The p-Block Elements Test

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The p-Block Elements Test - 57

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following metal on burining in mosist air does not give smell of ammonia?

A
$$Mg$$

B
$$Ca$$

C
$$K$$

D
$$Li$$
Question 2
1 / -0

In Haber process $$30L$$ of Dihydrogen and $$30L$$ of dinitrogen were taken for reaction which yielded only $$50\% $$of the expected product . what will be the composition of gaseous mixture under the aforesaid condition in the end?

A
$$20L$$ ammonia $$10L$$ nitrogen , $$30L$$ hydrogen

B
$$20L$$ ammonia , $$125L$$ nitrogen , $$25L$$ ydrogen

C
$$20L$$ ammonia , $$20L$$ nitrogen , $$20L$$hydrogen

D
$$10L$$ ammonia , $$25L$$nitrogen , $$15L$$ hydrogen

Solution

solution:
$$N_2+3H_2 \rightleftharpoons 2NH_3$$
30 30 0
30-x 30-x 2x

$$2x = 10$$

x = 5;
$$N_2 = 30 - 5 = 25 L$$

$$H_2 = 30 - 3X 5 = 15 L$$

30 L of H_2 will react with 10 L of N_2 to produce 20 L of $$NH_3$$ (Here $$H_2$$ is the limiting reactant). As the yield is only 50%, 15 L of $$H_2$$ will react with 5 L of $$N_2$$ to produce 10 L of $$NH_3$$
Reaction mixture will contain,
15 L of $$H_2$$ (30L-15L), and 10 L of $$NH_3$$
hence the correct opt: D
Question 3
1 / -0

Select the correct statement (s)

A
HCI and HF do not from free radical easily due to larger bond-energy

B
HI with a minimum bond energy is every reactive and instead from $${{\text{I}}_{\text{2}}}\,\,\,$$

C
Both (a) and (b)

D
None of the above

Solution

Bond dissociation energy with hydrogen decreased down the group so HI is will be the most reactive . HI is more reative because the bond dissociation energy is low due to difference in size of hydrogen and iodine so it can be easily breaked.
Question 4
1 / -0

Which of the following product (s) is/are obtained when $$C{I_2}O$$ reacts with $$N{H_3}$$ ?

A
$$N{O_2}$$

B
$${N_2}$$

C
$$NC{I_3}$$

D
$$N{H_4}CI$$

Solution

$$C{l_2}O + N{H_3} \to {N_2} + N{H_4}Cl + {H_2}O$$
Question 5
1 / -0

The most powerful oxidant among the following is:

A
$$HOCl$$

B
$$HClO_2$$

C
$$HClO_3$$

D
$$HClO_4$$

Solution

(B) $$HClO_2$$
Standard reduction potential is $$1.6$$V and is the highest strongest oxidant among all these.
Question 6
1 / -0

The compound that cannot be formed by Xenon is

A
$$XeO_3$$

B
$$XeF_4$$

C
$$XeCl_4$$

D
$$XeO_2F_2$$

Solution

Xenon will not form XeO₂F₂. because it doesnot exist
Question 7
1 / -0

The number of steps involved in the manufacturing of nitric acid through Ostwald's process is:

A
1

B
5

C
3

D
6

Solution

Ostwald process to manufacture nitric acid-

Step $$1$$: Catalytic oxidation of Ammonia

A mixture of dry air and dry ammonia in the ratio of $$10 : 1$$ by volume is compressed and then passed into a platinum gauze which acts as catalyst at about $$800 ℃$$.
$$4 N{H}_{3} + 5 {O}_{2} \longrightarrow 4 NO + 6 {H}_{2}O + \text{Heat}$$

Step $$2$$: Oxidation of nitric oxide

Nitric oxide combines with oxygen to form nitrogen dioxide at about 50°C.
$$2 NO + {O}_{2} \longrightarrow 2 N{O}_{2}$$

Step $$3$$: Absorption of nitrogen dioxide in water

The nitrogen dioxide and oxygen present in the air react with water to form nitric acid.
$$4 N{O}_{2} + 2 {H}_{2}O + {O}_{2} \longrightarrow \underset{\text{Nitric acid}}{4 HN{O}_{3}}$$

Nitric acid obtained is concentrated above $$50 \%$$. On further distillation, $$68 \%$$ nitric acid is produced.

Hence, the number of steps involved in the manufacturing of nitric acid is 3.
Question 8
1 / -0

Of the following species, one which is non-existent :

A
$$Xe{F_6}$$

B
$$Xe{F_5}\;$$

C
$$Xe{F_4}$$

D
$$Xe{F_2}$$

Solution

Xe forms only three flourides
$$XeF_2,XeF_4,XeF_6$$
so,$$XeF_5$$ is non-existent.
Question 9
1 / -0

$$N_{2}H_{4}+2H_{2}O_{4}\rightarrow 4H_{2}O_2$$.

This equation suggests the use of $$H_{2}O_{2}$$ as :

A
Bleaching agent

B
Antisepics

C
Restoring colour of old paintings

D
Rocket Propellant
Question 10
1 / -0

which of the following is not characteristic of interhalogen compounds?

A
They are more reactive than halogens.

B
They are covalent in nature.

C
They have low boiling point and are non-volatile.

D
they are diamagnetic.

Solution

Inter-halogen compounds have molecules that are covalent and diamagnetic in nature. The bonds formed more reactive than diatomic halogen bonds. They are volatile but not explosive. they have low boiling point.

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The p-Block Elements Test - 57

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The p-Block Elements Test - 57

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Question 1

$$Mg$$

$$Ca$$

$$K$$

$$Li$$

Question 2

$$20L$$ ammonia $$10L$$ nitrogen , $$30L$$ hydrogen

$$20L$$ ammonia , $$125L$$ nitrogen , $$25L$$ ydrogen

$$20L$$ ammonia , $$20L$$ nitrogen , $$20L$$hydrogen

$$10L$$ ammonia , $$25L$$nitrogen , $$15L$$ hydrogen

Solution

Question 3

HCI and HF do not from free radical easily due to larger bond-energy

HI with a minimum bond energy is every reactive and instead from $${{\text{I}}_{\text{2}}}\,\,\,$$

Both (a) and (b)

None of the above

Solution

Question 4

$$N{O_2}$$

$${N_2}$$

$$NC{I_3}$$

$$N{H_4}CI$$

Solution

Question 5

$$HOCl$$

$$HClO_2$$

$$HClO_3$$

$$HClO_4$$

Solution

Question 6

$$XeO_3$$

$$XeF_4$$

$$XeCl_4$$

$$XeO_2F_2$$

Solution

Question 7

1

5

3

6

Solution

Question 8

$$Xe{F_6}$$

$$Xe{F_5}\;$$

$$Xe{F_4}$$

$$Xe{F_2}$$

Solution

Question 9

Bleaching agent

Antisepics

Restoring colour of old paintings

Rocket Propellant

Question 10

They are more reactive than halogens.

They are covalent in nature.

They have low boiling point and are non-volatile.

they are diamagnetic.

Solution

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