The p-Block Elements Test

Result

The p-Block Elements Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

Apatite is an ore of:

A
fluorine.

B
chlorine.

C
bromine.

D
iodine.

Solution

Apatite is $$CaF_{2} \cdot 3Ca_{3}(PO_{4})_{2}$$
$$\therefore$$ It is an ore of fluorine with calcium.
Question 2
1 / -0

The common basic structural unit in silicates and silica is:

A
$$Si_2O_6^{4-}$$

B
$$SiO^{2-}_3$$

C
$$SiO^{4-}_4$$

D
$$Si_2O^{6-}_7$$

Solution

The common basic structural unit in silicates and silica is $$\displaystyle SiO^{4-}_4 $$ where each $$Si$$ atom is bonded tetrahedrally to four oxygen atoms. Si undergoes $$\displaystyle sp^3$$ hybridization.
Question 3
1 / -0

Which of the following forms vortex ring?

A
$$P_{2}O_{5}$$

B
$$PH_{3}$$

C
$$NH_{3}$$

D
$$P_{2}O_{10}$$

Solution

Phosphine forms vortex rings of $$P_{2}O_{5}$$ when it comes in contact of air. These rings are in form of white smoke.
$$\therefore$$ They are used in making smoke screens in warfare.
Question 4
1 / -0

In clatherates of xenon with water, the nature of bonding between xenon and water molecule is :

A
Dipole induced dipole interaction

B
Coordinate

C
Hydrogen bonding

D
Covalent

Solution

Clathrates are also known as cage compounds.
Noble gases are trapped within cavities of crystal lattices of certain organic and inorganic substances.
In xenon clathrates the xenon molecules occupy the gaps in the open hydrogen-bonded networks of the water molecules.
There is no conventional chemical bonding involved in such clathrates, as in the case of halogen clathrates
hence nature of bonding is Dipole induced dipole interaction.
Question 5
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Order of reactivity of halogen acids towards an alcohol is:

A
HCl > HBr > HI

B
HBr > HI > HCl

C
HI > HBr > HCl

D
HI > HCl > HBr

Solution

Longer the bond length, lesser will be dissociation energy and hence, more reactivity.
Among halogen acids bond length increases from HCl to HI.
$$\therefore$$ Order of reactivity of halogen acids towards alcohol is
$$HI > HBr > HCl$$
Question 6
1 / -0

Coordination number of aluminium is:

A
8

B
6

C
12

D
4

Solution

Write valence shell configuration of $$Al$$.
$$Al(13) = 1s^2, 2s^22p^6, 3s^23p^1$$
$$\therefore$$ It can have maximum coordination number as 6
Question 7
1 / -0

Which of the following is known as dead burnt?

A
Gypsum

B
Plaster of Paris

C
Anhydrite

D
None of these

Solution

$$\underset {gypsum}{CaSO_{4}\cdot 2H_{2}O} \xrightarrow {120^{\circ}} \underset {\text {Plaster of paris}}{CaSO_{4}\cdot 1/2H_{2}O} \xrightarrow {200^{\circ}C} \underset {\text {anhydrite or dead burnt plaster}}{CaSO_{4}}$$
The anhydrous $$CaSO_{4}$$ is called dead burnt plaster because it does not set like plaster of Paris when moistened with water.
Question 8
1 / -0

Which of the following correctly represents the variation of electronegativity (EN) with atomic number (Z) of group 13 elements?

A

B

C

D

Solution

The electronegativity of group 13 elements decreases from B to Al and then increases from Al to Tl.
Question 9
1 / -0

Electronegativity of group 13 elements follow the order :

A
B > Ca >Al > Th > In

B
B > Th> Ga > Al > In

C
B > Th> In > Ga > Al

D
B > Al > Th > In > Ga

Solution
Question 10
1 / -0

The stable bivalency of $$Pb$$ and trivalency of $$Bi$$ is :

A
due to d contraction in $$Pb$$ and $$Bi$$

B
due to relativistic contraction of the $$6s$$ orbitals of $$Pb$$ and $$Bi$$, leading to inert pair effect

C
due to screening effect

D
due to attainment of noble liquid configuration

Solution

$${ Pb=6s }^{ 2 }{ 6p }^{ 2 }\\ { Bi=6s }^{ 2 }{ 6p }^{ 3 }\\ $$
Due to inertness of $$ 6{s}^2 $$ electrons in Pb and Bi, they show bivalency and trivalent respectively instead of tetravalent and pentavalency.