The p-Block Elements Test

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The p-Block Elements Test - 46

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Question 1
1 / -0

Nitrogen can combine with:

A
only metals

B
only non-metals

C
both metals and non-metals

D
neither metals nor non-metals

Solution

Nitrogen can combine with both metals and non metals.
$$N_2+3H_2 \rightarrow NH_3$$
$$N_2+Li \rightarrow 2Li_3N$$
Question 2
1 / -0

Haber's process is a/an:

A
endothermic process

B
exothermic process

C
spontaneous with no heat liberation

D
None

Solution

Hint:

The process used for manufacturing ammonia from nitrogen and hydrogen is termed Haber’s process.

Explanation:

Part 1:

The chemical reaction involved in this process is $$N_2\left ( g \right )+3H_2\left ( g \right )\rightleftharpoons 2NH_3\left ( g \right )+\Delta H$$

Here, $$\Delta H=-92.4 \dfrac{kJ}{mol}$$

Part 2:

The important conditions necessary for this process are

The volume of gasses of nitrogen and hydrogen should maintain $$1:3$$ ratio.

The temperature and pressure should be above $$723K$$ and $$200 atm$$ respectively.

Iron is used as a catalyst and molybdenum as a promoter.

Final answer:

The negative sign of enthalpy in the chemical reaction signifies that the heat is released.

Therefore the correct answer is option B i.e., Haber’s process is an exothermic process.
Question 3
1 / -0

In the commercial preparation of dinitrogen by the liquefaction and fractional distillation,dinitrogen is obtained in:

A
liquid form

B
gaseous form

C
mixture of liquid and gases

D
none

Solution

In the commercial preparation of dinitrogen by the liquefaction and fractional distillation,dinitrogen is obtained in gaseous form.
Question 4
1 / -0

Compound formed on reaction of nitrogen with oxygen at very high temperature is:

A
$$NO_2$$

B
$$N_2O_4$$

C
$$NO$$

D
$$N_2O_3$$

Solution

$$N_2+O_2 \rightarrow NO$$
The reaction occurs at temperature greater than 2000 C.
Question 5
1 / -0

The single $$X-X$$ bond made by $$N$$ is weaker than that of $$P$$ because:

A
of repulsion between nitrogen nucleus due to smaller size of nitrogen atom

B
of repulsion between non-bonded electrons due to smaller size of nitrogen atom

C
of higher bond enthalpy of nitrogen

D
all of the above

Solution

The single $$N-N$$ bonds formed is weaker than single $$P-P$$ bonds because of lone pair repulsion on small nitrogen molecule.
Question 6
1 / -0

The multiple bonds that nitrogen forms with itself are:

A
$$d\pi-d\pi$$ bonds

B
$$d\pi-p\pi$$ bonds

C
$$p\pi-p\pi$$ bonds

D
none of the above

Solution

When bonding occurs between elements having their valence electron and vacant place in $$p$$ orbital then its called $$p\pi-p\pi$$ bonding. The nitrogen atom has valence electron in $$p$$ orbital so with itself form $$p\pi-p\pi$$ bonds.
Question 7
1 / -0

What results in the weak catenation property of nitrogen?

A
Smaller size

B
Repulsion between non-bonded electrons

C
Repulsion between protons

D
High electronegativity

Solution

The weak catenation property of nitrogen is due to lone pair repulsion of lone pair of electrons on nitrogen molecule.
Question 8
1 / -0

The complex formed in Brown Ring Test is ___________.

A
$$[Fe(H_2O)_4(NO)_2]^{2+}$$

B
$$[Fe(H_2O)_5(NO)]^{2+}$$

C
$$[Fe(H_2O)_3(NO)_3]^{2+}$$

D
$$[Fe(H_2O)_6(NO)]^{2+}$$

Solution

$$[Fe(H_2O)_5(NO)]SO_4$$ is the complex formed in brown ring test done to identify the presence of nirates.
Question 9
1 / -0

Urea on reacting with water gives:

A
$$NH_3$$

B
$$N_2O_3$$

C
$$HNO_2$$

D
$$HNO_3$$

Solution

$$NH_2CONH_2+H_2O \rightarrow 2NH_3+CO_2$$
Urea on reaction with water gives carbondioxide and ammonia.
Question 10
1 / -0

Inert gases such as helium behave like ideal gases over a wide range of temperature. However they condense into the solid state at very low temperature there is a:

A
Weak attractive forces between the atoms

B
Weak repulsive force between the atoms

C
Strong attractive force between the atoms

D
Strong repulsive force between the atome

Solution

Option (C) is correct.
Inert gases condenses into the solid state at very low temperature as there is strong attractive force between the atoms.
In solid state, vander waals attractive forces are predominant between the atoms. The attractive force increases with the size of the atom as a result of the increase in polarizability and the decrease in ionization potential.

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Re-Attempt Weekly Quiz Competition

The p-Block Elements Test - 46

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The p-Block Elements Test - 46

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SHARING IS CARING

Question 1

only metals

only non-metals

both metals and non-metals

neither metals nor non-metals

Solution

Question 2

endothermic process

exothermic process

spontaneous with no heat liberation

None

Solution

Question 3

liquid form

gaseous form

mixture of liquid and gases

none

Solution

Question 4

$$NO_2$$

$$N_2O_4$$

$$NO$$

$$N_2O_3$$

Solution

Question 5

of repulsion between nitrogen nucleus due to smaller size of nitrogen atom

of repulsion between non-bonded electrons due to smaller size of nitrogen atom

of higher bond enthalpy of nitrogen

all of the above

Solution

Question 6

$$d\pi-d\pi$$ bonds

$$d\pi-p\pi$$ bonds

$$p\pi-p\pi$$ bonds

none of the above

Solution

Question 7

Smaller size

Repulsion between non-bonded electrons

Repulsion between protons

High electronegativity

Solution

Question 8

$$[Fe(H_2O)_4(NO)_2]^{2+}$$

$$[Fe(H_2O)_5(NO)]^{2+}$$

$$[Fe(H_2O)_3(NO)_3]^{2+}$$

$$[Fe(H_2O)_6(NO)]^{2+}$$

Solution

Question 9

$$NH_3$$

$$N_2O_3$$

$$HNO_2$$

$$HNO_3$$

Solution

Question 10

Weak attractive forces between the atoms

Weak repulsive force between the atoms

Strong attractive force between the atoms

Strong repulsive force between the atome

Solution

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