d and f-Block Elements Test

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d and f-Block Elements Test - 13

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Weekly Quiz Competition

Question 1
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Only One Option Correct Type
Direction (Q. Nos. 1- 10) This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
Q.
Set of continuous atomic numbers of elements are present in the same group as well as same period.

A
56,57,58,59

B
88,89,90,91

C
101,102,103,104

D
68,69,70,71

Solution

Er (68), Tm (69), Yb (70) and Lu (71) elements are present in same group as well as same period.
Question 2
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The element with atomic number 62 belongs

A
III A group and 6th period

B
III B group and 7th period

C
IV B group and 7th period

D
III B group and 6th period

Solution

Samarium (Sm) (Z = 62) belongs to III B group and 6th period.
Question 3
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The radius of lanthanum is 1.06 Å. Then which of the following is closest to the radius of Lu3+ ion

A
1.06 Å

B
1.4 Å

C
1.6 Å

D
0.85 Å

Solution

Due to lanthanoid contraction.
Question 4
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+ 4 ion of which has half-filled 4f subshell

A
Tb

B
Lu

C
Gd

D
Eu

Solution

Tb (Z = 65) have [Xe] 4f⁹5d⁰6s² configuration. Tb⁴⁺ have half-filled 4f subshell, i.e. 4f⁷.
Question 5
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Lanthanide for which + II and + III oxidation states are common is

A
La

B
Nd

C
Ce

D
Eu

Solution

Europium (Eu) has [Xe] 4f⁷5d⁰6s² configuration. The oxidation state of + 2 and + 3 are exhibited by these elements, e.g. Eu²⁺ is[Xe] 4f⁷and Eu³⁺ is [Xe]4f⁶.
Question 6
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An alloy of lanthanide

A
misch metal

B
nichrome

C
wood metal

D
type metal

Solution

Misch metal is lanthanide metal (95%) and iron (5%) and traces of S, C, Ca, Al.
Question 7
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Across the lanthanide series, the basicity of the lanthanide hydroxides

A
increases

B
decreases

C
first increases and then decreases

D
first decreases and then increases

Solution

The basicity decreases from Ce(OH)₃ to Lu(OH)₃. Due to lanthanide contraction, the decrease in the size of the cation increases the covalent character between the lanthanide ion and hydroxide ion thereby, reducing the basic character of the lanthanide hydroxide.
Question 8
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Gadolinium belongs to 4f series. Its atomic number is 64. Which of the following is the correct electronic configuration of gadolinium?

A
[Xe] 4f⁷5d¹6s²

B
[Xe] 4f⁶5d²6s²

C
[Xe] 4f⁸6d²

D
[Xe] 4f⁹5s¹

Solution

Gadolinium (Gd) have [Xe] 4f⁷5d¹6s² configuration.
Question 9
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Lanthanides contraction causes

A
small increase in basic strength of oxides of lanthanides

B
great variation in chemical properties of lanthanides

C
small increase in electronegativities of lanthanides

D
small decrease in standard electrode potential value of lanthanides

Solution

Electronegativity increases as we move from left to right due to lanthanide contraction.
Question 10
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Which one of the following statement(s) is/are incorrect for lanthanides?

A
Lanthanide metal ions are mostly coloured

B
Lanthanides have high melting and boiling points

C
Lanthanides have low densities

D
Lanthanides have high ionisation energy

Solution

The hardness increases with increasing atomic number.
Question 11
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Comprehension Type
Direction (Q. Nos. 11 and 12) 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 12
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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 13
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Matching List Type
Direction (Q. No. 13) 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 14
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Statement Type
Direction (Q. Nos. 14 and 15) 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 Elements Test - 13

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d and f-Block Elements Test - 13

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Question 1

56,57,58,59

88,89,90,91

101,102,103,104

68,69,70,71

Solution

Question 2

III A group and 6th period

III B group and 7th period

IV B group and 7th period

III B group and 6th period

Solution

Question 3

1.06 Å

1.4 Å

1.6 Å

0.85 Å

Solution

Question 4

Tb

Lu

Gd

Eu

Solution

Question 5

La

Nd

Ce

Eu

Solution

Question 6

misch metal

nichrome

wood metal

type metal

Solution

Question 7

increases

decreases

first increases and then decreases

first decreases and then increases

Solution

Question 8

[Xe] 4f75d16s2

[Xe] 4f65d26s2

[Xe] 4f86d2

[Xe] 4f95s1

Solution

Question 9

small increase in basic strength of oxides of lanthanides

great variation in chemical properties of lanthanides

small increase in electronegativities of lanthanides

small decrease in standard electrode potential value of lanthanides

Solution

Question 10

Lanthanide metal ions are mostly coloured

Lanthanides have high melting and boiling points

Lanthanides have low densities

Lanthanides have high ionisation energy

Solution

Question 11

Sm2+ > Yb2+ > Eu2+

Eu2+ > Yb2+ >Sm2+

Eu2+ > Sm2+ > Yb2+

Sm2+ > Eu2+ > Yb2+

Solution

Question 12

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

[Xe] 4f⁷5d¹6s²

[Xe] 4f⁶5d²6s²

[Xe] 4f⁸6d²

[Xe] 4f⁹5s¹

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