The p-Block Elements Test

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The p-Block Elements Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

Aluminium exhibits +3 oxidation state. As we move down the group, +1 oxidation states get more stable. This is a consequence of :

A
increasing size of the atom

B
inert pair effect

C
electron deficient nature

D
$$p\pi $$-$$p\pi $$ bonding

Solution

Answer:- (B) inert pair effect
The atoms of group 13 elements have 3 valence electrons, two in s-subshell and one in p subshell.
The +1 oxidation state becomes more stable as we move down to the group due to inert pair effect. After the removal of 1 electron from p orbital, the remaining $$ns^2$$ electrons behave like stable noble gases due to inert pair effect and do not take part in compound formation. The two electrons present in the s-shell are strongly attracted by the nucleus and do not participate in bonding. This inert pair effect becomes more and more prominent on moving down the group.
Hence, the +1 oxidation state gets more stable as we move down the group.
Question 2
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Which is the most electropositive element?

A
Na

B
Cu

C
Cs

D
Ca

Solution

Metals are electropositive in nature as they easily lose the electron to form a cation. The element of group 1 have the highest metallic character and thus are the most electropositive elements. Also, similar to the metallic character, the electropositive character increases down the group. Thus Cs is most electropositive.
option C
Question 3
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Group $$13$$ elements show $$+1$$ and $$+3$$ oxidation states. Relative stability of $$+3$$ oxidation state may be given as :

A
$$Tl^{3+}>In^{3+}>Ga^{3+}>Al^{3+}>B^{3+}$$

B
$$B^{3+}>Al^{3+}>Ga^{3+}>In^{3+}>Tl^{3+}$$

C
$$Al^{3+}>Ga^{3+}>Tl^{3+}>In^{3+}>B^{3+}$$

D
$$Al^{3+}>B^{3+}>Ga^{3+}>Tl^{3+}>In^{3+}$$

Solution

Answer:- (B) $$B^{+3} > {Al}^{+3}> {Ga}^{+3}> {In}^{+3} > {Tl}^{+3}$$
Reason:-
As we move down the group the inert pair effect becomes more and more prominent which causes the decrease in stability of elements in +3 oxidation state.
Hence, (B) is correct answer.
Question 4
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Chemically borax is ?

A
sodium metaborate

B
sodium orthoborate

C
sodium tetraborate decahydrate

D
sodium hexaborate.

Solution

Answer:- (C) Sodium tetraborate decahydrate
Borax, also known as sodium perborate is an important boron compound, a mineral, and the salt of boric acid. It is usually a white powder consisting of soft colorless crystals that dissolve easily in water.
Chemical formula of borax is -
$${Na}_2B_4O_7.10H_2O$$ and called as sodium tetraborate decahydrate.
Question 5
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Aluminium oxide is not reduced by chemical reactions due to ?

A
its highly stable nature

B
its highly unstable nature

C
its amphoteric nature

D
its highly explosive nature

Solution

Answer:- its highly stable nature
Aluminium oxide is highly stable so extraction of aluminium is done by electrolytic process.
In this process, alumina $${Al}_2O_3$$ is used along with flourspar and cryolite.
Question 6
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Boron is unable to form $${ BF }_{ 6 }^{ 3- }$$ ions due to:

A
non-availability of d-orbitals

B
small size of boron atom

C
non-metallic nature

D
less reactivity towards halogens.

Solution

Boron does not have d orbitals. Because of this, it is not possible for the element to expand on its octet. This causes the maximum covalence of boron to be restricted to 4. Beyond this value, boron will become unstable and thus, cannot form $${BF_6}^{3-}$$ ions.

Hence, the correct answer is option $$\text{A}$$.
Question 7
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The tendency of group 14 elements to show +2 oxidation state increases in the order of:

A
$$C < Si < Sn < Pb < Ge$$

B
$$C < Si < Ge < Sn < Pb$$

C
$$Ge < Sn < Pb < C < Si$$

D
$$Pb < Sn < Ge < C < Si$$

Solution

The tendency to show $$+2$$ oxidation state increases as we move down the group. This is due to inert pair effect which causes the inability of $$ns^2$$ electrons of valence shell to participate in bonding.

Thus, the stability of elements in $$+2$$ oxidation state increases as we move down the group in the order $$C < Si < Ge < Sn < Pb$$.

Hence, the correct answer is option $$\text{B}$$.
Question 8
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The first member of the p-block elements differs from the remaining members of their corresponding groups due to :

A
small size and absence of d-orbitals

B
diagonal relationship with other elements

C
difference in ability to form double and triple bonds

D
high ionisation enthalpy

Solution

Answer:- (A) small size and absence of d-orbitals
Because of their smaller size and absence of d-orbitals the first member of p-block elements possess different properties from the remaining members of their corresponding group.
Thus, (A) is correct answer.
Question 9
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Which of the following bonds is shown in silicones?

A
$$O-\!\!\!\!(\underset{|}{ \overset{|}Si}-O-\!\!\!\!)_n$$

B

C
$$O-\underset{O}{\underset{|}{\overset{O}{\overset{|}{Si}}}}-O$$

D

Solution

Answer:-
Silicones are polymers, contains $${({R}_{2}Si{O})}_{n}$$ units as shown in fig.
Thus, (A) is correct answer.
Question 10
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Silicon is an important constituent of:

A
sand

B
atmosphere

C
plants

D
water bodies

Solution

Answer:- (A) sand
Silicon is the second most abundant element in the Earth's crust, making up $$25.7\%$$ of it by weight. It occurs in clay, feldspar, granite, quartz and sand, mainly in the form of silicon dioxide (also known as silica) and silicates (compounds containing silicon, oxygen and metals).

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Re-Attempt Weekly Quiz Competition

The p-Block Elements Test - 44

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The p-Block Elements Test - 44

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Question 1

increasing size of the atom

inert pair effect

electron deficient nature

$$p\pi $$-$$p\pi $$ bonding

Solution

Question 2

Na

Cu

Cs

Ca

Solution

Question 3

$$Tl^{3+}>In^{3+}>Ga^{3+}>Al^{3+}>B^{3+}$$

$$B^{3+}>Al^{3+}>Ga^{3+}>In^{3+}>Tl^{3+}$$

$$Al^{3+}>Ga^{3+}>Tl^{3+}>In^{3+}>B^{3+}$$

$$Al^{3+}>B^{3+}>Ga^{3+}>Tl^{3+}>In^{3+}$$

Solution

Question 4

sodium metaborate

sodium orthoborate

sodium tetraborate decahydrate

sodium hexaborate.

Solution

Question 5

its highly stable nature

its highly unstable nature

its amphoteric nature

its highly explosive nature

Solution

Question 6

non-availability of d-orbitals

small size of boron atom

non-metallic nature

less reactivity towards halogens.

Solution

Question 7

$$C < Si < Sn < Pb < Ge$$

$$C < Si < Ge < Sn < Pb$$

$$Ge < Sn < Pb < C < Si$$

$$Pb < Sn < Ge < C < Si$$

Solution

Question 8

small size and absence of d-orbitals

diagonal relationship with other elements

difference in ability to form double and triple bonds

high ionisation enthalpy

Solution

Question 9

$$O-\!\!\!\!(\underset{|}{ \overset{|}Si}-O-\!\!\!\!)_n$$

$$O-\underset{O}{\underset{|}{\overset{O}{\overset{|}{Si}}}}-O$$

Solution

Question 10

sand

atmosphere

plants

water bodies

Solution

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