Chemistry Test

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Chemistry Test - 13

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Weekly Quiz Competition

Question 1
4 / -1

Which of the following element exhibits \(+3\) oxidation State only?

A
Gallium

B
Thallium

C
Indium

D
Aluminium

Solution

Aluminium element exhibits \(+3\) oxidation State only.

Boron and aluminium exhibit oxidation state of \(+3\) only, while gallium, indium and thallium exhibit oxidation states of both \(+1\) and \(+3\). As we move down the group, the tendency to exhibit \(+3\) oxidation State decreases this occurs due to the inert pair effect.
Question 2
4 / -1

Chemical \(\mathrm{A}\) is used for water softening to remove temporary hardness, A reacts with \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) to generate caustic soda. When \(\mathrm{CO}_{2}\) is bubbled through \(\mathrm{A}\), it turns cloudy. What is the chemical formula of \(\mathrm{A}\)?

A
\(\mathrm {CaCO_{3}}\)

B
\(\mathrm {CaO}\)

C
\(\mathrm {Ca(OH)_{2}}\)

D
\(\mathrm {Ca(HCO_{3})_{2}}\)

Solution

The chemical \(\mathrm{A}\) is calcium hydroxide \(\left(\mathrm{Ca}(\mathrm{OH})_{2}\right) .\) It is used for water softening to remove temporary hardness. It reacts with sodium carbonate to form caustic soda.

\(\mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{Ca}(\mathrm{OH})_{2} \rightarrow \mathrm{CaCO}_{3} \downarrow+\mathrm{NaOH}\)

With carbon dioxide gas, it forms calcium carbonate which turns lime water milky.

\(\mathrm{Ca}(\mathrm{OH})_{2}+\mathrm{CO}_{2} \rightarrow \mathrm{CaCO}_{3} \downarrow+\mathrm{H}_{2} \mathrm{O}\)
Question 3
4 / -1

Which of the following electronic transitions in the line spectrum of \(He^{+}\) will have the same wavelength as that of the first line in the Lyman series of hydrogen atoms?

A

\(4 \leftarrow 2\)

B

\(2 \leftarrow 1\)

C

\(3 \leftarrow 1\)

D

\(4 \leftarrow 1\)

Solution

\(\bar{v}_{H}=R\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right] c m^{-1} \ldots(\mathrm{i})\)
\(\bar{v}_{H e^{+}}=R(2)^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right] \ldots .\) \((ii)\)
Now, \(\bar{v}_{H e^{+}}=R\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right]=R \times \frac{(2)^{2}}{(2)^{2}}\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right]\)
\(=R(2)^{2}\left[\frac{1}{2^{2}}-\frac{1}{4^{2}}\right] \ldots\) \((iii)\)
Comparing \((ii)\) and \((iii)\), it is clear that \(n_{2}=4, n_{1}=2\)
Hence the correct option is (A).
Question 4
4 / -1

In the presence of a catalyst, the heat evolved or absorbed during the reaction _______.

A
Increases

B
Decreases

C
Remains unchanged

D
May increase or decrease

Solution

In the presence of a catalyst, the heat evolved or absorbed during the reaction remains unchanged. The enthalpy change value of the reaction will not be affected by a catalyst. A catalyst will only lower the required activation energy for the reactions. Since it will lower the activation energy for both the forward and reverse reactions to the same extent, the net change in enthalpy is zero.
Question 5
4 / -1

In the electrolytic method of obtaining aluminium from purified bauxite, cryolite is added to the charge in order to:

A
Lower the melting point of bauxite

B
Dissolve bauxite and render it a conductor of electricity

C
Protect aluminium produced from oxygen

D
Minimise the heat loss due to radiation

Solution

In the electrolytic method of obtaining aluminium from purified bauxite, cryolite is added to the charge because it reduces the melting point of Bauxite (from \(1200^{\circ} \mathrm{C}\) to \(800^{\circ}-900^{\circ} \mathrm{C}\) ) and also it increases the electrical conductivity of the mixture.
Question 6
4 / -1

Which of the following is called the bitter of tin?

A
\(\mathrm{SnCl}_{2} .5 \mathrm{HO}\)

B
\(\mathrm{SnCl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}\)

C
\(\mathrm{SnCl} .5 \mathrm{H}_{2} \mathrm{O}\)

D
\(\mathrm{SnCl}_{2} .5 \mathrm{H}_{2} \mathrm{O}\)

Solution

\(\mathrm{SnCl}_{2} .5 \mathrm{H}_{2} \mathrm{O}\)is called the bitter of tin.

The compound Stannous chloride with \(5\) moles of water is called bitter of tin and it is used as a mordant in dyeing as it gives bright colours. Its chemical formula is given by \(\mathrm{SnCl}_{2} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) and it is a white crystalline solid.
Question 7
4 / -1

Aryl halides are less reactive towards nucleophilic substitution than alkyl halides. Which of the following factors is not responsible for this fact?

A
Formation of more stable carbonium ion

B
Resonance stabilisation

C
Inductive effect

D
sp²-hybridised carbon attached to the halogen

Solution

Aryl halides are less reactive towards nucleophilic substitution reaction as compared to alkyl halides due to resonance stabilization. In alkyl halides, the carbon atom is sp³-hybridized whereas, in aryl halides, it is sp²-hybridized and hence, more electronegative.
Question 8
4 / -1

Which of the following options is incorrect about physical adsorption?

A
It has low heat of adsorption

B
It is reversible

C
It does not form multimolecular layers

D
The state of adsorbate is the same as that in bulk

Solution

Physical adsorption is also known as physisorption. It is due to weak Van der Waals forces between adsorbate and adsorbent.

This type of adsorption is caused by physical forces.

Physisorption is a weak phenomenon.

This adsorption is a multi-layered process.

Physical adsorption is not specific and takes place all over the adsorbant.

Surface area, temperature, pressure, nature of adsorbate effects physisorption.

Energy for activation is low (20 – 40 kg/mol).
Question 9
4 / -1

Which of the following options is an example of salt gives an aqueous solution of pH less than 7?

A
Sodium carbonate

B
Sodium chloride

C
Ammonium chloride

D
Sodium bicarbonate

Solution

Ammonium Chloride is an example of a salt gives an aqueous solution of pH less than 7.

We know the reaction between Ammonium Chloride (NH₄Cl)

NH₄Cl + H₂O → NH₄OH + HCl

All the other salts when mixed with water give basic solution.
Question 10
4 / -1

Primary, secondary and tertiary amines can be identified by ________.

A
Hinsberg's reagent

B
Grignard's reagent

C
Fehling's solution

D
Tollen's reagent

Solution

The primary, secondary, and tertiary amines can be distinguished by Hinsberg's reagent.

Hinsberg reagent \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{2} \mathrm{Cl}\) reacts with primary amines and gives alkali soluble benzene sulphonamide, with secondary amine it gives alkali insoluble benzene sulphonamide, with tertiary amines it does not react.

Hinsberg's reagent (benzene sulphonyl chloride) is used to distinguish \(1^{\circ}, 2^{\circ}\) and \(3^{\circ}\) amines. With Hinsberg's reagent, \(1^{\circ}\) amines give a sulphonamide, soluble in alkali, \(2^{\circ}\) amines form a sulphonamide, insoluble in alkali, while \(3^{\circ}\) amines do not react at all.

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Chemistry Test - 13

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Chemistry Test - 13

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Question 1

Gallium

Thallium

Indium

Aluminium

Solution

Question 2

\(\mathrm {CaCO_{3}}\)

\(\mathrm {CaO}\)

\(\mathrm {Ca(OH)_{2}}\)

\(\mathrm {Ca(HCO_{3})_{2}}\)

Solution

Question 3

\(4 \leftarrow 2\)

\(2 \leftarrow 1\)

\(3 \leftarrow 1\)

\(4 \leftarrow 1\)

Solution

Question 4

Increases

Decreases

Remains unchanged

May increase or decrease

Solution

Question 5

Lower the melting point of bauxite

Dissolve bauxite and render it a conductor of electricity

Protect aluminium produced from oxygen

Minimise the heat loss due to radiation

Solution

Question 6

\(\mathrm{SnCl}_{2} .5 \mathrm{HO}\)

\(\mathrm{SnCl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}\)

\(\mathrm{SnCl} .5 \mathrm{H}_{2} \mathrm{O}\)

\(\mathrm{SnCl}_{2} .5 \mathrm{H}_{2} \mathrm{O}\)

Solution

Question 7

Formation of more stable carbonium ion

Resonance stabilisation

Inductive effect

sp2-hybridised carbon attached to the halogen

Solution

Question 8

It has low heat of adsorption

It is reversible

It does not form multimolecular layers

The state of adsorbate is the same as that in bulk

Solution

Question 9

Sodium carbonate

Sodium chloride

Ammonium chloride

Sodium bicarbonate

Solution

Question 10

Hinsberg's reagent

Grignard's reagent

Fehling's solution

Tollen's reagent

Solution

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sp²-hybridised carbon attached to the halogen