The p-Block Elements Test

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The p-Block Elements Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

Identify the compound $$A$$ in the reaction given below.
$$Na_{2}B_{4}O_{7}+2H_{2}O\rightarrow 2NaOH+A$$

A
Orthoboric acid

B
Metaboric Acid

C
Tetraboric Acid

D
Pyroboric acid

Solution

A reaction is given below:
$$Na_2B_4O_7+2H_2O\rightarrow 2NaOH+A$$
In the above reaction, $$A$$ is tetraboric acid.

The complete reaction is as follows:
$$ Na_{2} B_{4} O_{7} + 2 H_{2} O \rightarrow 2NaOH + H_2B_4O_7$$

Hence, option C is the correct answer.
Question 2
1 / -0

The aqueous solution of borax turns red litmus to:

A
blue

B
No Change

C
red

D
white

Solution

This is because of the alkanline nature of the aqueous solution of borax.
Question 3
1 / -0

Which of the following reaction will yield crystalline variety of boron?

A
$$ B_{2}O_{3}+2Al$$

B
$$BCl_{3}\: +H_{2}$$

C
$$B_{2}O_{3}+3Mg$$

D
$$2B+3MgO$$

Solution

Small amounts of crystalline Boron can be obtained by the following reaction using red hot W or Tantalum filament as catalyst.
$$2BX_{3} + 3H_{2} \rightarrow 2B + 6HX, X=Cl, Br$$
Question 4
1 / -0

The compound used in etching of glass is:

A
$$\mathrm{H}\mathrm{F}$$

B
$$\mathrm{H}\mathrm{Cl}\mathrm{O}_{4}$$

C
$$\mathrm{H}\mathrm{N}\mathrm{O}_{3}$$

D
$$\mathrm{H}_{2}\mathrm{S}\mathrm{O}_{4}$$

Solution

HF is used in etching of glass.

$$4\ HF+SiO_2\rightarrow Si\ F_4+2H_2O$$

HF reacts with $$SiO_2$$ present in glass.

Option A is correct.
Question 5
1 / -0

Which of the following represent prismatic borax?

A
$$Na_{2}B_{4}O_{7}.10H_{2}O$$

B
$$Na_{2}B_{4}O_{7}.5\, H_{2}\, O$$

C
$$Na_{2}B_{4}O_{7}.3H_{2}O$$

D
$$Na_{2}B_{4}O_{7}.4H_{2\, }O$$

Solution

Prismatic borax is $$Na_2B_4O_7. 10H_2O$$.
Question 6
1 / -0

Which metal burns in air at high temperature with
the evolution of much heat?

A
Cu

B
Al

C
Hg

D
Pb

Solution

Aluminium burns in air to give a mixture of Aluminium oxide and nitride. This is an exothermic process.
Question 7
1 / -0

The reducing character of Al, Ga, In, Tl increases in which of the following order?

A
$$Al< Tl< Ga> In$$

B
$$Ga< Al< Tl< In$$

C
$$Tl< In< Ga< Al$$

D
$$Al< Ga< In< Tl$$

Solution

Down the group, the reducing character decreases as electrode potential increases going down the group so reducing tendency decreases.
Hence, the order is Tl < In < Ga < Al.
Question 8
1 / -0

When HF reacts with glass, the compound formed is:

A
$${F}_{2}$$

B
$$NaF$$

C
$$Na_{3}AlF_{3}$$

D
$$Na_{2}SiF_{6}$$

Solution

When glass is attacked by $$HF$$, it forms $$Na_2SiF_6$$ and this property is used for the etching of glass.

The reaction is given below:
$$Na_2SiO_3 + 6HF \rightarrow Na_2SiF_6+ 3H_2O$$

Option D is correct.
Question 9
1 / -0

Which is true for an element R present in III A group of the periodic table?

A
It forms halide of type $$RX_{3}$$

B
It has oxidation state of +3

C
It forms oxide of type $$R_{2}O_{3}$$

D
.All the above

Solution

Given $$R$$ is an element in the $$IIIA$$ group of the periodic table, so it has a maximum oxidation state of $$+3$$

$$A)$$ As halide is monovalent, the element $$R$$ forms halides of the form $$R{ X }_{ 3 }$$

$$B)$$ It has an oxidation state of $$+3$$ because it has 3 valence electrons, so this option is true.

$$C)$$ As oxygen is divalent so the element $$R$$ forms oxides of the type $${ R }_{ 2 }{ O }_{ 3 }$$.

So, all the above options are true.
Question 10
1 / -0

Consider the reaction, $$LiH+AlH_{3}\rightarrow .....$$
Which is the incorrect statement about this reaction?

A
The product is a good reducing agent

B
$$AlH_{3}$$ acts as a lewis acid

C
There is decrease in the oxidation number of aluminium in the product

D
$$LiH$$ donates $$H^{-}$$ to $$AlH_{3}$$

Solution

There is no change in the oxidation state of Aluminium during the reaction
The reaction is as follow:
$$LiH + AlH_{3} \rightarrow LiAlH_{4}$$
The oxidation state of Al in both reactant as well as in product is -3.

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Re-Attempt Weekly Quiz Competition

The p-Block Elements Test - 22

Result

The p-Block Elements Test - 22

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SHARING IS CARING

Question 1

Orthoboric acid

Metaboric Acid

Tetraboric Acid

Pyroboric acid

Solution

Question 2

blue

No Change

red

white

Solution

Question 3

$$ B_{2}O_{3}+2Al$$

$$BCl_{3}\: +H_{2}$$

$$B_{2}O_{3}+3Mg$$

$$2B+3MgO$$

Solution

Question 4

$$\mathrm{H}\mathrm{F}$$

$$\mathrm{H}\mathrm{Cl}\mathrm{O}_{4}$$

$$\mathrm{H}\mathrm{N}\mathrm{O}_{3}$$

$$\mathrm{H}_{2}\mathrm{S}\mathrm{O}_{4}$$

Solution

Question 5

$$Na_{2}B_{4}O_{7}.10H_{2}O$$

$$Na_{2}B_{4}O_{7}.5\, H_{2}\, O$$

$$Na_{2}B_{4}O_{7}.3H_{2}O$$

$$Na_{2}B_{4}O_{7}.4H_{2\, }O$$

Solution

Question 6

Cu

Al

Hg

Pb

Solution

Question 7

$$Al< Tl< Ga> In$$

$$Ga< Al< Tl< In$$

$$Tl< In< Ga< Al$$

$$Al< Ga< In< Tl$$

Solution

Question 8

$${F}_{2}$$

$$NaF$$

$$Na_{3}AlF_{3}$$

$$Na_{2}SiF_{6}$$

Solution

Question 9

It forms halide of type $$RX_{3}$$

It has oxidation state of +3

It forms oxide of type $$R_{2}O_{3}$$

.All the above

Solution

Question 10

The product is a good reducing agent

$$AlH_{3}$$ acts as a lewis acid

There is decrease in the oxidation number of aluminium in the product

$$LiH$$ donates $$H^{-}$$ to $$AlH_{3}$$

Solution

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