The d- and f- Block Elements Test

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The d- and f- Block Elements Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

$$K_2Cr_2O_7$$ on heating with aqueous $$NaOH$$, gives:

A
$$CrO^{2-}_4$$

B
$$Cr(OH)_3$$

C
$$Cr_2O^{2-}_7$$

D
$$Cr(OH)_2$$

Solution

Reaction:
$$K_2Cr_2O_7 + 2NaOH \rightarrow K_2CrO_4 + Na_2CrO_4 + H_2O$$
Question 2
1 / -0

Following elements do not show the properties characteristic of d-block elements :

A
$$Cu, Ag, Au$$

B
$$Zn, Hg, Cd$$

C
$$Sc, Ti, V$$

D
$$Fe, Co, Ni$$

Solution

In the elements $$Zn, Hg, Cd, $$ the $$3d$$ shell is completely filled. Hence, they do not show the characteristic properties of transition elements.
For example, since no unpaired, electron is present, these compounds are colourless and diamagnetic.
Question 3
1 / -0

Which shows maximum magnetic moment among the bivalent ions of the first transition series ?

A
$$Fe^{2+}$$

B
$$Co^{2+}$$

C
$$Ni^{2+}$$

D
$$Mn^{2+}$$

Solution

As we know,
Magnetic moment $$\mu= \sqrt {n(n+2)}$$, n is no of unpaired electron.

So higher the number of unpaired electrons, the higher is the magnetic moment.

$$Fe^{2+} \rightarrow d^6\rightarrow$$ 4 unpaired electrons

$$Co^{2+} \rightarrow d^7\rightarrow$$ 3 unpaired electrons

$$Ni^{2+} \rightarrow d^8\rightarrow$$ 2 unpaired electrons

$$Mn^{2+} \rightarrow d^5 \rightarrow$$ 5 unpaired electrons so it will have the highest magnetic moment.
Question 4
1 / -0

What is the correct order of covalent size of the following ?
Nb ______ Ta

A
Nb = Ta

B
Nb < Ta

C
Nb > Ta

D
None of these.

Solution

The covalent radius, is a measure of the size of an atom that forms part of one covalent bond. Niobium $$Nb$$ (atomic number 41) and tantalum $$Ta$$ (atomic number 73) belongs to same group but both have almost equal covalent size due to lanthanoid contraction.

Hence, the correct option is $$A$$
Question 5
1 / -0

The correct order of ionic radii of $$Y^{3+}, La^{3+}, Eu^{3+}$$ and $$Lu^{3+}$$ is:
(Atomic number of Y=39, La=57, Eu=63, Lu=71)

A
$$Lu^{3+} < Eu^{3+} < La^{3+} < Y^{3+}$$

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

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

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

Solution

Due to lanthanide contraction size of lanthanides ions decreases from $$La^{3+}$$ to $$Lu^{3+}$$.

so order is $$Y^{3+} < Lu^{3+} < Eu^{3+} < La^{3+}$$

Hence, the correct option is $$D$$.
Question 6
1 / -0

To protect iron against corrosion, the most durable metal plating on it, is :

A
nickel plating

B
tin plating

C
copper plating

D
zinc plating

Solution

Hint: A thin layer of metal has been put to the outside of a substance and is known as plating.
Explanation:
Metal plating creates a sacrificial coating that shields the iron from corrosion. Instead of oxidizing iron, this metal layer oxidizes itself, protecting the iron.
For plating, a metal's oxidation should be preferred over iron, and it should therefore have a lower reduction potential than iron.
Zinc has a lower reduction potential than iron and thus it can be utilized for plating among the available possibilities. Galvanization is the name of the procedure.
Final Answer:
The correct answer is option $$(D)$$.
Question 7
1 / -0

$$CrO_3 +H_2O \rightarrow A \overset{OH^-}{\rightarrow}B$$
What are $$A$$ and $$B$$ in the above reaction?

A
$$H_2CrO_4,\ H_2Cr_2O_7$$

B
$$H_2Cr_2O_7,\ Cr_2O_3$$

C
$$CrO^{2-}_4,\ Cr_2O_7^{2-}$$

D
$$H_2Cr_2O_7,\ CrO_4^{2-}$$

Solution

The complete reaction is as follows:
$$CrO_3 +H_2O \rightarrow \underset{(A)}{H_2Cr_2O_7} \overset{OH^-}{\rightarrow}\underset{(B)}{CrO^{2-}_4}$$
Question 8
1 / -0

The atomic radii of cerium and promethium are quite similar because

A
they belong to the same period in periodic table

B
they have similar electronic configuration

C
f-orbitals have poor shielding effect

D
nuclear charge is higher on cerium than promethium

Solution

The most significant characteristic of lanthanides is the contraction of the atomic size with increase in atomic number called as lanthanide contraction. While moving along the lanthanides series from $$^{58}Ce$$ to $$^{71}Lu$$, a regular decrease in the size of the atom/ion and an increase in atomic number is observed. This decrease in size is called the lanthanide contraction. The decrease in size, though continuous, is not regular. Lanthanide contraction takes place due to the imperfect shielding of a $$4f$$- electron which increases by one more in the same sub-shell.
So, due to this poor shielding effect, $$Ce$$ and $$Pm$$ have similar radii.
Question 9
1 / -0

The transition metal used as a catalyst is :

A
nickel

B
platinum

C
cobalt

D
all of these

Solution

All are used as catalyst due to variable valency and large surface area.
Question 10
1 / -0

$$Fe$$ is made passive by:

A
$$dil.\ H_2SO_4$$

B
$$dil.\ HCl$$

C
$$conc. \ HNO_3$$

D
$$conc.\ H_2SO_4$$

Solution

The inertness exhibited by metals under conditions when chemical activity is to be expected is called chemical passivity.
When iron comes in contact with concentrated nitric acid for some time and then made to react with reagents, iron shows inertness it becomes inactive or passive for further reactions. Iron by treatment with concentrated nitric acid has lost its usual properties or it has been rendered inert or passive.
Reason: Formation of a thin layer of oxide upon the surface of metal, which protects the inner surface of metal to come in direct contact of air or other chemicals for reaction.

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Re-Attempt Weekly Quiz Competition

The d- and f- Block Elements Test - 24

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The d- and f- Block Elements Test - 24

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Question 1

$$CrO^{2-}_4$$

$$Cr(OH)_3$$

$$Cr_2O^{2-}_7$$

$$Cr(OH)_2$$

Solution

Question 2

$$Cu, Ag, Au$$

$$Zn, Hg, Cd$$

$$Sc, Ti, V$$

$$Fe, Co, Ni$$

Solution

Question 3

$$Fe^{2+}$$

$$Co^{2+}$$

$$Ni^{2+}$$

$$Mn^{2+}$$

Solution

Question 4

Nb = Ta

Nb < Ta

Nb > Ta

None of these.

Solution

Question 5

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

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

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

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

Solution

Question 6

nickel plating

tin plating

copper plating

zinc plating

Solution

Question 7

$$H_2CrO_4,\ H_2Cr_2O_7$$

$$H_2Cr_2O_7,\ Cr_2O_3$$

$$CrO^{2-}_4,\ Cr_2O_7^{2-}$$

$$H_2Cr_2O_7,\ CrO_4^{2-}$$

Solution

Question 8

they belong to the same period in periodic table

they have similar electronic configuration

f-orbitals have poor shielding effect

nuclear charge is higher on cerium than promethium

Solution

Question 9

nickel

platinum

cobalt

all of these

Solution

Question 10

$$dil.\ H_2SO_4$$

$$dil.\ HCl$$

$$conc. \ HNO_3$$

$$conc.\ H_2SO_4$$

Solution

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