The p-Block Elements Test

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The p-Block Elements Test - 63

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Question 1
1 / -0

Nitrogen reacts with a metal $$M$$ to produce a compound $$X$$. This compound $$X$$, when dissolved in water forms pungent gas $$Y$$ which can be identified easily by the formation of white fumes in the presence of $$HCl$$.

Identify the metal $$M$$.

A
Calcium

B
Sodium

C
Cobalt

D
Iron

Solution

Calcium metal reacts with nitrogen to form calcium nitride.

If we consider metal $$M$$ as calcium.

$$3\underset{M}{Ca}+N_2 \rightarrow Ca_3N_2.$$

The calcium nitride would react with water to form $$NH_3$$, which will react with $$HCl$$ to give white fumes of $$NH_4Cl$$ salt.

Thus, the metal $$M$$ is calcium.

Hence, option (A) is the right answer.
Question 2
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Ammonia gas is not prepared by :

A
reduction of sodium nitrite or sodium nitrate by the reaction of zinc dust and sodium hydroxide

B
hydrolysis of calcium cyanamide

C
heating ammonium chloride with slaked lime

D
heating of $$(NH_4)_2Cr_2O_7$$

Solution

(A) $$NO_2^-+7[H]^-\xrightarrow {Zn/NaOH} NH_3+2H_2O; NO_3^-+9[H]^-\xrightarrow {Zn/NaOH} NH_3+3H_2O$$
(B) $$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$$
(C) $$NH_4Cl+Ca(OH)_2\rightarrow 2NH_3+CaCl_2+2H_2O$$
(D) $$(NH_4)_2Cr_2O_7\rightarrow N_2+Cr_2O_3+4H_2O$$
Question 3
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The boiling points of neon and krypton are $$ -245.{ 9 }^{ \circ }C$$ and $$ -152.{ 9 }^{ \circ }C$$ respectively. Using these data estimate the boiling point of argon. (report answer as the negative of the closest integer)

A
$$-289^oC$$

B
$$-299^oC$$

C
$$-199^oC$$

D
$$-189^oC$$

Solution

The boiling point of argon can be roughly approximated as the average of the boiling points of neon and krypton.
$$\dfrac {-245.9 \: ^0C-152.9 \: ^0C}{2}=-199.4 \: ^0C$$
Hence, the boiling point of argon is approximately $$-199.4 \: ^0C$$.
Question 4
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Choose correct squence for the following species.

A
$$F_{2} > Cl_{2} > Br_{2} > I_{2} $$(Reducing power)

B
$$Cl_{2}< F_{2} < Br_{2} < I_{2}$$ (Reactivity)

C
$$I_{2} > Br_{2} > F_{2}> Cl_{2}$$(Bond length)

D
$$Cl_{2} > Br_{2} >F_{2} >I_{2}$$ (Bond dissociation enthalpy)

Solution

Bond dissociation enthalpy decreases down the group . Bond dissociation enthalpy is inversely proportional to the atomic radii. More the size less the attraction between atoms . But the bond dissociation enthalpy of fluorine is less than expected . There is interelectronic repulsion in fluorine atom due to its small size. Due to this repulsion bond dissociation enthalpy of fluorine is less than that of bromine and chlorine .
Question 5
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The high oxidising power of fluorine is due to its :

A
high electron affinity.

B
high heat of dissociation and low heat of hydration.

C
low heat of dissociation and high heat of hydration.

D
high heat of dissociation and high heat of hydration.

Solution

The atomisation energy of fluorine is abnormally low. This reflects the low bond enthalpy of fluorine. The main reason for such a high oxidising power of fluorine is the very high hydration enthalpy of the fluoride ion. This is because the ion is very small. There is a very strong attraction between the fluoride ions and water molecules. The stronger the attraction, the more heat is evolved when the hydrated ions are formed.
Hence, low heat of dissociation and high heat of hydration leads to high oxidising power of fluorine.
Question 6
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The steps for the manufacture of nitric acid by Ostwald process are given. Arrange them in sequence.
(a) Recirculation of nitric oxide
(b) Addition of $${H}_{2}O$$ to nitrogen dioxide
(c) Oxidation of nitric oxide
(d) Catalytic oxidation of ammonia

A
d c b a

B
d c a b

C
c d b a

D
c d a b

Solution

In ostwald process Ammonia is converted to nitric acid in 2 stages. It is oxidized (in a sense "burnt") by heating with oxygen in the presence of a catalyst to form nitric oxide and water. Nitric oxide is oxidized again to yield nitrogen dioxide. This gas is then readily absorbed by the water and forms nitric acid. Thus the steps of process are :
(d) Catalytic oxidation of ammonia
(c) Oxidation of nitric oxide
(b) Addition of $${H}_{2}O$$ to nitrogen dioxide
(a) Recirculation of nitric oxide
Question 7
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Directions For Questions

The chemistry of halogens is interesting. Halogens belong to the group 17 of the periodic table. The name Halogen was introduced by J.S.C. Schweiggen in 1811 to describe the property of chlorine (at that time unique among the elements) to combine directly with metals to give salts. Iodine is an essential trace element for humans and plays an important role in growth regulating hormone thyroxine, which is produced in the thyroid gland. Deficiency of iodine in diet leads to Goitre. In order to prevent this condition, about 0.01% $$NaI$$ is added to the table salt. Iodine is a lustrous solid and sublimes at ordinary temperature.

...view full instructions

The solubility of iodine in carbon tetrachloride is due to

A
Ionic solvent-solute interaction

B
Charge-transfer phenomenon

C
Van der Waals interaction

D
Dissociation of iodine

Solution

Both $$I_2$$ and $$CCl_4$$ are non polar in nature. The interaction between them is van der waal.
Question 8
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Atoms in a $${ P }_{ 4 }$$ molecule of white phosphorus are arranged regularly in which of 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 centre and corners of a tetrahedron

Solution

White phosphorus has the following structure
The structure shows that the four $$P$$ atoms in white phosphorus lie at the corners of a regular tetrahedron.
Question 9
1 / -0

How many oxides of monomeric nature can nitrogen form?

A
$$1$$

B
$$2$$

C
$$4$$

D
$$5$$

Solution

$$N$$ has 5 valence electron, therefore molecules with $$N_2O_x$$ will have even number of electrons and will be non radical. For monomeric oxide, $$N_2O$$ is the only monomeric nitrogen oxide.
Therefore option A is correct.
Question 10
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Which of the following gases has the greatest density at STP conditions?

A
$$Ne$$

B
$$Ar$$

C
$$He$$

D
$$Kr$$

E
$$Xe$$

Solution

The highest density among the inert gases is of Neon (Ne). This a factual data. Thus the highest density among given options is of $$Ne$$, as all the options are of inert gases.

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The p-Block Elements Test - 63

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The p-Block Elements Test - 63

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Question 1

Calcium

Sodium

Cobalt

Iron

Solution

Question 2

reduction of sodium nitrite or sodium nitrate by the reaction of zinc dust and sodium hydroxide

hydrolysis of calcium cyanamide

heating ammonium chloride with slaked lime

heating of $$(NH_4)_2Cr_2O_7$$

Solution

Question 3

$$-289^oC$$

$$-299^oC$$

$$-199^oC$$

$$-189^oC$$

Solution

Question 4

$$F_{2} > Cl_{2} > Br_{2} > I_{2} $$(Reducing power)

$$Cl_{2}< F_{2} < Br_{2} < I_{2}$$ (Reactivity)

$$I_{2} > Br_{2} > F_{2}> Cl_{2}$$(Bond length)

$$Cl_{2} > Br_{2} >F_{2} >I_{2}$$ (Bond dissociation enthalpy)

Solution

Question 5

high electron affinity.

high heat of dissociation and low heat of hydration.

low heat of dissociation and high heat of hydration.

high heat of dissociation and high heat of hydration.

Solution

Question 6

d c b a

d c a b

c d b a

c d a b

Solution

Question 7

Ionic solvent-solute interaction

Charge-transfer phenomenon

Van der Waals interaction

Dissociation of iodine

Solution

Question 8

At the corners of a cube

At the corners of an octahedron

At the corners of a tetrahedron

At the centre and corners of a tetrahedron

Solution

Question 9

$$1$$

$$2$$

$$4$$

$$5$$

Solution

Question 10

$$Ne$$

$$Ar$$

$$He$$

$$Kr$$

$$Xe$$

Solution

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