End to end overlap of orbitals takes place

Sidewise as well as end to end overlap of orbitals take place

Sidewise overlap of orbitals take place

MCQ Questions for Class 11 Medical Chemistry Chemical Bonding And Molecular Structure Quiz 10

The relatively high boiling point of

$HF$ is due to

0%
Hydrogen bonding
0%
Covalent bonding
0%
Unshared electron pair on F
0%
Being a halogen acid

Explanation

Since fluorine is more electronegative than nitrogen, H – F bonds are much more polar than N – H bonds. Due to the high partial charges on H and F in HF,

$\textbf{ a Hydrogen bond}$ between HF molecules is much stronger than that between NH3 molecules. The high boiling point of HF is thus due to the strong H-bond

The enolic form of acetone contains:

0%
9 sigma bonds, 1 pi bond and 2 lone pairs
0%
8 sigma bonds, 2 pi bond and 2 lone pairs
0%
10 sigma bonds, 1 pi bond and 1 lone pair
0%
9 sigma bonds, 2 pi bond and 1 lone pair

Explanation

$\quad { C }H_{ 3 }-\overset { \underset { | }{ :\overset { .. }{ O } -H } }{ { C } } =CH_{ 2 }$

It has

$9 \sigma , 1 \pi$ and 2 lone pairs.

Which of the following option is correctly matched

0%
$S_{3}O_{9}$ : contains no S-S linkage
0%
$S_{2}O^{2}_{6 }$ ; contains -O-O-linkage
0%
$(HPO_{3})_{3}$ ; contains P-P linkage
0%
$S_{2}O^{2-}_{8}$ ; contains S-S linkage

Explanation

$S_3O_9$ have contain S-S limkage

Water has high heat of vaporisation due to:

0%
Covalent bonding
0%
H-bonding
0%
Ionic bonding
0%
None of the above

Explanation

The energy required to break multiple

$\text{Hydrogen bonds}$ causes water to have a high heat of vaporization. A large amount of energy is needed to convert liquid water, where the molecules are attracted through their hydrogen bonds, to water vapor, where they are not.

Which contains strongest H-bond?

0%
$O-H...S$
0%
$S-H...O$
0%
$F-H...F$
0%
$F-H...O$

Explanation

The strongest hydrogen bond is in hydrogen fluoride because the power of hydrogen bond

$\propto$ electroneativity of atom.

Electronegativity

$\propto \dfrac {1}{atomic\ size}$

So fluorine has maximum electronegativity and minimum atomic size.

On the following hydrides which has the lowest boiling point?

0%
$NH_3$
0%
$PH_3$
0%
$SbH_3$
0%
$AsH_3$

Explanation

$NH_3>PH_3<AsH_3<SbH_3<BiH_3$

From

$PH_3 to BiH_3$ the boiling points increase with increase in molecular weights, due to the increase in Van der Waal's forces with molecular mass.

However, boiling point of

$NH_3>PH_3$ even though

$NH_3$ has lower molecular mass than

$PH_3$ , due to the intermolecular hydrogen bonding in Ammonia that is not present in

$PH_3$

and other hydrides in Group-15.

therefore Option B is a correct answer.

In which of the following compounds does hydrogen bonding occur?

0%
$SiH_4$
0%
$LiH$
0%
$HI$
0%
$NH_3$

Explanation

A hydrogen bond is the electrostatic attraction between polar groups that occurs when a hydrogen (H) atom bound to a highly electronegative atom such as nitrogen (N), oxygen (O) or fluorine (F) experiences attraction to some other nearby highly electronegative atom.

therefore,, only

${NH_3}$ shows occuring of Hydrogen bonding.

which of the following statements is false

0%
$H_{2}PO_{2}$ , $H_{3}PO_{3}$ , $H_{3}PO_{4}$ all are tribasic and reducing in nature
0%
Among anions, $NO_{3}^{-}$ , $SO^{2-}_{3}$ , $SO^{2-}_{4}$ , $S_{2}O_{4}^{2-}$ , $H_{2}SO_{4}$ , $SO^{2-}_{3}$ is basic reducing nature
0%
Number of anions $SO^{2-}_{3}$ , $SO^{2-}_{4}$ , $SO^{2-}_{4}$ , and $H_{2}SO_{4}$ , $SO^{2-}_{3}$
0%
Number of lone pair(s) electrons on Xe atoms in $XeF_{2}$ , $XeF_{4}$ , $XeF_{6}$ are 3,2,and 1 respectively.

Explanation

(C)

$SO^{2-}_{3}$ contains one lone pair of electrons and sulphur atoms.
1594680_1747872_ans_319cb78bc3b34b1692d9b52f2405cd81.png

An electrically charged atom or a group of atoms is known as:

0%
A meson
0%
A proton
0%
An ion
0%
A cyclotron

Explanation

Meson is any member of a family of subatomic particles composed of a quark and an antiquark.

proton is a positively charged elementary particle that is a fundamental constituent of all atomic nuclei.

ion is an atom or molecule with a net electric charge due to the loss or gain of one or more electrons.

A cyclotron is a class of devices that accelerates charged atomic or subatomic particles in a constant magnetic field.

Hence, option

$C$ is correct.

Which of the following is not correct

0%
A sigma bond is weaker than $\pi$ bond
0%
A sigma bond is stronger than $\pi$ bond
0%
A double bond is stronger than a single bond
0%
A double bond is shorter than a single bond

Explanation

Option (A), is not correct.

(A):

$\text{A sigma bond is weaker than pi bond.}$

A sigma bond is the end-to-end overlap of the bonding orbitals.whereas, a pi bond is the side-to-side overlap of (usually) two p-orbitals. This places the electron density above and below the plane of the nuclei with a node in the middle. The nodal region reduces the degree to which the orbitals overlap and therefore a pi bond is less strong than a sigma bond between the same two elements.

(B): A sigma bond is stronger than π bond.

(C): A(=) bond is stronger than a single bond. It is because there are more electrons to be shared by the two atoms, so the nuclei are more attracted. The bonds to be longer because there are more electrons and the electrons would repel each other more.

(D): A double bond is shorter than single bond. Another consequence of the presence of multiple bonds between atoms is the difference in the distance between the nuclei of the bonded atoms. Double bonds have shorter distances than single bonds, and triple bonds are shorter than double bonds.

The bond angle of water is

$104.5^{\circ}$ due to

0%
Repulsion between lone pair and bond pair
0%
$sp^3$ hybridization of O
0%
Bonding of $H_2O$
0%
Higher electronegativity of O

Explanation

The bond angle in

$H_2O$ is

$104.5^o$ due to the presence of two lone pair of electrons. This fact can be explained with the help of valence shell electron pair repulsion theory (

$VSEPR$ ).

Hence, option A is correct.

In the sigma bond

0%
Sidewise as well as end to end overlap of orbitals take place
0%
Sidewise overlap of orbitals take place
0%
End to end overlap of orbitals takes place
0%
None of the above

Explanation

Answer is Option C is a correct answer.

$Sigma$

$bond$ :

A covalent bond resulting from the formation of a molecular orbital by the

$\text{end-to-end overlap of atomic orbitals}$ , denoted by the symbol

$σ$ .

$N_2$ molecule, the atoms are bonded by:

0%
One $\sigma$ and Two $\pi$
0%
One $\sigma$ and one $\pi$
0%
Two $\sigma$ and one $\pi$
0%
Three $\sigma$ bonds.

Explanation

$N_2$ molecule contains 2 lone pairs and 1s and 2p bonds.

$:N≡N:$

A triple bond consist of

$\text{1 sigma and two pi bonds}$ . Lone pair refers to a pair of valence electrons that are not shared with another atom, it is also called non-bonding pair.

The number and type of bonds between two carbon atoms in calcium carbide are:

0%
One sigma, one pi
0%
One sigma, two pi
0%
Two sigma, one pi
0%
Two sigma, Two pi

The covalency of nitrogen in

$HNO_3$ is:

0%
0
0%
3
0%
4
0%
5

Explanation

The electronic configuration of N is

$1s^22s^2p^3$ . The

$1s$ orbital is completely filled and unavailable for bonding. That leaves the

$2^{nd}$ and

$2p$ orbitals. The formation of

$4$ bonds (

$covalency$ ) is explained by hybridation of the

$2s$ and

$2p$ orbital forming

$4sp^3$ orbitals.

Normally when the

$N$ atom bonds it forms 3 covalent bonds with one pair of non-bonding electrons. Hence,

$N$ normally has a covalency of 3 with an available pair of electrons. But, when a proton, or other group that is able to bond with a pair of electrons coordinates with the central

$N$ atom a 4th covalent bond is formed. Beyond the 4th bond there is no more capability of the N atom to form additional bonds so it can only have a maximum covalency of

$4$ .

Which can be described as a molecule with residual bonding capacity

0%
$BeCl_2$
0%
$NaCl$
0%
$CH_4$
0%
$N_2$

Explanation

Answer is option A

$BeCl_2$

At room temperature, it exists as a dimer (

$Be_2Cl_4$ ), so it has

$\text{residual bonding capacity}$ because although

$BeCl_2$ is a complete, neutral molecule, it can form bonds with another molecule to attain further stability.

The double bond between the two carbon atoms in ethylene consists of

0%
Two sigma bonds at right angles to each other
0%
One sigma bond and one pi bond
0%
Two pi bonds at right angles to each other
0%
Two pi bonds at an angle of $60^{\circ}$ to each other

Explanation

Therefore on seeing the figure we say that:

There is a

$\text{One sigma bond and one pi bond}$ .

Therefore Option B is the correct answer.

1671737_1752226_ans_c9fcb0f1cfe14b0f9c4e25938f45460a.jpg

Toluene has

0%
$6\sigma$ and $3\pi$ bond
0%
$9\sigma$ and $3\pi$ bond
0%
$9\sigma$ and $6\pi$ bond
0%
$15\sigma$ and $3\pi$ bond

Explanation

Hint: Toluene is a mono-substituted colorless liquid that consists of a methyl group attached to a benzene ring.

Step1: Chemical Formula of Toluene

The chemical formula of toluene is

$C_6H_5CH_3$ .

Step 2: Sigma and Pi bond

Sigma Bond: It is formed by the head-on overlapping of orbitals. It is denoted by

$\sigma$ .

Pi Bond: It is formed by lateral overlapping of orbitals. It is denoted by

$\pi$ .

From the structure of toluene, we can observe that the benzene ring comprises of

$12$

$\sigma$ bonds and

$3$

$\pi$ bonds, and the substituent

$-CH_3$ has three

$\sigma$ bonds.

Therefore, toluene has total

$15$ $\sigma$ bonds and $3$ $\pi$ bonds.

Final Step: Correct answer is Option(D)

Which compound has zero dipole moment?

The number of

$\sigma$ bonds in o-xylene is

0%
$6$
0%
$9$
0%
$12$
0%
$18$

Explanation

o-xylene has

$18\sigma$ bonds and

$3\pi$ bonds.
1677593_1761226_ans_ba4e38bcb21d477cb9200c3ae8043709.png

In which one of the following changes there are transfer of five electrons

0%
$Mn{O}_{4}^{-}\rightarrow {Mn}^{2+}$
0%
$Cr{O}_{4}^{2-}{Cr}^{3+}$
0%
$Mn{O}_{4}^{2-}Mn{O}_{2}$
0%
${Cr}_{2}{O}_{7}^{2-} 2{Cr}^{3+}$

Explanation

$\overset{+7}{Mn}{O}_{4}^{-}\rightarrow {Mn}^{2+}+5{e}^{-}$

Option A.

The type of bonding in

$HCl$ molecule is

0%
pure covalent
0%
polar covalent
0%
highly covalent
0%
$H-$ bonding

Explanation

$\mathrm{HCl}$ , Hydrogen atom shares an electron with Chorine atom and the bond is formed. So, the bond is covalent.

But according to Fazan's rule, a covalent bond is always not 100% covalent and an ionic bond is not 100% ionic.

Hence, Option "C" is the correct answer.

Peroxide bond is present in

0%
$MgO$
0%
$CaO$
0%
$Li_2O$
0%
$BaO_2$

Explanation

Oxide bond is present in

$MgO$ ,

$CaO$ &

$Li_2O$ , whereas peroxide bond is present in

$BaO_2$ .

Hence, Option "D" is the correct answer.

The number of

$\pi$ bonds in 3-hexyne-1-ene is

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

Explanation

$\overset{1}{CH_2}\overset{1\ \pi}=\overset{2}{CH}-\overset{3}{C}\overset{2\ \pi}{\equiv} \overset{4}{C}-\overset{5}{CH_2}-\overset{6}{CH_3}$

3-hexyne-1-ene

The number of

$\pi$ bonds in 3-hexyne-1-ene is 3.

Acetylene molecules contain

0%
$5\sigma$ bond
0%
$4\sigma$ bond and $1\pi$ bond
0%
$3\sigma$ and $2\pi$
0%
$3\sigma$ and $3\pi$

Explanation

The formula of acetylene is

$C_2H_2\ or \ H-C\equiv C-H$

$H\overset{\sigma}{-}C\overset{\sigma}{\equiv }C\overset{\sigma}{-}H$

$3\sigma$ and

$2\pi$ bonds are present in acetylene.

$PO_{4}^{3-}$ ion the formal charge on the oxygen atom of

$PO$ bond is

0%
$+1$
0%
$-1$
0%
$-0.75$
0%
$+0.75$

Explanation

The formal charge of an atom in a polyatomic molecule or ion may be defined as the difference between the number of valence electrons of that atom in an isolated or free state and the number of electrons assigned to that atom in the Lewis structure.

$\text{Formal charge of the atom in the molecule or ion = Number of valence electron in the free atom}$

$-$

$\text{(Number of lone pair electrons +}$

$\dfrac{1}{2}$

$\text{number of bonding electrons)}$

So formal charge on each

$O$ -atom:

$6-(6+\dfrac{1}{2}*2)=6-7=-1$

The enolic form of acetone contains

0%
$8\sigma$ bonds, $2\pi-$ bonds and $1$ lone pairs
0%
$9\sigma$ bonds, $1\pi-$ bonds and $2$ lone pairs
0%
$9\sigma$ bonds, $2\pi-$ bonds and $1$ lone pairs
0%
$10\sigma$ bonds, $1\pi-$ bonds and $1$ lone pairs

Explanation

It has

$9\sigma$ bonds,

$1\pi-$ bond and

$2$ lone pair (oxygen has two lone pairs).
1683958_1761349_ans_e1564ae0c9ab49b98577591f701d45dc.jpg

In which of the following combination hybridisation of central atom (*) does not change ?

0%
$H_{2}O + CO_{2}$
0%
$H_{3}BO_{3} + OH^{-}$
0%
$BF_{3} + NH_{3}$
0%
None of these

Number of S - S bond is

$H_{2}S_{n}O_{6}$ :

0%
n
0%
(n - 1)
0%
(n - 2)
0%
(n + 1)

A : tetracyanomethane
B : carbondioxide
C : benzene
D : 1, 3-butadiene
Ratio of

$\sigma$ and

$\pi$ bonds is in order :

0%
A = B < C < D
0%
A = B < D < C
0%
A = B = C= D
0%
C < D < A < B

In which of the following cases C-C bond length will be highest?

0%
$CH_3-CF_3$
0%
$FCH_2-CH_2F$
0%
$F_2CH-CHF_2$
0%
$CF_3-CF_3$

A diatomic molecule has a dipole moment of 1.2 D. If the bond length is

$1.0 \times 10^{-8} \ cm$ , what fraction of charge does exist each atom?

0%
0.1
0%
0.2
0%
0.25
0%
0.3

Explanation

Fraction of charge =

$\frac{1.2 \times 10^{-18}}{4.8 \times 10^{-10}\times 10^{-8}} = 0.25$

Correct statement(s) about dipole moment of

$R_3NO$ and

$R_3PO$ is/are :

0%
dipole moment of $R_3NO$ > dipole moment of $R_3PO$
0%
dipole moment of $R_3NO$ < dipole moment of $R_3PO$
0%
Experimental dipole moment of $R_3PO$ > theoretical dipole moment of $R_3PO$
0%
Experimental dipole moment of $R_3PO$ < theoretical dipole moment of $R_3PO$

Which of the following statements is incorrect for sigma and

$\pi$ - bonds formed between two carbon atoms?

0%
Sigma - bond is stronger than a $\pi$ - bond
0%
Bond energies of sigma and $\pi$ - bonds are of the order of 264 kJ/mol and 347 kJ/mol
0%
Free rotation of surrounding atoms about a sigma-bond is allowed but not in case of a $\pi$ - bond
0%
Sigma- bond determines the direction between carbon atoms but a $\pi$ - bond has no primary effect in this regard

How many lone pairs of electrons are in resonance of the given compound?

0%
$4$
0%
$2$
0%
$3$
0%
$6$

Explanation

Two (one of

$N$ -atom with double bond and one of

$O$ -atom with double bond) lone pairs of electrons are in resonance.

Which of the following compounds has zero dipole moment?

The total number of lone pairs in chlorate ion is:

0%
$8$
0%
$4$
0%
$6$
0%
$2$

Explanation

Correct option: $A$

Explanation:

When two atoms share their respective electron with each other to form a bond, the bond is called Covalent Bond.
The electron that takes part in bond formation is called a bond pair of electrons.
The electron present in the valence shell of the atom that does not take part in bond formation is called lone pair electrons.

The structure of chlorate ion is shown as;

From the structure it is clear that there are total of $7$ lone pairs on $O$ and $1$ lone pair on $Cl$ So, total of $8$ lone pairs are there

Correct Option: $A$ .

Give the correct order of initials T or F for following statements. Use T if statement is true and F if it is false :
(i) The order of the repulsion between different pair of electrons is

$l_{p} - l_{p} > l_{p}- b_{p} > b_{p}- b_{p}$
(ii) In general, as the number of lone pair of electrons on central atom increases, value of bond angle from normal bond angle also increases
(iii) The number of lone pair on O in

$H_{2}O$ is 2 while on N in

$NH_{3}$ is 1
(iv) The structures of xenon fluorides and xenon oxyfluorides could not be explained on the basis of VSEPR theory

0%
TTTF
0%
TFTF
0%
TFTT
0%
TFFF

Explanation

correct statements are

(ii) (F) In general, as the number of lone pair of electrons on the central atom increases, the value of bond angle from normal bond angle decreases due to

$lp - lp>lp - bp$

(iv) (F) Structures of xenon fluorides and xenon oxy fluoride are explained on the basis of VSEPR theory.

$SOBr_{2}$ S-O bond has a maximum bond length in comparison to S-O bond lengths in

$SOF_{2}$ and

$SOCl_{2}$ because in

$SOBr_{2}$ S-O bond has been formed by hybrid orbital containing less s-character.

Which of the following molecular species is/are having

$\pi_{2p}$ as H.O.M.O. (highest occupied molecular orbital):

0%
$N_2^-$
0%
$O_2^{2+}$
0%
$NO^+$
0%
$B^+_2$

Explanation

${O_2}^{2+}, NO^+, {B_2}^+$ have

$\pi_{2p}$ as HOMO

Select correct statements regarding

$\sigma$ and

$\pi$ -bonds :

0%
$\sigma$ -bond lies on the line joining the nuclei of bonded atoms
0%
$\pi$ -electron cloud lies on either side to the line joining the nuclei of bonded atoms
0%
$(2p\pi 3d\pi)$ Pi-bond is stronger than $(2p\pi - 3p\pi)$ Pi-bond
0%
$\sigma$ -bond has primary effect to decide direction of covalent bond, which $\pi$ -bond has no primary effect in direction of bond.

Explanation

All of the given options regarding

$\sigma$ and

$\pi$ bonds are correct.

How many sigma bonds are present in a molecule of diethyl either

$C_{2}H_{5}OC_{2}H_{5}$ ?

0%
14
0%
12
0%
8
0%
16

Which of the following molecules has as

$O-O$ bond?

0%
$H_2S_2O_8$
0%
$H_2S_2O_7$
0%
$H_2SO_5$
0%
$H_2S_2O_6$

Explanation

$H_2S_2O_8 \ and \ H_2SO_5$ has

$O-O$ bond.

Which of the following has no lone pair of electrons?

0%
$NO$
0%
$CO$
0%
$NH_3$
0%
$O_2$

How many sigma and pi bonds are present in toluene ?

0%
$10\sigma$ and $3\pi$ bonds
0%
$12\sigma$ and $3\pi$ bonds
0%
$15\sigma$ and $3\pi$ bonds
0%
$6\sigma$ and $3\pi$ bonds

Explanation

Toluene is

$C_{6}H_{5}CH_{3}.$

No. of

$\sigma$ bonds is

$15.$

No. of

$\pi$ bonds is

$3.$

1767668_1842048_ans_fdb5e328051e46578684da1a39f7fd22.png

State true or false:
The term chemical bond does not express the existence of strong forces of attraction between the atoms.

0%
True
0%
False

Explanation

$\text { False. } \\$

$\text { Chemical bonds are formed because of } \\$

$\text { strong forces of attraction which leads } \\$

$\text { to greater stability. } \\$

$\text { correct }-(B) \text { False. }$

${NO_{3}}^{-}$ ion, number of bond pair and lone pair of electrons on nitrogen atom are :

0%
$2,2$
0%
$3,2$
0%
$1,3$
0%
$4,0$

Explanation

According to the VSEPR theory,

$NO_3^-$ is sp² hybridized with 3 sigma bond pairs, 1 pi bond pair and no lone pair on the central atom as shown in the figure. Its shape is trigonal planar.

Hence, option D is correct.

The number of

$sp^{2}-sp^{2} \sigma$ bond in given compound is:

0%
$4$
0%
$5$
0%
$3$
0%
$6$

The molecule in which an atom is associated with more than

$8e^{-}s$ is known as hypervalent molecule. All hypervalent molecules must have

$d\pi - p\pi$ bonding but the molecule having back bonding need not have always

$d\pi - p\pi$ bonding.

Which of the following molecule is not having

$d\pi - p\pi$ bonding?

0%
$SO_{2}$
0%
$P_{4}O_{10}$
0%
$H_2O$
0%
$B_{3}N_{3}H_{6}$

Explanation

$SO_2, P_4O_{10}$ has

$d\pi - p \pi$ bonding.

No,

$d \pi - p \pi$ bonding exists in

$H_2O$ and

$B_3N_3H_6$ .

In water, oxygen has a complete octet after forming the two bonds with hydrogen.

$B_3N_3H_6$ , there is an absence of d - orbitals in boron and nitrogen. Dative bond is formed between N and B atoms using p-orbitals of both.

The ratio of number of lone pairs in the central atoms of

$NH_3$ ,

$H_2O$ and

$XeF_{2}$ is:

0%
$3:2:1$
0%
$2:1:3$
0%
$1:2:3$
0%
$2:3:1$

Explanation

$NH_3$ : 3 bond pairs and one lone pair because, out of 5 valence electrons of nitrogen, only 2 electrons are left to form a lone pair.

$H_2O$ : 2 bond pairs and two lone pairs because, out of 6 valence electrons of oxygen, 4 electrons are left to form 2 lone pairs.

$XeF_2$ : 2 bond pairs and three lone pairs because out of 8 valence electrons of xenon, 6 electrons are left to form 3 lone pairs.

So, the ratio of lone pairs is

$1:2:3$ .

In the analysis of an organic compound, it was found that it contains 7 carbon atoms, there are two C = C bonds and one

$C\equiv C$ bond. This compound is a hydrocarbon. Hydrocarbons have only carbon and hydrogen elements. On structural analysis it was found that it is covalent in nature and expected structure is given above. The ratio between the pure and hybrid orbitals is?

0%
$7:10$
0%
$8:7$
0%
$5:9$
0%
$1:2$

Explanation

$\displaystyle sp, sp^2$ and

$\displaystyle sp^3$ hybridization of

$C$ , the number of hybrid orbitals formed are

$2,3$ and

$4$ respectively and the number of pure orbitals are

$2,1$ and

$0$ respectively.
In this compound,

$2$

$\displaystyle sp$ hybridized,

$4$

$\displaystyle sp^2$ hybridized and

$1$

$\displaystyle sp^3$ hybridized

$C$ atom is present. The number of pure orbitals in these atoms are

$\displaystyle 4+4+0=8$
Also there are

$6\: H$ atoms which contribute

$6$ pure orbitals. Hence,the total number of pure orbitals are

$\displaystyle 8+6=14$
The number of hybrid orbitals are

$\displaystyle 4+12+4+0=20$
Hence, the ratio of pure orbitals to hybrid orbitals is

$\displaystyle 14: 20$ or

$\displaystyle 7:10$

CBSE Questions for Class 11 Medical Chemistry Chemical Bonding And Molecular Structure Quiz 10 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Chemical Bonding And Molecular Structure Quiz 10 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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