The p-Block Elements Test

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The p-Block Elements Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

Superhalogen is:

A
$$F_2$$

B
$$Cl_2$$

C
$$Br_2$$

D
$$I_2$$

Solution

Superhalogen is a halogen which has high electronegativity and high reactivity.
Since $$F_2$$ is highly reactive, it is a superhalogen. Also $$F_2$$ has the highest electronegativity.
Question 2
1 / -0

Which of the following bond has highest bond energy?

A
F-F

B
Cl-Cl

C
Br-Br

D
I-I

Solution

Correct Option: $$B$$

Explanation:

• $$Flourine$$ , $$Chlorine$$ , $$Bromine$$ and $$Iodine$$ are the group $$17$$ elements.

• Bond energy is the amount of energy needed to break the atoms involved in a molecular bond into free atoms.

• $$F-F$$ bond is the weak bond because the repulsion between the lone electron pairs of the two fluorine atoms is huge, Also the ionization energy of the $$Flourine$$ is very low and hence it can react easily with other elements and break the $$F-F$$ bond. The bond length of two $$F$$ atoms is more compared to the other group $$17$$ elements.

• $$Cl-Cl$$ bond is the strongest bond because of the smallest bond length compared to all the other halogen elements, Due to stronger bond and lesser repulsion, the bond length gets small.

• As we know that, The more the electrons are attracted, the stronger is the bond formed.

• The bond dissociation energy of $$Flourine$$ is however , lower than $$Chlorine$$ and $$Bromine$$ due to its small size its inter electronic repulsion is very high, so the bond dissociation enthalpy order as follows $$Cl_2$$ > $$Br_2$$ > $$F_2$$ > $$I_2$$ .

• $$I_2$$ has the smallest bond dissociation energy due to the largest size.
Thus, the correct option is B
Question 3
1 / -0

Fluorine differs from the other halogens due to :

A
Small size

B
Very high electronegativity

C
Non-availability of d-orbitals

D
All of these

Solution

Fluorine differs from other halogens in many respects.
1. Smallest size of (F) atom
2. Highest ionization energy and electronegativity of fluorine than other halogens.
3. Fluorine cannot expand their octet beyond 8 because of lack of d-orbital, but other halogens can expand their octet.

Hence, the correct option is D.
Question 4
1 / -0

Atoms in $$P_4$$ molecule of white phosphorus are arranged regularly in the following way :

A
at the corners of a cube.

B
at the corners of an octahedron.

C
at the corners of a tetrahedron.

D
at the center and corners of a tetrahedron.

Solution

Atoms in $$P_4$$ molecule of white phosphorus are arranged regularly at the corners of a tetrahedron such that each atom is single-bonded to other three.The $$P-P-P$$ bond angle is $$60^{0}$$.

Option (C) is correct.
Question 5
1 / -0

If atomic number of inert gas is $$Z$$, then ionic bond is formed between:

A
$$(Z - 1)$$ and $$(Z + 1)$$

B
$$(Z - 2)$$ and $$(Z + 2)$$

C
both (a) and (b)

D
none of these

Solution

If Z is the atomic number of the inert gas, then
$$\left.\begin{matrix}(z\,-\,1) &\Rightarrow &halogens (z\,+\,1)\,=\,alkali\,\, metals \\ (z\,-\,2)&\Rightarrow & oxygen family; (z\,+\,2)\,=\,alkaline\,\,earth\,\,metals\end{matrix}\right\}From \,\,ionic\,\, compound$$
Question 6
1 / -0

The following acids have been arranged in the order of decreasing acid strength. Identify the correct order: ClOH (I), BrOH (II), IOH (III).

A
$$I> II > III$$

B
$$II>I>III$$

C
$$III>II>I$$

D
$$I>III>II$$

Solution

For the same oxidation state, when the electronegativity of the halogen increases, the strength of the oxyacid increases.
$$HOCI > HOBr > HOI$$
Hence option A is correct.
Question 7
1 / -0

Which one of the following oxy acid of fluorine exists ?

A
$$HOF$$

B
$$HFO_{3}$$

C
$$HFO_{4}$$

D
$$HFO_{2}$$

Solution

In fluorine d-orbitals are absent so it shows only -1 oxidation state in oxo-acid,
but not + 3, + 5 or + 7.

That's why unlike other members (halogens) fluorine only forms one oxoacid HOF.

Hence, option A is correct.
Question 8
1 / -0

The statement, which prompted Neil Bartlett to prepare the first noble gas compound was :

A
$$Xe-F$$ bond has high bond energy

B
$$F_{2}$$ has exceptionally low bond energy

C
$$PtF_{6}$$ is a strong oxidant

D
$$O_{2}$$ molecule and Xe atom have very similar ionization energies.

Solution

Bartlett's main specialty was the chemistry of fluorine and of compounds containing fluorine. In 1962, Bartlett prepared the first noble gas compound, xenon hexafluoroplatinate. This contradicted established models of the nature of valency, as it was believed that all noble gases were entirely inert to chemical combination. He subsequently produced and reproduced several other fluorides of xenon.
The statement, which prompted Neil Bartlett to prepare the first noble gas compound was $$O_{2}$$ molecule and Xe atom have very similar ionization energies.
Question 9
1 / -0

The following acids have been arranged in the order of decreasing acid strength. Identify the correct order-
ClOH(I) BrOH(II) IOH(III)

A
I > II > III

B
II > I > III

C
III > II > I

D
I > III > II

Solution

Higher the I.E, higher the acidic strength of hypohalus acid (hydroxides)
As we move from top to bottom in halogens the I.E will be decreased,so the acidity will be decreased.
The correct oreder of acidity is I > II > III
Hence option A is correct.
Question 10
1 / -0

$$N_2(g) + 3H_2(g)\overset{Fe+Mo}{\rightleftharpoons}2NH_3(g)$$; Haber's process.
Mo is used as :

A
a catalyst

B
a catalytic promoter

C
an oxidising agent

D
a catalytic poison

Solution

$$NH_3$$ is manufactured by Haber's process.
$$N_2(g) + 3H_2(g)\overset{Fe+Mb}{\rightleftharpoons}2NH_3(g)$$
In this process, pure nitrogen and hydrogen gases react in the ratio of $$1:3$$ by volume in $$400^0C-500^0C$$ temperature, $$200-900$$ atm pressure in presence of iron as catalyst and Mo as promoter.

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The p-Block Elements Test - 36

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The p-Block Elements Test - 36

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Question 1

$$F_2$$

$$Cl_2$$

$$Br_2$$

$$I_2$$

Solution

Question 2

F-F

Cl-Cl

Br-Br

I-I

Solution

Question 3

Small size

Very high electronegativity

Non-availability of d-orbitals

All of these

Solution

Question 4

at the corners of a cube.

at the corners of an octahedron.

at the corners of a tetrahedron.

at the center and corners of a tetrahedron.

Solution

Question 5

$$(Z - 1)$$ and $$(Z + 1)$$

$$(Z - 2)$$ and $$(Z + 2)$$

both (a) and (b)

none of these

Solution

Question 6

$$I> II > III$$

$$II>I>III$$

$$III>II>I$$

$$I>III>II$$

Solution

Question 7

$$HOF$$

$$HFO_{3}$$

$$HFO_{4}$$

$$HFO_{2}$$

Solution

Question 8

$$Xe-F$$ bond has high bond energy

$$F_{2}$$ has exceptionally low bond energy

$$PtF_{6}$$ is a strong oxidant

$$O_{2}$$ molecule and Xe atom have very similar ionization energies.

Solution

Question 9

I > II > III

II > I > III

III > II > I

I > III > II

Solution

Question 10

a catalyst

a catalytic promoter

an oxidising agent

a catalytic poison

Solution

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