The p-Block Elements Test

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The p-Block Elements Test - 53

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following statements is not correct about $$XeF_{2}$$?

A
It can be obtained by direct reaction between $$F_{2}$$ and Xe at high pressure.

B
$$XeF_{2}$$ undergoes alkaline hydrolysis to give $$O_{2}$$ and Xe.

C
$$XeF_{2}$$ is a powerful reducing agent.

D
$$XeF_{2}$$ contains two bond pairs and three lone pairs.

Solution

$$XeF_2$$ is obtained by direct reaction of $$Xe$$ and $$F_2$$ at high pressure. It undergoes alkali hydrolysis to produce $$Xe$$ and $$O_2$$. $$XeF_2$$ is a powerful reducing agent where $$Xe^{+2}$$ can change to $$+6$$ state.
Question 2
1 / -0

Which of the following statements is not correct?

A
All the oxides of halogens are powerful oxidants.

B
The compounds of oxygen and fluorine are not called oxides but fluorides.

C
Oxygen fluorides form oxoacids.

D
In oxyhalides, bonds are mainly covalent due to small difference in electronegativity of oxygen and halogens.

Solution

Oxygen florides forms $$HF,O_2$$ on hydrolysis.
$$OF_2 + H_2O\rightarrow O_2 + 2 HF$$

$$2O_2F_2 + 2H2O\rightarrow O_2 + 4HF$$

so oxygen florides do not form forms hypohalous acids.
So the given statement oxygen florides forms oxoacids is wrong.
Hence option C is correct.
Question 3
1 / -0

Which factor is most responsible for the increase in boiling points of noble gases from $$He$$ to $$Xe$$?

A
Decrease in $$I.E.$$

B
Monoatomic nature

C
Decrease in polarisability

D
Increase in polarisability

Solution

Polarisability increases with increase in size. So, from $$He$$ to $$Xe$$ as size increases, polarisability increases.
Question 4
1 / -0

Which of the following substance will have highest boiling point?

A
$$He$$

B
$$CsF$$

C
$${ NH }_{ 3 }$$

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

Solution

Boiling point $$\times $$ Molar mass $$\times $$ Ionic character

which is highest for $$CsF$$
Question 5
1 / -0

Oxygen can be prepared by heating ___________.

A
potassium permanganate

B
methane

C
carbon

D
none of these

Solution

Oxygen can be prepared by heating potassium permanganate. On heating, potassium permanganate splits into potassium manganate and oxygen.
$$2KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2$$
Question 6
1 / -0

Which of the molecule does not exist?

A
$$ClF_3$$

B
$$ICl_3$$

C
$$BrF_3$$

D
$$ClI_3$$

Solution

$$ClI_3$$ does not exist as $$I$$ is bigger than $$Cl$$ so $$Cl$$ cannot have $$3I$$ as having $$3I$$ on $$Cl$$ will become crowded.
Question 7
1 / -0

The product which is not formed on the heating of potassium permanganate is ________.

A
$$K_2MnO_4$$

B
$$MnO_2$$

C
$$O_2$$

D
$$MnO_4$$

Solution

On heating potassium parmanganate we do not get $$MnO_4$$. All the other products are formed on heating potassium permanganate.
$$2KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2$$
Question 8
1 / -0

For the manufacture of ammonia by the reaction$$N_{2}+3H_{2} \rightleftharpoons 2NH_{3}+21.9\ K cal$$, the favorable conditions are:

A
low temperature, low pressure & catalyst

B
low temperature, high pressure & catalyst

C
high temperature, low pressure & catalyst

D
low temperature, high pressure

Solution

The equation representing habers process of $${ NH }_{ 3 }$$
synthesis is-$${ N }_{ 2 }(g)+{ 3H }_{ 2 }(g)\rightleftharpoons { 2NH }_{ 3 }(g)\\ $$
pressure: 150-200 atm
temperature : $$450℃-500℃$$
catalyst : Iron
Question 9
1 / -0

Among $$BF_{3},N(SiH_{3})_{3},N(CH_{3})_{3}, O(SiH_{3})_{2},CF_{3},CCl_{3}^{-}$$ the number of molecules or molecular ions which have back bonding is:

A
$$1$$

B
$$3$$

C
$$4$$

D
$$all$$

Solution

$$BF_{ 3 }$$ has trigonal planner structure in which B has empty P-orbital.Hue,F acts as lewis base and donater its lone pair to boron and it results in back bonding.
Question 10
1 / -0

Which of the following does not react with xenon to form compunds?

A
$$OSeF_{5}$$

B
$$OTeF_{5}$$

C
$$F_{2}$$

D
$$N_{2}$$

Solution

Xenon reacts directly with fluorine because fluorine is a very powerful oxidizing agent (hence it gets reduced and it gains an electron from Xenon). Xenon has larger radii; therefore the electron attraction to the nucleus is weaker in comparison to the smaller noble gases. Fluorine on the other hand is very tiny and highly electronegative, so it would steal an electron from fluorine, forming a compound.
Xenon does not react directly with oxygen. Xenon reacts with water molecules in order to form xenon oxides.

conclusion: hence option (D) is correct.

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Re-Attempt Weekly Quiz Competition

The p-Block Elements Test - 53

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The p-Block Elements Test - 53

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Question 1

It can be obtained by direct reaction between $$F_{2}$$ and Xe at high pressure.

$$XeF_{2}$$ undergoes alkaline hydrolysis to give $$O_{2}$$ and Xe.

$$XeF_{2}$$ is a powerful reducing agent.

$$XeF_{2}$$ contains two bond pairs and three lone pairs.

Solution

Question 2

All the oxides of halogens are powerful oxidants.

The compounds of oxygen and fluorine are not called oxides but fluorides.

Oxygen fluorides form oxoacids.

In oxyhalides, bonds are mainly covalent due to small difference in electronegativity of oxygen and halogens.

Solution

Question 3

Decrease in $$I.E.$$

Monoatomic nature

Decrease in polarisability

Increase in polarisability

Solution

Question 4

$$He$$

$$CsF$$

$${ NH }_{ 3 }$$

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

Solution

Question 5

potassium permanganate

methane

carbon

none of these

Solution

Question 6

$$ClF_3$$

$$ICl_3$$

$$BrF_3$$

$$ClI_3$$

Solution

Question 7

$$K_2MnO_4$$

$$MnO_2$$

$$O_2$$

$$MnO_4$$

Solution

Question 8

low temperature, low pressure & catalyst

low temperature, high pressure & catalyst

high temperature, low pressure & catalyst

low temperature, high pressure

Solution

Question 9

$$1$$

$$3$$

$$4$$

$$all$$

Solution

Question 10

$$OSeF_{5}$$

$$OTeF_{5}$$

$$F_{2}$$

$$N_{2}$$

Solution

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