The p-Block Elements Test

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The p-Block Elements Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

When borax is dissolved in water:

A
$$B(OH)_3$$ is formed only

B
$$[B(OH)_4]^-$$ formed only

C
both $$B(OH)_3$$ and $$[B(OH)_4]^-$$ are formed

D
$$[B_3O_3(OH)_4]^-$$ is formed only

Solution

Hint: Borax reacts to give an alkaline solution.
Correct Answer: Option C
Explanation:
The reaction of borax in water is as follows:

$${\left[ {{{\text{B}}_4}{{\text{O}}_7}} \right]^{2 - }} +\ 7{{\text{H}}_2}{\text{O}} \to 2\;{\text{B}}{({\text{OH}})_3} + 2{\left[ {\;{\text{B}}{{({\text{OH}})}_4}} \right]^ - }$$

Hence both $$B{(OH)}_3$$ and $$[B{(OH)}_4]^-$$ are formed.
Question 2
1 / -0

Which of the following molecule is /are chiral ?

A
I

B
II

C
III

D
IV

Solution

1 is chiral molecule because it is nonsuperimposable on its mirror imageand presence of an asymmetric carbon center
Question 3
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Which one of the following statements about $$H_{3}BO_{3}$$ is not correct?

A
It is a strong tribasic acid.

B
It is prepared by acidifying an aqueous solution of borax.

C
It has a layer structure in which planar BO, units are joined by hydrogen bonds.

D
It does not act as proton donor as it acts as a Lewis acid by accepting hydroxyl ions.

Solution

1. It is not a tribasic acid.
Option A is incorrect for $$H_3BO_3$$.

2. Aqueous solution, instead of donating the $$-OH$$ group it breaks the $$H-OH$$ bond of water and accepts the $$-OH$$ group from it leaving in the solution hence it is mono basic acid.
Question 4
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$$SiF_4$$ gets hydrolysed then which of the following is not expected to from?

A
$$SiO_2$$

B
$$Si(OH)_2F_2$$

C
$$H_2SiF_6$$

D
$$Si(OH)_4$$

Solution

Partial hydrolysis is not possible for $$SiF_{4}$$
$$4H_{2}O+SiF_{4}\longrightarrow Si(OH)_{4}+HF$$
$$H_{2}O+SiF_{4}\longrightarrow H_{2}SiO_{3}+H_{2}SiF_{6}$$
$$2H_{2}O+SiF_{4}\longrightarrow SiO_{2}+4HF$$
Hence $$Si(OH)_{2}F_{2}$$ is not possible
Question 5
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In boron atom screening on outermost electron is due to:

A
electrons of K shell only

B
all the electrons of K and L shell

C
two electrons of 1s and 2s each

D
only by electrons of L shell

Solution

Boron $$=1S^2 2S^2 2P^1$$
Screening effect :- Higher is the screening electron causes less attraction from the nucleus and can be easily removed, which leads to lower value of ionisation potential.
To Screening effect in Boron in Boron is due to two electrons of $$1S$$ and $$2S$$ subshells.
Question 6
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Elements of group 13 shows :

A
only +1 oxidation state

B
only +3 oxidation state

C
+1 and +3 oxidation states

D
+1, +2 and +3 oxidation states

Solution

Answer:- (C) +1 and +3 oxidation states
Reason:-
The element of group 13 has 3 electrons in their outermost orbit. The s-subshell contains 2 electrons whereas the p subshell contain only one electron. Thus, the elements of the boron family adopt +3 or +1 oxidation states. Due to stability and inert pair effect, the oxidation state decreases as the size increases. Hence the elements down the group show +1 oxidation state.
Hence, (C) is correct answer.
Question 7
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Electropositive character for the elements of group 13 follows the order ?

A
B > Al > Ga > In > Tl

B
B < Al < Ga < In < Tl

C
B < AL > Ga < In > Tl

D
B < Al > Ga > In > Tl

Solution

Answer:- (D) $$B<Al>Ga>In>Tl$$
Reason:-
The electropositive character of elements increases from boron to aluminium due to smaller size and high ionisation potential of B than Al.
However, the electropositive character decreases slowly from Gallium to Thallium due to extra d-electrons which exerts a very little shielding effect on outer electrons. Therefore, these electrons are more firmly held by the nucleus and hence there is a decrease in electropositive character.
Hence, (D) is correct answer.
Question 8
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In group 13, electronegativity first decreases from B to Al and then increase marginally down the group. This is because of :

A
non- metallic nature of B

B
discrepancies in atomic size of element

C
ability of B and Al to form $$p\pi $$ - $$p\pi $$ multiple bonds

D
irregular trend in electronegativity throughout the periodic table

Solution

Answer:- (B) discrepancies in atomic size of an element.
Reason:-
Electronegativity of an element depends upon
(a) Size of an atom
(b) The magnitude of a nuclear charge
On going from Al to Ga, The electrons start filling into the d-subshells. Since these intervening electrons do not screen the nuclear charge.
Hence, the electrons in Ga experience more force of attraction than in Al causing a decrease in atomic size and increase in electronegativity.
Thus, Electronegativity decreases from B to Al and then increases due to discrepancies of an atomic size of the elements.
Question 9
1 / -0

Which of the following metals does not show inert pair effect?

A
Thallium

B
Gallium

C
Indium

D
Aluminium
Solution
Answer:- (D) Aluminium
Inert pair effect means that the 2 s-electrons of the valence shell of heavier p-block elements form an inert pair and do not participate in bond formation.
Inert pair effect occurs due to the poor shielding effect of d and f orbitals.
Hence, due to small size and absence of d and f orbitals, Aluminium does not show inert pair effect.
Question 10
1 / -0

The reason behind the lower atomic radius of Ga as compared to Al is ?

A
poor screening effect of d-electrons for the outer electrons from the increased nuclear charge

B
increased force of attraction of increased nuclear charge on electrons

C
increased ionisation enthalpy of Ga as compared to Al

D
Anomalous behaviour of Ga

Solution

Answer:- (A) poor screening effect of d-electrons for the outer electrons from an increased nuclear charge.
Reason:-
The presence of additional 10d electrons in gallium offers poor screening effect for the outer electrons which increases the force of attraction between the outermost electrons and nuclear charge, due to which the atomic radius of gallium decreases.
Hence, Ga has a lower atomic radius as compared to Al.

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The p-Block Elements Test - 43

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The p-Block Elements Test - 43

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Question 1

$$B(OH)_3$$ is formed only

$$[B(OH)_4]^-$$ formed only

both $$B(OH)_3$$ and $$[B(OH)_4]^-$$ are formed

$$[B_3O_3(OH)_4]^-$$ is formed only

Solution

Question 2

I

II

III

IV

Solution

Question 3

It is a strong tribasic acid.

It is prepared by acidifying an aqueous solution of borax.

It has a layer structure in which planar BO, units are joined by hydrogen bonds.

It does not act as proton donor as it acts as a Lewis acid by accepting hydroxyl ions.

Solution

Question 4

$$SiO_2$$

$$Si(OH)_2F_2$$

$$H_2SiF_6$$

$$Si(OH)_4$$

Solution

Question 5

electrons of K shell only

all the electrons of K and L shell

two electrons of 1s and 2s each

only by electrons of L shell

Solution

Question 6

only +1 oxidation state

only +3 oxidation state

+1 and +3 oxidation states

+1, +2 and +3 oxidation states

Solution

Question 7

B > Al > Ga > In > Tl

B < Al < Ga < In < Tl

B < AL > Ga < In > Tl

B < Al > Ga > In > Tl

Solution

Question 8

non- metallic nature of B

discrepancies in atomic size of element

ability of B and Al to form $$p\pi $$ - $$p\pi $$ multiple bonds

irregular trend in electronegativity throughout the periodic table

Solution

Question 9

Thallium

Gallium

Indium

Aluminium

Solution

Question 10

poor screening effect of d-electrons for the outer electrons from the increased nuclear charge

increased force of attraction of increased nuclear charge on electrons

increased ionisation enthalpy of Ga as compared to Al

Anomalous behaviour of Ga

Solution

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