The d- and f- Block Elements Test

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The d- and f- Block Elements Test - 60

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Question 1
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Directions For Questions

Transition metal and their compounds are used as catalysts in industry and in biological system. For example, in the contact process, vanadium compounds in the +5 state ($$V_2O_5$$ or $$VO_3{^-}$$) are used to oxidise $$SO_2$$ to $$SO_3$$ : $$SO_2$$ + $$\displaystyle{\frac{1}{2}}O_2 \overset{V_2O_5} {\rightarrow}$$ $$SO_3$$. It is thought that the actual oxidation process takes place in two stages. In the first step, $$V^{5+}$$ in the presence of oxide ions converts $$SO_2$$ to $$SO_3$$. At the same time, $$V^{5+}$$ is reduced to $$V^{4+}$$. $$2V^{5+} + O^{2-} + SO_2 \rightarrow 2V^{4+} + SO_3$$. In the second step, $$V^{5+}$$ is regenerated from $$V^{4+}$$ by oxygen : $$2V^{4+} + \displaystyle{\frac{1}{2}O_2 \rightarrow} 2V^{5+} + O^{2-}$$. The overall process is, of course, the sum of these two steps: $$SO_2 + \displaystyle{\frac{1}{2}}O_2 \rightarrow SO_3$$.

...view full instructions

Transition metals and their compounds catalyse reactions because:

A
they have completely filled $$s$$ - subshell.

B
they have a comparable size due to poor shielding of d-subshell.

C
they introduce an entirely new reaction mechanism with lower activation energy.

D
they have variable oxidation states.

Solution

The catalytic activity of transition metals is attributed to the following reason. Because of their variable oxidation states transition metals sometimes form unstable intermediate compounds and provide a new path with a lower activation energy for the reaction.
Question 2
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Directions For Questions

Transition metal and their compounds are used as catalysts in industry and in biological system. For example, in the contact process, vanadium compounds in the +5 state ($$V_2O_5$$ or $$VO_3{^-}$$) are used to oxidise $$SO_2$$ to $$SO_3$$ : $$SO_2$$ + $$\displaystyle{\frac{1}{2}}O_2 \overset{V_2O_5} {\rightarrow}$$ $$SO_3$$. It is thought that the actual oxidation process takes place in two stages. In the first step, $$V^{5+}$$ in the presence of oxide ions converts $$SO_2$$ to $$SO_3$$. At the same time, $$V^{5+}$$ is reduced to $$V^{4+}$$. $$2V^{5+} + O^{2-} + SO_2 \rightarrow 2V^{4+} + SO_3$$. In the second step, $$V^{5+}$$ is regenerated from $$V^{4+}$$ by oxygen : $$2V^{4+} + \displaystyle{\frac{1}{2}O_2 \rightarrow} 2V^{5+} + O^{2-}$$. The overall process is, of course, the sum of these two steps: $$SO_2 + \displaystyle{\frac{1}{2}}O_2 \rightarrow SO_3$$.

...view full instructions

Which of the following ion involved in the above process will show paramagnetism?

A
$$V^{5+}$$

B
$$V^{4+}$$

C
$$O^{2-}$$

D
$$VO_3^-$$

Solution

As here in this process, the $$V^{5+}$$ first changes to $$V^{4+}$$ and then again, $$V^{4+}$$ changes to $$V^{5+}$$.
$$V^{4+}$$ has $$3d^1$$ configuration, so it has 1 unpaired electron, due to which this can show paramagnetism.
Whereas $$V^{5+}$$, has $$3d^0$$ configuration, hence, due to no unpaired electrons, it does not show paramagnetism.
In the same way, others also do not show paramagnetism.
Question 3
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Identify the order in which the spin only magnetic moment (in BM) increases for the following four ions:
(I) $$Fe^{2+}$$ (II) $$Ti^{2+}$$ (III) $$Cu^{2+}$$ (IV) $$V^{2+}$$

A
$$I, II, IV, III$$

B
$$IV, I, II, III$$

C
$$III, IV, I, II$$

D
$$III, II, IV, I$$

Solution

$$(I) Fe^{2+}$$ has $$3d^6$$ configuration.
Hence, it has 4 unpaired electrons.

$$(II) Ti^{2+} $$ has $$3d^2$$ configuration.
So, it has 2 unpaired electrons.

$$(III) Cu^{2+}$$ has $$3d^9$$ configuration.
So, it has 1 unpaired electron.

$$(IV) V^{2+}$$ has $$3d^3$$ configuration.
So, it has 3 unpaired electrons.

More the unpaired electrons more will be the spin only magnetic moment. So the order of increasing unpaired electrons and spin only magnetic moment of species is as follows:

$$(III)<(II)<(IV)<(I)$$
Question 4
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When $$MnO_2$$ is fused with $$KOH$$ and $$KNO_3$$, a coloured compound is formed, the product and its colour is:

A
$$K_2MnO_4, $$ Green

B
$$KMnO_4,$$ Purple

C
$$Mn_2O_3, $$ Blue

D
$$K_3MnO_2$$, Red

Solution

When $$MnO_2$$ is fused with $$KOH$$ and $$KNO_3$$ potassium magnate $$K_2MnO_4$$ is formed .$$K_2MnO_4$$ is green coloured
Question 5
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The bivalent metal ion having maximum paramagnetic behaviour among the first transition series elements is :

A
$$Mn^{2+}$$

B
$$Cu^{2+}$$

C
$$Sc^{2+}$$

D
$$Cu^+$$

Solution

$$Mn^{2+}$$ will have 5 unpaired electrons in its electronic configuration so it will have maximum paramagnetic behaviour.
Question 6
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Statement 1 : A piece of zinc placed in a blue copper nitrate solution will displace the copper from the solution, producing copper metal and a colorless $$\displaystyle { Zn }^{ 2+ }$$ solution.
Statement 2 : Copper is a much more active metal than zinc.

A
Both Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1.

B
Both Statement 1 and Statement 2 are correct and Statement 2 is not the correct explanation of Statement 1.

C
Statement 1 is correct but Statement 2 is not correct.

D
Statement 1 is not correct but Statement 2 is correct.

E
Both the Statement 1 and Statement 2 are not correct.

Solution

Any oxidizing agent (the species on the left of the half reaction) can oxidize any reducing agent (the species on the right of half reaction) that appears below it but cannot oxidize the specie located above it in the electrochemical series.
Cu is above Zn in the electrochemical series. Hence, Cu(II) can oxidize Zn.
Statement 1 : A piece of zinc placed in a blue copper nitrate solution will displace the copper from the solution, producing copper metal and a colorless Zn$$^{2+}$$ solution.
Statement 2 : Copper is a less active metal than zinc.
Statement 1 is correct but Statement 2 is not correct.
Hence, the option (C) is the correct answer.
Question 7
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The electrical conductivity of copper is due to which of the following?

A
Hydrogen bonding

B
Ionic bonding

C
Network bonding

D
London dispersion force

E
Metallic bonding

Solution

Metallic bonding allow copper to conduct electricity. Positively charged nuclei float in a sea of mobile electrons. The electrons are free to move from one nucleus to other. This helps in thermal and electrical conductivity.
Question 8
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When a brown compound of Mn(A) is treated with $$HCl$$, it gives a gas B. The gas B taken in excess reacts with $$NH_3$$ to give an explosive compound C. The compounds A, B and C are:

A
$$A=MnO_2, B=Cl_2, C=NCl_3$$

B
$$A=MnO, B=Cl_2, C=NH_4Cl$$

C
$$A=Mn_3O_4, B=Cl_2, C=NCl_3$$

D
$$A=MnO_3, B=Cl_2, C=NCl_2$$

Solution

When brown compound $$\displaystyle Mn_3O_4$$ is treated with $$HCl$$, it gives chlorine gas.

$$\displaystyle Mn_3O_4 + 8HCl \rightarrow 3MnCl_2+4H_2O+ Cl_2 \uparrow $$

Chlorine gas taken in excess reacts with ammonia to give an explosive compound $$\displaystyle NCl_3$$.
$$\displaystyle NH_3 + 3Cl_2(excess) \rightarrow NCl_3+3HCl$$

The compounds A, B and C are $$\displaystyle A=Mn_3O_4, B=Cl_2, C=NCl_3$$
Question 9
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Statement 1: Transition metal compounds generally exhibit bright colors.
Statement 2: The electrons in the partially filled d orbitals are easily promoted to excited states.

A
Both Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1.

B
Both Statement 1 and Statement 2 are correct and Statement 2 is not the correct explanation of Statement 1.

C
Statement 1 is correct but Statement 2 is not correct.

D
Statement 1 is not correct but Statement 2 is correct.

E
Both the Statement1 and Statement 2 are not correct.

Solution

Statement 1: Transition metal compounds generally exhibit bright colors.
This is correct statement.
Statement 2: The electrons in the partially filled d orbitals are easily promoted to excited states.
This is correct explanation .
As due to transition from the lower energy level to the higher energy level, the electrons absorbs energy and then emit energy when these come back to initial state. That energy fall in the visible light wavelength.
Question 10
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In a transition series, with the increase in atomic-number, the paramagnetism :

A
increases gradually

B
decreases gradually

C
first increases to a maximum and then decreases

D
first decreases to a minimum and then increases

Solution

For transition elements, with increase in atomic number electrons enter the empty $$d$$-orbitals. First they are singly occupied which increases paramagnetism until the point where all the $$d$$-orbitals are half-filled (point of maximum paramagnetism) and then they get paired which decreases paramagnetism. Hence, paramagnetism first increases till maximum then decreases.

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Re-Attempt Weekly Quiz Competition

The d- and f- Block Elements Test - 60

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The d- and f- Block Elements Test - 60

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Question 1

they have completely filled $$s$$ - subshell.

they have a comparable size due to poor shielding of d-subshell.

they introduce an entirely new reaction mechanism with lower activation energy.

they have variable oxidation states.

Solution

Question 2

$$V^{5+}$$

$$V^{4+}$$

$$O^{2-}$$

$$VO_3^-$$

Solution

Question 3

$$I, II, IV, III$$

$$IV, I, II, III$$

$$III, IV, I, II$$

$$III, II, IV, I$$

Solution

Question 4

$$K_2MnO_4, $$ Green

$$KMnO_4,$$ Purple

$$Mn_2O_3, $$ Blue

$$K_3MnO_2$$, Red

Solution

Question 5

$$Mn^{2+}$$

$$Cu^{2+}$$

$$Sc^{2+}$$

$$Cu^+$$

Solution

Question 6

Both Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1.

Both Statement 1 and Statement 2 are correct and Statement 2 is not the correct explanation of Statement 1.

Statement 1 is correct but Statement 2 is not correct.

Statement 1 is not correct but Statement 2 is correct.

Both the Statement 1 and Statement 2 are not correct.

Solution

Question 7

Hydrogen bonding

Ionic bonding

Network bonding

London dispersion force

Metallic bonding

Solution

Question 8

$$A=MnO_2, B=Cl_2, C=NCl_3$$

$$A=MnO, B=Cl_2, C=NH_4Cl$$

$$A=Mn_3O_4, B=Cl_2, C=NCl_3$$

$$A=MnO_3, B=Cl_2, C=NCl_2$$

Solution

Question 9

Both Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1.

Both Statement 1 and Statement 2 are correct and Statement 2 is not the correct explanation of Statement 1.

Statement 1 is correct but Statement 2 is not correct.

Statement 1 is not correct but Statement 2 is correct.

Both the Statement1 and Statement 2 are not correct.

Solution

Question 10

increases gradually

decreases gradually

first increases to a maximum and then decreases

first decreases to a minimum and then increases

Solution

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