The d- and f- Block Elements Test

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The d- and f- Block Elements Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following compounds is most widely used as an oxidizing agent in the volumetric analysis?

A
$$CeO_{2}$$

B
$$(NH_{4})_{2}[Ce(NO_{3})_{6}]$$

C
$$Ce(SO_{4})_{2}$$

D
$$CeF_{3}$$

Solution

Ceric sulphate $$Ce(SO_{4})_{2}$$ is most widely used as an oxidizing agent in the volumetric analysis.

Ce has +3 and +4 oxidation states. $$Ce^{+4}$$ readily gains an electron to form $$Ce^{+3}$$ and is reduced. Hence, it easily oxidizes other substances.
Question 2
1 / -0

Chemically Zinc group elements closely resemble ______.

A
I A group

B
II A group

C
III A group

D
IV A group

Solution

Zinc belongs to IIB and it closely resembles IIA because of the completely filled valence orbitals.
The reason is in IIA group all the elements have completely filled s-orbital while in IIB group the elements have completely filled s orbital along with d-orbital.
Question 3
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The correct order of ionic radii of $$Yb^{3+},La^{3+},Eu^{3+}$$ and $$Lu^{3+}$$ is:

A
$$Yb^{3+}> Lu^{3+}> Eu^{3+}> La^{3+}$$

B
$$Lu^{3+}< Yb^{3+}< Eu^{3+}< La^{3+}$$

C
$$Lu^{3+}< Eu^{3+}< La^{3+}< Yb^{3+}$$

D
$$La^{3+}< Eu^{3+}< Lu^{3+}< Yb^{3+}$$

Solution

In periodic table, as we move from left to right in any period, the atomic and ionic radii decreases. This is due to the increase in a nuclear charge in a period.
Thus the correct order of ionic radii in Lanthanoid is:
Ionic radii order: $$Lu^{3+}< Y^{3+}< Eu^{3+}< La^{3+}$$
Question 4
1 / -0

Differentiating electron in inner transition elements enters the ________ orbital.

A
s

B
p

C
d

D
f

Solution

In inner transition elements, the differentiating electron enters into f orbitals. So these elements are called f-block elements.

The ‘f’ – block elements are also known by the name of inner transition elements. In these elements, the last electron usually enters the penultimate i.e. (n – 2) f of the orbital.

Option D is correct.
Question 5
1 / -0

d -block elements form complexes because they have:

A
vacant orbitals

B
small sizes

C
higher nuclear charge

D
all of the above

Solution

Transition metals has incompletely filled d orbital. They have vacant orbitals to accept coordination bond. Complex compounbs are those in which the metal ion bind a number of anions or neutral molecules giving complex species with characteristics properties.
The transition metals form a large no. of complex compounds.
This is due to the comparatively smaller size of the metal ions,their high ionic charges and the availability of d orbitals for bond formation.
Question 6
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The elements which exhibit both vertical and horizontal similarities are _______.

A
inert gas elements

B
representative elements

C
rare elements

D
transition elements

Solution

Elements which exhibit both vertical and horizontal similarities are transition elements. They show vertical similarity because of same electronic configuration. They show horizontal similarity because of similar size.
Question 7
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Which of the following statement regarding transition elements is incorrect?

A
Their atoms contain partially filled d orbitals.

B
They are capable of showing variable valencies.

C
All of their ions are colourless.

D
They form complexes readily.

Solution

Properties of transition elements include:

  1) Have large charge/radius ratio;

2) Are hard and have high densities;

  3) Have high melting and boiling points;

4) Form compounds which are often paramagnetic;

  5) Show variable oxidation states;

6) Form coloured ions and compounds;

  7) Form compounds with profound catalytic activity;

8) Form stable complexes.
So, option C is an incorrect statement.
Hence option C is correct.
Question 8
1 / -0

Oil is converted into fat by using ____ as a catalyst.

A
Ni

B
Fe

C
Mn

D
$$V_{2}O_{5}$$

Solution

Nickel is used as catalyst in oil industry. It is used in catalytic hydrogenation.
Question 9
1 / -0

The following belongs to d-block but it is not transition element

A
Mn

B
Fe

C
Zn

D
Cr

Solution

Zinc is not a transition element due to complete d-orbital.
Question 10
1 / -0

Which of the following is not an interstitial compound?

A
$$TiC$$

B
$$MoC$$

C
$$Fe_{0.82}O$$

D
$$Cr_{2}O_{3}$$

Solution

$$TiC$$, $$MoC$$ and $$Fe_{0.82}O$$ are interstitial compounds. However, $$Cr_{2}O_{3}$$ is not an interstitial compound.
Interstitial compounds are those compounds which are formed when small atoms like $$H,\ C$$ or $$N$$ are trapped inside the interstitial spaces in the crystal lattice of metals. They are usually non-stoichiometric in nature and cannot be represented by a definite composition and are neither typically ionic nor covalent.

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The d- and f- Block Elements Test - 19

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The d- and f- Block Elements Test - 19

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SHARING IS CARING

Question 1

$$CeO_{2}$$

$$(NH_{4})_{2}[Ce(NO_{3})_{6}]$$

$$Ce(SO_{4})_{2}$$

$$CeF_{3}$$

Solution

Question 2

I A group

II A group

III A group

IV A group

Solution

Question 3

$$Yb^{3+}> Lu^{3+}> Eu^{3+}> La^{3+}$$

$$Lu^{3+}< Yb^{3+}< Eu^{3+}< La^{3+}$$

$$Lu^{3+}< Eu^{3+}< La^{3+}< Yb^{3+}$$

$$La^{3+}< Eu^{3+}< Lu^{3+}< Yb^{3+}$$

Solution

Question 4

s

p

d

f

Solution

Question 5

vacant orbitals

small sizes

higher nuclear charge

all of the above

Solution

Question 6

inert gas elements

representative elements

rare elements

transition elements

Solution

Question 7

Their atoms contain partially filled d orbitals.

They are capable of showing variable valencies.

All of their ions are colourless.

They form complexes readily.

Solution

Question 8

Ni

Fe

Mn

$$V_{2}O_{5}$$

Solution

Question 9

Mn

Fe

Zn

Cr

Solution

Question 10

$$TiC$$

$$MoC$$

$$Fe_{0.82}O$$

$$Cr_{2}O_{3}$$

Solution

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