p-Block Elements Test

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p-Block Elements Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

The correct order of increasing electron affinity of halogen of

A
I < Br < Cl

B
Br < I < Cl

C
Cl < Br < I

D
I < Cl < Br

Solution

Electron affinity decreases down the group as atomic size increases.
Question 2
1 / -0

Which is the correct arrangement of the compounds based on their bond strength?

A
HF > HCl > HBr > HI

B
HI > HBr > HCl > HF

C
HCl > HF > HBr > HI

D
HF > HBr > HCl > HI

Solution

Because electronegativity decreases down the group thereby e⁻ attracting tendency decreases down the group. F with high electronegativity will have strongest bond and I with least electronegativity will have least bond strength.
Question 3
1 / -0

The comparatively high boiling point of hydrogen fluoride is due to:

A
High reactivity of fluorine

B
Small size of hydrogen atom

C
Formation of hydrogen bonds

D
Small size of fluorine

Solution

F being highly electronegative has the ability to form hydrogen bonds due to which boiling point of HF is high.
Question 4
1 / -0

The high reactivity of fluorine is due to its _____ dissociation energy. Its shows only ______ oxidation state. It has_______electron affinity than chlorine. Among all hydrogen halides boiling point is highest for_______.

A
low, -1, lower, HF

B
high,+1, higher, HF

C
low,+1, lower, HCl

D
high,-1, higher, HF

Solution

F is having low dissociation energy as F being highly electronegative and small thereby repel each other, hence have low dissociation energy.
Shows −1 oxidation state as it can accept one e⁻ in order to complete its octet.
It has lower e⁻ affinity than chlorine due to its high electronegativity and small size it repels the new incoming e⁻.
HF has highest boiling point due to the strong bond formation.
Question 5
1 / -0

The property of halogens which is not correctly matched is

A
F > Cl > Br > I (Ionisation energy)

B
F > Cl > Br > I (Electronegativity)

C
I > Br > Cl > F (Density)

D
F > Cl > Br > I (Electron affinity)

Solution

Electron affinity of Cl is maximum. The correct trend is Cl > F > Br > I.
Question 6
1 / -0

Mark the correct statements about halogens.

A
Electron affinity of halogens is in the order F > Cl > Br > I

B
HF is the strongest hydrohalic acid

C
F₂ has lower bond dissociation energy than Cl₂

D
All halogens show variable oxidation states

Solution

F₂ has lower bond dissociation energy than Cl₂ due to its small size which results in interelectronic repulsion.
Question 7
1 / -0

What is the correct observation when Br₂ is treated with NaF, NaCl and Nal has taken in three test-tubes labeled as (X),(Y) and (Z)?

A
F₂ is liberated in (X) and Cl₂ in (Y)

B
Only I₂ is liberated in (Z)

C
Only Cl₂ is liberated in (Y)

D
Only F₂ is liberated in (X)

Solution

More reactive halogen can displace less reactive halogen from its salt solution. F₂ can displace Cl₂, Br₂, I₂ from their salt solutions while Cl₂ can displace Br₂ and I₂ from NaBr and Nal. Br₂ can displace only I₂ from Nal.
2NaI + Br₂ → 2NaBr + I₂
Question 8
1 / -0

The halogen that is most easily reduced is

A
F₂

B
Cl₂

C
Br₂

D
I₂

Solution

F₂is easily reduced as F⁻¹ is the strongest oxidising agent. Oxidising power can be explained by the fact that the ease with which a halogen [X] forms the halide ion H⁻ decreases down the group.
Question 9
1 / -0

Fluorine is the best oxidising agent because it has:

A
Highest electron affinity

B
Highest reduction potential

C
Highest oxidation potential

D
Lowest electron affinity

Solution

Fluorine is an element that has high electronegativity. Fluorine has low bond enthalpy and highest reduction potential so more ability to loose the electrons. Fluorine can also be easily reduced. Fluorine is thus one of the best oxidizing agents.
Question 10
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, O₂ on hydrolysis.
OF₂ + H₂O → O₂ + 2HF
2O₂F₂ + 2H₂O → O₂+ 4HF
so oxygen florides do not form forms hypohalous acids.
So the given statement oxygen florides forms oxoacids is wrong.

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p-Block Elements Test - 43

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p-Block Elements Test - 43

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Question 1

I < Br < Cl

Br < I < Cl

Cl < Br < I

I < Cl < Br

Solution

Question 2

HF > HCl > HBr > HI

HI > HBr > HCl > HF

HCl > HF > HBr > HI

HF > HBr > HCl > HI

Solution

Question 3

High reactivity of fluorine

Small size of hydrogen atom

Formation of hydrogen bonds

Small size of fluorine

Solution

Question 4

low, -1, lower, HF

high,+1, higher, HF

low,+1, lower, HCl

high,-1, higher, HF

Solution

Question 5

F > Cl > Br > I (Ionisation energy)

F > Cl > Br > I (Electronegativity)

I > Br > Cl > F (Density)

F > Cl > Br > I (Electron affinity)

Solution

Question 6

Electron affinity of halogens is in the order F > Cl > Br > I

HF is the strongest hydrohalic acid

F2​ has lower bond dissociation energy than Cl2​

All halogens show variable oxidation states

Solution

Question 7

F2​ is liberated in (X) and Cl2​ in (Y)

Only I2​ is liberated in (Z)

Only Cl2​ is liberated in (Y)

Only F2​ is liberated in (X)

Solution

Question 8

F2

Cl2

Br2

I2

Solution

Question 9

Highest electron affinity

Highest reduction potential

Highest oxidation potential

Lowest electron affinity

Solution

Question 10

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

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F₂ has lower bond dissociation energy than Cl₂

F₂ is liberated in (X) and Cl₂ in (Y)

Only I₂ is liberated in (Z)

Only Cl₂ is liberated in (Y)

Only F₂ is liberated in (X)

F₂

Cl₂

Br₂

I₂