NCERT Exemplar Class 12 Chemistry Solutions Chapter 8 The d and f block elements

NCERT Exemplar Class 12 Chemistry Solutions Chapter 8: MCQ (Type 1)

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

Metallic radii of some transition elements are given below. Which of these elements will have the highest density?

(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.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 8: MCQ (Type 2)

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.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 8: Short Answer Type

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

When ion is treated with , a white precipitate is formed. Explain the reaction with the help of the chemical equation.
Answer:

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

When a brown compound of manganese (A) is treated with it gives a gas (B). The gas taken in excess reacts with to give an explosive compound (C). Identify compounds A, B and C.
Answer:

A, B, C are shown below:

The reactions are explained as

Question:39

Although fluorine is more electronegative than oxygen, the ability of oxygen to stabilise higher oxidation states exceeds that of fluorine. Why?
Answer:

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

Although and ions have the same number of unpaired electrons the magnetic moment of is 3.87 B.M. and that of is 4.87 B.M. Why?
Answer:

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:

6^th period elements like have smaller sizes as compared to 7^th 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 4d and belongs to 5d transition series but it is quite difficult to separate them. Why?
Answer:

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

Although +3 oxidation states are the characteristic oxidation state of lanthanoids cerium shows +4 oxidation state also. Why?
Answer:

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 an orange solution containing ion is treated with an alkali, a yellow solution is formed and when ions are added to a yellow solution, an orange solution is obtained. Explain why does this happen?
Answer:

When is treated with an alkali :

When the yellow solution is treated with an acid, we get back the orange solution

Question:46

A solution of on reduction yields either colourless solution or a brown precipitate or a green solution depending on the pH of the solution. What different stages of the reduction do these represent and how are they carried out?
Answer:

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 second and third rows of transition elements resemble each other much more than they resemble the first row. Explain why?
Answer:

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

The halides of transition elements become more covalent with increasing oxidation state of the metal. Why?
Answer:

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 filling up of electrons in the atomic orbitals, the 4s orbital is filled before the 3d orbital but the reverse happens during the ionisation of the atom. Explain why?
Answer:

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 .

NCERT Exemplar Class 12 Chemistry Solutions Chapter 8: Matching Type

Question:52

Match the catalysts given in Column I with the processes given in Column II.

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.

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.

Answer:

(i c), (ii a), (iii b)

Question:55

Match the statements given in Column I with the oxidation states given in Column II.

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.

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.

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.

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

NCERT Exemplar Class 12 Chemistry Solutions Chapter 8: Assertion and Reason Type

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.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 8: Long Answer Type

Question:64

Identify A to E and also explain the reactions involved.

Answer:

Question:65

When a chromite ore (A) is fused with sodium carbonate in free excess of air and the product is dissolved in water, a yellow solution of compound (B) is obtained. After treatment of this yellow solution with sulphuric acid, compound (C) can be crystallized from the solution. When compound (C) is treated with , orange crystals of compound (D) crystallise out. Identify A to D and also explain the reactions.
Answer:

The compounds A, B, C and D are given as under:

Following are the reactions :

Question:66

When an oxide of manganese (A) is fused with in the presence of an oxidizing agent and dissolved in water, it gives a dark green solution of compound (B). Compound (B) disproportionates in neutral or acidic solution to give purple compound (C). An alkaline solution of compound (C) oxidises potassium iodide solution to a compound (D) and compound (A) is also formed. Identify compounds A to D and also explain the reactions involved.
Answer:

The compounds (i), (ii), (iii) and (iv) are given as under:

Question:67

On the basis of lanthanoid contraction, explain the following:
(a) Nature of bonding in and .
(b) Trends in the stability of oxo-salts of lanthanoids from .
(c) Stability of the complexes of lanthanoids.
(d) Radii of 4d and 5d-block elements.
(e) Trends in acidic character of lanthanoid oxides.
Answer:

(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) Answer the following questions:
(i) Which element of the first transition series has highest second ionization enthalpy?
(ii) Which element of the first transition series has highest third ionization enthalpy?
(iii) Which element of the first transition series has lowest enthalpy of atomization?
(b) Identify the metal and justify your answer:
(i) Carbonyl (ii)
Answer:

(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

Mention the type of compounds formed when small atoms like get trapped inside the crystal lattice of transition metals. Also give physical and Chemical characteristics of these compounds.
Answer:

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) 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.
Answer:

(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

A violet compound of manganese (A) decomposes on heating to liberate oxygen and compounds (B) and (C) of manganese are formed. Compound (C) reacts with in the presence of potassium nitrate to give compound (B). On heating compound (C) with cone. and , chlorine gas is liberated and a compound (D) of manganese along with other products is formed. Identify compounds A to D and also explain the reactions involved.
Answer:

The compounds A,B, C and D are given as :

The reactions are as