The p-Block Elements Test

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The p-Block Elements Test - 57

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Weekly Quiz Competition

Question 1
1 / -0

The nature of aqueous solution of borax is :

A
Acidic

B
Neutral

C
Amphoteric

D
Basic

Solution

When borax is dissolved in water it ionizes and hydrolysis of borate ion gives OH (the hydroxyl ion) and boric acid. Since boric acid is a weak acid and $$NaOH$$ is strong a alkali an aqueous solution of borax is alkaline in nature.
Question 2
1 / -0

Which of the following is an organo silicon polymer ?

A
Silica

B
Silicone

C
Silicone carbide

D
Silicic acid

Solution

Silicone is a polymeric organo silicon compound which is chemically inert and thermally stable.
hence B is correct answer
Question 3
1 / -0

$$SiCl_{4} \overset{H_2O}{\longrightarrow} X \overset{Heat}{\longrightarrow} Y \overset{NaOH}{\longrightarrow} Z$$
X, Y and Z in the above reaction are

A
$$X \rightarrow SiO_2 , Y \rightarrow Si , Z \rightarrow NaSi$$

B
$$X \rightarrow Si(OH)_4, Y \rightarrow SiO_2, Z \rightarrow Na_2SiO_3$$

C
$$X \rightarrow Si(OH)_4, Y \rightarrow Si, Z \rightarrow SiO_2$$

D
$$X \rightarrow SiO_2, Y \rightarrow SiCl_{4}, Z \rightarrow Na_2SiO_3$$

Solution
Question 4
1 / -0

Group $$13$$ elements show

A
only $$+1$$ oxidation state

B
only $$+3$$ oxidation state

C
$$+1$$ and $$+3$$ oxidation states

D
$$+1, +2$$ and $$+3$$ oxidation states.

Solution

Group -$$13$$ elements show $$+1$$ and $$+3$$ oxidation states. Tendency to show $$+3$$ oxidation state decreases down the group.
Question 5
1 / -0

Carborumdum is a:

A
molecular solid

B
covalent solid

C
ionic solid

D
amorphous solid

Solution

There are only two carbides that are considered completely covalent; they are formed with the two elements that are most similar to carbon in size and electronegativity, boron (B) and silicon (Si). silicon carbide (SiC) is known as carborundum and is prepared by the reduction of silicon dioxide (SiO₂) with elemental carbon in an electric furnace.
so option (b) is correct
Question 6
1 / -0

Which substance is having molecular solid :

A
graphite

B
$$C_{60}$$

C
gold

D
$$Ca_{3}(PO_{4})_{2}$$

Solution

Buckminsterfullerene is a type of fullerene with the formula C₆₀. It has a cage-like fused-ring structure that resembles a soccer ball, made of twenty hexagons and twelve pentagons. Each carbon atom has three bonds. It is a black solid that dissolves in hydrocarbon solvents to produce a violet solution.
Question 7
1 / -0

Which is the hardest compound of boron ?

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

B
$$BN$$

C
$$B_{4}C$$

D
$$B_{2}H_{6}$$

Solution

$$B_{4}C$$ is the hardest compound of boron .
Boron carbide $$B_{4}C$$ is a very hard substance like $$SiC.$$
Question 8
1 / -0

Choose the correct order of $$C - C$$ bond length in the given compounds:

A
Acetylene < ethylene < graphite < benzene < ethane

B
Acetylene < ethylene < benzene < graphite < ethane

C
Acetylene < graphite < ethylene < benzene < ethane

D
Acetylene < benzene < graphite < ethylene < ethane

Solution

thane contains carbon-carbon single bond.
Benzene is aromatic in nature. It has resonance. All the carbon-carbon bonds in benzene have partial double bond character.
Ethylene contains carbon-carbon double bond.
Acetylene contains carbon-carbon triple bond.
Hence, the correct order of carbon-carbon bond length is $$C_2H_6>C_6H_6>C_2H_4>C_2H_2$$.
Acetylene < benzene < graphite < ethylene < ethane
Question 9
1 / -0

The incorrect stability order of $$+3$$ and $$+1$$ states of $$13th$$ group elements (boron family) is:

A
$$Ga^{3+} < In^{3+} < Tl^{3+}$$

B
$$Tl^{+} > Tl^{3+}$$

C
$$Ga^{+} < ln^{+} < Tl^{+}$$

D
$$Ga^{3+} > Ga^{+}$$

Solution

stability is depend upon the stomic size in block top to bottam so the correvt order A is correct
Question 10
1 / -0

Correct order of bond angle in given species is:

A
$$SiO_{4}^{4-} > PCl_{3} > NCl_{3} > SbH_{3} > H_{2}Te$$

B
$$SiO_{4}^{4-} > NCl_{3} > PCl_{3} > SbH_{3} > H_{2}Te$$

C
$$SiO_{4}^{4-} > H_{2}Te > SbH_{3} > PCl_{3} > NCl_{3}$$

D
$$NCl_{3} > PCl_{3} > SiO_{4}^{4-} > SbH_{3} > H_{2}Te$$

Solution

$$\angle Cl - P - Cl$$ in $$PCl_{3} > \angle H - Sb - H$$ in $$SbH_{3},$$ as in $$SbH_{3},$$ bond pairs are formed by overlapping of almost pure p-orbitals.

$$\angle H - Sb - H$$ in $$SbH_{3} < \angle H - Te - H$$ in $$H_{2}Te$$ because two lone pairs are present on $$Te$$ while at $$Sb$$ there is one lone pair.

Hence, Option "B" is the correct answer.

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Re-Attempt Weekly Quiz Competition

The p-Block Elements Test - 57

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The p-Block Elements Test - 57

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Question 1

Acidic

Neutral

Amphoteric

Basic

Solution

Question 2

Silica

Silicone

Silicone carbide

Silicic acid

Solution

Question 3

$$X \rightarrow SiO_2 , Y \rightarrow Si , Z \rightarrow NaSi$$

$$X \rightarrow Si(OH)_4, Y \rightarrow SiO_2, Z \rightarrow Na_2SiO_3$$

$$X \rightarrow Si(OH)_4, Y \rightarrow Si, Z \rightarrow SiO_2$$

$$X \rightarrow SiO_2, Y \rightarrow SiCl_{4}, Z \rightarrow Na_2SiO_3$$

Solution

Question 4

only $$+1$$ oxidation state

only $$+3$$ oxidation state

$$+1$$ and $$+3$$ oxidation states

$$+1, +2$$ and $$+3$$ oxidation states.

Solution

Question 5

molecular solid

covalent solid

ionic solid

amorphous solid

Solution

Question 6

graphite

$$C_{60}$$

gold

$$Ca_{3}(PO_{4})_{2}$$

Solution

Question 7

$$B_{2}O_{3}$$

$$BN$$

$$B_{4}C$$

$$B_{2}H_{6}$$

Solution

Question 8

Acetylene < ethylene < graphite < benzene < ethane

Acetylene < ethylene < benzene < graphite < ethane

Acetylene < graphite < ethylene < benzene < ethane

Acetylene < benzene < graphite < ethylene < ethane

Solution

Question 9

$$Ga^{3+} < In^{3+} < Tl^{3+}$$

$$Tl^{+} > Tl^{3+}$$

$$Ga^{+} < ln^{+} < Tl^{+}$$

$$Ga^{3+} > Ga^{+}$$

Solution

Question 10

$$SiO_{4}^{4-} > PCl_{3} > NCl_{3} > SbH_{3} > H_{2}Te$$

$$SiO_{4}^{4-} > NCl_{3} > PCl_{3} > SbH_{3} > H_{2}Te$$

$$SiO_{4}^{4-} > H_{2}Te > SbH_{3} > PCl_{3} > NCl_{3}$$

$$NCl_{3} > PCl_{3} > SiO_{4}^{4-} > SbH_{3} > H_{2}Te$$

Solution

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