P-block Group 18 - Noble Gases(11th And 12th > Chemistry ) Questions and answers for exam Preparation

11th And 12th > Chemistry

P-BLOCK GROUP 18 - NOBLE GASES MCQs

Total Questions : 15 | Page 1 of 2 pages

Question 1.

Helium is the second most abundant element in the Universe. Yet, it is not the most abundant on Earth - Why?

Helium is highly reactive and constantly undergoes chemical transformation.
Helium atoms are small and cannot be retained by the Earth's gravitational field.
Both a and b are correct.
Helium is peculiarly attracted only to black holes.

Discuss Question

Answer: Option B. -> Helium atoms are small and cannot be retained by the Earth's gravitational field.
:
B

Helium accounts for up to 23 percent by mass of the Universe and the Sun. It is the most abundant element in the universe after hydrogen; Helium is rare in the atmosphere because its atoms are small and travel fast enough to escape the pull of Earth’s gravitational field. Do note that all the other noble gases occur in the atmosphere. There is no primordial Helium. In other words, there was no Helium from before the formation of Earth.

That leaves us with the question – where does the Helium on Earth come from? Helium comes into existence as $α$ particles from radioactive decay and tends to accumulate in natural gas deposits.

Question 2.

Xenon Fluorides are best stored in quartz containers – True or False?

True
False
Both a and b are correct.
Helium is peculiarly attracted only to black holes.

Discuss Question

Answer: Option B. -> False
:
B
Xenon Fluorides are stored in Nickel or Monel containers. They cannot be stored in glass containers or quartz containers as they react with

S i O_{2}

as:

2 X e F_{6} (s) + S i O_{2} (s) \to S i F_{4} (s) + 2 X e O F_{4} (s)

2 X e O F_{4} (s) + S i O_{2} (s) \to S i F_{4} (s) + 2 X e O_{2} F_{2} (s)

2 X e O_{2} F_{2} (s) + S i O_{2} (s) \to S i F_{4} (s) + 2 X e O_{3} (s)

Also, the first two reactions may be used to prepare Xenon oxyfluorides. Further, Xenon oxyfluorides can also be prepared by the controlled, partial hydrolysis of Xenon fluorides.
Hydrolysis of

X e F_{6}

is a multi-step process. Initially Xenon oxide tetrafluoride –

X e O F_{4}

is formed.

X e F_{6} (s) + H_{2} O (l) \to X e O F_{4} (l) + 2 H F (l)

Followed by

X e O F_{4} (l) + 2 H_{2} O (l) \to X e O_{3} (s) + 4 H F (l)

Question 3.

In which of the following compounds, the Fluorine atoms are placed axially in a trigonal bipyramidal geometry?

$X e O F_{2}$
$X e O F_{4}$
$X e F_{2}$
$X e O_{2} F_{2}$

Discuss Question

Answer: Option B. ->

X e O F_{4}

:
A, C, and D
We had earlier seen that

X e F_{2}

has a linear shape. The 3 lone pairs of electrons of central Xe atom lie on the plane perpendicular to the axial Xe – F bonds. The molecule has a trigonal bipyramidal geometry

In Which Of The Following Compounds, The Fluorine Atoms Are ...

In this molecule, if one oxygen atom takes the place of a lone pair, we get the geometry of

X e O F_{2}

. Do note that Xenon has a double bond with the oxygen atom. Let us look at the ground state and excited state:

Let us look at the geometry:

Now try to deduce the shape and geometry of

X e O_{2} F_{2}

:
A logical leap is to assume that the structure would be similar to replacing two of the three lone pairs with two oxygen atoms in the

X e F_{2}

structure. It is a good guess!

As for the geometry,

X e F_{2}

X e O F_{2}

and

X e O_{2} F_{2}

all have the same trigonal bipyramidal geometry.

X e F_{2}

has a linear shape.

X e O F_{2}

is T - Shaped.

X e O_{2} F_{2}

is see-saw shaped.
Earlier we had seen that the geometry of

X e F_{4}

has a square bipyramidal geometry and square planar shape:

In the above geometry, if we replace a lone pair with an Oxygen atom such that there is a double bond between Xe and O, then we get the geometry of

X e O F_{4}

. The four F atoms line in a plane perpendicular to the Xe – O axial sigma bond. This geometry is still the same as

X e F_{4}

but the shape will be square pyramidal!

So correct options are a c and d

Question 4.

$X e O_{3}$ is the product obtained by complete hydrolysis of which of the following compounds?

$X e F_{6}$
$X e F_{4}$
$X e F_{2}$
All of the above

Discuss Question

Answer: Option A. ->

X e F_{6}

:
A and B
Let us recall the hydrolysis reactions can we?

2 X e F_{2} (s) + 2 H_{2} O (l) \to 2 X e (g) + O_{2} (g) + 4 H F (l)

The above reaction easily disqualifies option c and d.

6 X e F_{4} (s) + 12 H_{2} O (l) \to 4 X e (g) + 3 O_{2} (g) + 2 X e O_{3} (s) + 24 H F (l)

X e F_{6} (S) + 3 H_{2} O (l) \to X e O_{3} (s) + 6 H F (l)

Hydrolysis of

X e F_{6}

is a multi-step process. Initially Xenon oxide tetrafluoride –

X e O F_{4}

is formed.

X e F_{6} (s) + H_{2} O (l) \to X e O F_{4} (l) + 2 H F (l)

Followed by

X e O F_{4} (l) + 2 H_{2} O (l) \to X e O_{3} (s) + 4 H F (l)

Question 5.

Which of the following can be termed as oxyfluorides of Xenon?

$X e O F_{2}$
$X e O F_{4}$
$O F_{2}$
$X e O_{2} F_{2}$

Discuss Question

Answer: Option A. ->

X e O F_{2}

:
A, B, and D

X e O F_{2}, X e O F_{4}

and

X e O_{2} F_{2}

are all well-known oxyfluorides of Xenon.

Question 6.

The shape of $X e O_{3}$ molecule is __

Discuss Question

Answer: Option A. ->

X e O F_{2}

:
Let us draw the Lewis structure for Xenon (VI) oxide or Xenon trioxide:

As we can see from the Lewis structure, there is one lone pair on the central atom and there are three atoms attached to the Xenon atom. Using VSEPR theory for a steric number = 3 bp + 1 lp = 4, we get a tetrahedral geometry and a pyramidal shape.

Question 7.

What happens when Xe (g) is heated with dioxygen gas in a sealed tube?

Only $X e O_{3}$ is formed
Initially $X e O_{3}$ is formed which slowly gets oxidized to $X e O_{4}$
A mixture of both $X e O_{3}$ and $X e O_{4}$ is obtained
No reaction

Discuss Question

Answer: Option D. -> No reaction
:
D
For the reaction,

X e (g) + O_{2} (g) \to xenon oxides Δ_{f} G^{\circ}

0

In other words, the standard change in free energy of the reaction is positive. Hence, this reaction is non-spontaneous. None of the Xenon oxides can be synthesized from this approach. So how can they be obtained?

Question 8.

Xenon forms different binary compounds with fluorine.

Discuss Question

Answer: Option D. -> No reaction
:
Xenon forms three fluorides:

X e (g) + F_{2} (g) 400^{\circ} C 1 a t m - ------ \to X e F_{2} (s)

{Xe in excess}

X e (g) + 2 F_{2} (g) 600^{\circ} C 6 a t m - ------ \to X e F_{4} (s) {X e : F_{2} = 1 : 1.5}

X e (g) + 3 F_{2} (g) 300^{\circ} C 50 a t m - ------- \to X e F_{6} (s) {X e : F_{2} = 1 : 20}

Question 9.

The Xenon Fluorides are very effective and clean Fluorinating agents. They also behave as good reducing agents – True or False?

True
False
A mixture of both $X e O_{3}$ and $X e O_{4}$ is obtained
No reaction

Discuss Question

Answer: Option B. -> False
:
B
Xenon Fluorides are clean fluorinating agents because usually they yield the fluorinated product and Xe gas. The required product can then easily be separated from Xe gas. In the following reaction,