an ionic compound consisting of $$Tl^{+}$$ and $$I_{3}^{-}$$ ions

an ionic compound consisting of $$Tl^{3+}$$ and three $$3I^{-}$$ ions

a mixture of $$TlI$$ and $$I_{2}$$

each silicon atom is surrounded by two oxygen atom and each oxygen atom is bonded to two silicon atoms

silicon atom is bonded to two oxygen atoms

there are double bonds between silicon and oxygen atoms

MCQ Questions for Class 11 Engineering Chemistry Some P-Block Elements Quiz 4

Which of the following represent prismatic borax?

0%
$Na_{2}B_{4}O_{7}.10H_{2}O$
0%
$Na_{2}B_{4}O_{7}.5\, H_{2}\, O$
0%
$Na_{2}B_{4}O_{7}.3H_{2}O$
0%
$Na_{2}B_{4}O_{7}.4H_{2\, }O$

Explanation

Prismatic borax is

$Na_2B_4O_7. 10H_2O$ .

$CaCO_3+SiO_2\rightarrow$ __________.

0%
$CaO+SiCO_3$
0%
$CaSiO_3+SiO_2$
0%
$CaSiO_3+CaO$
0%
$CaSiO_3+CO_2$

Explanation

The reaction in between calcium carbonate and silica is as follows :

$CaCO_3+SiO_2\rightarrow$

$CaSiO_3+CO_2$

$Ca_{2}B_{6}O_{11}+2Na_{2}CO_{3}\rightarrow X + 2CaCO_{3} + 2Na_{2}BO_{2}$ . The compound X in the above reaction is :

0%
$Na_{2}B_{4}O_{7}$
0%
$HBO_{2}$
0%
$H_{3}BO_{3}$
0%
$H_{2}B_{4}O_{7}$

Explanation

This is a method of preparation of borax. Colemanite powder is heated with

$Na_{2} CO_{3}$ solution giving a precipitate of

$CaCO_{3}$ . The filtrate is cooled and crystals of borax are precipitated.

In aluminate ion, the coordination number of aluminium is:

0%
2
0%
4
0%
6
0%
1

Explanation

Aluminate ion isolated is

$AlO_{2}^{-}$ .
When in solution, this is surrounded by 4 water molecules, 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 compound has 6 coordinate bonds.

Which of the following reacts with nitrogen when heated in air:

0%
Al
0%
C
0%
Na
0%
K

Explanation

$2Al + N_{2} \rightarrow 2AlN$

Thus, Al reacts with nitrogen when heated in air to form aluminium trinitride.

The reducing character of Al, Ga, In, Tl increases in which of the following order?

0%
$Al< Tl< Ga> In$
0%
$Ga< Al< Tl< In$
0%
$Tl< In< Ga< Al$
0%
$Al< Ga< In< Tl$

$Al$ reduces most of the metallic oxides due to its greater affinity for:

0%
oxygen
0%
metals
0%
electrons
0%
protons

Explanation

It reduces most oxides due to its affinity to oxygen

$2Al+\frac {3}{2}O_{2}\rightarrow Al_{2}O_{3}$

$\Delta G^{0}=-1582.3kJ/mole$

When HF reacts with glass, the compound formed is:

0%
${F}_{2}$
0%
$NaF$
0%
$Na_{3}AlF_{3}$
0%
$Na_{2}SiF_{6}$

Explanation

When glass is attacked by

$HF$ , it forms

$Na_2SiF_6$ and this property is used for the etching of glass.

The reaction is given below:

$Na_2SiO_3 + 6HF \rightarrow Na_2SiF_6+ 3H_2O$

Option D is correct.

Which is true for an element R present in III A group of the periodic table?

0%
It forms halide of type $RX_{3}$
0%
It has oxidation state of +3
0%
It forms oxide of type $R_{2}O_{3}$
0%
.All the above

Explanation

Given

$R$ is an element in the

$IIIA$ group of the periodic table, so it has a maximum oxidation state of

$+3$

$A)$ As halide is monovalent, the element

$R$ forms halides of the form

$R{ X }_{ 3 }$

$B)$ It has an oxidation state of

$+3$ because it has 3 valence electrons, so this option is true.

$C)$ As oxygen is divalent so the element

$R$ forms oxides of the type

${ R }_{ 2 }{ O }_{ 3 }$ .

So, all the above options are true.

Consider the reaction,

$LiH+AlH_{3}\rightarrow .....$
Which is the incorrect statement about this reaction?

0%
The product is a good reducing agent
0%
$AlH_{3}$ acts as a lewis acid
0%
There is decrease in the oxidation number of aluminium in the product
0%
$LiH$ donates $H^{-}$ to $AlH_{3}$

Explanation

There is no change in the oxidation state of Aluminium during the reaction

The reaction is as follow:

$LiH + AlH_{3} \rightarrow LiAlH_{4}$

The oxidation state of Al in both reactant as well as in product is -3.

Which amongst the following is also called as a sesqui oxide?

0%
$B_{2}O_{3}$
0%
$Al_{2}O_{3}$
0%
$Tl_{2}O_{3}$
0%
All of the above

Explanation

Sesquioxide is an oxide in which the ratio of atoms of oxygen and other atoms should be

$3:2$ like

$B_2O_3, Al_2O_3, Tl_2O_3$ .

Which of the following forms

$M_{2}O$ type of oxide?

0%
Al
0%
B
0%
Tl
0%
Ca

Explanation

$Tl^{+}$ (due to inert pair effect) is stable and forms compounds like

$Tl_{2}O$ .

Which of the following species cannot exist?

0%
$[BF_{6}]^{3-}$
0%
$[AlF_{6}]^{3-}$
0%
$[GaF_{6}]^{3-}$
0%
$[InF_{6}]^{3-}$

Explanation

$B^{3+}$ , being small in size, does not have empty d-orbitals like other atoms to accommodate 6 electron pairs of 6

$F^{-}$ ions.

Which of the following oxides dissolves in both hydrochloric acid and sodium hydroxide solution? (1993)

0%
Na
0%
MgO
0%
BaO
0%
$Al_{2}O_{3}$

Explanation

$Al_{2}O_{3}$ is an amphoteric oxide

Which metal burns in air at high temperature with
the evolution of much heat?

0%
Cu
0%
Al
0%
Hg
0%
Pb

Catenation i.e., linking of similar atoms depends on the size and electronic configuration of atoms. The tendency of catenation in Group 14 elements follows the order:

0%
C < Si < Ge < Sn
0%
C > Si > Ge > Sn
0%
Si > C > Sn > Ge
0%
Ge > Sn > Si > C

Explanation

Down the group, atomic size increases and electronegativity decreases. Therefore tendency of catenation decreases down the group as follows

$C > Si > Ge > Sn$ .

Which of the following oxidation states are the most characteristic for lead and tin respectively:

0%
$+2,+2$
0%
$+4, +2$
0%
$+2, +4$
0%
$+4, + 4$

Explanation

Lead and tin are in the fourth group of the periodic table,so we can expect an oxidation state of

$+4$ for both of them.But due to the reluctance of participation of electrons in the outermost

$s$ orbital in bonding i.e., inert pair effect which increases down the group.So, the oxidation states of lead and tin are

$+2$ and

$+4$ respectively which is also evident from their respective chemical reactions.

The repeating structural units in silicone is:

0%
$(R_2SiO-)$
0%
$SiO_{2}$
0%
$R-O-Si-\underset {\overset {|}R}O-$
0%
$(-R_2Si-O-O-R)$

The tendency of BF

$_{3}$ , BCl

$_{3}$ and BBr

$_{3}$ to behave as Lewis acid decreases in the sequence is:

0%
BCl $_{3}$ > BF $_{3}$ > BBr $_{3}$
0%
BBr $_{3}$ > BCl $_{3}$ > BF $_{3}$
0%
BBr $_{3}$ > BF $_{3}$ > BCl $_{3}$
0%
BF $_{3}$ > BCl $_{3}$ > BBr $_{3}$

Explanation

A molecule is said to be a Lewis acid if it can accept a lone pair.

This anomaly is explained on the basis of the relative tendency of the halogen atom to back donate its unutilised electrons to vacant p-orbital of boron atom. In BF3, boron has a vacant 2p-orbital and each flourine has fully filled unutilised 2p-orbitals. Fluorine transfers two electrons to vacant 2p-orbital of boron, thus forming pπ-pπ bond.

The tendency to back donate decreases from F to I as energy level difference between B and halogen atom increases from F to I.

So,the order of Lewis acidity is

$B{ Br }_{ 3 }>{ B }{ Cl }_{ 3 }>{ B }{ F }_{ 3 }$ .

Hence, the correct option is

$B$ .

The abundance of B, Al, Ga and Tl in earth crust follows the order:

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

What is the value of X in the given chemical formula of crystalline borax,

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

0%
5
0%
6
0%
7
0%
8

Explanation

Borax is white crystalline solid of formula

$Na_2B_4O_7.10H_2O$ which can also be written as

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

Which of the following is the electron deficient molecule?

0%
$C_{2}H_{6}$
0%
$B_{2}H_{6}$
0%
$SiH_{4}$
0%
$PH_{3}$

Explanation

$B_2H_6$ is the electron deficient molecule which completes its deficiency by forming Bridge or Banana
Bond. These bonds are also called $3e-3e$ bonds

Quartz is a crystalline variety of:

0%
Silicon carbide
0%
Sodium silicate
0%
Silica
0%
Silicon

Explanation

Quartz is a covalent crystal having a framework of silicates or silica, i.e. a three-dimensional network where all the four oxygen atoms of each of

$SiO_{4}$ tetrahedron are shared.

When borax is dissolved in water :

0%
$B(OH)_{3}$ is formed only
0%
$[B(OH)_{4}]^{-}$ is formed only
0%
both $B(OH)_3$ and $[B(OH)_{4}]^{-}$ are formed
0%
$[B_{3}O_{3}(OH)_{4}]^{-}$ is formed only

Explanation

Hint:

$Boric \ acid$ is a weak Lewis acid.

Explanation:

When

$borax$ is dissolved in water,

$sodium \ hydroxide$ and

$boric \ acid$ are formed:

$Na_2B_4O_7 + H_2O \rightarrow NaOH + H_3BO_3$

Since

$boric \ acid$ is a weak Lewis acid, it can accept an electron pair from a hydroxide ion.

$H_3BO_3 + OH^- \rightarrow [B(OH)_4]^-$

The correct answer is option $C$ .

The structure of silicon(IV) oxide belongs to the type:

0%
ionic lattice
0%
macromolecular, with a layer structure
0%
molecular lattice, with van der Waals forces among the molecules
0%
macromolecular, with a non-layer structure

Explanation

$Si(IV)$ oxide is macromolecular with a non layer structure similar to diamond.

$TlI_{3}$ is:

0%
an ionic compound consisting of $Tl^{3+}$ and three $3I^{-}$ ions
0%
a covalent compound
0%
an ionic compound consisting of $Tl^{+}$ and $I_{3}^{-}$ ions
0%
a mixture of $TlI$ and $I_{2}$

Explanation

${T}{l}{I}+{I}_{2}\rightarrow {T}{l}^{+}+{I}_{3}^{-}({T}{l}{I}_{3})$
So, it is an compound because of +1 oxidation state of Tl which is more ionic than +3 oxidation state.

Which of the following curves assures that the metal obtained by carbon reduction is in the vapour state?

Explanation

Since after boiling point of M plot

$C\rightarrow CO$ is below

$M\rightarrow MO$ , Hence metal can be obtained in vapour state.
1098562_73761_ans_26e2c9de527e40d384bb666d00c77fbc.C

Aluminium hydroxide is amphoteric.

0%
True
0%
False

Explanation

Amphoteric oxides are both acidic and basic in nature.

Aluminium hydroxide is amphoteric. It dissolves in acid forming

$\left[ Al\left( { H }_{ 2 }O \right) _{ 6 } \right] ^{ 3+ }$ . It also dissolves in strong alkali forming

$\left[ Al\left( OH \right) _{ 4 } \right] ^{ - }$

The gaseous product(s) expected at room temperature by reaction of sodium borohydride an boron trifluoride under anhydrous conditions is/are :

0%
$H_2$
0%
$B_2H_6$ and $H_2$
0%
$B_2H_6$
0%
$BH_2F$ and $H_2$

Explanation

$3Na[{ BH }_{ 4 }]+{ 4BF }_{ 3 }\longrightarrow { 2B }_{ 2 }{ H }_{ 6 }(g)+3Na[{ BF }_{ 4 }]$

$B_2H_6$ is the gaseous product

Name the type of the structure of silicate in which one oxygen atom of

$\left [ SiO_{4} \right ]^{4-}$ is shared:

0%
linear chain silicate
0%
sheet silicate
0%
pyrosilicate
0%
three dimensional silicate

Usually a disilicate share only one oxygen of silicate unit. But if in the disilicate, two

$O$ atoms are shared, then formula of its salt with potassium is :

0%
$K_6Si_2O_7$
0%
$K_{4}Si_2O_6$
0%
$K_2Si_2O_6$
0%
$K_8Si_2O_8$

Formula of Pyro silicic acid:

0%
$H_2Si_4O_9$
0%
$H_4Si_2O_6$
0%
$H_4Si_3O_8$
0%
$H_6Si_2O_7$

Explanation

Hydrated Silica is a form of silicon dioxide, which has a variable amount of water in the formula. It is also known as silicic acid, a term usually used for its form dissolved in water.

$H_6Si_2O_7$ is chemical formula of pyro silicic acid. Chemical Formula:

$SiO_{2n}H_2O$ .

Read the following short write-up and answer the questions at the end of it The name 'silica' covers an entire group of minerals, which have the general formula

$SiO_{2}$ , the most common of which is quartz. Quartz is a framework silicate with

$SiO_{4}$ tetrahedra arranged in spirals. The spirals can turn in a clockwise or anticlockwise direction- a feature that results in there being two mirror images, optically active, varieties of quartz.

The following pictures represent various silicate anions. Their formulae for silicon and oxygen respectively:

0%
$SiO_{3}^{2-}\:Si_{3}O_{7}^{2-}$
0%
$SiO_{4}^{4-}\:Si_{3}O_{10}^{8-}$
0%
$SiO_{4}^{2-}\:Si_{3}O_{9}^{2-}$
0%
$SiO_{3}^{4-}\:Si_{3}O_{7}^{8-}$

Explanation

The formulas are represented as

$SiO_{4}^{4-}\:Si_{3}O_{10}^{8-}$ .
Silicon is at the center of a tetrahedron and 4 oxygen atoms are present at four corners.
Hence,

$SiO_{4}^{4-}$ represents the formula of the silicate ion. When three tetrahedra are shown linked together, they share corner oxygen atoms. Hence, the formula of the ion becomes

$Si_{3}O_{10}^{8-}$ .

Borax is actually made of two tetrahedral and two triangular units joined together and should be written as:

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

Consider the following statements about borax.

0%
Each boron atom has four $B-O$ bonds
0%
Each boron atom has three $B-O$ bonds
0%
Two boron atoms have four $B-O$ bonds while other two have three $B-O$ bonds
0%
Each boron atom has one $-OH$ groups

Explanation

Structure of

$[B_{4}O_{5}(OH)_{4}]^{2-}$ is shown in figure. From figure it is evident that statement 3 and 4 are correct.

The possible oxidation state of Tl are:

0%
+1 and +2
0%
+2 and +3
0%
+1 and -1
0%
+1 and +3

Explanation

$Tl$ shows both +1 and +3 oxidation states. The stability of +1 oxidation state increases down the group

$Tl>In>Ga>Al$ and +3 oxidation state is highly oxidising in character.

In silicon dioxide :

0%
each silicon atom is surrounded by four oxygen atoms and each oxygen atom is bonded to two silicon atoms
0%
each silicon atom is surrounded by two oxygen atom and each oxygen atom is bonded to two silicon atoms
0%
silicon atom is bonded to two oxygen atoms
0%
there are double bonds between silicon and oxygen atoms

Explanation

Hint: In silicon dioxide, each $Si$ atom is tetrahedrally surrounded by $4$ $O$ atoms.

Correct Answer: Option

$A$

Explanation:

As we can see from the figure, each

$Si$ atom is surrounded by four oxygen atoms and each oxygen atom is bonded to two silicon atoms.

The repeating structural unit in silicone is:

0%
$R_2SiO$
0%
$SiO_2$
0%
$SiO_3$
0%
All of the above

Explanation

The repeating structural unit in silicone is

$R_2SiO$ .
Silicones are organosilicon polymers containing

$Si-O-Si$ linkages.

If we start with

$MeSiCl_{3}$ as the starting material, silicones formed is :

0%
0%
0%
both of the above
0%
none of the above

Explanation

The silicone obtained from

$MeSiCl_{3}$ will have the structure as represented by option B.
The repeating unit is

$MeSiO_3$ .

Which of the following are true about Silicones?

0%
They are formed by the hydrolysis of ${ R }_{ 2 }{ SiCl }_{ 2 }$
0%
They are polymer made by ${ R }_{ 2 }{ SiO }_{ 2 }$ units
0%
They are made up of ${ R }_{ 2 }{ SiO }_{ 2 }$ units
0%
They are macromolecules

Explanation

Silicones are the macromolecular polymers formed by the hydrolysis of

${ R }_{ 2 }{ SiCl }_{ 2 }$ .
The hydrolysis of

${ R }_{ 2 }{ SiCl }_{ 2 }$ will give dialkyl silanediol which on polymerization will give a linear polymer.
These polymers are made by

${ R }_{ 2 }{ SiO }_{ 2 }$ units.
Thus, the options A, B, and D are correct.

An inorganic compound (A) made of two most occurring elements into the earth crust, having a polymeric tetrahedral network structure. With carbon, compound (A) produces a poisonous gas (B) which is the most stable diatomic molecule. Compounds (A) and (B) will be :

0%
$SiO_{2},CO_{2}$
0%
$SiO_{2},CO$
0%
$SiC,CO$
0%
$SiO_{2},N_{2}$

Explanation

$SiO_2 + 2C \rightarrow Si + 2CO$
Compound A is silica (

$SiO_2$ ) made of

$Si$ and

$O_2$ , two most occuring elements into the earth crust, having a polymeric tetraheadral network structure.
Compound B is

$CO$ , a poisonous gas which is the most stable diatomic molecule.

$CO_2$ is gas, while

$SiO_2$ is solid because :

0%
$CO_2$ is a linear molecule, while $SiO_2$ is angular.
0%
van der Waals' forces are very strong in $SiO_2$ .
0%
$CO_2$ is covalent, while $SiO_2$ is ionic.
0%
$Si$ cannot form stable bonds with $O$ , hence $Si$ has to form a $3D$ lattice.

Explanation

$SiO_2$ is solid because

$Si$ cannot form stable bonds with

$O$ , hence

$Si$ has to form a

$3D$ lattice of

$SiO_2$ molecules and form solid structure.

Borax is actually made of two tetrahedral and two triangular units joined together and should be written as:

$\displaystyle Na_{2}[B_{4}O_{5}(OH)_{4}]\cdot 8H_{2}$ O
Consider the following statements about borax. Which of the following is the correct statement?

0%
Each boron atom has four $B-O$ bonds
0%
Each boron atom has three $B-O$ bonds while other two have three $B-O$ bonds
0%
Each boron atom has four $B-O$ bonds while other two have three $B-O$ bonds
0%
Each boron atom has one $-OH$ groups

${SiO_2}$ reacts with :

0%
${Na_2 CO_3}$
0%
${CO_2}$
0%
$HF$
0%
$HCl$

Explanation

$SiO_{2}(s)+4HF(\ell)\rightarrow SiF_{4}(\ell)+2H_{2}O(\ell)$

$SiO_{2}(s)+Na_{2}CO_{3}(s)\rightarrow Na_{2}SiO_{3}(s)+CO_{2}(g)$

Select the correct statement about silicates:

0%
Cyclic silicate having three Si atoms contains six Si---O---Si linkages
0%
${\displaystyle2\frac{1}{2}}$ ver oxygen atoms of phedron unit are shared in double chain silicate
0%
( ${Si_2O_5})^{2n-}_{n}$ is formula of double chain silicate
0%
${SiO^{4-}_{4}}$ units polymerize to form silicate because Si atom has less tendency to form ${\pi}$ - bond with oxygen

Explanation

From figure it can be seen that:
2(1/2) oxygen atoms of per tetrahedron unit are shared in double chain silicate.

${SiO^{4-}_{4}}$ units polymerize to form silicate because Si atom has less tendency to form

${\pi}$ - bond with oxygen.

$RCl \displaystyle \xrightarrow[Si]{cu-powder} R_{2}SiCl_{2} \xrightarrow{H_{2}O} R_{2}Si\left ( OH \right )_{2} \xrightarrow{condensation} A$
Compound ( A ) is :

0%
a linear silicone
0%
a chlorosilane
0%
a linear silane
0%
a network silane

Explanation

The hydrolysis of

${ R }_{ 2 }{ SiCl }_{ 2 }$ will give dialkyl silanediol which on polymerization will give a linear polymer.

$RCl \displaystyle \xrightarrow[Si]{Cu-powder} R_{2}SiCl_{2} \xrightarrow{H_{2}O} R_{2}Si\left ( OH \right )_{2} \xrightarrow{condensation} \underset {A,\ linear \ polymer} {-O-R_2Si-O-R_2Si-O-R_2Si-O-}$
Compound (A) is a linear silicone.

Silicate having one monovalent corner oxygen atom in each tetrahedron unit is:

0%
sheet silicate
0%
cyclic silicate
0%
single chain silicate
0%
double chain silicate

Explanation

The general formula of Sheet or Phyllo silicates is (Si

$_2$ O

$_5$ )n

$^{2n-}$ . Each SiO

$_4$ tetrahedron shares three oxygen atoms with others and thus by forming two-dimensional sheets. These silicates can be cleaved easily just like graphite.

Hence, the correct option is

$\text{A}$

Select correct statement:

0%
Oxides of carbon family $(MO_2)$ are all network solids with octahedral coordination.
0%
silicon dioxide(silica) is a network solid with tetrahedral coordination and is a giant molecule.
0%
$GeO_2, SnO_2$ and $PbO_2$ are all acidic oxides.
0%
None of the above.

Explanation

Silica has a three-dimensional network structure. In silica, silicon is

$sp^3$ -hybridized and is thus linked to four oxygen atoms and each oxygen atom is linked to two silicon atoms forming a three-dimensional giant molecule as shown in figure. This three-dimensional network structure imparts stability to

$SiO_2$ crystal and hence a large amount of energy is required to break the crystal resulting in high melting point.

Which of the following cuts ultraviolet rays?

0%
Soda glass.
0%
Crookes glass.
0%
Pyrex.
0%
None of these.

Stability of hydrides of carbon family is in order :

0%
$CH_4 > SiH_4 > GeH_4 > SnH_4 > PbH_4$
0%
$CH_4 < SiH_4 < GeH_4 < SnH_4 < PbH_4$
0%
$CH_4 > SnH_4 > GeH_4 > SiH_4 > PbH_4$
0%
None of these

Explanation

Hydrides of carbon are more stable since electronegativity values of carbon and hydrogen are nearly same.

Hydrides of silicon are less stable because in silanes hydrogen withdraw electrons from Si-H bond.

As we move down, electronegativity decreases so stability decreases.

$CH_4 > SiH_4 > GeH_4 > SnH_4 > PbH_4$ is correct order.

Silica will react in a similar way with metal sulphates (and Phosphates) as it does with carbonates, giving a solid residue and a gas.

$1$ mole of silica reacts with

$1$ mole of sodium sulphate,

$Na_2SO_4$ . Write the equation:

0%
$SiO_2(s)+Na_2SO_4(s)\rightarrow Na_2SiO_3(s)+SO_3(g)$
0%
$SiO_2(s)+Na_2SO_4(s)\rightarrow Na_2SiO_4(s)+SO_2(g)$
0%
$SiO_2(s)+Na_2SO_4(s)\rightarrow Na_2SiO_2(s)+SO_4(g)$
0%
None of these

Explanation

The balanced equation is as follows:

$SiO_2(s)+Na_2SO_4(s)\rightarrow Na_2SiO_3(s)+SO_3(g)$

CBSE Questions for Class 11 Engineering Chemistry Some P-Block Elements Quiz 4 - 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 4 - MCQExams.com

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Practice Class 11 Engineering Chemistry Quiz Questions and Answers

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