The p-Block Elements Test

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The p-Block Elements Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

The form of $$SiO_{2}$$ used as piezoelectric material :

A
cristobalite

B
quartz

C
kiesulguhr

D
silica gel

Solution

Quartz crystals have piezoelectric properties; they develop an electric potential upon the application of mechanical stress. An early use of this property of quartz crystals was in phonograph pickups. One of the most common piezoelectric uses of quartz today is as a crystal oscillator. The quartz clock is a familiar device using the mineral. The resonant frequency of a quartz crystal oscillator is changed by mechanically loading it, and this principle is used for very accurate measurements of very small mass changes in the quartz crystal microbalance and in thin film monitors.
Question 2
1 / -0

The material used in solar cell contains :

A
Cs

B
Si

C
Sn

D
T

Solution

Silicon is a semicoductor and is used in solar cells, sometimes doped.
Question 3
1 / -0

List -I List -2
A. The purest form of $$SiO_{2}$$ 1. Quartz
B. Carborundum 2. Graphite
C. Corundum 3. $$SiC$$
D. Hexagonal rings 4. $$ Al_{2} O_{3}$$
5. Diamond

Correct match with A B C D is:

A
1 3 4 2

B
3 1 4 2

C
1 3 2 4

D
3 1 4 3

Solution

Quartz is the crystalline form of $$SiO_{2}$$ and hence is the purest form.

Carborundum is the commercial name of $$SiC$$.

Corundum is an ore of Aluminium with the chemical formula $$Al_{2}O_{3}$$.

Graphite has a sheet-like structure, with each layer made of hexagonal rings.
Question 4
1 / -0

List -I List -2
A. Reactive form of Carbon 1. HF
B. Acid employed for 2. Diamond
etching of glass
C. Synthesis gas 3. Charcoal
D. Unreactive form of Carbon 4. $$CO H_{2}$$
The correct match with A B C D is ?

A
1 2 3 4

B
2 4 3 1

C
2 1 3 4

D
3 1 4 2

Solution

Charcoal is a reactive form of carbon.
Glass is etched by $$HF$$.
Water gas is also called synthesis gas.
Diamond is chemically inactive due to its presence of strong directional covalent C-C bonds.
Question 5
1 / -0

Which of the following does not exhibit inert pair effect?

A
Bi

B
Pb

C
B

D
Tl

Solution

Only high atomic weight elements in p block show inert pair effect. Here B is the first member of boron family and it does not have high atomic weight so it does not show inert pair effect.
Question 6
1 / -0

The most abundant metal among the following is:

A
$$Al$$

B
$$Ca$$

C
$$Fe$$

D
$$K$$

Solution

In the Earth's crust, aluminium is the most abundant ($$8.3\%$$ by weight) metallic element and the third most abundant of all elements (after oxygen and silicon).
Question 7
1 / -0

Among III-A group elements, the elements with highest and lowest I.P. values are:

A
B, Tl

B
B, In

C
B, Al

D
B, Ga

Solution

$$I.P.$$ does not smoothly decrease down $$III-A$$ group. There is an observed discontinuity in the $$I.P.$$ values between $$In$$ and $$Tl$$ due to non availability of f-electrons, which have low screening effect to compensate the increase in nuclear charge.
Question 8
1 / -0

Element with a giant molecular structure is:

A
B

B
Al

C
Ga

D
Tl

Solution

An element with a giant molecular structure is B. It exists in two allotropic forms.
(1) Grey black, non-metallic, high melting crystalline solid, chemically inert.
(2) Dark brown, amorphous powder, chemically active.
Also, Boron has icosahedron geometry which gives it a giant molecular structure.
Question 9
1 / -0

Assertion (A) : Thallium compounds are stable in +1 oxidation state.
Reason (R) : The $$6s^{2}$$ electrons in Tl show reluctance in participating in bond formation.

A
Both Assertion and Reason are true and Reason is the correct explanation of Assertion

B
Both Assertion and Reason are true and Reason is not the correct explanation of Assertion

C
Assertion is true but Reason is false

D
Assertion is false but Reason is true.

Solution

Inert pair effect. Because of poor shielding by f, Increased effective nuclear charge holds the 6s electrons tightly.
Question 10
1 / -0

The electro-positive character increases from B to Al and then decreases from Al to Tl because of:

A
decrease in ionization energy of the elements

B
increase in size of the elements

C
ineffective shielding of nuclear charge by d-electrons in case of Ga, In and Tl

D
decrease in electronegativity of the elements

Solution

The electro-positive character increases from B to Al and then decreases from Al to Tl because of ineffective shielding of nuclear charge by d-electrons in case of Ga, In and Tl.
Due to the small size of boron, the sum of its first three ionization enthalpies is very high. This prevents it to form +3 ions and forces it to form only covalent compounds.
But as we move from B to Al, the sum of the first three ionization enthalpies of Al considerably decreases and is, therefore, able to form $$Al^{3+}$$ ions. In fact, aluminium is a highly electropositive metal.
$$\text {However, down the group, due to the poor shielding effect of intervening d and f orbitals, the increased }$$
$$\text{ effective nuclear charge holds ns electrons tightly, restricting their participation in bonding}$$. As a result of this, an only p-orbital electron may be involved in bonding. In fact in Ga, In and Tl, both +1 and +3 oxidation states are observed.

Hence, the electro-positive character increases from B to Al and then decreases from Al to Tl.
Option C is correct.

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The p-Block Elements Test - 20

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The p-Block Elements Test - 20

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SHARING IS CARING

Question 1

cristobalite

quartz

kiesulguhr

silica gel

Solution

Question 2

Cs

Si

Sn

T

Solution

Question 3

1 3 4 2

3 1 4 2

1 3 2 4

3 1 4 3

Solution

Question 4

1 2 3 4

2 4 3 1

2 1 3 4

3 1 4 2

Solution

Question 5

Bi

Pb

B

Tl

Solution

Question 6

$$Al$$

$$Ca$$

$$Fe$$

$$K$$

Solution

Question 7

B, Tl

B, In

B, Al

B, Ga

Solution

Question 8

B

Al

Ga

Tl

Solution

Question 9

Both Assertion and Reason are true and Reason is the correct explanation of Assertion

Both Assertion and Reason are true and Reason is not the correct explanation of Assertion

Assertion is true but Reason is false

Assertion is false but Reason is true.

Solution

Question 10

decrease in ionization energy of the elements

increase in size of the elements

ineffective shielding of nuclear charge by d-electrons in case of Ga, In and Tl

decrease in electronegativity of the elements

Solution

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