The p-Block Elements Test

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The p-Block Elements Test - 55

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Weekly Quiz Competition

Question 1
1 / -0

Which compound of xenon is not possible ?

A
$$XeF_{2}$$

B
$$XeF_{4}$$

C
$$XeF_{5}$$

D
$$XeF_{6}$$

Solution

Xe has a complete filled 5p configuration. As a result when it undergoes bonding with an odd number (3 or 5) of F atoms it leaves behind one unpaired electron. This causes the molecule to become unstable. As a result $$XeF_3$$ and $$XeF_5$$ do not exist.
Hence option C is correct answer.
Question 2
1 / -0

Stability of tetrabromides follows the order :

A
$${ SeBr }_{ 4 } < { TeBr }_{ 4 } < { PoBr }_{ 4 }$$

B
$${ SeBr }_{ 4 } < { PoBr }_{ 4 } < { TeBr }_{ 4 }$$

C
$${PoBr }_{ 4 } < { TeBr }_{ 4 } < { SeBr}_{ 4 }$$

D
None of the these

Solution

Stability of higher oxidation state increase as we go down the group - Tetrabromides of sulphur and selenium. i.e,$$SBr_4,SeBr_4$$ are unstable where as $$TeBr_4, PoBr_4$$ are quite stable.

Stability : $$SeBr_4<TeBr_4<PoBr_4$$
Question 3
1 / -0

In Carius method, of estimation of halogens, $$250mg$$ of an organic compound gave $$141mg$$ of $$AgBr$$. The percentage of bromine in the compound is:
[Atomic mass $$Ag=108,Br=80$$ amu]

A
$$24$$

B
$$36$$

C
$$8$$

D
$$60$$

Solution

Hence, the correct option is $$A$$
Question 4
1 / -0

One of the following combinations which illustrates the law of reciprocal proportion?

A
N$$_{2}$$O$$_{3}$$,N$$_{2}$$O$$_{4}$$,N$$_{2}$$O$$_{5}$$

B
NaCl ,NaBr ,Nal

C
CS$$_{2}$$ ,CO$$_{2}$$, SO$$_{2}$$

D
MgO,Mg(OH)$$_{2}$$
Question 5
1 / -0

Which of the following alkyl halides is used as a methylating agent?

A
$${C}_{2}{H}_{5}Br$$

B
$${C}_{6}{H}_{5}Cl$$

C
$${CH}_{3}Cl$$

D
$${C}_{2}{H}_{5}Cl$$

Solution

methylation is a form of alkylation with a methyl group replacing a hydrogen atoms .
So, $$CH_3Cl$$ provides that methyl group wich replaces the hydrogen atom of the substrate.
Question 6
1 / -0

An alkyl halide reacts with alcoholic ammonia in a sealed tube, the product formed will be

A
a primary amine

B
a secondary amine

C
a tertiary amine

D
All of the above

Solution

When alkyl halide reacts with ammonia , it gives primary amine .
Reaction doesn't stop there and primary amine further gives secondary amine .
Secondary gives tertiary .
Hence mixture of three is formed .
Question 7
1 / -0

The correct order of stability is

A
$${ N }_{ 2 }>{ N }_{ 2 }^{ + }>{ N }_{ 2 }^{ - }$$

B
$${ N }_{ 2 }>{ N }^{ - }_{ 2 }>{ N }_{ 2 }^{ + }$$

C
$${ N }_{ 2 }^{ + }>{ N }^{ - }_{ 2 }>{ N }_{ 2 }$$

D
None of these

Solution

Bond order for $$N_2=3$$
Bond order for $$N_2^+=2.5$$
Bond order for $$N_2^-=2$$
Bond order $$\propto$$ Stability of molecule.
Therefore, the correct order of stability is:
$$N_2>N_2^+>N_2^-$$
Question 8
1 / -0

The electron affinity values for the halogens show the following trends

A
$$F< Cl> Br> I$$

B
$$F< Cl< Br< I$$

C
$$F> Cl> Br> I$$

D
$$F< Cl> Br< I$$

Solution

The electron affinity of an atom or molecule is defined as the amount of energy released or spent when an electron is added to a neutral atom or molecule in the gaseous state to form a negative ion.
Electron affinity of an elements depends on certain factor like
Number of protons and size of atom: Halogens are smaller atom compared to other elements in same horizontal lines in the modern Periodic table.

As fluorine sits atop chlorine in the periodic table, most people expect it to have the highest electron affinity, but this is not the case. Fluorine is a small atom with a small amount of space available in its 2p orbital. Because of this, any new electron trying to attach to fluorine experiences lower electron affinity from the electrons already living in the element's 2p orbital. Since chlorine's outermost orbital is a 3p orbital, there is more space, and the electrons in this orbital are inclined to share this space with an extra electron. Therefore, chlorine has a higher electron affinity than fluorine, and this orbital structure causes it to have the highest electron affinity of all of the elements.
Question 9
1 / -0

Less ionic compound among the following

A
NaF

B
KCl

C
KF

D
NaCl

Solution

The decreasing order of ionic character NaF > NaCl > NaBr > NaI.

Because among the halogens the electronegativity is more for Fluorine followed by Chlorine, Bromine, Dodine. According to Fajan's rule, the size should be small to have more ionic character. Thus NaF is more ionic. And NaCl is less ionic .

Hence , option D is correct .
Question 10
1 / -0

Which of the following has $$p\pi -d\pi $$ bonding?

A
$${ NO }_{ 3 }^{ - }$$

B
$${ CO }_{ 3 }^{ - }$$

C
$${ BO }_{ 3 }^{ - }$$

D
$${ SO }_{ 3 }^{ - }$$

Solution

$$p\pi-d\pi$$ bonding includes d-orbitals,if you observe the options only S belongs to $$3^{rd}$$ period and has d-orbitals(though unoccupied in ground state).
Other options $$C$$ and $$N$$ belongs to $$2^{nd}$$ period while B belongs to $$1^{st}$$ period and doesn't have d-orbitals.

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The p-Block Elements Test - 55

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The p-Block Elements Test - 55

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Question 1

$$XeF_{2}$$

$$XeF_{4}$$

$$XeF_{5}$$

$$XeF_{6}$$

Solution

Question 2

$${ SeBr }_{ 4 } < { TeBr }_{ 4 } < { PoBr }_{ 4 }$$

$${ SeBr }_{ 4 } < { PoBr }_{ 4 } < { TeBr }_{ 4 }$$

$${PoBr }_{ 4 } < { TeBr }_{ 4 } < { SeBr}_{ 4 }$$

None of the these

Solution

Question 3

$$24$$

$$36$$

$$8$$

$$60$$

Solution

Question 4

N$$_{2}$$O$$_{3}$$,N$$_{2}$$O$$_{4}$$,N$$_{2}$$O$$_{5}$$

NaCl ,NaBr ,Nal

CS$$_{2}$$ ,CO$$_{2}$$, SO$$_{2}$$

MgO,Mg(OH)$$_{2}$$

Question 5

$${C}_{2}{H}_{5}Br$$

$${C}_{6}{H}_{5}Cl$$

$${CH}_{3}Cl$$

$${C}_{2}{H}_{5}Cl$$

Solution

Question 6

a primary amine

a secondary amine

a tertiary amine

All of the above

Solution

Question 7

$${ N }_{ 2 }>{ N }_{ 2 }^{ + }>{ N }_{ 2 }^{ - }$$

$${ N }_{ 2 }>{ N }^{ - }_{ 2 }>{ N }_{ 2 }^{ + }$$

$${ N }_{ 2 }^{ + }>{ N }^{ - }_{ 2 }>{ N }_{ 2 }$$

None of these

Solution

Question 8

$$F< Cl> Br> I$$

$$F< Cl< Br< I$$

$$F> Cl> Br> I$$

$$F< Cl> Br< I$$

Solution

Question 9

NaF

KCl

KF

NaCl

Solution

Question 10

$${ NO }_{ 3 }^{ - }$$

$${ CO }_{ 3 }^{ - }$$

$${ BO }_{ 3 }^{ - }$$

$${ SO }_{ 3 }^{ - }$$

Solution

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