count.
Both arrangements can represent excited electronic configurations, provided their individual orbital occupancies satisfy the Pauli exclusion principle.
The presence of an electron in \(4s\) while \(3p\) remains incompletely filled is not, by itself, a violation of Pauli's principle.
It means the distribution is not the usual lowest-energy Aufbau arrangement.
The correct classification therefore requires electron counting before applying the ground-state filling rule.
312. A neutral atom has ground-state configuration \([Ar]\,3d^5\,4s^2\). It forms a \(2+\) ion by removal of both \(4s\) electrons. Assume the \(3d\) orbital arrangement follows Hund's rule and remains in its lowest-energy occupancy pattern.
Which option correctly compares the neutral atom and ion?
ⓐ. The neutral atom has seven unpaired electrons, while the ion has five.
ⓑ. The neutral atom has five unpaired electrons, while the ion has five; the ion retains a half-filled \(3d^5\) subshell.
ⓒ. The neutral atom has five unpaired electrons, while the ion has three because two \(3d\) electrons are removed.
ⓓ. Both species have zero unpaired electrons because the \(3d^5\) subshell is half-filled.
Correct Answer: The neutral atom has five unpaired electrons, while the ion has five; the ion retains a half-filled \(3d^5\) subshell.
Explanation: \(\textbf{Identify the neutral outer-electron distribution:}\)
The neutral atom has
\[
[Ar]\,3d^5\,4s^2
\]
The \(4s^2\) electrons are paired and do not contribute to the number of unpaired electrons.
The five electrons in \(3d\) occupy five degenerate spatial orbitals.
Hund's rule places them singly with parallel spins before any pairing occurs.
Therefore,
\[
N_{\mathrm{unpaired,neutral}}=5
\]
\(\textbf{Apply the specified electron removal:}\)
Formation of the \(2+\) ion removes the two \(4s\) electrons.
The remaining configuration is
\[
[Ar]\,3d^5
\]
The five \(3d\) electrons remain distributed singly across the five spatial orbitals.
Therefore,
\[
N_{\mathrm{unpaired,ion}}=5
\]
\(\textbf{Check the subshell's occupancy:}\)
A \(d\) subshell contains five spatial orbitals and can accommodate ten electrons.
With five electrons, it is exactly half-filled.
The unpaired-electron count is five because no orbital contains a pair in the usual Hund arrangement.
\(\textbf{Interpret the unchanged magnetic-electron count:}\)
The total electron count decreases by two during ionisation.
However, the removed electrons formed a paired \(4s^2\) set.
Their removal does not change the number of unpaired electrons in \(3d\).
Thus, a change in ionic charge need not alter the number of unpaired electrons.
The conclusion depends on both the subshell from which electrons are removed and the orbital occupancy of the remaining electrons.
\(\textbf{Final answer:}\)
\[
\boxed{N_{\mathrm{unpaired,neutral}}=N_{\mathrm{unpaired,ion}}=5}
\]
313. An atom contains six electrons in a set of five degenerate \(d\) orbitals. Four students propose arrangements that obey different combinations of the Pauli exclusion principle and Hund's rule. Each box represents one \(d\) spatial orbital.
| Arrangement | Orbital 1 | Orbital 2 | Orbital 3 | Orbital 4 | Orbital 5 |
|---|
| P | ↑↓ | ↑ | ↑ | ↑ | ↑ |
| Q | ↑↓ | ↑↓ | ↑ | ↑ | Empty |
| R | ↑↑ | ↑ | ↑ | ↑ | ↑ |
| S | ↑↓ | ↑ | ↑ | ↑ | ↓ |
Which assessment correctly identifies the preferred lowest-energy arrangement and the distinct defects in the other proposals?
ⓐ. Q is preferred; P violates Pauli, while R and S are excited configurations.
ⓑ. R is preferred because it maximises the number of electrons having parallel spins.
ⓒ. P is preferred; Q pairs prematurely, R violates Pauli, and S does not maximise parallel spins in singly occupied orbitals.
ⓓ. S is preferred because the paired orbital requires the remaining singly occupied orbitals to contain both spin orientations.
Correct Answer: P is preferred; Q pairs prematurely, R violates Pauli, and S does not maximise parallel spins in singly occupied orbitals.
Explanation: A \(d\) subshell contains five spatial orbitals.
Under Hund's rule, the first five electrons occupy different degenerate orbitals singly with parallel spins.
The sixth electron then pairs with one of these electrons.
Arrangement P follows this sequence.
It contains one paired orbital and four singly occupied orbitals having parallel spins.
Therefore, P is the preferred lowest-energy arrangement under the stated rules.
Arrangement Q contains two paired orbitals while another degenerate orbital remains empty.
Although the paired electrons have opposite spins and the total electron count is correct, this distribution does not satisfy Hund's preferred single-occupancy pattern.
Arrangement R places two electrons having the same spin projection in the same orbital.
Those two electrons would possess identical spatial quantum numbers and identical spin quantum numbers.
Therefore, R violates the Pauli exclusion principle.
Arrangement S has one paired orbital and four singly occupied orbitals.
However, one of the singly occupied electrons has a spin opposite to the other three.
This arrangement does not maximise the total spin multiplicity in the stated degenerate set.
The comparison illustrates that satisfying the Pauli exclusion principle is necessary but does not alone identify the Hund-rule ground-state arrangement.
Likewise, simply maximising the total number of upward arrows is not sufficient if two electrons are assigned identical quantum-number sets.
The correct configuration must satisfy electron count, Pauli occupancy and Hund's maximum-multiplicity requirement simultaneously.
314. An element has atomic number 24. A simplified application of the Aufbau filling sequence suggests \([Ar]\,3d^4\,4s^2\), whereas the observed neutral ground-state configuration is \([Ar]\,3d^5\,4s^1\).
Which option correctly identifies the element, compares the number of unpaired electrons in these configurations under Hund's rule and gives the relevant stabilisation explanation?
ⓐ. Chromium; four unpaired electrons in the simplified arrangement and six in the observed arrangement; the observed configuration is favoured by the extra stability associated with the half-filled \(3d^5\) subshell.
ⓑ. Chromium; six unpaired electrons in both arrangements; the change occurs because Pauli's principle forbids \(3d^4\).
ⓒ. Manganese; four unpaired electrons in the simplified arrangement and five in the observed arrangement; the change occurs because \(4s\) cannot accommodate two electrons.
ⓓ. Chromium; four unpaired electrons in the simplified arrangement and five in the observed arrangement; the observed state has a completely filled \(3d\) subshell.
Correct Answer: Chromium; four unpaired electrons in the simplified arrangement and six in the observed arrangement; the observed configuration is favoured by the extra stability associated with the half-filled \(3d^5\) subshell.
Explanation: \(\textbf{Identify the element:}\)
Atomic number 24 corresponds to chromium.
The argon core accounts for eighteen electrons.
The six remaining electrons are distributed between the \(3d\) and \(4s\) subshells.
\(\textbf{Evaluate the simplified arrangement:}\)
The simplified configuration is
\[
[Ar]\,3d^4\,4s^2
\]
Under Hund's rule, four \(3d\) electrons occupy four different \(d\) orbitals singly.
The \(4s^2\) electrons form a pair.
Thus,
\[
N_{\mathrm{unpaired,simplified}}=4
\]
\(\textbf{Evaluate the observed ground-state arrangement:}\)
Chromium has the observed configuration
\[
[Ar]\,3d^5\,4s^1
\]
The five \(3d\) electrons occupy all five \(d\) orbitals singly.
The remaining \(4s\) electron is also unpaired.
Hence,
\[
N_{\mathrm{unpaired,observed}}=5+1
\]
\[
N_{\mathrm{unpaired,observed}}=6
\]
\(\textbf{Explain the configuration exception:}\)
The \(3d\) and \(4s\) subshells are sufficiently close in energy for the total stability of alternative arrangements to matter.
The \(3d^5\) arrangement is half-filled and receives additional stabilisation associated with exchange energy and symmetrical electron distribution.
The resulting \([Ar]\,3d^5\,4s^1\) configuration is more stable for neutral chromium.
This does not imply that \(3d^4\) is forbidden by the Pauli exclusion principle.
It also does not establish a universal rule requiring electron promotion whenever half-filling can be imagined.
The exception must be understood in terms of the actual total-energy balance for the particular atom.
\(\textbf{Final answer:}\)
\[
\boxed{N_{\mathrm{unpaired,simplified}}=4,\qquad N_{\mathrm{unpaired,observed}}=6}
\]
315. Two neutral elements P and Q have atomic numbers 24 and 29, respectively. Their observed ground-state configurations are compared with the configurations predicted by a rigid filling sequence that does not account for exceptional subshell stability.
Consider the following statements.
Statement I: P has the observed configuration \([Ar]\,3d^5\,4s^1\).
Statement II: Q has the observed configuration \([Ar]\,3d^{10}\,4s^1\).
Statement III: P and Q exhibit their exceptional configurations for exactly the same reason: both achieve completely filled \(3d\) subshells.
Statement IV: The observed configurations reflect a balance of subshell energies and electronic stabilisation rather than an unrestricted rule that any electron may be promoted to create a filled or half-filled subshell.
Which statements are correct?
ⓐ. I, II and III only
ⓑ. II, III and IV only
ⓒ. I, III and IV only
ⓓ. I, II and IV only
Correct Answer: I, II and IV only
Explanation: Statement I is correct because atomic number 24 corresponds to chromium.
Its observed neutral ground-state configuration is \([Ar]\,3d^5\,4s^1\).
This arrangement contains a half-filled \(3d\) subshell.
Statement II is correct because atomic number 29 corresponds to copper.
Its observed neutral ground-state configuration is \([Ar]\,3d^{10}\,4s^1\).
Here the \(3d\) subshell is completely filled.
Statement III is incorrect because the two configurations have different \(3d\) occupancies.
Chromium has \(3d^5\), which is half-filled rather than completely filled.
Copper has \(3d^{10}\), which is completely filled.
The two elements therefore illustrate related but distinct forms of additional subshell stability.
Statement IV is correct because the energy differences between relevant \(3d\) and \(4s\) arrangements are small enough for electronic stabilisation effects to influence the observed ground state.
Exchange-energy considerations and symmetrical distributions contribute to the treatment of half-filled and completely filled subshells.
However, the simplified explanation must not be converted into a universal electron-promotion rule for all atoms.
A configuration exception is determined by the total energy of the electronic arrangement.
The usual Aufbau sequence remains useful for most configurations, but it must be applied with awareness of recognised exceptions.
Thus, Statements I, II and IV are correct.
316. Three species are considered:
P: \(\mathrm{Cu^+}\), atomic number 29;
Q: \(\mathrm{Zn^{2+}}\), atomic number 30;
R: neutral Zn, atomic number 30.
The neutral ground-state configurations of copper and zinc are \([Ar]\,3d^{10}\,4s^1\) and \([Ar]\,3d^{10}\,4s^2\), respectively. Electrons are removed from \(4s\) when the specified positive ions form.
Which statement correctly compares the configurations and unpaired-electron counts of P, Q and R?
ⓐ. P and Q are both \([Ar]3d^{10}\), while R is \([Ar]3d^{10}4s^2\); all three have zero unpaired electrons.
ⓑ. P has \([Ar]\,3d^9\,4s^1\) and two unpaired electrons; Q and R have no unpaired electrons.
ⓒ. P and Q both have \([Ar]\,3d^{10}\,4s^1\), while R has \([Ar]\,3d^{10}\,4s^2\).
ⓓ. All three species have identical electronic configurations because their \(3d\) subshells are completely filled.
Correct Answer: P and Q are both \([Ar]3d^{10}\), while R is \([Ar]3d^{10}4s^2\); all three have zero unpaired electrons.
Explanation: \(\textbf{Determine the copper-ion configuration:}\)
Neutral copper has
\[
[Ar]\,3d^{10}\,4s^1
\]
Formation of \(\mathrm{Cu^+}\) removes the single \(4s\) electron.
Therefore,
\[
\mathrm{Cu^+}:\quad[Ar]\,3d^{10}
\]
The \(3d\) subshell is completely filled, so it contains no unpaired electrons.
\(\textbf{Determine the zinc-ion configuration:}\)
Neutral zinc has
\[
[Ar]\,3d^{10}\,4s^2
\]
Formation of \(\mathrm{Zn^{2+}}\) removes both \(4s\) electrons.
Therefore,
\[
\mathrm{Zn^{2+}}:\quad[Ar]\,3d^{10}
\]
This species also has a completely filled \(3d\) subshell and no unpaired electrons.
\(\textbf{Compare with neutral zinc:}\)
Neutral zinc retains
\[
[Ar]\,3d^{10}\,4s^2
\]
The \(3d^{10}\) subshell is completely filled.
The \(4s^2\) electrons are paired.
Consequently, neutral zinc also has no unpaired electrons.
\(\textbf{Check electron counts and identities:}\)
Both \(\mathrm{Cu^+}\) and \(\mathrm{Zn^{2+}}\) contain 28 electrons.
They are isoelectronic and have the same stated configuration.
Neutral zinc contains 30 electrons and has two additional \(4s\) electrons.
Its absence of unpaired electrons does not make its configuration identical to those of the ions.
Likewise, a completely filled \(3d\) subshell does not require the atom and all its ions to have equal electron populations.
The result combines observed neutral configurations, electron removal, isoelectronic comparison and spin occupancy.
\(\textbf{Final answer:}\)
\[
N_{\mathrm{unpaired}}(\mathrm{Cu^+})=N_{\mathrm{unpaired}}(\mathrm{Zn^{2+}})=N_{\mathrm{unpaired}}(\mathrm{Zn})=0
\]
317. A hypothetical atom contains four electrons beyond a closed noble-gas core. Two arrangements are compared:
Arrangement P: \(ns^2\,np^2\), with the two \(np\) electrons occupying different orbitals with parallel spins.
Arrangement Q: \(ns^1\,np^3\), with the three \(np\) electrons occupying different orbitals with parallel spins.
The \(ns\) subshell lies below \(np\) in the ordinary Aufbau filling order. No special configuration exception is specified.
Which conclusion is justified?
ⓐ. Q must be the ground state because it contains a half-filled \(np\) subshell.
ⓑ. P is the ordinary ground state; Q is a Pauli-permitted promoted state with two additional unpaired electrons.
ⓒ. P violates Hund's rule because its two \(np\) electrons have parallel spins.
ⓓ. Q violates Pauli's principle because all four electrons are unpaired.
Correct Answer: P is the ordinary ground state; Q is a Pauli-permitted promoted state with two additional unpaired electrons.
Explanation: Arrangement P places two electrons in the lower-energy \(ns\) orbital and two in the \(np\) subshell.
The \(ns^2\) electrons are paired.
Under Hund's rule, the two \(np\) electrons occupy different degenerate orbitals with parallel spins.
P therefore contains two unpaired electrons.
Arrangement Q contains one electron in \(ns\) and three in \(np\).
The three \(np\) electrons occupy different orbitals with parallel spins, giving three unpaired electrons in that subshell.
The remaining \(ns\) electron is also unpaired.
Q therefore contains four unpaired electrons.
Both arrangements have the same total electron count.
The change from P to Q transfers one electron from the lower-energy \(ns\) subshell to the higher-energy \(np\) subshell.
Under the stated ordinary Aufbau conditions, this represents an excited arrangement.
The fact that \(np^3\) is half-filled does not automatically make Q the ground state.
The stabilisation associated with a half-filled subshell cannot be treated as an unrestricted rule that always overrides a lower-energy orbital occupancy.
Both P and Q satisfy the Pauli exclusion principle under the specified spin arrangements.
The correct interpretation therefore distinguishes electronic-state validity from lowest-energy configuration.
The unpaired-electron count rises from two to four even though no electron is added to the atom.
318. A neutral iron atom has ground-state configuration \([Ar]\,3d^6\,4s^2\). It forms \(\mathrm{Fe^{3+}}\) by successive removal of three electrons. Assume that the two \(4s\) electrons are removed before one \(3d\) electron and that the resulting \(3d\) occupancy follows Hund's rule.
Which option correctly gives the configuration of \(\mathrm{Fe^{3+}}\), its number of unpaired electrons and the change in that number relative to neutral iron?
ⓐ. \([Ar]\,3d^3\,4s^2\); three unpaired electrons; a decrease of one
ⓑ. \([Ar]\,3d^5\); five unpaired electrons; an increase of one
ⓒ. \([Ar]\,3d^5\); three unpaired electrons; a decrease of one
ⓓ. \([Ar]\,3d^6\,4s^1\); four unpaired electrons; no change
Correct Answer: \([Ar]\,3d^5\); five unpaired electrons; an increase of one
Explanation: \(\textbf{Identify the neutral occupancy:}\)
The neutral configuration is
\[
[Ar]\,3d^6\,4s^2
\]
The \(4s^2\) electrons are paired.
For \(3d^6\), Hund's rule gives one paired \(d\) orbital and four singly occupied orbitals.
Therefore, neutral iron has four unpaired electrons.
\(\textbf{Remove the first two electrons:}\)
The specified removal order takes the two electrons from \(4s\).
Thus,
\[
[Ar]\,3d^6\,4s^2\rightarrow[Ar]\,3d^6
\]
The resulting \(2+\) ion retains six \(3d\) electrons.
\(\textbf{Remove the third electron:}\)
The third electron is removed from \(3d\).
Therefore,
\[
[Ar]\,3d^6\rightarrow[Ar]\,3d^5
\]
The final species is \(\mathrm{Fe^{3+}}\).
\(\textbf{Apply Hund's rule to the final subshell:}\)
The five \(3d\) electrons occupy the five \(d\) orbitals singly with parallel spins.
Consequently,
\[
N_{\mathrm{unpaired,Fe^{3+}}}=5
\]
\(\textbf{Compare with the neutral atom:}\)
\[
\Delta N_{\mathrm{unpaired}}=5-4
\]
\[
\Delta N_{\mathrm{unpaired}}=+1
\]
\(\textbf{Interpret the change:}\)
Removal of electrons does not necessarily reduce the number of unpaired electrons.
The first two removed electrons form a \(4s\) pair and do not affect the initial four unpaired \(3d\) electrons.
Removal of the sixth \(3d\) electron then converts a paired \(3d^6\) occupancy into the half-filled \(3d^5\) arrangement.
The number of unpaired electrons increases by one.
This outcome depends on both the electron-removal order and Hund's rule for the remaining \(d\) electrons.
\(\textbf{Final answer:}\)
\[
\boxed{\mathrm{Fe^{3+}}:[Ar]\,3d^5,\qquad N_{\mathrm{unpaired}}=5,\qquad\Delta N_{\mathrm{unpaired}}=+1}
\]
319. The following electronic arrangements are proposed for neutral atoms with the indicated atomic numbers.
| Atom | Atomic number | Proposed ground-state configuration |
|---|
| P | 24 | \([Ar]\,3d^5\,4s^1\) |
| Q | 25 | \([Ar]\,3d^5\,4s^2\) |
| R | 29 | \([Ar]\,3d^9\,4s^2\) |
| S | 30 | \([Ar]\,3d^{10}\,4s^2\) |
Which option correctly identifies the proposals that match the actual neutral ground-state configurations and explains the remaining mismatch?
ⓐ. P and R only; Q and S require promotion to half-filled subshells.
ⓑ. Q and S only; P incorrectly leaves \(4s\) singly occupied.
ⓒ. P, Q and S only; R omits copper's observed completely filled \(3d^{10}\) subshell.
ⓓ. P, R and S only; Q must have \([Ar]\,3d^6\,4s^1\).
Correct Answer: P, Q and S only; R omits copper's observed completely filled \(3d^{10}\) subshell.
Explanation: Proposal P corresponds to chromium, atomic number 24.
Its observed ground-state configuration is \([Ar]\,3d^5\,4s^1\).
The half-filled \(3d^5\) arrangement contributes to its exceptional stability.
Therefore, P is correct.
Proposal Q corresponds to manganese, atomic number 25.
Its ordinary ground-state configuration is \([Ar]\,3d^5\,4s^2\).
The \(3d\) subshell is half-filled, and the \(4s\) subshell is fully occupied.
No additional rearrangement is required.
Therefore, Q is correct.
Proposal R corresponds to copper, atomic number 29.
The displayed configuration \([Ar]\,3d^9\,4s^2\) is the straightforward prediction obtained without accounting for copper's exceptional stability.
The observed neutral ground-state configuration is \([Ar]\,3d^{10}\,4s^1\).
Its completely filled \(3d\) subshell provides additional stabilisation.
Therefore, R does not match the actual ground state.
Proposal S corresponds to zinc, atomic number 30.
Its neutral ground-state configuration is \([Ar]\,3d^{10}\,4s^2\).
Both subshells are completely filled, so S is correct.
The comparison demonstrates that chromium and copper are recognised exceptions, but manganese and zinc must not be forced into unnecessary promoted configurations.
Half-filled and completely filled subshell stability must be considered together with actual orbital-energy differences.
The existence of one exception does not justify inventing an exception for every neighbouring element.
Thus, proposals P, Q and S are correct.
320. A neutral atom has the electronic configuration \([Ne]\,3s^2\,3p^5\). It gains one electron to form a stable negative ion. A second atom has the ground-state configuration \([Ne]\,3s^2\,3p^4\) and gains two electrons to form its corresponding \(2-\) ion.
Which option correctly compares the final electronic configurations, the \(3p\) orbital occupancies and the numbers of unpaired electrons?
ⓐ. Both ions attain \([Ne]\,3s^2\,3p^6\), contain three doubly occupied \(3p\) orbitals and have no unpaired electrons.
ⓑ. The first ion attains \(3p^6\), but the second retains \(3p^4\) because a \(p\) orbital cannot accept more than one electron.
ⓒ. Both ions attain \(3p^6\), but each must retain one unpaired electron because the added electrons have identical spins.
ⓓ. The first ion has \(3p^5\) and the second has \(3p^6\); only the second is a completely filled subshell.
Correct Answer: Both ions attain \([Ne]\,3s^2\,3p^6\), contain three doubly occupied \(3p\) orbitals and have no unpaired electrons.
Explanation: \(\textbf{Apply electron gain to the first atom:}\)
The first neutral configuration ends in
\[
3s^2\,3p^5
\]
Its \(3p\) subshell contains five electrons distributed among three spatial orbitals.
The sixth permitted electron completes that subshell.
Therefore,
\[
3p^5+e^-\rightarrow3p^6
\]
The resulting negative ion has the configuration
\[
[Ne]\,3s^2\,3p^6
\]
\(\textbf{Apply electron gain to the second atom:}\)
The second neutral configuration ends in
\[
3s^2\,3p^4
\]
Adding two electrons fills the two remaining available spin states in the \(3p\) subshell.
Thus,
\[
3p^4+2e^-\rightarrow3p^6
\]
Its \(2-\) ion also has
\[
[Ne]\,3s^2\,3p^6
\]
\(\textbf{Check orbital capacity and spin occupancy:}\)
A \(p\) subshell contains three spatial orbitals.
Each orbital can accommodate two electrons with opposite spins.
The completely filled \(3p^6\) arrangement is therefore
\[
[\uparrow\downarrow]\ [\uparrow\downarrow]\ [\uparrow\downarrow]
\]
All electrons in this subshell are paired.
The \(3s^2\) subshell is also paired.
Consequently, neither ion has an unpaired electron in its stated configuration.
\(\textbf{Interpret the comparison:}\)
The two ions begin from different neutral atoms and acquire different negative charges.
Nevertheless, both reach the same final electron configuration.
Their electronic configurations are identical, but their nuclear charges remain different.
The stable completely filled subshell results from adding electrons to previously unoccupied spin states, not from violating the Pauli exclusion principle.
The result combines electron counting, subshell capacity, ion formation and the stability of a completely filled electronic arrangement.
\(\textbf{Final answer:}\)
\[
\boxed{[Ne]\,3s^2\,3p^6,\qquad N_{\mathrm{unpaired}}=0\ \text{for both ions}}
\]
