MCQ Questions for Class 11 Medical Chemistry Chemical Bonding And Molecular Structure Quiz 2

The molecule having maximum number of sigma bonds in it is:

0%
Ethane
0%
Ethene
0%
Ethyne
0%
Sulphur hexafluoride

Explanation

Ethane,

${C}_{ 2 }{ H}_{6 }$ has seven sigma bonds.

Ethene

${C}_{ 2 }{H}_{4 }$ has five sigma bonds.

Ethyne

${C}_{ 2 }{H}_{2 }$ has three sigma bonds.

sulphur hexafluoride

$S{F }_{6 }$ has six sigma bonds.

The number of electron bond pairs involved in the formation of hydrogen cyanide molecule are:

0%
two
0%
eight
0%
three
0%
four

Explanation

$HCN$ contains

$4$ bonds (

$1\: C-H$ single bond and

$1$ triple bond between

$C$ and

$N$ ). Thus, total bond pairs are

$4$ .

Valence bond theory of Pauling and Slater accounts for the following characteristic of covalent bond:

0%
directional
0%
ionic
0%
strength
0%
hybrid

Covalency for central atom is maximum in :

0%
$BF_{3}$
0%
$SO_{2}Cl_{2}$
0%
$POCl_{3}$
0%
$BeCl_{2}$

Explanation

$\bf{Hint:-}$ Covalency denotes the number of bonds that can be formed by an atom in a molecule.

$\bf{Explanation-}$

When an element shares electrons with other atoms of the same or different elements to acquire stable electronic configuration then it is called covalency.

Covalency of

$S$ in

$SO_2Cl_2$ is

$6$ .

Covalency of

$B$ in

$BF_3$ is

$3$ .

Covalency of

$P$ in

$POCl_3$ is

$5$ .

Covalency of

$Be$ in

$BeCl_2$ is

$2$ .

$\bf{Conclusion-}$ Hence, covalency of central atom is maximum in

$SO_2Cl_2$ i.e

$6$ .

The number of electron pairs present in the valency shell of central atom in

$SF_{6}$ molecule are:

0%
$4$
0%
$6$
0%
$8$
0%
$7$

Explanation

$1$ electron pair for each bond and there are

$6\: S-F$ bonds in the given molecule.

Hybridisation involves:

0%
addition of an electron pair
0%
combination and redistribution of atomic orbitals
0%
removal of an electron pair
0%
separation of orbitals

Explanation

$\bf{Hint:-}$ Mixing of different energy orbitals to form new equal energy orbitals known as hybridization.

$\bf{Explanation-}$

$\bullet$ Hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals.

$\bullet$ Atomic orbitals with similar energy combine to give degenerated hybrid orbitals.

$\bullet$ The number of combining atomic orbitals is equal to the hybridised orbitals.

$\bullet$ They involve redistribution of electrons of combined atomic orbital to hybrid orbitals.

$\bf{Conclusion-}$ Hence, option

$B$ is correct.

Molecule having maximum number of lone pairs of electrons on central atom is:

0%
$PH_{3}$
0%
$H_{2}S$
0%
$CH_{4}$
0%
$BrF_{5}$

Explanation

$P$ belongs to group

$15$ and has

$5$ electrons in valence shell.

$3$ are involved in bonding with the

$3$

$H$ atoms.

So, lone pair

$= 1$ .

$S$ belongs to group

$16$ and has

$6$ electrons in valence shell.

$2$ are involved in bonding with the

$2$

$H$ atoms.

So, lone pairs

$= 2$ .

$C$ belongs to group

$14$ and has

$4$ electrons in valence shell.

$4$ are involved in bonding with the

$4$

$H$ atoms.

So, lone pair

$= 0$ .

$Br$ belongs to group

$17$ and has

$7$ electrons in valence shell.

$5$ are involved in bonding with the

$5$

$F$ atoms.

So, lone pair

$= 1$ .

The molecular weight of an oxide of Hydrogen is

$34$ . Therefore the number of covalent bonds in its molecule are:

0%
$4$
0%
$5$
0%
$2$
0%
$3$

Explanation

Oxide of hydrogen,

$H_2O$

$H_2O_2$
M.W-

$16$ M.W-

$34$
In

$H_2O_2$ , three covalent bonds are present.

$OF_{2}$ , number of bond pairs and lone pairs of electrons are respectively:

0%
$2, 6$
0%
$2, 8$
0%
$2, 10$
0%
$2, 9$

Explanation

Explanation:

For number of bond pair and lone pair identification we need to draw the structure of $OF_2$ , which is as follows:

It is clear from the structure that :There are two bond pairs of electrons which forms two $O-F$ bond pairs. $O$ atom has two lone pairs of electron. Each $F$ atom has 3 lone pairs of electrons. Thus in total there are 8 lone pairs of electrons.

Hence option is correct

Diethyl ether is mostly used in solvent extraction due to all of the following reasons except:

0%
Its solvation capacity is very high
0%
being inert, it does not react with most organic compound
0%
there are two lone pairs in it, therefore it acts as a strong nucleophile
0%
its boiling point is low, therefore, it can be easily separated by distillation

Explanation

$C_2H_5OC_2H_5$
There are two lone pairs in it but being inert, it does not react with most organic compound.

Among the following ions, the

$p\pi - d\pi$ overlap could be present in:

0%
$NO^{-}_{3}$
0%
$PO^{3-}_{4}$
0%
$CO^{2-}_{3}$
0%
$NO^{-}_{2}$

Explanation

$NO_3^-, CO_3^{2-}, NO_2^-$ only

$p-p$ orbitals are involved in bonding.
In

$PO_4^{3-}$ ,

$3d$ orbitals of phosphorus are also involved in bonding.

Which formulae does not correctly represents the bonding capacity of the atom involved?

According to VSEPR theory, the most probable shape of the molecule having

$4$ electron pairs in the outer shell of the central atom is:

0%
linear
0%
tetrahedral
0%
pyramidal
0%
octahedral

Explanation

Most probable shape of molecule having

$4$ bond pairs with no lone pairs is regular tetrahedron (tetrahedral) Ex:-

$CH_4, SiF_4$

The number of sigma bonds in

$1$ -butene is:

0%
$8$
0%
$10$
0%
$11$
0%
$12$

A molecule having 3 bond pairs and 2 lone pairs will have:

0%
T-shaped structure
0%
Trigonal planar structure
0%
Linear structure
0%
Square pyramidal structure

Explanation

For

$3$ bond pairs and

$2$ lone pairs, hybridization is

$sp^3d$ . Hence, it has trigonal bipyramidal geometry. The shape is called as T-shaped.

Example:

$ClF_3$ .

Which of the following unit conversion of dipole moment is correct?

0%
$1 D = 3.3356 Cm$
0%
$1D= 3.3356\times10^{-30}C m$
0%
$1 Cm = 3.3356 D$
0%
$1 C m = 3.3356\times10^{-30} D$

Explanation

One Debye,

$1D=10^{-18} esu\cdot cm$ esu. electrostatic unit
On coverting it in SI units,

$1D=3\cdot 33\times 10^{-30}$ coloumb metre.

Length of the hydrogen bond range from

$1.75A^{o}$ to _________.

0%
$2.5A^{o}$
0%
$2.75A^{o}$
0%
$2.6A^{o}$
0%
$3.2A^{o}$

Explanation

As H-bond is a weak bond compared to other covalent bonds, its bond length is a little bit more than covalent bonds. The range of the length of a hydrogen atom is

$1\cdot 75\overset {o}A$ to

$2\cdot 5\overset {o}A$

The percentage of s- character of the hybrid orbitals in ethane, ethene and ethyne are respectively:

0%
$50, 75, 100$
0%
$10, 20, 40$
0%
$25, 33, 50$
0%
$25, 50, 75$

Explanation

Ethane Ethene Ethyne

$CH_3 - CH_3$

$CH_2 = CH_2$

$HC = CH$

$SP^3 \,\,\,\,\, SP^3$

$SP^2 \,\,\,\,\, SP^2$

$SP \,\,\,\,\, SP$

$\% S = 25\%$

$\% S = 33.33 \%$

$\% S = 50 \%$

$\% P = 75\%$

$\% P = 66.67 \%$

$\% P = 50 \%$

Hence, option C is correct.

Which of the following does not contain any co-ordinate bond?

0%
$H_3O^+$
0%
$BF_4^-$
0%
$HF_2^-$
0%
$NH_4^+$

Compounds having equal number of hybrid orbitals among the following are:

0%
0%
0%
0%
none of these

Explanation

Compound A & C has equal no. of hybrid orbitals (as each carbon is

$sp^3$ hybridized and each compound has

$4$ carbon so total no. of

$sp^3$ hybridized orbitals are

$16$ ).

Option A and C are correct.

Which of the following is diamagnetic?

0%
${ H }_{ 2 }^+$
0%
$O_2$
0%
$H_2$
0%
${ H }_{ 2} ^-$

Explanation

There are no unpaired

$e^-$ in

$H_2$ so its diamagnetic.

In which of the compound given below contains more than one kind of hybridization (

$sp,sp^{2}$ ,

$sp^{3}$ ) for carbon:

0%
$CH_{3}CH_{2}CH_{2}CH_{3}$
0%
$CH_{3}-CH=CH-CH_{3}$
0%
$CH_{2}=CH-CH=CH_{2}$
0%
$H-C\equiv C-H$

Explanation

(i)

$\overset{sp^3}{CH_3}-\overset{sp^3}{CH_2}-\overset{sp^3}{CH_2}-\overset{sp^3}{CH_3}$

Only

$sp^3$ hybridized carbon.

(ii)

$\overset{sp^3}{CH_3}-\overset{sp^2}{CH}=\overset{sp^2}{CH}-\overset{sp^3}{CH_3}$

Both

$sp^2$ and

$sp^3$ hybridized carbon.

(iii)

$\overset{sp^2}{CH_2}=\overset{sp^2}{CH}-\overset{sp^2}{CH}=\overset{sp^2}{CH_2}$

Only

$sp^2$ hybridized carbon.

(iv)

$H- \underset{sp}{C}\equiv \underset{sp}{C}-H$

Only

$sp$ hybridized carbon.

Option B is correct.

The molecule in which xenon is surrounded by three lone pairs and two bond pairs is:

0%
$\mathrm{X}\mathrm{e}\mathrm{F}_{2}$
0%
$\mathrm{X}\mathrm{e}\mathrm{F}_{4}$
0%
$\mathrm{X}\mathrm{e}\mathrm{F}_{6}$
0%
$\mathrm{X}\mathrm{e}\mathrm{O}_{4}$

Explanation

The number of lone pair

$= 3$ and bond pair

$= 2$ in

$XeF_2$ .

Which one of the following molecular geometries (i.e. shapes) is not possible for the $sp^{3}d^{2}$ hybridization?

0%
Capped octahedral
0%
Octahedral
0%
Square planar
0%
Square pyramidal

Which of the following statements is not correct?

0%
$(s + p_y)$ produces sp hybrid orbitals which are lying in the yz plane
0%
$(s + p_y)$ produces sp hybrid orbitals which are lying the xz plane
0%
$(s + p_x + p_z)$ produces sp2 hybrid orbitals which are lying in the xz plane
0%
$(s p_y)$ produces sp hybrid orbitals which are lying along the y-axis

Explanation

$(S+P_y)$
Produces SP hybrid orbital which is orlinked in y- direction i.e. its nodal plane its xz plane.

Which of the following are isostructural ?

(I) $NO_{3}^{-}$

(II) $CO_{3}^{2-}$

(III) $CIO_{3}^{-}$

(IV) $SO_{3}$

(V) $XeO_{3}$

0%
(I) and (IV)
0%
(II) and (V)
0%
(III) and (IV)
0%
(IV) and (V)

The shape of

$XY_3$ molecule if there is no lone pair on the central atom is trigonal planar.

0%
True
0%
False

Explanation

It is true because

$X{ Y }_{ 3 }$ molecules if lone pair is present forms trigonal bi pyramidal due to the repulsion of lone pair and bond pair electrons, but if lone pair electrons are absent, then it forms trigonal planar structure.

The maximum covalency of oxygen in ozone is:

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

Which of these molecules have non-bonding electron pairs on the central atom?
I :

$SF_4$ II :

$ICl_3$ III:

$SO_2$

0%
II only
0%
I and II only
0%
I and III only
0%
I, II and III

Explanation

$SF_4$ molecule has one non-bonding electron pair on the central atom Sulfur.

$ICl_3$ molecule has two non-bonding electron pairs on the central atom Iodine.

$SO_2$ molecule has one non-bonding electron pair on the central atom Sulfur.

So the answer is D.

The compound with the highest degree of covalency is :

0%
NaCl
0%
MgO
0%
AgCl
0%
CsCl

Explanation

Correct Option : $C$

Hint: Compounds will pseudo noble gas configurations are the most covalent.

Explanation:

Fajan's rule states that the covalent character of an ionic compound depends on the polarisability of anion which increases its size and charge, and polarising power of cation, which increases with its charge, and decreases with size. In both cases, the charge is a more important factor than size.

However, if the cation has a pseudo noble gas configuration, then it will be the most covalent, and it supersedes the above-mentioned criterion. Pseudo noble gas configuration means

$d^{10}$ configuration.

Identify the noble gas and pseudo noble gas configurations.

$Na^+ \rightarrow noble \ gas \ configuration$

$Mg^{+2} \rightarrow noble \ gas \ configuration$

$Ag^+ \rightarrow \ pseudo \ noble \ gas \ configuration$

$Cs^+ \rightarrow \ noble \ gas \ configuration$

So, the order of covalent characters will be

$C>B>A>D$

Hence the compound with the highest degree of covalency is AgCl (Option $C$ )

Select the INCORRECT statement :

0%
$N_2O$ with sodium metal in liquid ammonia forms sodium azide and nitrogen gas is liberated.
0%
Ammonia is oxidized to nitrogen by dilute solution of sodium hypochlorite in presence of glue.
0%
Ammonium dichromate on heating decomposes to give nitrogen and a green coloured compound.
0%
CaNCN on hydrolysis produces a white precipitate and a gas which turns filter paper moistened with copper sulphate solution deep blue.

Explanation

$3 N_2O + 4 Na + NH_3\xrightarrow[]{NH_3(\ell)}NaN_3 + 3 NaOH + 2 N_2$

$\underbrace { 2NH_3 +NaOCl }_{ dilute \: solutions} \xrightarrow[]{glue} N_2H_4 + NaCl + H_2O$

$(NH_4)_2Cr_2O_7\xrightarrow[]{\Delta }N_2 + 4 H_2O + Cr_2O_3\downarrow (green)$

$CaCN_2 + 3 H_2O\longrightarrow CaCO_3\downarrow +2 NH_3$

$Cu^{2+} + 4 NH_3\longrightarrow [Cu(NH_3)_4]^{2+} (deep \: blue)$

Write the order of repulsions among the electron pairs.

Lone pair - lone pair > lone pair - bond pair > bond pair - bond pair

0%
True
0%
False

The number of non-bonded electrons present on the central atom in the molecule of ammonia is two.

0%
True
0%
False

Explanation

It is true because the central molecule has

$1$ atom.

$N$ in ammonia consists of one lone pair, which has

$2$ non bonded electrons.

How many unpaired electrons are present in

$N_2^+$ :

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

Explanation

According to MOT,

$N_2^+$ ion has one unpaired electron.

In the electronic structure of H

$_2$ SO

$_4$ , the total number of shared electrons is:

0%
$20$
0%
$16$
0%
$12$
0%
$8$

Explanation

In the electronic structure of sulphuric acid, total

$8$ bonds are present.
Each bond is formed by sharing of two electrons. Thus, total

$16$ electrons are shared.

Which metal(s) can be used to produce electrons b the photoelectric effect from given source of light?

0%
Barium only
0%
Barium or lithium
0%
Lithium, tantalum or tungsten
0%
Tungsten or tantalum

Explanation

$\Delta E=\frac {12400}{5400A^o}$

$\Delta 2.75 eV$
For photoelectric effect,

$\Delta E > W_0$ (work function).

The bond between two identical non metallic atoms has a pair of electron:

0%
unequally shared among the two atoms.
0%
equally shared among two atoms.
0%
equal transfer of electrons from one atom to another.
0%
shared by only one atom.

Explanation

Bonds between identical, non-metals, are purely covalent due to the same electronegativities of the bonded atoms.

Hence, the bonded atoms have a equal hold on the shared pair of electrons,

e.g. :

$F_2$ ,

$O_2$ ,

$N_2$ .

Hence, option (B) is the correct answer.

Which of the following compound has a directional bond?

0%
Magnesium chloride
0%
Solid neon
0%
Hydrogen sulphide
0%
Caesium fluoride

Linear overlapping of any two pure atomic p - orbitals lead to sigma bond formation.

0%
True
0%
False

Explanation

Linear overlapping of two pure atomic

${ P }_{ z }$ orbitals. lead to sigma bond formation.

This kind of overlapping takes place during the formation of

$Cl_2$ .

The Hydrogen bond is always formed between the molecules of the same substance.

0%
True
0%
False

A species

$H_2O^+$ pyramidal in shape. The number of lone pairs of electrons is:

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

Explanation

${ H }_{ 2 }{ O }^{ + }$

Since 'O' has 6 electrons in valence shell where 2 electron bonded with hydrogen and one electron is donated due to +ve charges. one electron is as non-bonding electron and thus there is one lone pair of electrons.

When a chemical bond is formed, there is decrease in :

0%
kinetic energy
0%
potential energy
0%
repulsive force
0%
attractive force

The formal charge of the

$O$ atoms in the ion is:

0%
$-2$
0%
$-1$
0%
$0$
0%
$+1$

Explanation

Formal charge of O atom

$=$ Number of valence electrons - Number of dots - Number of lines

$=6-4-2$

$=0$

The number of bonding pairs electron

$(X)$ and lone pairs electron

$(Y)$ around the central atom in the

${I}^-_3$ ion are

$X +Y$ :

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

Explanation

The number of bonding pairs electron

$(X)=2\times 2=4$

Total no. of lone pairs electron

$Y=3\times 3 \times 2=18$

$X+Y= 4+18=22$

Option A is the correct answer.

Which of the following molecules contain odd electron bonds?

0%
$H_2O$
0%
$NO$
0%
$NO_2$
0%
$ClO_2$

Explanation

$NO_2$

Nitrogen dioxide is the chemical compound with the formula

$NO_2$ . Again, nitrogen dioxide does not follow the octet rule for one of its atoms, namely nitrogen.

The total number of valence electrons is $5+2(6)=17$

There is the persistent radical character on nitrogen because it has an unpaired electron.

$ClO_2$

A Lewis structure shows $ClO_2$ has a total of $2\times 6+7=19$ valence electrons, or 9 electron pairs and a lone electron.

Cl is the central atom.

The molecule $ClO_2$ has an odd number of valence electrons.

$NO$

14 valence electrons in your trial structure.

The valence electrons you have available are: $1 N + 1 O = 1\times 5 + 1\times 6 = 11$ .

The molecule $NO$ , $NO_2$ and $ClO_2$ has an odd number of valence electrons.

The number of

$\sigma$ and

$\pi$ bonds in

$C_2H_2$ is:

0%
0 and 4
0%
2 and 2
0%
3 and 2
0%
4 and 2

Explanation

Acetylene form cylindrical

$\pi$ - electron cloud where nodal plane at one

$\pi$ - bond occupied by antinode of other

$\pi$ -bond.

It has 3

$\sigma$ -bond and 2

$\pi$ bond.

Which of the following would not have a dipole moment?
(i)

$CCl_4$ (iii)

$1, 2$ -dichloroacetylene
(ii) trans-

$1, 2$ -dichloroethene (iv)

$CH_2Cl_2$

0%
(i) and (iv)
0%
(i), (ii) and (iv)
0%
(i), (ii) and (iii)
0%
(ii) and (iii)

Explanation

In carbon tetrachloride, the resulting dipole moment of three

$C-Cl$ bonds is equal in magnitude and opposite in direction to the dipole moment of the fourth

$C-Cl$ bond. Hence the molecule has no net dipole moment.
In trans-

$1,2$ -dichloroethane, the dipole vectors of two

$C-Cl$ bonds are equal in magnitude and opposite in direction. Hence, they cancel each other.
Similarly the dipole vectors of two

$C-H$ bonds are equal in magnitude and opposite in direction. Hence, they cancel each other.
Thus the molecule has no net dipole moment.
In

$1, 2$ -dichloroacetylene, the dipole moment vectors of two

$C-Cl$ bonds are equal in magnitude and opposite in direction. Hence, they cancel each other. Hence, the molecule has no net dipole moment.
In dichloromethane, the resultant vector of dipole moments of two

$C-Cl$ bonds is not equal to the resultant vector of the dipole moments of two

$C-H$ bonds. Hence, the molecule has net dipole moment.

The heat of combustion of carbon to

$CO_2$ is –393.5 kJ/mol. The heat released upon formation of 35.2 g of

$CO_2$ from carbon and oxygen gas is:

0%
-315kJ
0%
+315kJ
0%
-630Kj
0%
+3.15kJ

Explanation

$\Delta H_f$ =

$-393.5kJ/mol$

Therefore heat released on formation of44 gm or 1 mole of

$CO_2$ =

$-393.5kJ/mol$

Heat released on formation of 35.2 gm of

$CO_2$ =

$\dfrac{-393.5kJ/mol}{44g}$

$\times35.2g$ =

$-315 kJ/mol$

Hence,option A is correct.

In which of the following mixtures, the London dispersion force acts as a major intermolecular force of attraction?

0%
Sodium chloride and water
0%
Cyclohexane and carbon tetrachloride
0%
Water and ethyl alcohol
0%
Benzene and acetone

Explanation

Cyclohexane

$(C_6H_{12})$ and carbon tetrachloride

$(CCl_4)$ both are non-polar compounds so there exists London dispersion forces between their constituent molecules.

Which compound have maximum dipole moment?

Explanation

Hint:

Mathematically, the product of the magnitude of the charge and the distance between the nucleus of both the charge is termed as dipole moment.

Explanation:

in compound A, the electron donor group destabilizes the stability of the resonant structure. thus have a low dipole moment.
In compound B, the electron acceptor group withdraws electron from the ring resulting lower dipole moment.
In compound C, it has one group which is electron donor $ONa^{+}$ and the other one is electron acceptor $NO_2^{-}$ .
So, here in this case the dipole moment is directed towards $NO_2$ as $ONa$ is electron-donating.
In compound D, both the groups are electron-withdrawing.

Final Answer:

Thus, from the resonant structures we conclude that the overall dipole moment is maximum in compound C

So, the correct answer is option C.

CBSE Questions for Class 11 Medical Chemistry Chemical Bonding And Molecular Structure Quiz 2 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Chemical Bonding And Molecular Structure Quiz 2 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.