401. The general electronic configuration used for actinoid atoms is:
ⓐ. \([\mathrm{Xe}]\,4f^{1-14}5d^{0-1}6s^2\)
ⓑ. \([\mathrm{Kr}]\,4d^{1-10}5s^2\)
ⓒ. \([\mathrm{Rn}]\,6f^{1-14}7d^{0-1}8s^2\)
ⓓ. \([\mathrm{Rn}]\,5f^{1-14}6d^{0-1}7s^2\)
Correct Answer: \([\mathrm{Rn}]\,5f^{1-14}6d^{0-1}7s^2\)
Explanation: Actinoids occur after radon, so their configurations are written using the \([\mathrm{Rn}]\) noble-gas core. The \(5f\) subshell is progressively occupied across the series. Depending on the individual element, zero or one electron may also occupy the \(6d\) subshell. The outer \(7s\) subshell commonly contains two electrons in the neutral atom. The expression is general because the exact configurations show irregularities. The xenon-core expression belongs to the lanthanoid series.
402. Assertion: Exact actinoid configurations show greater irregularity than a simple stepwise \(5f\)-filling pattern suggests.
Reason: The \(5f\), \(6d\), and \(7s\) subshells are close in energy.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: The general actinoid trend involves progressive filling of the \(5f\) subshell. However, the \(5f\), \(6d\), and \(7s\) energy levels lie relatively close together. Small energy differences can make more than one electron distribution competitive. Some neutral atoms therefore contain different numbers of \(5f\) and \(6d\) electrons than a mechanical filling rule would predict. This produces greater configurational irregularity across the series. The energy proximity stated in the Reason directly explains the Assertion.
403. An atom has the supplied configuration \([\mathrm{Rn}]\,5f^36d^17s^2\). Given that radon contains \(86\) electrons, the atomic number of the atom is:
ⓐ. \(90\)
ⓑ. \(92\)
ⓒ. \(91\)
ⓓ. \(94\)
Correct Answer: \(92\)
Explanation: \( \textbf{Electrons in the radon core:} \)
\[
86
\]
\( \textbf{Electrons in the \(5f\) subshell:} \)
\[
3
\]
\( \textbf{Electron in the \(6d\) subshell:} \)
\[
1
\]
\( \textbf{Electrons in the \(7s\) subshell:} \)
\[
2
\]
\( \textbf{Total electron count:} \)
\[
Z=86+3+1+2
\]
\[
Z=92
\]
\( \textbf{Classification check:} \) The radon core and occupied \(5f\) subshell are consistent with an actinoid atom.
\( \textbf{Final answer:} \) The atomic number is \(92\).
404. A learner rejects the configuration \([\mathrm{Rn}]\,5f^36d^17s^2\) because “the \(5f\) subshell must be completely filled before any electron can enter \(6d\).” The best correction is:
ⓐ. the configuration is impossible because an \(f\) subshell holds only three electrons
ⓑ. every actinoid must have the exact configuration \([\mathrm{Rn}]\,5f^n7s^2\)
ⓒ. a \(6d\) electron can occur only after both \(7s\) electrons are removed
ⓓ. the similar energies of \(5f\) and \(6d\) can permit \(6d^1\) before \(5f^{14}\)
Correct Answer: the similar energies of \(5f\) and \(6d\) can permit \(6d^1\) before \(5f^{14}\)
Explanation: Electron configurations are governed by relative orbital energies rather than by an inflexible subshell-completion rule. In actinoids, the \(5f\), \(6d\), and \(7s\) levels are close in energy. An arrangement containing a \(6d\) electron can therefore be energetically favourable even when \(5f\) is not full. An \(f\) subshell can hold fourteen electrons, not three. The general actinoid configuration explicitly allows \(6d^{0-1}\). The learner has applied a mechanical filling model where closely spaced energy levels must be considered.
405. Two supplied electronic configurations are \([\mathrm{Rn}]\,5f^46d^07s^2\) and \([\mathrm{Rn}]\,5f^36d^17s^2\). Which conclusion about their compatibility with the actinoid pattern is most appropriate?
ⓐ. Only the first can be an actinoid because the second contains a \(6d\) electron
ⓑ. Only the second can be an actinoid because the first has no \(6d\) electron
ⓒ. Both can fit the actinoid pattern because \(5f\) and \(6d\) occupancies may vary
ⓓ. Neither can be an actinoid because the \(7s\) subshell is occupied
Correct Answer: Both can fit the actinoid pattern because \(5f\) and \(6d\) occupancies may vary
Explanation: Both configurations use the radon core and contain electrons in the \(5f\) subshell. The general actinoid pattern permits either zero or one electron in \(6d\). Close energies of \(5f\) and \(6d\) make both distributions chemically plausible as supplied configurations. The presence of \(7s^2\) is also consistent with the general neutral-atom pattern. Actinoid classification does not require an identical \(6d\) occupancy for every member. Both configurations therefore belong within the allowed actinoid framework.
406. Radioactivity in the actinoid series is described by the statement:
ⓐ. Only actinoids with more than seven \(5f\) electrons are radioactive
ⓑ. Only synthetic actinoids are radioactive
ⓒ. Actinoids become non-radioactive after forming \(3+\) ions
ⓓ. All actinoids are radioactive
Correct Answer: All actinoids are radioactive
Explanation: Every member of the actinoid series has an unstable nucleus and is radioactive. This remains true whether the element occurs naturally or is produced synthetically. Radioactivity is a nuclear property and is not removed by ordinary ion formation. It is also not determined simply by whether the \(5f\) subshell is less than or more than half-filled. Different actinoids and isotopes can have very different half-lives. The universal feature across the series is radioactivity.
407. A student states, “All actinoids must be synthetic because all of them are radioactive.” The best correction is:
ⓐ. uranium and thorium occur naturally despite being radioactive
ⓑ. the statement is valid because no radioactive element can occur naturally
ⓒ. all actinoids are naturally abundant, and none has been produced artificially
ⓓ. uranium and thorium are naturally occurring because they are non-radioactive
Correct Answer: uranium and thorium occur naturally despite being radioactive
Explanation: Radioactive nuclei can exist naturally if their half-lives and geological history allow measurable amounts to remain. Uranium and thorium are important naturally occurring actinoids. Many heavier or later members have been produced mainly through nuclear reactions in laboratories or reactors. Both naturally occurring and synthetic actinoids are radioactive. Synthetic origin describes how an element is obtained, whereas radioactivity describes nuclear instability. The learner has incorrectly treated these two ideas as equivalent.
408. A mineral sample is found to contain naturally occurring members of the actinoid series. Which pair is most reasonably expected in such a sample?
ⓐ. Curium and californium
ⓑ. Uranium and thorium
ⓒ. Fermium and mendelevium
ⓓ. Nobelium and lawrencium
Correct Answer: Uranium and thorium
Explanation: Uranium and thorium are important naturally occurring actinoids found in minerals. Their long-lived isotopes allow appreciable quantities to persist in Earth materials. Many later actinoids are produced mainly by artificial nuclear processes. They often have shorter-lived isotopes and do not occur in ordinary minerals in comparable amounts. The question asks for the most reasonable naturally occurring pair rather than claiming that no trace of any other actinoid can ever exist. Uranium and thorium are therefore the appropriate choice.
409. An actinoid atom forms a \(3+\) ion, but its radioactive behaviour remains. This observation shows that:
ⓐ. removing three electrons makes the nucleus stable only after complex formation
ⓑ. radioactivity is nuclear and survives ordinary ion formation
ⓒ. the number of \(5f\) electrons alone determines the radioactive half-life
ⓓ. only neutral actinoid atoms can undergo radioactive decay
Correct Answer: radioactivity is nuclear and survives ordinary ion formation
Explanation: Formation of a \(3+\) ion changes the electronic configuration by removing electrons. Radioactive decay, however, originates from instability within the atomic nucleus. Ordinary oxidation, reduction, or complex formation does not change the numbers of protons and neutrons in the way required to eliminate that instability. An actinoid ion therefore remains radioactive even though its valence-electron arrangement differs from that of the neutral atom. Electron configuration affects chemical behaviour, whereas nuclear composition governs radioactivity.
410. A newly produced actinoid isotope decays so rapidly that only tiny amounts can be studied soon after its formation. Which inference is most reasonable?
ⓐ. its short half-life requires rapid study under controlled conditions
ⓑ. its short half-life implies a completely filled \(5f^{14}\) subshell
ⓒ. it should match the nearest natural actinoid in every chemical property
ⓓ. chemical bonding should greatly lengthen its nuclear half-life
Correct Answer: its short half-life requires rapid study under controlled conditions
Explanation: A short half-life means that the number of nuclei decreases rapidly with time. Such an isotope may not persist naturally in easily detectable quantities. It must often be produced through a nuclear reaction and investigated promptly. The half-life does not directly reveal the number of \(5f\) electrons. Chemical bonding changes the electron environment but does not normally eliminate nuclear radioactivity. The need for controlled preparation follows from the rapid nuclear decay described in the question.
411. A learner arranges actinoids by \(5f\)-electron count and concludes that radioactivity must increase regularly with each added \(5f\) electron. The best evaluation is:
ⓐ. valid, because electronic filling directly determines the rate of nuclear decay
ⓑ. valid, but only for actinoids that occur naturally in measurable quantities
ⓒ. invalid, because nuclear stability does not follow a simple \(5f\)-electron-count rule
ⓓ. invalid, because no member of the actinoid series contains \(5f\) electrons
Correct Answer: invalid, because nuclear stability does not follow a simple \(5f\)-electron-count rule
Explanation: The \(5f\)-electron count is an electronic property of the atom or ion. Radioactivity is governed by the composition and stability of the nucleus. Different isotopes of the same element can have very different half-lives even though their neutral electronic configurations are essentially the same. A regular electronic-filling sequence therefore cannot be used as a direct scale of radioactive decay rate. Actinoids do contain \(5f\) electrons, but those electrons are not the simple cause proposed by the learner. The conclusion confuses electronic structure with nuclear behaviour.
412. An early actinoid forms compounds in oxidation states \(+3\), \(+4\), \(+5\), and \(+6\), whereas a typical lanthanoid forms mainly the \(+3\) state. The wider range shown by the actinoid is best explained by:
ⓐ. large energy gaps between \(5f\), \(6d\), and \(7s\) prevent their joint participation
ⓑ. only \(7s\) electrons participate, fixing the oxidation state at \(+2\)
ⓒ. formation of \(5f^{14}\) is required before any higher oxidation state appears
ⓓ. similar \(5f\), \(6d\), and \(7s\) energies let several electrons bond
Correct Answer: similar \(5f\), \(6d\), and \(7s\) energies let several electrons bond
Explanation: Actinoid atoms contain valence electrons in the closely spaced \(5f\), \(6d\), and \(7s\) subshells. Different numbers of these electrons can participate in bond formation or ionisation. This makes several oxidation states accessible, particularly among the early actinoids. Lanthanoid \(4f\) electrons are more deeply buried and participate less readily in bonding. Lanthanoids consequently show a stronger predominance of the \(+3\) state. The wider actinoid oxidation-state range follows from the energetic and spatial accessibility of several subshells.
413. Which statement most accurately describes oxidation states in the actinoid series?
ⓐ. The \(+2\) state is the dominant state for every member of the series
ⓑ. The \(+3\) state is common, but higher states up to \(+7\) occur, especially among early members
ⓒ. The \(+7\) state becomes the only stable state toward the end of the series
ⓓ. The range becomes wider toward later members because their \(5f\) electrons are progressively less tightly held
Correct Answer: The \(+3\) state is common, but higher states up to \(+7\) occur, especially among early members
Explanation: The \(+3\) oxidation state occurs widely across the actinoid series. Early actinoids can also form accessible \(+4\), \(+5\), \(+6\), and \(+7\) states because \(5f\), \(6d\), and \(7s\) electrons can participate in bonding. As nuclear charge increases, the \(5f\) electrons become more tightly held. The range of stable oxidation states therefore generally narrows toward the later actinoids rather than becoming wider. Actinoid oxidation-state chemistry is variable, but it is not dominated by one universal state.
414. Examine the oxidation-state information below.
| Row | Species | Oxidation state of the actinoid |
|---|
| P | \(\mathrm{ThO_2}\) | \(+4\) |
| Q | \(\mathrm{PaF_5}\) | \(+5\) |
| R | \(\mathrm{UO_3}\) | \(+6\) |
| S | \(\mathrm{NpO_5^{3-}}\) | \(+7\) |
The consistent rows are:
ⓐ. P, Q, R and S
ⓑ. P and R only
ⓒ. Q and S only
ⓓ. P, Q and R only
Correct Answer: P, Q, R and S
Explanation: Oxygen has oxidation state \(-2\), so thorium is \(+4\) in \(\mathrm{ThO_2}\). Five fluoride ions contribute \(-5\), making protactinium \(+5\) in \(\mathrm{PaF_5}\). Three oxygen atoms contribute \(-6\), so uranium is \(+6\) in \(\mathrm{UO_3}\). For \(\mathrm{NpO_5^{3-}}\), let the oxidation state of neptunium be \(x\); then \(x+5(-2)=-3\), giving \(x=+7\). All four rows illustrate oxidation states accessible to early actinoids.
415. Assertion: Early actinoids generally show a wider range of oxidation states than later actinoids.
Reason: With increasing nuclear charge, the \(5f\) electrons become more tightly held and less readily available for bonding.
ⓐ. Assertion is true, but Reason is false
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Both Assertion and Reason are true, and Reason explains Assertion
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: In the early part of the series, several \(5f\), \(6d\), and \(7s\) electrons can participate in chemical bonding. This permits oxidation states extending from \(+3\) to relatively high values. As nuclear charge increases, the \(5f\) electrons are drawn closer to the nucleus. They become less available for removal or covalent participation. The later actinoids therefore show a stronger preference for the \(+3\) state. The Reason directly explains the trend stated in the Assertion.
416. A graph shows the number of commonly accessible oxidation states on the vertical axis and increasing atomic number across the actinoid series on the horizontal axis. The most reasonable overall pattern is:
ⓐ. a steady increase to the end of the series
ⓑ. a constant value because every actinoid behaves identically
ⓒ. many early oxidation states, narrowing among later members
ⓓ. zero oxidation states for early members and only \(+7\) for later members
Correct Answer: many early oxidation states, narrowing among later members
Explanation: Early actinoids have several valence subshells close enough in energy to participate in bonding. They consequently show numerous oxidation states. As atomic number increases, the \(5f\) electrons experience greater effective nuclear attraction. Their participation becomes less favourable, and the \(+3\) state becomes increasingly dominant. The graph should therefore begin with a relatively wide oxidation-state range and narrow later. Minor irregularities may occur, so the trend need not be perfectly smooth.
417. A valid comparison between lanthanoids and actinoids is:
ⓐ. Lanthanoids vary more because exposed \(4f\) electrons enter bonding readily
ⓑ. Actinoids vary more because \(5f\), \(6d\), and \(7s\) electrons bond more readily
ⓒ. Both series remain at \(+3\) because their \(f\) electrons never enter bonding
ⓓ. Actinoid states arise only through loss of the outer \(7s\) electrons
Correct Answer: Actinoids vary more because \(5f\), \(6d\), and \(7s\) electrons bond more readily
Explanation: Lanthanoids show a strong predominance of the \(+3\) oxidation state. Their \(4f\) electrons are relatively well shielded and participate only weakly in ordinary bonding. Actinoid \(5f\) orbitals extend farther from the nucleus and are closer in energy to \(6d\) and \(7s\). More electrons can therefore be involved in bond formation or removal. This produces a wider oxidation-state range among actinoids. Both series have exceptions, but their overall behaviours are clearly different.
418. Oxidation of uranium from \(\mathrm{U^{4+}}\) to \(\mathrm{UO_2^{2+}}\), where uranium is \(+6\), involves:
ⓐ. loss of two electrons per uranium atom
ⓑ. gain of two electrons per uranium atom
ⓒ. loss of six electrons per uranium atom
ⓓ. no electron transfer because oxygen is added
Correct Answer: loss of two electrons per uranium atom
Explanation: Uranium begins in oxidation state \(+4\) in \(\mathrm{U^{4+}}\). It has oxidation state \(+6\) in the uranyl ion. The oxidation-state increase is \(+6-(+4)=+2\). An increase in oxidation state corresponds to electron loss. Each uranium atom therefore loses two electrons during the oxidation. Addition of oxygen does not remove the need to account for the electron change.
419. A student claims, “Because \(+3\) is common among actinoids, species such as \(\mathrm{UF_6}\) and \(\mathrm{NpO_5^{3-}}\) cannot exist.” The best evaluation is:
ⓐ. correct, because actinoids can never exceed oxidation state \(+3\)
ⓑ. correct, because fluorine and oxygen have positive oxidation states in these species
ⓒ. incorrect, because only lanthanoids can form oxidation states above \(+3\)
ⓓ. incorrect, because early actinoids can form high oxidation states such as \(+6\) and \(+7\)
Correct Answer: incorrect, because early actinoids can form high oxidation states such as \(+6\) and \(+7\)
Explanation: Fluorine is \(-1\) in \(\mathrm{UF_6}\), so uranium is \(+6\). In \(\mathrm{NpO_5^{3-}}\), five oxygen atoms contribute \(-10\); the overall charge is \(-3\), so neptunium is \(+7\). These high oxidation states are accessible among early actinoids because \(5f\), \(6d\), and \(7s\) electrons can participate in bonding. The \(+3\) state is common but not exclusive. The formulas are therefore consistent with the variable oxidation-state chemistry of the actinoid series.
420. A set of radii for tripositive actinoid ions shows a gradual overall decrease as atomic number increases. The trend is identified as:
ⓐ. lanthanoid contraction caused by imperfect \(4f\) shielding
ⓑ. an oxidation-state effect because the ionic charge increases across the data set
ⓒ. actinoid contraction caused mainly by imperfect \(5f\) shielding
ⓓ. periodic expansion caused by addition of electrons to a new outer shell
Correct Answer: actinoid contraction caused mainly by imperfect \(5f\) shielding
Explanation: The ions all have the same \(+3\) charge, so the trend is not produced by a changing oxidation state. Across the actinoid series, nuclear charge increases while added \(5f\) electrons shield that increase incompletely. The effective nuclear attraction on the electron cloud therefore grows. The radii show an overall decrease known as actinoid contraction. Lanthanoid contraction is the analogous \(4f\)-series effect, not the name of the trend described here.
