Chemistry Test-7

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Chemistry Test-7

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Weekly Quiz Competition

Question 1
4 / -1

The order of basic strength in gas phase of the following compound would be

A
iv) < i) < iii) < ii)

B
i) < iv) < iii) < ii)

C
iv) < i) < ii) < iii)

D
ii) < iii) < i) < iv)

Solution

In decreasing order of basic strength in the gas phase,

In the gas phase, there is no hydrogen bonding. Therefore stabilization due to hydrogen bonding is not there. Thus, the only effect to determine the strength is the inductive effect. The +I effect increases with an increase in the alkyl group. Therefore the basic strength will be the highest in

(C₂H₅)₃N and least in NH₃. Therefore the increasing order of basic strength in the gas phase will be,
Question 2
4 / -1

The reaction: \(\mathrm{n}-\mathrm{BuBr}+\mathrm{KCN} \frac{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}-\mathrm{H}_{2} \mathrm{O}}{\rightarrow} \mathrm{n}-\mathrm{BuCN}\) is an example of:

A
Decarboxylation reaction

B
Elimination reaction

C
Condensation reaction

D
Nucleophile substitution

Solution

The reaction: \(n-\mathrm{BuBr}+\mathrm{KCN} \stackrel{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}-\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} n-\mathrm{BuCN}\) is an example of Nucleophile substitution..

This is an example of bimolecular nucleophilic substitution reaction \(\mathrm{SN}^{2}\). Cyanide ion acts as a nucleophile and bromide ion acts as a leaving group. It is one step reaction and the mechanism is as shown above.
Question 3
4 / -1

The general formula C_nH_2nO₂could be for open chain __________.

A
Diketones

B
Carboxylic acids

C
Diols

D
Dialdehydes

Solution

The general formula C_nH_2nO₂ could be for open chain carboxylic acids. For example: butanoic acid CH₃−CH₂−CH₂−COOH has formula C₄H₈O₂and decanoic acid CH₃−(CH₂)₈−COOH has formula C₁₀H₂₀O₂.
Question 4
4 / -1

Hybridisation shown by carbon and oxygen of –OH group in phenol are respectively

A
sp²,sp²

B
sp³, sp³

C
sp,sp²

D
sp²,sp³

Solution

The carbon atom attached to oxygen is sp² hybridized in phenol and O atom is sp³ hybridised as shown in the figure:
Question 5
4 / -1

The sp³ d² hybridisation of central atom of a molecule would lead to

A
Square planar geometry

B
Tetrahedral geometry

C
Trigonal bipyramidal geometry

D
Octahedral geometry

Solution

The sp³d² and d²sp³ hybridisation of central atom lead to octahedral geometry. Sp³d² hybridization has 1s, 3p and 2d orbitals, that undergo intermixing to form 6 identical sp³d² hybrid orbitals. These 6 orbitals are directed towards the corners of an octahedron. They are inclined at an angle of 90 degrees to one another.
Question 6
4 / -1

The coordination number of Cu in [Cu(H₂O)₄]²⁺complex is

A
2

B
1

C
3

D
4

Solution

The coordination number of Cu in [Cu(H₂O)₄]²⁺ complex is 4

Co-ordination number = Number of monodentate ligand attached to central metal ion = 4.

The total number of monodentate ligands attached to the central metal is known as the coordination number. So, in [Cu(H₂O)₄]²⁺ coordination number of Cu atoms in the complex is 4.
Question 7
4 / -1

When copper pyrites are roasted in excess of air, a mixture of Cu₂O+FeO is formed. FeO is present as impurities. This can be removed as slag during the reduction of Cu₂O. The flux added to form slag is ___________.

A
SiO₂ which is an acid flux

B
Lime stone, which is a basic flux

C
SiO₂, which is a basic flux

D
CaO, which is a basic flux

Solution

When copper pyrites are roasted in excess of air, a mixture of Cu₂O + FeO is formed. FeO is present as impurities. This can be removed as slag during the reduction of Cu₂O. The flux added to form slag is an acid flux.

The compounds which combine with impurities present in ore (at high temperature) and remove them as a fusible substance (slag) are known as flux. When basic impurities are present, an acidic flux is used and vice-versa.

During the roasting of copper pyrite ferrous oxide (basic impurity) combines with silica (acidic flux) SiO₂ and forms ferrous silicate. By this reaction, most of the iron is removed as slag.

FeO+SiO₂ ⟶ FeSiO₃
Question 8
4 / -1

The equivalent conductance of \(\mathrm{Ba}^{2+}\) and \(\mathrm{Cl}^{-}\) are respectively 127 and \(76 \mathrm{ohm}^{-1}\)\(\mathrm{cm}^{2} \mathrm{eq}^{-1}\) at infinite dilution. What will be the equivalent conductance of \(\mathrm{BaCl}_{2}\) at infinite dilution?

A

\(139.5 \mathrm{ohm}^{-1} \mathrm{~cm}^{2} \mathrm{eq}^{-1}\)

B

\(203 \mathrm{ohm}^{-1} \mathrm{~cm}^{2} \mathrm{eq}^{-1}\)

C

\(279 \mathrm{ohm}^{-1} \mathrm{~cm}^{2} \mathrm{eq}^{-1}\)

D

\(101.5 \mathrm{ohm}^{-1} \mathrm{~cm}^{2} \mathrm{eq}^{-1}\)

Solution

Kohlrausch law:
"At infinite dilution, when dissociation is complete, each ion makes a definite contribution towards equivalent conductance of the electrolyte irrespective of the nature of the ion with which it Is associated and the value of equivalent conductance at infinite dilution for any electrolyte is the sum of the contribution of its constituent ions", i.e., anions and cations.
The chemical equation is \(\mathrm{BaCl}_{2} \rightarrow \mathrm{Ba}^{2+}+2 \mathrm{Cl}\)
\(\Lambda_{\mathrm{BaCl}_{2}}^{0}=\Lambda_{\mathrm{Ba}^{2+}}^{0}+2 \mathrm{Cl}^{-}\)
\(\Rightarrow127+2 \times 76=279 \mathrm{ohm}^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
Equivalent conductivity \(=\frac{279}{2}=139.5 \mathrm{ohm}^{-1} \mathrm{~cm}^{2} \mathrm{eq}^{-1}\)
Question 9
4 / -1

A covalent bond between two atoms may be broken in different ways depending upon ____________.

A
Nature of given organic compound

B
Nature of attacking reagent

C
Reaction conditions

D
All ofthese

Solution

A covalent bond between two atoms may be broken in different ways depending upon nature of given organic compound, nature of attacking reagent and reaction conditions. They differ in their structure and properties. Covalent bonds consist of pairs of electrons shared by two atoms, and bind the atoms in a fixed orientation. Relatively high energies are required to break them (50 - 200 kcal/mol). Whether two atoms can form a covalent bond depends upon their electronegativity i.e. the power of an atom in a molecule to attract electrons to itself.
Question 10
4 / -1

In which among the following organic compound, carboxyl group is present?

A
Ethanol

B
Formaldehyde

C
Ethylene

D
Ethanoic acid

Solution

Carboxyl group:

It is defined as carbonyl and hydroxyl attached to a carbon atom, which means the carbon atom is double-bonded with oxygen and a single bond with the hydroxyl.

And Carboxyl acid is an organic compound that consists of a carboxyl group.

The general formula of the class is \(\mathrm{R-C(O)OH}\).

In this formula, \(\mathrm{R}\) is the alkyl or aryl group.

Carboxyl acids occur widely in nature.

However, the majority of the members of this group are manufactured synthetically.

The double bond presents in the structure of carboxyl acids play a very important role in the properties of the different compounds of carboxyl acids.

When a carbon compound is attached to the functional group −\(\mathrm{COOH}\) (carboxyl group) then the compound refers to as carboxyl acids.

The above structure formula of ethanoic acid is \(\mathrm{CH_{3}COOH}\), so it contains a carboxyl acid group.
Question 11
4 / -1

What is the percentage of oxygen in \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} ?\)

A
\(57.7 \%\)

B
\(56.1 \%\)

C
\(53.1 \%\)

D
\(52.6 \%\)

Solution

Given compound is Aluminium Sulphate \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\)
the Atomic mass of Aluminium \((\mathrm{Al})=27\).
The atomic mass of Sulphur \((S)=16 \times 2=32\)
The atomic mass of oxyen \((0)=8 \times 2=16\).
Total molar mass of the compound is \((27 \times 2)+(32 \times 3)+(16 \times 12)\)
\( \Rightarrow 342 \mathrm{gm} / \mathrm{mole}\).
Molar Mass of the oxygen \(=12 \times 16\)
\( \Rightarrow 192 \mathrm{gm} / \mathrm{mole}\).
Percentage of oxygen in total compound \(=(\frac{192 }{ 342}) \times 100 \)
\(\Rightarrow 56.140 \%\)
Therefore, the percentage of oxygen in Aluminium Sulphate is \(56.1 \%\)
Question 12
4 / -1

An ideal gas heat engine operates in Carnot's cycle between \(227^{\circ} \mathrm{C}\) and \(127^{\circ} \mathrm{C}\) It absorbs \(6 \times 10^{4} \mathrm{~J}\) at high temperature. The amount of heat converted into work is:

A
\(4.8 \times 10^{4} \times J\)

B
\(3.5 \times 10^{4} \times J\)

C
\(1.6 \times 10^{4} \times J\)

D
\(1.2 \times 10^{4} \times J\)

Solution

Given
\(T_{1}=227+273=500 \mathrm{~K}, T_{2}=127+273=400 \mathrm{~K}\),
Heat absorbed by the engine is \(Q_{1}=6 \times 10^{4} \mathrm{~J}\)
The efficiency of the heat engine is given by
\(\Rightarrow \eta=\frac{W}{Q_{1}}=1-\frac{T_{2}}{T_{1}} \)
\(\Rightarrow \frac{W}{6 \times 10^{4}}=1-\frac{400}{500} \)
\(\Rightarrow \frac{W}{6 \times 10^{4}}=1-\frac{4}{5} \)
\(\Rightarrow \frac{W}{6 \times 10^{4}}=\frac{1}{5} \)
\(\Rightarrow W=\frac{6 \times 10^{4}}{5}=1.2 \times 10^{4} \mathrm{~J}\)
Question 13
4 / -1

A substance 'A' decomposes by a first order reaction starting initially with \([\mathrm{A}]=\) \(2.00 \mathrm{M}\) and after \(200 \mathrm{~min}[\mathrm{~A}]=0.15 \mathrm{M}\). For this reaction what is the value of \(\mathrm{k}\)?

A
\(1.29 \times 10^{-2} \mathrm{~min}^{-1}\)

B
\(2.29 \times 10^{-2} \mathrm{~min}^{-1}\)

C
\(3.29 \times 10^{-2} \mathrm{~min}^{-1}\)

D
\(4.40 \times 10^{-2} \mathrm{~min}^{-1}\)

Solution

Given
\(\mathrm{A}(\mathrm{a})=2.00 \mathrm{M}, \mathrm{t}=200 \mathrm{~min}\) and \((\mathrm{a}-\mathrm{x})=0.15 \mathrm{M}\)
\(k=\frac{2.303}{t} \log \frac{a}{a-x}\)
\(=\frac{2.303}{200} \log \frac{2.00}{0.15}\)
\(=\frac{2.303}{200} \times(0.301+0.824)\)
\(=1.29 \times 10^{-2} \mathrm{~min}^{-1}\)
Question 14
4 / -1

Fr87 and Ra88, two radioactive elements belong to which of the following block of modern table ?

A
p-block Elements

B
s-block Elements

C
d-block Elements

D
f-block Elements
Solution
Fr87 and Ra88, two radioactive elements belong to s-block Elements of modern table.
Block division of Elements:
- The s-block elements having group 2 (alkaline earth metals) & group 1 (alkali metals).
- The p-block elements are groups 13-18 (basic metals, metalloids, nonmetals, halogens, and noble gases).
- The \(d\) and \(f\) Block elements in the groups of 3 to 11 are also called transition elements and inner transition elements respectively.
- 4 f and \(5 \mathrm{f}\) orbitals of f-block elements are steadily in the next two long periods.
- The position is highly thoughtful in the periodic table according to its nature & properties.
Question 15
4 / -1

For the same alkyl group, the boiling point of alkyl halides decrease in the order:

A
RI> RBr> RCl> RF

B
RI< RCl< RBr

C
RI< RBr< RCl< RF

D
RF< RBr< RCl< RI

Solution

The boiling points of alkyl halides decrease in the order RI> RBr> RCl> RF. This is because as the size of halogen increases, the magnitude of van der Waals forces increase, and hence the boiling point increase.
Straight chain alkyl halides have greater boiling point than their isomers. This is due to the reason that with branching, the surface area of the alkyl halide decreases and hence the magnitude of the van der Waal's force of attraction decreases. Therefore, CH₃CH₂CH₂CH₂Cl has highest boiling point.

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Chemistry Test-7

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Chemistry Test-7

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Question 1

iv) < i) < iii) < ii)

i) < iv) < iii) < ii)

iv) < i) < ii) < iii)

ii) < iii) < i) < iv)

Solution

Question 2

Decarboxylation reaction

Elimination reaction

Condensation reaction

Nucleophile substitution

Solution

Question 3

Diketones

Carboxylic acids

Diols

Dialdehydes

Solution

Question 4

sp2,sp2

sp3, sp3

sp,sp2

sp2,sp3

Solution

Question 5

Square planar geometry

Tetrahedral geometry

Trigonal bipyramidal geometry

Octahedral geometry

Solution

Question 6

2

1

3

4

Solution

Question 7

SiO2 which is an acid flux

Lime stone, which is a basic flux

SiO2, which is a basic flux

CaO, which is a basic flux

Solution

Question 8

\(139.5 \mathrm{ohm}^{-1} \mathrm{~cm}^{2} \mathrm{eq}^{-1}\)

\(203 \mathrm{ohm}^{-1} \mathrm{~cm}^{2} \mathrm{eq}^{-1}\)

\(279 \mathrm{ohm}^{-1} \mathrm{~cm}^{2} \mathrm{eq}^{-1}\)

\(101.5 \mathrm{ohm}^{-1} \mathrm{~cm}^{2} \mathrm{eq}^{-1}\)

Solution

Question 9

Nature of given organic compound

Nature of attacking reagent

Reaction conditions

All ofthese

Solution

Question 10

Ethanol

Formaldehyde

Ethylene

Ethanoic acid

Solution

Question 11

\(57.7 \%\)

\(56.1 \%\)

\(53.1 \%\)

\(52.6 \%\)

Solution

Question 12

\(4.8 \times 10^{4} \times J\)

\(3.5 \times 10^{4} \times J\)

\(1.6 \times 10^{4} \times J\)

\(1.2 \times 10^{4} \times J\)

Solution

sp²,sp²

sp³, sp³

sp,sp²

sp²,sp³

SiO₂ which is an acid flux

SiO₂, which is a basic flux