The p-Block Elements Test

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The p-Block Elements Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

Fluorine is highly reactive because:

A
F-F bond energy is high

B
F-F bond energy is low

C
it is gaseous at room temperature

D
F has a smaller size

Solution

$$F_{2}$$ is highly reactive because of its low bond energy. Also, it has the highest electronegativity.
Question 2
1 / -0

Assertion (A): Balloons made by nylon films are better for containing helium than the conventional rubber balloons.
Reason (R): R.M.S. velocity of helium is very high. So helium atoms can effuse out through rubber balloons.

A
Both $$A$$ and $$R$$ are correct and $$R$$ is the correct explanation of $$A$$.

B
Both $$A$$ and $$R$$ are correct and $$R$$ is not the correct explanation of $$A$$.

C
$$A$$ is true but $$R$$ is false.

D
$$A$$ is false but $$R$$ is true.

Solution

Root mean square velocity of helium gas is high due to its low mass.
$$RMS velocity \times \frac{1}{mass}$$
Therefore, nylon films are better for containing helium than the conventional rubber balloons.
Question 3
1 / -0

The atomic number of astatine which belongs to the sixth period is:

A
53

B
54

C
85

D
86

Solution

He (2)
F Ne (10)
Cl Ar (18)
Br kr (36)
I xe (54)
As Rn (86)
Since the position of astatine is just before the Rn.
so atomic no of astatine is 85
Question 4
1 / -0

What is the atomic number (Z) of the noble gas that reacts with fluorine?

A
54

B
10

C
18

D
2

Solution

He(2)
Ne(10)
Ar(18)
Kr(36)
Xe(54)

Xe reacts with $$F_{2}$$.

Its atomic no. is 54

Option A is correct.
Question 5
1 / -0

The inert gas which cannot be adsorbed over activated coconut charcoal?

A
Ne

B
Ar

C
He

D
Kr

Solution

Due to small size of helium, it can not be absorbed over activated coconut charcoal.

Option C is correct.
Question 6
1 / -0

Which of the following exhibits the weakest intermolecular forces?

A
$$\mathrm{H}_{2}\mathrm{O}$$

B
$$ \mathrm{N}\mathrm{H}_{3}$$

C
He

D
HCl

Solution

He has maximum ionization energy. So it has least tendency

to combine and has weakest intermolecular forces.
Question 7
1 / -0

The correct order of enthalpy of vaporisation of noble gases is :

A
$$ He < Ne < Ar < Kr < Xe $$

B
$$ He > Ne > Ar > Kr > Xe$$

C
$$ He < Ar < Kr < Ne < Xe $$

D
$$ He < Xe < Ar < Ne < Kr $$

Solution

Enthalpy of vaporization $$\alpha $$ Van der Waals forces $$\alpha$$ Molar mass
Thus, as molar mass increases, the enthalpy of vaporization increases.
Thus correct order of enthalpy of vapourisation of noble gases
$$Xe>Kr>Ar>Ne>He$$
Question 8
1 / -0

The atomic weight of noble gases is obtained by using the relationship :

A
Atomic weight = equivalent weight X valency

B
Atomic weight = equivalent weight/valency

C
At. weight = valency/equivalent weight

D
2 x V.D. = molecular weight = atomic weight

Solution

For Noble gases
$$\underset{weight}{Molecular} = \underset{weight}{Atomic}$$
$$\underset{Density}{Vapour} = \frac{Atomic weight}{2} $$
$$\therefore 2\times U.D = molecular weight$$
$$ = atomic weight $$
Question 9
1 / -0

Clathrates are_________

A
Inert gases

B
Inert gases $$+$$ organic or inorganic compounds

C
Halogen $$+$$ carbon black

D
Halogen $$+$$ transition elements

Solution

Clathrates compound are non-stoichiometric in nature.

They have a host molecule which could be a organic or inorganic compound while the guest molecule is inert gases.

Option B is correct.
Question 10
1 / -0

The most abundant element in the earth's crust among the following is_________.

A
$$\mathrm{P}$$

B
$$\mathrm{A}\mathrm{s}$$

C
$$\mathrm{S}\mathrm{b}$$

D
$$\mathrm{B}\mathrm{i}$$

Solution

The most abundant element in the earth's crust is phosphorus. It occurs in the minerals of the apatite family. An example is a fluorapatite $$ \displaystyle Ca_9[PO_4]_6 \cdot CaF_2$$
Phosphorus is the most abundant element of 15th group, accounting for 0.10% of the mass of the earth's crust.
Option A is correct.

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The p-Block Elements Test - 27

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The p-Block Elements Test - 27

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Question 1

F-F bond energy is high

F-F bond energy is low

it is gaseous at room temperature

F has a smaller size

Solution

Question 2

Both $$A$$ and $$R$$ are correct and $$R$$ is the correct explanation of $$A$$.

Both $$A$$ and $$R$$ are correct and $$R$$ is not the correct explanation of $$A$$.

$$A$$ is true but $$R$$ is false.

$$A$$ is false but $$R$$ is true.

Solution

Question 3

53

54

85

86

Solution

Question 4

54

10

18

2

Solution

Question 5

Ne

Ar

He

Kr

Solution

Question 6

$$\mathrm{H}_{2}\mathrm{O}$$

$$ \mathrm{N}\mathrm{H}_{3}$$

He

HCl

Solution

Question 7

$$ He < Ne < Ar < Kr < Xe $$

$$ He > Ne > Ar > Kr > Xe$$

$$ He < Ar < Kr < Ne < Xe $$

$$ He < Xe < Ar < Ne < Kr $$

Solution

Question 8

Atomic weight = equivalent weight X valency

Atomic weight = equivalent weight/valency

At. weight = valency/equivalent weight

2 x V.D. = molecular weight = atomic weight

Solution

Question 9

Inert gases

Inert gases $$+$$ organic or inorganic compounds

Halogen $$+$$ carbon black

Halogen $$+$$ transition elements

Solution

Question 10

$$\mathrm{P}$$

$$\mathrm{A}\mathrm{s}$$

$$\mathrm{S}\mathrm{b}$$

$$\mathrm{B}\mathrm{i}$$

Solution

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