The p-Block Elements Test

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The p-Block Elements Test - 19

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Question 1
1 / -0

Assertion (A) : Silica is less harder than diamond .
Reason (R) : Silica has only two dimensional network .

A
Both A and R are true and R is correct

explanation of A

B
Both A and R are true and R is not correct

explanation of A

C
A is true but R is false

D
A is false but R is true.

Solution

3 dimensional silicates exist. Eg: Quartz.
Question 2
1 / -0

Which of the following statements are true about quartz?
A. it is pure crystalline form of silica.
B. it is a tetrahedral polymer of $$SiO_{2}$$.
C. UV light can pass through quartz.

A
A and B are correct

B
B and C are correct

C
A and C are correct

D
all are correct

Solution

Various forms of silica ($$SiO_2$$) are quartz, sand, flint etc.
It is made up of a continuous framework of $$SiO_4$$ silicon-oxygen tetrahedra, with each oxygen being shared between two tetrahedra, giving an overall formula $$SiO_2$$.
Quartz is a good transmitter of UV rays. Most glasses are opaque to UV. The UV range is from 0.1 nm to about 400 nm (or from 100 $$A^o$$ to about 4000 $$A^o$$). The visible light wavelengths are longer, between 400 nm and 700 nm. Quartz will transmit UV from 180 nm to 400 nm. Since quartz is made of silicon dioxide, glasses made from a high proportion of this material will also readily transmit UV. One such glass is known as VYCOR, which is 96% silica.
So statement A and C are correct.
Question 3
1 / -0

Decreasing order of p-orbital character in the following is:
A) $$ SiO_{2}$$ B) $$ CO_{2}$$ C) Graphite

A
A>B>C

B
B>A>C

C
B>C>A

D
A>C>B

Solution

The more the number of p orbitals involved in hybridisation, the more is the p-character of the hybrid orbital and hence, more is the p-character of the bonds. The hybridisations are:
$$SiO_{2}$$ is $$sp^3$$ hybridised with 3 p-orbitals involved. $$CO_{2}$$ is $$sp$$ hybridised with 1 p-orbital involved. Graphite is $$sp^2$$ hybridised with 2 p-orbitals involved. Hence, p character is maximum in $$SiO_2$$, then graphite, and the least in $$CO_2$$.
Question 4
1 / -0

$$SiF_{4} + H_{2} O \rightarrow A \xrightarrow {1000^oC} B \xrightarrow {Na_{2}CO_{3}} C$$

In the above reaction, identify B and C respectively :

A
$$H_{4}SiO _{4}$$ and $$Na _{2} SiO_{3}$$

B
$$SiO _{2}$$ and $$Na _{2} S$$

C
$$SiO_{2}$$ and $$Na _{2}CO _{2}$$

D
$$SiO _{2}$$ and $$Na _{2} SiO_{3}$$

Solution

The reactions are as follows:
$$SiF_{4} + H_{2} O \rightarrow Si(OH)_4 \xrightarrow {1000^oC} SiO_2 \xrightarrow {Na_2CO_3} Na_2SiO_3$$

Hence, option D is the correct one.
Question 5
1 / -0

Arrange the group 14 elements in the order of decreasing ionisation energies.

A
C > Si > Ge > Pb > Sn

B
C > Si > Ge > Sn > Pb

C
C > Ge > Sn > Pb>Si

D
Si > C > Ge > Sn > Pb

Solution

As we move down the group, the ionisation energy decreases, due to increase in atomic size, thereby, increasing the distance of the electrons from the nucleus. The first ionisation energy of lead is slightly higher than that of tin. The reason is due to lanthanide contraction. The atomic size of tin and lead are very close, and hence, the charge density is more in case of lead, due to more charge. Due to higher charge density, the outer electrons are attracted more by the nucleus in lead and hence, the ionisation energy is more in the case of lead.
Question 6
1 / -0

A. Graphite acts as lubricant due to layer like structure.
B. The distance between two layers in graphite is $$1.42^{\circ}A$$.
C. Silica does not react with HCl, HBr and Hl.
Which of the above statement(s) is/are true?

A
A & C

B
A & B

C
only 'A'

D
A, B & C

Solution

A) Graphite has layer structure. Each layer is a planar sheet, composed of hexagonal rings of carbon atoms, with 3 electrons of each atom involved in strong single bonds with three adjacent atoms of hexagonal ring. The extra electron makes a very weak bond with the adjacent layer. Thus, one layer can slide over another and thus, it can be used as a lubricant.
B) The interlayer spacing is 3.35$$^0$$A, and not 1.42$$^0$$A. Since the layers are not joined by single bonds and 1.42$$^0$$A is the distance between two carbon atoms joined by single bond.
C) Silica is an acidic oxide and reacts only with bases and not with the acids like above.
Question 7
1 / -0

A. Silanes are good reducing agents.
B. $$SiO _{2}$$ is a giant tetrahedral polymer.
C. $$SnCl_{4}$$ acts as bronsted base.
The correct option is :

A
A and B are true

B
B and C are true

C
only C is true

D
all are true

Solution

Silanes are good reducing agents and $$SiO_2$$ shows giant tetrahedral structure.
$$SnCl_4 + 4H_2O \rightarrow Sn(OH)_4 + 4HCl$$
$$SnCl_4$$ acts as bronsted base. So all options are correct.
Question 8
1 / -0

STATEMENT-1 : $$SiO_{2}$$ is used as an acidic flux in metallurgy.
STATEMENT-2 : $$SiO_{2}$$ removes basic impurities forming a slag.

A
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

B
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

C
STATEMENT-1 is True, STATEMENT-2 is False

D
STATEMENT-1 is False, STATEMENT-2 is True

Solution

An acid flux is the chemicak substance which removes basic impurities.
$$SiO_2$$ is used as acidic flux in metallurgy.
$$SiO_2 + CaO \rightarrow CaSiO_3$$
Acidic Basic Slag
flux impurity
Question 9
1 / -0

Match list I and list II.
List -I      List -2
A)$$ SiCl_{4}$$ 1) $$ SiO_{2}$$
B) Tetrahedral   2) Acid-flux
C) Acheson's process   3) Lewis Acid
D) $$ SiO_{2}$$   4) Silicon
5)Graphite

A
1 2 4 3

B
5 1 2 4

C
3 1 5 2

D
3 2 4 1

Solution

$$SiCl_4$$ acts as lewis acid due to vacant d-orbitals.
$$SiO_2$$ has tetrahedral structure.
Acheson's process is one of the methods to prepare graphite.
$$SiO_2$$ is used as acid flux in metallurgy.
Question 10
1 / -0

The compounds of boron are :

A
mostly ionic

B
mostly covalent

C
both

D
crystalline

Solution

IE and electron affinity values of boron are too high to form ionic compounds.

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The p-Block Elements Test - 19

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The p-Block Elements Test - 19

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Question 1

Both A and R are true and R is correctexplanation of A

Both A and R are true and R is not correctexplanation of A

A is true but R is false

A is false but R is true.

Solution

Question 2

A and B are correct

B and C are correct

A and C are correct

all are correct

Solution

Question 3

A>B>C

B>A>C

B>C>A

A>C>B

Solution

Question 4

$$H_{4}SiO _{4}$$ and $$Na _{2} SiO_{3}$$

$$SiO _{2}$$ and $$Na _{2} S$$

$$SiO_{2}$$ and $$Na _{2}CO _{2}$$

$$SiO _{2}$$ and $$Na _{2} SiO_{3}$$

Solution

Question 5

C > Si > Ge > Pb > Sn

C > Si > Ge > Sn > Pb

C > Ge > Sn > Pb>Si

Si > C > Ge > Sn > Pb

Solution

Question 6

A & C

A & B

only 'A'

A, B & C

Solution

Question 7

A and B are true

B and C are true

only C is true

all are true

Solution

Question 8

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is False

STATEMENT-1 is False, STATEMENT-2 is True

Solution

Question 9

1 2 4 3

5 1 2 4

3 1 5 2

3 2 4 1

Solution

Question 10

mostly ionic

mostly covalent

both

crystalline

Solution

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Both A and R are true and R is correct

explanation of A

Both A and R are true and R is not correct

explanation of A