$$Na_{2}B_{4}O_{7}\, .\: 5H_{2}O$$

MCQ Questions for Class 11 Engineering Chemistry Some P-Block Elements Quiz 3

Hydrolysis of

$SiCl_{4}$ gives compound

$X$ and

$HCl$ . On heating to

$1000^{\circ}$

$C$ .

$X$ loses water and forms

$Y$ . Identify

$X$ and

$Y$ respectively.

0%
$SiO_{2}$ and $Si$
0%
$H_{4}SiO_{4}$ and $SiO_{2}$
0%
$SiO_{2}$ and $SiC$
0%
$H_{4}SiO_{4}$ and $Si$

Explanation

$H_4SiO_4 \xrightarrow {1000 \ ^oC} SiO_2 + H_2O$

On hydrolysis, all the bonds of

$SiCl_{4}$ are broken after which

$H^{+}$ forms bonds with

$Cl^{-}$ and

$OH^{-}$ forms bonds with

$Si^{4+}$ , thus forming

$H_{4}SiO_{4}$ and

$HCl$ .
On deep heating,

$H_{4}SiO_{4}$ loses two water molecules to form

$SiO_{2}$

Which of the following Si biocompatible

0%
Silicone
0%
Polythene
0%
Teflon
0%
Bakelite

Silica reacts with a hydride of superhalogen to form 'X'. On hydrolysis of 'X', another compound 'Y' is formed. 'Y' on heating at 1000

$^0$ C loses water to form 'Z'. The 'Z' can also be prepared by the reaction

0%
$SiO_{2}+4HF\rightarrow SiF_{2}+ 2H_{2}O$
0%
$Si+O_{2}\rightarrow SiO_{2}$
0%
$K_{2}SiF_{6}\rightarrow 6KF+Si$
0%
$Si+2NaOH+H_{2}\rightarrow Na_{2}SiO_{3}+2H_{2}$

Explanation

Silica reacts with hydride of superhalogen to form

$SiX_4$ which is compound X. It on hydrolysis gives

$Si(OH)_4$ which is compound Y. This on heating at

$1000^0C$ loses water to form

$SiO_2$ which is compound Z.

$SiO_2$ can also be prepared by the reaction of silicon with oxygen.

$Si+O_{2}\rightarrow SiO_{2}$

The high melting point of

$SiO_{2}$ is due to:

0%
$sp$ hybridised silicon atom
0%
tetrahedral polymer structure
0%
its solid state
0%
$Sp^{2}$ hybridised silicon atom

Which of the following bonds is present in silicone ?

0%
$Si-O-Si-O-Si$
0%
$Si-C-Si-O-Si$
0%
$Si-O-Si-C-Si$
0%
$Si-Si-Si-Si-Si$

Explanation

Explanation

Silicones belong to the organo-silicon polymer family since they are comprised entirely of carbon, silicon, and oxygen atoms.

The repeating unit on all silicones is $~{{R}_{2}}SiO$ , where R is the alkyl group.

Because silicon is a carbon family element, its atoms are tetravalent. Oxygen is a bivalent atom. Each silicon atom is connected to two oxygen atoms and two carbon atoms. In silicones, each silicon atom has two methyl groups attached to it.

As a result of the preceding construction, the bonding between silicon and oxygen atoms can be described as $Si-O-Si-O-Si$

Henceforth, the correct answer is option A.

$CCl_{4}$ does not act as lewis acid.
B. Silanes are strong reducing agents.
C. Crystalline silica will have Diamond like structure.
Which of the above statement(s) is/are true?

0%
A & B
0%
only 'A'
0%
only 'B'
0%
A, B & C

Explanation

$CCl_{4}$ is electron precise. Hence it is not a lewis acid.
Silanes possess

$Si-H$ bonds, in which the

$Si$ atom has a partial positive charge and the

$H$ atoms have a partial negative charge. Thus, silanes can donate a

$H^{-}$ ion, which can act as reducing agents.
Crystalline silica does have diamond like structure.

The most abundant metal among the following is:

0%
$Al$
0%
$Ca$
0%
$Fe$
0%
$K$

Explanation

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).

List -I List -2
Arrange properly.
A. Diamond Metal electrode
B. Graphite sp hybridized
C. Silica Acheson's process
D.

$CO_2$ Agate
Cutting of glass
The correct match is?

0%
2 1 3 4
0%
5 1 2 3
0%
5 3 4 2
0%
1 4 2 3

Explanation

Diamond is used for cutting glass as diamond is harder than glass.
Acheson process is used to prepare graphite from coke.

$CO_{2}$ is linear in shape and hence has

$sp$ hybridisation.
Silica is used as agate.

Option C is correct.

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

0%
Both A and R are true and R is correct

explanation of A
0%
Both A and R are true and R is not correct

explanation of A
0%
A is true but R is false
0%
A is false but R is true.

Decreasing order of p-orbital character in the following is:
A)

$SiO_{2}$ B)

$CO_{2}$ C) Graphite

0%
A>B>C
0%
B>A>C
0%
B>C>A
0%
A>C>B

Explanation

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

$CO_2$ .

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

In the above reaction, identify B and C respectively :

0%
$H_{4}SiO _{4}$ and $Na _{2} SiO_{3}$
0%
$SiO _{2}$ and $Na _{2} S$
0%
$SiO_{2}$ and $Na _{2}CO _{2}$
0%
$SiO _{2}$ and $Na _{2} SiO_{3}$

Explanation

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.

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

0%
C > Si > Ge > Pb > Sn
0%
C > Si > Ge > Sn > Pb
0%
C > Ge > Sn > Pb>Si
0%
Si > C > Ge > Sn > Pb

Which of the following is wrong about silica?

0%
Purest form of silica is know as quartz.
0%
Quartz is also know as rock crystal.
0%
Quartz contain hexagonal prisms.
0%
U.V. light cannot pass through quartz.

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 :

0%
A and B are true
0%
B and C are true
0%
only C is true
0%
all are true

Explanation

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.

STATEMENT-1 :

$SiO_{2}$ is used as an acidic flux in metallurgy.
STATEMENT-2 :

$SiO_{2}$ removes basic impurities forming a slag.

0%
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
0%
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1
0%
STATEMENT-1 is True, STATEMENT-2 is False
0%
STATEMENT-1 is False, STATEMENT-2 is True

Explanation

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

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?

0%
A & C
0%
A & B
0%
only 'A'
0%
A, B & C

Explanation

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.

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.

0%
A and B are correct
0%
B and C are correct
0%
A and C are correct
0%
all are correct

Explanation

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.

Anhydrous borax is:

0%
$Na_{2}B_{4}O_{7}$
0%
$Na_{2}B_{4}O_{7}\, .\: 5H_{2}O$
0%
$Na_{2}B_{4}O_{7}\,. 10H_{2}O$
0%
$Na_{2}B_{4}O_{7}.8H_{2}O$

Explanation

Anhydrous borax has formula

$Na_2B_4O_7$ .

Anhydrous means no water of crystallisation.

Option A is correct.

Borax is known as :

0%
Sodium borate.
0%
Sodium tetra borate.
0%
Disodium tetra borate.
0%
All

Explanation

Borax:

$Na_{2} [ B_{4} O_{5} (OH)_{4} ].8H_{2}O$

In the aluminothermic process, aluminium acts as:

0%
an oxidising agent
0%
a.flux
0%
a reducing agent
0%
a solder

Explanation

Aluminium gets oxidized.
The equation is,

$2Al+Fe_{2}O_{3}\rightarrow 2Fe_{(molten)}+Al_{2}O_{3}$

Borax is the sodium salt of:

0%
Ortho boric acid.
0%
Meta boric acid.
0%
Tetra boric acid.
0%
Pyro boric acid.

Explanation

Borax (

$Na_2B_4O_7.10H_2O$ ) is sodium salt of tetraboric acid (

$H_2B_4O_7$ ).

Match the List-I with List-II.

List -I List -II
a) Phosgene p) Sodium silicate
b) Water glass q) A poisonous gas
c) Carbon monoxide r) Fire extinguisher
d) Carbon dioxide s) Carbonyl ligand

0%
a-q; b-p; c-s; d-r
0%
a-r; b-q; c-r; d -p,q
0%
a-s; b-p; c-q; d-r,
0%
a-r; b-s; c-q; d-p

Explanation

Phosgene (

$COCl_2$ ) is a poisonous gas that was used during the first world war.

Sodium silicate (

$Na_2SiO_3$ ) is also called water glass.

Due to the presence of lone pair on the carbon atom,

$CO$ acts as ligand.

$CO_2$ is does not support combustion so it is used in fire extinguisher.

so option A is correct.

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

0%
1 2 4 3
0%
5 1 2 4
0%
3 1 5 2
0%
3 2 4 1

Explanation

$SiCl_4$ 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.

The form of

$SiO_{2}$ used as piezoelectric material :

0%
cristobalite
0%
quartz
0%
kiesulguhr
0%
silica gel

The material used in solar cell contains :

0%
Cs
0%
Si
0%
Sn
0%
T

List -I List -2
A. The purest form of

$SiO_{2}$ Quartz
B. Carborundum Graphite
C. Corundum

$SiC$
D. Hexagonal rings

$Al_{2} O_{3}$
Diamond

Correct match with A B C D is:

0%
1 3 4 2
0%
3 1 4 2
0%
1 3 2 4
0%
3 1 4 3

Explanation

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.

List -I List -2
A. Reactive form of Carbon HF
B. Acid employed for Diamond
etching of glass
C. Synthesis gas Charcoal
D. Unreactive form of Carbon

$CO H_{2}$
The correct match with A B C D is ?

0%
1 2 3 4
0%
2 4 3 1
0%
2 1 3 4
0%
3 1 4 2

Explanation

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.

Which of the following does not exhibit inert pair effect?

0%
Bi
0%
Pb
0%
B
0%
Tl

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

0%
B, Tl
0%
B, In
0%
B, Al
0%
B, Ga

Explanation

$I.P.$ does not smoothly decrease down in

$III-A$ group. The decrease from B to Al is associated with increase in size. The observed discontinuity in the

$I.P.$ values between

$Al$ and

$Ga$ , and between

$In$ and

$Tl$ are due to the inability of d- and f-electrons, which have low screening effect to compensate the increase in nuclear charge.

Element with a giant molecular structure is:

0%
B
0%
Al
0%
Ga
0%
Tl

Sources of Boron are:

0%
Kernite
0%
Borax
0%
Colemanite
0%
$B_{2}O_{3}$

The compounds of boron are :

0%
mostly ionic
0%
mostly covalent
0%
both
0%
crystalline

Which of the following is the correct formulae of
borax ?

0%
$Na[B_{4}O_{5}\left ( OH \right )_{4}].2H_{2}O$
0%
$Na[B_{4}O_{5}\left ( OH \right )_{4}].4H_{2}O$
0%
$Na[B_{4}O_{5}\left ( OH \right )_{4}].6H_{2}O$
0%
$Na_{2}[B_{4}O_{5}\left ( OH_{} \right )_{4}].8H_{2}O$

Bauxite is an oxide mineral of:

0%
barium
0%
boron
0%
bismuth
0%
aluminium

Explanation

Correct Answer: Option D

Explanation:

The members of Group 13 elements are:

Boron, Alumunium, Gallium, Indium and Thallium

Generally, the oxidation state of members of boron family is $+3$ and $+1$ . On moving down the group, the tendency to form $+1$ ion increases due to the inert pair effect.

Every element of the Boron family reacts with oxygen at high temperature forming trioxides, $M_2O_3$ .

$4M\left ( s \right ) + 6O_2\left ( g \right ) \rightarrow 2M_2O_3\left ( s \right )$

Bauxite is an oxide mineral of aluminium.

Chemical formula $Al_2O_3⋅2H_2O$ .

Hence Bauxite is an oxide mineral of Aluminium.

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.

0%
Both Assertion and Reason are true and Reason is the correct explanation of Assertion
0%
Both Assertion and Reason are true and Reason is not the correct explanation of Assertion
0%
Assertion is true but Reason is false
0%
Assertion is false but Reason is true.

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

0%
decrease in ionization energy of the elements
0%
increase in size of the elements
0%
ineffective shielding of nuclear charge by d-electrons in case of Ga, In and Tl
0%
decrease in electronegativity of the elements

Explanation

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.

Which of the following property does not support anamolous behaviour of boron?

0%
Small size
0%
High electronegativity
0%
High ionization energy
0%
Formation of trihalides

Explanation

Formation of trihalides does not support anomalous behaviour of boron as other elements of boron family also form trihalides such as

$\displaystyle AlCl_3$ .

Identify the compound

$A$ in the reaction given below.

$Na_{2}B_{4}O_{7}+2H_{2}O\rightarrow 2NaOH+A$

0%
Orthoboric acid
0%
Metaboric Acid
0%
Tetraboric Acid
0%
Pyroboric acid

Explanation

A reaction is given below:

$Na_2B_4O_7+2H_2O\rightarrow 2NaOH+A$

In the above reaction,

$A$ is tetraboric acid.

The complete reaction is as follows:

$Na_{2} B_{4} O_{7} + 2 H_{2} O \rightarrow 2NaOH + H_2B_4O_7$

Hence, option C is the correct answer.

The aqueous solution of borax turns red litmus to:

0%
blue
0%
No Change
0%
red
0%
white

Aqueous solution of borax is _______ in nature.

0%
Neutral
0%
Acidic
0%
Alkaline
0%
None

Which of the following reaction will yield crystalline variety of boron?

0%
$B_{2}O_{3}+2Al$
0%
$BCl_{3}\: +H_{2}$
0%
$B_{2}O_{3}+3Mg$
0%
$2B+3MgO$

Explanation

Small amounts of crystalline Boron can be obtained by the following reaction using red hot W or Tantalum filament as catalyst.

$2BX_{3} + 3H_{2} \rightarrow 2B + 6HX, X=Cl, Br$

Which of the following elements can present both in cation and anion of its salts?

0%
$B$
0%
$Al$
0%
$Ga$
0%
$Tl$

Explanation

Boron can only form cation of its salts.

$Al$ can form both cation and anion of its salts.

Example:

$Al^{3+}$ (cation)

$AlO_{3}^{3-}$ (Anion)

$Ga$ and

$Tl$ can only form cation of their salts.

An element R belongs to III A group. Which is true with respect to R?

0%
It is a gas at room temperature
0%
It has an oxidation state of +4
0%
It forms an oxide of the type $R_{2}\, O_{3}$
0%
It forms a halide of the type $Rx_{2}$

Explanation

An IIIA group element has valency 3. So they form an oxide of type

$R_{2}O_{3}$ .

Which of the following metal forms a protective oxide layer to prevent corrosion?

0%
$Cu$
0%
$Mg$
0%
$Al$
0%
$Mn$

Explanation

$Al_{2}O_{3}$ is formed which forms a protective layer on the metal

When borax is heated strongly it gives:

0%
$B_{2}O_{3}$
0%
$Na_{2}B_{4}O_{7}$
0%
$NaBO_{2}+B_{2}O_{3}$
0%
none of the above

Explanation

$NaBO_{2} + B_{2} O_{3}$ is the glassy bead that is formed on heating borax.

$AlO^{-}_{2}$ ion in aqueous solution exists as :

0%
$[Al\left ( OH \right )_{4}]^{-}$
0%
$[Al\left ( OH \right )_{4}\left ( H_{2}O \right )]^{-}$
0%
$[Al\left ( OH \right )_{4}\left ( H_{2}O \right )_{2}]^{-}$
0%
$[Al\left ( OH \right )_{6}]^{-}$

Explanation

Aluminate ion is isolated as

$AlO_{2}^{-}$ .
When in solution, this is surrounded by 4 water molecules, then two water molecules react with the oxygen to give overall 4

$OH^{-}$ groups. These four hydroxyl groups and two water molecules are held by coordinate bonds in the complex.
Hence, the formula becomes

$[Al(OH)_{4}(H_{2}O)_{2}]^{-}$ .

The valencies of nitrogen and boron in borazole are :

0%
$3,3$
0%
$4,4$
0%
$3,4$
0%
$4,3$

Oxidation of aluminium is :

0%
reversible
0%
exothermic
0%
endothermic
0%
slow reaction

Explanation

Al has a strong affinity for

$O_{2}$ . Oxidation of Al is a reaction of negative enthalpy(exothermic).

The inactiveness of aluminium in the presence of conc.

$HNO_{3}$ is due to the formation of a layer of :

0%
$AlN$
0%
$Al_{2}S_{3}$
0%
$Al_{4}C_{3}$
0%
$Al_{2}O_{3}$

Explanation

A passive layer of aluminium oxide is formed by the reaction of aluminium with concentrated nitric acid.

$2Al+6\underset {conc.}{HNO_{3}}\rightarrow Al_{2}O_{3}+6NO_{2}+3H_{2}O$

Therefore, option D is correct.

The compound used in etching of glass is:

0%
$\mathrm{H}\mathrm{F}$
0%
$\mathrm{H}\mathrm{Cl}\mathrm{O}_{4}$
0%
$\mathrm{H}\mathrm{N}\mathrm{O}_{3}$
0%
$\mathrm{H}_{2}\mathrm{S}\mathrm{O}_{4}$

Explanation

HF is used in etching of glass.

$4\ HF+SiO_2\rightarrow Si\ F_4+2H_2O$

HF reacts with

$SiO_2$ present in glass.

Option A is correct.

CBSE Questions for Class 11 Engineering Chemistry Some P-Block Elements Quiz 3 - MCQExams.com

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Practice Class 11 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Chemistry Some P-Block Elements Quiz 3 - MCQExams.com

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Practice Class 11 Engineering Chemistry Quiz Questions and Answers

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