101. Which factor is mainly responsible for the near-equality of \(4d\) and \(5d\) radii?
ⓐ. Strong shielding by \(4f\)-electrons
ⓑ. Complete absence of inner electrons in \(5d\) elements
ⓒ. Poor shielding by \(4f\)-electrons
ⓓ. Sudden increase in the size of the \(4d\) series
Correct Answer: Poor shielding by \(4f\)-electrons
Explanation: The \(4f\)-electrons are relatively ineffective at shielding the nucleus. Because of that, the effective nuclear charge felt by outer electrons increases strongly across the lanthanoid series. This contracts the atoms that appear after the lanthanoids, especially the \(5d\) elements. That poor shielding is the central reason their sizes become so close to those of the \(4d\) elements.
102. Which pair shows the most appropriate size relation?
ⓐ. \(Ti \approx Zr\)
ⓑ. \(Y \approx Sc\)
ⓒ. \(Zr < Ti\)
ⓓ. \(Zr \approx Hf\)
Correct Answer: \(Zr \approx Hf\)
Explanation: Titanium is a \(3d\) element and is notably smaller than zirconium, so \(Ti \approx Zr\) is not appropriate. Similarly, yttrium is larger than scandium. The important unusual relation is between zirconium and hafnium, which are nearly similar in size despite hafnium belonging to the \(5d\) series. This near-equality is one of the classic examples of lanthanoid contraction.
103. Assertion: The \(5d\) series should have been much larger than the \(4d\) series on simple down-group reasoning.
Reason: Lanthanoid contraction reduces the size of \(5d\) elements.
ⓐ. 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: On descending a group, one usually expects atomic size to increase because an additional shell is added. So without any special effect, \(5d\) elements should have been clearly larger than \(4d\) elements. However, lanthanoid contraction reduces the sizes of \(5d\) elements. That is exactly why the expected increase becomes much smaller than usual.
104. Which statement is false?
ⓐ. The \(3d\) series is generally smaller than the \(4d\) series.
ⓑ. The \(5d\) series is compressed by an effect linked to lanthanoids.
ⓒ. The similarity in size between \(4d\) and \(5d\) elements is due to poor shielding by \(4f\)-electrons.
ⓓ. The \(4d\) and \(5d\) series are nearly equal in size because \(4f\)-electrons shield very effectively.
Correct Answer: The \(4d\) and \(5d\) series are nearly equal in size because \(4f\)-electrons shield very effectively.
Explanation: The key point is that \(4f\)-electrons shield poorly, not effectively. Poor shielding allows a greater effective nuclear pull and causes contraction across the lanthanoids. This contraction reduces the size of the following \(5d\) elements. Therefore, the statement that attributes the effect to very effective shielding is incorrect.
105. Which sequence best represents the expected relative size of \(Ti\), \(Zr\), and \(Hf\)?
ⓐ. \(Ti > Zr > Hf\)
ⓑ. \(Zr > Hf > Ti\)
ⓒ. \(Ti < Zr \approx Hf\)
ⓓ. \(Ti \approx Zr < Hf\)
Correct Answer: \(Ti < Zr \approx Hf\)
Explanation: Titanium belongs to the \(3d\) series and is smaller than zirconium. Hafnium belongs to the \(5d\) series, so it might be expected to be much larger than zirconium, but lanthanoid contraction reduces that increase strongly. As a result, zirconium and hafnium are very close in size. This makes \(Ti < Zr \approx Hf\) the best relation.
106. Which statement best explains why \(Zr\) and \(Hf\) often resemble each other in chemical behaviour?
ⓐ. They have the same atomic number.
ⓑ. They have nearly similar radii.
ⓒ. They belong to different groups of the periodic table.
ⓓ. They have identical electron configurations.
Correct Answer: They have nearly similar radii.
Explanation: Chemical behaviour is strongly influenced by size, charge, and valence-shell pattern. Because zirconium and hafnium have very similar atomic and ionic radii, they often show comparable chemistry. Their similarity is striking enough to be a standard consequence of lanthanoid contraction. They are not identical in atomic number or full electron configuration, but their sizes are unusually close.
107. The near-equality of \(4d\) and \(5d\) radii is a consequence of
ⓐ. actinoid contraction
ⓑ. weak nuclear attraction in \(5d\) elements
ⓒ. addition of a new shell without any contraction
ⓓ. lanthanoid contraction
Correct Answer: lanthanoid contraction
Explanation: Before the \(5d\) series begins, the lanthanoid series is traversed. Across the lanthanoids, poor shielding by \(4f\)-electrons leads to a steady contraction in size. This reduces the size of the succeeding \(5d\) elements and brings them close to the \(4d\) elements. Hence the relevant effect here is lanthanoid contraction, not actinoid contraction.
108. Which statement best summarizes the size comparison among the \(3d\), \(4d\), and \(5d\) transition series?
ⓐ. \(3d\), \(4d\), and \(5d\) elements of the same group all have identical radii.
ⓑ. The size increases regularly from \(3d\) to \(4d\) to \(5d\) with no exception.
ⓒ. The \(4d\) series is smaller than the \(3d\) series, while the \(5d\) series is much larger than both.
ⓓ. The \(3d\) series is smaller; \(4d\) and \(5d\) are close because of lanthanoid contraction.
Correct Answer: The \(3d\) series is smaller; \(4d\) and \(5d\) are close because of lanthanoid contraction.
Explanation: Moving from \(3d\) to \(4d\), the size generally increases because of the extra shell. But the jump from \(4d\) to \(5d\) is not as large as simple group trends would suggest. Lanthanoid contraction compresses the \(5d\) series and makes it close in size to the \(4d\) series. That is the most useful broad comparison across these three transition series.
109. Across the first transition series, the first ionisation enthalpy generally
ⓐ. decreases sharply from \(Sc\) to \(Zn\)
ⓑ. shows a moderate overall increase
ⓒ. remains exactly constant for every element
ⓓ. rises very steeply as in the \(p\)-block
Correct Answer: shows a moderate overall increase
Explanation: As the atomic number increases across the first transition series, the nuclear charge also increases. That tends to hold the outer electrons more strongly, so more energy is generally needed to remove one electron. However, the increase is only moderate, not very steep. This happens because the added \(3d\)-electrons provide some shielding and soften the effect of the increasing nuclear charge.
110. Why does the ionisation enthalpy across transition elements not increase as sharply as it does across many \(p\)-block elements?
ⓐ. Transition elements have no increase in nuclear charge across the series.
ⓑ. Their outer electrons are always removed from a noble-gas core.
ⓒ. All transition elements have identical atomic radii.
ⓓ. Shielding by \(d\)-electrons and close orbital energies soften the rise.
Correct Answer: Shielding by \(d\)-electrons and close orbital energies soften the rise.
Explanation: The increasing nuclear charge does tend to raise ionisation enthalpy, but that is not the only effect operating across the transition series. The electrons entering the \(d\)-subshell provide some shielding, though not very effective. Also, the \(ns\) and \((n-1)d\) orbitals lie fairly close in energy. Because of these combined effects, the rise in ionisation enthalpy is smoother and less sharp than in many \(p\)-block sequences.
111. Which statement best compares the ionisation enthalpy of transition metals with that of alkali metals?
ⓐ. Transition metals generally have higher ionisation enthalpy than alkali metals.
ⓑ. Transition metals and alkali metals have nearly the same ionisation enthalpy in all cases.
ⓒ. Transition metals always have lower ionisation enthalpy than alkali metals.
ⓓ. Alkali metals have higher ionisation enthalpy because they lose electrons less easily.
Correct Answer: Transition metals generally have higher ionisation enthalpy than alkali metals.
Explanation: Alkali metals lose their single valence electron very easily, so their ionisation enthalpies are low. Transition metals hold their outer electrons more strongly because of higher effective nuclear attraction and more complex electronic structure. That makes electron removal less easy in general. So transition metals are usually less reactive toward simple electron loss than alkali metals.
112. Which statement about the ionisation enthalpy pattern in the first transition series is most accurate?
ⓐ. It rises smoothly with no irregularity at all.
ⓑ. It falls from left to right because \(d\)-electrons are added.
ⓒ. It generally increases, but not perfectly regularly.
ⓓ. It is unrelated to shielding and nuclear charge.
Correct Answer: It generally increases, but not perfectly regularly.
Explanation: A broad left-to-right increase is observed because nuclear charge increases across the series. Still, the pattern is not perfectly smooth, since shielding, subshell energies, and individual electronic arrangements also matter. This makes local irregularities possible. So the correct description is a general increase with some departures from strict regularity.
113. Which factor directly contributes to the moderate rise in ionisation enthalpy across the first transition series?
ⓐ. Decrease in nuclear charge from \(Sc\) to \(Zn\)
ⓑ. Increasing effective nuclear attraction
ⓒ. Sudden entry of electrons into a new principal shell
ⓓ. Permanent emptiness of the \(4s\) orbital in all atoms
Correct Answer: Increasing effective nuclear attraction
Explanation: Moving across the series, the number of protons in the nucleus increases. This raises the attraction between the nucleus and the electrons, especially the outer electrons. Although added \(d\)-electrons provide some shielding, the overall pull still becomes stronger. That is why ionisation enthalpy tends to rise.
114. Which statement is false?
ⓐ. Transition elements usually do not show as sharp an ionisation-enthalpy rise as many \(p\)-block elements.
ⓑ. The presence of \(d\)-electrons affects the ionisation-enthalpy trend.
ⓒ. Transition metals generally lose electrons less easily than alkali metals.
ⓓ. Ionisation enthalpy across the first transition series decreases steadily from \(Sc\) to \(Zn\).
Correct Answer: Ionisation enthalpy across the first transition series decreases steadily from \(Sc\) to \(Zn\).
Explanation: The first ionisation enthalpy does not show a steady decrease across the first transition series. Instead, it generally increases, though not in a perfectly uniform way. The other statements agree with the usual interpretation of the trend. So the steadily decreasing pattern is the incorrect one.
115. Assertion: Transition metals are generally less reactive than alkali metals toward simple loss of electrons.
Reason: Transition metals usually have higher ionisation enthalpies than alkali metals.
ⓐ. 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: Alkali metals have very low ionisation enthalpies, so they lose their outer electron readily and behave as highly reactive reducing metals. Transition metals generally require more energy for electron removal. Because of that, they are less reactive toward simple electron loss. The reason therefore explains the assertion directly.
116. Which feature makes the \(ns\) electrons of transition elements different from the valence electrons of many main-group elements?
ⓐ. They are always inner-shell electrons.
ⓑ. They are absent from all neutral transition-metal atoms.
ⓒ. They are close in energy to the \((n-1)d\) electrons.
ⓓ. They are always completely shielded from the nucleus.
Correct Answer: They are close in energy to the \((n-1)d\) electrons.
Explanation: In transition elements, the \(ns\) and \((n-1)d\) orbitals lie close in energy. This is one of the reasons their chemistry is rich and their ionisation-enthalpy trend is not extremely simple. The participation of both kinds of orbitals affects electron removal and bonding. That orbital closeness is a characteristic feature of transition-metal behaviour.
117. Which statement best explains why the first ionisation enthalpy of transition elements does not show a dramatic jump across the series?
ⓐ. Their nuclei become smaller from left to right.
ⓑ. \(d\)-electron shielding partly balances rising nuclear charge.
ⓒ. Every atom has the same number of effective valence electrons.
ⓓ. The \(4s\) orbital disappears after scandium.
Correct Answer: \(d\)-electron shielding partly balances rising nuclear charge.
Explanation: A stronger nucleus would by itself make electron removal steadily more difficult. But across the transition series, electrons are being added to the \(d\)-subshell, and these electrons contribute some shielding. That shielding is not strong enough to reverse the trend, but it does moderate it. As a result, the rise in ionisation enthalpy is gentler than one might otherwise expect.
118. Which comparison is most appropriate?
ⓐ. Transition elements are more reactive than alkali metals because they always have lower ionisation enthalpy.
ⓑ. Transition elements and alkali metals show identical ionisation-enthalpy trends across a period.
ⓒ. Transition elements show negligible ionisation enthalpy because \(d\)-electrons cancel nuclear charge completely.
ⓓ. Transition elements generally require more energy to remove an electron than alkali metals do.
Correct Answer: Transition elements generally require more energy to remove an electron than alkali metals do.
Explanation: Alkali metals have one outer electron that is relatively loosely held, so removing it requires comparatively little energy. In transition metals, the effective nuclear pull is greater and the electronic structure is less simple. That makes electron removal less easy in general. Hence transition elements usually have higher ionisation enthalpies than alkali metals.
119. Which statement about the role of shielding in transition elements is correct?
ⓐ. Poor \(d\)-electron shielding moderates but does not cancel the rise.
ⓑ. \(d\)-electrons shield so strongly that ionisation enthalpy falls sharply across the series.
ⓒ. \(d\)-electrons have no effect at all on ionisation enthalpy.
ⓓ. \(d\)-electrons completely remove the effect of increasing nuclear charge.
Correct Answer: Poor \(d\)-electron shielding moderates but does not cancel the rise.
Explanation: The \(d\)-electrons do provide shielding, but they are not especially effective at doing so. Therefore, the growing nuclear charge across the series still increases the attraction for the outer electrons. However, the shielding is enough to make the rise less sharp than it would otherwise be. This gives the characteristic moderate trend.
120. Which statement best summarizes the first ionisation-enthalpy trend in the first transition series?
ⓐ. It decreases because metallic character always increases across the series.
ⓑ. It remains fixed because the outer \(4s\) electron is identical in all cases.
ⓒ. A moderate overall increase with small shielding and orbital-energy irregularities.
ⓓ. It rises sharply with no exception because every added proton acts without any counter-effect.
Correct Answer: A moderate overall increase with small shielding and orbital-energy irregularities.
Explanation: The broad tendency is upward because increasing nuclear charge holds electrons more strongly. At the same time, shielding by \(d\)-electrons and the closeness of \(ns\) and \((n-1)d\) orbital energies prevent a steep or perfectly regular rise. This leads to a moderate increase with some local irregularity. That is the most accurate summary of the pattern.
