201. Xenon belongs to group 18 but forms fluorides \(\mathrm{XeF_2}\), \(\mathrm{XeF_4}\) and \(\mathrm{XeF_6}\) under suitable conditions. Which statement correctly interprets their oxidation states and the implication for noble-gas valency?
ⓐ. Xenon has oxidation state zero in all three because group-18 elements have completed outer shells.
ⓑ. Xenon has oxidation states \(+2,+4,+6\), respectively; the compounds demonstrate that closed-shell group membership does not imply absolute chemical inertness.
ⓒ. Xenon has oxidation states \(-2,-4,-6\), respectively, because it belongs to the same period as strongly electropositive metals.
ⓓ. Xenon has oxidation states \(+2,+4,+6\), respectively; therefore all noble gases must form analogous stable hexafluorides.
Correct Answer: Xenon has oxidation states \(+2,+4,+6\), respectively; the compounds demonstrate that closed-shell group membership does not imply absolute chemical inertness.
Explanation: \(\textbf{Identifying the periodic classification:}\)
Xenon belongs to group 18 and has the outer configuration \(5s^2\,5p^6\).
Its valence shell is complete.
This contributes to its high ionization enthalpy and relatively low chemical reactivity under ordinary conditions.
Nevertheless, a completed valence shell does not guarantee that no compounds can form under any conditions.
\(\textbf{Examining xenon difluoride:}\)
Fluorine has oxidation state \(-1\) in its compounds.
For \(\mathrm{XeF_2}\), let xenon's oxidation state be \(x\).
Then:
\[
x+2(-1)=0
\]
\[
x=+2
\]
\(\textbf{Examining xenon tetrafluoride:}\)
For \(\mathrm{XeF_4}\):
\[
y+4(-1)=0
\]
\[
y=+4
\]
\(\textbf{Examining xenon hexafluoride:}\)
For \(\mathrm{XeF_6}\):
\[
z+6(-1)=0
\]
\[
z=+6
\]
\(\textbf{Interpreting the compounds:}\)
Xenon can exhibit several positive oxidation states in suitable fluorides.
The existence of these compounds illustrates variation in chemical reactivity within the noble-gas family.
It does not imply that neon, helium and all other noble gases must form corresponding stable fluorides.
Differences in atomic size, electron accessibility and the chemical environment remain important.
\(\textbf{Distinguishing valency from group number:}\)
The ordinary low reactivity of a noble gas is associated with its closed-shell configuration.
It should not be converted into a universal rule that every group-18 element has permanently zero combining capacity.
The correct option identifies the formal oxidation states and recognises that noble-gas inertness is not absolute.
202. Nitrogen occurs as \(\mathrm{N_2}\), containing an N≡N triple bond, while chlorine occurs as \(\mathrm{Cl_2}\), containing a Cl–Cl single bond. Both elements also form hydrogen compounds, \(\mathrm{NH_3}\) and \(\mathrm{HCl}\), respectively.
Which statement correctly distinguishes covalency from oxidation state in these four species?
ⓐ. Nitrogen has covalency 3 and oxidation state 0 in \(\mathrm{N_2}\); chlorine has covalency 1 and oxidation state 0 in \(\mathrm{Cl_2}\).
ⓑ. Nitrogen has covalency 0 in \(\mathrm{N_2}\) and 3 in \(\mathrm{NH_3}\), because oxidation state is zero only when no covalent bond exists.
ⓒ. Chlorine has covalency 0 in \(\mathrm{Cl_2}\) and 1 in \(\mathrm{HCl}\), because bonds between identical atoms are excluded from covalency.
ⓓ. Nitrogen has oxidation state \(+3\) in \(\mathrm{N_2}\) and \(-3\) in \(\mathrm{NH_3}\), whereas chlorine has oxidation state \(+1\) in \(\mathrm{Cl_2}\) and \(-1\) in \(\mathrm{HCl}\).
Correct Answer: Nitrogen has covalency 3 and oxidation state 0 in \(\mathrm{N_2}\); chlorine has covalency 1 and oxidation state 0 in \(\mathrm{Cl_2}\).
Explanation: \(\textbf{Examining the elemental molecules:}\)
Nitrogen and chlorine belong to groups 15 and 17, respectively.
An \(\mathrm{N_2}\) molecule contains three shared electron pairs between its nitrogen atoms.
Thus, each nitrogen atom has covalency three in the stated bonding description.
A \(\mathrm{Cl_2}\) molecule contains one shared electron pair, giving each chlorine atom covalency one.
However, atoms present in their free elemental forms have oxidation state zero.
Therefore, the oxidation state of nitrogen in \(\mathrm{N_2}\) and chlorine in \(\mathrm{Cl_2}\) is zero.
\(\textbf{Comparing the hydrogen compounds:}\)
In ammonia, nitrogen forms three N–H bonds and retains covalency three.
Since hydrogen is assigned oxidation state \(+1\), nitrogen has oxidation state \(-3\).
In hydrogen chloride, chlorine forms one H–Cl bond and has oxidation state \(-1\).
Its covalency remains one.
\(\textbf{Resolving the distinction:}\)
Covalency concerns the covalent bonds associated with an atom in a specified structure.
Oxidation state concerns formal assignment of bonding electrons.
A bond between identical atoms contributes to covalency but does not produce unequal oxidation-state assignments.
Therefore, nonzero covalency is entirely compatible with oxidation state zero.
The first option correctly separates these descriptors without confusing elemental identity with chemical bonding.
203. The sulfate ion, \(\mathrm{SO_4^{2-}}\), contains sulfur directly connected to four oxygen atoms. The sulfite ion, \(\mathrm{SO_3^{2-}}\), contains sulfur directly connected to three oxygen atoms. Assume oxygen has oxidation state \(-2\) in both ions.
Which statement correctly compares the formal oxidation states of sulfur and the numbers of oxygen atoms directly attached to it?
ⓐ. Sulfur has oxidation state \(+4\) in both ions, while the number of attached oxygen atoms changes from three to four.
ⓑ. Sulfur has oxidation state \(+6\) in both ions, because its group number fixes a single positive oxidation state.
ⓒ. Sulfur has oxidation states \(+6\) and \(+4\) in sulfate and sulfite, respectively, while the corresponding oxygen-neighbour counts are four and three.
ⓓ. Sulfur has oxidation states \(+4\) and \(+6\) in sulfate and sulfite, respectively, while the corresponding oxygen-neighbour counts are four and three.
Correct Answer: Sulfur has oxidation states \(+6\) and \(+4\) in sulfate and sulfite, respectively, while the corresponding oxygen-neighbour counts are four and three.
Explanation: \(\textbf{Determining the sulfate oxidation state:}\)
The oxidation-state sum in a polyatomic ion equals its net charge.
Let the sulfur oxidation state in sulfate be \(x\).
For \(\mathrm{SO_4^{2-}}\):
\[
x+4(-2)=-2
\]
\[
x-8=-2
\]
\[
x=+6
\]
\(\textbf{Determining the sulfite oxidation state:}\)
Let the sulfur oxidation state in sulfite be \(y\).
For \(\mathrm{SO_3^{2-}}\):
\[
y+3(-2)=-2
\]
\[
y-6=-2
\]
\[
y=+4
\]
\(\textbf{Interpreting the bonding information:}\)
In sulfate, sulfur is directly connected to four oxygen atoms.
In sulfite, sulfur is directly connected to three oxygen atoms.
These counts are structural coordination numbers, not oxidation states.
Both ions can be represented using suitable resonance descriptions.
The oxidation-state calculations do not require selecting one particular resonance contributor.
\(\textbf{Connecting the result with periodic behaviour:}\)
Sulfur belongs to group 16 and can exhibit several oxidation states.
Its group position does not force every compound to contain sulfur in the same formal oxidation state.
The addition of an oxygen atom while retaining the same overall ionic charge changes sulfur's oxidation state in this comparison.
Thus, sulfate contains sulfur in the \(+6\) state with four oxygen neighbours, while sulfite contains sulfur in the \(+4\) state with three.
204. Carbon dioxide and carbon tetrachloride contain carbon in its \(+4\) oxidation state. In the conventional molecular structures, carbon dioxide has two C=O double bonds, whereas carbon tetrachloride has four C–Cl single bonds.
Which statement correctly distinguishes carbon's oxidation state, covalency and number of directly attached atoms?
ⓐ. Both have carbon oxidation state \(+4\) and covalency 4, but carbon has two directly attached atoms in \(\mathrm{CO_2}\) and four in \(\mathrm{CCl_4}\).
ⓑ. Both have carbon oxidation state \(+4\), but covalency is 2 in \(\mathrm{CO_2}\) and 4 in \(\mathrm{CCl_4}\), because a double bond counts as one shared electron pair.
ⓒ. Carbon has covalency 4 in both molecules, but oxidation states \(+2\) and \(+4\) because the carbon dioxide molecule contains only two oxygen atoms.
ⓓ. Both have four directly attached atoms around carbon because each contains four shared electron pairs.
Correct Answer: Both have carbon oxidation state \(+4\) and covalency 4, but carbon has two directly attached atoms in \(\mathrm{CO_2}\) and four in \(\mathrm{CCl_4}\).
Explanation: \(\textbf{Comparing the bonding arrangements:}\)
Carbon belongs to group 14 and possesses four valence electrons.
In carbon dioxide, its conventional structure contains two C=O double bonds.
Each double bond contains two shared electron pairs.
Thus, carbon participates in four shared electron pairs and has covalency four.
However, only two oxygen atoms are directly attached to carbon.
\(\textbf{Examining carbon tetrachloride:}\)
In \(\mathrm{CCl_4}\), carbon forms four ordinary C–Cl single bonds.
These provide four shared electron pairs and connect carbon to four chlorine atoms.
Its covalency is consequently four, but its number of directly attached atoms is also four.
\(\textbf{Determining oxidation states:}\)
In carbon dioxide, oxygen has oxidation state \(-2\).
Carbon must therefore have oxidation state \(+4\) to give a neutral molecule.
In carbon tetrachloride, each chlorine has oxidation state \(-1\).
The four chlorine atoms again require carbon to have oxidation state \(+4\).
\(\textbf{Distinguishing three descriptors:}\)
Oxidation state is a formal electron-assignment quantity.
Covalency in these conventional structures counts the shared electron pairs associated with carbon.
The number of directly attached atoms is a different structural count.
A double bond changes the relationship between these latter quantities.
Therefore, equal oxidation states and equal covalencies do not require identical numbers of neighbouring atoms.
205. Tin forms the chlorides \(\mathrm{SnCl_2}\) and \(\mathrm{SnCl_4}\), while lead commonly forms \(\mathrm{PbCl_2}\) and can form \(\mathrm{PbCl_4}\). Both elements belong to group 14.
Consider the following statements.
Statement I: The oxidation state of the metal is \(+2\) in the dichlorides and \(+4\) in the tetrachlorides.
Statement II: The relative stability of the \(+2\) state increases from tin to lead.
Statement III: The \(+2\) compounds show that tin and lead have only two valence electrons in their neutral atoms.
Statement IV: Their common group membership is compatible with different relative stabilities of the same oxidation states.
Which combination is correct?
ⓐ. Statements I, II and III only
ⓑ. Statements II, III and IV only
ⓒ. Statements I, III and IV only
ⓓ. Statements I, II and IV only
Correct Answer: Statements I, II and IV only
Explanation: Statement I is correct because chlorine is assigned oxidation state \(-1\) in these compounds.
Two chlorine atoms require a metal oxidation state of \(+2\) in each dichloride.
Four chlorine atoms require oxidation state \(+4\) in each tetrachloride.
Statement II is correct because the lower \(+2\) oxidation state becomes relatively more stable on descending from tin to lead.
This is associated with the increasing significance of the inert-pair effect among heavier p-block elements.
The outer \(ns^2\) electron pair becomes relatively less available for bonding.
Statement III is incorrect because the neutral atoms of both tin and lead possess the characteristic group-14 valence configuration \(ns^2np^2\).
Each has four valence electrons.
The occurrence of the \(+2\) state does not remove two electrons from the neutral atom's underlying valence-electron count.
Rather, the lower state is associated with reduced participation of the outer s-electron pair.
Statement IV is correct because group placement expresses a recurring valence-shell pattern.
It does not require every oxidation state to have equal stability in all members of the group.
The relative stability of compounds is also influenced by bonding and the detailed electronic environment.
Thus, the observations are consistent with periodic classification.
Only Statements I, II and IV are correct.
206. An unknown third-period element \(X\) forms a covalent hydride \(\mathrm{H_2X}\) and a neutral oxide \(\mathrm{XO_3}\). In the hydride, hydrogen is assigned oxidation state \(+1\); in the oxide, oxygen is assigned \(-2\).
Which statement correctly identifies \(X\) and explains the differing oxidation states?
ⓐ. \(X=\mathrm{Si}\); its oxidation states are \(-2\) and \(+6\), consistent with group 14.
ⓑ. \(X=\mathrm{Cl}\); its oxidation states are \(-2\) and \(+6\), consistent with group 17.
ⓒ. \(X=\mathrm{P}\); its oxidation states are \(+2\) and \(+6\), consistent with group 15.
ⓓ. \(X=\mathrm{S}\); its oxidation states are \(-2\) and \(+6\), consistent with group 16.
Correct Answer: \(X=\mathrm{S}\); its oxidation states are \(-2\) and \(+6\), consistent with group 16.
Explanation: \(\textbf{Using the hydride formula:}\)
The compound \(\mathrm{H_2X}\) is electrically neutral.
Each hydrogen is assigned oxidation state \(+1\).
Let the oxidation state of \(X\) be \(a\).
Therefore:
\[
2(+1)+a=0
\]
\[
a=-2
\]
\(\textbf{Using the oxide formula:}\)
The compound \(\mathrm{XO_3}\) is also neutral.
Each oxygen atom has oxidation state \(-2\).
Let the oxidation state of \(X\) be \(b\).
Then:
\[
b+3(-2)=0
\]
\[
b=+6
\]
\(\textbf{Identifying the periodic element:}\)
Sulfur belongs to group 16 of the third period.
Its characteristic valence configuration is \(3s^2\,3p^4\).
It forms hydrogen sulfide, \(\mathrm{H_2S}\), with sulfur in oxidation state \(-2\).
It also forms sulfur trioxide, \(\mathrm{SO_3}\), with sulfur in oxidation state \(+6\).
Both formulas therefore match the supplied information.
\(\textbf{Interpreting the change:}\)
The oxidation-state difference arises from the different bonding partners.
Hydrogen is less electronegative than sulfur, while oxygen is more electronegative.
Formal electron assignment consequently gives sulfur negative and positive oxidation states in the two compounds.
The change does not alter sulfur's group membership or its neutral valence-electron configuration.
\(\textbf{Final interpretation:}\)
Combining the two formula constraints identifies sulfur more reliably than using either formula as an isolated clue.
The same element can exhibit different characteristic oxidation states while retaining one fixed position in the periodic table.
207. Magnetite has the empirical formula \(\mathrm{Fe_3O_4}\). It may be described in terms of one iron(II) centre and two iron(III) centres per formula unit. Assume oxygen has oxidation state \(-2\).
Which statement correctly combines the average oxidation state of iron with the numbers of the two formal iron oxidation states?
ⓐ. Average oxidation state \(+\frac83\); one \(\mathrm{Fe^{2+}}\) and two \(\mathrm{Fe^{3+}}\) centres
ⓑ. Average oxidation state \(+\frac83\); two \(\mathrm{Fe^{2+}}\) and one \(\mathrm{Fe^{3+}}\) centre
ⓒ. Average oxidation state \(+3\); one \(\mathrm{Fe^{2+}}\) and two \(\mathrm{Fe^{3+}}\) centres
ⓓ. Average oxidation state \(+2\); two \(\mathrm{Fe^{2+}}\) and one \(\mathrm{Fe^{3+}}\) centre
Correct Answer: Average oxidation state \(+\frac83\); one \(\mathrm{Fe^{2+}}\) and two \(\mathrm{Fe^{3+}}\) centres
Explanation: \(\textbf{Calculating the average oxidation state:}\)
The oxide contains three iron atoms and four oxygen atoms per empirical formula unit.
Let the average oxidation state of iron be \(x\).
For a neutral compound:
\[
3x+4(-2)=0
\]
\[
3x-8=0
\]
\[
x=\frac83
\]
The value \(+\frac83\) is an average oxidation state.
It does not require every iron centre to carry an individual oxidation state of \(+\frac83\).
\(\textbf{Resolving the mixed-valence composition:}\)
The question specifies iron(II) and iron(III) as the constituent formal oxidation states.
Let their numbers per formula unit be \(a\) and \(b\), respectively.
The total number of iron centres gives:
\[
a+b=3
\]
Electrical neutrality requires:
\[
2a+3b=8
\]
Substituting \(a=3-b\):
\[
2(3-b)+3b=8
\]
\[
6+b=8
\]
\[
b=2,\qquad a=1
\]
\(\textbf{Checking the result:}\)
One iron(II) contribution gives \(+2\).
Two iron(III) contributions give \(+6\).
Their total positive oxidation-state contribution is \(+8\), balancing the four oxide ions.
\(\textbf{Chemical interpretation:}\)
The empirical formula alone gives the average oxidation state.
Additional chemical information is needed to assign the individual oxidation states to particular centres.
The given mixed-valence description supplies that information.
\(\textbf{Final answer:}\)
\[
\overline{\mathrm{OS}}(\mathrm{Fe})=+\frac83;\quad n(\mathrm{Fe^{2+}})=1,\quad n(\mathrm{Fe^{3+}})=2
\]
208. The chromate ion, \(\mathrm{CrO_4^{2-}}\), and dichromate ion, \(\mathrm{Cr_2O_7^{2-}}\), are related by the following acid-dependent equilibrium:
\[
\mathrm{2CrO_4^{2-}+2H^+\rightleftharpoons Cr_2O_7^{2-}+H_2O}
\]
Assume oxygen has oxidation state \(-2\). Which conclusion correctly characterises this interconversion?
ⓐ. Chromium is oxidised from \(+4\) to \(+6\), so the forward process is a redox reaction.
ⓑ. Chromium is reduced from \(+6\) to \(+3\), while oxygen is oxidised from \(-2\) to zero.
ⓒ. Chromium remains at oxidation state \(+6\) in both ions, so the stated interconversion is not a redox reaction.
ⓓ. Chromium remains at oxidation state \(+3\) in both ions, but hydrogen ions serve as oxidising agents.
Correct Answer: Chromium remains at oxidation state \(+6\) in both ions, so the stated interconversion is not a redox reaction.
Explanation: \(\textbf{Calculating chromium's state in chromate:}\)
Let the oxidation state of chromium in \(\mathrm{CrO_4^{2-}}\) be \(x\).
The ionic charge is \(-2\).
Therefore:
\[
x+4(-2)=-2
\]
\[
x=+6
\]
\(\textbf{Calculating chromium's state in dichromate:}\)
Let each chromium atom have oxidation state \(y\).
Charge balance gives:
\[
2y+7(-2)=-2
\]
\[
2y-14=-2
\]
\[
2y=12
\]
\[
y=+6
\]
\(\textbf{Examining the complete reaction:}\)
Chromium has oxidation state \(+6\) on both sides.
Oxygen retains its assigned oxidation state of \(-2\).
Hydrogen is assigned \(+1\) in both the hydrogen ions and water.
No element undergoes an oxidation-state change.
\(\textbf{Interpreting the equilibrium:}\)
The reaction involves the acid-dependent interconversion of two oxygen-containing chromium(VI) ions.
It changes their compositions and the distribution of atoms among the species.
Nevertheless, it does not require electron transfer between oxidising and reducing agents.
Thus, a change in the formula of a chromium-containing ion does not necessarily indicate oxidation or reduction.
The deciding test is the change, if any, in assigned oxidation states.
The correct conclusion is that the interconversion is not a redox reaction.
209. The oxide \(\mathrm{CrO_3}\) and the oxide \(\mathrm{Cr_2O_3}\) contain the same transition-metal element but exhibit different acid-base characters. Under ordinary classification, \(\mathrm{CrO_3}\) is acidic, whereas \(\mathrm{Cr_2O_3}\) is amphoteric.
Which statement correctly relates these classifications to chromium's oxidation states?
ⓐ. Chromium has oxidation state \(+3\) in both oxides, so the difference must arise from a change in the number of oxygen atoms alone.
ⓑ. Chromium has oxidation states \(+6\) and \(+3\), respectively; the more highly oxidised oxide has the more pronounced acidic character.
ⓒ. Chromium has oxidation states \(+6\) and \(+3\), respectively; the more highly oxidised oxide must be more strongly basic.
ⓓ. Chromium has oxidation state \(+6\) in both oxides because the metal belongs to group 6.
Correct Answer: Chromium has oxidation states \(+6\) and \(+3\), respectively; the more highly oxidised oxide has the more pronounced acidic character.
Explanation: \(\textbf{Assigning chromium's state in the first oxide:}\)
Oxygen has oxidation state \(-2\) in \(\mathrm{CrO_3}\).
The compound is neutral.
Therefore:
\[
x+3(-2)=0
\]
\[
x=+6
\]
Chromium is present in oxidation state \(+6\).
\(\textbf{Assigning chromium's state in the second oxide:}\)
In \(\mathrm{Cr_2O_3}\), the oxidation-state sum gives:
\[
2y+3(-2)=0
\]
\[
2y=6
\]
\[
y=+3
\]
Thus, chromium has oxidation state \(+3\) in this oxide.
\(\textbf{Comparing the chemical characters:}\)
Chromium trioxide is an acidic oxide and is associated with chromium(VI) oxyacid chemistry.
Chromium(III) oxide exhibits amphoteric behaviour.
The difference is consistent with the broad tendency for certain transition-metal oxides to become more acidic as the formal oxidation state increases.
\(\textbf{Recognising the scope of the trend:}\)
This relationship is a useful qualitative tendency rather than an exception-free classification rule for every oxide.
Acid-base character also depends on bonding and the actual chemical species.
The group number of chromium does not force oxidation state \(+6\) in every compound.
The two oxides therefore demonstrate variable oxidation states accompanied by different chemical behaviour.
The second option correctly connects the supplied oxide classifications with chromium's formal electron assignments.
210. Consider hydrogen peroxide, \(\mathrm{H_2O_2}\), and the superoxide ion, \(\mathrm{O_2^-}\). The two oxygen atoms are equivalent within each species for the purpose of assigning average oxidation states.
Which statement correctly compares the oxidation state of oxygen and the O–O bonding units?
ⓐ. Oxygen has oxidation state \(-2\) in both species, and both contain the oxide ion \(\mathrm{O^{2-}}\).
ⓑ. Oxygen has oxidation state \(-1\) in hydrogen peroxide and \(-\frac12\) in superoxide; both contain an O–O bonded unit.
ⓒ. Oxygen has oxidation state \(-\frac12\) in hydrogen peroxide and \(-1\) in superoxide; only hydrogen peroxide contains an O–O bond.
ⓓ. Oxygen has oxidation state zero in hydrogen peroxide and \(-2\) in superoxide because both oxygen atoms are identical.
Correct Answer: Oxygen has oxidation state \(-1\) in hydrogen peroxide and \(-\frac12\) in superoxide; both contain an O–O bonded unit.
Explanation: \(\textbf{Examining hydrogen peroxide:}\)
Hydrogen peroxide has the conventional connectivity \(\mathrm{H-O-O-H}\).
Each hydrogen atom is assigned oxidation state \(+1\).
Let the oxidation state of each oxygen atom be \(x\).
For the neutral molecule:
\[
2(+1)+2x=0
\]
\[
x=-1
\]
Thus, oxygen has oxidation state \(-1\) in hydrogen peroxide.
\(\textbf{Examining the superoxide ion:}\)
The superoxide ion contains two oxygen atoms and has total charge \(-1\).
Let their common average oxidation state be \(y\).
Therefore:
\[
2y=-1
\]
\[
y=-\frac12
\]
\(\textbf{Connecting the formulas with oxygen bonding:}\)
Hydrogen peroxide contains an O–O single bond in its conventional structure.
The superoxide ion also contains an O–O bonded unit, although its bonding is not identical to an ordinary peroxide single bond.
The different overall charges produce different average oxidation-state assignments.
\(\textbf{Distinguishing oxidation state from bond order:}\)
The oxidation state of oxygen is not determined merely by counting how many oxygen atoms appear in the formula.
It is obtained from the total charge and the assignments for any other constituent elements.
The fractional value in superoxide is a formal average oxidation state.
It does not mean that the ion contains half an electron or half an oxygen atom.
The correct comparison is \(-1\) for peroxide oxygen and \(-\frac12\) for superoxide oxygen.
211. A transition-metal oxide has the formula \(\mathrm{M_2O_3}\), while a second oxide of the same metal has formula \(\mathrm{MO_2}\). Oxygen has oxidation state \(-2\) in both.
A student argues that the metal must move to a different group because its oxidation state changes.
Which conclusion is correct?
ⓐ. The metal changes from oxidation state \(+3\) to \(+4\), but its atomic number and periodic position remain unchanged.
ⓑ. The metal changes from oxidation state \(+2\) to \(+4\), so its group number increases by two.
ⓒ. The metal changes from oxidation state \(+3\) to \(+2\), but its group number remains unchanged.
ⓓ. The metal has oxidation state \(+4\) in both compounds because the second formula has two oxygen atoms.
Correct Answer: The metal changes from oxidation state \(+3\) to \(+4\), but its atomic number and periodic position remain unchanged.
Explanation: \(\textbf{Examining the first oxide:}\)
The formula \(\mathrm{M_2O_3}\) contains two metal atoms and three oxygen atoms.
Let the metal oxidation state be \(x\).
Charge balance gives:
\[
2x+3(-2)=0
\]
\[
2x=6
\]
\[
x=+3
\]
\(\textbf{Examining the second oxide:}\)
For \(\mathrm{MO_2}\), let the metal oxidation state be \(y\).
Then:
\[
y+2(-2)=0
\]
\[
y=+4
\]
\(\textbf{Distinguishing chemical state from elemental identity:}\)
The metal exhibits oxidation states \(+3\) and \(+4\) in the two compounds.
Different oxidation states can arise from different formal distributions of bonding electrons.
They do not require the nucleus to gain or lose protons.
The atomic number of the metal remains fixed.
Consequently, the element retains the same periodic group and period.
\(\textbf{Interpreting transition-metal behaviour:}\)
Variable oxidation states are common among several transition metals because electrons associated with their outer s- and nearby d-subshells can participate in bonding.
The oxide formulas provide evidence of the formal oxidation states but do not by themselves determine the element's atomic number.
Thus, one metal can form both oxides without changing its position in the periodic table.
The first option correctly separates oxidation-state variation from periodic classification.
212. Two unknown oxides, \(P\) and \(Q\), are examined. Oxide \(P\) is formed by a third-period element.
Oxide \(P\) reacts with excess dilute hydrochloric acid and also dissolves in sufficiently concentrated aqueous sodium hydroxide under suitable conditions.
Oxide \(Q\) does not exhibit ordinary acidic or basic oxide behaviour toward water, dilute acid or dilute alkali under the stated test conditions.
Which assignment is consistent with the observations?
ⓐ. \(P=\mathrm{MgO}\), a basic oxide; \(Q=\mathrm{SiO_2}\), a neutral oxide
ⓑ. \(P=\mathrm{Al_2O_3}\), an amphoteric oxide; \(Q=\mathrm{CO_2}\), an acidic oxide
ⓒ. \(P=\mathrm{SiO_2}\), an amphoteric oxide; \(Q=\mathrm{SO_3}\), a neutral oxide
ⓓ. \(P=\mathrm{Al_2O_3}\), an amphoteric oxide; \(Q=\mathrm{NO}\), a neutral oxide
Correct Answer: \(P=\mathrm{Al_2O_3}\), an amphoteric oxide; \(Q=\mathrm{NO}\), a neutral oxide
Explanation: \(\textbf{Interpreting the behaviour of oxide P:}\)
An amphoteric oxide reacts with suitable acids and bases.
Aluminium oxide is an important amphoteric oxide.
With hydrochloric acid, it forms aluminium chloride and water.
Under suitable alkaline conditions, it forms soluble aluminium-containing species.
Therefore, the observations for \(P\) are consistent with \(\mathrm{Al_2O_3}\).
Magnesium oxide is predominantly basic and does not satisfy the same broad amphoteric behaviour.
Silicon dioxide is classified as an acidic oxide rather than an amphoteric oxide.
\(\textbf{Interpreting the behaviour of oxide Q:}\)
Neutral oxides do not exhibit ordinary acidic or basic oxide behaviour.
Nitric oxide, \(\mathrm{NO}\), is classified as a neutral oxide.
Under the stipulated tests, it does not behave as an ordinary acid anhydride or basic oxide.
Sulfur trioxide, by contrast, is strongly acidic.
\(\textbf{Applying the period and reaction conditions:}\)
Oxide \(P\) must originate from a third-period element; aluminium satisfies this restriction.
No period restriction is imposed on the element forming oxide \(Q\).
Nitric oxide, \(\mathrm{NO}\), is a neutral oxide of the second-period element nitrogen and agrees with the stated behaviour of \(Q\).
Carbon dioxide, \(\mathrm{CO_2}\), is an acidic oxide, so the alternative pairing it with amphoteric \(\mathrm{Al_2O_3}\) fails the observation for \(Q\).
The observed amphoteric and neutral properties, together with the period requirement for \(P\), uniquely identify the fourth assignment.
213. A sample of sodium oxide is reacted completely with water, while a second sample of dichlorine heptoxide is also reacted completely with water:
\[
\mathrm{Na_2O+H_2O\rightarrow2NaOH}
\]
\[
\mathrm{Cl_2O_7+H_2O\rightarrow2HClO_4}
\]
If \(0.15\ \mathrm{mol}\) of each oxide reacts independently, what quantities of hydroxide and acid are formed, and what does their character indicate about the parent elements?
ⓐ. \(0.15\ \mathrm{mol}\) NaOH and \(0.15\ \mathrm{mol}\) HClO₄; both oxides are basic.
ⓑ. \(0.15\ \mathrm{mol}\) NaOH and \(0.30\ \mathrm{mol}\) HClO₄; sodium oxide is basic and dichlorine heptoxide is neutral.
ⓒ. \(0.30\ \mathrm{mol}\) NaOH and \(0.15\ \mathrm{mol}\) HClO₄; sodium oxide is amphoteric and dichlorine heptoxide is acidic.
ⓓ. \(0.30\ \mathrm{mol}\) NaOH and \(0.30\ \mathrm{mol}\) HClO₄; sodium oxide is basic and dichlorine heptoxide is acidic.
Correct Answer: \(0.30\ \mathrm{mol}\) NaOH and \(0.30\ \mathrm{mol}\) HClO₄; sodium oxide is basic and dichlorine heptoxide is acidic.
Explanation: \(\textbf{Identifying the first reaction ratio:}\)
Sodium is a highly electropositive group-1 metal.
Its normal oxide is basic and reacts with water to form sodium hydroxide.
The balanced equation is:
\[
\mathrm{Na_2O+H_2O\rightarrow2NaOH}
\]
One mole of sodium oxide produces two moles of sodium hydroxide.
For the given sample:
\[
n(\mathrm{NaOH})=2n(\mathrm{Na_2O})
\]
\[
n(\mathrm{NaOH})=2(0.15)
\]
\[
n(\mathrm{NaOH})=0.30\ \mathrm{mol}
\]
\(\textbf{Examining dichlorine heptoxide:}\)
Chlorine belongs to group 17 and is a non-metal.
Dichlorine heptoxide is an acidic oxide corresponding to perchloric acid.
Its water reaction is:
\[
\mathrm{Cl_2O_7+H_2O\rightarrow2HClO_4}
\]
Again, one mole of oxide produces two moles of the corresponding product.
Therefore:
\[
n(\mathrm{HClO_4})=2n(\mathrm{Cl_2O_7})
\]
\[
n(\mathrm{HClO_4})=2(0.15)
\]
\[
n(\mathrm{HClO_4})=0.30\ \mathrm{mol}
\]
\(\textbf{Interpreting the periodic chemistry:}\)
Sodium oxide produces an alkali, consistent with the metallic character of sodium.
Dichlorine heptoxide produces an oxyacid, consistent with the acidic oxide behaviour of a non-metal in a high oxidation state.
Equal quantities of products arise here from the identical stoichiometric coefficients, not from identical chemical character.
\(\textbf{Final answer:}\)
\[
\boxed{n(\mathrm{NaOH})=0.30\ \mathrm{mol},\quad n(\mathrm{HClO_4})=0.30\ \mathrm{mol}}
\]
214. The following third-period oxides are considered under suitable reaction conditions:
| Oxide | Representative chemical behaviour |
|---|
| \(\mathrm{Na_2O}\) | Reacts with water to form sodium hydroxide |
| \(\mathrm{Al_2O_3}\) | Reacts with suitable acids and strong bases |
| \(\mathrm{SiO_2}\) | Reacts with suitable strong bases but not ordinarily with water |
| \(\mathrm{SO_3}\) | Reacts with water to form sulfuric acid |
Which sequence correctly classifies these oxides and explains the broad progression?
ⓐ. Basic → Amphoteric → Acidic → Acidic; the pattern is consistent with decreasing metallic character across the representative-element region.
ⓑ. Basic → Basic → Amphoteric → Acidic; the pattern is consistent with increasing metallic character.
ⓒ. Amphoteric → Basic → Acidic → Acidic; the pattern requires identical valence-electron counts.
ⓓ. Basic → Amphoteric → Neutral → Acidic; silicon dioxide is neutral because it does not react directly with ordinary water.
Correct Answer: Basic → Amphoteric → Acidic → Acidic; the pattern is consistent with decreasing metallic character across the representative-element region.
Explanation: \(\textbf{Identifying sodium oxide:}\)
Sodium belongs to group 1 and is highly electropositive.
Its oxide reacts with water to form sodium hydroxide.
It is therefore classified as a basic oxide.
\(\textbf{Identifying aluminium oxide:}\)
Aluminium lies farther to the right in the third period.
Its oxide reacts with suitable acids and sufficiently strong bases.
This dual acid-base behaviour makes \(\mathrm{Al_2O_3}\) amphoteric.
\(\textbf{Identifying silicon dioxide:}\)
Silicon is a metalloid in group 14.
Silicon dioxide reacts with suitable basic substances to produce silicates.
It is classified as an acidic oxide.
Its limited reaction with ordinary water does not make it neutral.
An acidic oxide need not dissolve or react rapidly with water under every condition.
\(\textbf{Identifying sulfur trioxide:}\)
Sulfur is a group-16 non-metal.
Sulfur trioxide reacts with water to form sulfuric acid.
It is therefore an acidic oxide.
\(\textbf{Connecting the sequence with periodicity:}\)
Metallic character generally decreases from left to right across the representative-element portion of a period.
Correspondingly, oxide character progresses broadly from basic toward acidic.
The change is not necessarily represented by a distinct category at every successive element.
The correct sequence is basic, amphoteric, acidic and acidic.
It is consistent with the general change from strongly metallic to non-metallic character across the third period.
215. Three gaseous elements belong to the same period. Their selected properties are:
| Element | First ionization enthalpy (\(\mathrm{kJ\,mol^{-1}}\)) | First electron-gain enthalpy (\(\mathrm{kJ\,mol^{-1}}\)) | Outer configuration |
|---|
| \(P\) | 496 | \(-53\) | \(3s^1\) |
| \(Q\) | 1251 | \(-349\) | \(3s^2\,3p^5\) |
| \(R\) | 1521 | \(+96\) | \(3s^2\,3p^6\) |
Which statement correctly predicts their characteristic ordinary chemical roles?
ⓐ. \(P\) readily accepts electrons, \(Q\) is chemically inert and \(R\) readily loses electrons.
ⓑ. \(P\) is chemically inert, \(Q\) readily loses electrons and \(R\) behaves as a strongly reactive halogen.
ⓒ. \(P\) and \(Q\) both readily lose electrons, whereas \(R\) readily accepts electrons because its outer shell is complete.
ⓓ. \(P\) readily loses an electron, \(Q\) readily accepts an electron in suitable reactions and \(R\) exhibits low ordinary reactivity.
Correct Answer: \(P\) readily loses an electron, \(Q\) readily accepts an electron in suitable reactions and \(R\) exhibits low ordinary reactivity.
Explanation: \(\textbf{Identifying element P:}\)
The configuration \(3s^1\) identifies sodium.
Its relatively low first ionization enthalpy is consistent with easy removal of its outer electron.
The resulting sodium ion possesses the stable neon configuration.
Therefore, sodium characteristically behaves as an electropositive metal and reducing agent.
\(\textbf{Identifying element Q:}\)
The configuration \(3s^2\,3p^5\) identifies chlorine.
Its strongly negative first electron-gain enthalpy indicates favourable electron addition to an isolated gaseous atom.
The resulting chloride ion has a completed third principal shell.
Chlorine consequently participates readily as an oxidising agent in suitable electron-transfer reactions.
\(\textbf{Identifying element R:}\)
The configuration \(3s^2\,3p^6\) identifies argon.
Its completed valence shell contributes to a high first ionization enthalpy.
Its positive electron-gain enthalpy indicates that addition of another electron is unfavourable.
Argon therefore exhibits very low ordinary chemical reactivity.
\(\textbf{Integrating the properties:}\)
The three elements occupy different characteristic regions of the same period.
Their electron-loss and electron-gain tendencies differ because of their valence-shell configurations.
The data illustrate why ordinary chemical reactivity does not increase monotonically with atomic number.
Reactive electropositive metals occur near the left side of a period.
Reactive non-metals occur toward the right, but the noble gases provide an important closed-shell qualification.
The second option correctly identifies these contrasting chemical roles.
216. A mixture contains \(0.20\ \mathrm{mol}\) of chlorine molecules and \(0.30\ \mathrm{mol}\) of bromide ions in aqueous solution. The reaction proceeds according to:
\[
\mathrm{Cl_2+2Br^-\rightarrow2Cl^-+Br_2}
\]
Assume the reaction goes to completion and no side reactions occur. What amount of bromine is formed, and what periodic property explains why chlorine oxidises bromide under these conditions?
ⓐ. \(0.20\ \mathrm{mol}\); bromine has a greater first ionization enthalpy than chlorine.
ⓑ. \(0.30\ \mathrm{mol}\); chlorine's first electron-gain enthalpy is positive, so it cannot be reduced.
ⓒ. \(0.10\ \mathrm{mol}\); bromine is a stronger oxidising agent because it has a larger atomic radius.
ⓓ. \(0.15\ \mathrm{mol}\); chlorine is a stronger oxidising agent than bromine in the stated aqueous displacement reaction.
Correct Answer: \(0.15\ \mathrm{mol}\); chlorine is a stronger oxidising agent than bromine in the stated aqueous displacement reaction.
Explanation: \(\textbf{Identifying the reaction stoichiometry:}\)
The balanced equation shows that one mole of chlorine molecules reacts with two moles of bromide ions.
It produces one mole of bromine molecules and two moles of chloride ions.
\(\textbf{Checking the reactant requirements:}\)
The initial amount of chlorine is:
\[
n(\mathrm{Cl_2})=0.20\ \mathrm{mol}
\]
The initial amount of bromide is:
\[
n(\mathrm{Br^-})=0.30\ \mathrm{mol}
\]
To consume all the chlorine, the required bromide amount would be:
\[
n_{\mathrm{required}}(\mathrm{Br^-})=2(0.20)
\]
\[
n_{\mathrm{required}}(\mathrm{Br^-})=0.40\ \mathrm{mol}
\]
Only \(0.30\ \mathrm{mol}\) of bromide is available.
Therefore, bromide is the limiting reactant.
\(\textbf{Calculating bromine formation:}\)
Two moles of bromide produce one mole of bromine.
Thus:
\[
n(\mathrm{Br_2})=\frac{n(\mathrm{Br^-})}{2}
\]
\[
n(\mathrm{Br_2})=\frac{0.30}{2}
\]
\[
n(\mathrm{Br_2})=0.15\ \mathrm{mol}
\]
\(\textbf{Connecting the reaction with periodic behaviour:}\)
Chlorine and bromine belong to group 17.
Under the stated aqueous conditions, chlorine has the greater oxidising tendency.
It therefore accepts electrons from bromide ions, converting them into elemental bromine.
The reaction illustrates the general decrease in halogen oxidising strength down the group.
\(\textbf{Final answer:}\)
\[
\boxed{n(\mathrm{Br_2})=0.15\ \mathrm{mol}}
\]
217. Carbon monoxide and carbon dioxide are both oxides of carbon. Under ordinary oxide classification, carbon monoxide is neutral, whereas carbon dioxide is acidic.
Which statement correctly connects their acid-base characters with the oxidation states of carbon?
ⓐ. Carbon has oxidation state \(+2\) in CO and \(+4\) in CO₂; the higher oxidation state is associated with acidic oxide behaviour in this comparison.
ⓑ. Carbon has oxidation state \(+4\) in both because the element belongs to group 14; the difference arises entirely from molecular mass.
ⓒ. Carbon has oxidation state zero in CO because neutral oxides contain elements exclusively in the zero oxidation state.
ⓓ. Carbon has oxidation states \(+2\) and \(+4\), respectively, but carbon monoxide must also be acidic because it contains oxygen.
Correct Answer: Carbon has oxidation state \(+2\) in CO and \(+4\) in CO₂; the higher oxidation state is associated with acidic oxide behaviour in this comparison.
Explanation: \(\textbf{Determining carbon's state in monoxide:}\)
Oxygen ordinarily has oxidation state \(-2\).
Let the oxidation state of carbon in CO be \(x\).
For the neutral molecule:
\[
x+(-2)=0
\]
\[
x=+2
\]
\(\textbf{Determining carbon's state in dioxide:}\)
For CO₂:
\[
y+2(-2)=0
\]
\[
y=+4
\]
\(\textbf{Comparing oxide behaviour:}\)
Carbon monoxide is classified as a neutral oxide.
It does not exhibit ordinary acidic or basic oxide behaviour under the conventional classification.
Carbon dioxide is an acidic oxide.
It can react with suitable bases and forms carbonic acid in equilibrium with water.
\(\textbf{Distinguishing charge from oxide classification:}\)
Both molecules are electrically neutral.
Nevertheless, the carbon atoms possess nonzero oxidation states in both compounds.
A neutral oxide is not an oxide whose constituent atoms all have oxidation state zero.
Rather, the term refers to its ordinary acid-base behaviour.
\(\textbf{Interpreting the oxidation-state relationship:}\)
Carbon dioxide contains carbon in the higher oxidation state.
Its acidic character contrasts with the neutral character of carbon monoxide.
This is consistent with the broader tendency of many non-metal and high-oxidation-state oxides to show acidic behaviour.
However, oxidation state alone should not be treated as an infallible universal predictor of oxide acidity.
The first option correctly combines oxidation-state assignment with the experimentally established oxide classifications.
218. An unknown element \(X\) belongs to the third period. It forms an oxide \(\mathrm{X_2O_3}\) that dissolves in suitable acidic and strongly alkaline media, and a chloride \(\mathrm{XCl_3}\) in which chlorine has oxidation state \(-1\).
Which conclusion is consistent with all three observations?
ⓐ. \(X\) is magnesium; the chloride formula proves that group-2 elements commonly exhibit oxidation state \(+3\).
ⓑ. \(X\) is phosphorus; its trioxide is amphoteric because phosphorus belongs to the p-block.
ⓒ. \(X\) is aluminium; its common oxidation state is \(+3\), and its oxide is amphoteric.
ⓓ. \(X\) is sulfur; its chloride and oxide formulas establish that it belongs to group 13.
Correct Answer: \(X\) is aluminium; its common oxidation state is \(+3\), and its oxide is amphoteric.
Explanation: \(\textbf{Interpreting the oxide stoichiometry:}\)
The unknown oxide has formula \(\mathrm{X_2O_3}\).
Assigning oxygen oxidation state \(-2\), let the oxidation state of \(X\) be \(x\).
Then:
\[
2x+3(-2)=0
\]
\[
2x=6
\]
\[
x=+3
\]
\(\textbf{Checking the chloride formula:}\)
The chloride has formula \(\mathrm{XCl_3}\).
Each chlorine atom contributes oxidation state \(-1\).
Therefore:
\[
y+3(-1)=0
\]
\[
y=+3
\]
Both formulas are consistent with an element that commonly exhibits oxidation state \(+3\).
\(\textbf{Using the oxide's acid-base behaviour:}\)
The oxide dissolves in suitable acids and sufficiently strong bases.
This is characteristic amphoteric behaviour.
Aluminium oxide, \(\mathrm{Al_2O_3}\), is a well-established amphoteric oxide.
Magnesium oxide is basic, while ordinary phosphorus oxides are acidic rather than amphoteric.
\(\textbf{Applying the period constraint:}\)
Aluminium has atomic number 13 and belongs to period 3, group 13.
Its valence configuration is \(3s^2\,3p^1\).
Its common \(+3\) oxidation state and the formulas \(\mathrm{Al_2O_3}\) and \(\mathrm{AlCl_3}\) satisfy the supplied constraints.
\(\textbf{Final interpretation:}\)
The oxide formula alone determines a formal oxidation state but does not uniquely identify an element.
The chloride formula confirms the same oxidation state without resolving every possible chemical identity.
The amphoteric oxide behaviour and third-period position provide the decisive additional evidence.
Thus, the unknown element is aluminium.
219. Two separate beakers contain excess water at the same temperature. Exactly \(0.10\ \mathrm{mol}\) of sodium is added to the first beaker and \(0.10\ \mathrm{mol}\) of potassium to the second. Assume both metals react completely, all the hydrogen is collected, and no side reactions occur.
Potassium is observed to react more vigorously than sodium. Which conclusion correctly distinguishes the rate-related observation from the final amounts of hydrogen produced?
ⓐ. Potassium reacts more vigorously, but both metals ultimately produce \(0.05\ \mathrm{mol}\) of hydrogen.
ⓑ. Potassium reacts more vigorously and ultimately produces twice as much hydrogen because it has a lower first ionization enthalpy.
ⓒ. Sodium reacts more vigorously, but both metals ultimately produce \(0.10\ \mathrm{mol}\) of hydrogen.
ⓓ. Potassium reacts more vigorously and produces \(0.10\ \mathrm{mol}\) of hydrogen, whereas sodium produces \(0.05\ \mathrm{mol}\).
Correct Answer: Potassium reacts more vigorously, but both metals ultimately produce \(0.05\ \mathrm{mol}\) of hydrogen.
Explanation: \(\textbf{Identifying the periodic relationship:}\)
Sodium and potassium belong to group 1 and possess the characteristic outer electronic configuration \(ns^1\).
Each commonly loses one electron to form a monovalent cation.
Potassium lies below sodium and has a larger atomic radius.
Its outer electron experiences greater shielding and is generally easier to remove.
These electronic factors contribute to its more vigorous reaction with water under comparable conditions.
\(\textbf{Examining sodium's reaction:}\)
The balanced equation is:
\[
\mathrm{2Na+2H_2O\rightarrow2NaOH+H_2}
\]
Two moles of sodium produce one mole of hydrogen.
Therefore:
\[
n(H_2)_{\mathrm{Na}}=\frac{n(\mathrm{Na})}{2}
\]
\[
n(H_2)_{\mathrm{Na}}=\frac{0.10}{2}
\]
\[
n(H_2)_{\mathrm{Na}}=0.05\ \mathrm{mol}
\]
\(\textbf{Examining potassium's reaction:}\)
Potassium reacts according to:
\[
\mathrm{2K+2H_2O\rightarrow2KOH+H_2}
\]
The metal-to-hydrogen stoichiometric ratio remains \(2:1\).
Hence:
\[
n(H_2)_{\mathrm{K}}=\frac{0.10}{2}
\]
\[
n(H_2)_{\mathrm{K}}=0.05\ \mathrm{mol}
\]
\(\textbf{Distinguishing rate from yield:}\)
The two metals need not consume equal amounts per second merely because their balanced equations have the same stoichiometric coefficients.
Reaction vigour concerns how rapidly the reaction proceeds under the specified conditions.
Final hydrogen yield, when reaction is complete, follows from the initial amount of limiting reactant and the balanced equation.
Thus, potassium can react more vigorously without producing more hydrogen from an equal molar quantity.
\(\textbf{Final answer:}\)
\[
\boxed{n(H_2)_{\mathrm{Na}}=n(H_2)_{\mathrm{K}}=0.05\ \mathrm{mol}}
\]
220. Two hypothetical samples contain pure isotopic forms of elemental lithium, \({}^{6}\mathrm{Li}\) and \({}^{7}\mathrm{Li}\). Each sample has a mass of \(42\ \mathrm{g}\) and reacts completely with excess water according to the ordinary lithium–water reaction.
For this calculation, take the molar masses of the two isotopic forms as exactly \(6\) and \(7\ \mathrm{g\,mol^{-1}}\), respectively.
Which statement correctly compares the amounts of hydrogen produced and the periodic positions of the two lithium isotopes?
ⓐ. Both produce \(3.0\ \mathrm{mol}\) of hydrogen because identical atomic numbers guarantee identical molar amounts in equal masses.
ⓑ. The isotope of mass number 7 produces more hydrogen because it contains an additional neutron.
ⓒ. The isotope of mass number 6 produces hydrogen in the ratio \(7:6\) relative to the isotope of mass number 7, although both occupy the same periodic position.
ⓓ. The isotope of mass number 6 produces hydrogen in the ratio \(6:7\) relative to the isotope of mass number 7, because its atoms are lighter.
Correct Answer: The isotope of mass number 6 produces hydrogen in the ratio \(7:6\) relative to the isotope of mass number 7, although both occupy the same periodic position.
Explanation: \(\textbf{Identifying the chemical similarity:}\)
Both isotopes contain three protons and therefore have atomic number 3.
Their neutral atoms possess the same ordinary electronic configuration, \(1s^2\,2s^1\).
Both consequently occupy group 1 and period 2.
The additional neutron in \({}^{7}\mathrm{Li}\) changes its mass but not its ordinary valence-electron count.
Both isotopes follow the same reaction stoichiometry with water.
\(\textbf{Writing the governing reaction:}\)
\[
\mathrm{2Li+2H_2O\rightarrow2LiOH+H_2}
\]
Two moles of lithium atoms produce one mole of hydrogen molecules.
\(\textbf{Calculating hydrogen from lithium-6:}\)
The initial amount is:
\[
n({}^{6}\mathrm{Li})=\frac{m}{M}
\]
\[
n({}^{6}\mathrm{Li})=\frac{42\ \mathrm{g}}{6\ \mathrm{g\,mol^{-1}}}
\]
\[
n({}^{6}\mathrm{Li})=7\ \mathrm{mol}
\]
Therefore:
\[
n(H_2)_6=\frac{7}{2}=3.5\ \mathrm{mol}
\]
\(\textbf{Calculating hydrogen from lithium-7:}\)
\[
n({}^{7}\mathrm{Li})=\frac{42\ \mathrm{g}}{7\ \mathrm{g\,mol^{-1}}}
\]
\[
n({}^{7}\mathrm{Li})=6\ \mathrm{mol}
\]
Hence:
\[
n(H_2)_7=\frac{6}{2}=3.0\ \mathrm{mol}
\]
\(\textbf{Comparing the yields:}\)
\[
\frac{n(H_2)_6}{n(H_2)_7}=\frac{3.5}{3.0}
\]
\[
\frac{n(H_2)_6}{n(H_2)_7}=\frac76
\]
\(\textbf{Interpreting the result:}\)
Equal masses of different isotopes need not contain equal numbers of atoms.
The lighter isotope supplies more moles of lithium in the same sample mass.
It therefore produces more hydrogen even though the ordinary chemical reaction and periodic classification are unchanged.
The difference arises from molar amount, not from a change in group, valency or reaction stoichiometry.
\(\textbf{Final answer:}\)
\[
\boxed{n(H_2)_6:n(H_2)_7=7:6}
\]
