101. A data passage describes a transition series in which melting points rise toward the middle, one central element lies unexpectedly below neighbouring values, and the values fall strongly near the \(d^{10}\) end. This pattern shows that:
ⓐ. atomic mass alone determines the melting point throughout a transition series
ⓑ. a broad metallic-bonding trend is modified by configuration-dependent anomalies
ⓒ. the middle of every transition series must contain its lowest-melting element
ⓓ. a completely filled \(d\) subshell invariably strengthens metallic bonding
Correct Answer: a broad metallic-bonding trend is modified by configuration-dependent anomalies
Explanation: The rise toward the middle is consistent with stronger metallic bonding when several valence electrons can participate. The unexpectedly low central value represents an anomaly caused by element-specific electronic or structural factors. The sharp fall near the \(d^{10}\) end reflects weaker participation of filled \(d\) electrons in metallic cohesion. Atomic mass alone cannot reproduce these features. The passage therefore supports a broad chemical trend without requiring every point to lie on a smooth curve.
102. Transition metals generally have high densities because they commonly possess:
ⓐ. high atomic masses and compact atomic volumes
ⓑ. low atomic masses and very large atomic radii
ⓒ. loosely packed metallic lattices with large atomic volumes
ⓓ. a small number of occupied electron shells
Correct Answer: high atomic masses and compact atomic volumes
Explanation: Density depends on the mass contained within a given volume. Transition-metal atoms often have substantial atomic masses, while their atomic radii change only moderately across a series. Their metallic structures also pack atoms relatively closely. The combination of increasing mass and modest atomic volume produces high densities. Density is not determined by atomic mass alone, because lattice packing and atomic size also affect the occupied volume.
103. Use the graph description below. The vertical axis represents density and the horizontal axis moves across much of a transition series. Curve P generally rises with small irregularities. Curve Q falls steadily to zero. Curve R remains exactly constant, while Curve S alternates between extremely high and negligible values. The most suitable curve is:
ⓐ. Curve R
ⓑ. Curve S
ⓒ. Curve Q
ⓓ. Curve P
Correct Answer: Curve P
Explanation: Atomic mass generally increases across a transition series. Atomic radius decreases initially and then changes only slightly, so atomic volume does not rise in proportion to mass. Density therefore tends to increase across much of the series. Crystal structures and individual atomic volumes introduce irregularities, so a perfectly smooth line is not expected. A generally rising curve with modest deviations captures the combined mass and volume effects.
104. Equal-mass samples of transition metals X and Y have densities \(7.2\,g\,cm^{-3}\) and \(9.0\,g\,cm^{-3}\), respectively. Together, the two samples occupy \(45.0\,cm^3\). The mass of each sample and the sample occupying the larger volume are:
ⓐ. \(144\,g\); sample X
ⓑ. \(180\,g\); sample Y
ⓒ. \(180\,g\); sample X
ⓓ. \(360\,g\); sample Y
Correct Answer: \(180\,g\); sample X
Explanation: \( \textbf{Let the mass of each sample be:} \)
\[
m\,g
\]
\( \textbf{Volume relation:} \)
\[
V=\frac{m}{\rho}
\]
\( \textbf{Volume of sample X:} \)
\[
V_X=\frac{m}{7.2}
\]
\( \textbf{Volume of sample Y:} \)
\[
V_Y=\frac{m}{9.0}
\]
\( \textbf{Combined-volume condition:} \)
\[
\frac{m}{7.2}+\frac{m}{9.0}=45.0
\]
\( \textbf{Simplification:} \)
\[
m\left(\frac{5}{36}+\frac{4}{36}\right)=45.0
\]
\[
\frac{m}{4}=45.0
\]
\[
m=180\,g
\]
\( \textbf{Individual volumes:} \)
\[
V_X=\frac{180}{7.2}=25.0\,cm^3
\]
\[
V_Y=\frac{180}{9.0}=20.0\,cm^3
\]
\( \textbf{Final answer:} \) Each sample has a mass of \(180\,g\), and sample X occupies the larger volume because it has the lower density.
105. Later \(5d\) transition metals can possess exceptionally high densities mainly because they combine:
ⓐ. large atomic masses and compact atomic sizes
ⓑ. small atomic masses with expanded electron clouds
ⓒ. loose metallic packing with enlarged atomic volumes
ⓓ. atomic volumes increasing faster than atomic masses
Correct Answer: large atomic masses and compact atomic sizes
Explanation: The \(5d\) elements contain heavy nuclei and therefore have large atomic masses. Lanthanoid contraction keeps many of their atomic radii smaller than a simple extra-shell argument would predict. Their atoms can consequently pack substantial mass into a relatively small volume. This produces some of the highest densities among metallic elements. The result depends on both mass and compactness rather than on atomic mass alone.
106. Metal P has relative atomic mass \(100\) and relative atomic volume \(20.0\), while metal Q has relative atomic mass \(120\) and relative atomic volume \(22.0\). Metal R has relative atomic mass \(150\) and the same relative density as Q. Using
\[
\rho\propto\frac{\text{relative atomic mass}}{\text{relative atomic volume}},
\]
the required relative atomic volume of R and the density order are:
ⓐ. \(25.0\); \(\rho_R\gt\rho_Q\gt\rho_P\)
ⓑ. \(27.5\); \(\rho_P\gt\rho_Q=\rho_R\)
ⓒ. \(27.5\); \(\rho_Q=\rho_R\gt\rho_P\)
ⓓ. \(33.0\); \(\rho_P=\rho_Q\gt\rho_R\)
Correct Answer: \(27.5\); \(\rho_Q=\rho_R\gt\rho_P\)
Explanation: \( \textbf{Relative density of P:} \)
\[
\rho_P\propto\frac{100}{20.0}=5.00
\]
\( \textbf{Relative density of Q:} \)
\[
\rho_Q\propto\frac{120}{22.0}=\frac{60}{11}
\]
\( \textbf{Condition for R:} \)
\[
\rho_R=\rho_Q
\]
Let the relative atomic volume of R be \(V_R\).
\[
\frac{150}{V_R}=\frac{120}{22.0}
\]
\( \textbf{Solve for \(V_R\):} \)
\[
V_R=\frac{150\times22.0}{120}
\]
\[
V_R=27.5
\]
\( \textbf{Density comparison:} \)
\[
\rho_Q=\rho_R=\frac{60}{11}\approx5.45
\]
\[
\rho_P=5.00
\]
\( \textbf{Final answer:} \) R requires a relative atomic volume of \(27.5\), and the order is \(\rho_Q=\rho_R\gt\rho_P\). Equal atomic masses are not required for two metals to have equal densities because atomic volume also contributes.
107. Transition elements commonly exhibit variable oxidation states because:
ⓐ. only electrons belonging to the inner noble-gas core participate in bonding
ⓑ. the \((n-1)d\) subshell remains completely filled in all their compounds
ⓒ. the outer \(ns\) electrons are too strongly bound to be removed during oxidation
ⓓ. the \(ns\) and \((n-1)d\) electrons have similar energies and can both enter bonding
Correct Answer: the \(ns\) and \((n-1)d\) electrons have similar energies and can both enter bonding
Explanation: The outer \(ns\) and penultimate \((n-1)d\) subshells of transition elements are relatively close in energy. Electrons from either set may therefore participate in bonding or be removed during oxidation. Different numbers of available electrons can be involved in different compounds, producing several oxidation states for the same element. Main-group elements often show a narrower range because their valence subshells are more clearly separated from the inner electrons. Variable oxidation state is thus rooted in accessible valence-electron distributions rather than in metallic character alone.
108. Assertion: Transition elements near the middle of a series generally show a wider range of oxidation states than elements near either end.
Reason: Near the middle, a larger number of \(ns\) and \((n-1)d\) electrons can participate in bonding without the \(d\) subshell being initially empty or completely filled.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Early transition elements have relatively few \(d\) electrons available, while elements near the end approach a filled \(d^{10}\) subshell. Middle members possess several \(d\) electrons together with outer \(s\) electrons that may participate in bonding. This permits access to several electron-loss or electron-sharing patterns. The oxidation-state range consequently becomes broad near the middle of the series. The Reason directly links the number of available valence electrons with the observed variability.
109. A metal atom has the valence configuration \((n-1)d^5ns^2\). Compared with an element having \((n-1)d^{10}ns^2\), the first metal is more likely to show several oxidation states because it:
ⓐ. contains no \((n-1)d\) valence electrons
ⓑ. has several \(d\) and \(s\) electrons available for bonding
ⓒ. must lose all seven valence electrons in every compound
ⓓ. possesses an inert \(d^5\) subshell under all conditions
Correct Answer: has several \(d\) and \(s\) electrons available for bonding
Explanation: The \(d^5s^2\) atom has seven electrons outside the noble-gas core that may participate to different extents in bonding. It is not necessary for all seven to be removed in every compound. Different bonding environments can stabilise several oxidation states. A \(d^{10}s^2\) element has a filled \(d\) subshell, and its common chemistry often involves mainly the two outer \(s\) electrons. Half-filled stability influences particular states but does not make the \(d^5\) configuration chemically inert.
110. A learner states, “Variable oxidation states arise only because transition metals can lose their outer \(ns\) electrons.” The best correction is:
ⓐ. the statement is incorrect because only noble-gas-core electrons are removed
ⓑ. the statement is correct because \(d\) electrons never take part in bonding
ⓒ. the statement is incomplete because \(d\) electrons may also be removed
ⓓ. the statement is correct only for elements near the middle of a series
Correct Answer: the statement is incomplete because \(d\) electrons may also be removed
Explanation: Loss of the outer \(ns\) electrons commonly produces the \(+2\) state in many first-row transition elements. Higher oxidation states require participation or removal of additional \((n-1)d\) electrons. The closeness in energy of the two subshells makes this possible. Restricting oxidation to the \(ns\) electrons would predict only a narrow set of states and would fail to explain compounds containing metals in high oxidation states. Core electrons remain largely unaffected in ordinary transition-metal chemistry.
111. A group-\(6\) transition element forms compounds in the \(+2\), \(+3\), and \(+6\) oxidation states. The number of \(d\) electrons remaining in these states is respectively:
ⓐ. \(d^6,\ d^5,\ d^0\)
ⓑ. \(d^4,\ d^3,\ d^0\)
ⓒ. \(d^2,\ d^3,\ d^6\)
ⓓ. \(d^4,\ d^5,\ d^6\)
Correct Answer: \(d^4,\ d^3,\ d^0\)
Explanation: \( \textbf{Group number:} \) The element belongs to group \(6\).
\( \textbf{For oxidation state } +2\textbf{:} \)
\[
d=6-2=4
\]
\( \textbf{For oxidation state } +3\textbf{:} \)
\[
d=6-3=3
\]
\( \textbf{For oxidation state } +6\textbf{:} \)
\[
d=6-6=0
\]
\( \textbf{Resulting sequence:} \)
\[
d^4,\ d^3,\ d^0
\]
\( \textbf{Chemical meaning:} \) Increasing oxidation state removes additional valence electrons and lowers the \(d\)-electron count.
\( \textbf{High-state check:} \) The \(+6\) state leaves no \(d\) electrons and is therefore \(d^0\).
\( \textbf{Final answer:} \) The three configurations are \(d^4,\ d^3,\) and \(d^0\).
112. A metal forms a low oxidation-state chloride and a much higher oxidation-state oxoanion. The high oxidation state is more readily stabilised in the oxoanion because:
ⓐ. oxygen forms strong bonds to highly charged metal centres
ⓑ. the metal gains \(d\) electrons as its oxidation state increases
ⓒ. oxoanions contain no electronegative atoms
ⓓ. high oxidation states are stable only in elemental metals
Correct Answer: oxygen forms strong bonds to highly charged metal centres
Explanation: Highly electronegative oxygen stabilises a strongly positive metal centre through strong metal-oxygen bonding. Multiple bonding and charge delocalisation within an oxoanion can further support the high oxidation state. A simple chloride may stabilise lower states more effectively under comparable conditions. Increasing oxidation state removes electrons rather than adding \(d\) electrons. High oxidation states are therefore commonly encountered in oxides and oxoanions rather than as isolated aqueous cations.
113. A graph plots the maximum oxidation state of the first transition-series elements from \(\mathrm{Sc}\) to \(\mathrm{Zn}\). The expected broad shape is:
ⓐ. a steady decrease from scandium to zinc with no intermediate maximum
ⓑ. a horizontal line at \(+2\) for every member of the series
ⓒ. a minimum near manganese followed by a steady increase toward zinc
ⓓ. a rise to a maximum near manganese followed by a decline toward zinc
Correct Answer: a rise to a maximum near manganese followed by a decline toward zinc
Explanation: The highest oxidation state generally increases from scandium toward manganese as more \(d\) electrons become available for bonding. Manganese reaches the broad maximum of \(+7\). Beyond manganese, increasingly filled \(d\) subshells make removal or sharing of all available electrons less favourable. The maximum oxidation state consequently declines toward zinc. The graph is a broad trend and does not imply that every intermediate state is equally stable for every element.
114. The sequence of highest oxidation states for \(\mathrm{Sc}\), \(\mathrm{Ti}\), \(\mathrm{V}\), \(\mathrm{Cr}\), and \(\mathrm{Mn}\) is:
ⓐ. \(+3,\ +4,\ +5,\ +6,\ +7\)
ⓑ. \(+2,\ +3,\ +4,\ +5,\ +6\)
ⓒ. \(+3,\ +3,\ +4,\ +5,\ +5\)
ⓓ. \(+1,\ +2,\ +3,\ +4,\ +5\)
Correct Answer: \(+3,\ +4,\ +5,\ +6,\ +7\)
Explanation: Scandium, titanium, vanadium, chromium, and manganese belong to groups \(3\), \(4\), \(5\), \(6\), and \(7\), respectively. Up to manganese, the highest oxidation state broadly corresponds to the group number. The sequence therefore rises by one unit at each step. These highest states are often stabilised in compounds containing oxygen or fluorine. The pattern changes after manganese, where the maximum accessible state begins to decrease.
115. The oxidation state of chromium in \(\mathrm{CrO_4^{2-}}\) and manganese in \(\mathrm{MnO_4^-}\) is respectively:
ⓐ. \(+4\) and \(+6\)
ⓑ. \(+6\) and \(+6\)
ⓒ. \(+6\) and \(+7\)
ⓓ. \(+7\) and \(+6\)
Correct Answer: \(+6\) and \(+7\)
Explanation: \( \textbf{For chromate:} \) Let the oxidation state of chromium be \(x\).
\[
x+4(-2)=-2
\]
\[
x-8=-2
\]
\[
x=+6
\]
\( \textbf{For permanganate:} \) Let the oxidation state of manganese be \(y\).
\[
y+4(-2)=-1
\]
\[
y-8=-1
\]
\[
y=+7
\]
\( \textbf{Trend connection:} \) Chromium and manganese reach the high states expected for groups \(6\) and \(7\).
\( \textbf{Final answer:} \) The oxidation states are \(+6\) in \(\mathrm{CrO_4^{2-}}\) and \(+7\) in \(\mathrm{MnO_4^-}\).
116. Assertion: The highest oxidation state generally corresponds to the group number for the early first-row transition elements up to manganese.
Reason: All available \(ns\) and \((n-1)d\) valence electrons may participate in bonding in their highest oxidation states.
ⓐ. Assertion is true, but Reason is false
ⓑ. Both Assertion and Reason are true, and Reason explains Assertion
ⓒ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Scandium through manganese belong to groups \(3\) through \(7\). Their highest oxidation states broadly increase from \(+3\) to \(+7\). In these states, the available outer \(s\) and developing \(d\) electrons are involved in bonding or formally removed. The total number of such electrons corresponds approximately to the group number. Oxygen and fluorine often stabilise these highly oxidised arrangements through strong bonding.
117. Two elements lie on opposite sides of manganese in the first transition series. Element X is an early member whose maximum oxidation state equals its group number, while element Y is a later member approaching a filled \(d\) subshell. The more reasonable comparison is:
ⓐ. Y should generally show a higher maximum oxidation state than X
ⓑ. X may sustain high states, but Y shows fewer common states
ⓒ. both must show only the \(+2\) state
ⓓ. neither can form compounds with oxygen
Correct Answer: X may sustain high states, but Y shows fewer common states
Explanation: Early elements up to manganese can use a large fraction of their available \(d\) and \(s\) electrons in bonding. Their maximum oxidation states therefore often reach the group number. Beyond manganese, increasing \(d\)-electron occupancy makes removal or sharing of many electrons progressively less favourable. Later elements usually show a narrower set of common states centred more strongly on lower values. Oxygen-containing compounds can still form for both regions, but the accessible oxidation-state ranges differ.
118. A sample contains equal molar amounts of chromium in \(\mathrm{Cr^{3+}}\) and \(\mathrm{CrO_4^{2-}}\). What is the average oxidation state of chromium in the sample?
ⓐ. \(+3.0\)
ⓑ. \(+4.0\)
ⓒ. \(+4.5\)
ⓓ. \(+6.0\)
Correct Answer: \(+4.5\)
Explanation: \( \textbf{First chromium species:} \)
\[
\mathrm{Cr^{3+}}\Rightarrow \text{oxidation state}=+3
\]
\( \textbf{Second chromium species:} \) In \(\mathrm{CrO_4^{2-}}\), let the oxidation state be \(x\).
\[
x+4(-2)=-2
\]
\[
x=+6
\]
\( \textbf{Equal-amount condition:} \) The two chromium populations contribute equally to the average.
\( \textbf{Average oxidation state:} \)
\[
\frac{(+3)+(+6)}{2}
\]
\[
\frac{9}{2}=+4.5
\]
\( \textbf{Interpretation:} \) The fractional result is an average over two species, not the oxidation state of an individual chromium atom.
\( \textbf{Final answer:} \) The average oxidation state of chromium is \(+4.5\).
119. The statement that an oxidation state is “stable” means that the state:
ⓐ. must possess either a half-filled or completely filled \(d\) subshell
ⓑ. can never undergo oxidation or reduction under any conditions
ⓒ. is always the most common state in every compound and solvent
ⓓ. is energetically favoured under the specified chemical environment
Correct Answer: is energetically favoured under the specified chemical environment
Explanation: Oxidation-state stability is always linked to a particular chemical environment. Hydration enthalpy, lattice energy, bond strength, ligand type, and electronic configuration may all contribute. A state stable in a solid oxide may be less stable as a simple ion in water. Half-filled, empty, or filled subshells can provide useful stabilisation, but none acts as an absolute rule. Redox conditions can also shift an element from one stable state to another. The word “stable” therefore describes a favourable energetic balance under stated conditions.
120. Assertion: \(\mathrm{Mn^{2+}}\) is often relatively stable in aqueous chemistry.
Reason: \(\mathrm{Mn^{2+}}\) has a half-filled \(3d^5\) configuration.
ⓐ. Assertion is true, but Reason is false
ⓑ. Both Assertion and Reason are true, but Reason does not contribute to the Assertion
ⓒ. Assertion is false, but Reason is true
ⓓ. Both Assertion and Reason are true, and Reason contributes to the explanation
Correct Answer: Both Assertion and Reason are true, and Reason contributes to the explanation
Explanation: Neutral manganese has the configuration \([\mathrm{Ar}]\,3d^54s^2\). Removal of the two \(4s\) electrons gives \(\mathrm{Mn^{2+}}\) with configuration \([\mathrm{Ar}]\,3d^5\). The half-filled \(d^5\) arrangement provides favourable exchange-energy stabilisation. Hydration of the ion and the surrounding chemical environment also affect its relative stability. The Reason is therefore chemically relevant without being the only contributing factor. Treating \(d^5\) stability as an absolute rule would oversimplify the aqueous redox behaviour.
