The d and f-Block Elements
Multiple Choice Questions (MCQs)
1. Electronic configuration of a transition element
(a) 25
(b) 26
(c) 27
(d) 24
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Answer
(b) Electronic configuration of
It repersents the total number of
Therefore, atomic number of
Hence, option (b) is correct.
-
Option (a) 25: If the atomic number of element
were 25, then in its +3 oxidation state, it would have lost 3 electrons. The electronic configuration of would be . Removing 3 electrons would result in , not . -
Option (c) 27: If the atomic number of element
were 27, then in its +3 oxidation state, it would have lost 3 electrons. The electronic configuration of would be . Removing 3 electrons would result in , not . -
Option (d) 24: If the atomic number of element
were 24, then in its +3 oxidation state, it would have lost 3 electrons. The electronic configuration of would be . Removing 3 electrons would result in , not .
2. The electronic configuration of
(a)
(b)
(c)
(d) Stability of
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Answer
(a)
-
(b)
is less stable: This is incorrect because is actually more stable than due to the greater effective nuclear charge in , which allows it to hold its electrons more tightly despite having one less electron than . -
(c)
and are equally stable: This is incorrect because is more stable than . The effective nuclear charge in makes it more stable despite the configuration, compared to the configuration of . -
(d) Stability of
and depends on nature of copper salts: This is incorrect because the intrinsic stability of over is due to the effective nuclear charge and not dependent on the nature of the copper salts.
3. Metallic radii of some transition elements are given below. Which of these elements will have highest density?
Element | ||||
---|---|---|---|---|
Metallic radii/pm | 126 | 125 | 125 | 128 |
(a)
(b)
(c) Co
(d)
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Answer
(d) On moving left to right along period, metallic radius decreases while mass increases. Decrease in metallic radius coupled with increase in atomic mass results in increase in density of metal.
Hence, among the given four choices
-
(a) Fe: Iron (Fe) has a larger metallic radius (126 pm) compared to Co and Ni, and it is positioned to the left of Cu in the periodic table. This means it has a lower atomic mass and a larger atomic radius, resulting in a lower density compared to Cu.
-
(b) Ni: Nickel (Ni) has a smaller metallic radius (125 pm) compared to Cu, but it is positioned to the left of Cu in the periodic table. Although it has a smaller radius, its atomic mass is lower than that of Cu, leading to a lower density.
-
(c) Co: Cobalt (Co) has a metallic radius of 125 pm, similar to Ni, and is also positioned to the left of Cu in the periodic table. Like Ni, it has a lower atomic mass than Cu, resulting in a lower density.
4. Generally, transition elements form coloured salts due to the presence of unpaired electrons. Which of the following compounds will be coloured in solid state?
(a)
(b)
(c)
(d)
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Answer
(b) Transition elements form coloured salt due to the presence of unpaired electrons. In
-
(a)
: Silver (Ag) in is in the +1 oxidation state, which has a completely filled d^10 configuration. There are no unpaired electrons, so it does not exhibit color. -
(c)
: Zinc (Zn) in is in the +2 oxidation state, which has a completely filled d^10 configuration. There are no unpaired electrons, so it does not exhibit color. -
(d)
: Copper (Cu) in is in the +1 oxidation state, which has a completely filled d^10 configuration. There are no unpaired electrons, so it does not exhibit color.
5. On addition of small amount of
(a)
(b)
(c)
(d)
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Answer
(a) On addition of
-
(b)
: This compound is manganese dioxide, which is a solid and not a green oily liquid. It is also not highly explosive in nature. -
(c)
: This compound is manganese(II) sulfate, which is a solid and typically forms a pink or pale red solution in water. It is not a green oily liquid and is not highly explosive. -
(d)
: This compound is manganese(III) oxide, which is a solid and not a green oily liquid. It is also not highly explosive in nature.
6. The magnetic nature of elements depends on the presence of unpaired electrons. Identify the configuration of transition element, which shows highest magnetic moment.
(a)
(b)
(c)
(d)
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Thinking Process
This problem is based on calculation of magnetic moment can be done as Magnetic moment
Answer
(b) Greater the number of unpaired electron, higher will be its value of magnetic moment. Since,
-
(a)
: This configuration has 3 unpaired electrons. The magnetic moment is calculated as: Since it has fewer unpaired electrons than , its magnetic moment is lower. -
(c)
: This configuration has 2 unpaired electrons. The magnetic moment is calculated as: Since it has fewer unpaired electrons than , its magnetic moment is lower. -
(d)
: This configuration has 2 unpaired electrons. The magnetic moment is calculated as: Since it has fewer unpaired electrons than , its magnetic moment is lower.
7. Which of the following oxidation state is common for all lanthanoids?
(a) +2
(b) +3
(c) +4
(d) +5
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Answer
(b) Lanthanoids show common oxidation state of +3 . Some of which also show +2 and +4 stable oxidation state alongwith +3 oxidation state. These are shown by those elements which by losing 2 or 4 electrons acquire a stable configuration of
-
(a) +2: While some lanthanoids can exhibit a +2 oxidation state, it is not common to all lanthanoids. The +2 state is typically less stable and is observed in only a few specific lanthanoids such as Eu and Yb.
-
(c) +4: The +4 oxidation state is also not common to all lanthanoids. It is observed in a limited number of lanthanoids like Ce and Tb, where the elements achieve a stable electronic configuration by losing four electrons.
-
(d) +5: The +5 oxidation state is extremely rare and not observed in lanthanoids. Lanthanoids typically do not achieve this high oxidation state due to the high energy required to remove five electrons.
8. Which of the following reactions are disproportionation reactions?
(i)
(ii)
(iii)
(iv)
(a) (i)
(b) (i), (ii) and (iii)
(c) (ii), (iii) and (iv)
(d) (i) and (iv)
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Answer
(a) The reaction in which oxidation as well as reduction, occur upon same atom simultaneously is known as disproportionation reaction.
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-
Option (b) (i), (ii) and (iii):
- Reaction (ii) is not a disproportionation reaction because it involves the reduction of
to and the oxidation of to , but these changes do not occur on the same atom simultaneously. - Reaction (iii) is not a disproportionation reaction because it involves the decomposition of
into , , and , but there is no simultaneous oxidation and reduction of the same species.
- Reaction (ii) is not a disproportionation reaction because it involves the reduction of
-
Option (c) (ii), (iii) and (iv):
- Reaction (ii) is not a disproportionation reaction because it involves the reduction of
to and the oxidation of to , but these changes do not occur on the same atom simultaneously. - Reaction (iii) is not a disproportionation reaction because it involves the decomposition of
into , , and , but there is no simultaneous oxidation and reduction of the same species. - Reaction (iv) is not a disproportionation reaction because it involves the reduction of
to and the oxidation of to , but these changes do not occur on the same atom simultaneously.
- Reaction (ii) is not a disproportionation reaction because it involves the reduction of
-
Option (d) (i) and (iv):
- Reaction (iv) is not a disproportionation reaction because it involves the reduction of
to and the oxidation of to , but these changes do not occur on the same atom simultaneously.
- Reaction (iv) is not a disproportionation reaction because it involves the reduction of
9. When
(a)
(b) reaction is exothermic
(c)
(d)
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Answer
(d) When
Reduction half
Oxidation half
Overall equation
End point of this reaction Colourless to light pink
-
(a)
is formed as the product: The formation of as a product does not explain the change in reaction rate. The decolourisation becoming instantaneous is due to the catalytic effect of , not the production of . -
(b) reaction is exothermic: While the reaction may release heat, the exothermic nature of the reaction does not account for the initial slowness followed by a rapid decolourisation. The key factor is the autocatalytic role of
. -
(c)
catalyses the reaction: is actually a reactant in this reaction, not a catalyst. The catalyst in this reaction is , which is produced during the reaction and speeds up the process.
10. There are 14 elements in actinoid series. Which of the following elements does not belong to this series?
(a)
(b)
(c)
(d) Fm
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Answer
(c)
- (a)
(Uranium) is incorrect because it belongs to the actinoid series with an atomic number of 92. - (b)
(Neptunium) is incorrect because it belongs to the actinoid series with an atomic number of 93. - (d)
(Fermium) is incorrect because it belongs to the actinoid series with an atomic number of 100.
Q.11
(a)
(b)
(c)
(d)
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Answer
The reaction of
5 moles of
-
Option (b)
is incorrect because it suggests that 1 mole of ions would react with moles of , which is not supported by the stoichiometry of the balanced chemical equation. The correct stoichiometry indicates that 1 mole of ions reacts with moles of . -
Option (c)
is incorrect because it overestimates the amount of needed to react with 1 mole of ions. According to the balanced equation, only moles of are required for 1 mole of ions. -
Option (d)
is incorrect because it underestimates the amount of needed to react with 1 mole of ions. The balanced chemical equation shows that moles of are necessary for 1 mole of ions.
12. Which of the following is amphoteric oxide?
(a)
(b)
(c)
(d)
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Answer
(a)
Note In lower oxides, the basic character is predominant while in higher oxides, the acidic character is predominant.
-
Option (b)
: Both and are not amphoteric oxides. They are acidic oxides because they react with bases to form salts and water but do not react with acids. -
Option (c)
: is a basic oxide, not amphoteric. It reacts with acids to form salts and water but does not react with bases. is amphoteric, but the presence of makes this option incorrect. -
Option (d)
: is a basic oxide, not amphoteric. It reacts with acids to form salts and water but does not react with bases. is amphoteric, but the presence of makes this option incorrect.
13. Gadolinium belongs to
(a)
(c)
(b)
(d)
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Answer
(a) Gadolinium belongs to
It has extra stability due to half-filled
-
Option (b):
is incorrect because it does not account for the extra stability provided by the half-filled subshell. Gadolinium has 7 electrons in the subshell, not 6. -
Option (c):
is incorrect because it suggests that there are 8 electrons in the subshell and 2 electrons in the subshell, which does not match the actual electron count for gadolinium. Additionally, it omits the electrons. -
Option (d):
is incorrect because it suggests that there are 9 electrons in the subshell and 1 electron in the subshell, which is not the correct electron configuration for gadolinium. The subshell should not be involved, and the subshell should have 7 electrons.
14. Interstitial compounds are formed when small atoms are trapped inside the crystal lattice of metals. Which of the following is not the characteristic property of interstitial compounds?
(a) They have high melting points in comparison to pure metals
(b) They are very hard
(c) They retain metallic conductivity
(d) The are chemically very reactive
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Answer
(d) Interstitial compounds are formed when small atoms are trapped inside the crystal lattice of metals. Some of their important characteristics are as follows
(i) They are very hard and rigid.
(ii) They have high melting point which are higher than those of the pure metals.
(iii) They show conductivity like that of the pure metal.
(iv) They acquire chemical inertness.
-
(a) They have high melting points in comparison to pure metals: This is incorrect because interstitial compounds indeed have high melting points, often higher than those of the pure metals due to the strong bonding between the metal atoms and the small atoms trapped in the lattice.
-
(b) They are very hard: This is incorrect because interstitial compounds are known for their hardness and rigidity, which is a result of the small atoms fitting into the interstices of the metal lattice and strengthening the overall structure.
-
(c) They retain metallic conductivity: This is incorrect because interstitial compounds retain the metallic conductivity of the pure metals, as the presence of small atoms in the lattice does not significantly disrupt the flow of electrons.
15. The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of
(a)
(b)
(c)
(d)
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Answer
(b) The magnetic moment is associated with its spin angular momentum and orbital angular momentum.
Spin only magnetic moment value of
Hence, magnetic moment
-
Option (a)
: This value is incorrect because it corresponds to a different number of unpaired electrons. For a ion, which has 3 unpaired electrons, the correct calculation yields , not . -
Option (c)
: This value is incorrect because it does not match the result of the spin-only magnetic moment formula for 3 unpaired electrons. The correct calculation is , not . -
Option (d)
: This value is incorrect because it also does not match the result of the spin-only magnetic moment formula for 3 unpaired electrons. The correct calculation is , not .
Q.16
(a)
(b)
(c)
(d)
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Answer
(c)
Reaction
or,
-
(a)
: In an alkaline medium, is a strong oxidizing agent and it oxidizes iodide ions ( ) to higher oxidation states. Molecular iodine ( ) is typically formed in acidic or neutral conditions, not in alkaline conditions. -
(b)
: The oxidation state of iodine in is +1. However, in the presence of a strong oxidizing agent like in an alkaline medium, iodide ions are oxidized to a higher oxidation state than +1, specifically to +5 in . -
(d)
: The oxidation state of iodine in is +7. While is a strong oxidizing agent, in an alkaline medium, it typically oxidizes iodide ions to the +5 oxidation state, forming , rather than the +7 oxidation state in .
17. Which of the following statements is not correct?
(a) Copper liberates hydrogen from acids
(b) In its higher oxidation states, manganese forms stable compounds with oxygen and fluorine
(c)
(d)
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Answer
(a) Copper lies below hydrogen in the electrochemical series and hence does not liberate
Other three options (
-
(b) In its higher oxidation states, manganese forms stable compounds with oxygen and fluorine: This statement is correct. Manganese in higher oxidation states (like +4, +6, and +7) forms stable compounds such as ( \text{MnO}_2 ), ( \text{MnO}_3 ), and ( \text{MnF}_4 ).
-
(c) ( \text{Mn}^{3+} ) and ( \text{Co}^{3+} ) are oxidising agents in aqueous solution: This statement is correct. Both ( \text{Mn}^{3+} ) and ( \text{Co}^{3+} ) have a tendency to gain electrons and get reduced, thus acting as oxidizing agents in aqueous solution.
-
(d) ( \text{Ti}^{2+} ) and ( \text{Cr}^{2+} ) are reducing agents in aqueous solution: This statement is correct. Both ( \text{Ti}^{2+} ) and ( \text{Cr}^{2+} ) have a tendency to lose electrons and get oxidized, thus acting as reducing agents in aqueous solution.
18. When acidified
(a)
(b)
(c)
(d)
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Answer
(c) When acidified
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-
Option (a)
: This option is incorrect because is being oxidized in the reaction with acidified . Oxidation involves an increase in the oxidation state, not a reduction to elemental tin ( ), which would be a decrease in the oxidation state. -
Option (b)
: This option is incorrect because the oxidation state of tin in this reaction changes from +2 to +4. There is no intermediate oxidation state of +3 involved in this specific redox reaction. -
Option (d)
: This option is incorrect because would represent a reduction in the oxidation state from +2 to +1. However, in the reaction with acidified , is oxidized, meaning its oxidation state increases, not decreases.
19. Highest oxidation state of manganese in fluoride is
(a) fluorine is more electronegative than oxygen
(b) fluorine does not possess
(c) fluorine stabilises lower oxidation state
(d) in covalent compounds, fluorine can form single bond only while oxygen forms double bond
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Answer
(d) Highest oxidation state of manganese in fluoride is
-
(a) Fluorine is more electronegative than oxygen, but this does not directly explain why manganese achieves a higher oxidation state in oxides compared to fluorides. The ability to form multiple bonds (as in the case of oxygen) is more relevant to the stabilization of higher oxidation states.
-
(b) The fact that fluorine does not possess
orbitals is not relevant to the oxidation state of manganese. The oxidation state is determined by the ability of the element to stabilize different charges, which is more influenced by bonding characteristics rather than the presence of orbitals in fluorine. -
(c) While fluorine can stabilize lower oxidation states due to its high electronegativity, this does not explain why manganese achieves a higher oxidation state in oxides. The key factor is the ability of oxygen to form multiple bonds, which stabilizes higher oxidation states of manganese.
20. Although zirconium belongs to
(a) both belong to
(b) both have same number of electrons
(c) both have similar atomic radius
(d) both belong to the same group of the Periodic Table
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Answer
(c) Due to lanthanoide contraction,
-
(a) Both belong to
-block: While it is true that both zirconium and hafnium belong to the -block, this is not the primary reason for their similar physical and chemical properties. Many elements in the -block do not exhibit such close similarities in properties. -
(b) Both have same number of electrons: Zirconium and hafnium do not have the same number of electrons. Zirconium has 40 electrons, while hafnium has 72 electrons. The number of electrons is not the reason for their similar properties.
-
(d) Both belong to the same group of the Periodic Table: Although zirconium and hafnium do belong to the same group (Group 4) of the Periodic Table, this alone does not account for their nearly identical physical and chemical properties. The primary reason is the lanthanoid contraction, which results in their nearly identical atomic and ionic radii.
21. Why is
(a) Both
(b)
(c)
(d)
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Answer
(b)
-
(a) Both
and act as oxidising agents: This is incorrect because is not an oxidizing agent; it is a strong acid that can be oxidized by . -
(c)
is a weaker oxidising agent than : This is incorrect because is a much stronger oxidizing agent compared to . -
(d)
acts as a reducing agent in the presence of : This is incorrect because is a strong oxidizing agent and does not act as a reducing agent in the presence of .
Multiple Choice Questions (More Than One Options)
22. Generally transition elements and their salts are coloured due to the presence of unpaired electrons in metal ions. Which of the following compounds are coloured?
(a)
(b)
(c)
(d)
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Answer
is not coloured because titanium in is in the +4 oxidation state, which has a electronic configuration, meaning there are no unpaired electrons to cause colour. is not coloured because copper in is in the +1 oxidation state, which has a electronic configuration, meaning all electrons are paired and there are no unpaired electrons to cause colour.
23. Transition elements show magnetic moment due to spin and orbital motion of electrons. Which of the following metallic ions have almost same spin only magnetic moment?
(a)
(b)
(c)
(d)
Show Answer
Answer
Electronic configuration of
Electronic configuration of
Electronic configuration of
Electronic configuration of
Hence, it is clearly seen that both
-
: The electronic configuration of is , which means it has 4 unpaired electrons. This is different from the 3 unpaired electrons in and . -
: The electronic configuration of is , which means it has 5 unpaired electrons. This is different from the 3 unpaired electrons in and .
24. In the form of dichromate,
(a)
(b)
(c) Higher oxidation states of heavier members of group-6 of transition series are more stable.
(d) Lower oxidation states of heavier members of group-6 of transition series are more stable.
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Answer
In d-block elements, for heavier elements, the higher oxidation states are more stable. Hence,
-
(a)
is more stable than and : This is incorrect because, in reality, and are more stable than . The stability of higher oxidation states increases as we move down the group in the periodic table, making and more stable than . -
(d) Lower oxidation states of heavier members of group-6 of transition series are more stable: This is incorrect because, for heavier elements in the d-block, the higher oxidation states are generally more stable. Therefore, the statement that lower oxidation states are more stable for heavier members of group-6 is not accurate.
25. Which of the following actinoids show oxidation states upto +7 ?
(a) Am
(b)
(c)
(d)
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Answer
The oxidation states of the following actinoids are
(a) Americium
(b) Plutonium
(c) Uranium (
(d) Neptunium (
-
Americium (Am): Americium does not show an oxidation state of +7. The highest oxidation state it exhibits is +6.
-
Uranium (U): Uranium does not show an oxidation state of +7. The highest oxidation state it exhibits is +6.
26. General electronic configuration of actinoids is
(a)
(b)
(c) Pu (Atomic number. 94)
(d) Am (Atomic number. 95)
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Answer
(
General electronic configuration of actinoids is
-
Pu (Atomic number 94): Plutonium typically has the electronic configuration [Rn] 5f^6 7s^2. It does not have an electron in the 6d orbital.
-
Am (Atomic number 95): Americium typically has the electronic configuration [Rn] 5f^7 7s^2. It does not have an electron in the 6d orbital.
27. Which of the following lanthanoids show +2 oxidation state besides the characteristic oxidation state +3 of lanthanoids?
(a)
(b)
(c)
(d) Ho
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Answer
(a) Cerium
(b) Europium
(c) Ytterbium
(d) Holmium
-
Cerium (Ce): Cerium typically exhibits oxidation states of +3 and +4. It does not commonly show a +2 oxidation state.
-
Holmium (Ho): Holmium predominantly exhibits an oxidation state of +3 and does not commonly show a +2 oxidation state.
28. Which of the following ions show higher spin only magnetic moment value?
(a)
(b)
(c)
(d)
Show Answer
Answer
As,
Crystal field splitting energy (CFSE) is high in
-
: This ion has a single unpaired electron in the 3d orbital ( configuration). The spin-only magnetic moment is directly related to the number of unpaired electrons. With only one unpaired electron, has a lower spin-only magnetic moment compared to ions with more unpaired electrons. -
: This ion has a configuration. Due to the high crystal field splitting energy (CFSE) in , the electrons pair up in the orbitals, resulting in no unpaired electrons ( ). Therefore, has a very low or zero spin-only magnetic moment.
29. Transition elements form binary compounds with halogens. Which of the following elements will form
(a)
(b) Co
(c)
(d)
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Answer
(a, b)
Transition elements such as
-
Cu: Copper typically forms compounds in the +1 and +2 oxidation states, such as
and . The +3 oxidation state is less stable for copper, making unlikely to form. -
Ni: Nickel commonly forms compounds in the +2 oxidation state, such as
. The +3 oxidation state is not stable for nickel, hence is not typically formed.
30. Which of the following will not act as oxidising agents?
(a)
(b)
(c)
(d)
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Answer
A species can act as oxidising agent only when metal is present in high oxidation state but lower oxidation state show stability. As higher oxidations states of
-
(a)
: Chromium trioxide ( ) contains chromium in the +6 oxidation state, which is a high oxidation state. Chromium in this state can readily accept electrons and be reduced to a lower oxidation state, making it a strong oxidizing agent. -
(d)
: Chromate ion ( ) also contains chromium in the +6 oxidation state. Similar to , chromium in this state can act as an oxidizing agent by accepting electrons and being reduced to a lower oxidation state.
31. Although +3 is the characteristic oxidation state for lanthanoids but cerium also shows +4 oxidation state because
(a) it has variable ionisation enthalpy
(b) it has a tendency to attain noble gas configuration
(c) it has a tendency to attain
(d) it resembles
Show Answer
Answer
Electronic configuration of
Therefore, electronic configuration of
Thus, it has a tendency to attain noble gas configuration and attain
-
(a) it has variable ionisation enthalpy: This option is incorrect because the primary reason for cerium showing a +4 oxidation state is not due to variable ionisation enthalpy but rather its ability to achieve a stable electronic configuration.
-
(d) it resembles
: This option is incorrect because the resemblance to is not a significant factor in cerium exhibiting a +4 oxidation state. The key reasons are related to its electronic configuration and stability.
Short Answer Type Questions
32. Why does copper not replace hydrogen from acids?
Show Answer
Answer
Copper not replace hydrogen from acids because
33. Why
Show Answer
Answer
Negative values of
Hence,
34. Why first ionisation enthalpy of
Show Answer
Answer
Ionisation enthalpy of
35. Transition elements show high melting points. Why?
Show Answer
Answer
Transition elements show high melting point, due to involvement of greater number of electrons in the interatomic bonding from
36. When
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Answer
When
(In this reaction,
37. Out of
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Answer
Among
38. When a brown compound of manganese
Show Answer
Thinking Process
This problem is based on the properties of
Answer
(i)
(ii)
39. Although fluorine is more electronegative than oxygen, but the ability of oxygen to stabilise higher oxidation states exceeds that of fluorine. Why?
Show Answer
Answer
Oxygen can form multiple bonds with metals, while fluorine can’t form multiple bond with metals. Hence, oxygen has more ability to stabilize higher oxidation state rather than fluorine
40. Although
Show Answer
Answer
Magnetic moment of any metal ion can be decided on the basis of spin as well as orbital contribution of electron. Due to symmetrical electronic configuration, there is no orbital contribution in
41. Ionisation enthalpies of
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Answer
In the beginning, when 5 -orbitals begin to be occupied, they will penetrate less into the inner core of electrons. The
Therefore, outer electrons are less firmly held and they are available for bonding in the actinoids
42. Although
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Answer
Separation of
43. Although +3 oxidation states is the characteristic oxidation state of lanthanoids but cerium shows +4 oxidation state also. Why?
Show Answer
Answer
It is due to the fact that after losing one more electron Ce acquires stable
44. Explain why does colour of
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Answer
When oxalic acid is added to acidic solution of
45. When orange solution containing
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Answer
When orange solution containing
when
Q.46
Show Answer
Answer
Oxidising behaviour of
In acidic medium (
In alkaline medium
In neutral medium
47. The second and third rows of transition elements resemble each other much more than they resemble the first row. Explain, why?
Show Answer
Answer
Due to lanthanoid contraction, the atomic radii of the second and third row transition elements is almost same. So, they resemble each other much more as compared to first row elements and show similar character.
48.
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Answer
49. The halides of transition elements become more covalent with increasing oxidation state of the metal. Why?
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Thinking Process
This problem is based on concept of Fajan’s rule and its application.
Answer
As the oxidation state increases, size of the ion of transition element decreases. As per Fajan’s rule, as the size of metal ion decreases, covalent character of the bond formed increases.
Therefore, the halide of transition elements become more covalent with increasing oxidation state of the metal.
50. While filling up of electrons in the atomic orbitals, the
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Answer
During filling up of electrons follow
51. Reactivity of transition elements decreases almost regularly from Se to Cu. Explain.
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Answer
Reactivity of transition elements depends mostly upon their ionisation enthalpies. As we move from left to right in the periodic table (
Hence, their reactivity decreases almost regularly from
Matching The Columns
52. Match the catalysts given in Column I with the processes given in Column II.
Column I (Catalyst) |
Column II (Process) |
||
---|---|---|---|
A. | 1. | Ziegler-Natta catalyst | |
B. | 2. | Contact process | |
C. | 3. | Vegetable oil to ghee | |
D. | Finely divided iron | 4. | Sandmeyer reaction |
E. | 5. | Haber’s process | |
6. | Decomposition of |
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Answer
A.
B.
C.
E.
Catalyst | Process | |
---|---|---|
A. | Vegetable oil to ghee | |
B. | Sandmeyer reaction | |
C. | Contact process |
|
D. | Finely divided iron | Haber’s process |
E. |
53. Match the compounds/elements given in Column I with uses given in Column II.
Column I (Compound/element) |
Column II (Use) |
||
---|---|---|---|
A. | Lanthanoid oxide | 1. | Production of iron alloy |
B. | Lanthanoid | 2. | Television screen |
C. | Misch metall | 3. | Petroleum cracking |
D. | Magnesium based alloy is constituent of | 4. | Lanthanoid metal + iron |
E. | Mixed oxides of lanthanoids are employed | 5. | Bullets |
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Answer
A.
B.
C.
D.
E.
Compound /Element | Use | |
---|---|---|
A. | Lanthanoid oxide | Television screen |
B. | Lanthanoid | Production of iron alloy |
C. | Misch metall | Lanthanoid metal + iron |
D. | Magnesium based alloy is constitute of | Bullets |
E | Mixed oxides of lanthanoids are employed | Petroleum cracking |
54. Match the properties given in Column I with the metals given in Column II.
Column I (Property) |
Column II (Metal) |
||
---|---|---|---|
A. | An element which can show +8 oxidation state | 1. | |
B. | oxidation state |
2. | |
C. | 3. | Os | |
4. |
Show Answer
Answer
A.
B.
C.
A. Osmium is an element which can show +8 oxidation state.
B.
C.
55. Match the statements given in Column I with the oxidation states given in Column II.
Column I | Column II | ||
---|---|---|---|
A. | Oxidation state of |
1. | +2 |
B. | Most stable oxidation state of |
2. | +3 |
C. | Most stable oxidation state of |
3. | +4 |
D. | Characteristic oxidation state of lanthanoids is | 4. | +5 |
5. | +7 |
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Answer
A.
B.
C.
D.
A. Oxidation state of
B. Most stable oxidation state of
C. Most stable oxidation state of
D. Characteristic oxidation state of lanthanoids is +3 .
56. Match the solutions given in Column I and the colours given in Column II.
Column I (Aqueous solution of salt) |
Column II (Colour) |
||
---|---|---|---|
A. | 1. | Green | |
B. | 2. | Light pink | |
C. | 3. | Blue | |
D. | 4. | Pale green | |
E. | 5. | Pink | |
6. | Colourless |
Show Answer
Answer
A.
B.
C.
D.
E.
Aqueous solution of salt | Colour | |
---|---|---|
A. | Pale green | |
B. | Green | |
C. | Light pink | |
D. | Pink | |
E. | Colourless |
57. Match the property given in Column I with the element given in Column II.
Column I (Property) |
Column II (Element) |
||
---|---|---|---|
A. | Lanthanoid which shows +4 oxidation state | 1. | Pm |
B. | Lanthanoid which can show +2 oxidation state |
2. | |
C. | 3. | Lu | |
D. | Lanthanoid which has configuration in +3 oxidation state |
4. | |
E. | Lanthanoid which has configuration in +3 oxidation state |
5. | Gd |
Show Answer
Answer
A.
B.
C.
D.
E.
A. Lanthanoid which shows +4 oxidation state is cerium.
B. Lanthanoid which can show +2 oxidation state is europium.
C. Radioactive lanthanoid is promethium. It is the only synthetic (man-made) radioactive lanthanoid.
D. Lanthnoid which has
E. Lanthanoid which has
58. Match the properties given in Column I with the metals given in Column II.
Column I (Property) |
Column II (Metal) |
||
---|---|---|---|
A. | Element with highest second ionisation enthalpy | 1. | |
B. | Element with highest third ionisation enthalpy | 2. | |
C. | 3. | ||
D. | Element with highest heat of atomisation | 4. | |
5. |
Show Answer
Answer
A.
B.
C.
D.
A.
B.
C. Metal carbonyl with formula
D. Nickel is the element with highest heat of atomisation.
Assertion and Reason
In the following questions a statement of assertion (A) followed by a statement of reason
(a) Both assertion and reason are true, and reason is the correct explanation of the assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is not true but reason is true.
(d) Both assertion and reason are false.
59. Assertion
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Answer
(a) Assertion and reason both are correct and reason is the correct explanation of assertion.
Copper (II) iodide
60. Assertion (A) Separation of
Show Answer
Answer
(b) Assertion and reason are true but reason is not correct explanation of assertion.
Separation of
61. Assertion (A) Actinoids form relatively less stable complexes as compared to lanthanoids.
Reason (R) Actinoids can utilise their 5 f orbitals alongwith
Show Answer
Answer
(c) Assertion is not true but reason is true.
Actinoids form relatively more stable complexes as compared to lanthanoids because of actinoids can utilise their
62. Assertion (A) Cu cannot liberate hydrogen from acids.
Reason (R) Because it has positive electrode potential.
Show Answer
Answer
(a) Assertion and reason both are correct and reason is correct explanation of assertion. Cu can not liberate hydrogen from acids because it has positive electrode potential. Metals having negative value of electrode potential liberate
63. Assertion (A) The highest oxidation state of osmium is +8 .
Reason (R) 0smium is a 5d-block element.
Show Answer
Answer
(b) Assertion and reason both are correct but reason is not the correct explanation of assertion.
The highest oxidation state of osmium is +8 due to its ability to expand their octet by using its all 8 electrons ( 2 from 6 s and 6 from
Long Answer Type Questions
64. Identify A to E and also explain the reaction involved.
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Show Answer
Answer
The substances from
Reactions:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
65. When a chromite ore
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Answer
The given process is the preparation method of potassium dichromate from chromite ore.
(i)
(ii)
(iii)
66. When an oxide of manganese
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Thinking Process
This problem based on the concept of preparation and properties of potassium permanganate.
Answer
It is the method of preparation of potassium permanganate (purple).
Thus,
67. On the basis of lanthanoid contraction, explain the following:
(i) Nature of bonding in
(ii) Trends in the stability of oxo salts of lanthanoids from La to Lu.
(iii) Stability of the complexes of lanthanoids.
(iv) Radii of
(v) Trends in acidic character of lanthanoid oxides.
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Answer
(i) As the size decreases covalent character increases. Therefore,
(ii) As the size decreases from La to Lu, stability of oxo salts also decreases.
(iii) Stability of complexes increases as the size of lanthanoids decreases.
(iv) Radii of
(v) Acidic character of oxides increases from La to Lu.
68. (a) Answer the following questions
(i) Which element of the first transition series has highest second ionisation enthalpy?
(ii) Which element of the first transition series has highest third ionisation enthalpy?
(iii) Which element of the first transition series has lowest enthalpy of atomisation?
(b) Identify the metal and justify your answer.
(i) Carbonyl
(ii)
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Answer
(a) (i)
(ii) Zinc, because of electronic configuration of
(iii) Zinc, because of it has completely filled
(b) (i) Carbonyl
According to EAN rule, the effective number of a metal in a metal carbonyl is equal to the atomic number of nearest inert gas EAN is calculated as
EAN
So, the metal is
(ii)
In
Let us assume that oxidation state of
or,
69. Mention the type of compounds formed when small atoms like
Show Answer
Answer
When small atoms like
(a) Such compounds are called interstitial compounds.
(b) Their characteristic properties are;
(i) They have high melting points, higher than those of pure metals.
(ii) They are very hard.
(iii) They retain metallic conductivity.
(iv) They are chemically inert.
70. (a) Transition metals can act as catalysts because these can change their oxidation state. How does Fe (III) catalyse the reaction between iodide and persulphate ions?
(b) Mention any three processes where transition metals act as catalysts.
Show Answer
Answer
(a) Reaction between iodide and persulphate ions is
(b) (i) Vanadium (
(ii) Finely divided iron in Haber’s process in conversion of
(iii)
71. A violet compound of manganese
Show Answer
Thinking Process
This problem is based on preparation and properties of
Answer
Since, compound
Thus,