The p-Block Elements Test

Result

The p-Block Elements Test - 9

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Aluminium is usually found in $$+3$$ oxidation state. In contrast, thallium exists in $$+1$$ and $$+3$$ oxidation states. This is due to :

A
lanthanoid contraction

B
lattice effect

C
diagonal relationship

D
inert pair effect

Solution

Inert pair effect is the prominent character of the p-block element.

In this, the high molecular weight element of the group show a lower oxidation state.

This is because on going down the group, the shielding effect increases, but d-subshell and f-subshell show poor shielding effect.

The high molecular weight members of p-block groups contain d-subshell and f-subshell.

Because of this, they show a lower oxidation state.

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

Correct statements among a to d regarding silicones are:

(a) They are polymers with hydrophobic character.

(b) They are biocompatible.

(c) In general, they have high thermal stability and low dielectric strength.

(d) Usually, they are resistant to oxidation and used as greases.

A
(a), (b) and (c) only.

B
(a), and (b) only.

C
(a), (b), (c) and (d).

D
(a), (b) and (d) only.

Solution

Silicones are the polymer of silicon-containing $$[-(R)_3Si-O-]$$ linkage.

It is a polymer with hydrophobic character thus used in making water-resistant seals.

They are biocompatible.

They have high thermal stability and low dielectric strength which is not correct.

They are also resistant to oxidation. Because of there wax-like texture, they are also used in greases.

Hence, the correct option is D.
Question 3
1 / -0

Among the following substituted silanes the one which will give rise to cross linked silicone polymer on hydrolysis is:

A
$$\mathrm{R}_{2}\mathrm{S}\mathrm{i}\mathrm{C}1_{2}$$

B
$$\mathrm{R}_{3}\mathrm{S}\mathrm{i}\mathrm{C}1$$

C
$$\mathrm{R}_{4}\mathrm{S}\mathrm{i}$$

D
$$\mathrm{R}\mathrm{S}\mathrm{i}\mathrm{C}1_{3}$$

Solution

Trichlorosilane is the basic ingredient used in the production of purified cross linked silicon polymer. In water, it rapidly decomposes to produce a silicone polymer while giving off hydrochloric acid.
$$RSiCl_3 + H_2O \rightarrow RSi(OH)_3 + HCl$$
Linear and cyclic silicones are produced by the reaction of water with organochlorosilanes of the general formula, $$R_2SiCl_2$$ followed by a polymerization reaction that occurs by the elimination of a molecule of water from two hydroxyl groups of adjacent $$R_2Si(OH)_2$$ molecules.
Question 4
1 / -0

The amorphous form of silica is:

A
Quartz

B
Kieselguhr

C
Cristobalite

D
Tridymite

Solution

Kieselguhr is amorphous form of silica.
Question 5
1 / -0

Which one of the following elements is unable to form $$MF_{6}^{3-}$$ ion?

A
$$B$$

B
$$Ga$$

C
$$In$$

D
$$Al$$

Solution

B is unable to form $$MF_{6}^{3-}$$ ion.
This is because $$\displaystyle B^{3+}$$ has no vacant d orbitals. It has only on s and 3 p orbitals in the valence shell.
Hence, it can accept 4 electron pairs and can form $$\displaystyle BF_4^-$$.
therefore option A is correct.
Question 6
1 / -0

Which of the following is incorrect statement ?

A
$$PbF_4$$ is covalent in nature

B
$$SiCl_4$$ is easily hydrolysed

C
$$GeX_4 (X=F,Cl,Br,I)$$ is more stable than $$GeX_2$$

D
$$SnF_4$$ is ionic in nature

Solution

$$PbF_4$$ and $$SnF_4$$ are ionic in nature.

Hence option A is the correct answer.
Question 7
1 / -0

How many electrons can fit in the orbital for which $$n = 3$$ and $$l = 1$$?

A
10

B
14

C
2

D
6

Solution

$$n= 3$$ and $$l= 1$$

$$\Rightarrow 3p$$ orbital so $$3p$$ orbital can accommodate 6 electrons.
Question 8
1 / -0

The stability of $$+1$$ oxidation state among $$Al$$, $$Ga$$, $$In$$ and $$Tl$$ increases in the sequence:

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

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

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

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

Solution

The stability of $$+1$$ oxidation state among $$Al$$, $$ Ga$$, $$In$$ and $$Tl$$ increases in the sequence.
$$\displaystyle Al < Ga < In < Tl $$.
This is due to inert pair effect due to which, on moving down the group, the stability of $$+3$$ oxidation state decreases while the stability of $$+1$$ oxidation state increases.
Question 9
1 / -0

The element $$Z = 114$$ has been discovered recently. It will belong to which of the following family/ group and electronic configuration?

A
Nitrogen family, $$[Rn] 5f^{14} 6d^{10} 7s^{2} 7p^{6}$$

B
Halogen family, $$[Rn] 5f^{14} 6d^{10} 7s^{2} 7p^{5}$$

C
Carbon family, $$[Rn] 5f^{14} 6d^{10} 7s^{2} 7p^{2}$$

D
Oxygen family, $$[Rn] 5f^{14} 6d^{10} 7s^{2} 7p^{4}$$

Solution

The element $$Z=114$$ has been discovered recently.

It will belong to carbon family and its electronic configuration is $$\displaystyle [Rn] 5f^{14} 6d^{10} 7s^{2} 7p^{2}$$. It belongs to seventh period and fourteenth group.

Hence, the correct option is $$C$$.
Question 10
1 / -0

What is the change in oxidation number of carbon in the following reaction?
$$CH_4 (g) + 4Cl_2 (g)\longrightarrow CCl_4 (l) + 4HCl(g)$$

A
$$0$$ to $$+ 4$$

B
$$-4$$ to $$+ 4$$

C
$$0$$ to $$-4$$

D
$$+4$$ to $$+ 4$$

Solution

$$\overset {-4} CH_4 (g) + 4Cl_2 (g)\longrightarrow \overset {+4} CCl_4 (l) + 4HCl(g)$$

The oxidation state of $$H$$ and $$Cl$$ are +1 and -1 respectively. Therefore the oxidation state of C in methane is $$-4$$ and that in carbon tetrachloride is $$+4$$.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

The p-Block Elements Test - 9

Result

The p-Block Elements Test - 9

Score

-

Rank

-

SHARING IS CARING

Question 1

lanthanoid contraction

lattice effect

diagonal relationship

inert pair effect

Solution

Question 2

(a), (b) and (c) only.

(a), and (b) only.

(a), (b), (c) and (d).

(a), (b) and (d) only.

Solution

Question 3

$$\mathrm{R}_{2}\mathrm{S}\mathrm{i}\mathrm{C}1_{2}$$

$$\mathrm{R}_{3}\mathrm{S}\mathrm{i}\mathrm{C}1$$

$$\mathrm{R}_{4}\mathrm{S}\mathrm{i}$$

$$\mathrm{R}\mathrm{S}\mathrm{i}\mathrm{C}1_{3}$$

Solution

Question 4

Quartz

Kieselguhr

Cristobalite

Tridymite

Solution

Question 5

$$B$$

$$Ga$$

$$In$$

$$Al$$

Solution

Question 6

$$PbF_4$$ is covalent in nature

$$SiCl_4$$ is easily hydrolysed

$$GeX_4 (X=F,Cl,Br,I)$$ is more stable than $$GeX_2$$

$$SnF_4$$ is ionic in nature

Solution

Question 7

10

14

2

6

Solution

Question 8

$$Tl < In < Ga < Al$$

$$In < Tl < Ga < Al$$

$$Ga < In < Al < Tl$$

$$Al < Ga < In < Tl$$

Solution

Question 9

Nitrogen family, $$[Rn] 5f^{14} 6d^{10} 7s^{2} 7p^{6}$$

Halogen family, $$[Rn] 5f^{14} 6d^{10} 7s^{2} 7p^{5}$$

Carbon family, $$[Rn] 5f^{14} 6d^{10} 7s^{2} 7p^{2}$$

Oxygen family, $$[Rn] 5f^{14} 6d^{10} 7s^{2} 7p^{4}$$

Solution

Question 10

$$0$$ to $$+ 4$$

$$-4$$ to $$+ 4$$

$$0$$ to $$-4$$

$$+4$$ to $$+ 4$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.