equal sharing of electrons takes place between two atoms

transfer of electrons takes place.

one atom acts as donor and other acts as acceptor

the electrons are shared by only one atom

when two orbitals directly overlap but there is only one bonding interaction.

when two orbitals directly overlap and there are two bonding interaction.

when two valence electrons combine into one orbital.

when two atoms overlap and the valence electrons bond.

the creation of an S shaped orbital.

MCQ Questions for Class 11 Medical Chemistry Chemical Bonding And Molecular Structure Quiz 4

State True or False.

All the orbitals in a set of hybridised orbitals have the same shape and energy

0%
True
0%
False

Which condition don't favours the bond formation:

0%
maximum attraction and maximum potential energy.
0%
minimum attraction and minimum potential energy.
0%
minimum potential energy and maximum attraction.
0%
minimum attraction and maximum potential energy.

In the electronic configurations of elements

$A\;and\;B$ are

$1s^2,\,2s^2\;2p^6,\,3s^1\;and\;1s^2,\,2s^2\;2p^6,\,3s^2\;3p^4$ respectively, then the formula of the compound formes by the combination of these elements will be

0%
$AB$
0%
$AB_3$
0%
$AB_2$
0%
$A_2B$

Explanation

Electronic configuration of

$A$ is

$1s^2\,2s^2\,2p^6\,3s^1$
Electronic configuration of

$B$ is

$1s^2\,2s^2\,2p^6\,3s^2\,3p^4$
Valency of

$A=1$
Valency of

$B=8-6=2$

$\therefore$ The compound formed by

$A\;\&\;B$ is
1313633_329861_ans_06f34f4359e6433f9e993bbab0147700.bmp

Which reaction involves a change in the electron-pair geometry for the underlined element?

0%
$\underline{B}F_3\, +\, F^-\, \rightarrow\, {\underline{B}F_4}^-$
0%
$\underline{N}H_3\, +\, H^+\, \rightarrow\, \underline{N}H_4^+$
0%
$2\underline{S}O_2\, +\, O_2\, \rightarrow\, 2\underline{S}O_3$
0%
$H_2\underline{O}\, +\, H^+\, \rightarrow\, H_3\underline{O}^+$

Explanation

$BF_3$ is sp

$_2$ hybridised and trigonal planar in shape while

$BF_4^-$ is sp

$_3$ hybridised and tetrahedral in shape .

So both have different electron-pair geometry .

Bond-breaking is an _________________process.

0%
endothermic
0%
exothermic
0%
neither exothermic nor endothermic
0%
either exothermic or endothermic

In a triple bond, there is sharing of:

0%
$3$ electrons
0%
$4$ electrons
0%
$6$ electrons
0%
none of these

Explanation

A triple bond is formed by sharing of

$6$ electrons.

Consider the following three halides:
(i)

$CH_3-CH_2-Cl$
(ii)

$CH_2=CH-Cl$
(iii)

$C_6H_5-Cl$
Arrange C-Cl bond length of these compounds in decreasing order.

0%
i > ii > iii
0%
i > iii > ii
0%
iii > ii > i
0%
ii > iii > i

Explanation

Bond length is the average distance between nuclei of two bonded atoms in a molecule. Alkenes have more of s character so they tightly hold the electronegative chloride ions towards itself making C-Cl bond length shorter than alkyl chlorides. In aryl halides, the ring produces strain on Cl atom because of resonance and shows double bond character so it has shortest C-Cl bond length. Hence, the correct order is as follows:

$CH_3-CH_2-Cl(i)\ >\ CH_2=CH-Cl(ii)\ >\ C_6H_5-Cl(iii)$

Which of the following has least polarity in bond?

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

Explanation

Hint: Bond polarity is directly proportional to the electronegativity difference between two atoms.

Correct option :

$D$

Explanation of Correct Option:

The ability of an atom in a molecule to attract shared electrons is called electronegativity. The higher the electronegativity of an atom, the greater its ability to attract shared electrons.

It is a measure of an atom's ability to attract shared electrons to itself and

bond polarity increases with an increasing difference in electronegativity.

Now Let's discuss trends of electronegativity in the periodic table

On the periodic table, electronegativity generally increases as you move from left to right across a period and decreases as you move down a group.

Hence on basis of trends of electronegativity order of electronegativities is as follows:

$\mathrm{F}>\mathrm{O}>\mathrm{N}>\mathrm{Cl}>\mathrm{Br}>\mathrm{I}>\mathrm{S}>\mathrm{C}>\mathrm{H}$

Now clearly the difference in electronegativities is minimum between sulphur and hydrogen as they are very near.

Therefore,

$\mathrm{H}-\mathrm{S}$ bond will show minimum polar character.

Explanation of Incorrect Options:

As other options have more electronegativity difference, therefore, their bond polarity will be more than the

$H-S$ bond.

In covalency,

0%
transfer of electrons takes place.
0%
one atom acts as donor and other acts as acceptor
0%
the electrons are shared by only one atom
0%
equal sharing of electrons takes place between two atoms

A sigma bond is formed by the overlapping of:

0%
$s-s$ orbital alone
0%
$s$ and $p$ orbital alone
0%
$s-s,s-p$ or $p-p$ orbital along internuclear axis
0%
$p-p$ orbital along the sides

Explanation

A sigma bond is formed by the overlapping of

$s-s,s-p$ or

$p-p$ orbital along inter-nuclear axis. On the other hand, the lateral overlap of

$p-p$ orbitals gives a pi bond.

In which of the following species, each atom carries same number of lone pair electrons on it?

0%
$XeO_4^{2-}$
0%
$XeF_2$
0%
$XeO_6^{4-}$
0%
$O_3$

Explanation

$Xe$ and

$F$ have

$3$ lone pairs on it in

$XeF_2$

In which of the following molecules partial charge separation does not take place?

0%
Chlorine
0%
Hydrochloric acid
0%
Water
0%
Ammonia

How many lone pair of electrons are present on the central atom of CH

$_4$ , NH

$_3$ , PCl

$_3$ and PCl

$_5$ molecules?

0%
$0, 1, 1, 0$
0%
$0, 1, 1, 1$
0%
$0, 0, 1, 1$
0%
$0, 0, 0, 1$

Explanation

Methane

$(CH_4)$ does not have lone pair of electrons.

Ammonia

$(NH_3)$ has one lone pair of electron on nitrogen atom.

$PCl_3$ has one lone pair of electron on phosphorous atom.

$PCl_5$ does not have lone pair of electrons.

So answer is A.

Which of the following molecules does not have a lone pair of electrons?

0%
$HCl$
0%
$CO_2$
0%
$CH_4$
0%
$NH_3$

Explanation

$CH_4$ does not has a lone pair because every electron in valence shell of carbon atoms is shared with Hydrogen atoms.

The total number of electrons involved in the formation of CH

$_4$ molecule are:

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

Which of the following species has four lone pair of electrons?

0%
$OH^-$
0%
$Cl^-$
0%
$O^-$
0%
$I$

Explanation

$Cl^-$ among the species has four lone pairs of electrons.

In which of the following molecules, is the covalent bond most polar?

0%
HI
0%
HBr
0%
HCl
0%
H $_2$

The nature of the chemical bond in MgO is:

0%
electrovalent
0%
covalent
0%
dative
0%
polar covalent

Explanation

$Mg$ is a metal and completes its octet by the loss of

$e^-$ . Oxygen is a non-metal and completes its octet by gain of

$e^-$ . So, the formation of

$MgO$ involves a transfer of an electron from metal to non-metal. Hence, the bonding is electrovalent.

The charge in coulombs on

$N^{3-}$ ion is:

0%
$4.80\times 10^{-19}$ C
0%
$1.60\times 10^{-19}$ C
0%
$-3$ C
0%
$1.8\times 10^{-24}$ C

Explanation

Charge on one electron

$= 1.6 \times { 10 }^{ -19 }\ C$

${ N }^{ 3- }$ has

$-3$ charge on it.

Charge on three electron

$=3 \times 1.6 \times { 10 }^{ -19 }C=4.8 \times { 10 }^{ -19 }C$

Which of the following pairs of compounds will not form H-bonds with each other?

0%
Ethanol-methanol
0%
Ethanol-water
0%
Chloroform-acetone
0%
Carbon tetrachloride-acetone

Explanation

$CCl_4$ and

$(CH_3)_2C = O$ will not form H-bonds with each other. Both acetone and carbon tetrachloride are non-polar covalent compounds which are not miscible with each other and do not form hydrogen bonds whereas the others are polar and form H-bonds. In option C, the O of acetone and H of chloroform can form hydrogen bond but it is not that stable as other molecules with N,O and F.

Among the following mixture, maximum dipole-dipole interaction is present in:

0%
benzene and ethanol
0%
acetonitrile and acetone
0%
$KCl$ and water
0%
benzene and carbon tetra chloride

Explanation

Acetonitrile ( $CH_2=CHCN$ ) and acetone ( $CH_3COCH_3$ ) both are polar molecules.So they show more dipole-dipole interactions.

Hence option B is correct.

The structure of the tetrathionate ion is:

Explanation

The tetrathionate anion,

$S_4O_6^{2-}$ , is a sulfur oxoanion derived from the compound tetrathionic acid,

$H_2S_4O_6$ . Two of the sulfur atoms present in the ion are in oxidation state 0 and two are in oxidation state +5.

Hence, option C is correct.

What type of bond is formed if atom looses one or more valence electrons to give positive metal ions?

0%
Metallic
0%
Hydrogen bonding
0%
Polar covalent
0%
Ionic
0%
Covalent

$1-butene$ number of

$\sigma$ - bonds is:

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

Explanation

The first bond between any two atoms is

$\sigma$ and rest are

$\pi$ bonds. Draw structure of

$butene-1$ and count

$\sigma$ bonds in it.

$\therefore$ It has

$11 \sigma$ and

$1 \pi$ bond.
1142014_460283_ans_1b82d8e256b44c00a4c36ee432ca7ce0.jpg

The maximum covalency exhibited by an element isThe number of unpaired electrons in the ground state of an atom is:

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

Explanation

Sulphur can show a maximum valency of 6 because of vacant d-orbital. Sulphur atom has a configuration of

$[Ne] 3s^23p^4$

In-ground state, sulphur atom has 2 unpaired electrons.

Bonding that holds a sample of barium iodide (

$BaI_2$ ) together is :

0%
metallic bonding
0%
non-polar covalent bonding
0%
hydrogen bonding
0%
ionic bonding
0%
polar covalent bonding

Explanation

Bonding that holds a sample of barium iodide,

$BaI_2$ , together is ionic bonding.

An ionic compound is formed between a metal (Ba) and a non-metal (iodine).

Hence, the correct option is D.

In which of the following pairs, the central atoms have the same number of lone pairs of electrons?

0%
$PCl_{5}, \ BrF_{5}$
0%
$XeF_{2}, \ {ICl}$
0%
$XeF_{4},\ {ClO_{4}}^{-}$
0%
$SCl_{4}, \ CH_{4}$

Explanation

Explanation:

A lone pair refers to a pair of valence electrons that are not shared with another atom in a covalent bond and is sometimes called an unshared pair

Option A:

$PCl_5$ has

$0$ lone pair, as all-electron groups are bonding pairs, so the structure is designated as

$AX_5$

whereas in

$BrF_5$ there are

$7$ electrons in

$Br$ valence shell from which

$5$ electrons went to make bond with

$F$ , while the remaining two makes a lone pair. and thus only

$1$ lone pair exist in

$BrF_5$

Option B:

$XeF_2$ there are two pairs of bonded electrons and

$3$ lone pairs of electrons whereas in

$ICl$ number of lone pairs are

$3$ and one pair of bonded electrons

Option C:

$XeF_4$ there are 3 lone pairs present on each

$F$ atom and two lone pairs present on

$Xe$ . So, the total number of lone pairs are: 4 x 3 + 2 = 14

whereas in

$ClO_4^-$ there are nine lone pairs on atoms. There are three oxygen atoms that are connected through double bonds to the chlorine atom. Each of those oxygen atoms has two lone pairs and 3 in one.

Option D: In

$SCl_4$

$1$ lone pair is present.

$S$ has 10 valence electrons in

$SCl_4$ from 4 bonds and 1 lone pair.

whereas in

$CH_4$ all are bond pair

$0$ lone pair. If these are all bond pairs the molecular geometry is tetrahedral

Hence option B is correct

Which one of the following is correct regarding

$\sigma$ molecular orbital ?

0%
It is formed by the partial overlap of atomic orbitals at right angle to inter-nuclear axis.
0%
The overlapping region is very less.
0%
It is a very weak bond
0%
The rotation along the inter nuclear axis is symmetric.

Explanation

$\displaystyle \sigma$ molecular orbital
(A) It is formed by the overlap of atomic orbitals in the direction of inter-nuclear axis.
(B) The overlapping region is significant.
(C) It is a strong bond
(D) The rotation along the inter nuclear axis is symmetric.

The molecules having the same number of lone pair as

$Xe$ are:

$(i) { Xe } {O}_{3}$

$(ii) \ { Xe } O{F}_{4}$

$(iii)\ { Xe } {F}_{6}$

0%
$(i)$ and $(ii)$
0%
$(i)$ and $(iii)$
0%
$(ii)$ and $(iii)$
0%
$(i),\ (ii)$ and $(iii)$

Dipole moment of is

$1.5\;D$ . The dipole moment of is

0%
$1.5\;D$
0%
$2.25\;D$
0%
$1\;D$
0%
$3\;D$

Explanation

Given, for the molecule $\mu_1=1.5\,D$

For $\mu=0$ (as it is symmetrical)

Hence, for $\mu$ will be $1.5\,D$ .

1295376_471358_ans_78030195272944b6957dfdaa6fc7a2a5.png

The valency of nitrogen in

$NH_{3}$ is :

0%
$3$
0%
$4$
0%
$5$
0%
$1$

Explanation

Structure of

$N{H}_{3}$

(Image )

$3$ bp and

$1$ lp around N

Since N-atom has formed

$3$ bonds.

Its valency is

$3$

Which compound is not paired with its correct name?

0%
$FeCl_2$ / iron(II) chloride
0%
$K_2O$ / potassium oxide
0%
$NO_2$ / nitrogen dioxide
0%
$PCl_3$ / potassium trichloride
0%
$NH_4Cl$ / ammonium chloride

Explanation

$PCl_3$ / potassium trichloride represents incorrect pair.

$\displaystyle PCl_3$ is phosphorous trichloride.
The elemental symbol P represents phosphorous and not potassium.

A sigma bond forms :

0%
when two orbitals directly overlap but there is only one bonding interaction.
0%
when two orbitals directly overlap and there are two bonding interaction.
0%
when two valence electrons combine into one orbital.
0%
when two atoms overlap and the valence electrons bond.
0%
the creation of an S shaped orbital.

The number of bonds predicted for

$N_{2}$ :

0%
zero
0%
one
0%
two
0%
three
0%
four

Explanation

The number of bonds predicted for

$N_2$ molecule is three.

Each N atom has 5 valence electrons. It needs 3 more electrons to complete its octet. It shares 3 electrons to form 3 chemical bonds. Thus each N atom has completed its octet.

$\displaystyle : N \equiv N :$

Hence the correct option is D.

Which of the following is/has the weakest attractive forces?

0%
Vander Waals
0%
Coordinate covalent bonding
0%
Covalent bonding
0%
Polar covalent bonding
0%
Ionic bonding

Dipoles are found in:

0%
Water
0%
Hydrogen bromide
0%
Iron
0%
Argon
0%
Sodium chloride

Non-polar molecule having polar bonds is:

0%
phosphorus trichloride (PCl $_3$ )
0%
sulfur dioxide (SO $_2$ )
0%
water (H $_2$ O)
0%
hydrogen chloride (HCl)
0%
carbon dioxide (CO $_2$ )

Explanation

Non-polar molecule having polar bonds is carbon dioxide,

$\displaystyle CO_2$ .

$\displaystyle C=O$ bonds are polar as two atoms (C and O) significantly differ in electronegativity.
Carbon dioxide is a linear molecule,

$\displaystyle O=C=O$
The individual bond dipoles cancel each other.
Hence, the molecule on the whole is non polar.

Which of the following ions or molecules has a nonzero dipole moment?

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

Explanation

Option D is the correct answer.

Sulphur dioxide is a bent molecule. An image has been attached to make the molecule look clear. The angle between the bonds is about 119 degrees. Oxygen is more electronegative in nature than Suphur, hence

$SO_2$ has a dipole moment which is towards the oxygen atom.

Which one among the following is not paramagnetic?

0%
$Be^-$
0%
$Ne^{2+}$
0%
$Cl^-$
0%
$As^+$

Explanation

$\displaystyle Cl^-$ ion is not paramagnetic. It is dimagnetic as all electrons are paired. It has electronic configuration

$\displaystyle [Ne] 3s^23p^6$

Among the following identify the molecules with

$sp^2 - p$

$\sigma$ bonds.

0%
$BeCl_2$
0%
$C_2H_2$
0%
$BCl_3$
0%
$BF_3$

Explanation

The molecules with

$\displaystyle sp^2 - p \: \sigma$ bonds are (C)

$\displaystyle BCl_3$ and (D)

$\displaystyle BF_3$ . These molecules are planar. The central

$\displaystyle B$ atom is

$\displaystyle sp^2$ hybridised. These molecules have trigonal planar geometry.

Note:

$\displaystyle BeCl_2$ has

$\displaystyle sp - p \: \sigma$ bonds.

$\displaystyle C_2H_2$ has

$\displaystyle sp-s \: \sigma$ bonds.

Number of lone pairs and bond pairs in ammonia molecule is ___________.

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

Statement

$1$ : The molecule

$SO_{2}$ has a net dipole.
Statement

$2$ : Oxygen has the higher electronegativity than sulfur.

0%
Both Statement $1$ and Statement $2$ are correct and Statement $2$ is the correct explanation of Statement $1$ .
0%
Both Statement $1$ and Statement $2$ are correct, but Statement $2$ is NOT the correct explanation of Statement 1.
0%
Statement $1$ is correct, but Statement $2$ is not correct.
0%
Statement $1$ is not correct, but Statement $2$ is correct.

Explanation

The

$SO_2$ molecule has bent structure. This is because in general, electrons want to stay as far apart from each other as possible.

Oxygen has a stronger pull on the electron in the bond betweem

$S$ and

$O$ , the result is a dipole moment, which can be seem more easily using vectors.

1305217_525871_ans_85dcc65c38cd482b90780d5524ed2980.PNG

The correct order of strength of bonds is:

0%
$s-s>s-p>p-p$
0%
$s - s < s - p < p - p$
0%
$s-p>s-s>p-p$
0%
$s-s>p-p>s-p$

Explanation

The correct order of strength of bonds is

$\displaystyle s−s>s−p>p−p$
The extent of overlap is higher for

$\displaystyle s$ orbital than for

$\displaystyle p$ orbital. Higher is the extent of overlap, stronger is the bond formed.

The bonds present in $N_2O_5$ are :

0%
only ionic
0%
covalent and coordinate
0%
only covalent
0%
covalent and ionic

Explanation

Since nitrogen and oxygen are non-metals, the bonds between them are NOT ionic.

The structure of $N_2O_5$ is given above.

There are both covalent and coordinate covalent bonds.

Hence correct option is B.

Using formal charge, which of the following would be the least likely structure for dinitrogen monoxide based on the knowledge that oxygen is NOT the central atom?

0%
0%
0%
0%
All three structures are equally likely as drawn.

The polarity of a molecular compound results from an imbalance of the sharing electrons in the bond based on electronegativity differences and atomic structure.
Which of the following gases would be the least polar?

0%
Nitrogen monoxide
0%
Ammonia
0%
Ethane $(CH_{3}CH_{3})$ .
0%
Hydrogen cyanide.

Use the figure below to answer the question that follows.
For which of the molecules in the figure does the central atom, or atoms, have

$sp^{2}$ hybridization?

0%
III only
0%
IV only
0%
III and IV only
0%
II and IV only

Explanation

$N$ in

$NO_2^-$ is having

$sp^2$ hybridization.

As hybridization =

$\cfrac{1}{2}$ [number of valence electrons in central atom - number of monovalent atom attached - cationic charge + anionic charge]

$\cfrac{1}{2}[5+1] = 3 = sp^2$

Main axis of diatomic molecule is

$Z$ . The orbitals

$p_x$ and

$p_y$ overlap to form :

0%
$\pi$ -molecular orbital
0%
$\sigma$ -molecular orbital
0%
$\delta$ -molecular orbital
0%
no bond is formed

Explanation

If the main axis of diatomic molecule is Z axis, then the orbitals

$P_x$ and

$P_y$ overlap to form

$\pi$ bond.

Which of the following structures has a dipole moment in terms of atom and unbonded electron pair placement from a Lewis Dot structure view?

0%
$IF_{4}^{1-}$
0%
$XeF_{4}$
0%
$IF_{4}^{1+}$
0%
$NH_{4}^{1+}$

Explanation

${ IF }_{ 4 }^{ + }$ has a net dipole towards

$F$ as fluorine is more electronegative than iodine. Thus, fluorine pulls the

${ e }^{ - }$ density towards itself.

Correct order of $CO$ bond length is:

0%
$CO^+ >{ C }_{ 2 }{ O }_{ 4 }^{ 2- } >CO$
0%
$CO^+ > CO > { C }_{ 2 }{ O }_{ 4 }^{ 2- }$
0%
${ C }_{ 2 }{ O }_{ 4 }^{ 2- }> CO^+ > CO$
0%
${ C }_{ 2 }{ O }_{ 4 }^{ 2- }> CO> CO^+$

CBSE Questions for Class 11 Medical Chemistry Chemical Bonding And Molecular Structure Quiz 4 - 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 4 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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