Chemistry Test

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Chemistry Test - 7

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following represents the correct order of increasing first ionization enthalpy for Ca, Ba, S, Se and Ar.

A
\(\mathrm{Ca}<\mathrm{S}<\mathrm{Ba}<\mathrm{Se}<\mathrm{A}\)

B
\(\mathrm{S}<\mathrm{Se}<\mathrm{C} \mathrm{a}<\mathrm{Ba}<\mathrm{Ar}\)

C

D

Solution
Question 2
1 / -0

Which among the given haloalkanes develop colour when exposed to light?

A
\(\mathrm{R}-\mathrm{Br}\) and \(\mathrm{R}-\mathrm{Cl}\)

B
\(\mathrm{R-I}\) and \(\mathrm{R-F}\)

C
\(\mathrm{R}-\mathrm{Cl}\) and \(\mathrm{R}-\mathrm{Br}\)

D
\(\mathrm{R}-\mathrm{Br}\) and \(\mathrm{R}-\mathrm{I}\)
Solution
Haloalkanes of the form\(\mathrm{R}-\mathrm{Br}\) and \(\mathrm{R}-\mathrm{I}\)develop colour when exposed to light.
- As we progress down the periodic table from fluorine to iodine, molecular size increases.
- As a result, we also see an increase in bond length.
- As the bond length increases, upon irradiation of light, after absorption the compound emit light in visible region.
- Therefore, \(\mathrm{R}-\mathrm{Br}\) and \(\mathrm{R}-\mathrm{I}\) bond exhibit color when exposed to light.
Question 3
1 / -0

Find out the major products from the following reaction sequence.

A

B

C

D

Solution
Question 4
1 / -0

The atomic numbers of elements \(X, Y\) and \(Z\) are \(19\), \(21\) and \(25\) respectively. The order of number of electrons present in the M shells of these elements ______.

A
\(Z>X>Y\)

B
\(X>Y>Z\)

C
\(Z>Y>X\)

D
\(Y>Z>X\)

Solution

\(\begin{array}{cccccc}X^{19} & : & 1 s^{2} & 2 s^{2} 2 p^{6} & 3 s^{2} 3 p^{6} & 4 s^{1} \\ & & K & L & M & N \\ Y^{21} & : & 1 s^{2} & 2 s^{2} 2 p^{6} & 3 s^{2} 3 p^{6} 3 d^{1} & 4 s^{1} \\ & & K & L & M & N \\ Z^{25} & : & 1 s^{2} & 2 s^{2} 2 p^{6} & 3 s^{2} 3 p^{6} 3 d^{5} & 4 s^{1} \\ & & K & L & M & N\end{array}\)
Electrons in \(M\) shell of these elements are
\(Z^{25}(13)>Y^{21}(9)>X^{19}(8)\)
Question 5
1 / -0

Crystalline solids are _________ in nature.

A
pseudo solids

B
anisotropic

C
isotropic

D
short range order
Solution
Crystalline solids are anisotropic in nature.
- It is because the arrangement of constituent particles is regular and ordered along all the directions.
- Therefore, the value of any physical property (electrical resistance or refractive index) would be different along each direction.
- That is, some of their physical index show different values when measured along different directions in the same crystal.
Question 6
1 / -0

Which of the following is a closed system?

A
Pressure cooker

B
Rocket engine during propulsion

C
Tea placed in a steel kettle

D
Jet engine

Solution

A pressure cookeris a closed system.
There will be no interactions with the surroundings. For example, the contents of a pressure cooker on a stove with its lid tightly closed and the whistle in position is a closed system as no mass can enter or leave the pressure cooker, but heat can be transferred to it.
Question 7
1 / -0

Determine the empirical formula of an oxide of iron which has \(69.9 \%\) iron and \(30.1 \%\) dioxygen by mass.

A
\(\mathrm{Fe}_2 \mathrm{O}_3\)

B
\(\mathrm{Fe}_3 \mathrm{O}_2\)

C
\(\mathrm{Fe}_3 \mathrm{O}_4\)

D
None of these

Solution

Given that:
The iron oxide has \(69.9 \%\) iron and \(30.1 \%\) dioxygen by mass.
Thus, \(100 \mathrm{~g}\) of iron oxide contains \(69.9 \mathrm{~g}\) iron and \(30.1 \mathrm{~g}\) dioxygen.
The number of moles of iron present in \(100 \mathrm{~g}\) of iron oxide are:
\(=\frac{\text { Mass }}{\text { Atomic mass }} \)
\( =\frac{69.9}{55.8} \)
\(=1.25\)
The number of moles of dioxygen present in \(100 \mathrm{~g}\) of iron oxide are:
\(=\frac{\text { Mass }}{\text { Atomic mass }} \)
\(=\frac{30.1}{32} \)
\(=0.94\)
The ratio of the number of oxygen atoms to the number of iron atoms present in one formula unit of iron oxide is:
\(=\frac{2 \times 0.94}{1.25} \)
\(=1.5: 1 \)
\(=3: 2\)
Therefore, the formula of the iron oxide is \(\mathrm{Fe}_2 \mathrm{O}_3\).
Question 8
1 / -0

The ligand \(N \left( CH _{2} CH _{2} NH _{2}\right)_{3}\):

A
Bidentate

B
Tridentate

C
Tetradentate

D
Pentadentate

Solution

Number of donor atoms in \(N \left( CH _{2} CH _{2} NH _{2}\right)_{3}\) is four so it is a tetradentate ligand. Each nitrogen atom is a donor atom.
Question 9
1 / -0

Among the following compounds, the one which can be dehydrated most easily is:

A
\(\mathrm{CH}_{3}-\mathrm{CH}_{2}-{\underset{{\underset {\mathrm{OH}} |}} {\overset { \overset{\mathrm{CH}_{3}}{\|}} {\mathrm{C}}}}-\mathrm{CH}_{2} \mathrm{CH}_{3}\)

B
\(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-{\underset {\underset {\mathrm{OH}} {|}} {\mathrm{CH}}}-\mathrm{CH}_{3}\)

C
\(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{OH}\)

D
\(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{CH}_{2}-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}_{3}\)

Solution

The compound which can be easily dehydrated will form most stable carbocation after removal of \(\mathrm{H}_{2} \mathrm{O}\) molecule.
Since stability of carbocation follows 3 degree> 2 degree> 1 degree
So, compound \(A\) is most easily dehydrated.
Question 10
1 / -0

Across the lanthanide series, the basicity of the lanthanide hydroxides?

A
Increases

B
Decreases

C
First increase and then decrease

D
First decrease and then increase

Solution

Across the lanthanide series the size of the cations decreases thereby decreasing ionic character and increasing covalent character of hydroxides of lanthanides. Thus, the basicity of hydroxides decreases.