In this question paper, some of the important one mark, two and five marks questions from the chapter The d- and f- Block Elements are covered. The questions are prepared from the book back and previous year questions.
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Questions:

Generic 2

6 x 2 = 12

Give reasons:
(i) Mn shows the highest oxidation state of +7 with oxygen but with fluorine, it shows the highest oxidation state of +4.
(ii) Transition metals show variable oxidation states.
(iii) Actinoids show irregularities in their electronic configurations.

Why are Mn²⁺ compounds more stable than Fe²⁺ compounds towards oxidation to their +3 state ?

Write a note on lanthanoid contraction.

Why Zr and Hf or Nb and Ta exhibit similar properties ?

Or Zirconium (atomic number 40) and hafnium (atomic number 72) occur together in minerals and they exhibit similar properties. Give reasons.

Why does copper not replace hydrogen from acids?

Most of the transition metal ions exhibit characterisitic in colours in aqueous solutions. Give reason.

Generic 3

5 x 3 = 15

Zinc is a transition element and has many useful applications. The presence of zinc in trace amounts is essential in humans and many animals.
Answer the following questions:
(i) What is the role of zinc in the body of humans and animals?
(ii) A compound of zinc is used as a rodent poison. Name the compound.
(iii) Name the compound of zinc used in paints.
(iv) Is \(ZnSO_{ 4 }(aq)\) coloured or colourless?

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

Why do the transition elements exhibit higher enthalpies of atomisation?

10.

Identify the first row transition metal ions which have outer electronic configurations of 3d⁴ and 3d⁶ and describe their oxidation states.

11.

How would you account for the following?
(I) The E^o value for the Mn³⁺ /Mn²⁺couple is much more positive than that for the Cr³⁺/Cr²⁺ couple or Fe ³⁺/Fe ²⁺ couple.
(ii) The highest oxidation state of a metal is exhibited in its oxide or fluoride.
(iii) The atomic radii ofthe metals ofthe third (Sd) series of transition elements are virtually the same as those of the corresponding members of the second (4d) series.

Multiple Choice Question

10 x 1 = 10

12.

Mohr salt, FeSO₄, (NH₄)₂SO₄. 6 H₂ O, is preferred over FeSO₄. 7 H₂O for standardization of KMnO₄ solution because.

(a)

Mohr salt is a double salt while ferrous sulphate is a single salt.

(b)

Mohr salt is not hygroscopic but FeSO₄. 7 H₂O is hygroscopic

(c)

Mohr salt contains only ferrous ions whereas ferrous sulphate contains some ferric ions.

(d)

Mohr salt solution can be titrated even in the absence of H₂SO₄

13.

When pyrolusite id fused with KOH and KCIO₃, we get

(a)

KMnO₄

(b)

K₂MnO₄

(c)

Both KMnO₄ and K₂MnO₄

(d)

None of these

14.

When manganous salt is fused with a mixture of KNO₃ and solid NaOH, the oxidation number of Mn changes from +2 to

(a)

(b)

(c)

(d)

15.

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 coniguration

(c)

it has a tendency to attain f^o configuration

(d)

it resembles Pb⁴⁺

16.

Which of the following reactions are disproportionation reactions?
\((i)\ { Cu }^{ + }\longrightarrow { Cu }^{ 2+ }+{ Cu }\)
\((ii)\ { 3MnO }_{ 4 }^{ - }+{ 4H }^{ + }\longrightarrow { 2MnO }_{ 4 }^{ - }+{ MnO }_{ 2 }+{ 2H }_{ 2 }O\)
\((iii)\ { 2KMnO }_{ 4 }\longrightarrow { K }_{ 2 }{ MnO }_{ 4 }+{ MnO }_{ 2 }+{ O }_{ 2 }\)
\((iv)\ { 2MnO }_{ 4 }^{ - }+{ 3Mn }^{ 2+ }+{ 2H }_{ 2 }O\longrightarrow { 5MnO }_{ 2 }+{ 4H }^{ + }\)

(a)

(i),(ii)

(b)

(i),(ii),(iii)

(c)

(ii),(iii),(iv)

(d)

(i),(iv)

17.

Electronic configuration of a transition element X in +3 oxidation state is [Ar] 3d⁵. What is its atomic number ?

(a)

(b)

(c)

(d)

18.

KMnO₄, oxidation number of Mn is

(a)

+ 2

(b)

+ 4

(c)

+ 6

(d)

+ 7

19.

Which metal has highest density ?

(a)

(b)

(c)

(d)

20.

Which element does not show variable oxidation state ?

(a)

(b)

(c)

(d)

21.

Which of the following is not a transition element ?

(a)

(b)

(c)

(d)

Fill up / 1 Marks

1 x 1 = 1

22.

In the first transition series, the maximum number of oxidation states is shown by _______________.

Generic 5

2 x 5 = 10

23.

Lokesh is a social worker. A milkman in the village has been complaining that a factory in his nearby area dumps chemical waste in his field which has become a major cause of decreasing productivity. Lokesh visited that place and found after analysis that the major waste was potassium permanganate which is being absorbed by the soil. He advised the factory people that they should treat potassium permanganate solution before dumping it into the drain.
Comment in brief:
(i) About the environmental values displayed by Lokesh.
(ii) Write balanced chemical equations for the two reactions showing oxidizing nature of potassium permanganate.

24.

(a) Give reasons for the following:
(i) \(\\ Mn^{ 3+ }\) is a good oxidising agent.
(ii) \(E°_{ M^{ 2+ }/M }\) values are not regular for first row transition metals (3d series).
(iii) Although 'F' is more electronegative than 'O' the highest Mn fluoride is \(MnF_{ 4 }\),
(b) Complete the following equations:
(i) \(2CrO_{ 4 }^{ 2- }+2H^{ + }\longrightarrow \)
(ii) \(KMnO_{ 4 }\overset { Heat }{ \longrightarrow } \)

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Answers Key:

Generic 2

6 x 2 = 12

(i) Manganese shows the highest oxidation state of +7 with oxygen but +4 with fluorine. This is because oxygen has a tendency to form multiple bonds and hence stabilize the high oxidation states.
(ii) The transition elements exhibit variable oxidation states. The variable oxidation states of transition metals are due to the participation of ns and (n - 1) d-electrons. This is because of the very small difference between the energies of (n - 1) d and ns orbitals. For the first five elements, the minimum oxidation state is equal to the number of electrons in the 4s orbitals and the other oxidation states are equal to the sum of 4s and some of the 3d-electrons.
The highest oxidation state is equal to the sum of 4s and 3d electrons. For the remaining elements, the minimum oxidation state is equal to electrons in 4s-orbitals and the maximum oxidation state is not equal to the sum of 4s and 3d electrons. In general, the oxidation state increases up to the middle and then decreases.
(iii) The actinoids involve the filling of 5f-subshell. From thorium (Z = 90) onwards, 5f orbitals get progressively filled. But the energy of 5f and 6d subshells are almost equal and therefore, there are uncertainties regarding the filling of 5f and 6d subshells. The electrons may enter either of these two subshells. Thus, there are irregularities in the electronic configurations of actinoids

Electronic configuration of Mn²⁺ is 3d 5 which is half-filled and hence stable. Therefore, 3rd ionization enthalpy is very high, i.e.,a 3rd electron cannot be lost easily. In the case of Fe²⁺, an electronic configuration is 3d₆. Hence, it can lose one electron easily to give the stable configuration 3d₅.

(a) Lanthanoid contraction :
In the lanthanoids, there is a regular decrease in the size of atoms and ions with an increase in atomic number. This decrease of ionic and atomic radii in the lanthanoid elements is called lanthanoid contraction. For example, the ionic radii decrease from Ce³⁺(111pm) to Lu³⁺ (93 pm).
The cause of lanthanoid contraction:
As we move through the lanthanoid series, 4f-electrons are being added, one at each step.
The mutual shielding effect of electrons is very little, even smaller than that of d-electrons. This is due to -the shape of the f-orbitals. The nuclear charge, however, increases by one at each step. Hence, the inward pull experienced by the 4f-electrons increases. This causes a reduction in the size of the entire 4f' shell. The sum of the successive reductions gives the total lanthanoid contraction. Consequences of lanthanoid contraction. The important consequences of lanthanoid contraction are:
(i) It has a very important effect on the relative properties of the elements which follow the lanthanoids with those which proceed the lanthanoids. We know that there is a regular increase in size as we go from Sc to Y to La in group 3. Similarly, we expect normal, increase in size in other groups from
\(Ti\rightarrow Zr\rightarrow Hf\)
\(V\rightarrow Nb\rightarrow Ta\)
However after lanthanoids, the increase in radii from second to third transition series vanishes. Consequently, the pairs of elements Zr-Hf, Nb- Ta, M~W, etc. have almost similar sizes. Therefore, the properties of these elements are similar.
(ii) There is a small but steady increase in the standard electrode potential values (EO) for the process

Due to lanthanoid contraction, Hf (Z = 72) has size similar to thet of Zr (Z = 40). Hence, their properties are similar and therefore, occur together. For the same reason, Nb and Ta have similar size and hence similar properties.

It is because copper is less reactive than hydrogen and has +ve value of reaction potential.

It is due to presence of unpaired electrons, they undergo d-d transition by absorbing visible light and radiate complementary colour.

Generic 3

5 x 3 = 15

(i) Zinc is needed in our diet for digestion of proteins. There are more than 20 zinc-containing enzymes in the human body which are responsible for proper absorption of CO2 by red blood cells in muscles and other tissues and for maintaining proper pH. Some of the zinc-containing enzymes play an important role in the digestion of proteins by animals.
(ii) Zinc phosphide is used as a rodent poison.
(iii) Zinc oxide (white zinc) is used in paints.
(iv) ZnSO₄(aq) is colorless because it has filled d-subshell.

This is because of ability to form multiple bonds.

Because of large number of unpaired electrons in their atoms they have stronger interatomic interaction and hence stronger bonding between atoms resulting in higher enthalpies of atomisation.

10.

Cr²⁺ has electronic configuration 3d⁴ Chromium also shows +3 and +6 oxidation states. Fe²⁺ has electronic configuration 3d⁶. Iron has oxidation states +2 and +3.

11.

(i) It is because Mn²⁺ is more stable than Mn³⁺ due to stable half filled 3d5 configuration, whereas Cr³⁺(t_2g³) and Fe³ (3d⁵) are more stable than Cr²⁺ sg and Fe²⁺ respectively.
(ii) It is because oxygen and fluorine are strong oxidising agents, highly electronegative, small size and can provide energy for formation of transition metal ion in higher oxidation state.
(iii) It is due to lanthanoid contraction which is due to poor shielding effect off-electrons.

Multiple Choice Question

10 x 1 = 10

12.

13.

(b)
K₂MnO₄

14.

15.

16.

(a)
(i),(ii)

17.

(b)
26

18.

(d)
+ 7

19.

(b)
Os

20.

(a)
Sc

21.

(d)
Pb

Fill up / 1 Marks

1 x 1 = 1

22.

()
manganese

Generic 5

2 x 5 = 10

23.

(i) Lokesh knows about the soil pollution caused by chemical industrial waste. His responsibility towards society and environment and social awareness are displayed here.
(ii) 10FeSO₄ + 2KMn0₄ + 8H₂SO₄ \(\longrightarrow \)K₂SO₄ + 2MnSO₄ + 5 Fe₂(SO₄)₃ + 8H₂O
5H₂C₂O₄ + 2KMnO₄ + 3H₂SO₄ \(\longrightarrow \)+ K₂SO₄ + 2MnSO₄ + 8H₂O + 10CO₂

24.

(i) Mn³⁺ (3d4) on changing to Mn²⁺ (3d5) becomes stable, half filled configuration has extra stability. Therefore, Mn³⁺can be easily reduced and acts as a good oxidizing agent.
(ii) EO(M₂+/M) values are not regular in the first transition series metals because of irregular variation of ionization enthalpies (IE₁ + IE₂) and the sublimation energies.
(iii) Among transition elements, the bonds formed in +2 and +3 oxidation states are mostly ionic. The compounds formed in higher oxidation states are generally formed by sharing of d-electrons.
(a) The transition elements show variable oxidation states because their atoms can lose a different number of electrons. This is due to the participation of inner (n - 1) d-electrons in addition to outer electrons because the energies of the ns and (n - 1) d-subshells are almost equal.
(i) Manganese
(ii) Scandium
(b) The steady decrease in atomic and ionic sizes of lanthanide elements with increasing atomic number is called lanthanide contraction. In the lanthanoids, there is a regular decrease in the size of atoms and ions with an increase in atomic number. For example, the ionic radii decrease from Ce³⁺(111 pm) to Lu³⁺ (93 pm). Cause of lanthanoid contraction. As we move through the lanthanoid series, 4f-electrons are being added, one at each step. The mutual shielding effect of electrons is very little, even smaller than that of d-electrons. This is due to the shape of f-orbitals.
The nuclear charge, however, increases by one at each step. Hence, the inward pull experienced by the 4f-electrons increases. This causes a reduction in the size of the entire 4f' shell. The sum of the successive reductions gives the total lanthanoid contraction.
The important alloy is misch metal.