NCERT Exemplar Class 11 Chemistry Solutions Chapter 4 Chemical Bonding and Molecular Structure 2021

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Q: What all topics are covered in this chapter?

Different topics are covered in this chapter i.e., VSEPR theory, Lewis structures, The concept of hybridization, Hydrogen bonding, etc.

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It is very important to understand NCERT Exemplar Class 11 Chemistry Solutions Chapter 4 as it will help you to score the required marks in the examination.

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There are a total of 40 questions in this chapter which are answered by the experts.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 4 Chemical Bonding and Molecular Structure

Edited By Sumit Saini | Updated on Sep 10, 2022 04:46 PM IST

NCERT Exemplar Class 11 Chemistry solutions chapter 4 Chemical Bonding and Molecular Structure focuses on the chemical bonding and molecular structure. It helps understanding the several fundamental concepts in the field of chemistry. This chapter of NCERT Class 11 Chemistry Solutions deals with different topics that are related to chemical bonding and molecular structure. Class 11 Chemistry Exemplar solutions chapter 4 also aids students in solving different questions and finding the appropriate answers accordingly.
Also, check NCERT Solutions for Class 11 for other subjects.

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NCERT Exemplar Class 11 Chemistry solutions chapter 4: MCQ (Type 1)
NCERT Exemplar Class 11 Chemistry Solutions Chapter 4: MCQ (Type 2)
NCERT Exemplar Class 11 Chemistry Solutions Chapter 4: Short Answer Type
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NCERT Exemplar Class 11 Chemistry Solutions Chapter 4: Assertion and Reason Type
NCERT Exemplar Class 11 Chemistry Solutions Chapter 4: Long Answer Type
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NCERT Exemplar Class 11 Chemistry Solutions Chapter Wise

NCERT Exemplar Class 11 Chemistry solutions chapter 4: MCQ (Type 1)

Question:1

Isostructural species are those which have the same shape and hybridization. Among the given species identify the isostructural pairs.
(i) $\left [ NF_{3}\; and\; BF_{3} \right ]$
(ii) $\left [ BF{_{4}}^{-}\; and\; NH_{4}^{+} \right ]$
(iii) $\left [ BCl_{3}\; and\; BrCl_{3} \right ]$
(iv) $\left [ NH_{3}\; and\; NO{_{3}}^{-} \right ]$
Answer:

The answer is the option (ii) $\left [ BF{_{4}}^{-}\; and\; NH_{4}^{+} \right ]$
Explanation: The pair above is isostructural because both the molecules $\left [ BF{_{4}}^{-}\; and\; NH_{4}^{+} \right ]$ are sp3 hybridized and their shape is tetrahedral.

Question:2

Polarity in a molecule and hence the dipole moment depends primarily on electronegativity of the constituent atoms and shape of a molecule. Which of the following has the highest dipole moment?
(i) $CO_{2}$
(ii) $HI$
(iii) $H_{2}O$
(iv) $SO_{2}$
Answer:

The answer is the option (iii) $H_{2}O$
Explanation: $H_{2}O$ has the highest dipole moment because it is highly electronegative and has two lone pair on it.

Question:3

The types of hybrid orbitals of nitrogen in $NO{_{2}}^{+},NO{_{3}}^{-}$ and $NH{_{4}}^{+}$ respectively are expected to be
(i) $sp, sp^{3} \; and\; sp^{2}$
(ii) $sp, sp^{2} \; and\; sp^{3}$
(iii) $sp^{2}, sp \; and\; sp^{3}$
(iv) $sp^{2}, sp^{3} \; and\; sp$
Answer:

The answer is the option (ii) $sp, sp^{2} \; and\; sp^{3}$
Explanation: $NO{_{2}}^{+}$ is sp hybridized while $NO{_{3}}^{-}$ is sp²hybridised and $NH{_{4}}^{+}$ is $Sp^{3}$ hybridized.

Question:4

Hydrogen bonds are formed in many compounds e.g., $H_{2}O, HF,NH_{3}.$ The boiling point of such compounds depends to a large extent on the strength of hydrogen bond and the number of hydrogen bonds. The correct decreasing order of the boiling points of above compounds is :
(i) $HF>H_{2}O>NH_{3}$
(ii) $H_{2}O>HF>NH_{3}$
(iii) $NH_{3}>HF>H_{2}O$
(iv) $NH_{3}>H_{2}O>HF$

Answer:

The answer is the option (ii) $H_{2}O>HF>NH_{3}$
Explanation: The factors that affect the strength of hydrogen bond are electronegativity, size of atom and number of hydrogen bonds per mol. Usually, F is the most electronegative element, but, in H₂O, there are more H-bonds, and their size is smaller than N.

Question:5

In $PO{_{4}}^{3-}$ ion the formal charge on the oxygen atom of $P-O$ bond is
(i) + 1
(ii) – 1
(iii) – 0.75
(iv) + 0.75

Answer:

The answer is the option (ii) -1
Explanation: Formal charge in an atom of a polyatomic molecule or ion can also 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.
Formula:
The formal charge of the atom in the molecule or ion = (Number of the valence electron in the free atom) – (Number of lone pair electrons + ½ number of bonding electrons)
Therefore, the formal charge on each O-atom
= 6 – (6+ 1/2 × 2) = 6 – 7 = -1

Question:6

In $NO{_{3}}^{-}$ ion, the number of bond pairs and lone pairs of electrons on nitrogen atom are
(i) 2, 2
(ii) 3, 1
(iii) 1, 3
(iv) 4, 0
Answer:

The answer is the option (iv) 4, 0
Explanation: N, viz., the central atom of the molecule is surrounded by 2 covalent bonds with 1 oxygen atom and 2 coordinate covalent bonds with 2 oxygen atoms. Therefore, there are 4 bond pairs and no lone pair on the nitrogen atom in $NO{_{3}}^{-}$
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Question:7

Which of the following species has tetrahedral geometry?
(i) $BH{_{4}}^{-}$
(ii) $NH{_{2}}^{-}$
(iii) $CO{_{3}}^{2-}$
(iv) $H_{3}O^{+}$
Answer:

The answer is the option (i) $BH{_{4}}^{-}$
Explanation: 4 bond pairs surround Boron here.
In $BH{_{4}}^{-}$ , there are 4 bond pair and no lone pair.
Its sp³ hybridized, and therefore has tetrahedral geometry
$CO{_{3}}^{-2}$ is triangular planar.

Question:8

Number of $\pi$ bonds and σ bonds in the following structure is–

1660043565379
(i) 6, 19
(ii) 4, 20
(iii) 5, 19
(iv) 5, 20
Answer:

The answer is the option (iii) 5, 19
Explanation: Here, each C atom is sp² hybridized and surrounded by 3 sigma bonds and 1 pi bond between two C atoms.
8 C – H + 11 C – C bonds = 19 bonds.
There are $5\pi$ - bonds and 19 bonds.

Question:9

Which molecule/ion out of the following does not contain unpaired electrons?
(i) $N{_{2}}^{+}$
(ii) $O_{2}$
(iii) $O{_{2}}^{2-}$
(iv) $B_{2}$
Answer:

The answer is the option (iii) $O{_{2}}^{2-}$
Explanation:
The molecular orbital configuration of $O{_{2}}^{2-}$ ion is

$KK\sigma (2s)^{2}\sigma *(2s)^{2}\sigma (2p_{z})^{2}\pi (2p_{x})^{2}\pi (2p_{y})^{2}\pi *(2p_{x})^{2}\pi *(2p_{y})^{2}$

Where, $KK$ represents non-bonding molecular orbital of 1s orbital; thus, $O{_{2}}^{2-}$ possesses no unpaired electrons.

Question:10

In which of the following molecule/ion, all the bonds are not equal?
(i) $XeF_{4}$
(ii) $BF{_{4}}^{-}$
(iii) $C_{2}H_{4}$
(iv) $SiF_{4}$
Answer:

The answer is the option (iii) $C_{2}H_{4}$
Explanation: In $C_{2}H_{4}$ all the bonds are not equal because here, each C atom is surrounded by 3 $\sigma$ bonds and $1\pi$ bond.

Question:11

In which of the following substances will hydrogen bond be strongest?
(i) $HCl$
(ii) $H_{2}O$
(iii) $HI$
(iv) $H_{2}S$
Answer:

The answer is the option (ii) $H_{2}O$
Explanation: The hydrogen bond will be the strongest in $H_{2}O$ because compared to the other elements Oxygen is more electronegative as well as it's small in size.
capture-11

Question:12

If the electronic configuration of an element is $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{2} 4s^{2},$ the four electrons involved in chemical bond formation will be_____.
(i) $3p^{6}$
(ii) $3p^{6},4s^{2}$
(iii) $3p^{6},3d^{2}$
(iv) $3d^{2},4s^{2}$
Answer:

The answer is the option (iv) $3d^{2},4s^{2}$
Explanation: In transition elements ns electrons and (n-1)d electrons participate, while in bonding, electrons and valence electrons from penultimate shell participate. Therefore, the bond formation will be $3d^{2},4s^{2}$

Question:13

Which of the following angle corresponds to sp² hybridization?
(i) 90°
(ii) 120°
(iii) 180°
(iv) 109°
Answer:

The answer is the option (ii) 120°
Explanation: Triangular planar is forming an angle of 120°with each other as a result of the sp² hybridization geometry.

Question:14

The electronic configurations of three elements A, B and C are given below.
Answer the questions from 14 to 17 on the basis of these configurations.
A $1s^{2} 2s^{2} 2p^{6}$
B $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$
C $1 s^{2}2s^{2}2p^{6}3s^{2}3p^{5}$
Stable form of A may be represented by the formula
(i) A
(ii) A₂
(iii) A₃
(iv) A₄
Answer:

The answer is the option (i) A
Explanation: The formula will be A only, as the octet of A is complete. It has an atomic number of 10.

Question:15

The electronic configurations of three elements A, B and C are given below.
Answer the questions from 14 to 17 on the basis of these configurations.
A $1s^{2} 2s^{2} 2p^{6}$
B $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$
C $1 s^{2}2s^{2}2p^{6}3s^{2}3p^{5}$
Stable form of C may be represented by the formula
(i) C
(ii) C₂
(iii) C₃
(iv) C₄
Answer:

The answer is the option (ii) C₂
Explanation:
Bond order $=\frac{1}{2}[6-2]=2;$ ; hence stable form of c i.e. dichlorine (Cl₂) may be represented as C₂.

Question:16

The electronic configurations of three elements A, B and C are given below.
Answer the questions from 14 to 17 on the basis of these configurations.
A $1s^{2} 2s^{2} 2p^{6}$
B $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$
C $1 s^{2}2s^{2}2p^{6}3s^{2}3p^{5}$
The molecular formula of the compound formed from B and C will be
(i) BC
(ii) B₂C
(iii) BC₂
(iv) BC₃
Answer:

The answer is the option (iv) BC₃
Explanation: Here, B represents phosphorus(P) and C represents Chlorine(Cl). Therefore, the compound formed will be PCl₃, i.e., BC₃.

Question:17

The electronic configurations of three elements A, B and C are given below.
Answer the questions from 14 to 17 on the basis of these configurations.
A $1s^{2} 2s^{2} 2p^{6}$
B $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$
C $1 s^{2}2s^{2}2p^{6}3s^{2}3p^{5}$
The bond between B and C will be
(i) Ionic
(ii) Covalent
(iii) Hydrogen
(iv) Coordinate
Answer:

The answer is the option (ii) Covalent
Explanation: The compounds B and C are non-metals; hence, they will form a covalent bond between them.

Question:18

Which of the following order of energies of molecular orbitals of $N_{2}$ is correct?
(i) $(\pi 2py)<(\sigma 2pz)<(\pi ^{*}2px)\approx (\pi ^{*}2py)$
(ii) $(\pi 2py)>(\sigma 2pz)>(\pi ^{*}2px)\approx (\pi ^{*}2py)$
(iii) $(\pi 2py)<(\sigma 2pz)>(\pi ^{*}2px)\approx (\pi ^{*}2py)$
(iv) $(\pi 2py)>(\sigma 2pz)<(\pi ^{*}2px)\approx (\pi ^{*}2py)$
Answer:

The answer is the option (i) $(\pi 2py)<(\sigma 2pz)<(\pi ^{*}2px)\approx (\pi ^{*}2py)$
Explanation: Molecules like B₂, C₂ & N₂ have 1 to 3 electrons in p orbital energy of $\sigma$ 2p molecular orbital are greater than that of $\pi2p_{x} \; and \; \pi 2p_{y}$ molecular orbitals.

Question:19

Which of the following statement is not correct from the view point of molecular orbital theory?
(i) Be₂ is not a stable molecule.
(ii) He₂ is not stable but $He{_{2}}^{+}$ is expected to exist.
(iii) Bond strength of N₂ is maximum amongst the homonuclear diatomic molecules belonging to the second period.
(iv) The order of energies of molecular orbitals in N₂ molecule is
$\sigma 2s<\sigma ^{*}2s<\sigma 2p_{z}<(\pi 2px=\pi 2pr)<(\pi ^{*}2px=\pi ^{*}2py)<\sigma ^{*}2pz$
Answer:

The answer is the option (iv) The order of the energies of molecular orbitals in N₂ is
$\sigma 2s<\sigma ^{*}2s<\sigma 2p_{z}<(\pi 2px=\pi 2pr)<(\pi ^{*}2px=\pi ^{*}2py)<\sigma ^{*}2pz$

Question:20

Which of the following options represents the correct bond order :
(i) $O{_{2}}^{-}>O_{2}>O{_{2}}^{+}$
(ii) $O{_{2}}^{-}<O_{2}<O{_{2}}^{+}$
(iii) $O{_{2}}^{-}>O_{2}<O{_{2}}^{+}$
(iv) $O{_{2}}^{-}<O_{2}>O{_{2}}^{+}$
Answer:

The answer is the option (ii) $O{_{2}}^{-}<O_{2}<O{_{2}}^{+}$
Explanation: N_b electrons > N_a electrons as well as, it depends on the bond order.
B.O. = ½(N_b – N_a)

Question:21

The electronic configuration of the outer most shell of the most electronegative element is
(i) $2s^{2}2p^{5}$
(ii) $3s^{2}3p^{5}$
(iii) $4s^{2}4p^{5}$
(iv) $5s^{2}5p^{5}$
Answer:

The answer is the option (i) $2s^{2}2p^{5}$
Explanation: Elements of group 17 has $ns^{2}np^{5}$ electronic configuration. Electronegativity decreases down the group. The elements of group 17 are the most electronegative.

Question:22

Amongst the following elements whose electronic configurations are given below, the one having the highest ionization enthalpy is
(i) $[Ne]3s^{2}3p^{1}$
(ii) $[Ne]3s^{2}3p^{3}$
(iii) $[Ne]3s^{2}3p^{2}$
(iv) $[Ar]3d^{10}4s^{2}4p^{3}$
Answer:

The answer is the option (ii) $[Ne]3s^{2}3p^{3}$
Explanation: As compared to partially filled orbital, half-filled p-orbital is more stable. Hence opt(ii).

NCERT Exemplar Class 11 Chemistry Solutions Chapter 4: MCQ (Type 2)

Question:23

Which of the following have identical bond order?
(i) $CN^{-}$
(ii) $NO^{+}$
(iii) $O{_{2}}^{-}$
(iv) $O{_{2}}^{2-}$
Answer:

The answer is the option (i) $CN^{-}$ and (ii) $NO^{+}$
Explanation: They have identical bond order because in both of the species number of electrons in N_b = number of electrons in N_ab, i.e. 14. Whereas the number of electrons in $O{_{2}}^{-}$ is 17 and in $O{_{2}}^{2-}$ is 18.

Question:24

Which of the following attain the linear structure:
(i) $BeCl_{2}$
(ii) $NCO^{+}$
(iii) $NO_{2}$
(iv) $CS_{2}$
Answer:

The answer is the option (i) $BeCl_{2}$ and (iv) $CS_{2}$
Explanation: Both of the choices have the central atom sp hybridized; therefore both have a linear structure.

Question:25

CO is isoelectronic with
(i) $NO^{+}$
(ii) $N_{2}$
(iii) $SnCl_{2}$
(iv) $NO{_{2}}^{-}$
Answer:

The answer is the option (i) $NO^{+}$ and (ii) $N_{2}$
Explanation: In both the species have the same number of electrons but a different number of protons. Hence, they are isoelectronic with CO.

Question:26

Which of the following species have the same shape?
(i) $CO_{2}$
(ii) $CCl_{4}$
(iii) $O_{3}$
(iv) $NO{_{2}}^{-}$
Answer:

The answer is the option (iii) $O_{3}$ and (iv) $NO{_{2}}^{-}$
capture-26
Explanation: In both the species the central atom has the same hybridized state and geometry. Hence, they have the same shape.

Question:27

Which of the following statements are correct about $CO{_{3}}^{2-}$ ?
(i) The hybridization of central atom is sp³.
(ii) Its resonance structure has one C–O single bond and two C=O double bonds.
(iii) The average formal charge on each oxygen atom is 0.67 units.
(iv) All C–O bond lengths are equal.
Answer:

The answer is the option (iii) The average formal charge on each oxygen atom is 0.67 units and (iv) All C-O bond lengths are equal.
Explanation: In resonating, structures bonds are not fixed, and as a result, all bond lengths are equal.

Question:28

Dimagnetic species are those which contain no unpaired electrons. Which among the following are dimagnetic?
(i) $N_{2}$
(ii) $N{_{2}}^{2-}$
(iii) $O_{2}$
(iv) $O{_{2}}^{2-}$
Answer:

The answer is the option (i) $N_{2}$ and (iv) $O{_{2}}^{2-}$
Explanation: Both of them have no unpaired electrons; therefore, they are diamagnetic.

Question:29

Species having same bond order are :
(i) $N_{2}$
(ii) $N{_{2}}^{-}$
(iii) $F{_{2}}^{+}$
(iv) $O{_{2}}^{-}$
Answer:

The answer is the option (iii) $F{_{2}}^{+}$ and (iv) $O{_{2}}^{-}$
Explanation: They have the same order because, Both the species have the same number of electrons in bonding molecular orbital as well as in antibonding orbital.

Question:30

Which of the following statements are not correct?
(i) NaCl being an ionic compound is a good conductor of electricity in the solid state.
(ii) In canonical structures there is a difference in the arrangement of atoms.
(iii) Hybrid orbitals form stronger bonds than pure orbitals.
(iv) VSEPR Theory can explain the square planar geometry of $XeF_{4}.$
Answer:

The answer is the option (i) NaCl being an ionic compound is a good conductor of electricity in solid-state and (ii) In canonical structures, there is a difference in the arrangement of atoms
Explanation: In (i) ions are not free in solid-state but are strongly bonded through electrostatic forces, and in (ii) it does not show canonical structures being ionic compounds.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 4: Short Answer Type

Question:31

Explain the non-linear shape of $H_{2}S$ and non-planar shape of $PCl_{3}$ using valence shell electron pair repulsion theory.
Answer:

In $H_{2}S$ as well as in $PCl_{3}$ , lone pairs are present along with bond pairs around the central atom. Now, according to VSEPR, theory L.P-L.P > L.P-B.P > B.P-B.P. Therefore, $H_{2}S$ is non-linear, and $PCl_{3}$ is non-planar.

Question:32

Using molecular orbital theory, compare the bond energy and magnetic character of $O{_{2}}^{+}$ and $O{_{2}}^{-}$ species.
Answer:

The molecular orbital configuration of $O{_{2}}^{+}$ and $O{_{2}}^{-}$ has been specified as follows: -
$O^{2+}(15):\sigma (1s)^{2}\; \sigma ^{*}(1s)^{2}\; \sigma (2s)^{2}\sigma ^{*}(2s)^{2}\; \sigma (2p_{z})^{2}\pi ( 2p_{x}^{2}=\pi\ 2p_{y}^{2})\pi ^{*}2p_{x}^{1}$
$O^{2+}(15):\sigma (1s)^{2}\; \sigma ^{*}(1s)^{2}\; \sigma (2s)^{2}\sigma ^{*}(2s)^{2}\; \sigma (2p_{z})^{2}\pi ( 2p_{x}^{2}=\pi\ 2p_{y}^{2})\(pi ^{*}2p_{x}^{2}\pi ^{*}2p_{y}^{1})$
Therefore, the bond order for $O{_{2}}^{+}= 10-5/2 = 2.5$
And the bond order for $O{_{2}}^{-}= 10-7/2 = 1.5$
We know that as per the molecular orbital theory, greater is the bond order, greater is the bond energy. Therefore, $O{_{2}}^{+}$ is more stable than $O{_{2}}^{-}$ .

Question:33

Explain the shape of $BrF_{5}.$
Answer:

The central atom in $BrF_{5}.$ is bromine. Its hybridization is $sp^{3}d^{2}$ .
In total, there are 7 valence electrons in a Br atom out of which 5 valence electrons are used for making a pair with the F atoms, and two valence electrons are used for making a lone pair of electrons. We know that the bond pair and lone pair repel each other. Hence, the shape is square pyramidal.
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Question:34

Structures of molecules of two compounds are given below :
capture-34
(a) Which of the two compounds will have intermolecular hydrogen bonding and which compound is expected to show intramolecular hydrogen bonding.
(b) The melting point of a compound depends on, among other things, the extent of hydrogen bonding. On this basis, explain which of the above two compounds will show a higher melting point.
(c) The solubility of compounds in water depends on the power to form hydrogen bonds with water. Which of the above compounds will form a hydrogen bond with water easily and be more soluble in it?
Answer:

(a) $NO_{2}$ and $OH$ groups are close together compared to the other compound; therefore, compound (I) will form an intramolecular hydrogen bonding, whereas compound (II) will show intermolecular bonding.
(b) Compound (II) forms intermolecular H-bonds, joining, more and more molecules are joined together through H-bonds; hence it will have a higher melting point.
(c) Compound (I) will not be able to form hydrogen bonds with water due to intramolecular H-bonding, thus, will be less soluble in it while compound (II) will be able to form H-bond with water more easily and will be soluble in water.

Question:35

Why does the type of overlap given in the following figure not result in bond formation?
capture-35
Answer:

No bond formation occurs because the same charges repel as well as there is zero overlappings here.

Question:36

Both these compounds have different structures due to the different state of hybridisation of the central atom i.e., phosphous and iodine.
PCl₅:sp³d
lF₅:sp³d²

Answer:

In , $PCl_{5}$ P is surrounded by 5 bond pairs and no lone pairs, whereas in $IF_{5}$ , iodine atom is surrounded by 5 bond pairs and one lone pair; therefore the one lone pair in $IF_{5}$ makes its shape different.

Question:37

In both water and dimethyl ether $(CH_{3} - \ddot{O}- CH_{3}),$ oxygen atom is central atom, and has the same hybridization, yet they have different bond angles. Which one has greater bond angle? Give reason.
Answer:

Because of more repulsion between bond pairs of $CH_{3}$ groups attached in ether than between bond pair of hydrogen atoms attached to oxygen in the water. Dimethyl ether will have a larger bond angle. The carbon of $CH_{3}$ in ether is attached to three hydrogen atoms through bonds, and electron pairs of these bonds add to the electronic charge density on the carbon atom. Hence, the repulsion between two $-CH_{3}$ groups will be more than that between two hydrogen atoms.

Question:38

Write Lewis structure of the following compounds and show formal charge on each atom. $HNO_{3},NO_{2},H_{2}SO_{4}$
Answer:

The formula for calculation of the formal charge is as follows: -
Formal charge = ½ [total no: of bonding or shared electrons]
Oxygen with single bond will have formal charge of = 6-6-2/2 = -1
Oxygen with double bond will have formal charge of = 6-4-4/2 = 0
Therefore, nitrogen will have a formal charge of = 5-2-6/2 = 0
Oxygen 1 and 4 will have a formal charge = 6-4-4/2= 0
Oxygen 2 and 3 will have a formal charge = 6-4-4/2=0
Hydrogen 1 and 2 will have a formal charge = 1-0-2/2=0
Sulfur will have a formal charge =6-0-12/2 = 0
i) HNO₃
capture-38-1

ii) NO₂

capture-38-2

iii) H₂SO₄

capture-38-3

Question:39

The energy of $\sigma 2p_{z}$ molecular orbital is greater than $\pi 2p_{x}$ and $\pi 2p_{y}$ molecular orbitals in nitrogen molecule. Write the complete sequence of energy levels in the increasing order of energy in the molecule. Compare the relative stability and the magnetic behaviour of the following species :
$N_{2},N_{2}^{+},N_{2}^{-},N{_{2}}^{2+}$
Answer:

We know that the general sequence of the energy level of the molecular orbital can be written as,
$\sigma 1s<\sigma ^{*}1s<\sigma 2s<\sigma ^{*}2s<\pi 2px=\pi 2py<\sigma 2pz$
$N2\; \sigma 1s^{2}<\sigma ^{*}1s^{2}<\sigma 2s^{2}<\sigma ^{*}2s^{2}<\pi 2p^{2}x=\pi 2p^{2}y<\sigma 2p^{2}z$
$N2^{+}\; \sigma 1s^{2}<\sigma ^{*}1s^{2}<\sigma 2s^{2}<\sigma ^{*}2s^{2}<\pi 2p^{2}x=\pi 2p^{2}y<\sigma 2p^{1}z$
$N2^{-}\; \sigma 1s^{2}<\sigma ^{*}1s^{2}<\sigma 2s^{2}<\sigma ^{*}2s^{2}<\pi 2p^{2}x=\pi 2p^{2}y\sigma 2p^{2}z\sigma 2p^{2}x$
$N{_{2}}^{2+}\; \sigma 1s^{2}<\sigma ^{*}1s^{2}<\sigma 2s^{2}<\sigma ^{*}2s^{2}<\pi 2p^{2}x=\pi 2p^{2}y$
We are also aware that, Bond order = ½ [electrons in BMO – electrons in ABMO]
In case of $N_{2}=10-\frac{4}{2}=3$
Hence, the bond order for $N{_{2}}^{+}=9-\frac{4}{2}=2.5$
Hence, the bond order for $N{_{2}}^{-}=10-\frac{5}{2}=2.5$
Hence, the bond order for $N{_{2}}^{2+}=8-\frac{4}{2}=2$
Therefore, we can conclude that the order of stability is:
$N2>N{_{2}}^{-}>N{_{2}}^{+}>N{_{2}}^{2+}$

Question:40

What is the effect of the following processes on the bond order in $N_{2}$ and $O_{2}$ ?
(i) $N_{2}\rightarrow N{_{2}}^{+}+e^{-}$
(ii) $O_{2}\rightarrow O{_{2}}^{+}+e^{-}$

Answer:

(i) $N_{2}$ possesses a total of 14 electrons. When one electron is donated, these electrons are removed from the Bonding molecular orbital. Therefore, the BO for $N_{2}=3$
(ii) $O_{2}$ possesses a total of 16 electrons, out of which 8 electrons are in the molecular orbitals and 4 are in the antibonding molecular orbitals. Hence, the BO for $O_{2}=2$

Question:41

Give reasons for the following :
(i) Covalent bonds are directional bonds while ionic bonds are nondirectional.
(ii) The water molecule has a bent structure whereas carbon dioxide molecule is linear.
(iii) Ethyne molecule is linear.
Answer:

(i) A covalent bond is formed by the overlapping of the atomic orbitals, therefore, it’s a directional bond. Ionic bond is formed by the transference of electrons and electrostatic field of an ion is there which is non-directional, hence it’s a non-directional bond. Positive ion is surrounded by number of anions in any direction depending upon the size and similarly, negative ion is surrounded by no. of cations in any directional depending upon the size.
(ii) In $H_{2}O$ , oxygen atom is sp³ hybridized and we already know that it is surrounded by two lone pairs and two bonded pairs. There are four sp³ hybrid orbitals which give distorted tetrahedral geometry in which the two corners are occupied by hydrogen atoms and other two by lone pairs. The bond angle is reduced to 104.5° and 109.5°due to greater repulsive forces between 1p-1p and molecules acquire a bent or V-shaped structure. Carbon atom is sp hybridized in $CO_{2}$ and the two sp hybrid orbitals forms linear shape, i.e., an angle of 180° and hence are oriented in opposite direction.
(iii) Both the carbon atoms are sp hybridized in ethyne molecule, and there are two unhybridized orbitals. Hybridized orbitals of both the carbon atoms are directed in opposite directions forming an angle of 180°, and hence, a linear structure is formed.

Question:42

What is an ionic bond? With two suitable examples explain the difference between an ionic and a covalent bond?
Answer:

Ionic bond: The bond which is formed by the transferring electrons from one atom to other atom completely, and as a result positive and negative ions are formed in this bond. Here, the ions are held together through electrostatic force of attraction. e.g.,
The formation of calcium fluoride,
$Ca\rightarrow Ca^{2+}+2e^{-}$
$[Ar]4s^{2}[Ar]$
$F+e^{-}\rightarrow F^{-}$
$[He]2s^{2}2p^{5}[He]2s^{2}2p^{6}\; or[Ne]$
$Ca^{2+}+2F^{-}\rightarrow CaF_{2}\; or\; Ca^{2+}(F^{-})_{2}$
The information of $NaCl$
$Na\rightarrow Na^{+}+e^{-}$
$[Ne]3s^{2}3p^{5}[Ne]3s^{2}3p^{6}\; or[Ar]$
$Na^{+}+Cl^{-}\rightarrow NaCl\; or\; Na^{+}Cl^{-}$
Covalent bond: The bond which is formed between the two atoms of non-metals by mutual sharing of electron between them is called covalent bond, e.g., the formation of chlorine molecule can be explained as
aerfed
Also formation of HCl:
wefsadw

Question:43

Arrange the following bonds in order of increasing ionic character giving a reason
$N-H, F-H, C-H \; and\; O-H$
Answer:

Ionic character of the bond depends on the electronegativity.
Ionic character $\alpha$ to the difference of electronegativity.
In the following bonds, hydrogen is in all but have different electronegativity.
$N-H, F-H, C-H \; and\; O-H$
Increasing order of electronegativity is:
$C<N<O<F$
Therefore, ionic character of the bond in increasing order will be-
$C-H < N-H < O-H < F-H$

Question:44

Explain why $CO{_{3}}^{2-}$ ion cannot be represented by a single Lewis structure. How can it be best represented?
Answer:

In ion, $CO{_{3}}^{2-}$ there are three C-O bonds, and all have same bond length.
The reason is that they show resonating structures not only one structure. All the properties are explained below by the Lewis structure.
q44-image

Question:45

Predict the hybridization of each carbon in the molecule of the organic compound given below. Also indicate the total number of sigma and pi bonds in this molecule.
q45-image
Answer:

In the given structure, two carbon atoms are sp hybridized and linked through triple bond, two carbon atoms are sp² hybridized and linked through double bonds to O atoms and one is sp³ hybridized and that is linked to two C atoms through single bonds and to two H atoms through single bonds, hence, there are total 5 carbon atoms.
Therefore it is clear that in the molecule, there are 11 $\sigma$ bonds and 4 $\pi$ bonds.

Question:46

Group the following as linear and non-linear molecules :
$H_{2}O, HOCl, BeCl_{2}, Cl_{2}O$
Answer:

Due to lp-lp repulsion, all the molecules except $BeCl_{2}$ are non-linear. It is because in all other molecules, central atom O is surrounded by two lone pairs.
The structure of $BeCl_{2}$ is-
Cl-Be-Cl

Question:47

Elements X, Y and Z have 4, 5 and 7 valence electrons respectively.
(i) Write the molecular formula of the compounds formed by these elements individually with hydrogen.
(ii) Which of these compounds will have the highest dipole moment?
Answer:

(i) The molecular formula of the compounds formed by X, Y and Z with hydrogen atom are XH₄, YH₃ and H-Z.
(ii) X, Y and Z have 4, 5 and 7 electrons respectively. These elements belong to second period and 14^th, 15^th and 17^th groups. Electronegativity of elements increase across the period from group 1 to group17. Hence, H-Z will have the highest dipole moment.

Question:48

Draw the resonating structure of
(i) Ozone molecule
(ii) Nitrate ion
Answer:

(i) Resonating structure of Ozone $(O)_{3}:$ :
q49-image
(ii) Resonating structure of $NO{_{3}}^{-}:$
q49-image-2

Question:49

Predict the shapes of the following molecules based on hybridization.
$BCl_{3}, CH_{4}, CO_{2}, NH_{3}$

Answer:

(i) In $BCl_{3}$ , Boron is sp² hybridized and thus, the shape of $BCl_{3}$ is trigonal planar.
(ii) The shape of $CH_{4}$ is tetrahedral, carbon is sp³ hybridized.
(iii) The shape of $NH_{3}$ molecule is pyramidal, nitrogen atom is sp³ hybridized nut lone pair on N gives pyramidal shape.
(iv) $CO_{2}$ is linear because carbon is sp hybridized.

Question:50

All the C—O bonds in carbonate ion $(CO2^{-3})$ are equal in length. Explain
Answer:

In $(CO2^{-3})$ ion, carbon is bonded to 3 oxygen atoms. It is bonded to 2 oxygen atoms by double bond. Here, the bonds are not fixed and show resonance therefore, all C-O bonds are equal in length.

Question:51

What is meant by the term average bond enthalpy? Why is there a difference in bond enthalpy of O—H bond in ethanol $(C_{2}H_{5}OH)$ and water?
Answer:

(i) Average bond enthalpy is obtained by dividing total bond dissociation enthalpy by the number of bonds broken.
(ii) All the identical bonds in a molecule do not have the same bond enthalpies, e.g., in water $(H_{2}O)$ , there are two O-H bonds but breaking of first O-H bond, the second O-H bond undergoes some change because of the charge.
Therefore, in polyatomic molecules average bond enthalpies is used and calculated by dividing total bond dissociation enthalpy by the number of bonds broken.
$H_{2}O\; (g)\; \rightarrow H(g)+OH(g);\Delta _{a}H{_{1}}^{o}=502\; kJ\; mol^{-1}$
$OH\; (g)\; \rightarrow H(g)+O(g);\Delta _{b}H{_{2}}^{o}=427\; kJ\; mol^{-1}$
Therefore, Average O-H bond enthalpy $= 502+\frac{427}{2} = 464.5\; kJ\; mol-1$
The bond enthalpies of O-H in $C_{2}H{_{5}}^{2}OH.H_{2}O$ are different because of different electronic environment around oxygen atom.
$CH_{3}CH_{2}OH\; H-O-H$
In ethanol, -OH is attached to carbon and in water O-H is attached to hydrogen atom.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 4: Matching Type

Question:52

Match the species in Column I with the type of hybrid orbitals in Column II.

Column I	Column II
(i) $SF_{4}$	(a) sp³d²
(ii) $if_{5}$	(b) d²sp³
(iii) $NO{_{2}}^{+}$	(c) sp³ d
(iv) $NH_{4}^{+}$	(d) sp³
	(e) sp

Answer:

(i)→ (c); (ii) →(a); (iii) →(e); (iv) →(d)
Explanation:

Column I	Column II
SF₄	S is surrounded by 4 bond pairs and 1 lone pair (sp³d hybridization)
IF₅	I is surrounded by 5 bond pairs and 1 lone pair (sp³d² hybridization)
NO₂⁺	N has 2 bond pairs and no lone pair (sp hybridization)
NH₄⁺	N has 4 bond pairs and no lone pair (sp³hybridization)

Question:53

Match the species in Column I with the geometry/shape in Column II.

Column I	Column II
(i) $H_{3}O^{+}$	(a) Linear
(ii) $HC\equiv CH$	(b) Angular
(iii) $ClO{_{2}}^{-}$	(c) Tetrahedral
(iv) $NH{_{4}}^{+}$	(d) Trigonal bipyramidal
–	(e) Pyramidal

Answer:

(i) →(e); (ii) →(a); (iii) →(b); (iv) →(c)
Explanation :

Column I	Column II
1. $H_{3}O^{+}$	Oxygen has 3 bond pairs and 1 lone pair (pyramidal shape)
2. $HC\equiv CH$	Linear sp hybridization
$ClO{_{2}}^{-}$	Cl has 2 bond pairs and two lone pairs (Angular shape)
$NH{_{4}}^{+}$	N has 4 bond pairs and no lone pair (tetrahedral)

Question:54

Match the species in Column I with the bond order in Column II.

Column I	Column II
(i) NO	(a) 1.5
(ii) CO	(b) 2.0
(iii) $O{_{2}}^{-}$	(c) 2.5
(iv) $O_{2}$	(d) 3.0

Answer:

(i) →(c); (ii) →(d); (iii) →(a); (iv) →(b)
Explanation:

Column I	Column II
NO	$NO(7+8=15\; electrons)$ $\sigma 1s^{2}<\sigma ^{}1s^{2}<\sigma 2s^{2}<\sigma ^{}2s^{2}<\sigma 2pz^{2}<[\pi 2px^{2}=\pi 2px^{2}]<[\pi^{} 2px^{1}=\pi^{} 2px]<\sigma ^{}2p_{z}$ Bond order = $(N_{b}-N_{a})/2=^{}(\frac{(10-5)}{2})=2.5$
CO	$CO (6+8 = 14\; electrons)$ $\sigma 1s^{2}<\sigma ^{}1s^{2}<\sigma 2s^{2}<\sigma ^{}2s^{2}<[\pi ^{}2p{_{x}}^{2}=\pi wp{_{x}}^{2}]\sigma 2pz{_{z}}^{2}<[\pi^{} 2px=\pi ^{}2px]<\sigma ^{}2p$ Bond order = $\frac{(N_{b}-N_{a})}{2}=^{*}(\frac{(10-4)}{2})=3$
$O{_{2}}^{-}$	$(8+8+1=17\; electrons)$ $\sigma 1s^{2}<\sigma ^{}1s^{2}<\sigma 2s^{2}<\sigma ^{}2s^{2}<[\pi 2p{_{x}}^{2}=\pi 2p{_{z}}^{2}]<\sigma 2p{_{z}}^{2}<[\pi^{}2p{_{x}}^{1}=\pi ^{}2p{_{z}}^{1}]<\sigma ^{*}2p$ Bond order = $\frac{(N_{b}-N_{a})}{2}=(\frac{(10-7)}{2})=1.5$
$O_{2}$	$(8+8=16\; electrons)$ $\sigma 1s^{2}<\sigma ^{}1s^{2}<\sigma 2s^{2}<\sigma ^{}2s^{2}<[\pi 2p{_{z}}^{2}=\pi 2p{_{z}}^{2}]<\sigma 2p{_{z}}^{2}<[\pi^{}2p{_{x}}^{1}=\pi ^{}2p{_{x}}^{1}]<\sigma ^{*}2p$ Bond order = $\frac{(N_{b}-N_{a})}{2}=(\frac{(10-6)}{2})=2$

Question:55

Match the items given in Column I with examples given in Column II.

Column I	Column II
(i) Hydrogen bond	(a) C
(ii) Resonance	(b) LiF
(iii) Ionic solid	(c) H₂
(iv) Covalent solid	(d) HF
	(e) O₃

Answer:

(i) →(d); (ii) →(e); (iii) →(b); (iv) →(a)

Question:56

Match the shape of molecules in Column I with the type of hybridization in Column II.

Column I	Column II
(i) Tetrahedral	(a) sp²
(ii) Trigonal	(b) sp
(iii) Linear	(c) sp³

Answer:

(i) →(c);(ii) →(a); (iii) →(b)

NCERT Exemplar Class 11 Chemistry Solutions Chapter 4: Assertion and Reason Type

Question:57

In the following questions, a statement of Assertion (A) followed by a statement of the reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Sodium chloride formed by the action of chlorine gas on sodium metal is a stable compound.
Reason (R): This is because sodium and chloride ions acquire octet in sodium chloride formation.
(i) A and R both are correct, and R is the correct explanation of A.
(ii) A and R both are correct, but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A and R both are false.
Answer:

The answer is the option (i) A and R both are correct, and R is the correct explanation of A
Explanation:
$Na+Cl\rightarrow NaCl$
2,8,1 2,8,7 (2,8 X 2,8,8)
In NaCl, Sodium and Chlorine ions have complete octet. Therefore, NaCl is a stable compound.

Question:58

In the following questions, a statement of Assertion (A) followed by a statement of the reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A) : Though the central atom of both NH₃ and H₂O molecules are sp³ hybridized, yet H–N–H bond angle is greater than that of H–O–H.
Reason (R) : This is because nitrogen atom has one lone pair and oxygen atom has two lone pairs.
(i) A and R both are correct, and R is the correct explanation of A.
(ii) A and R both are correct, but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A and R both are false.
Answer:

The answer is the option (i) A and R both are correct, and R is the correct explanation of A.
Explanation:
$H_{2}O$ has two lone pairs but NH3 has one lone pair. Hence, $H_{2}O$ .

Question:59

In the following questions, a statement of Assertion (A) followed by a statement of the reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Among the two O–H bonds in H₂O molecule, the energy required to break the first O–H bond and the other O–H bond is the same.
Reason (R) : This is because the electronic environment around oxygen is the same even after breakage of one O–H bond.
(i) A and R both are correct, and R is correct explanation of A.
(ii) A and R both are correct, but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A and R both are false
Answer:

The answer is the option (iv) A and R both are false.
Explanation:
Correct assertion: The bond enthalpy in H-O-H is not same for both the O-H bonds.
Correct reason: Because electronic charge on oxygen atom is different after breaking of one O-H bond.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 4: Long Answer Type

Question:60

i) Discuss the significance/ applications of dipole moment.
(ii) Represent diagrammatically the bond moments and the resultant dipole moment in $CO_{2}, NF_{3}$ and $CHCl_{3}$ .
Answer:

(i) The formula of calculating dipole moment is,
Dipole moment $(\mu )$ = charge(Q) × distance of separation(r)
Dipole moment is expressed in Debye units (D).
Significance of Dipole moment is:

Helps predicting polar and non-polar nature of compounds
Percentage of ionic character can be calculated by the formula:

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% of ionic character = $(\mu )$ observed × 100/ $\mu$ , ionic

It helps to know the symmetry of the molecule.

Symmetrical molecules have zero dipole moment, although they have two or more polar bonds. For example,
The dipole moment in case of $BeF_{2}$ is zero. This is because the two equal bond dipoles point in opposite directions and cancel the effect of each other.
(ii)
q60-image

Question:61

Use the molecular orbital energy level diagram to show that $N_{2}$ would be expected to have a triple bond, $F_{2}$ a single bond and $Ne_{2}$ no bond.
Answer:

The general sequence of the energy level of molecular orbitals for nitrogen is:
$\sigma 1s<\sigma ^{*}1s<\sigma 2s<\sigma ^{*}2s<\pi 2p_{x}=\pi 2p_{y}<\sigma 2p_{z}$
The molecular orbitals in the sequence of their energy levels for the given molecules have hereby been described below:
$N_{2}\sigma 1s^{2}\; \sigma^{*} 1s^{2}\; \sigma 2s^{2}\sigma^{*} 2s^{2}\pi 2p{_{x}}^{2}=\pi 2p{_{y}}^{2}\sigma 2p{_{z}}^{2}$
Now, we know that the,
Bond order for $N_{2}=\frac{8-2}{2}=3$
Now, as the bond order is 3, it means that $N_{2}$ will have a triple bond.
capture-61-1
The molecular orbital of Fluorine has been described below: -
$F_{2}= \sigma 1s^{2}, \sigma ^{*} 1s^{2}, \sigma 2s^{2}, \sigma ^{*} 2s^{2}, \sigma \; 2px^{2}, \pi 2px^{2} = \pi 2py^{2}$
The bond order of $F_{2}=\frac{10-8}{2}=1$
Thus, we can imply that when the bond order is 1, the number of bonds must also be 1.
q61-image-1
$Ne_{2}= \sigma 1s^{2}, \sigma ^{*} 1s^{2}, \sigma 2s^{2}, \sigma ^{*} 2s^{2}, \sigma \; 2px^{2}, \pi 2px^{2} = \pi 2py^{2},\pi ^{*}2p{_{x}}^{2}=\pi ^{*}2p{_{y}}^{2}$
Now, we can calculate the bond order of $Ne_{2}=\frac{10-10}{2}=0$
q61-image-2
Thus, $Ne_{2}$ has no bond.

Question:62

Briefly describe the valence bond theory of covalent bond formation by taking an example of hydrogen. How can you interpret energy changes taking place in the formation of dihydrogen?
Answer:

Valence bond theory was introduced by Heilter and London in 1927. It was developed by Pauling. It is based on knowledge of atomic orbitals, electronic configuration of elements, overlapping of atomic orbitals and hybridization of orbitals.
Let us consider two hydrogen atoms A and B. Let’s assume that they are approaching each other and let their nuclei be N_A and N_B and electrons present in them be e_A and e_B.
Now let us imagine that the two atoms are at a large distance from each other, hence, there is no interaction between them. Now think that they are coming close to each other, and new attractive and repulsive forces begin to operate.
Now we know that attractive forces arises between:

Nucleus of one atom and its own electron that is $N_{A}-e_{A}$ and $N_{B}-e_{b}$ .
Nucleus of one atom and electron of other atom, i.e., $N_{A}-e_{B},N_{B}-e_{A}$ , NB-eA. Similarly, repulsive forces arise between

(a) Electrons of two atoms like $e_{A}-e_{B}$
(b) Nuclei of two atoms $N_{A}-N_{B}$ .
It has been found that the magnitude of the new attractive force is more than the new repulsive forces. As a result, two atoms approach each other and potential energy decreases. Eventually the net force of attraction balances the force of repulsion and system acquires minimum energy. At this point two hydrogen atoms are said to be bonded together to form a stable molecule having bond length 74pm.
Since the energy is released when the bond is formed between two hydrogen atoms, the hydrogen molecule is more stable than that of the isolated hydrogen atoms. The energy so released is called the bond enthalpy, which is corresponding to minimum in the curve depicted in figure. Conversely, 435.8kJ of energy is required to dissociate one mole of H₂ molecule.
$H_{2}(g)+435.8 \; kJ\; mol^{-1}\rightarrow H(g)+H(g).$
The potential energy curve for the formation of H₂ molecule as a function of internuclear distance of H atoms. The minimum in the curve corresponds to the most stable state of H₂.

Question:63

Describe hybridization in the case of $PCl_{5}$ and $SF_{6}$ . The axial bonds are longer as compared to equatorial bonds in $PCl_{5}$ whereas in $SF_{6}$ both axial bonds and equatorial bonds have the same bond length. Explain.
Answer:

The hybridization of the S in $SF_{6}$ is $sp^{3}d^{2}$ and the hybridization of the P in $PCl_{5}$ is $sp^{3}d.$
We know that $PCl_{5}$ has trigonal bipyramidal geometry. As such, the axial bonds are slightly longer as compared to the equatorial bonds. This is a result of the axial bonds experiencing greater repulsion from other bonds as compared to the equatorial bonds.
We know that $SF_{6}$ has an octahedral structure. As such, all the bonds possess the same length this is because all bonds experience similar repulsion from other bonds.

Question:64

(i)Discuss the concept of hybridization. What are its different types in a carbon atom?
(ii) What is the type of hybridization of carbon atoms marked with star.
capture-64
Answer:

(i) The process of hybridization can be defined as the process that involves the intermixing of the orbitals with slightly different or similar energy levels in order to form new orbitals that have similar shapes and energy levels. These orbitals are known as the hybrid orbitals.
The valence shell of an atom is always hybridized. Additionally, the merging orbitals are of nearly similar energy or same energy. These orbitals do not form pie bonds.
(a) sp hybridization: Carbon compounds with triple bond i.e. $C\equiv C$ have this type of hybridization.
(b) sp² hybridization: Carbon compounds with a double bond i.e. $C= C$ have this type of hybridization.
(c) sp³ hybridization: Carbon compounds with single bonds i.e. $C - C$ have this type of hybridization.
(ii)
(a)
capture-64-1
(b)
capture-64-2
(c)
capture-64-3
(d)
capture-64-4
(e)
capture-64-5

Question:65

Comprehension given below is followed by some multiple choice questions. Each question has one correct option. Choose the correct option.
Molecular orbitals are formed by the overlap of atomic orbitals. Two atomic orbitals combine to form two molecular orbitals called bonding molecular orbital (BMO) and anti bonding molecular orbital (ABMO). Energy of anti bonding orbital is raised above the parent atomic orbitals that have combined and the energy of the bonding orbital is lowered than the parent atomic orbitals. Energies of various molecular orbitals for elements hydrogen to nitrogen increase in the order :
$\sigma 1s < \sigma^{*}1s <\sigma\: 2s <\sigma^{*}\: 2s < (\pi \: 2p_{x} = \pi \: 2p_{y}) < \sigma 2p_{z} < \sigma 2p_{z} < (\pi ^{*}\: 2p_{x} = \pi ^{*} \: 2p_{y}) < \sigma ^{*}2p_{z}:$ and for oxygen and fluorine order of energy of molecular orbitals is given below :
$\sigma 1s < \sigma^{*}1s <\sigma\: 2s <\sigma^{*} 2s< \sigma 2p_{z} < \sigma 2p_{z} < (\pi \: 2p_{x} = \pi \: 2p_{y}) < (\pi ^{*}\: 2p_{x} = \pi ^{*} \: 2p_{y})< \sigma ^{*}2p_{z}$ Different atomic orbitals of one atom combine with those atomic orbitals of the second atom which have comparable energies and proper orientation. Further, if the overlapping is head on, the molecular orbital is called ‘Sigma’, $(\sigma)$ and if the overlap is lateral, the molecular orbital is called ‘pi’, $(\pi)$ . The molecular orbitals are filled with electrons according to the same rules as followed for filling of atomic orbitals. However, the order for filling is not the same for all molecules or their ions. Bond order is one of the most important parameters to compare the strength of bonds.
Which of the following statements is correct?
(i) In the formation of dioxygen from oxygen atoms 10 molecular orbitals will be formed.
(ii) All the molecular orbitals in the dioxygen will be completely filled.
(iii) Total number of bonding molecular orbitals will not be same as total number of anti bonding orbitals in dioxygen.
(iv) Number of filled bonding orbitals will be same as number of filled anti bonding orbitals
Answer:

(i) In the formation of dioxygen from oxygen atoms 10 molecular orbitals will be formed.
The above statement is the correct answer.
There are unpaired electrons in two molecular orbitals in the dioxygen. Hence, the statement (ii) is not correct.
The total number of antibonding orbitals in dioxygen is equal to the total number of bonding molecular orbitals. Therefore, the statement (iii) is not correct.
The number of filled antibonding orbitals is not always going to be the same as the number of filled bonding orbitals. hence, the statement (iv) is not correct.

Question:66

Which of the following molecular orbitals has a maximum number of nodal planes?
(i) $\sigma ^{*}1s$
(ii) $\sigma ^{*}2p_{z}$
(iii) $\pi 2p_{x}$
(iv) $\pi^{*} 2p_{y}$
Answer:

The answer is the option (ii) The molecular orbital that has the maximum no. of nodal planes is $\sigma ^{*}2p_{z}$

Question:67

Which of the following pair is expected to have the same bond order?
(i) $O_{2},N_{2}$
(ii) $O{_{2}}^{+},N{_{2}}^{-}$
(iii) $O{_{2}}^{-},N{_{2}}^{+}$
(iv) $O{_{2}}^{-},N{_{2}}^{-}$
Answer:

The answer is the option (ii) $O{_{2}}^{+},N{_{2}}^{-}$ ; because the Isoelectronic species have a similar bond order.
Option $I-O_{2}$ has 16 electrons and N₂ has 14 electrons.
Option II - Both $O_{2}^{+}$ and $N_{2}^{-}$ has 15 electrons.
Option III - $O_{2}^{-}$ has 17 electrons and $N_{2}^{+}$ has 13 electrons.

Option IV - $O_{2}^{-}$ has 17 electrons and $N_{2}^{-}$ has 15 electrons

Question:68

In which of the following molecules, $\sigma 2p_{z}$ molecular orbital is filled after $\pi 2p_{x}$ and $\pi 2p_{y}$ molecular orbitals?
(i) $O_{2}$
(ii) $Ne_{2}$
(iii) $N_{2}$
(iv) $F_{2}$
Answer:

The answer is the option (iii) N₂
The first atomic orbitals on two atoms form two molecular orbitals designated as
$\sigma 1s\; and\; \sigma ^{*}1s$
Similarly, the 2s and 2p atomic orbitals give rise to the eight molecular orbitals given below
Antibonding $MOs\; \sigma^{*}2s\; \sigma^{*}2p_{z}\pi ^{*}2p_{x}\; \pi ^{*}2p_{y}$
Bonding $Mos\; \sigma 2s\; \sigma2p_{z}\; \pi 2p_{x}\; \pi 2p_{y}$
For molecules such as B₂, C₂, N₂, etc. it is observed that the increasing order of energies of various molecular orbitals is -
Here, the important characteristic feature is that energy of $\sigma 2p_{z}$ molecular orbital is higher than that of $\pi 2p_{x}$ and $\pi 2p_{y}$ molecular orbitals

The students can use NCERT Exemplar Class 11 Chemistry Solutions Chapter 4 pdf download to save the copy of solutions in their device. class 11 Chemistry NCERT exemplar solutions chapter 4 will help the students for understanding the topics and concepts in a better manner. Thus they can solve any questions with confidence.

Topics and Sub-topics of NCERT Exemplar Class 11 Chemistry Solutions Chapter 4

Here is The List of Topics Included in NCERT Exemplar Solutions for Class 11 Chemistry Chapter 4:

VSEPR theory
Lewis structures
Valence bond theory
The polar character of covalent bonds
The concept of hybridization
The molecular orbital theory of homonuclear diatomic molecules
Hydrogen bonding

What will students learn in NCERT Exemplar Class 11 Chemistry Solutions Chapter 4?

The NCERT Exemplar Class 11 Chemistry chapter 4 solutions will help the students in getting a better understanding of the topics and the concepts that are included in the book.
NCERT Class 11 Chemistry Book is equipped with all the subject details with the probable questions from the exam.
In NCERT Exemplar Class 11 Chemistry solutions chapter 4, The students will learn about how to deal with a chemical bond and its long-lasting attractive force.
It’s working on the atoms, ions, compounds, etc. as a group in a chemical reaction. It is also explained in several theories which include valence shell electron pair repulsion theory, etc.

NCERT Exemplar Class 11 Chemistry Solutions Chapter Wise

Chapter 1 Some Basic Concepts of Chemistry

Chapter 2 Structure of Atom

Chapter 3 Classification of Elements and Periodicity in Properties

Chapter 5 States of Matter

Chapter 6 Thermodynamics

Chapter 7 Equilibrium

Chapter 8 Redox Reactions

Chapter 9 Hydrogen

Chapter 10 The s-Block Elements

Chapter 11 The p-Block Elements

Chapter 12 Organic Chemistry – Some Basic Principles and Techniques

Chapter 13 Hydrocarbons

Chapter 14 Environmental Chemistry

Important topics to cover for exams from NCERT Exemplar Class 11 Chemistry solutions chapter 4 Chemical Bonding and Molecular Structure

Given below are some of the important topics that the students will learn from this chapter:

The lessons and topics will help you to learn about the chemical bond and the reaction. A chemical bond has been explained in the several theories and also these bonds which will gain stability that may result in the formation of electrostatic force.
Class 11 NCERT Exemplar Chemistry solutions chapter 4 explains the chemical bond and its importance and its usefulness. The solutions incorporate simple methods that are explained for easy understanding of concepts.
The students will also learn about the theories that are explained by different theorists in the book. It involves different concepts that will develop more understanding among the students.

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2. What all topics are covered in this chapter?

Different topics are covered in this chapter i.e., VSEPR theory, Lewis structures, The concept of hybridization, Hydrogen bonding, etc.

3. How important it is to understand this chapter?

It is very important to understand NCERT Exemplar Class 11 Chemistry Solutions Chapter 4 as it will help you to score the required marks in the examination.

4. How many questions are there in this chapter?

There are a total of 40 questions in this chapter which are answered by the experts.

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Exemplar Class 11 Chemistry Solutions Chapter 4 Chemical Bonding and Molecular Structure

NCERT Exemplar Class 11 Chemistry solutions chapter 4: MCQ (Type 1)

NCERT Exemplar Class 11 Chemistry Solutions Chapter 4: MCQ (Type 2)

NCERT Exemplar Class 11 Chemistry Solutions Chapter 4: Short Answer Type

NCERT Exemplar Class 11 Chemistry Solutions Chapter 4: Matching Type

NCERT Exemplar Class 11 Chemistry Solutions Chapter 4: Assertion and Reason Type

NCERT Exemplar Class 11 Chemistry Solutions Chapter 4: Long Answer Type

Here is The List of Topics Included in NCERT Exemplar Solutions for Class 11 Chemistry Chapter 4:

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