The p-Block Elements Test

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The p-Block Elements Test - 5

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Weekly Quiz Competition

Question 1
1 / -0

Catenation i.e., linking of similar atoms depends on size and electronic configuration of atoms. The tendency of catenation in Group 14 elements follows the order:

A
Ge > Sn > Si > C

B
C > Si > Ge ≈ Sn

C
C > Si > Ge > Sn

D
Si > C > Sn > Ge

Solution

Catenation property decreases down the group as size of the atom increases.
Question 2
1 / -0

Silicon has a strong tendency to form polymers like silicones. The chain length of silicone polymer can be controlled by adding

A
Me₃SiCl

B
Me₂Si Cl₂

C
Me₄Si

D
MeSiCl₃

Solution

Me₃SiCl has only one Cl and (can react once) to forms only dimer thus it is used as chain controlling agent (terminate the polymerisation process).
Question 3
1 / -0

In the structure of diborane

A
2 boron atoms and 4 terminal hydrogen atoms lie in the same plane and 2 bridging hydrogen atoms lie in the perpendicular plane.

B
All hydrogen atoms lie in one plane and boron atoms lie in a plane perpendicular to this plane

C
4 bridging hydrogen atoms and boron atoms lie in one plane and two terminal hydrogen atoms lie in a plane perpendicular to this plane.

D
All the atoms are in the same plane.

Solution

B₂H₆ has 4 terminal H attached to B atoms ans all these lie I the same plane. Also in diborane we have 2 bridging hydrogen(3 centre 2 electron ) which lie perpendicular to the plane.
Question 4
1 / -0

Quartz is extensively used as a piezoelectric material, it contains ___________.

A
Sn

B
Si

C
Ti

D
Pb

Solution

Quartz contains Si. [Piezoelectric materials produces electricity when pressure is applied]
Question 5
1 / -0

The linear shape of CO₂ is due to _________.

A
sp₂ hybridisation of carbon

B
sp₃ hybridisation of carbon

C
dsp² hybridisation of carbon

D
pπ– pπ bonding between carbon and oxygen

Solution

In CO₂ (O = C = O ) hybridisation of C is sp so due to pЛ –pЛ bonding it is linear. Further carbon is sp hybridised, and have linear structure.
Question 6
1 / -0

The reason for small radius of Ga compared to Al is _______.

A
presence of higher orbitals

B
decrease in nuclear charge

C
poor screening effect of d and f orbitals

D
higher atomic number

Solution

The presence of additional 10d- electrons offer poor screening effect for the outer electrons from the increased nuclear charge in gallium.
[Electronic configuration of Ga is [Ar] 3d¹⁰ 4s² 4p¹, that of aluminum is [Ne]3s²3p¹]
Question 7
1 / -0

Cement, the important building material is a mixture of oxides of several elements. Besides iron and sulphur, oxides of elements of which of the group (s) are present in the mixture?

A
Groups 2, 13 and 14

B
groups 2 and 14

C
groups 2 and 13

D
group 2

Solution

Main Composition of Portland cement are: lime (CaO) (group 2), silica (SiO ₂)(group 14) and alumina (Al ₂O ₃).(group 13)

Cement is a heavy powder of calcium aluminates and silicates
Question 8
1 / -0

Dry ice is

A
Solid CO₂

B
Solid NH₃

C
Solid N₂

D
Solid SO₂

Solution

Dry ice is solid CO₂.

Solid CO₂ sublimates. Thus it does not wet anything and called dry ice.
Question 9
1 / -0

In trivalent state, for example, the trichlorides, being covalent are hydrolysed in water form

A
trigonal structure

B
tetrahedral structure

C
planar structure

D
pentagonal structure

Solution

Trihalides hydrolysis to form tetrahedral structure with four covalent bonds.

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Re-Attempt Weekly Quiz Competition

The p-Block Elements Test - 5

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The p-Block Elements Test - 5

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Question 1

Ge > Sn > Si > C

C > Si > Ge ≈ Sn

C > Si > Ge > Sn

Si > C > Sn > Ge

Solution

Question 2

Me3SiCl

Me2Si Cl2

Me4Si

MeSiCl3

Solution

Question 3

2 boron atoms and 4 terminal hydrogen atoms lie in the same plane and 2 bridging hydrogen atoms lie in the perpendicular plane.

All hydrogen atoms lie in one plane and boron atoms lie in a plane perpendicular to this plane

4 bridging hydrogen atoms and boron atoms lie in one plane and two terminal hydrogen atoms lie in a plane perpendicular to this plane.

All the atoms are in the same plane.

Solution

Question 4

Sn

Si

Ti

Pb

Solution

Question 5

sp2 hybridisation of carbon

sp3 hybridisation of carbon

dsp2 hybridisation of carbon

pπ– pπ bonding between carbon and oxygen

Solution

Question 6

presence of higher orbitals

decrease in nuclear charge

poor screening effect of d and f orbitals

higher atomic number

Solution

Question 7

Groups 2, 13 and 14

groups 2 and 14

groups 2 and 13

group 2

Solution

Question 8

Solid CO2

Solid NH3

Solid N2

Solid SO2

Solution

Question 9

trigonal structure

tetrahedral structure

planar structure

pentagonal structure

Solution

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Me₃SiCl

Me₂Si Cl₂

Me₄Si

MeSiCl₃

sp₂ hybridisation of carbon

sp₃ hybridisation of carbon

dsp² hybridisation of carbon

Solid CO₂

Solid NH₃

Solid N₂

Solid SO₂