The p-Block Elements Test

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The p-Block Elements Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

The tendency of BF$$_{3}$$, BCl$$_{3}$$ and BBr$$_{3}$$ to behave as Lewis acid decreases in the sequence is:

A
BCl$$_{3}$$ > BF$$_{3}$$ > BBr$$_{3}$$

B
BBr$$_{3}$$ > BCl$$_{3}$$ > BF$$_{3}$$

C
BBr$$_{3}$$ > BF$$_{3}$$ > BCl$$_{3}$$

D
BF$$_{3}$$ > BCl$$_{3}$$ > BBr$$_{3}$$

Solution

A molecule is said to be a lewis acid if it can accept a lone pair.

When the halide is more electronegative the interelectronic repulsion increases making the molecule difficult to accept a lone pair.

Size of halide also behaves in the same way. Also the degree of hydrolysis increases from fluorine to iodine,so the lewis acidic character increases in the same order.

So,the order of lewis acidity is $$B{ Br }_{ 3 }>{ B }{ Cl }_{ 3 }>{ B }{ F }_{ 3 }$$.

Hence, the correct option is $$B$$.
Question 2
1 / -0

What is the value of X in the given chemical formula of crystalline borax, $$Na_{2}[B_{4}O_{5}(OH)_{4}].XH_{2}O$$?

A
5

B
6

C
7

D
8

Solution

Borax is white crystalline solid of formula $$Na_2B_4O_7.10H_2O$$ which can also be written as $$Na_{2}[B_{4}O_{5}(OH)_{4}].8H_{2}O$$.
Question 3
1 / -0

Which of the following is the electron deficient molecule?

A
$$C_{2}H_{6}$$

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

C
$$SiH_{4}$$

D
$$PH_{3}$$

Solution

$$B_2H_6$$ is the electron deficient molecule which completes its deficiency by forming Bridge or Banana
Bond. These bonds are also called $$3e-3e$$ bonds
Question 4
1 / -0

When borax is dissolved in water :

A
$$B(OH)_{3}$$ is formed only

B
$$[B(OH)_{4}]^{-}$$ is formed only

C
both $$B(OH)3$$ and $$[B(OH)_{4}]^{-}$$ are formed

D
$$[B_{3}O_{3}(OH)_{4}]^{-}$$ is formed only

Solution

Hint: $$Boric \ acid$$ is a weak Lewis acid.

Explanation:

When $$borax$$ is dissolved in water, $$sodium \ hydroxide$$ and $$boric \ acid$$ are formed:

$$Na_2B_4O_7 + H_2O \rightarrow NaOH + H_3BO_3$$

Since $$boric \ acid$$ is a weak Lewis acid, it can accept an electron pair from a hydroxide ion.

$$H_3BO_3 + OH^- \rightarrow [B(OH)_4]^-$$

The correct answer is option $$C$$.
Question 5
1 / -0

The structure of silicon(IV) oxide belongs to the type:

A
ionic lattice

B
macromolecular, with a layer structure

C
molecular lattice, with van der Waals forces among the molecules

D
macromolecular, with a non-layer structure

Solution

$$Si(IV)$$ oxide is macromolecular with a non layer structure similar to diamond.
Question 6
1 / -0

The gaseous product(s) expected at room temperature by reaction of sodium borohydride an boron trifluoride under anhydrous conditions is/are :

A
$$H_2$$

B
$$B_2H_6$$ and $$H_2$$

C
$$B_2H_6$$

D
$$BH_2F$$ and $$H_2$$

Solution

$$3Na[{ BH }_{ 4 }]+{ 4BF }_{ 3 }\longrightarrow { 2B }_{ 2 }{ H }_{ 6 }(g)+3Na[{ BF }_{ 4 }]$$
$$B_2H_6$$ is the gaseous product
Question 7
1 / -0

The possible oxidation state of Tl are:

A
+1 and +2

B
+2 and +3

C
+1 and -1

D
+1 and +3

Solution

$$Tl$$ shows both +1 and +3 oxidation states. The stability of +1 oxidation state increases down the group $$Tl>In>Ga>Al$$ and +3 oxidation state is highly oxidising in character.
Question 8
1 / -0

The repeating structural unit in silicone is:

A
$$R_2SiO$$

B
$$SiO_2$$

C
$$SiO_3$$

D
All of the above

Solution

The repeating structural unit in silicone is $$R_2SiO$$.
Silicones are organosilicon polymers containing $$Si-O-Si$$ linkages.
Question 9
1 / -0

Formula of Pyro silicic acid:

A
$$ H_2Si_4O_9 $$

B
$$H_4Si_2O_6$$

C
$$H_4Si_3O_8$$

D
$$H_6Si_2O_7$$

Solution

Hydrated Silica is a form of silicon dioxide, which has a variable amount of water in the formula. It is also known as silicic acid, a term usually used for its form dissolved in water. $$H_6Si_2O_7$$ is chemical formula of pyro silicic acid. Chemical Formula: $$SiO_{2n}H_2O$$.
Question 10
1 / -0

$$CO_2$$ is gas, while $$SiO_2$$ is solid because :

A
$$CO_2$$ is a linear molecule, while $$SiO_2$$ is angular.

B
van der Waals' forces are very strong in $$SiO_2$$.

C
$$CO_2$$ is covalent, while $$SiO_2$$ is ionic.

D
$$Si$$ cannot form stable bonds with $$O$$, hence $$Si$$ has to form a $$3D$$ lattice.

Solution

$$SiO_2$$ is solid because $$Si$$ cannot form stable bonds with $$O$$, hence $$Si$$ has to form a $$3D$$ lattice of $$SiO_2$$ molecules and form solid structure.

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Re-Attempt Weekly Quiz Competition

The p-Block Elements Test - 24

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The p-Block Elements Test - 24

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Question 1

BCl$$_{3}$$ > BF$$_{3}$$ > BBr$$_{3}$$

BBr$$_{3}$$ > BCl$$_{3}$$ > BF$$_{3}$$

BBr$$_{3}$$ > BF$$_{3}$$ > BCl$$_{3}$$

BF$$_{3}$$ > BCl$$_{3}$$ > BBr$$_{3}$$

Solution

Question 2

5

6

7

8

Solution

Question 3

$$C_{2}H_{6}$$

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

$$SiH_{4}$$

$$PH_{3}$$

Solution

Question 4

$$B(OH)_{3}$$ is formed only

$$[B(OH)_{4}]^{-}$$ is formed only

both $$B(OH)3$$ and $$[B(OH)_{4}]^{-}$$ are formed

$$[B_{3}O_{3}(OH)_{4}]^{-}$$ is formed only

Solution

Question 5

ionic lattice

macromolecular, with a layer structure

molecular lattice, with van der Waals forces among the molecules

macromolecular, with a non-layer structure

Solution

Question 6

$$H_2$$

$$B_2H_6$$ and $$H_2$$

$$B_2H_6$$

$$BH_2F$$ and $$H_2$$

Solution

Question 7

+1 and +2

+2 and +3

+1 and -1

+1 and +3

Solution

Question 8

$$R_2SiO$$

$$SiO_2$$

$$SiO_3$$

All of the above

Solution

Question 9

$$ H_2Si_4O_9 $$

$$H_4Si_2O_6$$

$$H_4Si_3O_8$$

$$H_6Si_2O_7$$

Solution

Question 10

$$CO_2$$ is a linear molecule, while $$SiO_2$$ is angular.

van der Waals' forces are very strong in $$SiO_2$$.

$$CO_2$$ is covalent, while $$SiO_2$$ is ionic.

$$Si$$ cannot form stable bonds with $$O$$, hence $$Si$$ has to form a $$3D$$ lattice.

Solution

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