501. Carbon and silicon both belong to group 14 and form the covalent tetrafluorides \(\mathrm{CF_4}\) and \(\mathrm{SiF_4}\).
Under suitable conditions, silicon tetrafluoride accepts additional fluoride ions:
\[
\mathrm{SiF_4+2F^-\rightarrow[SiF_6]^{2-}}
\]
An analogous ordinary stable hexafluorocarbon(IV) ion is not formed by simple fluoride addition to CF₄ under corresponding conditions.
Consider the following statements.
Statement I: Silicon has oxidation state \(+4\) in both SiF₄ and \(\mathrm{[SiF_6]^{2-}}\).
Statement II: The number of directly bonded fluorine neighbours around silicon increases from four to six.
Statement III: Silicon must change from group 14 to group 16 when its coordination number increases.
Statement IV: The greater size and different bonding possibilities of third-period silicon help accommodate a higher coordination number than is ordinarily available to second-period carbon.
Which combination is correct?
ⓐ. Statements I, II and III only
ⓑ. Statements I, III and IV only
ⓒ. Statements I, II and IV only
ⓓ. Statements II, III and IV only
Correct Answer: Statements I, II and IV only
Explanation: Statement I is correct because fluorine is assigned oxidation state \(-1\) in the specified compounds.
In neutral SiF₄, the four fluorine atoms contribute total oxidation state \(-4\).
Silicon therefore has oxidation state \(+4\).
In \(\mathrm{[SiF_6]^{2-}}\), the six fluorine atoms contribute \(-6\).
Silicon must again contribute \(+4\) to give the ion its total charge of \(-2\).
Statement II is correct because the initial tetrafluoride contains four silicon–fluorine connections.
The hexafluorosilicate ion contains six directly attached fluorine atoms.
Its silicon coordination number is consequently six.
Statement III is incorrect because coordination number does not determine an element's periodic group.
Silicon retains atomic number 14 and remains in group 14 in both species.
Statement IV is correct because carbon belongs to period 2 and is considerably smaller than silicon.
Ordinary second-period carbon bonding is constrained by its compact valence-shell electron arrangement.
Third-period silicon can accommodate different higher-coordination bonding environments under suitable conditions.
The specified fluoride-addition reaction illustrates this difference between corresponding group members.
The reaction does not require oxidation of silicon from \(+4\) to \(+6\).
Only Statements I, II and IV are correct.
502. Four isolated gaseous species are arranged into two isoelectronic pairs:
Pair I: \(\mathrm{O^-(g)}\) and \(\mathrm{F(g)}\).
Pair II: \(\mathrm{O^{2-}(g)}\) and \(\mathrm{F^-(g)}\).
The transformations connecting the pairs are:
\[
\mathrm{O^-(g)+e^-\rightarrow O^{2-}(g)},\qquad\Delta H=+744\ \mathrm{kJ\,mol^{-1}}
\]
\[
\mathrm{F(g)+e^-\rightarrow F^-(g)},\qquad\Delta H=-328\ \mathrm{kJ\,mol^{-1}}
\]
Assume the ordinary isolated-species electronic-size convention applies.
Which statement correctly combines the electron counts, ionic-size comparisons and electron-addition energetics?
ⓐ. Pair I contains nine electrons per species and Pair II contains ten; oxygen species are larger within both pairs because oxygen has the smaller nuclear charge, but only fluorine's stated electron addition is exothermic.
ⓑ. Both pairs contain ten electrons per species; fluorine species are larger because fluorine has the greater nuclear charge.
ⓒ. Pair I contains nine electrons per species and Pair II contains ten; the oxygen species are smaller in both pairs because oxygen has the less positive nucleus.
ⓓ. Both electron-addition reactions are endothermic because they involve species that already have at least seven valence electrons.
Correct Answer: Pair I contains nine electrons per species and Pair II contains ten; oxygen species are larger within both pairs because oxygen has the smaller nuclear charge, but only fluorine's stated electron addition is exothermic.
Explanation: \(\textbf{Counting electrons in Pair I:}\)
Oxygen has atomic number 8.
The singly negative oxygen ion contains nine electrons.
Neutral fluorine has atomic number 9 and also contains nine electrons.
The two species are therefore isoelectronic.
\(\textbf{Comparing their electronic extents:}\)
Oxygen has eight protons.
Fluorine has nine.
For the common nine-electron population, fluorine exerts stronger nuclear attraction.
Therefore:
\[
r(\mathrm{O^-})\gt r(\mathrm F)
\]
\(\textbf{Counting electrons in Pair II:}\)
The oxide ion contains ten electrons.
Fluoride also contains ten electrons.
These two ions share the neon electronic configuration.
Their nuclei still contain eight and nine protons, respectively.
Thus:
\[
r(\mathrm{O^{2-}})\gt r(\mathrm{F^-})
\]
\(\textbf{Interpreting the different electron-addition stages:}\)
The oxygen transformation begins with an already negatively charged ion.
The incoming electron experiences substantial electrostatic repulsion.
The supplied positive enthalpy means the process is endothermic.
The fluorine transformation begins with a neutral atom.
Addition of an electron completes its outer p-subshell.
The supplied negative enthalpy means energy is released.
\(\textbf{Integrating size and energy:}\)
The two pairs demonstrate the influence of nuclear charge at constant electron count.
The electron-addition reactions demonstrate the influence of initial charge and final orbital occupancy.
An isoelectronic size comparison does not determine the sign of an electron-gain enthalpy.
The first option correctly distinguishes these connected but different electronic effects.
503. Carbon forms methanol, methanal, methanoic acid and carbon dioxide.
For this question, use the conventional structures:
Methanol, CH₃OH: carbon has three C–H single bonds and one C–O single bond.
Methanal, HCHO: carbon has two C–H single bonds and one C=O double bond.
Methanoic acid, HCOOH: carbon has one C–H single bond, one C=O double bond and one C–O single bond; the latter oxygen also carries an O–H bond.
Carbon dioxide, CO₂: carbon has two C=O double bonds.
For oxidation-state calculations, take H as \(+1\) and O as \(-2\). Count one single bond as bond order 1 and one double bond as bond order 2.
Which statement correctly compares carbon's oxidation states and its total bond-order sum?
ⓐ. Carbon has oxidation states \(-4,-2,0,+2\), respectively, and its total bond-order sum decreases from four to two.
ⓑ. Carbon has oxidation states \(-2,0,+2,+4\), respectively, while its total bond-order sum remains four in every specified structure.
ⓒ. Carbon has oxidation state \(+4\) in all four compounds because it belongs to group 14 and has four valence electrons.
ⓓ. Carbon has oxidation states \(-2,0,+2,+4\), respectively, and the number of directly attached atoms remains four in every structure.
Correct Answer: Carbon has oxidation states \(-2,0,+2,+4\), respectively, while its total bond-order sum remains four in every specified structure.
Explanation: \(\textbf{Identifying the common periodic feature:}\)
Carbon belongs to group 14 and has four valence electrons.
The specified structures use the ordinary tetravalent bonding arrangements of carbon.
However, four valence electrons do not force carbon to have oxidation state \(+4\) in every compound.
The formal oxidation state also depends on the bonded atoms.
\(\textbf{Calculating carbon's state in methanol:}\)
For neutral CH₃OH, there are four hydrogen atoms and one oxygen atom.
\[
x+4(+1)+(-2)=0
\]
\[
x=-2
\]
\(\textbf{Calculating carbon's state in methanal:}\)
For HCHO:
\[
y+2(+1)+(-2)=0
\]
\[
y=0
\]
\(\textbf{Calculating carbon's state in methanoic acid:}\)
For HCOOH:
\[
z+2(+1)+2(-2)=0
\]
\[
z=+2
\]
\(\textbf{Calculating carbon's state in carbon dioxide:}\)
For CO₂:
\[
w+2(-2)=0
\]
\[
w=+4
\]
\(\textbf{Checking the carbon bond-order sums:}\)
In methanol, carbon has four single bonds.
In methanal, it has two single bonds and one double bond.
In methanoic acid, it has two single bonds and one double bond.
In carbon dioxide, it has two double bonds.
The carbon bond-order sum is therefore four in every case.
\(\textbf{Distinguishing bonds from neighbouring atoms:}\)
The number of directly attached atoms does not remain constant.
Carbon is attached to four atoms in methanol, three in methanal, three in methanoic acid and two in carbon dioxide.
A double bond involves one neighbouring atom but contributes two to the bond-order sum.
\(\textbf{Chemical interpretation:}\)
The oxidation state changes progressively from \(-2\) to \(+4\).
The total carbon bond-order sum remains four.
This demonstrates why oxidation state, coordination number and conventional bonding capacity must not be treated as interchangeable.
The second option correctly integrates all three descriptors.
504. Two isolated gaseous metal atoms, one sodium and one potassium, each donate one electron to two isolated gaseous halogen atoms, one fluorine and one chlorine.
The following data are supplied:
| Atomic process | Enthalpy change (\(\mathrm{kJ\,mol^{-1}}\)) |
|---|
| \(\mathrm{Na(g)\rightarrow Na^+(g)+e^-}\) | \(+496\) |
| \(\mathrm{K(g)\rightarrow K^+(g)+e^-}\) | \(+419\) |
| \(\mathrm{F(g)+e^-\rightarrow F^-(g)}\) | \(-328\) |
| \(\mathrm{Cl(g)+e^-\rightarrow Cl^-(g)}\) | \(-349\) |
Two alternative pairings are considered.
Path I: Sodium transfers its electron to fluorine, while potassium transfers its electron to chlorine.
Path II: Sodium transfers its electron to chlorine, while potassium transfers its electron to fluorine.
In each path, one mole of each metal atom and one mole of each halogen atom participate. No ionic lattice, hydration or molecular bond formation is included.
Which statement correctly compares the two total enthalpy changes?
ⓐ. Path I requires \(21\ \mathrm{kJ\,mol^{-1}}\) more energy because fluorine has the less negative electron-gain enthalpy.
ⓑ. Path II releases \(77\ \mathrm{kJ\,mol^{-1}}\) more energy because potassium has the lower first ionization enthalpy.
ⓒ. Both paths require \(+238\ \mathrm{kJ}\) for the specified quantities, although their individual electron-transfer enthalpies differ.
ⓓ. Both paths have total enthalpy changes of \(-238\ \mathrm{kJ}\) because two halogen atoms acquire completed p-subshells.
Correct Answer: Both paths require \(+238\ \mathrm{kJ}\) for the specified quantities, although their individual electron-transfer enthalpies differ.
Explanation: \(\textbf{Identifying the invariant processes:}\)
Each path begins with the same four types of isolated gaseous neutral atoms.
Each ends with the same four ionic species:
\[
\mathrm{Na^+,\ K^+,\ F^-,\ Cl^-}
\]
The electron donors and acceptors are unchanged.
Only the description of which donor supplies which acceptor differs.
Enthalpy is a state function, so the total must depend on the initial and final states rather than the chosen pairing description.
\(\textbf{Calculating Path I — Sodium to fluorine:}\)
\[
\Delta H_{\mathrm{Na\rightarrow F}}=496-328
\]
\[
\Delta H_{\mathrm{Na\rightarrow F}}=+168\ \mathrm{kJ}
\]
\(\textbf{Path I — Potassium to chlorine:}\)
\[
\Delta H_{\mathrm{K\rightarrow Cl}}=419-349
\]
\[
\Delta H_{\mathrm{K\rightarrow Cl}}=+70\ \mathrm{kJ}
\]
Therefore:
\[
\Delta H_{\mathrm I}=168+70
\]
\[
\Delta H_{\mathrm I}=+238\ \mathrm{kJ}
\]
\(\textbf{Calculating Path II — Sodium to chlorine:}\)
\[
\Delta H_{\mathrm{Na\rightarrow Cl}}=496-349
\]
\[
\Delta H_{\mathrm{Na\rightarrow Cl}}=+147\ \mathrm{kJ}
\]
\(\textbf{Path II — Potassium to fluorine:}\)
\[
\Delta H_{\mathrm{K\rightarrow F}}=419-328
\]
\[
\Delta H_{\mathrm{K\rightarrow F}}=+91\ \mathrm{kJ}
\]
Therefore:
\[
\Delta H_{\mathrm{II}}=147+91
\]
\[
\Delta H_{\mathrm{II}}=+238\ \mathrm{kJ}
\]
\(\textbf{Explaining the invariant total:}\)
Both calculations contain the same two ionization contributions and the same two electron-gain contributions.
Their sum is:
\[
496+419-328-349=+238\ \mathrm{kJ}
\]
The individual pairings have different enthalpy changes.
However, these differences cancel when both electron transfers are included.
The positive result also shows that the stipulated isolated gas-phase ion formation is endothermic.
Formation of stable ionic solids would introduce additional thermochemical contributions absent from this comparison.
\(\textbf{Final answer:}\)
\[
\boxed{\Delta H_{\mathrm I}=\Delta H_{\mathrm{II}}=+238\ \mathrm{kJ};\qquad\Delta H_{\mathrm I}-\Delta H_{\mathrm{II}}=0}
\]
505. Chlorine and bromine are consecutive members of group 17, with chlorine occupying period 3 and bromine occupying period 4. Bromine forms the bromate ion, \(\mathrm{BrO_3^-}\), in which oxygen is assigned oxidation state \(-2\). Consider the formal oxidation state of bromine in this ion, the relative atomic radii of neutral chlorine and bromine atoms, and their first electron-gain enthalpies. Which combination is consistent with periodic trends?
ⓐ. Bromine has oxidation state \(+5\); bromine has the smaller atomic radius; chlorine has the more negative first electron-gain enthalpy.
ⓑ. Bromine has oxidation state \(+6\); bromine has the larger atomic radius; chlorine has the more negative first electron-gain enthalpy.
ⓒ. Bromine has oxidation state \(+5\); bromine has the larger atomic radius; bromine has the more negative first electron-gain enthalpy.
ⓓ. Bromine has oxidation state \(+5\); bromine has the larger atomic radius; chlorine has the more negative first electron-gain enthalpy.
Correct Answer: Bromine has oxidation state \(+5\); bromine has the larger atomic radius; chlorine has the more negative first electron-gain enthalpy.
Explanation: \(\textbf{Determining bromine's formal oxidation state:}\)
Bromine belongs to group 17 and has seven electrons in its outermost shell.
Group membership does not require bromine to exhibit oxidation state \(-1\) in every compound or ion.
The bromate ion, \(\mathrm{BrO_3^-}\), carries an overall charge of \(-1\).
Oxygen is assigned oxidation state \(-2\) in this ion.
Let the oxidation state of bromine be \(x\).
The sum of oxidation states must equal the ionic charge:
\[
x+3(-2)=-1
\]
\[
x-6=-1
\]
\[
x=+5
\]
Therefore, bromine has formal oxidation state \(+5\) in bromate.
This positive oxidation state is a formal electron-accounting assignment and does not imply the existence of an isolated \(\mathrm{Br^{5+}}\) ion.
\(\textbf{Comparing atomic radii:}\)
Chlorine has the outer electronic configuration \(3s^2\,3p^5\).
Bromine has the outer electronic configuration \(4s^2\,4p^5\).
Moving from chlorine to bromine adds a higher principal electron shell.
The increased shell occupancy and shielding outweigh the increased nuclear charge in determining the general down-group size trend.
Consequently, bromine has a larger atomic radius than chlorine.
\(\textbf{Comparing first electron-gain enthalpies:}\)
Both chlorine and bromine are halogens and have a strong tendency to acquire an electron to complete their outer p-subshells.
The relevant gaseous electron-addition processes are:
\[
\mathrm{Cl(g)+e^-\rightarrow Cl^-(g)}
\]
\[
\mathrm{Br(g)+e^-\rightarrow Br^-(g)}
\]
Both first electron-gain enthalpies are negative because energy is released in these electron-addition processes.
Chlorine is smaller than bromine, so the incoming electron is added to a region experiencing comparatively stronger effective nuclear attraction.
The first electron-gain enthalpy of chlorine is therefore more negative than that of bromine.
This comparison is consistent with the general decrease in the tendency to gain electrons while moving down group 17 from chlorine to bromine.
\(\textbf{Distinguishing the two chemical contexts:}\)
The \(+5\) oxidation state concerns bromine within an oxygen-containing polyatomic ion.
First electron-gain enthalpy concerns electron addition to an isolated neutral gaseous atom.
These quantities describe different processes and must not be equated.
A positive formal oxidation state in bromate does not prevent a neutral bromine atom from acquiring an electron exothermically.
\(\textbf{Final inference:}\)
Bromine has oxidation state \(+5\) in \(\mathrm{BrO_3^-}\), its neutral atom is larger than chlorine, and chlorine has the more negative first electron-gain enthalpy.
The combined result follows from charge balance, electronic configuration, shielding and the down-group trend in electron-gain enthalpy.
506. Chlorine and bromine belong to group 17 and have neutral valence configurations \(ns^2np^5\). They form the oxygen-containing ions \(\mathrm{ClO_3^-}\) and \(\mathrm{BrO_4^-}\). Assume oxygen has oxidation state \(-2\) in both ions.
Which statement correctly assigns the central-element oxidation states and interprets their relation to periodic classification?
ⓐ. Chlorine is \(+5\) and bromine is \(+7\); the two group-17 elements may exhibit different positive oxidation states in their oxyanions.
ⓑ. Chlorine is \(+6\) and bromine is \(+8\); each central element has an oxidation state equal to twice the oxygen count.
ⓒ. Chlorine is \(+5\) and bromine is \(+5\); shared group membership forces equal oxidation states.
ⓓ. Chlorine is \(-1\) and bromine is \(-1\); the ion charge directly gives each central atom's oxidation state.
Correct Answer: Chlorine is \(+5\) and bromine is \(+7\); the two group-17 elements may exhibit different positive oxidation states in their oxyanions.
Explanation: Both central atoms belong to the halogen family and ordinarily have seven outer-shell electrons.
Group membership explains this shared valence configuration, not a single fixed oxidation state in all compounds.
The sum of formal oxidation states in a monovalent anion must equal \(-1\).
For chlorate, write:
\[
x+3(-2)=-1
\]
\[
x=+5
\]
For perbromate, the corresponding relation is:
\[
y+4(-2)=-1
\]
\[
y=+7
\]
The extra oxygen in the bromine oxyanion changes the formal charge balance.
The positive values do not represent free \(\mathrm{Cl^{5+}}\) or \(\mathrm{Br^{7+}}\) ions.
They are formal oxidation states assigned within oxygen-containing covalent species.
Thus elements in the same group can show a related range of oxidation states without having identical values in every compound.
507. Xenon has the neutral ground-state configuration:
\[
[\mathrm{Kr}]\,4d^{10}\,5s^2\,5p^6
\]
Although xenon belongs to group 18, it forms fluorides such as \(\mathrm{XeF_2}\), \(\mathrm{XeF_4}\) and \(\mathrm{XeF_6}\) under suitable conditions.
Consider the following statements.
Statement I: Xenon has oxidation states \(+2,+4,+6\) in the three specified fluorides, respectively.
Statement II: Xenon's periodic classification changes from group 18 to group 16 when it forms XeF₆.
Statement III: The existence of xenon fluorides shows that a completed neutral-atom valence shell does not make compound formation impossible under every condition.
Statement IV: The conventional number of Xe–F bonds differs among the three compounds, while xenon's atomic number remains unchanged.
Which combination is correct?
ⓐ. Statements I, II and III only
ⓑ. Statements II, III and IV only
ⓒ. Statements I, II and IV only
ⓓ. Statements I, III and IV only
Correct Answer: Statements I, III and IV only
Explanation: Statement I is correct because fluorine is assigned oxidation state \(-1\) in its compounds.
In XeF₂, two fluorine atoms contribute \(-2\), giving xenon oxidation state \(+2\).
In XeF₄, xenon has oxidation state \(+4\).
In XeF₆, xenon has oxidation state \(+6\).
Statement II is incorrect because an element's periodic group is determined by its atomic number and characteristic electronic configuration.
Xenon retains atomic number 54 in all its compounds.
Formation of a higher oxidation state does not move an element to a different periodic group.
Statement III is correct because xenon possesses a completed \(5s^2\,5p^6\) valence shell in its neutral ground state.
This contributes to its low ordinary reactivity.
However, xenon can form stable compounds with highly electronegative fluorine under suitable conditions.
Its noble-gas classification therefore does not imply absolute chemical inertness.
Statement IV is correct because the specified molecular formulas contain two, four and six Xe–F bonds in their conventional bonding descriptions.
The corresponding xenon oxidation states also differ.
Nevertheless, covalency, oxidation state and atomic number remain distinct descriptors.
The number of protons in xenon is unchanged by chemical bonding.
Only Statements I, III and IV are correct.
508. Magnesium chloride and aluminium chloride contain the same monovalent anion but different cations.
For a simplified polarisation model, define the indicator:
\[
P=\frac{|z|}{r^2}
\]
where \(z\) is the cation charge number and \(r\) is its effective radius under one comparable ionic-radius convention.
| Cation | Charge number | Radius (pm) |
|---|
| \(\mathrm{Mg^{2+}}\) | 2 | 72 |
| \(\mathrm{Al^{3+}}\) | 3 | 54 |
Which statement correctly compares the indicators and their qualitative significance for chloride bonding?
ⓐ. Aluminium(III) has \(1.50\) times the indicator of magnesium(II) because only the charge difference matters.
ⓑ. Magnesium(II) has \(2.67\) times the indicator of aluminium(III) because the larger ionic radius increases polarisation.
ⓒ. Aluminium(III) has approximately \(2.67\) times the indicator of magnesium(II); its greater charge and smaller radius both favour stronger polarisation of chloride.
ⓓ. The indicators are equal because magnesium and aluminium occupy the same period and both form chlorides.
Correct Answer: Aluminium(III) has approximately \(2.67\) times the indicator of magnesium(II); its greater charge and smaller radius both favour stronger polarisation of chloride.
Explanation: \(\textbf{Identifying the competing cation properties:}\)
Magnesium and aluminium are neighbouring third-period metals.
Magnesium commonly forms a divalent cation.
Aluminium commonly forms a trivalent cation.
The supplied aluminium ion has both the greater positive charge and the smaller radius.
Both differences favour a stronger polarising influence in the specified model.
\(\textbf{Writing the ratio:}\)
\[
R=\frac{P_{\mathrm{Al^{3+}}}}{P_{\mathrm{Mg^{2+}}}}
\]
Using the given definition:
\[
R=\frac{3/54^2}{2/72^2}
\]
Rearranging:
\[
R=\frac32\left(\frac{72}{54}\right)^2
\]
\[
R=\frac32\left(\frac43\right)^2
\]
\[
R=\frac32\times\frac{16}{9}
\]
\[
R=\frac83
\]
\[
R\approx2.67
\]
\(\textbf{Interpreting chloride polarisation:}\)
A positively charged cation can distort the electron cloud of a neighbouring anion.
A greater positive charge increases the electrostatic influence.
A smaller effective radius concentrates that charge within a smaller region.
Aluminium(III) is consequently predicted to polarise chloride more strongly.
Greater polarisation favours an increased covalent contribution to bonding.
\(\textbf{Connecting the result with periodicity:}\)
The characteristic cation charges reflect the group positions of magnesium and aluminium.
The ionic-radius difference is also consistent with their different nuclear charges and electron-removal states.
However, the calculated ratio is not a quantitative ratio of actual covalent percentages in MgCl₂ and AlCl₃.
Actual bonding depends on the complete electronic and structural environment.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{P_{\mathrm{Al^{3+}}}}{P_{\mathrm{Mg^{2+}}}}=\frac83\approx2.67}
\]
509. Nitrogen and phosphorus are group-15 elements. Consider the compounds \(\mathrm{NO_2}\), \(\mathrm{N_2O_5}\), \(\mathrm{PCl_3}\) and \(\mathrm{H_3PO_4}\).
Assign oxygen oxidation state \(-2\), hydrogen \(+1\) in the acid, and chlorine \(-1\) in phosphorus trichloride. Which sequence gives the formal oxidation states of N, N, P and P in the stated order?
ⓐ. \(+2,+5,+3,+3\)
ⓑ. \(+4,+5,+3,+5\)
ⓒ. \(+4,+10,+3,+5\)
ⓓ. \(+4,+5,-3,+5\)
Correct Answer: \(+4,+5,+3,+5\)
Explanation: The characteristic outer-shell pattern of nitrogen and phosphorus is \(ns^2np^3\).
This group relationship allows several positive oxidation states depending on the atoms bonded to the central element.
For a neutral compound, the sum of formal oxidation states must be zero.
In \(\mathrm{NO_2}\), the two oxygens contribute \(-4\), so nitrogen is \(+4\).
In \(\mathrm{N_2O_5}\), five oxygens contribute \(-10\).
Two nitrogen atoms must therefore contribute \(+10\), giving \(+5\) per nitrogen.
In \(\mathrm{PCl_3}\), three chlorine atoms contribute \(-3\), giving phosphorus \(+3\).
For \(\mathrm{H_3PO_4}\), the three hydrogens contribute \(+3\) and the four oxygens contribute \(-8\).
Therefore the phosphorus oxidation state \(x\) satisfies \(3+x-8=0\), giving \(x=+5\).
The coefficient of an element in a molecular formula must not be confused with its oxidation state.
The oxidation-state assignments depend on charge bookkeeping, not merely on the elements being in group 15.
The second sequence satisfies all four independent balances.
510. Magnesium forms the halides \(\mathrm{MgF_2}\) and \(\mathrm{MgI_2}\).
Assume comparable solid-state bonding conditions. Fluoride is much smaller and less readily polarised than iodide, while the magnesium cation has the same charge in both compounds.
Consider the following statements.
Statement I: MgI₂ is expected to have a greater covalent contribution than MgF₂ because the larger iodide electron cloud is more readily polarised.
Statement II: The greater covalent contribution expected in MgI₂ necessarily means that its Mg–I internuclear distance is shorter than the Mg–F distance.
Statement III: Magnesium retains oxidation state \(+2\) in both compounds, although their degrees of ionic and covalent character may differ.
Statement IV: Relative anion polarizability can affect bonding character without changing the periodic group or characteristic charge of the metal.
Which combination is correct?
ⓐ. Statements I, II and III only
ⓑ. Statements II, III and IV only
ⓒ. Statements I, II and IV only
ⓓ. Statements I, III and IV only
Correct Answer: Statements I, III and IV only
Explanation: Statement I is correct because fluoride and iodide belong to group 17.
Iodide has a larger and more diffuse electronic distribution.
Its electron cloud is therefore more readily distorted by a nearby positive ion.
Under the stated conditions, magnesium(II) can induce a greater covalent contribution in MgI₂ than in MgF₂.
Statement II is incorrect because covalent character and internuclear distance are not identical quantities.
The iodide ion is substantially larger than fluoride.
Its greater size generally favours a larger Mg–I separation under comparable structural conditions.
Increased polarisation does not require the bond to become shorter than one involving a much smaller anion.
Statement III is correct because magnesium belongs to group 2.
Its characteristic cation is \(\mathrm{Mg^{2+}}\).
Fluoride and iodide each have formal charge \(-1\) in the specified compounds.
Charge balance therefore gives MgF₂ and MgI₂ with magnesium in oxidation state \(+2\).
Statement IV is correct because polarizability depends on the electronic extent and distribution of the anion.
Changing fluoride to iodide changes the bonding environment without changing magnesium's atomic number or group.
The comparison illustrates why ionic radii, formal charges, polarisation and covalent contribution must be considered together.
Only Statements I, III and IV are correct.
511. Carbon belongs to group 14 and exhibits both negative and positive oxidation states in compounds. Consider \(\mathrm{CH_4}\), \(\mathrm{CO}\) and \(\mathrm{CO_2}\), with hydrogen assigned \(+1\) and oxygen \(-2\).
Carbon monoxide is conventionally classified as a neutral oxide, whereas carbon dioxide is an acidic oxide. Which statement correctly combines oxidation-state calculation and oxide classification?
ⓐ. Carbon has oxidation states \(-4,+2,+4\), respectively; the two carbon oxides differ both in formal carbon oxidation state and ordinary oxide character.
ⓑ. Carbon has oxidation states \(+4,-2,-4\), respectively; an acidic oxide must contain carbon in a negative oxidation state.
ⓒ. Carbon has oxidation state \(+4\) in all three compounds because it belongs to group 14.
ⓓ. Carbon has oxidation states \(-4,+4,+2\), respectively; carbon monoxide is acidic and carbon dioxide neutral.
Correct Answer: Carbon has oxidation states \(-4,+2,+4\), respectively; the two carbon oxides differ both in formal carbon oxidation state and ordinary oxide character.
Explanation: Carbon's characteristic outer configuration is \(ns^2np^2\), with four valence electrons.
The presence of four valence electrons does not force carbon to have oxidation state \(+4\) in every compound.
In methane, four hydrogen atoms are assigned \(+1\), so the carbon oxidation state is \(-4\).
In carbon monoxide, the oxygen atom is assigned \(-2\), giving carbon oxidation state \(+2\).
In carbon dioxide, the two oxygen atoms together contribute \(-4\), giving carbon oxidation state \(+4\).
Carbon monoxide is a neutral oxide under the conventional oxide classification.
Carbon dioxide is an acidic oxide and can react with bases under suitable conditions.
The oxidation-state difference is one aspect of their distinct chemical behaviour, but it is not by itself a universal formula for predicting oxide character.
Both compounds are oxides of the same group-14 element, so periodic position alone does not make their oxide behaviour identical.
The central atom's bonding environment and the supplied classifications must also be considered.
Thus the correct option links formal electron bookkeeping with the appropriate qualitative reactivity distinction.
512. The following metallic radii and first ionization enthalpies are supplied for three consecutive group-1 elements:
| Element | Metallic radius (pm) | First ionization enthalpy (\(\mathrm{kJ\,mol^{-1}}\)) |
|---|
| K | 231 | 419 |
| Rb | 248 | 403 |
| Cs | 265 | 376 |
A student argues that equal radius increases between consecutive members require equal decreases in first ionization enthalpy. Which conclusion follows?
ⓐ. The claim is correct; each ionization-enthalpy decrease is exactly \(17\ \mathrm{kJ\,mol^{-1}}\).
ⓑ. The claim is incorrect; both radius increments are \(17\ \mathrm{pm}\), but the ionization-enthalpy decreases are \(16\) and \(27\ \mathrm{kJ\,mol^{-1}}\).
ⓒ. The claim is correct; the first ionization enthalpy must be directly proportional to metallic radius within group 1.
ⓓ. The claim is incorrect because atomic radius decreases as additional shells are occupied down group 1.
Correct Answer: The claim is incorrect; both radius increments are \(17\ \mathrm{pm}\), but the ionization-enthalpy decreases are \(16\) and \(27\ \mathrm{kJ\,mol^{-1}}\).
Explanation: \(\textbf{Read the two periodic trends separately:}\)
The supplied radii increase from potassium to rubidium to caesium.
The first ionization enthalpies decrease along the same sequence.
Additional principal shells and shielding help explain these general directions.
\(\textbf{Calculate the consecutive radius changes:}\)
\[
\Delta r_{\mathrm{K\rightarrow Rb}}=248-231=17\ \mathrm{pm}
\]
\[
\Delta r_{\mathrm{Rb\rightarrow Cs}}=265-248=17\ \mathrm{pm}
\]
\(\textbf{Calculate the corresponding enthalpy changes:}\)
\[
\Delta IE_{\mathrm{K\rightarrow Rb}}=403-419=-16\ \mathrm{kJ\,mol^{-1}}
\]
\[
\Delta IE_{\mathrm{Rb\rightarrow Cs}}=376-403=-27\ \mathrm{kJ\,mol^{-1}}
\]
The ionization-enthalpy changes are not equal, despite the equal reported radius increments.
Radius and first ionization enthalpy respond to several interacting factors, including shielding and effective nuclear attraction.
A general inverse trend does not establish a simple numerical proportionality.
\(\textbf{Final answer:}\)
\[
\boxed{\Delta r=(17,17)\ \mathrm{pm};\quad \Delta IE=(-16,-27)\ \mathrm{kJ\,mol^{-1}}}
\]
513. A text-described graph shows the first ionization enthalpies of four third-period elements in increasing atomic-number order.
The horizontal axis lists Mg, Al, P and S. The vertical axis shows first ionization enthalpy in \(\mathrm{kJ\,mol^{-1}}\). The plotted values are Mg: 738, Al: 578, P: 1012 and S: 1000. The points are joined only to guide the eye, not to imply continuous variation between different elements.
Two local decreases appear even though effective nuclear attraction generally increases across the period. Which explanation accounts for both decreases?
ⓐ. Mg to Al reflects loss from a more weakly bound \(3p\) electron, while P to S reflects pairing within the \(3p\) subshell.
ⓑ. Mg to Al reflects removal from an inner shell, while P to S reflects a decrease in nuclear charge.
ⓒ. Both decreases arise because Al and S possess one fewer occupied principal shell than their predecessors.
ⓓ. Both decreases arise because the third period is ordered by atomic mass rather than atomic number.
Correct Answer: Mg to Al reflects loss from a more weakly bound \(3p\) electron, while P to S reflects pairing within the \(3p\) subshell.
Explanation: \(\textbf{Establish the configurations:}\)
Magnesium has outer configuration \(3s^2\), whereas aluminium has \(3s^2\,3p^1\).
Phosphorus has outer configuration \(3s^2\,3p^3\), whereas sulfur has \(3s^2\,3p^4\).
All four elements occupy the third principal period.
\(\textbf{Interpret the first local decrease:}\)
The outer \(3p\) electron in aluminium is removed more readily than a \(3s\) electron in magnesium under the relevant atomic conditions.
The graph accordingly falls from \(738\) to \(578\ \mathrm{kJ\,mol^{-1}}\).
This does not imply that aluminium has a smaller nuclear charge.
\(\textbf{Interpret the second local decrease:}\)
The three \(3p\) electrons of phosphorus occupy different p-orbitals with parallel spins.
Sulfur's fourth \(3p\) electron must pair with one already present in an orbital.
The associated electron-electron repulsion makes electron removal from sulfur comparatively easier.
Thus the graph falls slightly from \(1012\) to \(1000\ \mathrm{kJ\,mol^{-1}}\).
Subshell energy and electron pairing explain the two deviations from the broad left-to-right increase.
The graph does not establish a continuously decreasing or mass-based periodic trend.
514. Carbon and silicon belong to group 14 and form the oxides CO₂ and SiO₂, respectively.
Observation I: Carbon dioxide consists of discrete molecular units and reacts with water to establish an acidic aqueous system.
Observation II: Silicon dioxide consists of an extended covalent network and does not readily react with water under ordinary conditions.
Observation III: Under suitable heating conditions, silicon dioxide reacts with strong alkali to form silicate.
Both oxides contain the central element in formal oxidation state \(+4\).
A student concludes that SiO₂ must be a neutral oxide because it does not behave like CO₂ when mixed with water.
Which statement correctly evaluates the conclusion?
ⓐ. The conclusion is correct because only discrete molecular oxides can exhibit acidic behaviour.
ⓑ. The conclusion is incorrect because SiO₂ is amphoteric, while CO₂ is basic, despite both having central elements at \(+4\).
ⓒ. The conclusion is correct because silicon has fewer valence electrons than carbon and therefore cannot form an acidic oxide.
ⓓ. Both oxides are acidic under suitable conditions, despite their differences in structure and reactions with water.
Correct Answer: Both oxides are acidic under suitable conditions, despite their differences in structure and reactions with water.
Explanation: \(\textbf{Identifying the group similarity:}\)
Carbon and silicon belong to group 14.
Their neutral atoms have the characteristic valence configuration \(ns^2np^2\).
Both can form oxides in which the central element has oxidation state \(+4\).
Nevertheless, their structures and physical properties need not be identical.
\(\textbf{Examining carbon dioxide:}\)
CO₂ consists of discrete covalent molecules.
Each carbon atom is bonded to two oxygen atoms in the ordinary molecular structure.
Carbon dioxide reacts with water to establish an acidic solution through the corresponding carbonic-acid system.
It is therefore classified as an acidic oxide.
\(\textbf{Examining silicon dioxide:}\)
Silicon dioxide forms an extended three-dimensional covalent network.
Each silicon atom is connected through oxygen atoms within the network.
This structure contributes to its high thermal stability and resistance to rapid reaction with water.
The absence of ready water reaction does not establish neutral oxide character.
\(\textbf{Applying the supplied alkali test:}\)
Silicon dioxide reacts with strong alkali under suitable conditions.
Formation of a silicate is characteristic of its acidic oxide behaviour.
Therefore, SiO₂ is not classified as neutral merely because it does not produce a simple acidic solution upon ordinary mixing with water.
\(\textbf{Interpreting the periodic relationship:}\)
Carbon's compact second-period orbitals permit strong multiple bonding in discrete CO₂ molecules.
Silicon more commonly forms extended Si–O bonding networks in its dioxide.
The difference in bonding arrangements influences their physical states and reaction behaviour.
\(\textbf{Final interpretation:}\)
Oxidation state alone does not specify molecular structure.
Direct reaction with water alone does not determine oxide classification.
The fourth option correctly distinguishes structural differences from the shared acidic character of the two oxides.
515. Oxygen, sulfur and selenium occupy successive positions down group 16. The following approximate properties refer to their isolated neutral atoms:
| Element | Outer configuration | First electron-gain enthalpy (\(\mathrm{kJ\,mol^{-1}}\)) |
|---|
| O | \(2s^2\,2p^4\) | \(-141\) |
| S | \(3s^2\,3p^4\) | \(-200\) |
| Se | \(4s^2\,4p^4\) | \(-195\) |
Which conclusion explains why the electron-gain trend is not simply more exothermic for the smallest atom?
ⓐ. Oxygen has the most exothermic first electron gain because its radius is smallest.
ⓑ. Sulfur's roomier \(3p\) subshell reduces incoming-electron repulsion relative to oxygen; the data then show a small decline in exothermicity from sulfur to selenium.
ⓒ. Selenium has the most exothermic first electron gain because it possesses the greatest number of occupied shells.
ⓓ. All three first electron-gain enthalpies must be equal because they have the same characteristic valence configuration.
Correct Answer: Sulfur's roomier \(3p\) subshell reduces incoming-electron repulsion relative to oxygen; the data then show a small decline in exothermicity from sulfur to selenium.
Explanation: All three elements have characteristic outer configuration \(ns^2np^4\) and belong to group 16.
First electron-gain enthalpy measures the signed enthalpy change for adding an electron to a neutral gaseous atom.
Each of the supplied values is negative, so the corresponding first electron addition is exothermic.
Sulfur's value of \(-200\ \mathrm{kJ\,mol^{-1}}\) is more negative than oxygen's \(-141\ \mathrm{kJ\,mol^{-1}}\).
The additional electron enters the very compact \(2p\) region of oxygen, where electron-electron repulsion is relatively severe.
The larger \(3p\) region of sulfur moderates that crowding effect.
This accounts for a recognised deviation from the simple expectation that the smallest atom must release the most energy on electron addition.
Selenium has an even larger valence shell, but greater distance and shielding also affect the incoming electron.
Its listed value of \(-195\) is slightly less negative than sulfur's \(-200\ \mathrm{kJ\,mol^{-1}}\).
The trend therefore reflects competing effects of effective nuclear attraction, electron repulsion, shell size and shielding.
The numerical data support sulfur as the most exothermic of the three cases supplied.
The result cannot be obtained reliably from atomic radius or group membership alone.
516. Sodium and magnesium are neighbouring period-3 metals with ordinary oxidation states \(+1\) and \(+2\), respectively. Under suitable conditions they form the normal oxides \(\mathrm{Na_2O}\) and \(\mathrm{MgO}\).
In separate idealised preparations, exactly \(0.200\ \mathrm{mol}\) of each metal is converted completely into its normal oxide using oxygen gas. No peroxide, superoxide or other product forms. Which statement correctly compares the required amounts of oxygen gas?
ⓐ. Sodium requires \(0.100\ \mathrm{mol}\) and magnesium \(0.050\ \mathrm{mol}\) of \(\mathrm{O_2}\).
ⓑ. Sodium requires \(0.100\ \mathrm{mol}\) and magnesium \(0.100\ \mathrm{mol}\) of \(\mathrm{O_2}\).
ⓒ. Sodium requires \(0.050\ \mathrm{mol}\) and magnesium \(0.100\ \mathrm{mol}\) of \(\mathrm{O_2}\).
ⓓ. Sodium requires \(0.200\ \mathrm{mol}\) and magnesium \(0.400\ \mathrm{mol}\) of \(\mathrm{O_2}\).
Correct Answer: Sodium requires \(0.050\ \mathrm{mol}\) and magnesium \(0.100\ \mathrm{mol}\) of \(\mathrm{O_2}\).
Explanation: \(\textbf{Determine the characteristic oxide formulas:}\)
Sodium belongs to group 1 and normally gives an oxidation state of \(+1\).
Two \(\mathrm{Na^+}\)-type contributions balance one oxide ion of charge \(-2\), producing \(\mathrm{Na_2O}\).
Magnesium belongs to group 2 and normally gives an oxidation state of \(+2\).
A one-to-one ratio with oxide is therefore consistent with \(\mathrm{MgO}\).
\(\textbf{Write the balanced formation relations:}\)
\[
4\mathrm{Na}+\mathrm{O_2}\rightarrow2\mathrm{Na_2O}
\]
\[
2\mathrm{Mg}+\mathrm{O_2}\rightarrow2\mathrm{MgO}
\]
\(\textbf{Calculate oxygen use in each independent process:}\)
\[
n(\mathrm{O_2})_{\mathrm{Na}}=0.200\times\frac14=0.050\ \mathrm{mol}
\]
\[
n(\mathrm{O_2})_{\mathrm{Mg}}=0.200\times\frac12=0.100\ \mathrm{mol}
\]
The oxygen molecules supply two oxygen atoms each, which is why the required oxygen amount differs from the amount of oxide formula units.
The assumption excludes sodium peroxide and other oxygen-rich products; without it, the stoichiometry might change.
The result links group position, usual oxidation state, oxide formula and balanced composition.
\(\textbf{Final answer:}\)
\[
\boxed{n(\mathrm{O_2})_{\mathrm{Na}}=0.050\ \mathrm{mol},\quad n(\mathrm{O_2})_{\mathrm{Mg}}=0.100\ \mathrm{mol}}
\]
517. Three neighbouring third-period elements X, Y and Z have characteristic valence configurations:
| Element | Valence configuration | Representative oxide |
|---|
| X | \(3s^2\,3p^1\) | \(\mathrm{X_2O_3}\) |
| Y | \(3s^2\,3p^2\) | \(\mathrm{YO_2}\) |
| Z | \(3s^2\,3p^3\) | \(\mathrm{Z_2O_5}\) |
A student argues that the oxide of X must be strongly basic simply because X is metallic, and that the oxides of Y and Z must have the same chemical behaviour because their central atoms both belong to the p-block. Which correction is justified?
ⓐ. X is Al with amphoteric oxide; Y is Si and Z is P with acidic oxides, whose reactions with water need not be identical.
ⓑ. X is magnesium with a basic oxide; Y is aluminium and Z is silicon, both with amphoteric oxides.
ⓒ. X is aluminium with an acidic oxide; Y and Z are silicon and phosphorus, both having predominantly basic oxides.
ⓓ. X is silicon with a neutral oxide; Y is phosphorus and Z is sulfur, both with amphoteric oxides.
Correct Answer: X is Al with amphoteric oxide; Y is Si and Z is P with acidic oxides, whose reactions with water need not be identical.
Explanation: \(\textbf{Use the electronic configurations:}\)
The configuration \(3s^2\,3p^1\) identifies aluminium in group 13.
The configuration \(3s^2\,3p^2\) identifies silicon in group 14.
The configuration \(3s^2\,3p^3\) identifies phosphorus in group 15.
They occur consecutively in the third period.
\(\textbf{Check oxidation states and oxide formulas:}\)
For \(\mathrm{X_2O_3}\), the aluminium oxidation state is \(+3\).
For \(\mathrm{YO_2}\), silicon is in oxidation state \(+4\).
For \(\mathrm{Z_2O_5}\), phosphorus is in oxidation state \(+5\).
\(\textbf{Evaluate oxide character:}\)
Aluminium oxide is amphoteric and can show both acidic and basic reactions under appropriate conditions.
Silicon dioxide is an acidic oxide, although it does not ordinarily react readily with water to form a simple acid.
Phosphorus(V) oxide is also classified as acidic.
The broad progression is consistent with the reduction in metallic character across the period.
Being metallic does not make every metal oxide strongly basic, and shared block membership does not guarantee identical reaction details.
Thus the first option correctly integrates configuration, oxidation state and oxide character.
518. Nitrogen and phosphorus belong to group 15. Under ordinary conditions, elemental nitrogen commonly exists as N₂, whereas one important form of elemental phosphorus contains P₄ molecules.
Molecule I, N₂: two nitrogen atoms are joined by a strong nitrogen–nitrogen triple bond.
Molecule II, P₄: four phosphorus atoms occupy the vertices of a tetrahedron. Each phosphorus atom is directly bonded to the other three phosphorus atoms by ordinary P–P single bonds.
Consider the following statements.
Statement I: Nitrogen and phosphorus have oxidation state zero in their respective elemental molecules.
Statement II: Each phosphorus atom in the specified P₄ molecule has three directly bonded phosphorus neighbours.
Statement III: Because P₄ contains more atoms per molecule than N₂, the P₄ molecule must be less reactive under every chemical condition.
Statement IV: The strong N≡N bond and the strained P–P bonding arrangement in tetrahedral P₄ help explain important differences in their chemical reactivities.
Which combination is correct?
ⓐ. Statements I, II and III only
ⓑ. Statements I, III and IV only
ⓒ. Statements I, II and IV only
ⓓ. Statements II, III and IV only
Correct Answer: Statements I, II and IV only
Explanation: Statement I is correct because both N₂ and P₄ are elemental forms.
The formal oxidation state of an element in its uncombined elemental form is zero.
This remains true even when different numbers of atoms occur in the elemental molecule.
Statement II is correct because the supplied P₄ structure is tetrahedral.
Every vertex is connected to the other three vertices.
Each phosphorus atom therefore has three directly bonded phosphorus neighbours.
The molecule contains six P–P edges in total.
Statement III is incorrect because molecular atom count does not by itself determine chemical reactivity.
A larger elemental molecule is not necessarily less reactive.
Bond strengths, molecular geometry and the energies of accessible products are also important.
Statement IV is correct because molecular nitrogen contains a particularly strong triple bond.
Breaking or substantially rearranging this bond often requires considerable energy.
Tetrahedral P₄ contains comparatively weaker P–P single bonds.
Its small bond angles also create substantial structural strain.
These features contribute to the different chemical behaviour of this phosphorus allotrope.
\(\textbf{Periodic interpretation:}\)
Nitrogen and phosphorus have the same number of neutral-atom valence electrons.
Nevertheless, their different atomic sizes and orbital-overlap characteristics favour different elemental bonding arrangements.
The example shows why common group membership does not require identical molecular structures or reactivities.
Only Statements I, II and IV are correct.
519. Aluminium and gallium occupy group 13 in periods 3 and 4, respectively. The following approximate data use consistent atomic-radius and first-ionization conventions:
| Element | Valence configuration | Atomic radius (pm) | \(IE_1\) (\(\mathrm{kJ\,mol^{-1}}\)) |
|---|
| Al | \(3s^2\,3p^1\) | 143 | 578 |
| Ga | \(4s^2\,4p^1\) | 135 | 579 |
A student predicts that gallium must have a much larger radius and substantially lower ionization enthalpy solely because it lies one period below aluminium. Which explanation best addresses the supplied observations?
ⓐ. Gallium's filled \(3d^{10}\) electrons provide almost perfect shielding, so its fourth-shell radius should greatly exceed aluminium's.
ⓑ. Gallium's smaller measured radius proves that its \(4p^1\) electron is actually in the same principal shell as aluminium's \(3p^1\) electron.
ⓒ. Gallium's intervening \(3d^{10}\) electrons shield imperfectly, helping its greater effective nuclear attraction counter the expected increase in radius down the group.
ⓓ. The similar ionization enthalpies prove that the increased nuclear charge and fourth-shell occupation have no opposing effects.
Correct Answer: Gallium's intervening \(3d^{10}\) electrons shield imperfectly, helping its greater effective nuclear attraction counter the expected increase in radius down the group.
Explanation: Aluminium and gallium both have three characteristic valence electrons of the pattern \(ns^2np^1\).
They therefore belong to group 13 in the p-block.
Gallium's electron configuration includes a filled inner \(3d^{10}\) subshell absent from aluminium.
Although gallium occupies the higher outer principal shell \(n=4\), these inner d-electrons do not shield nuclear charge as efficiently as a simple shell-counting rule might suggest.
The outer electrons can consequently experience a relatively strong effective nuclear attraction.
The supplied radius values show gallium to be slightly smaller rather than substantially larger.
The first ionization enthalpies are also nearly equal in the supplied data.
These observations qualify, rather than overturn, the broad increase of atomic radius and decrease of ionization enthalpy down many main-group families.
An empirical periodic trend is not a strict numerical law applying without exceptions.
Gallium remains in the p-block because its characteristic differentiating electron enters a p-subshell.
The presence of inner d-electrons does not reclassify it as a d-block element.
The third option is therefore consistent with both periodic placement and shielding effects.
520. Chromium and manganese occupy groups 6 and 7 in the fourth-period d-block. Consider their representative high-oxidation-state oxides \(\mathrm{CrO_3}\) and \(\mathrm{Mn_2O_7}\), assigning \(-2\) to oxygen.
A student reasons that the two oxides must have the same central-element oxidation state because their parent elements are neighbours. Which statement corrects this error?
ⓐ. Chromium is \(+3\) and manganese is \(+7\); the central oxidation state is numerically equal to the oxygen subscript.
ⓑ. Chromium is \(+6\) and manganese is \(+7\); the formulas are consistent with different high positive oxidation states in neighbouring d-block elements.
ⓒ. Chromium is \(+6\) and manganese is \(+14\); each manganese atom carries half the total oxygen charge only if it is monatomic.
ⓓ. Chromium is \(+3\) and manganese is \(+2\); d-block elements cannot reach positive oxidation states higher than three.
Correct Answer: Chromium is \(+6\) and manganese is \(+7\); the formulas are consistent with different high positive oxidation states in neighbouring d-block elements.
Explanation: The two elements lie in adjacent groups of the d-block and can exhibit variable oxidation states.
The oxidation-state assignment must be made for the specific compound formula.
In neutral \(\mathrm{CrO_3}\), three oxide contributions total \(-6\).
The chromium centre must therefore contribute \(+6\).
In neutral \(\mathrm{Mn_2O_7}\), seven oxygen atoms contribute \(-14\).
The two manganese atoms together must contribute \(+14\).
Each manganese atom therefore has a formal oxidation state of \(+7\).
These values are consistent with characteristic high oxidation states found in appropriate oxides of chromium and manganese.
The two neutral atoms have different groups and different available valence-electron arrangements.
Neighbouring positions in a period do not imply identical positive oxidation states.
The total oxidation-state sum and number of atoms must both be respected.
High positive oxidation states are not the same as simple isolated gaseous ions with charges \(+6\) and \(+7\).
The second option correctly combines formula charge balance with periodic variation.
