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Question:1
Electronic configuration of a transition element X in +3 oxidation state is . What is its atomic number?
(i) 25
(ii) 26
(iii) 27
(iv) 24
Answer:
The answer is the option (ii).
Electronic configuration of a transition element in
oxidation state is
.
Atomic number = 26
Question:2
The electronic configuration of is
whereas that of
is
. Which of the following is correct?
(i) is more stable
(ii) is less stable
(iii) and
are equally stable
(iv) Stability of and
depends on the nature of copper salts
Answer:
The answer is the option (i). Thestability of is more due to greater effective nuclear charge of
Question:3
Element Metallic radii/pm |
|
|
|
|
(i)
(ii)
(iii)
(iv)
Answer:
The answer is the option (iv). In a periodic table, moving across the period from left to right the atomic radii of element decreases. This is the reason behind the increase of density as we move rightwards in a period.
Amongst all the options, lies to the right most side of Periodic Table and has the highest density.
Question: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?
(i)
(ii)
(iii)
(iv)
Answer:
The answer is the option (ii) has one unpaired electron due to which it forms coloured salts in solid state.
Question:5
On addition of a small amount of to concentrated
, a green oily compound is obtained which is highly explosive. Identify the compound from the following.
(i)
(ii)
(iii)
(iv)
Answer:
The answer is the option (i).
Thus, we get which is highly explosive in nature
Question:6
The magnetic nature of elements depends on the presence of unpaired electrons. Identify the configuration of transition element, which shows the highest magnetic moment.
(i)
(ii)
(iii)
(iv)
Answer:
The answer is the option (ii).The value of magnetic moment is directly proportional to the number of unpaired electrons. Therefore, has maximum number of unpaired electrons and highest magnetic moment.
Question:7
Which of the following oxidation state is common for all lanthanoids?
(i) +2
(ii) +3
(iii) +4
(iv) +5
Answer:
The answer is the option (ii). +3 oxidation state is common for all lanthanoids. Although sometimes +2 and +4 ions are also obtained in solution or solid compound.
Question:8
Which of the following reactions are disproportionation reactions?
(a)
(b)
(c)
(d)
(i) a, b
(ii) a, b, c
(iii) b, c, d
(iv) a, d
Answer:
The answer is the option (i). Disproportionation reaction is where an element is simultaneously oxidised and reduced.
Question:9
When solution is added to the oxalic acid solution, the decolourisation is slow in the beginning but becomes instantaneous after some time because
(i) is formed as the product.
(ii) The reaction is exothermic.
(iii) catalyses the reaction.
(iv) acts as an autocatalyst.
Answer:
The answer is the option (iv). Acidified oxidises Oxalate ion to
on addition and
acts as an auto catalyst. This is why decolorisation is slow in the beginning but becomes instantaneous after some time.
Reduction half
Oxidation half
Overall equation
End point of this reaction colourless to light pink.
Question:10
There are 14 elements in the actinoid series. Which of the following elements does not belong to this series?
(i)
(ii)
(iii)
(iv)
Answer:
The answer is the option (iii). Tm (Thulium) that has an atomic number of 69 is a lanthanoid (4f) series.
Question:11
acts as an oxidising agent in acidic medium. The number of moles of
that will be needed to react with one mole of sulphide ions in acidic solution is
(i)
(ii)
(iii)
(iv)
Answer:
The answer is the option (i).
For 5 moles of S the number of moles of
For 1 mole of S the number of moles of
Question:12
Which of the following is amphoteric oxide?
(i)
(ii)
(iii)
(iv)
Answer:
The answer is the option (i). Since they react with both acids and bases, and
are amphoteric oxides.
Also, the basic character is predominant in lower oxides whereas the acidic character is predominant in higher oxides
Question:13
Gadolinium belongs to 4f series. Its atomic number is 64. Which of the following is the correct electronic configuration of gadolinium?
(i)
(ii)
(iii)
(iv)
Answer:
The answer is the option (i) is the correct electronic configuration of gadolinium that belongs to 4f series.
Question: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?
(i) They have high melting points in comparison to pure metals.
(ii) They are very hard.
(iii) They retain metallic conductivity.
(iv) They are chemically very reactive.
Answer:
The answer is the option (iv). Interstitial compounds are not chemically very active. Instead, they are inert in nature.
Question:15
The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of ion is___________.
(i) 2.87 B.M.
(ii) 3.87 B.M.
(iii) 3.47 B.M.
(iv) 3.57 B.M
Answer:
The answer is the option (ii). The determination of magnetic moment is done by the number of unpaired electrons. The is associated with its spin angular and orbital angular momentum.
Spin only magnetic moment value of ion is
Hence, magnetic moment
Question:16
acts as an oxidising agent in alkaline medium. When alkaline
is treated with
, iodide ion is oxidised to ____________.
(i)
(ii)
(iii)
(iv)
Answer:
The answer is the option (iii). On treating alkaline with
, we get
Question:17
Which of the following statements is not correct?
(i) Copper liberates hydrogen from acids.
(ii) In its higher oxidation states, manganese forms stable compounds with oxygen and fluorine.
(iii) are oxidising agents in aqueous solution.
(iv) are reducing agents in aqueous solution.
Answer:
The answer is the option (i). There is no liberation of hydrogen from acids when copper is added.
Question:18
When acidified solution is added to
salts then
changes to
(i)
(ii)
(iii)
(iv)
Answer:
The answer is the option (iii). On addition of acidified (Acidified potassium dichromate) solution to
salt,
changes to
.
The reaction is given below
Question:19
Highest oxidation state of manganese in fluoride is but highest oxidation state in oxides is
because ____________.
(i) fluorine is more electronegative than oxygen.
(ii) fluorine does not possess d-orbitals.
(iii) fluorine stabilises lower oxidation state.
(iv) in covalent compounds, fluorine can form a single bond only while oxygen forms double bond.
Answer:
The answer is the option (iv). This is because in covalent compounds, fluorine can form single bond only whereas oxygen has the ability to form multiple bonds and therefore it forms double bond
Question:20
Although Zirconium belongs to the 4d transition series and Hafnium to 5d transition series even then they show similar physical and chemical properties because___________.
(i) both belong to d-block.
(ii) both have the same number of electrons.
(iii) both have a similar atomic radius.
(iv) both belong to the same group of the periodic table
Answer:
The answer is the option (iii). Zirconium and hafnium show similar physical and chemical properties due to the almost identical radii of and
. Similar atomic radius results from lanthanoid contraction.
Question:21
Why is not used to make the medium acidic in oxidation reactions of
in an acidic medium?
(i) Both and
act as oxidising agents.
(ii) oxidises
into
which is also an oxidising agent.
(iii) is a weaker oxidising agent than
.
(iv) acts as a reducing agent in the presence of
.
Answer:
The answer is the option (ii). is oxidised by
into
which is also an oxidising agent. Since hydrochloric acid is oxidised to chlorine, permanganate titrations in their presence are unsatisfactory.
Question: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?
(i)
(ii)
(iii)
(iv)
Answer:
The answer is the option (i, ii) Amongst the given options, the compounds and
are coloured due to the presence of unpaired electrons in metal ions.
Question:23
Transition elements show a magnetic moment due to spin and orbital motion of electrons. Which of the following metallic ions have almost the same spin only magnetic moment?
(i)
(ii)
(iii)
(iv)
Answer:
The answer is the option (i, iv) The metallic ions and
have same number of unpaired electrons which is 3. Thus, they have similar spin only magnetic moment.
Question:24
In the form of dichromate, is a strong oxidising agent in acidic medium but
in
and
in
are not because ___________.
(i) is more stable than
and
.
(ii) and
are more stable than
.
(iii) Higher oxidation states of heavier members of group-6 of transition series are more stable.
(iv) Lower oxidation states of heavier members of group-6 of transition series are more stable.
Answer:
The answer is the option (ii, iii) In d-block elements heavier elements exhibit higher oxidation states in their stable form. Like in group 6, and
are more stable than
. This is why, in the form of dichromate
is a stronger oxidizing agent in acidic medium whereas
in
and
in
are not.
Question:25
Which of the following actinoids show oxidation states up to +7?
(i)
(ii)
(iii)
(iv)
Answer:
The answer is the option (ii, iv) oxidation state of and
is +7 as well.
Question:26
General electronic configuration of actinoids is . Which of the following actinoids have one electron in
orbital?
(i) (Atomic no. 92)
(ii) (Atomic no.93)
(iii) (Atomic no. 94)
(iv) (Atomic no. 95)
Answer:
The answer is the option (i, ii) have one electron in
orbital.
Question:27
Which of the following lanthanoids show +2 oxidation state besides the characteristic oxidation state +3 of lanthanoids?
(i)
(ii)
(iii)
(iv)
Answer:
The answer is the option (ii, iii)
(a) Cerium
Electronic Configuration
Oxidation state of
(b) Europium
Electronic configuration =
Oxidation state of
(c) Ytterbium
Electronic Configuration
Oxidation state of
(d) Holmium
Electronic Configuration
Oxidation state of
Question:28
Which of the following ions show higher spin only magnetic moment value?
(i)
(ii)
(iii)
(iv)
Answer:
The answer is the option (ii, iii). and
will show higher values of spin only magnetic moment.
Question:29
Transition elements form binary compounds with halogens. Which of the following elements will form type compounds?
(i)
(ii)
(iii)
(iv)
Answer:
The answer is the option (i, ii). Due to higher lattice energy in cobalt can form halides like
type of compounds. On the other hand, chromium can form
due to higher bond enthalpy.
Question:30
Which of the following will not act as oxidising agents?
(i)
(ii)
(iii)
(iv)
Answer:
The answer is the option (ii, iii) For a species to act as an oxidizing agent the metal should be in a higher oxidation state whereas stability is exhibited by its lower oxidation state. Since, higher oxidations states of are more stable; therefore, they do not act as oxidizing agents.
Question:31
Although +3 is the characteristic oxidation state for lanthanoids but cerium also shows +4 oxidation state because ___________.
(i) it has variable ionisation enthalpy
(ii) it has a tendency to attain noble gas configuration
(iii) it has a tendency to attain configuration
(iv) it resembles
Answer:
The answer is the option (ii, iii). The extra stability of empty, half-filled or completely filled f subshell gives rise to this irregularity. The noble gas configuration of favours its formation.
Question:32
Why does copper not replace hydrogen from acids?
Answer:
The standard reduction potential value of copper is positive. Therefore obeying the reactivity series of metals, it does not liberate hydrogen from acids.
Question:33
Why values for
and
are more negative than expected?
Answer:
values of
are more negative than expected because of the stability they have due to the half-filled d subshell
and completely-filled
configuration respectively.
Question:34
Why first ionisation enthalpy of is lower than that of
?
Answer:
Given below is the electronic configuration of chromium and zinc
has half-filled d orbital which is very stable and results in lower Ionization enthalpy. Similarly, the value for
is higher because its electron comes out from completely filled 4s orbital.
Question:35
Transition elements show high melting points. Why?
Answer:
Transition metals have a high melting point because of higher degree of metallic bonds formed. In addition to the ns electrons, the d electrons also contribute to the metallic bonding.
Question:36
Iodide ions reduce the ions :
Question:37
Out of and
, which is more stable and why?
Answer:
Due to a high hydration enthalpy, exhibits higher stability
.
Question:38
A, B, C are shown below:
The reactions are explained as
Question:39
Fluorine can form a single bond as it has a single unpaired electron. However, Oxygen
can form double bonds and can thus stabilize higher oxidation states.
Question:40
Unlike ,
has a symmetrical electronic configuration, due to which the magnetic moment is higher for Cobalt ion.
Question:41
Ionisation enthalpies of are higher than
. Why?
Answer:
6th period elements like have smaller sizes as compared to 7th period elements like
. As the size of atom increases, the removal of electron becomes easier (increased distance between outermost electrons and nucleus).
Question:42
Although belongs to the 4f series and
to the 5f series, they have nearly the same size, because of which they exhibit very similar properties.
Question:43
Lanthanoids generally lose 3 electrons from to exhibit +3 oxidation state. However, Cerium has a configuration
, and gains additional stability by losing
electrons as well. It is the case because on losing 4 electrons it has only completely filled orbitals
Question:44
Explain why the colour of disappears when oxalic acid is added to its solution in acidic medium.
Answer:
In an acidic medium, acts as an oxidising agent and itself converts to
, which is colourless.
Question:45
When is treated with an alkali :
When the yellow solution is treated with an acid, we get back the orange solution
Question:46
The oxidising nature of depends on the pH of the solution. They change to colourless manganous ion in acidic medium.
An alkaline solution will turn green because of the formation of manganate
In a neutral solution they will leave a brown precipitate, Manganese Oxide
Question:47
The f-orbital contributes extraordinarily little to shielding, because of which the effective nuclear charge increases and the size decreases. Due to this the second and third row transition elements have similar atomic radii and resemble each other much more than the third-row elements.
Question:48
of
is + 0.34V while that of
is – 0.76V. Explain.
Answer:
has to lose an electron from a fully filled
orbital to become
. This isn’t balance by the hydration enthalpy giving
a positive
value. On the other hand,
forms a relatively more stable
with an electronic configuration of
, which is completely filled, it has a lower ionization enthalpy than
, but nearly the same hydration energy and a negative
value.
Question:49
With an increase in the oxidation state, the element’s charge increases and size decreases. Fajan’s rule states that the smaller the ion, the more the bond will exhibit a covalent nature
Question:50
While electrons are filled according to rule (which is lower for
than
, they leave according to the ionisation enthalpy. 4s electrons, being further from nucleus, are more loosely held and are removed.
Question:51
Reactivity of transition elements decreases almost regularly from . Explain.
Answer:
Effective nuclear charge increases as we move along the period from left to right. Due to this, there is a decrease in size as well. Therefore, the electrons will be held more tightly and removing them from the outermost shell will be difficult. Thus, the ionization enthalpy also increases and reactivity decreases. is more reactive than
.
Question:52
Match the catalysts given in Column I with the processes given in Column II.
Column I (Catalyst) | Column II (Process) |
(i) | (a) Zieglar Natta catalyst |
(ii) | (b) Contact process |
(iii) | (c) Vegetable oil to ghee |
(iv) Finely divided iron | (d) Sandmeyer reaction |
(v) | (e) Haber's Process |
(f) Decomposition of |
Answer:
(i c), (ii
d), (iii
b), (iv
e), (v
a)
Question:53
Match the compounds/elements given in Column I with uses given in Column II.
Column I (Compound/element) | Column II (Use) |
(i) Lanthanoid oxide | (a) Production of iron alloy |
(ii) Lanthanoid | (b) Television screen |
(iii)Misch metal | (c) Petroleum cracking |
(iv) Magnesiumbased alloy is constituents of | (d) Lanthanoid metal+iron |
(v) Mixed oxides of lanthanoids are employed | (e) Bullets |
(d) In X-ray screen |
Answer:
(i b), (ii
a), (iii
d), (iv
e), (v
c)
Question:54
Match the properties given in Column I with the metals given in Column II.
Column I (Property) | Column II (Metal) |
(i) An element which can show +8 oxidation state | (a) |
(ii) 3d block element that can show upto +7 oxidation state | (b) |
(iii)3d block element with highest melting point | (c) |
(d) |
Answer:
(i c), (ii
a), (iii
b)
Question:55
Match the statements given in Column I with the oxidation states given in Column II.
Column I | Column II |
(i) Oxidation state of | (a) +2 |
(ii) Most stable oxidation state of | (b) +3 |
(iii) Most stable oxidation state of | (c) +4 |
(iv) Characteristic oxidation state of lanthanoids is | (d) +5 |
(e) +7 |
Answer:
(i c), (ii
a), (iii
e), (iv
b)
Question:56
Match the solutions given in Column I and the colours given in Column II.
Column I (Aqueous solution of salt) | Column II (Colour) |
(i) | (a) Green |
(ii) | (b) Light pink |
(iii) | (c) Blue |
(iv) | (d) Pale green |
(v) | (e) Pink |
(f) Colourless |
Answer:
(i d), (ii
a), (iii
b), (iv
e), (v
f)
Question:57
Match the property given in Column I with the element given in Column II.
Column I (Property) | Column II (Element) |
(i) Lanthanoid which shows +4 oxidation state | (a) |
(ii) Lanthanoid which can show +2 oxidation state | (b) |
(iii) Radioactive lanthanoid | (c) |
(iv) Lanthanoid which has | (d) |
(v) Lanthanoid which has | (e) |
(f) |
Answer:
(i) → (b) (ii) → (d) (iii) → (a) (iv) → (e) (v) → (c)
(i) Cerium has oxidation state +4
(ii) Europium has oxidation state of +2
(iii) Promethium is a man-made radioactive lanthanoid.
(iv) Gadolinium has electronic configuration in +3 oxidation state is.
Question:58
Match the properties given in Column I with the metals given in Column II.
Column I (Property) | Column II (Metal) |
(i) Element with highest second ionisation enthalpy | (a) |
(ii) Element with highest third ionisation enthalpy | (b) |
(iii) M in | (c) |
(iv) Element with highest heat of atomisation | (d) |
(e) |
Answer:
(i c), (ii
d), (iii
b), (iv
a)
i) has a configuration
, making removal of the next electron a challenge
ii) Similar to ,
has a configuration
iii)
iv) Energy of atomization is highest for Nickel
Question:59
In the following question, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: iodide is not known.
Reason: oxidises I– to iodine.
(i) Both assertion and reason are true, and reason is the correct explanation of the assertion.
(ii) Both assertion and reason are true but reason is not the correct explanation of assertion.
(iii) Assertion is not true but reason is true.
(iv) Both assertion and reason are false.
Answer:
The answer is the option (i). Iodide gets oxidised in presence of to Iodine.
Question:60
In the following question, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Separation of is difficult.
Reason: Because lie in the same group of the periodic table.
(i) Both assertion and reason are true, and reason is the correct explanation of the assertion.
(ii) Both assertion and reason are true but reason is not the correct explanation of assertion.
(iii) Assertion is not true but reason is true.
(iv) Both assertion and reason are false.
Answer:
The answer is the option (ii). Due to their similar sizes, it is difficult to separate
Question:61
In the following question, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Actinides form relatively less stable complexes as compared to lanthanoids.
Reason: Actinides can utilise their 5f orbitals along with 6d orbitals in bonding but lanthanoids do not use their 4f orbital for bonding.
(i) Both assertion and reason are true, and reason is the correct explanation of the assertion.
(ii) Both assertion and reason are true but reason is not the correct explanation of assertion.
(iii) Assertion is not true but reason is true.
(iv) Both assertion and reason are false.
Answer:
The answer is the option (iii). Actinoids form more stable complex compounds than Lanthanoids.
Question:62
In the following question, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion : cannot liberate hydrogen from acids.
Reason: Because it has positive electrode potential.
(i) Both assertion and reason are true, and reason is the correct explanation of the assertion.
(ii) Both assertion and reason are true but reason is not the correct explanation of assertion.
(iii) Assertion is not true but reason is true.
(iv) Both assertion and reason are false.
Answer:
(i) has a higher electrode potential than hydrogen because of which it can’t liberate Hydrogen in acidic solutions.
Question:63
In the following question, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: The highest oxidation state of osmium is +8.
Reason: Osmium is a 5d-block element
(i) Both assertion and reason are true, and reason is the correct explanation of the assertion.
(ii) Both assertion and reason are true but reason is not the correct explanation of assertion.
(iii) Assertion is not true but reason is true.
(iv) Both assertion and reason are false.
Answer:
The answer is the option (ii). Osmium can use all its 8 electrons to expand its octet and show the oxidation state of +8.
Question:64
Identify A to E and also explain the reactions involved.
Answer:
Question:65
The compounds A, B, C and D are given as under:
Following are the reactions :
Question:66
The compounds (i), (ii), (iii) and (iv) are given as under:
Question:67
(a). Covalent character of bond increases as the size of atom decreases. La is larger than Lu (Lanthanide contraction), hence, displays more ionic behaviour, but
displays more covalent behaviour.
(b). Decrease in size of atoms, also leads to decrease in stability of the oxo-salts.
(c). Decrease in size of atoms, corresponds with production of lower stable compounds.
(d). 4d and 5d block elements in the same column have nearly equal radii.
(e) There is an increase in the acidic nature of oxides as the size decreases.
Question:68
(a)
(i)Copper has an electronic configuration & the second electron, which has to be removed from a fully filled d-orbital leads to a very high ionization enthalpy.
(ii) Zinc has an electronic configuration of & the third electron, which has to be removed from a fully filled d-orbital leads to a very high ionization enthalpy.
(iii) With no unpaired electrons, Zinc has the lowest energy of atomization
(b)
(i)
(ii) (oxidation state of +7)
Question:69
Interstitial compounds are formed when crystal lattice of larger transition metals trap smaller atoms like Hydrogen, Carbon and Nitrogen. These compounds have a higher melting points than pure metals, are very hard, conduct electricity and are chemically inert.
Question:70
(a) Reaction between iodide and persulphate ions is :
Role of ions :
(b) acts as a catalyst :
(ii) Vanadium Oxide acts as a catalyst in contact process :
(iii) Finely divided iron acts as a catalyst in Haber's process :
Question:71
The compounds A,B, C and D are given as :
The reactions are as
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This chapter deals with elements of d and f blocks of the modern periodic table. It is essentially the study of matter and its properties. The solutions clear all fundamental concepts. The inner d orbits of group 3 to 13 are progressively filled. Overall, NCERT exemplar Class 12 Chemistry solutions chapter 8 along with providing the right information and interesting for the students who wish to learn about the different aspects of the block elements.
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- NCERT exemplar Class 12 Chemistry chapter 8 solutions provides all the related information about the d and f block elements.
- This chapter provides all the information about the different phases of the block elements. It also gives you the required information for a clear and deep understanding for the students.
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | The d and f block elements |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 | |
Chapter 15 | |
Chapter 16 |
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Also Check NCERT Books and NCERT Syllabus here:
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The experts in this field use all the correct information and provide information with their practical knowledge to solve the questions.
The students who are preparing for their board exams or the students who are preparing for the entrance exams can benefit from this.
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hope this helps.
Hello Akash,
If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.
You can get the Previous Year Questions (PYQs) on the official website of the respective board.
I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.
Thank you and wishing you all the best for your bright future.
Hello student,
If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:
Remember
, these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.
I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.
Good luck with your studies!
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