p-Block Elements Test

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p-Block Elements Test - 5

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Weekly Quiz Competition

Question 1
1 / -0

Graphite is a good conductor of electricity because of

A
weak van der Waals' force between its layers

B
covalent bonding between the carbon atoms of its layers

C
delocalised electrons in its each layer

D
sp hybridisation of carbon atom in its each layer

Solution

Each carbon atom in graphite is sp² hybridized and forms three covalent bonds to create a lattice of hexagonal rings arranged in layers. Therefore each carbon atom in graphite is left with one free electron. These free electrons act as a charge carriers across the graphite crystal and make it a good conductor of electricity.
Question 2
1 / -0

How many carbon atoms join to form spherical molecules in buckminsterfullerene?

A
50

B
55

C
60

D
62

Solution

Buckminsterfullerene is a spherical fullerene molecule with the formula C₆₀. It is a cage-like fused-ring structure which resembles an association football made of twenty hexagons and twelve pentagons, with a carbon atom at the vertices of each polygon and a bond along each polygon edge.
Question 3
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The order of increasing acidic strengths of BF₃, BCl₃ and BBr₃ is:

A
BBr₃ < BCl₃ < BF₃

B
BF₃ < BCl₃ < BBr₃

C
BBr₃ < BF₃ < BCl₃

D
BCl₃ < BBr₃ < BF₃

Solution

The order of increasing acidic strengths of BF₃, BCl₃ and BBr₃ is: BF₃ < BCl₃ < BBr₃.
The extent of back bonding is maximum in BF₃ and minimum in BBr₃.
Question 4
1 / -0

The following question has four choices out of which ONLY ONE is correct.

Detect the wrong statement regarding boron and silicon.

A
Both BCl₃ and SiCl₄ are hydrolysed by water.

B
Elemental boron and silicon are semiconductors.

C
Both of B and Si produce oxides soluble in NaOH.

D
Both BF₃ and SiF₄are Lewis bases.

Solution

Both BF₃ and SiF₄ are Lewis acids. BF₃is electron deficient and SiF₄has empty d-orbitals where it can accommodate electron pairs.
Question 5
1 / -0

Tin plague means

A
tin plating

B
conversion of white tin to grey tin

C
emission of sound while bending a tin plate

D
conversion of stannous salts to stannic salts

Solution

Tin pest or tin plague is an autocatalytic, allotropic transformation of the element tin, which causes deterioration of tin objects at low temperatures. Tin pest has also been called tin disease, tin blight or tin leprosy.

This happens at very low temperatures, so white tin loses its crystalline structure and gets converted into grey powdered form.
Question 6
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The unique ability of carbon to form C = C, C ≡ C, C = S and C ≡ N bonds is due to its ability to form

A
pπ – pπ bonding

B
pπ – dπ bonding

C
dπ – dπ bonding

D
None of these

Solution

The unique ability of carbon to form C = C, C ≡ C, C = S and C ≡ N bonds is due to its ability to form pπ – pπ bonding on account of its small size.
Question 7
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Property of catenation is the strongest in

A
C

B
Si

C
N

D
O

Solution

Bond strength of M – M bond and tendency for catenation decrease in the order:
C – C > Si – Si > Ge – Ge > Sn – Sn
The greater the bond energy, the greater the strength of the bond.
Question 8
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Which of the following reactions results in the formation of water gas?

A
CO + N₂

B
CO + CO₂ + CH₄

C
CO₂ + N₂

D
C + H₂O

Solution

Water-gas is a mixture of carbon monoxide (CO) and hydrogen (H₂) produced by passing steam over red-hot coke using the endothermic reaction
C + H₂O → CO + H₂
Question 9
1 / -0

Directions For Questions

The following question has four choices, out of which ONLY ONE is correct.

...view full instructions

Which of the following statements is correct?

A
Unlike carbon, silicon cannot form pπ - pπ bonds with oxygen atoms due to its larger size (117 pm) compared to C (77 pm).

B
In SiO₂, Si atoms use sp hybrid orbitals to bond with each other, and also with oxygen.

C
Si=O bonds are stronger than Si-O bonds and C=O bonds are stronger than C-O bonds.

D
CO₂and SiO₂both exist as discretemolecules.

Solution

The statement in option (1) is true. Owing to its large size, silicon cannot form pπ - pπ bonds with oxygen like carbon.

Consequently, SiO₂ forms a three-dimensional network structure, while carbon dioxide forms discrete linear molecules.

The hybridisation of Si in SiO₂ is sp³.

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p-Block Elements Test - 5

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p-Block Elements Test - 5

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Question 1

weak van der Waals' force between its layers

covalent bonding between the carbon atoms of its layers

delocalised electrons in its each layer

sp hybridisation of carbon atom in its each layer

Solution

Question 2

50

55

60

62

Solution

Question 3

BBr3 < BCl3 < BF3

BF3 < BCl3 < BBr3

BBr3 < BF3 < BCl3

BCl3 < BBr3 < BF3

Solution

Question 4

Both BCl3 and SiCl4 are hydrolysed by water.

Elemental boron and silicon are semiconductors.

Both of B and Si produce oxides soluble in NaOH.

Both BF3 and SiF4 are Lewis bases.

Solution

Question 5

tin plating

conversion of white tin to grey tin

emission of sound while bending a tin plate

conversion of stannous salts to stannic salts

Solution

Question 6

pπ – pπ bonding

pπ – dπ bonding

dπ – dπ bonding

None of these

Solution

Question 7

C

Si

N

O

Solution

Question 8

CO + N2

CO + CO2 + CH4

CO2 + N2

C + H2O

Solution

Question 9

Unlike carbon, silicon cannot form pπ - pπ bonds with oxygen atoms due to its larger size (117 pm) compared to C (77 pm).

In SiO2, Si atoms use sp hybrid orbitals to bond with each other, and also with oxygen.

Si=O bonds are stronger than Si-O bonds and C=O bonds are stronger than C-O bonds.

CO2 and SiO2 both exist as discrete molecules.

Solution

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BBr₃ < BCl₃ < BF₃

BF₃ < BCl₃ < BBr₃

BBr₃ < BF₃ < BCl₃

BCl₃ < BBr₃ < BF₃

Both BCl₃ and SiCl₄ are hydrolysed by water.

Both BF₃ and SiF₄are Lewis bases.

CO + N₂

CO + CO₂ + CH₄

CO₂ + N₂

C + H₂O

In SiO₂, Si atoms use sp hybrid orbitals to bond with each other, and also with oxygen.

CO₂and SiO₂both exist as discretemolecules.