Classification of Elements and Periodicity in Properties MCQs with Answers – Part 5 (NEET Chemistry)

Explanation: Tin and lead belong to group 14. Their neutral valence configurations have the characteristic pattern \(ns^2np^2\). Both therefore possess four valence electrons. Nevertheless, they can exhibit more than one positive oxidation state. In heavier p-block elements, the outer \(ns^2\) pair can become comparatively less available for bonding. This tendency is known as the inert-pair effect. It contributes to the increased relative stability of lower oxidation states down certain p-block groups. For group 14, the lower state is \(+2\), while the higher characteristic state is \(+4\). Tin(II) can undergo oxidation to tin(IV) in suitable reactions. It can therefore serve as a reducing agent. Lead(IV) can undergo reduction to the relatively stable lead(II) state. It can therefore serve as an oxidising agent. These observations do not mean that every reaction involving Sn(II) or Pb(IV) must proceed spontaneously under all conditions. The actual direction also depends on the other reactants and their chemical environments. The fourth option correctly connects the relative oxidation-state stability with the opposing redox roles.

Explanation: \(\textbf{Interpreting the first displacement:}\) Metal M displaces metal N from a solution containing \(\mathrm{N^{2+}}\). M therefore acts as the stronger reducing agent under the stated conditions. Its atoms lose electrons while \(\mathrm{N^{2+}}\) gains electrons. The observation gives the relative order: \[ M\gt N \] in aqueous reducing strength. \(\textbf{Interpreting the second displacement:}\) Metal N deposits P from a solution containing \(\mathrm{P^{2+}}\). Therefore: \[ N\gt P \] Combining the two observations: \[ M\gt N\gt P \] \(\textbf{Checking the third observation:}\) Metal P does not displace M from its divalent ionic solution. This is consistent with P being the weaker reducing agent. The third test supports rather than contradicts the inferred order. \(\textbf{Identifying the metals:}\) The ordinary aqueous reducing-strength order for the specified metals is: \[ \mathrm{Mg\gt Zn\gt Cu} \] Thus: \[ M=\mathrm{Mg},\qquad N=\mathrm{Zn},\qquad P=\mathrm{Cu} \] \(\textbf{Evaluating the ionization-enthalpy shortcut:}\) Zinc has the greatest supplied first ionization enthalpy. Copper's first ionization enthalpy is nevertheless only slightly greater than magnesium's. These values alone do not reproduce the complete aqueous displacement order. Metal-ion formation in water involves additional thermodynamic contributions, including the starting metal state and hydration of the resulting ions. First ionization enthalpy refers only to removal of an electron from an isolated gaseous atom. \(\textbf{Final interpretation:}\) Direct displacement observations provide evidence about the complete reactions under the stated conditions. They should not be replaced by a ranking based on one isolated atomic property. The second option correctly identifies the metals and explains why the two kinds of data must be distinguished.

Explanation: \(\textbf{Establishing the periodic positions:}\) Sodium, magnesium, aluminium and silicon occupy consecutive groups across period 3. Their characteristic valence-electron counts increase from one to four. Metallic character generally decreases across this sequence. The nature of their chloride bonding changes accordingly. \(\textbf{Examining sodium and magnesium chlorides:}\) Sodium is strongly electropositive and commonly forms \(\mathrm{Na^+}\). Magnesium commonly forms \(\mathrm{Mg^{2+}}\). Their chlorides have substantial ionic character and consist of extended ionic solids. When molten, sufficiently mobile ions can carry electrical charge. \(\textbf{Examining aluminium chloride:}\) Aluminium belongs to group 13. Its trichloride exhibits appreciable covalent character. The electron-deficient aluminium centre can participate in bridging interactions that produce associated species under suitable conditions. Its behaviour therefore cannot be inferred by treating every formula unit as an isolated set of simple ions. \(\textbf{Examining silicon tetrachloride:}\) Silicon belongs to group 14. It forms molecular SiCl₄ with four covalent Si–Cl bonds. The molecule is electrically neutral. Pure molecular SiCl₄ does not possess the freely moving ions characteristic of a molten ionic chloride. \(\textbf{Connecting bonding with periodicity:}\) Increasing nuclear charge and changes in valence configuration modify the tendency toward electron loss and sharing across the period. The chloride stoichiometry reflects characteristic combining capacities. It does not alone determine whether the substance is an ionic lattice, an associated covalent species or a discrete molecule. The third option correctly identifies both the periodic progression and the structural limitation of formula-based predictions.

Explanation: Statement I is correct because chlorine, bromine and iodine belong to the same periodic group. Their valence orbitals occupy progressively higher principal shells down the group. The H–X bond length generally increases. The overlap between the larger halogen valence orbitals and hydrogen's compact \(1s\) orbital becomes less effective. This contributes to the weakening of the H–X bond. Statement II is correct because the oxidation of iodide to iodine is generally easier than oxidation of chloride to chlorine under comparable aqueous redox conditions. Hydrogen iodide can therefore exhibit stronger reducing behaviour than hydrogen chloride in suitable reactions. This reducing tendency concerns electron transfer from the iodide-containing species. Statement III is incorrect because reducing strength does not determine aqueous acid strength directly. Aqueous acid dissociation concerns proton transfer and stabilisation of the resulting solvated species. Hydrogen iodide is ordinarily a strong aqueous acid. Its greater reducing tendency does not imply that it must be a weaker acid than HCl. Statement IV is correct because the three properties refer to different chemical processes. Bond dissociation enthalpy concerns cleavage of a specified bond. Acid strength concerns equilibrium proton donation. Reducing tendency concerns electron donation in an appropriate redox reaction. The relevant energy balances are therefore not identical. Only Statements I, II and IV are correct.

Explanation: \(\textbf{Determining sulfur's initial oxidation state:}\) Oxygen has oxidation state \(-2\) in SO₂. Therefore: \[ x+2(-2)=0 \] \[ x=+4 \] \(\textbf{Determining sulfur's state after oxidation:}\) For SO₃: \[ y+3(-2)=0 \] \[ y=+6 \] Sulfur is oxidised from \(+4\) to \(+6\). Oxygen gas acts as the oxidising agent in the first reaction. \(\textbf{Stage 1 — Identifying the limiting reactant:}\) The first equation requires two moles of SO₂ per mole of O₂. To consume all SO₂: \[ n(O_2)_{\mathrm{required}}=\frac{0.100}{2} \] \[ n(O_2)_{\mathrm{required}}=0.050\ \mathrm{mol} \] Only \(0.040\ \mathrm{mol}\) oxygen is available. Oxygen is limiting. \(\textbf{Calculating sulfur trioxide production:}\) \[ n(SO_3)=2n(O_2) \] \[ n(SO_3)=2(0.040) \] \[ n(SO_3)=0.080\ \mathrm{mol} \] \(\textbf{Determining residual sulfur dioxide:}\) The amount of SO₂ consumed equals the amount of SO₃ formed. Therefore: \[ n(SO_2)_{\mathrm{remaining}}=0.100-0.080 \] \[ n(SO_2)_{\mathrm{remaining}}=0.020\ \mathrm{mol} \] \(\textbf{Stage 2 — Hydration of sulfur trioxide:}\) The second balanced equation converts one mole of SO₃ into one mole of sulfuric acid. Thus: \[ n(H_2SO_4)=n(SO_3) \] \[ n(H_2SO_4)=0.080\ \mathrm{mol} \] \(\textbf{Checking oxidation-state conservation in hydration:}\) Sulfur remains at oxidation state \(+6\) during conversion of SO₃ into H₂SO₄. The second stage is a hydration reaction rather than an additional sulfur oxidation. The two stages must therefore be interpreted differently. \(\textbf{Final answer:}\) \[ \boxed{n(H_2SO_4)=0.080\ \mathrm{mol},\qquad n(SO_2)_{\mathrm{remaining}}=0.020\ \mathrm{mol};\quad \mathrm{S}:+4\rightarrow+6} \]

Explanation: \(\textbf{Establishing the electron-transfer processes:}\) Sodium belongs to group 1 and commonly forms \(\mathrm{Na^+}\). Magnesium belongs to group 2 and commonly forms \(\mathrm{Mg^{2+}}\). Both metals can reduce hydrogen ions to elemental hydrogen under the stated conditions. However, the balanced equations require different amounts of metal per mole of hydrogen. \(\textbf{Experiment I — Sodium reaction capacity:}\) Two moles of sodium consume two moles of HCl. Thus, the sodium-to-acid ratio is \(1:1\). The acid required to consume all sodium would be: \[ n(HCl)_{\mathrm{required,I}}=0.200\ \mathrm{mol} \] Only \(0.150\ \mathrm{mol}\) HCl is available. Acid is limiting. \(\textbf{Calculating the sodium-experiment products:}\) Two moles of HCl produce one mole of H₂. \[ n(H_2)_{\mathrm I}=\frac{0.150}{2} \] \[ n(H_2)_{\mathrm I}=0.075\ \mathrm{mol} \] The sodium consumed equals the acid amount: \[ n(Na)_{\mathrm{consumed}}=0.150\ \mathrm{mol} \] Therefore: \[ n(Na)_{\mathrm{remaining}}=0.200-0.150 \] \[ n(Na)_{\mathrm{remaining}}=0.050\ \mathrm{mol} \] \(\textbf{Experiment II — Magnesium reaction capacity:}\) One mole of magnesium consumes two moles of HCl. Complete reaction of the magnesium would require: \[ n(HCl)_{\mathrm{required,II}}=2(0.100) \] \[ n(HCl)_{\mathrm{required,II}}=0.200\ \mathrm{mol} \] Again, acid is limiting. \(\textbf{Calculating the magnesium-experiment products:}\) \[ n(H_2)_{\mathrm{II}}=\frac{0.150}{2} \] \[ n(H_2)_{\mathrm{II}}=0.075\ \mathrm{mol} \] The magnesium consumed is: \[ n(Mg)_{\mathrm{consumed}}=\frac{0.150}{2} \] \[ n(Mg)_{\mathrm{consumed}}=0.075\ \mathrm{mol} \] Therefore: \[ n(Mg)_{\mathrm{remaining}}=0.100-0.075 \] \[ n(Mg)_{\mathrm{remaining}}=0.025\ \mathrm{mol} \] \(\textbf{Interpreting the invariant hydrogen yield:}\) The two experiments contain the same limiting amount of HCl. Each mole of H₂ requires two moles of hydrogen ions. Consequently, both produce the same amount of hydrogen despite the metals having different valencies. The amounts of metal consumed differ because one sodium atom contributes one electron, whereas one magnesium atom contributes two. \(\textbf{Final answer:}\) \[ \boxed{n(H_2)_{\mathrm I}=n(H_2)_{\mathrm{II}}=0.075\ \mathrm{mol};\quad n(Na)_{\mathrm{remaining}}=0.050,\quad n(Mg)_{\mathrm{remaining}}=0.025\ \mathrm{mol}} \]

Explanation: Statement I is correct because magnesium is an elemental metal on the reactant side. Its initial oxidation state is zero. In magnesium oxide, oxygen has oxidation state \(-2\), so magnesium has oxidation state \(+2\). Magnesium therefore loses electrons and undergoes oxidation. Statement II is correct because oxygen has oxidation state \(-2\) in carbon dioxide. Carbon consequently has oxidation state \(+4\). In the elemental carbon product, its oxidation state is zero. Carbon is therefore reduced. Statement III is incorrect because an acidic oxide is not necessarily a reducing agent. Acid-base oxide classification and electron-transfer behaviour describe different chemical properties. In the supplied reaction, carbon dioxide accepts electrons indirectly through the reduction of its carbon centre. It acts as an oxidising agent, while magnesium acts as the reducing agent. Statement IV is correct because carbon dioxide does not necessarily remain chemically unreactive toward every burning material. Magnesium has a sufficiently strong tendency to form its stable oxide that it can reduce carbon dioxide under suitable conditions. This explains why a substance useful for extinguishing some fires may fail for a strongly reactive metal. The example also illustrates the danger of treating a general application of a chemical as an exception-free reactivity rule. Only Statements I, II and IV are correct.

Explanation: \(\textbf{Stage I — Establishing the precipitation requirement:}\) Aluminium belongs to group 13 and commonly exhibits oxidation state \(+3\). Each aluminium(III) ion requires three hydroxide ions to form aluminium hydroxide. The required hydroxide amount is: \[ n(OH^-)_{\mathrm{precipitation}}=3n(Al^{3+}) \] \[ n(OH^-)_{\mathrm{precipitation}}=3(0.100) \] \[ n(OH^-)_{\mathrm{precipitation}}=0.300\ \mathrm{mol} \] The supplied \(0.370\ \mathrm{mol}\) hydroxide is sufficient to precipitate all the aluminium initially. The precipitate formed at this stage is: \[ n(\mathrm{Al(OH)_3})_{\mathrm{initial}}=0.100\ \mathrm{mol} \] \(\textbf{Stage II — Determining the excess hydroxide:}\) After complete precipitation: \[ n(OH^-)_{\mathrm{excess}}=0.370-0.300 \] \[ n(OH^-)_{\mathrm{excess}}=0.070\ \mathrm{mol} \] The second equation requires one additional hydroxide ion per aluminium hydroxide formula unit. Therefore: \[ n(\mathrm{Al(OH)_3})_{\mathrm{dissolved}}=0.070\ \mathrm{mol} \] The amount of aluminate species formed is: \[ n(\mathrm{[Al(OH)_4]^-})=0.070\ \mathrm{mol} \] \(\textbf{Determining the remaining precipitate:}\) \[ n(\mathrm{Al(OH)_3})_{\mathrm{remaining}}=0.100-0.070 \] \[ n(\mathrm{Al(OH)_3})_{\mathrm{remaining}}=0.030\ \mathrm{mol} \] \(\textbf{Checking aluminium conservation:}\) The aluminium present in the precipitate and dissolved complex totals: \[ 0.030+0.070=0.100\ \mathrm{mol} \] This equals the original aluminium amount. \(\textbf{Chemical interpretation:}\) Aluminium hydroxide can first precipitate and subsequently dissolve when additional hydroxide is supplied. This behaviour reflects amphoterism rather than an oxidation-state change. Aluminium remains at oxidation state \(+3\) in both products. A larger hydroxide supply would be required to dissolve all the precipitate under the stipulated sequence. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{Al(OH)_3})_{\mathrm{remaining}}=0.030\ \mathrm{mol},\qquad n(\mathrm{[Al(OH)_4]^-})=0.070\ \mathrm{mol}} \]

Explanation: \(\textbf{Identifying iron's periodic position:}\) Iron belongs to the first transition series. Its ground-state configuration is \([\mathrm{Ar}]\,3d^6\,4s^2\). Iron commonly exhibits oxidation states \(+2\) and \(+3\). These states cannot be chosen solely from the neutral atom's group position. The reacting chemical environment matters. \(\textbf{Examining the hydrochloric acid reaction:}\) Metallic iron begins with oxidation state zero. In FeCl₂, the two chloride ions each have oxidation state \(-1\). Iron consequently has oxidation state \(+2\). The hydrogen ions of HCl have oxidation state \(+1\). They are reduced to elemental hydrogen with oxidation state zero. Thus, iron acts as the reducing agent and hydrogen ions as the oxidising species. \(\textbf{Examining the chlorine reaction:}\) Iron again begins at oxidation state zero. The three chlorine atoms in FeCl₃ each have oxidation state \(-1\). Iron therefore acquires oxidation state \(+3\). Elemental chlorine begins with oxidation state zero and is reduced to chloride. Chlorine gas acts as the oxidising agent. \(\textbf{Reconciling the different products:}\) The first equation transfers two electrons per iron atom to hydrogen ions. The second transfers three electrons per iron atom to chlorine. The balanced reactions therefore lead to different iron oxidation states. The fact that both products contain chloride does not imply that iron must have the same charge in each compound. \(\textbf{Final interpretation:}\) Variable oxidation states are an important feature of transition-element chemistry. A characteristic oxidation state does not specify which product will form under every set of reaction conditions. The third option correctly connects the different electron acceptors with the resulting iron compounds.

Explanation: The total oxidation-state sum in a polyatomic ion equals its stated net charge. The oxide contribution must be counted separately for each oxygen atom. For \(\mathrm{SO_3^{2-}}\), the sulfur state \(x\) satisfies \(x+3(-2)=-2\). Therefore \(x=+4\) for sulfur in sulfite. For \(\mathrm{SO_4^{2-}}\), the corresponding relation is \(x+4(-2)=-2\). Therefore the sulfur state in sulfate is \(+6\). For \(\mathrm{ClO_3^-}\), use \(y+3(-2)=-1\). The chlorine oxidation state is \(y=+5\). The equal net charges of sulfite and sulfate do not force equal central-atom oxidation states. An additional oxygen can be accompanied by a different formal positive state at the central atom. Sulfur and chlorine are neighbouring third-period non-metals that can exhibit several positive oxidation states in oxygen-containing species. These formal states should not be mistaken for actual isolated-ion charges carried by the central atoms.

Explanation: \(\textbf{Identifying the periodic positions:}\) Element W has the outer configuration of aluminium. Element X corresponds to silicon. Element Y corresponds to phosphorus. Element Z corresponds to chlorine. All four belong to period 3. Their respective group positions are 13, 14, 15 and 17. \(\textbf{Checking the oxide evidence:}\) Aluminium forms amphoteric Al₂O₃. Silicon forms the acidic network oxide SiO₂. These observations agree with the supplied identities. \(\textbf{Checking the orbital occupancy evidence:}\) Phosphorus has the configuration \([\mathrm{Ne}]\,3s^2\,3p^3\). Its three p-electrons occupy different orbitals according to Hund's rule. It therefore has three unpaired p-electrons. Chlorine has \(3p^5\), one electron short of a completed p-subshell. It can accept an electron to form chloride under suitable conditions. \(\textbf{Interpreting the fifth observation:}\) Silicon begins with \(3p^2\). Electron addition can produce the relatively stable \(3p^3\) arrangement. Phosphorus already begins with \(3p^3\). Adding one electron produces \(3p^4\), introducing a paired-electron orbital. The pairing interaction reduces the favourability of electron addition. The first electron-gain enthalpy of phosphorus is consequently less negative than that of silicon. \(\textbf{Integrating the conclusion:}\) The property exception follows from the starting and final p-subshell occupancies. It does not contradict the increasing nuclear charge across the period. The other observations independently establish the correct elemental identities. The second option correctly connects the electronic exception with the multi-property classification.

Explanation: \(\textbf{Controlling the ionic charge:}\) Magnesium and calcium belong to group 2. Both commonly form divalent cations. Their charges therefore have the same magnitude. The supplied model assigns a square dependence on radius. \(\textbf{Applying the polarising-field relation:}\) \[ P\propto\frac{|z|}{r^2} \] For magnesium: \[ P_{\mathrm{Mg^{2+}}}\propto\frac{2}{72^2} \] For calcium: \[ P_{\mathrm{Ca^{2+}}}\propto\frac{2}{100^2} \] \(\textbf{Calculating the comparison ratio:}\) \[ R=\frac{P_{\mathrm{Mg^{2+}}}}{P_{\mathrm{Ca^{2+}}}} \] \[ R=\frac{2/72^2}{2/100^2} \] \[ R=\left(\frac{100}{72}\right)^2 \] \[ R=\frac{10000}{5184} \] \[ R\approx1.93 \] \(\textbf{Interpreting the bonding tendency:}\) The smaller magnesium ion has greater charge density. It consequently exerts stronger polarising influence on the chloride electron cloud. Greater anion polarisation favours a larger covalent contribution to the interaction. This is consistent with Fajans' qualitative rules. \(\textbf{Connecting the result with periodicity:}\) Calcium has an additional occupied principal shell compared with magnesium. Its divalent ion is therefore larger under the stated convention. The difference in size modifies the polarising influence despite their common ionic charge. The computed ratio is a property of the stipulated field model, not a direct numerical ratio of percentage covalent character. \(\textbf{Final answer:}\) \[ \boxed{\frac{P_{\mathrm{Mg^{2+}}}}{P_{\mathrm{Ca^{2+}}}}\approx1.93} \]

Explanation: \(\textbf{Comparing the electronic structures:}\) Lithium, sodium and potassium each possess one valence s-electron. Their outer configurations are \(2s^1\), \(3s^1\) and \(4s^1\), respectively. Moving down the group introduces additional occupied principal shells. Their atomic radii therefore increase. \(\textbf{Explaining the melting-temperature trend:}\) Metallic bonding involves attraction between the positive metal-ion cores and delocalised valence electrons. The greater separation and changing charge density associated with larger alkali-metal atoms generally weaken metallic cohesion down the group. The supplied melting temperatures consequently decrease. This is a property of the condensed metallic solids. \(\textbf{Explaining first ionization enthalpy:}\) First ionization removes an electron from an isolated gaseous atom. The outer electron becomes farther from the nucleus and more strongly shielded on descending the group. Consequently, the first ionization enthalpy decreases. This is consistent with an increasing tendency toward electron loss in ordinary chemical reactions. \(\textbf{Reconciling water reactivity:}\) Potassium generally reacts more vigorously with water than lithium under comparable ordinary conditions. This behaviour is influenced by chemical energetics and reaction conditions. It does not require potassium to possess stronger metallic cohesion. Indeed, the lower melting temperature can influence the physical behaviour of a metal during an exothermic reaction. \(\textbf{Final interpretation:}\) Melting temperature measures a condensed-phase physical transition. First ionization enthalpy measures gas-phase electron removal. Observable chemical reaction vigour depends on the full reaction pathway. The third option correctly separates these properties while connecting them to the shared down-group size trend.

Explanation: \(\textbf{Relating the formulas to periodic groups:}\) Representative group-1 elements possess one outer valence electron. Group-2 elements possess two. Group-13 elements possess three. Group-14 elements possess four. The stated oxide formulas are consistent with these characteristic positive oxidation states. \(\textbf{Checking the oxide charge balances:}\) In \(\mathrm{M_2O}\), two M atoms collectively balance oxygen's \(-2\) contribution. Each M atom therefore has oxidation state \(+1\). In \(\mathrm{NO}\), N balances one oxide oxygen and has oxidation state \(+2\). In \(\mathrm{P_2O_3}\), the two P atoms collectively balance \(-6\), giving \(+3\) each. In \(\mathrm{QO_2}\), Q balances \(-4\) and therefore has oxidation state \(+4\). \(\textbf{Connecting the progression with oxide chemistry:}\) Strongly electropositive metals near the left side of a representative period generally form basic oxides. Moving rightward, metallic character decreases. Some intermediate metallic oxides exhibit amphoteric behaviour. Oxides of more non-metallic elements often exhibit acidic character. The precise classification of an individual oxide nevertheless depends on its actual chemical reactions. \(\textbf{Recognising the limitation:}\) The central-element oxidation states increase across the stipulated formula sequence. However, oxide character is not determined by the oxidation-state number alone. The change in metallic and non-metallic character provides the broader periodic explanation. The first option correctly identifies the formal states and the qualitative direction of oxide-character change.

Explanation: Statement I is correct because both O₂ and O₃ are elemental forms of oxygen. The oxidation state of an element in its uncombined elemental form is zero. This remains true whether its elemental molecules are diatomic or triatomic. Statement II is correct because chemical oxidising strength depends on the energetics of the complete reaction. Ozone and dioxygen have different molecular bonding arrangements. Their reactant and product energy relationships consequently differ. Ozone can therefore exhibit a stronger oxidising tendency under suitable conditions. Statement III is incorrect because atomic number describes the number of protons in one atom. Every oxygen atom has atomic number 8. Changing the number of oxygen atoms in a molecule does not change their individual nuclear charges. Statement IV is correct because allotropy changes the arrangement of atoms within an elemental substance. It can therefore change physical and chemical properties. However, each oxygen atom retains the same elemental identity and periodic position. The comparison also illustrates why formal oxidation state alone cannot rank the oxidising strengths of different elemental substances. Both allotropes begin with oxygen at oxidation state zero but need not have identical reaction energetics. Only Statements I, II and IV are correct.

Explanation: \(\textbf{Identifying copper's oxidation states:}\) Oxygen has oxidation state \(-2\) in both ordinary copper oxides. For \(\mathrm{Cu_2O}\), let the oxidation state of copper be \(x\). \[ 2x+(-2)=0 \] \[ x=+1 \] Thus, \(\mathrm{Cu_2O}\) contains copper in oxidation state \(+1\). For \(\mathrm{CuO}\), the corresponding charge balance gives: \[ x+(-2)=0 \] \[ x=+2 \] Therefore, copper is oxidised from \(+1\) to \(+2\) during the reaction. Molecular oxygen is reduced from oxidation state zero to \(-2\), so it acts as the oxidising agent. \(\textbf{Determining the limiting reactant:}\) The balanced equation gives: \[ 2\ \mathrm{mol\ Cu_2O}\longrightarrow4\ \mathrm{mol\ CuO} \] for every \(1\ \mathrm{mol}\) of oxygen consumed. The oxygen required to oxidise all the initial copper(I) oxide is: \[ n(\mathrm{O_2})_{\mathrm{required}}=\frac{0.120}{2} \] \[ n(\mathrm{O_2})_{\mathrm{required}}=0.060\ \mathrm{mol} \] However, only \(0.040\ \mathrm{mol}\) of oxygen is available. Since the available quantity is smaller than the required quantity, oxygen is the limiting reactant. \(\textbf{Calculating the unreacted copper(I) oxide:}\) From the balanced equation: \[ n(\mathrm{Cu_2O})_{\mathrm{consumed}}=2n(\mathrm{O_2}) \] Substituting the available oxygen quantity: \[ n(\mathrm{Cu_2O})_{\mathrm{consumed}}=2(0.040) \] \[ n(\mathrm{Cu_2O})_{\mathrm{consumed}}=0.080\ \mathrm{mol} \] The unreacted quantity of copper(I) oxide is: \[ n(\mathrm{Cu_2O})_{\mathrm{remaining}}=0.120-0.080 \] \[ n(\mathrm{Cu_2O})_{\mathrm{remaining}}=0.040\ \mathrm{mol} \] \(\textbf{Calculating the copper(II) oxide formed:}\) One mole of oxygen produces four moles of CuO. Therefore: \[ n(\mathrm{CuO})=4n(\mathrm{O_2}) \] \[ n(\mathrm{CuO})=4(0.040) \] \[ n(\mathrm{CuO})=0.160\ \mathrm{mol} \] \(\textbf{Determining the Cu(I):Cu(II) atomic ratio:}\) Each formula unit of \(\mathrm{Cu_2O}\) contains two Cu(I) atoms. Each formula unit of \(\mathrm{CuO}\) contains one Cu(II) atom. The amount of copper atoms remaining in oxidation state \(+1\) is: \[ n(\mathrm{Cu(I)})=2(0.040)=0.080\ \mathrm{mol} \] The amount of copper atoms present in oxidation state \(+2\) is: \[ n(\mathrm{Cu(II)})=1(0.160)=0.160\ \mathrm{mol} \] Hence: \[ \mathrm{Cu(I):Cu(II)}=0.080:0.160 \] \[ \mathrm{Cu(I):Cu(II)}=1:2 \] The conservation check gives \(0.080+0.160=0.240\ \mathrm{mol}\) of copper atoms, equal to the initial \(2(0.120)=0.240\ \mathrm{mol}\). \(\textbf{Connecting oxidation states with d-electron configurations:}\) Neutral copper has the ground-state configuration: \[ \mathrm{Cu}:[\mathrm{Ar}]\,3d^{10}\,4s^1 \] Formation of \(\mathrm{Cu^+}\) involves removal of the outer \(4s\) electron: \[ \mathrm{Cu^+}:[\mathrm{Ar}]\,3d^{10} \] Formation of \(\mathrm{Cu^{2+}}\) involves removal of one further electron from the \(3d\) subshell: \[ \mathrm{Cu^{2+}}:[\mathrm{Ar}]\,3d^9 \] Therefore, Cu(I) has a completely filled d-subshell, whereas Cu(II) has an incompletely filled d-subshell. The existence of the \(3d^9\) Cu(II) state is consistent with copper's transition-element character, despite its filled d-subshell in the neutral atom. \(\textbf{Final inference:}\) Oxygen is limiting, the final Cu(I):Cu(II) atomic ratio is \(1:2\), and isolated \(\mathrm{Cu^{2+}}\) has the \(3d^9\) configuration. The result simultaneously follows from oxidation-state assignment, limiting-reactant analysis, copper-atom conservation and d-block electronic configuration.

Explanation: Beryllium belongs to group 2 and period 2. Aluminium belongs to group 13 and period 3. They therefore occupy diagonally related positions. Beryllium commonly exhibits oxidation state \(+2\), while aluminium commonly exhibits \(+3\). Their chloride formulas are correspondingly BeCl₂ and AlCl₃. Both metal centres can exert substantial polarising influence because of their relatively small sizes and positive charges. This contributes to appreciable covalent character in their chlorides. The electron-deficient bonding arrangements of these compounds can also permit interaction with electron-pair donors. Water can therefore participate in hydrolysis under suitable conditions. Sodium chloride, in contrast, is predominantly ionic under ordinary conditions. Its dissolution in water is not equivalent to the characteristic hydrolysis of electron-deficient covalent chlorides. The resemblance between beryllium and aluminium does not require identical formulas or identical formal charges. Diagonal relationships reflect similarities in electronic and chemical behaviour that can coexist with different group positions. The second option correctly identifies the chemical basis of the resemblance without equating hydrolysis with oxidation or chlorine liberation.

Explanation: \(\textbf{Examining chlorine's valence occupancy:}\) Chlorine belongs to group 17. Its neutral atom contains seven valence electrons. The outer \(3p\) subshell has one vacancy. Addition of an electron produces chloride with the argon electronic configuration. This is consistent with chlorine's favourable first electron-gain enthalpy. \(\textbf{Examining argon's valence occupancy:}\) Argon belongs to group 18. Its outer \(3s^2\,3p^6\) arrangement is completely filled. The corresponding shell closure contributes to argon's low ordinary chemical reactivity. An incoming electron cannot occupy an additional position within the already completed \(3p\) subshell. It would have to enter a higher-energy available orbital region. This makes ordinary first electron addition to isolated gaseous argon energetically unfavourable. \(\textbf{Comparing ionization:}\) Argon's completed outer-shell configuration also contributes to strong resistance against first electron removal. Chlorine has an incompletely filled p-subshell and a lower first ionization enthalpy. Argon's greater nuclear charge is relevant, but shell occupancy is also essential. \(\textbf{Interpreting the periodic boundary:}\) The general increase in non-metallic electron-accepting tendency toward the right of a representative period does not continue without qualification into the noble-gas column. The completed shell produces an important exception to any proposed monotonic reactivity trend. The third option correctly reconciles chlorine's electron-gain tendency with argon's closed-shell resistance.

Explanation: \(\textbf{Identifying the shared group characteristic:}\) Group-1 metals have one valence s-electron. Their atomic radii generally increase down the group. The first ionization enthalpy generally decreases as the outer electron occupies progressively larger and more shielded shells. This favours easier gas-phase electron removal. \(\textbf{Examining Condition II:}\) The specified measurement begins with isolated gaseous neutral atoms. It ends with isolated gaseous monovalent cations and electrons. It excludes the cohesive energy of the metallic solid and the subsequent interactions of ions with water. The first ionization enthalpy is therefore a precisely defined atomic property. \(\textbf{Examining Condition I:}\) The water reaction begins with condensed metallic pieces. It involves formation of aqueous metal-containing products and hydrogen gas. The observable intensity depends on the reaction rate, local heating, surface condition and transport of reactants and products. Equal-mass pieces also need not contain equal amounts of metal atoms. The measured vigour therefore does not represent only one isolated electron-removal step. \(\textbf{Resolving the proposed equivalence:}\) A lower first ionization enthalpy contributes to greater electron-loss tendency. However, it does not provide a direct numerical scale for reaction vigour. Thermodynamic favourability and kinetic rate are distinct concepts. The relative hydration of the product ions can also modify the overall energy balance. The second option correctly distinguishes the relevant physical and chemical processes.

Explanation: \(\textbf{Identifying the element from configuration:}\) The outer configuration \(3s^2\,3p^2\) contains four valence electrons. An element with four valence electrons in period 3 belongs to group 14. It is silicon. \(\textbf{Examining the chloride:}\) Silicon commonly exhibits covalency four. It forms SiCl₄, a molecular covalent compound under ordinary conditions. Each silicon atom is bonded to four chlorine atoms. The formula is therefore consistent with its group-based valence configuration. \(\textbf{Examining the oxide:}\) Silicon dioxide has the empirical formula SiO₂. It forms an extended covalent network rather than isolated molecular units analogous to SiCl₄. In SiO₂, oxygen has oxidation state \(-2\). Silicon consequently has formal oxidation state \(+4\). \(\textbf{Interpreting the reaction with alkali:}\) The supplied reaction shows that silicon dioxide can react with strong alkali to form a silicate. This is characteristic of acidic oxide behaviour under suitable conditions. The absence of rapid dissolution in water does not make SiO₂ a neutral oxide. \(\textbf{Integrating the structural information:}\) The chloride and oxide contain silicon in a positive formal oxidation state. However, the actual bonding arrangements differ. An empirical formula cannot by itself determine whether a compound is molecular or an extended network. The fourth option correctly uses electronic configuration, chemical formula and oxide reaction behaviour to identify the element.