101. The following first ionization enthalpies are supplied for two pairs of neighbouring Period 2 and Period 3 elements.
| Element | Outermost configuration | First ionization enthalpy \((\mathrm{kJ\,mol^{-1}})\) |
|---|
| N | \(2s^22p^3\) | 1402 |
| O | \(2s^22p^4\) | 1314 |
| P | \(3s^23p^3\) | 1012 |
| S | \(3s^23p^4\) | 1000 |
Although oxygen has a higher atomic number than nitrogen and sulphur has a higher atomic number than phosphorus, each Group 16 element has a slightly lower first ionization enthalpy than the preceding Group 15 element. Which explanation correctly accounts for both comparisons?
ⓐ. Group 16 elements have one fewer valence electron than Group 15 elements, making electron removal easier.
ⓑ. Paired-electron repulsion in \(np^4\) can ease ionization relative to the half-filled \(np^3\) arrangement.
ⓒ. Group 15 elements possess completely filled \(p\) subshells, whereas Group 16 elements possess empty \(p\) subshells.
ⓓ. Electron removal from Group 16 requires disruption of an inner \(s\) subshell, whereas Group 15 loses only outermost \(p\) electrons.
Correct Answer: Paired-electron repulsion in \(np^4\) can ease ionization relative to the half-filled \(np^3\) arrangement.
Explanation: \(\textbf{Identify the expected periodic trend:}\)
First ionization enthalpy generally increases from left to right across a period because effective nuclear charge tends to increase.
Consequently, one might initially expect oxygen to have a higher first ionization enthalpy than nitrogen.
The same simple argument would suggest that sulphur should exceed phosphorus.
However, the supplied data show the reverse order for both pairs.
\(\textbf{Examine the Group 15 configurations:}\)
Nitrogen has the outermost configuration \(2s^22p^3\).
Phosphorus has the corresponding configuration \(3s^23p^3\).
A \(p\) subshell contains three degenerate orbitals.
According to Hund's rule, the three \(p\) electrons occupy these orbitals singly with parallel spins before pairing occurs.
This produces a relatively stable half-filled \(p\) subshell.
\(\textbf{Examine the Group 16 configurations:}\)
Oxygen has \(2p^4\), while sulphur has \(3p^4\).
In each case, one of the three \(p\) orbitals contains a pair of electrons.
The two electrons sharing the same orbital experience appreciable electron–electron repulsion.
Removal of one electron from this paired arrangement reduces that repulsion.
The resulting \(np^3\) arrangement is also relatively stable.
Consequently, the energy required to remove the first electron can be slightly lower than that required for the preceding Group 15 element.
\(\textbf{Interpret the exception correctly:}\)
The general increase in nuclear attraction across a period remains important.
However, subshell occupancy and electron-pair repulsion modify the expected trend.
The observed reversal is therefore not evidence that oxygen or sulphur has fewer valence electrons.
Group 15 elements possess five valence electrons, whereas Group 16 elements possess six.
The correct explanation combines the stability of a half-filled \(p\) subshell with the repulsion experienced by paired \(p\) electrons.
Thus, both N/O and P/S comparisons illustrate the same electronic-origin exception to the general ionization-enthalpy trend.
102. Nitrogen occurs in both \(\mathrm{NH_4^+}\) and \(\mathrm{NO_3^-}\). In the usual bonding descriptions, nitrogen is directly attached to four hydrogen atoms in ammonium and three oxygen atoms in nitrate. Which statement correctly compares nitrogen's formal oxidation states and coordination numbers?
ⓐ. Oxidation states are +4 and +3; coordination numbers are four and three, respectively.
ⓑ. Oxidation states are -3 and +3; coordination numbers are four and three, respectively.
ⓒ. Oxidation states are +3 and +5; coordination numbers are three and four, respectively.
ⓓ. Oxidation states are -3 and +5; coordination numbers are four and three, respectively.
Correct Answer: Oxidation states are -3 and +5; coordination numbers are four and three, respectively.
Explanation: \(\textbf{Determine the oxidation state in ammonium:}\)
Nitrogen is more electronegative than hydrogen.
Hydrogen is therefore assigned oxidation state +1 in the ammonium ion.
Let nitrogen's oxidation state be \(x\).
Since the total ionic charge is +1:
\[
x+4(+1)=+1
\]
\[
x+4=1
\]
\[
x=-3
\]
Thus, nitrogen has oxidation state -3 in \(\mathrm{NH_4^+}\).
\(\textbf{Determine the oxidation state in nitrate:}\)
Oxygen is assigned oxidation state -2.
For the nitrate ion:
\[
y+3(-2)=-1
\]
\[
y-6=-1
\]
\[
y=+5
\]
Nitrogen therefore has oxidation state +5 in \(\mathrm{NO_3^-}\).
\(\textbf{Determine the coordination numbers:}\)
Coordination number in this molecular-connectivity comparison refers to the number of atoms directly bonded to the central atom.
In ammonium, nitrogen is directly bonded to four hydrogen atoms.
Its coordination number is therefore four.
In nitrate, nitrogen is directly bonded to three oxygen atoms.
Its coordination number is three.
The resonance description of nitrate does not change the number of directly attached oxygen atoms.
\(\textbf{Resolve the apparent contradiction:}\)
Ammonium has the larger coordination number but the lower nitrogen oxidation state.
Nitrate has the smaller coordination number but the higher nitrogen oxidation state.
This is possible because oxidation state depends on formal electron assignment, whereas coordination number depends on molecular connectivity.
The positive charge on ammonium also does not imply that nitrogen itself must have a positive oxidation state.
Likewise, the negative charge on nitrate does not require a negative oxidation state for nitrogen.
The correct comparison is therefore nitrogen -3 and +5, with coordination numbers four and three, respectively.
103. Hydrogen peroxide undergoes decomposition under suitable conditions according to:
\[
\mathrm{2H_2O_2\longrightarrow 2H_2O+O_2}
\]
Which statement correctly classifies the reaction and identifies the oxidation-state changes of oxygen?
ⓐ. It is disproportionation; oxygen changes from -1 to both -2 and 0.
ⓑ. It is comproportionation; oxygen changes from -2 and 0 to -1.
ⓒ. It is oxidation only; every oxygen atom changes from -1 to 0.
ⓓ. It is reduction only; every oxygen atom changes from -1 to -2.
Correct Answer: It is disproportionation; oxygen changes from -1 to both -2 and 0.
Explanation: \(\textbf{Determine the initial oxidation state:}\)
Hydrogen peroxide contains an O–O peroxide linkage.
Hydrogen is assigned oxidation state +1.
For the neutral molecule:
\[
2(+1)+2x=0
\]
\[
2x=-2
\]
\[
x=-1
\]
Therefore, oxygen has oxidation state -1 in hydrogen peroxide.
\(\textbf{Determine the product oxidation states:}\)
In water, hydrogen has oxidation state +1.
The oxidation state of oxygen is:
\[
2(+1)+y=0
\]
\[
y=-2
\]
In elemental molecular oxygen, \(\mathrm{O_2}\), the oxidation state of oxygen is zero.
\(\textbf{Identify both electron-transfer directions:}\)
Some oxygen atoms change from -1 in hydrogen peroxide to -2 in water.
This decrease represents reduction.
Each such oxygen atom formally gains one electron.
Other oxygen atoms change from -1 to zero in molecular oxygen.
This increase represents oxidation.
Each such oxygen atom formally loses one electron.
\(\textbf{Verify electron conservation:}\)
The balanced reaction contains four oxygen atoms on each side.
Two oxygen atoms finish in water and collectively gain two electrons.
The other two finish in molecular oxygen and collectively lose two electrons.
Thus:
\[
\text{Electrons gained}=2
\]
\[
\text{Electrons lost}=2
\]
The electron balance is satisfied.
\(\textbf{Classify the reaction:}\)
Disproportionation occurs when the same element in one initial oxidation state undergoes both oxidation and reduction.
Oxygen initially exists only in oxidation state -1 in the reactant.
It produces oxygen in two different oxidation states, -2 and zero.
Therefore, hydrogen peroxide undergoes disproportionation.
This demonstrates that an intermediate oxidation state can permit a species to behave simultaneously as an oxidizing and reducing agent in an overall reaction.
The classification follows from both oxidation-state directions, not merely from the formation of two products.
104. The first electron-gain enthalpy of oxygen is negative, while the second electron-gain enthalpy for the process \(\mathrm{O^-(g)+e^-\rightarrow O^{2-}(g)}\) is strongly positive. Nevertheless, stable ionic oxides containing oxide ions can form with sufficiently electropositive metals. Which explanation correctly reconciles these facts?
ⓐ. The second electron-gain enthalpy of oxygen becomes negative whenever the incoming electron enters the \(2p\) subshell.
ⓑ. Stable oxides form because isolated gaseous \(\mathrm{O^{2-}}\) ions are necessarily more stable than isolated gaseous oxygen atoms.
ⓒ. Adding an electron to \(\mathrm{O^-}\) is unfavourable, but lattice stabilization can favour formation of oxide salts.
ⓓ. The existence of stable ionic oxides proves that oxidation state -2 is impossible for oxygen in covalent compounds.
Correct Answer: Adding an electron to \(\mathrm{O^-}\) is unfavourable, but lattice stabilization can favour formation of oxide salts.
Explanation: \(\textbf{Examine the first electron attachment:}\)
A neutral oxygen atom has the ground-state configuration \(1s^22s^22p^4\).
It can accept one electron into its incompletely filled \(2p\) subshell.
The first electron-gain process is:
\[
\mathrm{O(g)+e^-\rightarrow O^-(g)}
\]
This process has a negative electron-gain enthalpy, indicating release of energy.
The resulting \(\mathrm{O^-}\) ion has the configuration \(1s^22s^22p^5\).
\(\textbf{Examine the second electron attachment:}\)
A further electron can formally complete the \(2p\) subshell.
However, the incoming electron must approach an ion already carrying a negative charge.
Strong electrostatic repulsion opposes this process.
Therefore, the second electron-gain enthalpy is positive despite the formation of a filled-shell electronic configuration.
The process is:
\[
\mathrm{O^-(g)+e^-\rightarrow O^{2-}(g)}
\]
The closed-shell configuration of the product does not automatically make every isolated electron-attachment step exothermic.
\(\textbf{Explain the stability of ionic oxides:}\)
Formation of a stable ionic solid involves more than electron attachment to an isolated gaseous oxygen species.
Metal atom ionization, electron transfer and lattice formation all contribute to the overall energetic balance.
When oppositely charged ions assemble into a crystal lattice, substantial energy may be released.
Strong electrostatic attraction between metal cations and oxide ions can compensate for energetically unfavourable individual steps.
Consequently, compounds containing oxide ions can be stable even though the isolated second electron attachment is endothermic.
\(\textbf{Distinguish atomic and bulk stability:}\)
The energetics of an isolated gaseous ion are not identical to those of the same formal ion within a crystal.
Likewise, the oxidation state -2 does not prove that a compound must be ionic.
Oxygen can also have oxidation state -2 in covalent compounds such as water.
Thus, the existence of stable ionic oxides is consistent with a positive second electron-gain enthalpy.
The decisive principle is that compound stability depends on the complete energetic balance, not a single atomic electron-gain process.
105. Water and hydrogen fluoride both exhibit appreciable intermolecular hydrogen bonding. Their normal boiling points are approximately \(100^\circ\mathrm{C}\) and \(19.5^\circ\mathrm{C}\), respectively. HF has a very polar H–F bond, yet water has the much higher boiling point. Which explanation best accounts for this observation?
ⓐ. Water has a higher boiling point because its O–H covalent bonds must be broken completely during vaporization.
ⓑ. HF cannot form hydrogen bonds because fluorine possesses only one unpaired electron in its neutral atomic state.
ⓒ. Water has a higher boiling point because oxygen is more electronegative than fluorine.
ⓓ. Water can form a more extensive three-dimensional hydrogen-bonding network, so individual bond polarity alone does not determine the boiling point.
Correct Answer: Water can form a more extensive three-dimensional hydrogen-bonding network, so individual bond polarity alone does not determine the boiling point.
Explanation: \(\textbf{Identify the common intermolecular interaction:}\)
Both water and hydrogen fluoride contain hydrogen bonded to a small, highly electronegative atom.
Their molecules can consequently participate in intermolecular hydrogen bonding.
Hydrogen bonding is substantially stronger than the ordinary dispersion interactions characteristic of many small non-polar molecules.
However, its overall contribution to a liquid depends on the number, arrangement and strength of interactions between molecules.
\(\textbf{Examine the water molecule:}\)
Oxygen belongs to Group 16 and has six valence electrons.
A water molecule contains two O–H bonds and two lone pairs on oxygen.
Its two hydrogen atoms can participate as hydrogen-bond donors.
Its oxygen lone pairs can participate as hydrogen-bond acceptors.
This combination permits formation of an extensive three-dimensional intermolecular network.
Considerable energy is required to disrupt sufficient portions of this network during vaporization.
\(\textbf{Examine hydrogen fluoride:}\)
Fluorine belongs to Group 17 and has seven valence electrons.
In HF, fluorine forms one bond with hydrogen and retains three lone pairs.
HF exhibits strong intermolecular hydrogen bonding.
However, each molecule contains only one hydrogen atom capable of acting as a conventional hydrogen-bond donor.
Its associated molecular arrangements are therefore different from the extensive network possible in water.
The stronger polarity of an individual H–F bond does not ensure that the total intermolecular cohesion of liquid HF exceeds that of water.
\(\textbf{Connect the explanation with boiling:}\)
Boiling principally requires overcoming intermolecular attractions rather than breaking the ordinary O–H or H–F covalent bonds.
Water's extensive hydrogen-bonding network contributes to its significantly higher boiling point.
This comparison illustrates why individual bond polarity, molecular mass and the mere presence of hydrogen bonding are insufficient when considered separately.
The molecular arrangement and collective strength of intermolecular interactions must also be evaluated.
106. Consider chlorine in three different species: an isolated ground-state chlorine atom, a ground-state \(\mathrm{Cl_2}\) molecule and a chloride ion, \(\mathrm{Cl^-}\). Which statement correctly compares their unpaired-electron behaviour under ordinary ground-state conditions?
ⓐ. The isolated chlorine atom is paramagnetic, whereas ground-state \(\mathrm{Cl_2}\) and \(\mathrm{Cl^-}\) are diamagnetic.
ⓑ. All three species are paramagnetic because chlorine belongs to Group 17.
ⓒ. The isolated chlorine atom and chloride ion are diamagnetic, whereas \(\mathrm{Cl_2}\) is paramagnetic.
ⓓ. All three species are diamagnetic because chlorine has seven valence electrons.
Correct Answer: The isolated chlorine atom is paramagnetic, whereas ground-state \(\mathrm{Cl_2}\) and \(\mathrm{Cl^-}\) are diamagnetic.
Explanation: \(\textbf{Examine the isolated atom:}\)
Chlorine has atomic number \(17\) and the ground-state configuration \([\mathrm{Ne}]\,3s^23p^5\).
The three \(p\) orbitals are occupied according to Hund's rule.
In the \(3p^5\) arrangement, two orbitals contain electron pairs while one orbital contains a single electron.
Therefore, an isolated ground-state chlorine atom possesses one unpaired electron.
A species with an unpaired electron is paramagnetic.
\(\textbf{Examine the chlorine molecule:}\)
Two chlorine atoms combine to form an ordinary \(\mathrm{Cl_2}\) molecule.
Each chlorine atom contributes one electron to the shared electron pair forming the Cl–Cl covalent bond.
In the usual closed-shell ground-state bonding description, the electrons are paired.
Consequently, ground-state molecular chlorine is diamagnetic.
The presence of an unpaired electron in each separated atom does not require an unpaired electron to remain after the atoms form a stable covalent molecule.
\(\textbf{Examine the chloride ion:}\)
A neutral chlorine atom can acquire one electron to form \(\mathrm{Cl^-}\).
The ion then possesses eighteen electrons and has the argon configuration.
Its outermost \(3p\) subshell is completely filled.
All its electrons are paired in the ground state.
Therefore, the chloride ion is diamagnetic.
\(\textbf{Interpret the comparison:}\)
Magnetic behaviour depends on the actual electronic arrangement of the species being considered.
The same element can consequently show different electron-pairing behaviour in its isolated atomic, molecular and ionic forms.
Group membership identifies the neutral atom's general valence configuration but does not directly determine the magnetic behaviour of every species containing that element.
The correct comparison is one paramagnetic atomic species and two diamagnetic species.
107. The following approximate properties are supplied for fluorine, chlorine and iodine.
| Element | First ionization enthalpy \((\mathrm{kJ\,mol^{-1}})\) | Normal boiling point (K) |
|---|
| F | 1681 | 85 |
| Cl | 1251 | 239 |
| I | 1008 | 457 |
Which interpretation correctly reconciles the opposing trends in these two properties?
ⓐ. Both trends are controlled only by nuclear charge, so one of the supplied data sets must be incorrect.
ⓑ. First ionization enthalpy decreases because outer-electron attraction generally weakens, while boiling point increases because molecular dispersion interactions strengthen.
ⓒ. First ionization enthalpy decreases because the outermost valence-electron count decreases, while boiling point increases because the molecules become ionic.
ⓓ. First ionization enthalpy decreases because the X–X covalent bond strengthens, while boiling point increases because electron-gain enthalpy becomes more negative.
Correct Answer: First ionization enthalpy decreases because outer-electron attraction generally weakens, while boiling point increases because molecular dispersion interactions strengthen.
Explanation: \(\textbf{Interpret the ionization-enthalpy trend:}\)
Fluorine, chlorine and iodine all belong to Group 17.
Their general valence-shell configuration is \(ns^2np^5\).
Therefore, each neutral atom possesses seven valence electrons.
Down the group, the outermost electrons occupy successively higher principal shells.
Increasing atomic size and shielding generally reduce the attraction experienced by the outermost electrons.
Consequently, the first ionization enthalpy decreases from fluorine to iodine.
This trend concerns electron removal from isolated gaseous atoms.
\(\textbf{Interpret the boiling-point trend:}\)
Elemental fluorine, chlorine and iodine occur as homonuclear diatomic molecules under ordinary conditions.
These molecules are non-polar because the two bonded atoms have identical electronegativities.
Their intermolecular attractions are dominated by London dispersion forces.
Down the group, the number of electrons and the spatial extent of the molecular electron cloud increase.
The electron cloud consequently becomes more polarizable.
Temporary dipoles can induce stronger attractions between neighbouring molecules.
Thus, more energy is required to separate the molecules during vaporization.
The normal boiling point therefore increases from fluorine to iodine.
\(\textbf{Reconcile the opposing directions:}\)
First ionization enthalpy measures the energy needed to remove an electron from an isolated atom.
Boiling point reflects the energetics of separating molecules from a condensed phase.
These processes depend on different aspects of electronic structure.
Increasing electron-cloud extent can make the outermost electron easier to remove while simultaneously strengthening dispersion interactions between molecules.
The two trends are therefore compatible rather than contradictory.
The increase in boiling point also does not require the X–X covalent bond to become stronger.
Ordinary boiling leaves the intramolecular covalent bonds intact.
The correct explanation distinguishes atomic electron removal from intermolecular attraction.
108. Potassium and argon have atomic numbers \(19\) and \(18\), respectively. Their first ionization enthalpies are approximately \(419\,\mathrm{kJ\,mol^{-1}}\) and \(1521\,\mathrm{kJ\,mol^{-1}}\). The second ionization enthalpy of potassium is approximately \(3052\,\mathrm{kJ\,mol^{-1}}\). Which conclusion correctly reconciles the three values?
ⓐ. The first and second ionization enthalpies of potassium must be equal because both processes remove one electron.
ⓑ. Argon has a higher first ionization enthalpy than the second ionization enthalpy of potassium because neutral argon is a noble gas.
ⓒ. The second ionization of potassium removes another \(4s\) electron, making it easier than the first ionization of argon.
ⓓ. \(\mathrm{K^+}\) and Ar are isoelectronic, but \(\mathrm{K^+}\) has the greater nuclear charge, making further electron removal more difficult.
Correct Answer: \(\mathrm{K^+}\) and Ar are isoelectronic, but \(\mathrm{K^+}\) has the greater nuclear charge, making further electron removal more difficult.
Explanation: \(\textbf{Establish the neutral configurations:}\)
Argon has atomic number \(18\) and the electronic configuration \([\mathrm{Ne}]\,3s^23p^6\).
Its outermost principal shell is completely filled.
Potassium has atomic number \(19\) and the configuration \([\mathrm{Ar}]\,4s^1\).
It therefore possesses one electron outside an argon-like inner core.
\(\textbf{Interpret potassium's first ionization:}\)
The first ionization process removes the outermost \(4s\) electron.
\[
\mathrm{K(g)\rightarrow K^+(g)+e^-}
\]
The resulting ion has eighteen electrons.
Its electronic configuration is:
\[
\mathrm{K^+:[Ar]}
\]
Removal of the outer \(4s\) electron produces a closed-shell configuration and requires comparatively little energy.
Thus, the first ionization enthalpy of potassium is only approximately \(419\,\mathrm{kJ\,mol^{-1}}\).
\(\textbf{Interpret argon's first ionization:}\)
Argon already possesses a closed-shell configuration.
Removal of an electron disrupts its completely filled \(3p\) subshell.
The process consequently requires substantially more energy than removal of potassium's outer \(4s\) electron.
\(\textbf{Examine potassium's second ionization:}\)
The second ionization process is:
\[
\mathrm{K^+(g)\rightarrow K^{2+}(g)+e^-}
\]
The electron must now be removed from the completed third principal shell of \(\mathrm{K^+}\).
This is fundamentally different from removal of the original \(4s\) electron.
Furthermore, \(\mathrm{K^+}\) and neutral argon are isoelectronic.
Both contain eighteen electrons, but potassium has nineteen protons, whereas argon has eighteen.
The greater nuclear charge of \(\mathrm{K^+}\) attracts the same number of electrons more strongly.
Consequently, removal of an electron from \(\mathrm{K^+}\) requires more energy than removal of an electron from neutral argon.
\(\textbf{Verify the ordering:}\)
The supplied values give:
\[
419\lt 1521\lt 3052
\]
Therefore:
\[
IE_1(\mathrm{K})\lt IE_1(\mathrm{Ar})\lt IE_2(\mathrm{K})
\]
The ordering is consistent with outer-shell removal, closed-shell stability and the nuclear-charge comparison of isoelectronic species.
It demonstrates why successive ionization enthalpies depend on the actual electronic state of the species, not merely on the fact that one electron is removed in each step.
The correct conclusion is that \(\mathrm{K^+}\) and Ar are isoelectronic, but electron removal from \(\mathrm{K^+}\) is more difficult because of its greater nuclear charge.
109. An unknown Group 13 element X forms a normal oxide \(\mathrm{X_2O_3}\). The oxide contains \(25.6\%\) oxygen by mass. The possible elements are boron, aluminium, gallium and indium, with approximate atomic masses \(10.8\), \(27.0\), \(69.7\) and \(114.8\,\mathrm{g\,mol^{-1}}\), respectively. Which identification and oxide classification are consistent with the composition and general Group 13 chemistry?
ⓐ. Aluminium; amphoteric oxide
ⓑ. Gallium; amphoteric oxide
ⓒ. Indium; predominantly basic oxide
ⓓ. Boron; acidic oxide
Correct Answer: Gallium; amphoteric oxide
Explanation: \(\textbf{Extract the composition information:}\)
The oxide has the general formula \(\mathrm{X_2O_3}\), indicating two atoms of X and three oxygen atoms per formula unit.
Oxygen has an atomic mass of approximately \(16.0\,\mathrm{g\,mol^{-1}}\).
The mass contributed by oxygen in one mole of the oxide is:
\[
m_{\mathrm O}=3(16.0)=48.0\,\mathrm{g}
\]
Let the atomic mass of X be \(M\).
The molar mass of the oxide is:
\[
M_{\mathrm{oxide}}=2M+48.0
\]
\(\textbf{Apply the percentage-composition relation:}\)
\[
\%\mathrm O=\frac{48.0}{2M+48.0}\times100
\]
Substituting the supplied oxygen percentage:
\[
25.6=\frac{48.0}{2M+48.0}\times100
\]
\[
0.256(2M+48.0)=48.0
\]
\[
2M+48.0=187.5
\]
\[
2M=139.5
\]
\[
M=69.75\,\mathrm{g\,mol^{-1}}
\]
The calculated atomic mass agrees with gallium, whose supplied atomic mass is approximately \(69.7\,\mathrm{g\,mol^{-1}}\).
\(\textbf{Connect the identification with oxide behaviour:}\)
Gallium belongs to Group 13 and commonly exhibits oxidation state +3.
Its normal oxide therefore has the formula \(\mathrm{Ga_2O_3}\).
Gallium is metallic, but gallium(III) oxide exhibits amphoteric behaviour under suitable conditions.
It can participate in reactions with appropriate acids and strong bases.
The oxide is not classified as exclusively basic merely because gallium is metallic.
The composition establishes the elemental identity, while the acid–base classification requires knowledge of Group 13 chemical behaviour.
Thus, the correct identification is gallium with an amphoteric oxide.
110. The following radii are supplied for aluminium and gallium. Both ionic radii refer to six-coordinate trivalent ions.
| Element | Neutral-atom covalent radius | Trivalent ionic radius |
|---|
| Al | \(143\,\mathrm{pm}\) | \(54\,\mathrm{pm}\) |
| Ga | \(135\,\mathrm{pm}\) | \(62\,\mathrm{pm}\) |
Suppose the polarizing powers of \(\mathrm{Al^{3+}}\) and \(\mathrm{Ga^{3+}}\) towards the same anion are compared using their supplied ionic radii. Which prediction is justified?
ⓐ. \(\mathrm{Ga^{3+}}\) has greater polarizing power because neutral gallium has the smaller covalent radius.
ⓑ. Both ions have identical polarizing power because they belong to Group 13 and possess equal charges.
ⓒ. \(\mathrm{Ga^{3+}}\) has greater polarizing power because the larger ionic radius allows a more concentrated positive charge.
ⓓ. \(\mathrm{Al^{3+}}\) has greater polarizing power because its equal positive charge is concentrated within the smaller ionic radius.
Correct Answer: \(\mathrm{Al^{3+}}\) has greater polarizing power because its equal positive charge is concentrated within the smaller ionic radius.
Explanation: \(\textbf{Identify the relevant comparison:}\)
Polarizing power describes the ability of a cation to distort the electron cloud of a neighbouring anion.
For ions with comparable electronic environments, greater positive charge generally increases polarizing power.
For cations carrying the same positive charge, smaller ionic size generally produces stronger polarization.
Both aluminium(III) and gallium(III) carry charge +3.
Therefore, their relative ionic sizes are particularly important in this comparison.
\(\textbf{Apply the supplied ionic radii:}\)
The ionic radius of \(\mathrm{Al^{3+}}\) is \(54\,\mathrm{pm}\).
The corresponding radius of \(\mathrm{Ga^{3+}}\) is \(62\,\mathrm{pm}\).
Thus, aluminium(III) has the smaller radius under the specified coordination convention.
Its positive charge is consequently concentrated within a smaller region.
The smaller cation can exert a stronger distorting influence on the electron cloud of the same nearby anion.
Therefore, \(\mathrm{Al^{3+}}\) is expected to have greater polarizing power.
\(\textbf{Resolve the neutral-atom comparison:}\)
The neutral-atom covalent radii show the opposite ordering.
Neutral gallium is slightly smaller than neutral aluminium.
That difference is influenced by the relatively poor shielding provided by gallium's filled \(3d^{10}\) subshell.
However, the electronic structures change when the atoms form trivalent ions.
Aluminium(III) has a neon-like configuration, whereas gallium(III) retains a filled \(3d^{10}\) subshell beyond an argon core.
The neutral-atom radius ordering therefore cannot be transferred automatically to their ionic polarizing powers.
The correct prediction must use the radii of the actual charged species being compared.
This example illustrates how the same pair of Group 13 elements can exhibit different size orderings in neutral and ionic forms, leading to different chemical consequences.
111. Aluminium and thallium belong to Group 13. Aluminium commonly forms compounds containing Al(III), whereas Tl(I) is generally more stable than Tl(III). Consider their simple sulphides, with sulphur assigned oxidation state \(-2\). Which pair of formulas is consistent with these oxidation-state preferences and electrical neutrality?
ⓐ. \(\mathrm{Al_2S_3}\) and \(\mathrm{Tl_2S}\)
ⓑ. \(\mathrm{AlS}\) and \(\mathrm{TlS}\)
ⓒ. \(\mathrm{Al_2S}\) and \(\mathrm{Tl_2S_3}\)
ⓓ. \(\mathrm{AlS_2}\) and \(\mathrm{Tl_3S_2}\)
Correct Answer: \(\mathrm{Al_2S_3}\) and \(\mathrm{Tl_2S}\)
Explanation: \(\textbf{Establish the group similarity:}\)
Aluminium and thallium both possess the general valence-shell configuration \(ns^2np^1\).
Therefore, their neutral atoms contain three valence electrons.
However, identical valence-electron counts do not require the same preferred oxidation state.
Aluminium commonly exhibits the +3 state.
Thallium generally favours the +1 state because the inert-pair effect reduces the tendency of its outermost \(6s^2\) pair to participate in bonding.
\(\textbf{Determine the aluminium sulphide formula:}\)
Aluminium has oxidation state +3, while sulphur has oxidation state \(-2\).
Electrical neutrality requires the total positive and negative oxidation-state contributions to cancel.
The least common multiple of three and two is six.
Two aluminium atoms contribute:
\[
2(+3)=+6
\]
Three sulphur atoms contribute:
\[
3(-2)=-6
\]
The simplest neutral formula is:
\[
\mathrm{Al_2S_3}
\]
\(\textbf{Determine the thallium(I) sulphide formula:}\)
Thallium is assigned oxidation state +1.
Each sulphur atom contributes \(-2\).
Two thallium atoms are therefore required for one sulphur atom.
\[
2(+1)+(-2)=0
\]
The corresponding formula is:
\[
\mathrm{Tl_2S}
\]
\(\textbf{Interpret the chemical significance:}\)
The different formulas do not mean that aluminium and thallium belong to different groups.
They arise because different oxidation states have been selected for the two elements.
The Group 13 valence framework permits both +1 and +3 states, but their relative stability changes down the group.
This distinction is essential when predicting formulas of compounds containing heavier p-block elements.
The correct pair is \(\mathrm{Al_2S_3}\) and \(\mathrm{Tl_2S}\).
112. Carbon has the ground-state valence-shell configuration \(2s^22p^2\). According to Hund's rule, an isolated ground-state carbon atom has two unpaired valence electrons, yet carbon commonly forms four covalent bonds in compounds such as methane. Which explanation correctly reconciles these observations?
ⓐ. Carbon possesses four unpaired electrons in its isolated ground state because the filled \(2s\) subshell contains two unpaired electrons.
ⓑ. Carbon must form only two covalent bonds because the number of unpaired electrons in its ground state permanently determines its covalency.
ⓒ. Bond formation can involve redistribution and hybridization of valence orbitals, allowing four bonds despite the two unpaired electrons in the isolated ground-state atom.
ⓓ. Carbon forms four bonds by acquiring four electrons and converting into a free \(\mathrm{C^{4-}}\) ion before every covalent reaction.
Correct Answer: Bond formation can involve redistribution and hybridization of valence orbitals, allowing four bonds despite the two unpaired electrons in the isolated ground-state atom.
Explanation: \(\textbf{Examine the isolated atomic configuration:}\)
Carbon belongs to Group 14 and has four valence electrons.
Its ground-state electronic configuration is:
\[
\mathrm{1s^22s^22p^2}
\]
The two \(2s\) electrons occupy the same orbital with opposite spins.
They therefore form an electron pair rather than two unpaired electrons.
The two \(2p\) electrons occupy separate degenerate \(p\) orbitals according to Hund's rule.
Thus, the isolated ground-state carbon atom has two unpaired valence electrons.
\(\textbf{Distinguish atomic configuration from molecular bonding:}\)
The number of unpaired electrons in an isolated atom does not always directly equal the number of covalent bonds formed in a molecule.
During bond formation, the electronic arrangement is influenced by interactions with neighbouring atoms.
For a conventional valence-bond description of methane, promotion and hybridization provide four suitable bonding orbitals.
Four equivalent C–H sigma bonds are formed in a tetrahedral arrangement.
The energy required for electronic rearrangement is compensated by the energy released during bond formation.
\(\textbf{Interpret the Group 14 consequence:}\)
Carbon's tetravalency is consistent with its four available valence electrons.
However, tetravalency does not require four unpaired electrons to exist in the isolated atomic ground state.
It also does not imply formation of free \(\mathrm{C^{4-}}\) ions during ordinary covalent bonding.
The distinction between atomic ground-state occupancy and bonding-state orbital organization is central to understanding carbon's chemistry.
It prevents the incorrect use of Hund's-rule electron counting as an absolute limit on molecular covalency.
113. Diamond and graphite are two elemental forms of carbon. Both contain carbon atoms in oxidation state zero, yet diamond is an electrical insulator while graphite conducts electricity along its layers. Which explanation correctly connects their structures with this difference?
ⓐ. Graphite has delocalized electrons in layers; diamond has a three-dimensional localized covalent network.
ⓑ. Graphite conducts because its layers contain freely moving carbon cations, whereas diamond contains neutral carbon atoms.
ⓒ. Diamond and graphite should have identical conductivity because oxidation state and elemental composition are identical.
ⓓ. Diamond conducts poorly because its atoms have fewer valence electrons than graphite atoms.
Correct Answer: Graphite has delocalized electrons in layers; diamond has a three-dimensional localized covalent network.
Explanation: \(\textbf{Identify the common chemical feature:}\)
Diamond and graphite are allotropes of carbon.
They consist entirely of carbon atoms and therefore contain carbon in oxidation state zero.
Every carbon atom has four valence electrons.
The difference between their properties cannot be explained by different atomic numbers or valence-electron counts.
\(\textbf{Examine diamond:}\)
In diamond, each carbon atom forms four sigma bonds with neighbouring carbon atoms.
These bonds create an extended three-dimensional tetrahedral network.
The valence electrons are strongly associated with the covalent bonding framework.
There are no readily mobile electronic charge carriers under ordinary conditions.
Diamond therefore behaves as an electrical insulator.
\(\textbf{Examine graphite:}\)
In graphite, each carbon atom is bonded to three neighbouring carbon atoms within a planar layer.
Three valence electrons participate in the sigma-bonding framework.
The remaining valence-electron contribution forms a delocalized electronic system extending through the layer.
These delocalized electrons permit electrical conduction predominantly along the graphite layers.
The layers themselves are held together by comparatively weak interactions.
\(\textbf{Draw the chemical inference:}\)
The different conductivity arises from the arrangement and availability of valence electrons in the two structures.
Identical elemental composition and oxidation state do not require identical electronic behaviour.
Carbon's ability to form different stable covalent networks is therefore an important consequence of its bonding versatility.
The correct explanation must distinguish localized tetrahedral bonding from layered bonding with electronic delocalization.
114. Carbon monoxide and carbon dioxide both contain carbon and oxygen, but their chemical classifications differ. Carbon monoxide is a neutral oxide, whereas carbon dioxide is acidic. Which conclusion correctly combines their oxidation states and oxide behaviour?
ⓐ. Carbon has oxidation state zero in CO and +4 in \(\mathrm{CO_2}\); the zero state makes carbon monoxide neutral.
ⓑ. Carbon has oxidation state +4 in both CO and \(\mathrm{CO_2}\); oxide classification changes only because the molecular masses differ.
ⓒ. Carbon has oxidation state +2 in both CO and \(\mathrm{CO_2}\); carbon dioxide is acidic because its coordination number is greater.
ⓓ. Carbon has oxidation state +2 in CO and +4 in \(\mathrm{CO_2}\); different oxidation states and bonding arrangements accompany different oxide classifications.
Correct Answer: Carbon has oxidation state +2 in CO and +4 in \(\mathrm{CO_2}\); different oxidation states and bonding arrangements accompany different oxide classifications.
Explanation: \(\textbf{Determine the oxidation state in carbon monoxide:}\)
Oxygen is assigned oxidation state \(-2\) in ordinary carbon monoxide.
For the neutral molecule:
\[
x+(-2)=0
\]
\[
x=+2
\]
Therefore, carbon has oxidation state +2 in CO.
\(\textbf{Determine the oxidation state in carbon dioxide:}\)
Each oxygen atom contributes \(-2\).
For the neutral molecule:
\[
y+2(-2)=0
\]
\[
y=+4
\]
Thus, carbon has oxidation state +4 in \(\mathrm{CO_2}\).
\(\textbf{Relate oxidation state to chemical behaviour:}\)
Carbon monoxide is classified as a neutral oxide because it does not exhibit the characteristic acid–base behaviour of ordinary acidic or basic oxides.
Carbon dioxide is an acidic oxide.
It reacts with suitable bases to form carbonate or hydrogencarbonate species.
The two molecules also have different bonding arrangements and electronic structures.
However, oxidation state alone is not a universal rule for predicting oxide acidity.
The observed classifications arise from the actual chemical behaviour of each oxide.
\(\textbf{Identify the broader Group 14 principle:}\)
Carbon can exhibit different positive oxidation states while remaining a non-metallic element.
A lower oxidation state does not automatically establish basic oxide behaviour.
Likewise, a higher oxidation state does not independently prove that every oxide must be acidic.
The important conclusion is that formal oxidation state, bonding and observed acid–base behaviour must be considered together.
The correct comparison is carbon +2 in neutral CO and carbon +4 in acidic \(\mathrm{CO_2}\).
115. Phosphorus forms two representative oxides with molecular formulas \(\mathrm{P_4O_6}\) and \(\mathrm{P_4O_{10}}\). Oxygen has oxidation state \(-2\) in both compounds. Which statement correctly compares phosphorus oxidation states and the broad acid–base character of these oxides?
ⓐ. Phosphorus is +3 in both oxides, and both are neutral because phosphorus is non-metallic.
ⓑ. Phosphorus is +3 in \(\mathrm{P_4O_6}\) and +5 in \(\mathrm{P_4O_{10}}\); both are acidic oxides.
ⓒ. Phosphorus is +5 in \(\mathrm{P_4O_6}\) and +3 in \(\mathrm{P_4O_{10}}\); both are amphoteric oxides.
ⓓ. Phosphorus is +3 in \(\mathrm{P_4O_6}\) and +5 in \(\mathrm{P_4O_{10}}\); both are basic oxides.
Correct Answer: Phosphorus is +3 in \(\mathrm{P_4O_6}\) and +5 in \(\mathrm{P_4O_{10}}\); both are acidic oxides.
Explanation: \(\textbf{Establish the Group 15 valence framework:}\)
Phosphorus has five valence electrons and belongs to Group 15.
Its general outermost configuration is \(ns^2np^3\).
It can exhibit different positive oxidation states depending on the compound.
The common positive states include +3 and +5.
\(\textbf{Determine the oxidation state in the first oxide:}\)
In \(\mathrm{P_4O_6}\), six oxygen atoms contribute a total oxidation-state sum of:
\[
6(-2)=-12
\]
The four phosphorus atoms must collectively contribute +12.
\[
4x-12=0
\]
\[
4x=12
\]
\[
x=+3
\]
Thus, phosphorus has oxidation state +3 in \(\mathrm{P_4O_6}\).
\(\textbf{Determine the oxidation state in the second oxide:}\)
In \(\mathrm{P_4O_{10}}\), ten oxygen atoms contribute \(-20\).
\[
4y+10(-2)=0
\]
\[
4y=20
\]
\[
y=+5
\]
Therefore, phosphorus has oxidation state +5 in \(\mathrm{P_4O_{10}}\).
\(\textbf{Interpret oxide character:}\)
Phosphorus is a non-metal.
Both representative oxides exhibit acidic behaviour.
The fact that their central atoms have different oxidation states does not require one oxide to become basic or amphoteric.
This is consistent with the broad acidic nature of oxides of non-metallic Group 15 elements.
The comparison also shows why the number of valence electrons must not be confused with the oxidation state in every compound.
The correct conclusion combines the +3/+5 oxidation-state distinction with the acidic character of both oxides.
116. White phosphorus and red phosphorus are two allotropes of the same element. White phosphorus consists of discrete tetrahedral \(\mathrm{P_4}\) molecules and is comparatively reactive, whereas red phosphorus possesses a more extensively connected structure and is considerably less reactive under ordinary conditions. Which explanation best accounts for their different behaviour?
ⓐ. White phosphorus contains phosphorus in oxidation state +4, whereas red phosphorus contains phosphorus in oxidation state zero.
ⓑ. Red phosphorus contains fewer valence electrons per phosphorus atom, making its bonds stronger.
ⓒ. Bond strain in tetrahedral \(\mathrm{P_4}\) makes white phosphorus more reactive than red phosphorus.
ⓓ. White phosphorus is more reactive because it contains nitrogen-like \(\mathrm{P\equiv P}\) triple bonds, while red phosphorus contains only single bonds.
Correct Answer: Bond strain in tetrahedral \(\mathrm{P_4}\) makes white phosphorus more reactive than red phosphorus.
Explanation: \(\textbf{Identify the common elemental character:}\)
White phosphorus and red phosphorus are allotropes of phosphorus.
Both consist entirely of phosphorus atoms.
Therefore, phosphorus has oxidation state zero in both elemental forms.
Each phosphorus atom also retains the same atomic number and valence-electron count.
Their different chemical behaviour cannot be attributed to different elemental identities or formal oxidation states.
\(\textbf{Examine white phosphorus:}\)
White phosphorus contains discrete tetrahedral \(\mathrm{P_4}\) molecules.
Each phosphorus atom is directly bonded to three other phosphorus atoms.
The P–P–P bond angles within the tetrahedron are approximately \(60^\circ\).
These angles produce a strained bonding arrangement.
The molecular structure consequently contains a relatively high-energy bonding framework.
This contributes to the greater reactivity of white phosphorus.
\(\textbf{Examine red phosphorus:}\)
Red phosphorus possesses a more extensively connected structure.
Its bonding arrangement avoids much of the severe strain characteristic of the isolated tetrahedral \(\mathrm{P_4}\) molecule.
Consequently, red phosphorus is generally less reactive and more stable under ordinary conditions.
\(\textbf{Connect structure with allotropy:}\)
Allotropy refers to the existence of an element in different structural forms.
Different allotropes can have different physical and chemical properties despite possessing identical elemental composition.
The important difference here is the connectivity and strain of the phosphorus bonding framework.
The conclusion does not require different oxidation states or different valence-electron counts.
Thus, the greater reactivity of white phosphorus is consistent with its strained tetrahedral molecular structure.
117. Ammonia, \(\mathrm{NH_3}\), and nitrogen trifluoride, \(\mathrm{NF_3}\), both contain nitrogen bonded to three atoms and possess a lone pair on nitrogen. Under ordinary comparable conditions, ammonia is a much stronger Lewis base than nitrogen trifluoride. Which explanation correctly accounts for this difference?
ⓐ. Nitrogen trifluoride has no nitrogen lone pair because fluorine removes all five nitrogen valence electrons.
ⓑ. Ammonia has four nitrogen–hydrogen bonds, whereas nitrogen trifluoride has only three nitrogen–fluorine bonds.
ⓒ. Nitrogen trifluoride is the stronger base because fluorine increases the electron density available for nitrogen-to-acceptor donation.
ⓓ. The strongly electron-withdrawing fluorine atoms reduce the availability of nitrogen's lone pair for donation compared with ammonia.
Correct Answer: The strongly electron-withdrawing fluorine atoms reduce the availability of nitrogen's lone pair for donation compared with ammonia.
Explanation: \(\textbf{Establish the common nitrogen framework:}\)
Nitrogen belongs to Group 15 and possesses five valence electrons.
In both ammonia and nitrogen trifluoride, nitrogen forms three ordinary covalent bonds.
The remaining two valence electrons are represented by a lone pair.
Both molecules therefore possess the electronic requirement for possible Lewis-base behaviour.
Their molecular geometries are trigonal pyramidal in the usual bonding description.
However, identical numbers of bonding pairs and lone pairs do not guarantee identical basicity.
\(\textbf{Examine ammonia:}\)
Hydrogen is less electronegative than nitrogen.
In ammonia, the bonding environment allows the nitrogen lone pair to act effectively as an electron-pair donor.
The lone pair can be donated to a suitable Lewis acid.
Ammonia therefore exhibits substantial Lewis basicity.
\(\textbf{Examine nitrogen trifluoride:}\)
Fluorine is considerably more electronegative than nitrogen.
The three fluorine atoms withdraw electron density through the N–F bonds.
This strongly electron-withdrawing environment reduces the electron-donating tendency of the nitrogen centre.
Consequently, the nitrogen lone pair is much less effective as a donor than the lone pair in ammonia.
\(\textbf{Interpret the distinction:}\)
Lewis basicity depends on the energetic availability of an electron pair, not merely on whether a lone pair exists.
The chemical environment surrounding the donor atom can alter its basic strength substantially.
The comparison also demonstrates why identical central-atom coordination numbers do not require identical chemical reactivity.
Thus, ammonia is the stronger Lewis base even though both molecules contain one nitrogen lone pair.
118. Potassium forms three oxygen-containing compounds with formulas \(\mathrm{K_2O}\), \(\mathrm{K_2O_2}\) and \(\mathrm{KO_2}\). Potassium has oxidation state +1 in all three. The oxygen species are oxide, peroxide and superoxide, respectively. Which sequence correctly gives the average oxidation state of oxygen in these compounds?
ⓐ. \(-2,-1,-\frac12\)
ⓑ. \(-2,-\frac12,-1\)
ⓒ. \(-1,-2,-\frac12\)
ⓓ. \(-1,-\frac12,-2\)
Correct Answer: \(-2,-1,-\frac12\)
Explanation: \(\textbf{Apply the electrical-neutrality principle:}\)
Potassium is an alkali metal and has oxidation state +1 in the three supplied compounds.
For every neutral formula unit, the sum of the oxidation-state contributions must be zero.
The oxygen oxidation state can therefore be determined independently for each formula.
\(\textbf{Examine potassium oxide:}\)
In \(\mathrm{K_2O}\), two potassium atoms contribute +2.
The single oxygen atom must contribute \(-2\).
\[
2(+1)+x=0
\]
\[
x=-2
\]
This is the usual oxide oxidation state of oxygen.
\(\textbf{Examine potassium peroxide:}\)
In \(\mathrm{K_2O_2}\), two potassium atoms again contribute +2.
The two oxygen atoms together must contribute \(-2\).
\[
2(+1)+2y=0
\]
\[
y=-1
\]
The peroxide unit contains an O–O bond and has total charge \(-2\).
Each equivalent oxygen atom is therefore assigned oxidation state \(-1\).
\(\textbf{Examine potassium superoxide:}\)
In \(\mathrm{KO_2}\), one potassium atom contributes +1.
The two oxygen atoms together must contribute \(-1\).
\[
(+1)+2z=0
\]
\[
2z=-1
\]
\[
z=-\frac12
\]
Thus, the average oxidation state per oxygen atom is \(-\frac12\).
\(\textbf{Interpret the fractional value:}\)
The fractional oxidation state does not imply that an isolated oxygen atom carries exactly half an electron as a physical charge.
It is a formal average arising from the two-oxygen superoxide unit, \(\mathrm{O_2^-}\).
This differs from the peroxide ion, \(\mathrm{O_2^{2-}}\), and the monatomic oxide ion, \(\mathrm{O^{2-}}\).
The three compounds show why oxygen cannot be assigned oxidation state \(-2\) without examining the actual chemical formula and bonding context.
The required sequence is \(-2,-1,-\frac12\).
119. Two solid samples of elemental sulphur are known to contain the same cyclic \(\mathrm{S_8}\) molecules but different crystalline arrangements. Their melting temperatures and other physical properties are not identical. Which interpretation is consistent with allotropy and the bonding of sulphur?
ⓐ. The samples must contain sulphur in different oxidation states because their melting temperatures differ.
ⓑ. Different crystal packing can produce different bulk physical properties even when the molecular composition is the same.
ⓒ. The samples must have different numbers of valence electrons per sulphur atom because their structures are different.
ⓓ. Different melting temperatures prove that one sample contains ionic S–S bonds while the other contains metallic S–S bonds.
Correct Answer: Different crystal packing can produce different bulk physical properties even when the molecular composition is the same.
Explanation: \(\textbf{Identify the molecular similarity:}\)
Both samples contain elemental sulphur.
Each sulphur atom has six valence electrons and oxidation state zero in the elemental form.
The stated molecules have the same composition, \(\mathrm{S_8}\).
In these cyclic molecules, neighbouring sulphur atoms are connected through covalent S–S bonds.
Therefore, the molecular formula and elemental oxidation state are the same for both samples.
\(\textbf{Distinguish molecular structure from crystal structure:}\)
A molecular formula identifies the numbers and kinds of atoms in a molecule.
It does not completely specify how large numbers of those molecules are arranged in a solid.
Different crystalline arrangements can have different molecular packing and intermolecular interactions.
Consequently, they may exhibit different physical properties.
Rhombic and monoclinic sulphur provide familiar examples of crystalline allotropes involving cyclic \(\mathrm{S_8}\) molecules.
Their structures differ in the arrangement of these molecules within the crystals.
\(\textbf{Connect packing with melting:}\)
Melting involves disruption of sufficient interactions within the solid to produce a liquid phase.
The energy required for this change can depend on the crystalline arrangement.
Thus, two solids containing the same type of molecule need not possess identical melting behaviour.
The difference does not require changes in sulphur's atomic number, valence-electron count or oxidation state.
Nor does it establish that one allotrope must involve ionic or metallic S–S bonding.
This example illustrates why bulk physical properties depend on structural organization in addition to the bonding within individual molecules.
The correct conclusion is that different crystal packing can produce different physical behaviour despite identical molecular composition.
120. Hydrogen sulphide reacts with bromine under suitable conditions according to the balanced equation:
\[
\mathrm{H_2S+Br_2\longrightarrow S+2HBr}
\]
Which statement correctly identifies the oxidation-state changes and the functions of the reactants?
ⓐ. Sulphur changes from +2 to zero, while bromine changes from zero to +1; \(\mathrm{H_2S}\) is the oxidizing agent.
ⓑ. Sulphur changes from \(-2\) to zero, while bromine changes from zero to \(-1\); \(\mathrm{Br_2}\) is the reducing agent.
ⓒ. Sulphur changes from \(-2\) to zero, while bromine changes from zero to \(-1\); \(\mathrm{H_2S}\) is the reducing agent.
ⓓ. Sulphur changes from zero to \(-2\), while bromine changes from \(-1\) to zero; \(\mathrm{Br_2}\) is the oxidizing agent.
Correct Answer: Sulphur changes from \(-2\) to zero, while bromine changes from zero to \(-1\); \(\mathrm{H_2S}\) is the reducing agent.
Explanation: \(\textbf{Determine the initial oxidation states:}\)
In hydrogen sulphide, hydrogen has oxidation state +1.
For the neutral molecule:
\[
2(+1)+x=0
\]
\[
x=-2
\]
Therefore, sulphur begins in oxidation state \(-2\).
In elemental bromine, \(\mathrm{Br_2}\), each bromine atom has oxidation state zero.
\(\textbf{Determine the product oxidation states:}\)
Elemental sulphur has oxidation state zero.
In hydrogen bromide, hydrogen has oxidation state +1 and bromine has oxidation state \(-1\).
Therefore, sulphur changes from \(-2\) to zero, while bromine changes from zero to \(-1\).
\(\textbf{Identify oxidation and reduction:}\)
The increase in sulphur's oxidation state represents oxidation.
The formal electron-loss process is:
\[
\mathrm{S(-II)\longrightarrow S(0)+2e^-}
\]
Bromine undergoes reduction.
Each bromine atom gains one electron, giving two electrons gained per \(\mathrm{Br_2}\) molecule.
\[
\mathrm{Br_2+2e^-\longrightarrow 2Br^-}
\]
\(\textbf{Verify the electron balance:}\)
One sulphur atom formally loses two electrons.
The two bromine atoms collectively gain two electrons.
The electron transfers are therefore balanced.
\(\textbf{Interpret the chemical roles:}\)
A reducing agent is the species that undergoes oxidation.
Hydrogen sulphide is oxidized and therefore acts as the reducing agent.
Bromine accepts electrons and acts as the oxidizing agent.
This reaction illustrates the ability of a Group 16 hydride containing sulphur in a low oxidation state to behave as a reducing agent.
It also illustrates the electron-accepting tendency of a Group 17 molecular halogen.
The conclusion requires identifying the oxidation states of both reacting elements rather than assuming that the molecule containing hydrogen must be the oxidizing agent.
Thus, hydrogen sulphide is the reducing agent and molecular bromine is the oxidizing agent.
