d and f Block Test

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d and f Block Test - 6

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Weekly Quiz Competition

Question 1
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One or More than One Options Correct Type

Direction (Q. Nos. 1-5) This section contains 5 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.

Q.

Which of the following statements are incorrect regarding lanthanides?

A
All ions having unpaired electrons are coloured

B
The first and second ionisation energies are comparable with calcium

C
Ce⁴⁺ is a strong reducing agent

D
Option A,C are Correct

Solution

Ce³⁺ (4f¹ ) and Yb³⁺ (f¹³ ) are colourless because they do not absorb in the visible region. The E °value for Ce⁴⁺ /Ce³⁺ is + 1.74 V which suggest it is strong oxidising agent.
Question 2
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Observe the following statements. Which is correct statement regarding lanthanides?

A
Cerium shows + 4 and + 3 oxidation states

B
Element with atomic numbers 66 and 99 belong to f-block

C
Lanthanides belong to 3rd group and 7th period

D
Monazite is the mineral that constitutes the major source of the lanthanides

Solution

The correct answer is option B
Ce shows +4 oxidation state in aq.solution
Ce (58) to lu (71)- lanthanide
Th (90) to Lr (103)- actinoids.
Question 3
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The colourless ions among the lanthanides are

A
La³⁺

B
Gd³⁺

C
Lu³⁺

D
All are correct

Solution

La³⁺ , Gd³⁺ and Lu³⁺ have empty (4f⁰ ) half-filled (4f⁷ ) and full-filled (4f¹⁴ ) orbitals. Ce³⁺ (4f¹ ) is colourless because it does not absorb in the visible region.
Question 4
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The correct statements among the following are

A
Reducing power decreases from La to Lu

B
La to Lu E °values become less negative

C
Promethium is artificial radioactive element

D
All are correct

Solution

Basic strength decreases from La(OH)₃ to Lu (OH)₃ ,
Question 5
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Which of the following pairs have same atomic size?

A
Zr and Hf

B
Nb and Ta

C
Mo and W

D
All are correct

Solution

The pair of elements Zr - Hf, Nb - Ta, Mo - W possess almost the same size. After the lanthanoids the increase in radii from second to third transition series almost vanishes.
Question 6
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Comprehension Type

Direction : This section contains a paragraph, describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).

Passage

The observed oxidation state of lanthanides either in solution or in insoluble compounds form tripositive cations. Based on the ionisation enthalpies and hydration energy factors, it is concluded that tripositive species is more stable than di or tetrapositive species in aqueous solution. In general, ions having half-filled (Tb⁴⁺ , Eu²⁺ , Gd³⁺ ) or completely filled 4f orbitals (Yb²⁺ , Lu³⁺ ) or noble gas configuration (La³⁺ , Ce⁴⁺ ) are stable. This is supported by their SRP values. Ions with + 4 oxidation state act as oxidising and ions having + 2 oxidation state act as reducing agents. Ions having same number of unpaired electrons in f-orbitals possess same colour.

Q.

Observe the following equations.
Sm³⁺ (aq) + e^- →Sm²⁺ (aq), E °= - 1.55 V
Eu³⁺ (aq) + e^- →Eu²⁺ (aq), E °= - 0.43 V
Yb³⁺ (aq) + e^- →Yb²⁺ (aq), E °= - 1.55V

Based on the above equations, the correct reducing strength is in the order of

A
Sm²⁺ >Yb²⁺ >Eu²⁺

B
Eu²⁺ >Yb²⁺ >Sm²⁺

C
Eu²⁺ >Sm²⁺ >Yb²⁺

D
Sm²⁺ >Eu²⁺ >Yb²⁺

Solution

More negative the SRP value, stronger is the reducing agent.
Question 7
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The observed oxidation state of lanthanides either in solution or in insoluble compounds form tripositive cations. Based on the ionisation enthalpies and hydration energy factors, it is concluded that tripositive species is more stable than di or tetrapositive species in aqueous solution. In general, ions having half-filled (Tb⁴⁺ , Eu²⁺ , Gd³⁺ ) or completely filled 4f orbitals (Yb²⁺ , Lu³⁺ ) or noble gas configuration (La³⁺ , Ce⁴⁺ ) are stable. This is supported by their SRP values. Ions with + 4 oxidation state act as oxidising and ions having + 2 oxidation state act as reducing agents. Ions having same number of unpaired electrons in f-orbitals possess same colour.

Q. Which is correct statement?

A
Ce⁴⁺ is more stable than Ce³⁺ in solution

B
Ce³⁺ is more sable than Ce⁴⁺ in solution

C
Ce⁴⁺ and Ce³⁺ are equally stable

D
Ce⁴⁺ has xenon configuration

Solution

Ce⁴⁺ is [Xe] 4f⁰ , it tends to revert to more stable oxidation state of +3 by gain of an electron. That is why, in aqueous solution, Ce⁴⁺ is good oxidising agent.
Question 8
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Matching List Type

Direction : Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d), out of which one is correct.

Q.

Match the Column I with Column II and mark the correct option from the codes given below.

A
a

B
b

C
c

D
d

Solution

(i) →(p.t) (ii) →(r) (iii) →(q,s) (iv) →(q,s)
Question 9
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One Integer Value Correct Type

Direction : This section contains 5 questions. When worked out will result in an integer from 0 to 9 (both inclusive).

Q.

Number of species that have half-filled 4f subshell among Gd, Lu, Eu, Yb, Eu²⁺ Gd³⁺ ,Eu³⁺ ,Ce⁴⁺ ,Tb⁴⁺ .

A
5

B
4

C
3

D
2

Solution

Gd: [Xe] 4f⁷ 5d¹ 6s² ,
Lu: [Xe] 4f¹⁴ 5d¹ 6s² ,
Eu: [Xe] 4f⁷ 5d⁰ 6s² ,
Yb: [Xe] 4f¹⁴ 5d⁰ 6s² ,
Eu²⁺ : [Xe] 4f⁷ 5d⁰ 6s⁰ ,
Gd³⁺ : [Xe] 4f⁷ 5d⁰ 6s⁰ ,
Eu³⁺ : [Xe] 4f⁶ 5d⁰ 6s⁰ ,
Tb⁴⁺ : [Xe] 4f⁷ 5d⁰ 6s⁰ ,
Ce⁴⁺ : [Xe] 4f⁰ 5d⁰ 6s⁰ ,
Question 10
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Number of unpaired electrons in ytterbium is/are

A
1

B
2

C
5

D
0

Solution

Ytterbium has configuration [Xe] 4f¹⁴ 5d⁰ 6s²
Question 11
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Number of diamagnetic lanthanide ions among

La³⁺ , Ce³⁺ Ce⁴⁺ , Gd³⁺ ,Yb²⁺ ,Yb³⁺ ,Lu²⁺ , Lu³⁺ are

A
4

B
3

C
2

D
1

Solution

La³⁺ , Ce⁴⁺ , Yb²⁺ and Lu³⁺ are diamagnetic in nature.
Question 12
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Number of good reducing agents among

Ce⁴⁺ , Eu²⁺ , Sm²⁺ ,Yb²⁺ ,Tb⁴⁺ ,La³⁺ Ce³⁺ are

A
1

B
2

C
3

D
4

Solution

Eu²⁺ ,Sm²⁺ ,Yb²⁺ are reducing agents.
Question 13
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The highest oxidation state shown by any transition element is

A
5

B
8

C
3

D
0

Solution

The highest oxidation state shown by the transition element is Ru (+8) in 4d series and Os in 5d series.
Question 14
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Statement Type

Direction : This section is based on Statement I and Statement II. Select the correct answer from the codes given below.

Q.

Statement I : The second and third rows of transition elements resemble each other much more than they resemble the first row.

Statement II : Due to lanthanide contraction, the atomic radii of the second and the third row transition elements.

A
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

B
Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I

C
Statement I is correct but Statement II is incorrect

D
Statement II is correct but Statement I is incorrect

Solution

Both statements I and II are correct. Statement II correctly explains the statement I.
Question 15
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Statement I : Many trivalent lanthanide ions are coloured both in solid state and in aqueous solution.

Statement II : Colour of these ions is due to the presence of f-electrons.

A
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

B
Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I

C
Statement I is correct but Statement II is incorrect

D
Statement II is correct but Statement I is incorrect

Solution

Many trivalent lanthanide ions are coloured due to the presence of f-electrons, in solid state as well as in aqueous solutions.

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d and f Block Test - 6

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d and f Block Test - 6

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Question 1

All ions having unpaired electrons are coloured

The first and second ionisation energies are comparable with calcium

Ce4+ is a strong reducing agent

Option A,C are Correct

Solution

Question 2

Cerium shows + 4 and + 3 oxidation states

Element with atomic numbers 66 and 99 belong to f-block

Lanthanides belong to 3rd group and 7th period

Monazite is the mineral that constitutes the major source of the lanthanides

Solution

Question 3

La3+

Gd3+

Lu3+

All are correct

Solution

Question 4

Reducing power decreases from La to Lu

La to Lu E °values become less negative

Promethium is artificial radioactive element

All are correct

Solution

Question 5

Zr and Hf

Nb and Ta

Mo and W

All are correct

Solution

Question 6

Sm2+ >Yb2+ >Eu2+

Eu2+ >Yb2+ >Sm2+

Eu2+ >Sm2+ >Yb2+

Sm2+ >Eu2+ >Yb2+

Solution

Question 7

Ce4+ is more stable than Ce3+ in solution

Ce3+ is more sable than Ce4+ in solution

Ce4+ and Ce3+ are equally stable

Ce4+ has xenon configuration

Solution

Question 8

a

b

c

d

Solution

Question 9

5

4

3

2

Solution

Question 10

1

2

5

0

Solution

Question 11

4

3

2

1

Solution

Question 12

1

2

3

4

Solution

Ce⁴⁺ is a strong reducing agent

La³⁺

Gd³⁺

Lu³⁺

Sm²⁺ >Yb²⁺ >Eu²⁺

Eu²⁺ >Yb²⁺ >Sm²⁺

Eu²⁺ >Sm²⁺ >Yb²⁺

Sm²⁺ >Eu²⁺ >Yb²⁺

Ce⁴⁺ is more stable than Ce³⁺ in solution

Ce³⁺ is more sable than Ce⁴⁺ in solution

Ce⁴⁺ and Ce³⁺ are equally stable

Ce⁴⁺ has xenon configuration