201. Using the atomic masses \(\mathrm{Ti}=48\) and \(\mathrm{C}=12\), the percentage by mass of carbon in the non-stoichiometric carbide \(\mathrm{TiC_{0.5}}\) is closest to:
ⓐ. \(10.0\%\)
ⓑ. \(12.5\%\)
ⓒ. \(11.1\%\)
ⓓ. \(20.0\%\)
Correct Answer: \(11.1\%\)
Explanation: \( \textbf{Average formula:} \)
\[
\mathrm{TiC_{0.5}}
\]
\( \textbf{Mass contribution from titanium:} \)
\[
1\times48=48
\]
\( \textbf{Mass contribution from carbon:} \)
\[
0.5\times12=6
\]
\( \textbf{Total average formula mass:} \)
\[
48+6=54
\]
\( \textbf{Mass-percentage relation:} \)
\[
\%\mathrm{C}=\frac{\text{mass of carbon}}{\text{total mass}}\times100
\]
\( \textbf{Substitution:} \)
\[
\%\mathrm{C}=\frac{6}{54}\times100
\]
\( \textbf{Calculation:} \)
\[
\%\mathrm{C}=11.11\ldots\%
\]
\( \textbf{Rounded result:} \)
\[
\%\mathrm{C}\approx11.1\%
\]
\( \textbf{Final answer:} \) Carbon constitutes approximately \(11.1\%\) of the carbide by mass.
202. Which atom is least likely to occupy an ordinary interstitial site in a transition-metal lattice because of its comparatively large size?
ⓐ. \(\mathrm{Na}\)
ⓑ. \(\mathrm{H}\)
ⓒ. \(\mathrm{C}\)
ⓓ. \(\mathrm{N}\)
Correct Answer: \(\mathrm{Na}\)
Explanation: Interstitial sites are small gaps between the much larger host-metal atoms. Hydrogen, carbon, and nitrogen are sufficiently small to enter such spaces in many transition-metal lattices. Sodium atoms are considerably larger and cannot ordinarily fit into the same gaps without severely disrupting the structure. A larger metal atom is more likely to participate through substitution or a separate phase. Atomic size is therefore central to identifying possible interstitial occupants.
203. A learner states, “A non-stoichiometric interstitial compound is merely an uncontrolled mixture with no organised metal lattice.” The best evaluation is:
ⓐ. correct, because a fractional composition cannot occur in an ordered crystalline solid
ⓑ. incorrect, because an ordered metal lattice can have variably occupied interstices
ⓒ. correct, because every non-stoichiometric solid must have an amorphous structure
ⓓ. incorrect only because every interstitial site must remain fully occupied
Correct Answer: incorrect, because an ordered metal lattice can have variably occupied interstices
Explanation: Non-stoichiometry does not require loss of crystalline order. The host metal atoms can retain a regular lattice. Variability may arise because some interstitial sites are occupied while others remain vacant. The average composition then need not be a simple whole-number ratio. Such a solid can still have reproducible structural and physical properties despite its variable occupancy.
204. Transition metals readily form alloys with one another mainly because many of them have:
ⓐ. completely different crystal structures and extremely different atomic sizes
ⓑ. no metallic bonding in the pure state
ⓒ. similar atomic radii and compatible metallic structures
ⓓ. identical atomic numbers
Correct Answer: similar atomic radii and compatible metallic structures
Explanation: Atoms of similar size can replace one another in a metallic lattice without causing excessive distortion. Compatible crystal structures further support mixing in the solid state. The resulting material may form a substitutional solid solution. Transition metals often satisfy these size and structural conditions. Alloy formation does not require the elements to have identical atomic numbers or identical electronic configurations.
205. A substitutional solid-solution alloy is produced when:
ⓐ. small non-metal atoms occupy only the gaps between host atoms
ⓑ. atoms of one metal replace some atoms of another metal in the lattice
ⓒ. the components remain as separate macroscopic metal layers
ⓓ. every atom loses all of its valence electrons to form a molecular gas
Correct Answer: atoms of one metal replace some atoms of another metal in the lattice
Explanation: In a substitutional alloy, atoms of the added metal occupy ordinary lattice positions. They replace a fraction of the host-metal atoms rather than fitting only into interstitial gaps. Similar atomic radii make this replacement structurally favourable. The resulting material can remain a single metallic solid solution. This arrangement differs from an interstitial alloy, in which much smaller atoms occupy spaces between host atoms.
206. A corrosion-resistant material is required for kitchen equipment and chemical-processing vessels. The most suitable choice is stainless steel because alloying iron with elements such as chromium and nickel:
ⓐ. improves corrosion resistance and metallic strength
ⓑ. converts the entire material into a brittle molecular crystal
ⓒ. removes all mobile electrons from the metal
ⓓ. makes the alloy chemically identical to pure iron
Correct Answer: improves corrosion resistance and metallic strength
Explanation: Stainless steel is an iron-based alloy commonly containing chromium and often nickel. Chromium promotes formation of a protective surface layer that resists further corrosion. Nickel can improve structural stability and mechanical performance. The alloy retains the useful strength and workability of a metallic material. Its improved behaviour arises because alloying changes composition and microstructure rather than leaving the iron chemically unchanged.
207. Nichrome is well suited for electrical heating elements because it combines:
ⓐ. very low resistance together with a low melting point and rapid oxidation
ⓑ. appreciable resistance together with thermal and oxidation resistance
ⓒ. complete electrical insulation together with rapid high-temperature corrosion
ⓓ. high molecular volatility together with very low mechanical strength
Correct Answer: appreciable resistance together with thermal and oxidation resistance
Explanation: Nichrome is an alloy based mainly on nickel and chromium. Its relatively high electrical resistance causes electrical energy to be converted efficiently into heat. It can tolerate elevated temperatures without melting or weakening rapidly. Chromium also improves resistance to oxidation at the hot surface. These properties make nichrome more suitable for heating coils than a highly conductive pure metal.
208. Assertion: An alloy may be harder than either of the pure metals used to form it.
Reason: Atoms of different sizes can distort the lattice and hinder the movement of layers or defects through the solid.
ⓐ. Assertion is true, but Reason is false
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is false, but Reason is true
ⓓ. Both Assertion and Reason are true, and Reason explains Assertion
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Pure metal atoms are often arranged in relatively regular layers that can slide under an applied force. Introducing atoms of another size distorts this regular arrangement. The distortion obstructs easy movement of dislocations and reduces layer slippage. Greater force is then needed to deform the material. This microscopic effect explains why alloying commonly increases hardness and strength.
209. A \(250\,g\) stainless-steel sample contains \(18\%\) chromium and \(8\%\) nickel by mass. Pure chromium is added until chromium forms \(20\%\) of the new total mass. The mass of chromium added and the new nickel percentage are:
ⓐ. \(6.25\,g\) and \(7.80\%\)
ⓑ. \(5.00\,g\) and \(7.84\%\)
ⓒ. \(6.25\,g\) and \(8.00\%\)
ⓓ. \(12.50\,g\) and \(7.62\%\)
Correct Answer: \(6.25\,g\) and \(7.80\%\)
Explanation: \( \textbf{Initial chromium mass:} \)
\[
m_{\mathrm{Cr}}=0.18\times250
\]
\[
m_{\mathrm{Cr}}=45.0\,g
\]
\( \textbf{Initial nickel mass:} \)
\[
m_{\mathrm{Ni}}=0.08\times250
\]
\[
m_{\mathrm{Ni}}=20.0\,g
\]
Let the mass of pure chromium added be \(x\,g\).
\( \textbf{New chromium fraction:} \)
\[
\frac{45.0+x}{250+x}=0.20
\]
\( \textbf{Solve the composition equation:} \)
\[
45.0+x=50.0+0.20x
\]
\[
0.80x=5.00
\]
\[
x=6.25\,g
\]
\( \textbf{New total mass:} \)
\[
250+6.25=256.25\,g
\]
The nickel mass remains \(20.0\,g\) because only chromium was added.
\( \textbf{New nickel percentage:} \)
\[
\%\mathrm{Ni}=\frac{20.0}{256.25}\times100
\]
\[
\%\mathrm{Ni}=7.80\%
\]
\( \textbf{Final answer:} \) The sample requires \(6.25\,g\) of chromium, and its new nickel content is \(7.80\%\). Adding one component increases the total mass and therefore decreases the mass percentages of components whose masses remain unchanged.
210. As the oxidation state of a transition metal increases, the acid-base character of its oxides generally changes from:
ⓐ. acidic to neutral to basic
ⓑ. amphoteric to basic to neutral
ⓒ. basic to amphoteric to acidic
ⓓ. acidic to amphoteric to basic
Correct Answer: basic to amphoteric to acidic
Explanation: In a low oxidation state, the metal-oxygen bond has greater ionic character. Such oxides commonly react with acids and behave as basic oxides. At an intermediate oxidation state, an oxide may react with both acids and bases and therefore becomes amphoteric. A high oxidation state increases the covalent character of the metal-oxygen bonds. The corresponding oxide then commonly behaves as an acidic oxide or as the anhydride of an oxoacid.
211. Examine the classifications of chromium oxides.
| Row | Oxide | Chromium oxidation state | Acid-base character |
|---|
| P | \(\mathrm{CrO}\) | \(+2\) | Basic |
| Q | \(\mathrm{Cr_2O_3}\) | \(+3\) | Amphoteric |
| R | \(\mathrm{CrO_3}\) | \(+6\) | Acidic |
The correctly classified rows are:
ⓐ. P, Q and R
ⓑ. P and Q only
ⓒ. Q and R only
ⓓ. P and R only
Correct Answer: P, Q and R
Explanation: Chromium is in the \(+2\) state in \(\mathrm{CrO}\), and this low-state oxide is basic. In \(\mathrm{Cr_2O_3}\), chromium is \(+3\), and the oxide shows amphoteric behaviour. Chromium reaches \(+6\) in \(\mathrm{CrO_3}\), which is an acidic oxide. The sequence demonstrates increasing acidic character with increasing oxidation state. All three rows therefore agree with the general transition-metal oxide trend.
212. An oxide of a transition metal reacts with both a strong acid and a strong alkali. The oxide should be classified as:
ⓐ. exclusively basic
ⓑ. amphoteric
ⓒ. exclusively acidic
ⓓ. chemically inert
Correct Answer: amphoteric
Explanation: An amphoteric oxide displays both acidic and basic behaviour. It reacts with acids as though it were a base and with strong alkalis as though it were an acid. Intermediate oxidation-state oxides of transition metals commonly show this dual behaviour. \(\mathrm{Cr_2O_3}\) is a standard example. Reaction with only an acid would suggest basic character, while reaction with only an alkali would suggest acidic character.
213. A graph plots acidic character of the oxides of one transition metal against the metal oxidation state. The most reasonable graph would:
ⓐ. fall continuously as the oxidation state increases
ⓑ. remain horizontal for all oxidation states
ⓒ. alternate regularly between maximum acidity and maximum basicity
ⓓ. rise overall, possibly through an amphoteric region
Correct Answer: rise overall, possibly through an amphoteric region
Explanation: Low oxidation-state oxides commonly have relatively ionic metal-oxygen bonds and show basic behaviour. As oxidation state rises, the metal centre exerts a stronger polarising effect and the bonding becomes more covalent. Intermediate oxides may react with both acids and bases. High oxidation-state oxides commonly behave as acidic oxides. The graph should therefore show increasing acidic character rather than a constant or decreasing trend.
214. The expected order of increasing acidic character is:
ⓐ. \(\mathrm{MnO}\lt\mathrm{MnO_2}\lt\mathrm{Mn_2O_7}\)
ⓑ. \(\mathrm{Mn_2O_7}\lt\mathrm{MnO_2}\lt\mathrm{MnO}\)
ⓒ. \(\mathrm{MnO_2}\lt\mathrm{Mn_2O_7}\lt\mathrm{MnO}\)
ⓓ. \(\mathrm{MnO}\lt\mathrm{Mn_2O_7}\lt\mathrm{MnO_2}\)
Correct Answer: \(\mathrm{MnO}\lt\mathrm{MnO_2}\lt\mathrm{Mn_2O_7}\)
Explanation: Manganese is in oxidation states \(+2\), \(+4\), and \(+7\) in the three oxides. \(\mathrm{MnO}\) is a low oxidation-state oxide and is predominantly basic. \(\mathrm{MnO_2}\) has intermediate character and can show amphoteric behaviour. \(\mathrm{Mn_2O_7}\) contains manganese in its highest oxidation state and is strongly acidic. The acidic character consequently increases in the stated order.
215. A student states, “Because transition elements are metals, all their oxides must be basic.” The best correction is:
ⓐ. all transition-metal oxides are neutral because oxygen has oxidation state \(-2\)
ⓑ. only oxides containing a \(d^{10}\) metal centre can be acidic
ⓒ. transition-metal oxides are acidic only when dissolved in organic solvents
ⓓ. basic at low states, amphoteric at intermediate states, acidic at high states
Correct Answer: basic at low states, amphoteric at intermediate states, acidic at high states
Explanation: Metallic character alone does not determine the acid-base behaviour of every oxide of an element. A transition metal can occur in several oxidation states, and the nature of its bonding with oxygen changes accordingly. Low-state oxides generally have greater ionic character and are basic. Intermediate-state oxides may be amphoteric. High-state oxides have greater covalent character and commonly behave as acidic oxides.
216. A transition metal X forms \(\mathrm{XO}\), \(\mathrm{X_2O_3}\), and \(\mathrm{XO_3}\). Experiments show that \(\mathrm{XO}\) neutralises acids, \(\mathrm{X_2O_3}\) reacts with both acids and alkalis, and \(\mathrm{XO_3}\) forms an oxoacid with water. The observations are best explained by:
ⓐ. decreasing oxidation state producing increasing acidic character
ⓑ. greater oxidation state increases covalent and acidic character
ⓒ. identical bonding in all three oxides
ⓓ. a change in the number of protons in X between the oxides
Correct Answer: greater oxidation state increases covalent and acidic character
Explanation: The oxidation states of X are \(+2\), \(+3\), and \(+6\) in the three oxides. The reaction of \(\mathrm{XO}\) with acids identifies it as a basic oxide. Reaction of \(\mathrm{X_2O_3}\) with both acids and alkalis indicates amphoteric behaviour. Formation of an oxoacid from \(\mathrm{XO_3}\) identifies it as an acidic oxide. The sequence reflects a shift from more ionic to more covalent metal-oxygen bonding as the oxidation state rises.
217. Chromium(VI) oxide reacts with hydroxide ions according to:
\[
\mathrm{CrO_3+2OH^-\rightarrow CrO_4^{2-}+H_2O}.
\]
The amount of \(\mathrm{OH^-}\) required to react completely with \(0.25\,mol\) of \(\mathrm{CrO_3}\) is:
ⓐ. \(0.125\,mol\)
ⓑ. \(0.25\,mol\)
ⓒ. \(0.50\,mol\)
ⓓ. \(1.00\,mol\)
Correct Answer: \(0.50\,mol\)
Explanation: \( \textbf{Balanced equation:} \)
\[
\mathrm{CrO_3+2OH^-\rightarrow CrO_4^{2-}+H_2O}
\]
\( \textbf{Stoichiometric relation:} \)
\[
1\,mol\ \mathrm{CrO_3}:2\,mol\ \mathrm{OH^-}
\]
\( \textbf{Amount of chromium(VI) oxide:} \)
\[
n(\mathrm{CrO_3})=0.25\,mol
\]
\( \textbf{Required hydroxide amount:} \)
\[
n(\mathrm{OH^-})=2n(\mathrm{CrO_3})
\]
\( \textbf{Substitution:} \)
\[
n(\mathrm{OH^-})=2\times0.25
\]
\( \textbf{Calculation:} \)
\[
n(\mathrm{OH^-})=0.50\,mol
\]
\( \textbf{Acid-base interpretation:} \) \(\mathrm{CrO_3}\) consumes hydroxide because it is an acidic oxide.
\( \textbf{Final answer:} \) Complete reaction requires \(0.50\,mol\) of \(\mathrm{OH^-}\).
218. The reaction
\[
\mathrm{CrO+2H^+\rightarrow Cr^{2+}+H_2O}
\]
demonstrates that \(\mathrm{CrO}\) behaves as:
ⓐ. an oxidising agent only
ⓑ. an acidic oxide
ⓒ. a neutral oxide
ⓓ. a basic oxide
Correct Answer: a basic oxide
Explanation: A basic oxide reacts with an acid or with hydrogen ions to form a salt-like metal ion and water. In the given reaction, oxide oxygen combines with \(2\mathrm{H^+}\) to produce \(\mathrm{H_2O}\). Chromium remains in oxidation state \(+2\), so the reaction is not being classified primarily as a redox process. Consumption of hydrogen ions reveals the basic character of \(\mathrm{CrO}\). This behaviour is consistent with chromium being in a relatively low oxidation state.
219. Oxygen and fluorine are especially effective at stabilising high oxidation states of transition metals because they:
ⓐ. have low electronegativities and donate electrons completely to the metal
ⓑ. are high polarizability spreads charge away from the metal centre
ⓒ. prevent the formation of strong metal-ligand bonds
ⓓ. are highly electronegative and form strong bonds to oxidised metals
Correct Answer: are highly electronegative and form strong bonds to oxidised metals
Explanation: A metal in a high oxidation state has a strong tendency to attract electron density. Highly electronegative oxygen and fluorine form strong bonds with such a metal centre. These bonds lower the energy of the high-state compound and make it accessible. Oxygen can additionally support strong multiple-bond character in oxides and oxoanions. High oxidation states are therefore frequently found in oxides, oxoanions, and fluorides rather than as bare monatomic ions.
220. Manganese in oxidation state \(+7\) is stable in \(\mathrm{MnO_4^-}\), but a simple aqueous \(\mathrm{Mn^{7+}}\) ion is not common. The best explanation is:
ⓐ. manganese has oxidation state \(+2\) in permanganate
ⓑ. oxygen reduces manganese to its elemental state
ⓒ. strong \(\mathrm{Mn-O}\) bonding and charge delocalisation stabilise \(+7\)
ⓓ. a monatomic \(7+\) ion would have lower charge density than permanganate
Correct Answer: strong \(\mathrm{Mn-O}\) bonding and charge delocalisation stabilise \(+7\)
Explanation: A bare \(\mathrm{Mn^{7+}}\) ion would have an exceptionally high positive charge concentrated on a small centre. Such a species would be extremely unfavourable in ordinary aqueous solution. In permanganate, manganese is strongly bonded to four oxygen atoms. The bonding framework distributes electron density and stabilises the formal \(+7\) oxidation state. The oxidation state remains \(+7\), but it is supported by the complete oxoanion rather than by an isolated monatomic ion.
