d and f-Block Elements Test

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d and f-Block Elements Test - 4

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Question 1
1 / -0

The equivalent weight of MnSO₄ is ___ of its molecular weight when it is converted to MnO₂.

A
half

B
one-fourth

C
twice

D
thrice

Solution

Oxidation state of Mn in MnSO₄ = +2
Oxidation state of Mn in MnO₂ = +4
The increase in oxidation state is 2. Therefore, the equivalent weight of MnSO₄ is half of its molecular weight when it is converted to MnO₂ .
Question 2
1 / -0

Given below are some points of comparison between actinoids and lanthanoids. Choose the incorrect statements from among them.

a. Actinoids show greater number of oxidation states than lanthanoids.
b. The chemistry of actinoids elements is not so smooth as that of lanthanoids.
c. Most of the cations of actinoids and lanthanoids are coloured.

A
a and b

B
b and c

C
a and c

D
None of these

Solution

All the given statements are correct.
a. Actinoids show greater number of oxidation states than lanthanoids. This is because the 5f orbitals extend further from the nucleus than the 4f orbitals.
b. The chemistry of actinoids elements is not so smooth as that of lanthanoids.
c. Most of the cations of actinoids and lanthanoids are coloured.
Question 3
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Which of following is used in the preparation of the optical glass of the camera having a high refractive index?

A
Ceric sulphate

B
CeO₂

C
Lanthanum oxide

D
Gadolinium sulphate

Solution

Lanthanum oxide is added to the glass of optical instruments to raise the refractive index of glass and to increase the stability.
Neodymium and praseodymium oxides are used for making coloured glasses for goggles.
Question 4
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Directions For Questions

The following question has four choices, out of which ONLY ONE is correct.

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The correct order of ionic radii of Yb³⁺, La³⁺, Eu³⁺ and Lu³⁺ is:

A
La³⁺ < Eu³⁺ < Lu³⁺ < Yb³⁺

B
Yb³⁺ < La³⁺ < Eu³⁺ < Lu³⁺

C
Lu³⁺ < Yb³⁺ < Eu³⁺ < La³⁺

D
Lu³⁺ < Eu³⁺ < La³⁺ < Yb³⁺

Solution

In the lanthanide series, there is a regular decrease in the atomic as well as ionic radii with the increase in the atomic number. This is referred to as the lanthanide contraction.

Although the atomic radii do show some irregularities in the trend, the ionic radii of the trivalent ions (M³⁺) decrease with the increase in the atomic number, i.e. from La⁺³ (106 pm) to Lu³⁺ (86 pm).
Question 5
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In which of the following pairs of elements, does the element with the higher atomic number have a lower ionization energy as compared to the preceding element?

A
Co, Ni

B
Cr, Mn

C
Mn, Fe

D
Cu, Zu

Solution

The ionization energy values increase across the period due to increase in the effective nuclear charge.
However, even though Ni has a higher atomic number than cobalt, it has a lower ionization energy.
This may probably be, because Co with a slightly higher atomic mass has a slightly stronger pull on the valence electrons.
Question 6
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Directions For Questions

The following question has four choices, out of which ONLY ONE is correct.

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For Cr, the oxidation states (+ 2), (+ 3) and (+ 4) are well known, but the most stable oxidation state amongst these is

A
+2

B
+3

C
+4

D
Both +2 and +4

Solution

Cr⁺³ is the most stable oxidation state of Cr as the trivalent cation of chromium has a stable electronic configuration of half filled t₂g orbitals.
In this state, chromium is neither oxidizing nor reducing.
Question 7
1 / -0

The values of the reduction potentials of Cr³⁺ and Mn³⁺ are -0.41V and +1.51V, respectively. Choose the option that is inferred from the given factual information?

A
Cr³⁺ is less stable than Cr²⁺

B
Mn³⁺ is a good oxidizing agent

C
Cr²⁺ is a good oxidizing agent

D
Mn³⁺ is more stable than Mn²⁺

Solution

Mn²⁺ has a stable configuration of half filled d- orbitals and Mn³⁺ is readily reduced to Mn²⁺.

Therefore, Mn³⁺is a good oxidizing agent.

The other statements are wrong as Cr⁺³ is more stable than Cr⁺² and as a result Cr⁺² is a strong reducing agent.
Question 8
1 / -0

H₂O₂gives blue colour on treatment with acidified solution of potassium dichromate. This is due to the formation of

A
Chromium trioxide

B
Chromium(III) oxide

C
Chromium(VI) peroxide

D
Chromic acid

Solution

K₂Cr₂O₇+ H₂SO₄ + 4H₂O₂ → 2CrO₅ (chromic peroxide) + K₂SO₄ + 5H₂O
The blue colour due to chromic peroxide fades away gradually due to decomposition of CrO₅ to Cr³⁺ ions and oxygen.

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d and f-Block Elements Test - 4

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d and f-Block Elements Test - 4

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Question 1

half

one-fourth

twice

thrice

Solution

Question 2

a and b

b and c

a and c

None of these

Solution

Question 3

Ceric sulphate

CeO2

Lanthanum oxide

Gadolinium sulphate

Solution

Question 4

La3+ < Eu3+ < Lu3+ < Yb3+

Yb3+ < La3+ < Eu3+ < Lu3+

Lu3+ < Yb3+ < Eu3+ < La3+

Lu3+ < Eu3+ < La3+ < Yb3+

Solution

Question 5

Co, Ni

Cr, Mn

Mn, Fe

Cu, Zu

Solution

Question 6

+2

+3

+4

Both +2 and +4

Solution

Question 7

Cr3+ is less stable than Cr2+

Mn3+ is a good oxidizing agent

Cr2+ is a good oxidizing agent

Mn3+ is more stable than Mn2+

Solution

Question 8

Chromium trioxide

Chromium(III) oxide

Chromium(VI) peroxide

Chromic acid

Solution

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CeO₂

La³⁺ < Eu³⁺ < Lu³⁺ < Yb³⁺

Yb³⁺ < La³⁺ < Eu³⁺ < Lu³⁺

Lu³⁺ < Yb³⁺ < Eu³⁺ < La³⁺

Lu³⁺ < Eu³⁺ < La³⁺ < Yb³⁺

Cr³⁺ is less stable than Cr²⁺

Mn³⁺ is a good oxidizing agent

Cr²⁺ is a good oxidizing agent

Mn³⁺ is more stable than Mn²⁺