Chemistry Test 211

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Chemistry Test 211

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Weekly Quiz Competition

Question 1
4 / -1

Gadolinium has a low value of third ionisation enthalpy because of

A
small size

B
high exchange enthalpy

C
high electronegativity

D
high basic character

Solution

Electronic configuration of Gadolinium

Gd :- [Xe] 4f⁷5d¹6s²

In case of 3rd ionisation enthalpy electron will be removed from 5d and resultant configuration will be [Xe]4f f^⊤ that is stable electronic configuration as it will have high exchange energy, hence less energy will be required to remove 3^rd electron.
Question 2
4 / -1

If radius of second Bohr orbit of the He⁺ion is 105.8pm, what is the radius of third Bohr orbit of Li²⁺ ion?

A
158.7pm

B
15.87pm

C
1.587pm

D
158.7Å

Solution
Question 3
4 / -1

Given below are two statements:

Statement I: Cr2⁺ is oxidising and Mn³⁺ is reducing in nature.

Statement II: Sc³⁺ compounds are repelled by the applied magnetic field.

In the light of the above statements, choose the most appropriate answer from the options given below:

A
Statement I is incorrect but Statement II is correct

B
Both Statement I and Statement II are correct

C
Both Statement I and Statement II are incorrect

D
Statement I is correct but Statement II is incorrect

Solution

Cr⁺²: [Ar]3d⁴

Cr⁺² is reducing as its configuration changes from d⁴ to d³(t_2g³)

Mn⁺³ is oxidising in nature.

Mn⁺²: [Ar] 3d⁵ (extra stability)

Statement I is incorrect

Sc⁺³: [Ar]

diamagnetic - repelled by magnetic field.

Statement (II) is correct.
Question 4
4 / -1

Which of the following reactions is the metal displacement reaction? Choose the right option

A

B

C

D

Solution

Both reactions (1) and (4) are examples of decomposition reactions.

Reactions (2) and (3), both are examples of displacement reactions, while reaction (2) is an example of metal displacement reaction.
Question 5
4 / -1

Identify the incorrect statement.

A
The transition metals and their compounds are known for their catalytic activity due to their ability to adopt multiple oxidation states and to form complexes.

B
Interstitial compounds are those that are formed when small atoms like H,C or N are trapped inside the crystal lattices of metals

C
The oxidation states of chromium in CrO₄²⁻ and Cr₂O₇²⁻ are not the same.

D
Cr²⁺(d⁴) is a stronger reducing agent than Fe₂₊(d⁶) in water.

Solution

Oxidation state of Cr in CrO₄²⁻ and Cr²O₇²⁻ is +6 i.e. oxidation states are same.
Question 6
4 / -1

Which one of the following ions exhibits d−d transition and paramagnetism as well?

A
CrO₄²⁻

B
Cr₂O₇²⁻

C
MnO₄⁻

D
MnO₄²⁻

Solution

In MnO₄²⁻, one unpaired electron (n) is present in d -orbital so, d−d transition is possible.
Question 7
4 / -1

Name the gas that can readily decolourise acidified KMnO4 solution.

A
SO₂

B
NO₂

C
P₂O₅

D
CO₂

Solution

SO₂ readily decolourises pink violet colour of acidified KMnO₄ solution.
Question 8
4 / -1

The reason for greater range of oxidation states in actinoids is attributed to

A
actinoid contraction

B
5f,6d and 7s levels having comparable energies

C
4f and 5d levels being close in energies

D
the radioactive nature of actinoids.

Solution

Actinoids have a greater range of oxidation states due to comparable energies of 5f,6d and 7s orbitals. Hence, all their electrons can take part in bond formation.
Question 9
4 / -1

Which one of the following statements is correct when SO₂ is passed through acidified K₂Cr₂O₇ solution?

A
SO₂ is reduced.

B
Green Cr₂(SO₄)₃ is formed.

C
The solution turns blue.

D
The solution is decolourised.

Solution
Question 10
4 / -1

The catalytic activity of transition metals and their compounds is ascribed mainly to

A
their magnetic behaviour

B
their unfilled d-orbitals

C
their ability to adopt variable oxidation states

D
their chemical reactivity

Solution

The catalytic activity of transition metals and their compounds is ascribed mainly to their ability to adopt variable oxidation states.