The p-Block Elements Test

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The p-Block Elements Test - 45

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Question 1
1 / -0

Which of the following properties correctly explain the compound $$SiO_2$$?

A
Linear, basic

B
Tetrahedral, acidic

C
Tetrahedral, basic

D
Linear, acidic

Solution

Silicon dioxide, $$SiO_2$$ is one of the most common minerals in the earth. It has a tetrahedral arrangement with one silicon-bonded to four oxygen atoms. Some oxygen atoms will be bonded to two silicon atoms so that two tetrahedral are joined at a corner (bridging atoms). The orientation can be random, leading to an amorphous structure.

Silicon dioxide has no basic properties as it doesn't contain oxide ions and it doesn't react with acids. Instead, it is very weakly acidic, reacting with strong bases.

Thus, $$Si{O}_{2}$$ has a tetrahedral structure and is acidic in nature.

Hence, the correct answer is option $$\text{B}$$.
Question 2
1 / -0

Which of the following ions is the most stable?

A
$$Sn^{2+}$$

B
$$Ge^{2+}$$

C
$$Si^{2+}$$

D
$$Pb^{2+}$$
Solution
Hint: These compounds belong to the carbon family having a covalency of $$4$$.
Correct Answer: Option D
Explanation:
- The common oxidation states exhibited by this family are $$+4$$ and $$+2$$.
- The tendency to show $$+2$$ oxidation state increases in down the group.
- Due to the inert pair effect, down the group the tendency to form ions with $$+2$$ oxidation state increases from $$Ge$$ to $$Pb$$.
- The stability order is in increasing order of atomic number.
Hence, the correct answer is $$Pb^{2+}.$$
Question 3
1 / -0

The shapes and hybridisation of $$ BF_3 \,\,and \,\,{ BH }_{4 }^{- }$$ respectively are:

A
$$BF_3$$ - Trigonal, $$sp^2$$ hybridisation ;

$${ BH }_{4 }^{- }$$ - Square planar, $$sp^3$$ hybridisation

B
$$BF_3$$ - Triangular, $$sp^3$$d hybridisation ;

$${ BH }_{4 }^{- }$$ - Hexagonal, $$sp^3$$d hybridisation

C
$$BF_3$$ - Trigonal, $$sp^2$$ hybridisation ;

$${ BH }_{4 }^{- }$$ - Tetrahedral, $$sp^3$$ hybridisation

D
$$BF_3$$ - Tetrahedral, $$sp^3$$ hybridisation;

$${ BH }_{4 }^{- }$$ - Tetrahedral, $$sp^3$$ hybridisation.

Solution

$$BF_3 \; \rightarrow$$ As a result of its small size and high electronegativity, boron tends to form monomeric covalent halides. These halides have a planar triangular geometry. This triangular shape is formed by the overlap of three $$sp^2$$ hybridised orbitals of boron with the sp orbitals of three halogen atoms. Boron is $$sp^2$$ hybridised in $$BF_3$$.

$${BH_{4}}^{-} \; \rightarrow$$ Boron-hydride ion $$({BH{_4}}^{–})$$ is formed by $$sp^3$$ hybridisation of boron orbitals. Therefore, it is tetrahedral in structure.

Hence, the correct answer is option $$\text{C}$$.
Question 4
1 / -0

An oxide X in its normal form is almost non-reactive due to very high X - O bond enthalpy. It resists the attack by halogens, hydrogen and most of acids and metals even at elevated temperatures. It is only attacked by HF and NaOH. The oxide X is :

A
$$SiO_2$$

B
$$CO_2$$

C
$$SnO_2$$

D
$$PbO_2$$

Solution

Since X is non-reactive and it cannot be attacked by the halogens, hydrogen, and acids as well as metals. Even at high-temperature reaction is not happening. But exceptionally it reacts with HF and NaOH. Thus, X will be $$SiO_{2}$$.

It doesn't react because of the difficulty of breaking up its giant covalent structure. It is insoluble in the water and organic solvents.
The correct option is A.
Question 5
1 / -0

Calcium magnesium silicate is:

A
asbestos

B
litmus

C
prussian blue

D
slate

Solution

Asbestos is the other name of Calcium Magnesium Silicate.$$(CaMgO_4Si)$$
Asbestos materials have been used for insulation in buildings and in various products such as roofing materials, water supply lines and clutches, brake linings, gaskets and pads for automobiles.
Hence option A is correct answer.
Question 6
1 / -0

Topaz, opal, agate, and onyx are precious and beautiful stones found in nature. All these are different forms of:

A
silica

B
soda

C
carbon

D
lime
Solution
- Topaz is a rare silicate mineral with a chemical composition of $$Al_2SiO_4(FOH)_2$$. It usually forms in fractures and cavities of igneous rocks such as pegmatite and rhyolite, late in their cooling history.
- Opal is a non-crystalline form of the mineral silica and is formed from amorphous "balls" or "lumps" of silica. So the opal is a very unique gem compared with others, like diamonds or a ruby which is formed crystalline. This is the reason why variation of colours are in opals.
- Agate is a banded microcrystalline form of the mineral Quartz, and does not occur in visible crystals. It occurs in nodules, in massive form, as botryoidal, mammilary, and stalactitic formations, as smooth rounded pebbles, as amygdules, and as the linings of geodes.
- Onyx is formed of bands of chalcedony in alternating colors. It is cryptocrystalline, consisting of fine intergrowths of the silica minerals quartz and moganite. Its bands are parallel to one another, as opposed to the more chaotic banding that often occurs in agates.
- Hence option A is correct answer.
Question 7
1 / -0

Which of the following is not matched correctly with its use?

A
piezoelectric material - Quartz

B
Ion-exchangers - Graphite

C
Filtration plants - silica

D
Electrical insulators - Silicones

Solution

Ion exchange is an exchange of ions between two electrolytes or between an electrolyte solution and a complex. Zeolites are used as ion-exchangers in softening of hard water.

Hence, option $$B$$ is not correctly matched.
Question 8
1 / -0

Borax (sodium tetraborate) has many uses, some of which are mentioned here, except?

A
For giving a glaze to linen, in laundering

B
In bleaching of paper pulp

C
Making heat-resistant

D
As a flux in welding and soldering

Solution

Borax also known as sodium borate, Borax today is still used for many cleaning purposes as well as many other tasks. You can use it to make homemade laundry detergent, add a cleaning boost to any other detergent and even keep ants and other pests away. It is also used for giving a glaze to linen, in laundering,Making heat-resistant,As a flux in welding and soldering.
Hence option B is not the use of Borax.
Question 9
1 / -0

Silica gel is a very useful:

A
raw material for soap

B
drying agent

C
wetting agent

D
solvent

Solution

Silica gel's high specific surface area allows it to absorb water readily, making it useful as a desiccant (drying agent).
Question 10
1 / -0

Spaceships have to be air-tight because in space there is no air, only vaccum. Which of the following compounds has been used for air-tight closing of the door (hatch) in lunar modules?

A
Chlorides

B
Silicones

C
Bakelite

D
Hydrocarbons

Solution

Silicones are used for air tight closing of door in lunar modules

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The p-Block Elements Test - 45

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The p-Block Elements Test - 45

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Question 1

Linear, basic

Tetrahedral, acidic

Tetrahedral, basic

Linear, acidic

Solution

Question 2

$$Sn^{2+}$$

$$Ge^{2+}$$

$$Si^{2+}$$

$$Pb^{2+}$$

Solution

Question 3

$$BF_3$$ - Trigonal, $$sp^2$$ hybridisation ;$${ BH }_{4 }^{- }$$ - Square planar, $$sp^3$$ hybridisation

$$BF_3$$ - Triangular, $$sp^3$$d hybridisation ;$${ BH }_{4 }^{- }$$ - Hexagonal, $$sp^3$$d hybridisation

$$BF_3$$ - Trigonal, $$sp^2$$ hybridisation ;$${ BH }_{4 }^{- }$$ - Tetrahedral, $$sp^3$$ hybridisation

$$BF_3$$ - Tetrahedral, $$sp^3$$ hybridisation;$${ BH }_{4 }^{- }$$ - Tetrahedral, $$sp^3$$ hybridisation.

Solution

Question 4

$$SiO_2$$

$$CO_2$$

$$SnO_2$$

$$PbO_2$$

Solution

Question 5

asbestos

litmus

prussian blue

slate

Solution

Question 6

silica

soda

carbon

lime

Solution

Question 7

piezoelectric material - Quartz

Ion-exchangers - Graphite

Filtration plants - silica

Electrical insulators - Silicones

Solution

Question 8

For giving a glaze to linen, in laundering

In bleaching of paper pulp

Making heat-resistant

As a flux in welding and soldering

Solution

Question 9

raw material for soap

drying agent

wetting agent

solvent

Solution

Question 10

Chlorides

Silicones

Bakelite

Hydrocarbons

Solution

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$$BF_3$$ - Trigonal, $$sp^2$$ hybridisation ;

$${ BH }_{4 }^{- }$$ - Square planar, $$sp^3$$ hybridisation

$$BF_3$$ - Triangular, $$sp^3$$d hybridisation ;

$${ BH }_{4 }^{- }$$ - Hexagonal, $$sp^3$$d hybridisation

$$BF_3$$ - Trigonal, $$sp^2$$ hybridisation ;

$${ BH }_{4 }^{- }$$ - Tetrahedral, $$sp^3$$ hybridisation

$$BF_3$$ - Tetrahedral, $$sp^3$$ hybridisation;

$${ BH }_{4 }^{- }$$ - Tetrahedral, $$sp^3$$ hybridisation.