p-Block Elements Test

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p-Block Elements Test - 4

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following compounds is/are obtained by heating boric acid?

a. HBO₂b. H₂B₄O₇c. B₂O₃

A
a only

B
a and b

C
a and c

D
All of the above
Question 2
1 / -0

The maximum oxidation state shown by any p-block element is

A
the sum of the s electrons

B
the sum of the p-electrons

C
half of the sum of the s and p-electrons

D
the sum of the s and p-electrons

Solution

Maximum oxidation state of any p block element is the number of ionisable electrons which is equal to the sum of s and p electrons present in valence shell. For example valance shell electronic configuration of group 15 element is ns²np³ so max oxidation state is +5
Question 3
1 / -0

As we move from B to Al in the p-block elements the sum of the first three ionisation enthalpies

A
considerably decreases

B
remains the same

C
slightly decreases

D
increases

Solution

On moving from B to Al size of the atom increases. So ionisation enthalpy decreases. Thus as we move from B to Al in the p-block elements the sum of the first three ionisation enthalpies considerably decreases.
This is the reason aluminium forms Al3+ ion.
Question 4
1 / -0

Directions: In the given question, two statements are provided which are labeled as Assertion (A) and Reason (R). Read both of them carefully and select the alternative that most suitably follows the statements.

Assertion (A): Diamond is a bad conductor of electricity.
Reason (R): All C-C bond lengths in diamond are of 154 pm.

A
Both (A) and (R) are true and (R) is the correct reason for (A).

B
Both (A) and (R) are true and (R) is not the correct reason for (A).

C
Both (A) and (R) are false.

D
(A) is false and (R) is true
Question 5
1 / -0

There are six groups of p-block elements in the periodic table numbering from

A
14 to 19

B
11 to 16

C
13 to 18

D
12 to 17

Solution

After two s block and 10 d block, p block elements starts from group 13 to group 18.
Question 6
1 / -0

One of the following p-block elements has unusually low melting point

A
Indium

B
aluminium

C
gallium

D
boron

Solution

Ga does not do not follow any of the simple crystal structures. And two nearest Ga atoms are bonded by covalent bond which leads to low melting point.
Question 7
1 / -0

The stability of the dihalides of Si, Ge, Sn and Pb increases steadily in the sequence

A
SiX₂ << GeX₂ << PbX₂ << SnX₂

B
SiX₂<< GeX₂ << SnX₂ << PbX₂

C
PbX₂ << SnX₂ << GeX₂ << SiX₂

D
GeX₂ << SiX₂ << SnX₂ << PbX₂
Question 8
1 / -0

The valence shell electronic configuration of p-block elements is ns2np1-6. But this is not true in the case of

A
nitrogen

B
Helium

C
carbon

D
Boron

Solution

He has 1s² configuration. It is a noble gas.
Question 9
1 / -0

Aluminium should not be washed with materials containing washing soda as

A
washing soda is expensive

B
washing soda is easily decomposed

C
washing soda reacts with aluminium to form insoluble oxide

D
washing soda forms soluble metal aluminate
Question 10
1 / -0

Graphite is a smooth lubricant that is extremely difficult to melt. The reason for this anomalous behaviour is that graphite

A
has carbon atoms arranged in large plates of rings of strongly bonded carbon atoms with weak interpolate bonds

B
is a non-crystalline substance

C
is an allotrope of carbon

D
has molecules of variable molecular masses like polymers
Question 11
1 / -0

Which of these is not a monomer for a high molecular mass silicon polymer?

A
Me₃SiCl

B
PhSiCl₃

C
MeSiCl₃

D
Me₂SiCl₂
Question 12
1 / -0

Which of the following is the correct increasing order of oxidizing power?

A
CO₂ < SiO₂ < SnO₂< PbO₂

B
SiO₂< CO₂ < PbO₂ < SnO₂

C
SnO₂< PbO₂ < SiO₂< CO₂

D
PbO₂< SnO₂ < CO₂ < SiO₂
Question 13
1 / -0

Which of the following is not isostructural with SiCl₄?

A
SO₄^2-

B
NH₄⁺

C
PO₄^3-

D
SCl₄
Question 14
1 / -0

The tendency of BF₃, BCl₃ and BBr₃ to behave as Lewis acids decreases in the sequence

A
BCl₃ > BF₃ > BBr₃

B
BBr₃ > BCl₃ > BF₃

C
BBr₃ > BF₃ > BCl₃

D
BF₃ > BCl₃ > BBr₃
Question 15
1 / -0

Which of the following statements are correct?

a. Fullerenes have dangling bonds.
b. Fullerenes are cage-like molecules.
c. Graphite is thermodynamically the most stable allotrope of carbon.
d. The successive layers in graphite are held together by weak forces of attraction.

A
a and b

B
b and c

C
c and d

D
b, c and d
Question 16
1 / -0

PbCl₄ exists while PbBr₄ and PbI₄ do not because

A
bromine and iodine are unable to oxidize Pb²⁺ to Pb⁴⁺

B
Br^- and I^- ions are bigger in size

C
bromine and iodine are more electronegative in nature

D
chlorine is a gas
Question 17
1 / -0

A colourless aqueous solution, on adding water and on heating, gave a white precipitate. This precipitate, when reacted with NH₄Cl and NH₄OH in excess, resulted in dissolution of some of the precipitate and a gelatinous precipitate is obtained. What is the hydroxide formed in aqueous solution?

A
Zn(OH)₂

B
Mg(OH)₂

C
Ca(OH)₂

D
Al(OH)₃
Question 18
1 / -0

If CO₂ is passed through water, which of the following species will be present in water?

A
H₂CO₃, CO₂, CO₃^2-, HCO₃^-

B
HCO₃^-, CO₃^2-

C
CO₃^2-, CO₂

D
H₂CO₃, CO₂
Question 19
1 / -0

Which of the following anions is present in the chain structure of silicates?

A
(Si₂O₅)_n^2-

B
(SiO₃)_n^2-

C
SiO₄^4-

D
Si₂O₇^6-
Question 20
1 / -0

Borax is prepared by treating colemanite with

A
NaNO₃

B
NaCl

C
Na₂CO₃

D
NaHCO₃
Question 21
1 / -0

The variation in properties in a group of p-block of elements is due to

A
difference in inner core of the electronic configuration

B
difference in oxidation states

C
difference in electronegativities

D
difference in ionisation enthalpies

Solution

The difference in properties of p block element is because of presence of inner lying d and f electrons which have poor shielding effect.
Question 22
1 / -0

Among the following which is the strongest oxidizing agent?

A
Br₂

B
Cl₂

C
F₂

D
I₂

Solution

Because of its highest electronegativity it (F2) is very strongest oxidizing agent.
Question 23
1 / -0

The stability of + 1 oxidation state increases in the sequence

A
Tl  <  In  <  Ga  < Al

B
In  <  Tl  <  Ga  <  Al

C
Ga <  In <  Al < Tl

D
Al  <  Ga  <  In  <  Tl
Question 24
1 / -0

The sum of the s and p-electrons is called

A
the most stable valence number of p-block elements

B
inert pair of p-block elements

C
the group oxidation state of p-block elements

D
the most unstable valence number of p-block elements

Solution

Group oxidation state is the highest oxidation state exhibited by the elements present in any particular group.
For p block elements it is the sum of the s and p-electrons.
Question 25
1 / -0

Which of the following does not exist in free form?

A
BF₃

B
BCl₃

C
BBr₃

D
BH₃
Question 26
1 / -0

Which of the following statements is not correct?

A
PbCl₄ is less stable than SnCl₄ but PbCl₂ is more stable than SnCl₂.

B
PbO₂ is a stronger oxidant than SnO₂.

C
(SiH)₃N is a stronger base than (CH₃)₃N.

D
Sn(II) is a reducing agent but Pb(II) is not.
Question 27
1 / -0

Inert pair effect of p-block elements is due to the

A
the group oxidation state of p-block elements

B
the most stable valence number of p-block elements

C
the most unstable valence number of p-block elements

D
Inertness of the inner ns electrons because of ineffective shielding

Solution

The inert pair effect is nothing but the extra stability concept. Generally P-block elements shows variable valiancy which is equal to the maximum oxidation (Group oxidation state) state of that group of p-block elements, p block elements may show other oxidation state i.e.+2.

Group oxidation state +4 is obtained when electrons are lost from both ns and np levels,

+2 oxidation state is obtained while only np level electrons are lost due to extra stability and poor shielding of ns electron which are called inert pair. The effect is the inert effect
Question 28
1 / -0

Which of the following statements is correct?

A
BCl₃ and AlCl₃ are both Lewis acids and BCl₃ is stronger than AlCl₃.

B
BCl₃ and AlCl₃ are both Lewis acids and AlCl₃ is stronger than BCl₃.

C
BCl₃ and AlCl₃ are equally strong Lewis acids.

D
Neither BCl₃ nor AlCl₃ is a Lewis acid.
Question 29
1 / -0

The correct order of C-O bond length among CO, CO₃^2- and CO₂ is

A
CO < CO₂ < CO₃^2-

B
CO₂ < CO₃^2- < CO

C
CO < CO₃^2- < CO₂

D
CO₃^2- < CO₂< CO
Question 30
1 / -0

Boron forms only covalent compounds and not +3 ions because

A
the sum of its first three ionization enthalpies is very high.

B
the sum of its first two ionization enthalpies is very low.

C
the sum of its first two ionization enthalpies is very high.

D
the sum of its first three ionization enthalpies is very low.

Solution

Due to smaller size and high nuclear charge B has very high ionization energy value so does not show +3 oxidation state.

Further according to Fajan's rule, cation with small size and high positive charge leads to covalent bond.

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p-Block Elements Test - 4

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Question 1

a only

a and b

a and c

All of the above

Question 2

the sum of the s electrons

the sum of the p-electrons

half of the sum of the s and p-electrons

the sum of the s and p-electrons

Solution

Question 3

considerably decreases

remains the same

slightly decreases

increases

Solution

Question 4

Both (A) and (R) are true and (R) is the correct reason for (A).

Both (A) and (R) are true and (R) is not the correct reason for (A).

Both (A) and (R) are false.

(A) is false and (R) is true

Question 5

14 to 19

11 to 16

13 to 18

12 to 17

Solution

Question 6

Indium

aluminium

gallium

boron

Solution

Question 7

SiX2 << GeX2 << PbX2 << SnX2

SiX2 << GeX2 << SnX2 << PbX2

PbX2 << SnX2 << GeX2 << SiX2

GeX2 << SiX2 << SnX2 << PbX2

Question 8

nitrogen

Helium

carbon

Boron

Solution

Question 9

washing soda is expensive

washing soda is easily decomposed

washing soda reacts with aluminium to form insoluble oxide

washing soda forms soluble metal aluminate

Question 10

has carbon atoms arranged in large plates of rings of strongly bonded carbon atoms with weak interpolate bonds

is a non-crystalline substance

is an allotrope of carbon

has molecules of variable molecular masses like polymers

Question 11

Me3SiCl

PhSiCl3

MeSiCl3

Me2SiCl2

Question 12

CO2 < SiO2 < SnO2 < PbO2

SiO2 < CO2 < PbO2 < SnO2

SnO2 < PbO2 < SiO2 < CO2

PbO2 < SnO2 < CO2 < SiO2

Question 13

SO42-

NH4+

PO43-

SCl4

Question 14

BCl3 > BF3 > BBr3

SiX₂ << GeX₂ << PbX₂ << SnX₂

SiX₂<< GeX₂ << SnX₂ << PbX₂

PbX₂ << SnX₂ << GeX₂ << SiX₂

GeX₂ << SiX₂ << SnX₂ << PbX₂

Me₃SiCl

PhSiCl₃

MeSiCl₃

Me₂SiCl₂

CO₂ < SiO₂ < SnO₂< PbO₂

SiO₂< CO₂ < PbO₂ < SnO₂

SnO₂< PbO₂ < SiO₂< CO₂

PbO₂< SnO₂ < CO₂ < SiO₂

SO₄^2-

NH₄⁺

PO₄^3-

SCl₄

BCl₃ > BF₃ > BBr₃

BBr₃ > BCl₃ > BF₃

BBr₃ > BF₃ > BCl₃

BF₃ > BCl₃ > BBr₃

bromine and iodine are unable to oxidize Pb²⁺ to Pb⁴⁺

Br^- and I^- ions are bigger in size

Zn(OH)₂

Mg(OH)₂

Ca(OH)₂

Al(OH)₃

H₂CO₃, CO₂, CO₃^2-, HCO₃^-

HCO₃^-, CO₃^2-

CO₃^2-, CO₂

H₂CO₃, CO₂

(Si₂O₅)_n^2-

(SiO₃)_n^2-

SiO₄^4-

Si₂O₇^6-

NaNO₃

Na₂CO₃

NaHCO₃

Br₂

Cl₂

F₂

I₂

BF₃

BCl₃

BBr₃

BH₃

PbCl₄ is less stable than SnCl₄ but PbCl₂ is more stable than SnCl₂.

PbO₂ is a stronger oxidant than SnO₂.

(SiH)₃N is a stronger base than (CH₃)₃N.