Chemistry Test

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Chemistry Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

How many geometrical isomers are possible in the following two alkenes?

(i)CH₃−CH=CH−CH=CH−CH₃

(ii)CH₃−CH=CH−CH=CH−Cl

A
4 and 4

B
4 and 3

C
3 and 3

D
3 and 4

Solution

When the ends of alkene containing n double bonds are different the number of geometrical isomers is \(2^{n}\) thus for the given Number of geometrical isomers \(=2^{2}=4\)

When the ends of alkene containing n double bonds are same, then the number of geometrical isomers \(=2^{n-1}+2^{p-1}\) where \(p=\frac{n}{2}\) for even \(n\) and \(\frac{n+1}{2}\) for odd \(n\) Number of geometrical isomers \(=2^{2-1}+2^{\frac{2}{2}-1}=3\)

Hence option D is the correct answer.
Question 2
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Alkaline earth's metals are denser than alkali metals because metallic bonding in alkaline earth's metal is:

A
stronger

B
weaker

C
volatile

D
not present

Solution

Alkaline earth metals (ns²) are denser than alkali metals (ns¹) because metallic bonding in alkaline earth metals is stronger than alkali metals due to presence of two electrons in valence shell as compared to one electron in alkali metals.

Hence option A is the correct answer.
Question 3
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The elements of group 1 are called alkali metals because :

A
their oxides from acidic solutions on treating with water.

B
their peroxides from alkine solutions on treating with water.

C
their oxides and hydroxides from alkine solutions on treating with water.

D
their hydroxides from acidic solutions on treating with water.

Solution

The elements of group 1 are called alkali metals because their oxides and hydroxides form alkaline solutions on treating with water. Therefore, the solution becomes basic or alkaline due to the formation of hyroxide ions.

Hence option C is the correct answer.
Question 4
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The characteristic not related to alkali metal is:

A
High ionisation

B
Their ions are isoelectronic with noble gases

C
Low melting point

D
Low electronegativity

Solution

Alkali metals have low ionisation energy. They possess minimum value of ionisation energy in their period.

Hence option A is the correct answer.
Question 5
1 / -0

Among the following the least thermally stable is:

A
K₂CO₃

B
Na₂CO₃

C
BaCO₃

D
Li₂CO₃

Solution

Thermal stability of carbonates increases on moving down a group. Thus, lithium carbonate, Li₂CO₃Li₂CO₃ is least thermally stable.

\(L i_{2} C O_{3} \stackrel{\triangle}{\rightarrow} L i_{2} O+C O_{2} \uparrow\)

Hence option D is the correct answer.
Question 6
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Which one of the following conformer of 3-amino-2-butanol is most stable?

A

B

C

D

Solution

Conformer (A) of 3-amino-2-butanol is most stable due to intramolecular hydrogen bonding between NH₂NH₂and OHOH, and there is less steric repulsion between the two methyl group as they are at opposite angles (farthest possible angle) to each other.

Hence option A is the correct answer.
Question 7
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The correct IUPAC name of the given compound is:

A
N-dimethyl 1-methyl-1-ethyl propane

B
3-methyl-3(N, N-dimethyl amino) pentane

C
N-dimethyl amino-3-methyl pentane

D
N, N, 3-trimethylpentane-3-amine

Solution

The correct IUPAC name for the given compound is N, N, 3-tri methyl pentane-3-amine.

The parent compound contains 5 carbon atoms and an amino group at third carbon atom.

Hence, it is named pentane-3-amine.

There are three methyl groups as substituents.

Hence option D is the correct answer.
Question 8
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Which of the following elements does not show +4 oxidation state?

A
Zr

B
Pt

C
La

D
Ti

Solution

La does not show +4 oxidation state.

It shows only +3 oxidation state by losing two 6s and one 5d electron.

Hence option C is the correct answer.
Question 9
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A burning strip of Mg is introduced into a jar containing a gas. After some time, the walls of the container are coated with carbon. The gas in the container is :

A
O₂

B
N₂

C
CO₂

D
H₂O

Solution

Magnesium burns in carbon dioxide to form magnesium oxide and carbon.

2Mg+CO₂→2MgO+C

Hence option C is the correct answer.
Question 10
1 / -0

The IUPAC name of the compound CH₃CH=CHC≡CH is:

A
pent-4-yn-2-ene

B
pent-3-en-1-yne

C
pent-2-en-4-yne

D
pent-1-yn-3-ene

Solution

The IUPAC name of the compound CH₃CH=CHC≡CH is pent-3-en-1-yne.

C⁵H₃C⁴H=C³HC²≡C¹H

Pent-3-en-1-yne

Fact : If a molecule contains both carbon-carbon double or triple bonds,

The numbering is done from the end which is nearer to the == bond followed by the ≡ bond, and according to the lowest sum of locant rule.

The numbering or sum rule will follow the alphabetical order of the substituent.

However, if the sum of numbers turns out to be the same starting from either of the carbon chain, then the lowest number is given to the C=C double bond.

Hence option B is the correct answer.