CHAPTER -1

Chemical bonding and shape of molecules

A. Very short Answer Questions:

1.

Difference between sigma (σ) and pi (π) bonds.

 

Sigma (σ) bond	Pie (π) bond
1. Sigma bond is formed by the overlapping of half field orbital of one atom with half field orbital of another atom along the intermolecular axis.	1. Pie bond is formed by the side wiseoverlapping of half field orbital of another atom perpendicular to the intermolecular axis.
2. Overlapping takes between s – s orbital, s – p orbital and p – p orbital.	2. Overlapping takes place between only p – p orbital (py – py), (p2 – p2)
3. Sigma molecular orbital is symmetrical about the intermolecular axis.	3. π molecular orbital is discontinuous because two pie bonds are formed above the intermolecular axis.

 

Sigma bond is stronger as the overlapping takes place to larger extent whereas in pie bond,the overlapping takes place in side wise direction.,so, the extent of overlapping is less. Hence, π bond is weak.

 

2.

 Types of hybridization:

(i) Sp – hybridization.

(ii) Sp² hybridization.

(iii) Sp³ hybridization.

(iv) dsp² hybridization.

(v) dsp³ hybridization.

(vi) d²sp³ hybridization.

 

a. XeF₂

It has linear shape.

Its bond angle is 180°.

It is SP hybridization type.

 

 

b. CIF₃

It is SP² hybridization.

It bond angle is 120°.

It geometrical shape is triangonal plane.

 

 

c. PCl₅

It is sp³d type of hybridization.

Its bond angle is 120° and 90°.

It has trigonal bipyramidal shape.

 

3.

 

 

 

 

In water molecule, there are two lone pairs of electrons in oxygen. In NH₃ molecule, there is one lone pair of electron in nitrogen. According to VSEPR theory we have, lone – pair – lone pair repulsion > bond pair – bond pair repulsion.Since, the no.of lone pairs are more in oxygen, the O – H bond pairs come more closer than N – H bond pairs in NH₃. Thus, bond angle in H₂O (1o4°.27’) is less than in NH₃$\left( {{{107}^{\rm{o}}}{{48}^{\rm{'}}}} \right)$.

 

4.

Energy level diagram of O₂ molecule.

 

Bond order of O₂ = $\frac{1}{2}$(8 – 4) = $\frac{4}{2}$ = 2

It has paramagnetic nature.

 

5.

 It states that molecules take up the shape that minimizes the repulsion between the bonding and non – bonding pairs of electron.

 

6.${\rm{C}}{{\rm{H}}_4},{\rm{\: N}}{{\rm{H}}_3},{\rm{\: }}{{\rm{H}}_2}0$all involve four pairs of electrons around the central atom and should have a tetrahedral  structure like ${\rm{C}}{{\rm{H}}_4}{\rm{\: }}$with bond angle 109.5, but as the lone pair-bond pair repulsion> bond pair-bond pair repulsion, the bond angle in NH₃ and H₂O is decreased to 107^o48^’ and 104.5^o respectively.

 

7.

 The geometrical arrangement of Sp³, Sp² and Sp hybrid orbitals are tetrahedral with bond angles 109.5°, triangular planer with bond angle 120° and linear in structure with bond angle 180° respectively.

 

8.

It is defined as half of the difference between the number of electrons in bonding molecular orbital and the no. of electron in antibonding molecular orbitals.

The bond order of O₂ is:

${\rm{B}}{{\rm{O}}_{{{\rm{O}}_2}}} = \frac{1}{2}\left( {{{\rm{N}}_{\rm{b}}} - {{\rm{N}}_{\rm{a}}}} \right)$

= $\frac{1}{2}$(10 – 6)

= $\frac{4}{2}$ = 2

Similarly the bond order of H₂ is:

${\rm{B}}{{\rm{O}}_{{{\rm{H}}_2}}} = \frac{1}{2}\left( {{{\rm{N}}_{\rm{b}}} - {{\rm{N}}_{\rm{a}}}} \right)$

= $\frac{1}{2}\left( {10 - 4} \right)$

= $\frac{6}{2}$

= 3.

 

9.

The electronic configuration of fluorine (Z = s) is 1s², 2s², 2p⁵.

So,

Or, F₂ = (6ls²)(6 * 1s)²(62s)²(6 * 2s)²(62p_x)²(π2p_y)²(π²p_z)²(π * 2P_y)²(π * 2p_z)²

So, bond order in F₂ = $\frac{1}{2}$

= $\frac{1}{2}\left( {10 - 8} \right)$

= $\frac{1}{2}{\rm{*}}2$ = 1

Similarly, the electronic configuration of N₂ in;

N₂: (6l²)(6 * ls)²(62s)²(62p_x)²(π2p_y)²(π2p_z)²

So, Bond order = $\frac{{{{\rm{N}}_{\rm{b}}} - {{\rm{N}}_{\rm{a}}}}}{2}$

= $\frac{{10 - 4}}{2} = \frac{6}{2}$ = 3.

It is clear that bond order in N₂ molecule is 3 and F₂ molecule is 1. Since, greater the value of bond order, stronger is the bond.

10. The three rules for linear combination of atomic orbitals are:

(i) The atomic orbitals must be roughly of the same energy. This is important when considering overlap between two different types of atoms.

(ii) The orbitals must overlap one another as much as possible. This implies that the atoms must be close enough for effective overlap and that the radial distribution functions of the two atoms must be similar at this distance.

(iii) In order to produce bonding and antibonding molecular orbitals, either the symmetry of the two atomic orbitals must remain unchanged when rotated about the internuclear line, or both atomic orbitals must change symmetry in an identical manner.

 

11.

The three hybrid orbitals that may be formed by an atom with s and p orbitals are: BeF₂, BeH₂, BeCl₂, etc.

 

12.

 Covalent compound have weak intermolecular force, that is to say, the force between molecules is very weak. The consequences of this is that it becomes easy to move individual molecules away from other molecules as the bonding between molecule is weak, hence, it needs little energy to break. Since, temperature is proportional to energy, which means the temperature required to melt the compound is low. Hence, the covalent compound has lower melting point than ionic compound.

 

13.

 The geometry of following molecules are as follows:

(i) BS₃ – Sp₂ hybridization with bond angle 120°

(ii) CCl₄ – sp³ hybridization with bond angle 109°.28’

(iii) SF₆ – sp³ hybridization, octahedial geometry with bond angle 90°.

(iv) PCl– sp³ hybridization, trigonal bipyramidal geometry with bond angle 120°

(v) NH₃ – sp³ hybridization, pyramidal geometry with bond angle 109.5° to 107.48°.

(vi) XeF₄ – Its geometry is square planer.

 

14.

 It is a type of atomic orbital that results when two or more atomic orbitals of an isolated atom mix (the numbers of hybrid orbitals on a covalently bonded atom is equal to the no. of atomic orbital used to form the hybrid orbital.)

The type of hybridization involved in CO₂ is sp² hybridization and BF₃ in sp² hybridization.

 

15.

 The bond order of O₂ is 2 whereas, of O₂⁺ ion is $\frac{5}{2}$ so it is clear that higher bond orders are dissociated with higher bond strength and shorter bond length. We know,

Bond order $ \propto $$\frac{1}{{{\rm{strength\: length}}}}$

So, the bond length of above molecule and ion is O₂⁺ > O₂.

Therefore, bond length in O₂⁺ ion is greater than that in O₂ molecule.

 

16.

 

17.

 In O₂ molecule, atoms are held by a double covalent bond. It may be noticed that it contains two unpaired electron. So, O₂molecule has paramagnetic nature. But in case of N₂ molecule, atoms are held by triple bond. Hence, the nitrogen molecule is highly stable and is confirmed by its high bond dissociation energy and small bond length. Since, there are no unpaired electron in any orbital, N₂ molecule is diamagnetic.

 

18.

 The shape of following molecule are:

(i) The shape of CO₂ is linear as there is sp² hybridization.

(ii)  The shape of NH₃ is pyramidal with bond angle 107^oas there is sp³ hybridization.

 

19.

In O₂ molecule, atoms are held by a double covalent bond. It may be noticed that it contains two unpaired electron. So, O₂molecule has paramagnetic nature. But in case of N₂ molecule,atom are held by triple bond. Hence, the nitrogen molecule is highly stable and is confirmed by its high bond dissociation energy and small bond length. Since, there are no unpaired electron in any orbital N₂ molecule is diamagnetic

 

20.

 The molecule orbital structure of H₂ and He²⁺ are:

Or, H₂ = (61s)²

Or, H_e⁺² = (6ls)²(6 * ls)²

 

21.

It is defined as the intermixing of atomic orbital having similar energy content to form a new set of orbitals called hybrid orbitals.

E.g: CH₄

₆C = 1s²,2s²,2p⁶

Ground state: ₆C

Excited state = ₆C

CH₄formation

 

22.

The shape of following molecules are:

(i) BF₃

It has sp² hybridization with trigonal plane geometry and bond angle 120°.

 

(ii) H₂O

It has disordered tetrahedral or V – shaped geometry with bond angle 104.5° with sp³ hybridization.

 

(iii) SF₆

It has sp³ hybridization with bond angle 90° and octahedral geometry.

 

23.

 The pair of species which have higher energy are:

(i) Fe, Fe₂⁺

Fe has bond orders 1 and F₂⁺ has 1.5. Hence, Fe⁺ has higher bond energy.

(ii) Be₂, Be⁺₂.

 Be⁺₂ has bond order greater that Be₂. Hence, Be⁺₂ has higher bond energy.

 

24.

It is due to the presence of lone pair of electrons in NF_3. As the lone pair bond pair repulsion is more than bond pair bond pair repulsion, NF₃ is compressed and has a bond angle of 103^o rather than 120^o .

 

25.

 

 

 

26.

It is a type of atomic orbital that results when two or more atomic orbitals of an isolated atom mix (the numbers of hybrid orbitals on a covalently bonded atom is equal to the no. of atomic orbital used to form the hybrid orbital.)

The type of hybridization involved in CO₂ is sp² hybridization and BF₃ in sp² hybridization

The orbital picture of hydrocarbon (Ethane C₂H₆) moving tetrahedral structure in:

Fig: Orbital picture of ethane.

 

27.

 In water molecule, there are two lone pair of electrons in oxygen. Due to small size of oxygen, the lone pair – lone pair and the bond – pair – bond – pair repulsion is large and as a result, the electrons tend to push the molecules apart. However,  the sulphur atom is greater in size and the lone pair tends to be far apart, thereby considerably reducing the repulsions and also the bond angle that is the reason why H₂O bond angle is greater than H₂S.

 

28.

The mode of hybridization in:

(i) C of C₂H₄ – sp² hybridization.

(ii) B of BF₃ – sp² hybridization.

 

29.

The mode of hybridization of the central atom whose molecular geometry is tetrahedral is sp³ hybridization.

E.g. CH₄,CCl₄, etc.

 

 

30.

The example of the following:

(i) Tetrahedral hybridization – CH₄, CCl₄, etc.

(ii) Trigonal hybridization – BF₃, BCl₃, etc.

 

B. Short answer question:

1.

It is the process when s – orbital and two p – orbitals (p_x and p_y) are mixed together to form a new set of the orbitals known as sp² hybridization.

E.g:

Formation of BH₃ (Boron trihydride):

(i) Configuration of boron B(Y = 5) is 1s², 2s², 2p¹

When Boron ‘B’ is in excited state.

(ii) Configuration is 1s²,2s¹,2p²

n = 3.

So, boron undergoes sp² hybridization (one 2s² two 2p orbitals are mixed together to form 36p² hybrid orbitals which are situated at the corners of the triangle making an angle of + 20°. The electronic configuration is:

 

(iii) Hybrid orbitals of B:

Now in the formation of BH₃, the s orbital of the hydrogen atom overlaps axially with each sp² hybrid orbitals of Boron atoms forming H – B sigma bonds. So, BH₃ molecule has bond angle of +20°. In this structure the four atoms lie in one plane. So, it is also known as trigonal planar geometry.

 

2.

State of hybridization in ethylen molecule.

In the case of ethylene the sp² hybrid orbital of one carbon atom overlaps basically with similar sp² hybrid orbital of another C – atom, forming C – C sigma bond. The remaining two sp² hybrid orbital of each carbon overlaps with its orbital of two hydrogen atoms forming two C – H sigma bonds.

 

3.

Shape of BeF₂ by using VSEPR theory:

BeF₂: In BeF₂, Be is the central atom. The electronic configuration of Be is 1s², 2s². It consists two valance electrons. The two valance electrons forms two bond pairs with two fluorine atoms. Now, the cetral Be is surrounded by two bond pairs which is situated in linear manner making an angle of 180°. So, BeF₂ molecule has linear geometry with (FBeF = 180°) FBeF bond angle is 180°.

The lewis dot structure is:

 

Electron pair = 2

Bond pair = 2

Lone pair = 0.

 

BF₃: In BF₃, boron B is the central atom. The electronic configuration of B(2 = 5) is 1s²,2s²,2p¹. It consists of three electron pairs with three fluorine atoms. The central boron atom is surrounded by three bond pairs. So, three bond pairs adopt a trigonal structure which is situated in one plane. So, BF₃ has trigonal geometry with FBF bond angle equal to 120°. The lewis structure is:

 

Election pairs = 3.

Bond pairs = 3

Lone pairs = 0

 

4.

Difference between bonding and antibonding.

Bonding molecular orbital	Anti – bonding orbital
1. The formation of addition overlap of atomic orbitals.	The formation by subtraction overlap of atomic orbitals.
2. It has more electron density in the region between the two nuclei and this accounts for the stability of the bond.	2. It has less electron density in the region between the two nuclei and this leads to the instability of the bond.
3. It may or may not have a nodal plane.	3. It always has a nodal plane between the nuclei.
4. It possesses lower energy than the atomic orbital from which it is formed.	4. Electron in antibonding molecular orbital contributes to repulsion between two atoms.
5. Electron in a bonding molecular orbital contributes to attraction between two atoms.	5. Electron in antibonding molecular orbital contributes to repulsion between two atoms.
6. These are designed as σ, π, δ etc.	6. These are designed as σ+, π+, δ+ etc

 

5.

Features of VSEPR theory:

(i) In polyatomic covalent compound, there is a neutral atom to which other atoms are bonded together.

(ii) The central atom contains electron pair which may be bond pair or lone pair.

(iii) The electron pairs repel each other so that they are situated as far as possible so as to minimize the repulsion between them.

(iv) The electron pairs are situated in fixed position ion in the space giving a definite geometrical shape.

(v) The geometry of the electron pairs gives the geometry of the compound.

 

6.

NH₃:

In NH₃ molecule, there is one lone pair of electron in nitrogen. According to VSEPR theory we have, lone – pair – lone pair repulsion > bond pair – bond pair repulsion.sp³ hybridization, pyramidal geometry with bond angle 109.5° to 107.48°.

 

H₂O:

In water molecule, there are two lone pairs of electrons in oxygen. In NH₃ molecule, there is one lone pair of electron in nitrogen. According to VSEPR theory we have, lone – pair – lone pair repulsion > bond pair – bond pair repulsion.Since, the no.of lone pairs are more in oxygen, the O – H bond pairs come more closer than N – H bond pairs in NH₃. Thus, bond angle in H₂O (140°.27’) is less than in NH₃(10’48°).

 

 

7.

 Geometry of SF₆:

In SF₆ sulphur (S) is the central atom. The central atom sulphur has 1s², 2s², 2p⁶, 3s², 3p⁴ configuration. It has six election in valence shell which forms six bond pairs with six fluorine atoms, the central sulphur atom is surrounded by six bond pairs which are situated at the size corners of the hexagons. In such a way that the four electron pairs are situated at the four corners of the square making an angle of 90°.The remaining two bond pairs are situated above and below the plane of the source making angle of 90° with plane. So, SF₆ has octahedral geometry with CF_SF bond angle is equal to 90°.

 

Geometry of PCl₅:

1. It is sp³d type of hybridization.

2. Its bond angle is 120° and 90°.

3. It is triangonal bipyramidal.

 

8.

 

 

Sigma (σ) bond	Pie (π) bond
1. Sigma bond is formed by the overlapping of half field orbital of one atom with half field orbital of another atom along the intermolecular axis.	1. Pie bond is formed by the side wise of half field orbital of another atom perpendicular to the intermolecular axis.
2. Overlapping takes place between s – s orbital, s – p orbital and p – p orbital.	2. Overlapping takes place between only p – p orbital (py – py), (p2 – p2)
3. Sigma molecular orbital is symmetrical about the intermolecular axis.	3. π molecular orbital is discontinuous because two pie bonds are formed above the intermolecular axis.

 

 

9.

Geometry of methane CH₄.

In CH₄ carbon is the neutral atom. The electronic configuration of carbon (z = 6) is 1s²,2s²,2p². It consists of four valence electrons which forms four bond pairs with four hydrogen atoms. Now, the central carbon atom is surrounded by four electron pairs, which are situated at the four corners of the regular tetrahedral. Hence, CH₄ molecule has tetrahedral geometry with HC + l bond angle is 109.5.

Str.

 

C. Long answer question:

1.

Hybridization is the process of mixing of dissimilar atomic orbitals of same atoms giving rise to equal number of a new set of orbitals having same energy contents.

 

Importance of hybridization:

(i) Hybridization is the process of mixing of energies not the electron.

(ii) The hybrid orbitals have shape and same energy.

(iii) The number of Hybrid orbital formed is equal to the number of atomic orbitals mixing together.

(iv) The hybrid orbitals are situated in fire position in the space giving a definite geometrical shape.

(v) Only the bigger lobe takes part in overlapping.

 

VSEPR theory is applied to predict the shape of simple molecules or ions in which the atoms are bonded by single bond. This theory is based upon the fact that, the direction of bonds around the central atom in polyatomic molecule depends upon the interaction of electron pairs of the central atom will arrange themselves in space in such a way that their minimum repulsion between them so that the molecule will have definite geometry with minimum energy and maximum stability.

 

2.

 The process of mixing of dissimilar atomic orbitals of same atoms giving rise to equal number of a new set of orbitals having same energy is known as hybrid orbital.

 

Fig:

 

The molecular orbital configuration and bond of O₂ and F₂ are as follows:

O₂ = KK(62S)²(6 * 2s)²(62p_x)²(π2p_y)²( π2p_z)²(π * 2p_y)¹(π * p_z)¹

Bond angle = $\frac{1}{2}\left( {8 - 3} \right)$ = $\frac{5}{2}$ = 2.$\frac{1}{2}$.

F₂ = σ²₁σ₁²σ²₂σ²₂σ_px²π²p_yπ²p_zπ²p_yπ²p_z

Bond angle = $\frac{1}{2}.10 - \frac{1}{2}.8$ = 1

Here, O₂is paramagnetic compared to F₂.

 

 

3.

Valence bond theory of H₂ molecule:

When two hydrogen atoms are at large separation from each other, so that there is no orbital overlap, the total energy is equal to the sum of the energies to two H – atoms. When two atoms come close to each other then attractive and repulsive force began to operate. As the attractive force exceeds the repulsive force, orbitals start to overlap and potential energy of the system begins to decrease. At a certain distance between the two atoms, the attractive and repulsive interaction because each other and energy of the system attains minimum value (which is 433) or H₂ molecule. The internuclear separation at which the orbitals overlap to the proper extent and possessing minimum energy called bond energy.

Limitations of valence bond theory:

1. This theory doesn’t take into account the effect of lone pair electrons during overlap between two orbitals. It can’t explain formation of co – ordinate covalent bond.

2. This theory doesn’t explain magnetic properties of compounds. For eg, it can’t explain paramagnetic behavior of oxygen molecule.

3. This theory fails to explain bonding in electron deficient compounds.

 

4.

 Difference between bond theory and molecular orbital theory:

Valence bond theory	Molecules orbital theory
1. In valence bond theory, two atomic orbitals give an inter atomic orbital obtained by space filling of two unpaired electrons one being in each of the two atomic orbitals.	1. In molecular orbital theory, molecular orbitals are formed by linear combinations of atomic orbitals (LACO) method.
2. The resulting molecule, consists of atoms and retain their individual character.	2. Atomic orbitals of the resulting molecule lose their individual identities.
3. Atomic orbitals are monocentric.	3. Molecular orbitals are poly – centric.

 

 

The molecules contain more than one electron, the exact wave function for molecules orbitals cannot be obtained. For obtaining the wave function for molecular orbital approximately applied is that the molecular orbitals are linear combination of atomic orital. This approximate method is known as linear combination of atomic orbital (LCAO) method.

The electronic configuration of oxygen is 1s²,2s²,2p⁴. Total electrons in O₈ molecule is 16. The electronic configuration of O₂ molecule, therefore is:

O₂ : (61s)²(6 * 1s)²(62s)²(6 * 2s)²(62p_x)²(π2p_y)²(π2p_z)²(π2p_y)¹(π * 2p_z)¹

Bond orders = $\frac{{{{\rm{N}}_{\rm{b}}} - {{\rm{N}}_{\rm{a}}}}}{2}$

= $\frac{{\left( {10 - 6} \right)}}{2}$ = $\frac{4}{2}$ = 2.

In O₂ molecule, atoms are held by a double covalent bond. It maybe noticed that it contains two unpaired electron in π * 2p_y and π * 2p_z molecular orbitals. So, O₂ molecule has paramagnetic nature.

 

5.

a.

 VSEPR model:

It states that molecule takes up the shape that minimizes the repulsions between the bonded and non – bonded pairs of electron.

The above conclusion is derived from following postulates:

(i) The no. of valence shell electron pair present in the central atom and their orientation in the space around the central atom determines the geometry or shape of the molecule.

(ii) Electrons are negatively charged particles, and hence electron – pair try to repel with each other to remain as far apart as possible to gain stability with minimum potential energy.

(iii) Lone – pair electrons exert a greater repulsive force on the bond pairs than one bond pair repels with another bond pair in Non – bonded electron repel more strongly than bondedelectrons.

(iv) If a central atom in a molecule has only one bond pair it has regular geometry and if the central atom has more lone pair, molecule gets distorted to same extent giving rise to irregular geometry to the molecule. Thus, the presence of lone pair of electron causes the decrease in the bond angle and hence causes the certain distortion in regular geometry of the molecule than we except.

(v) Greater the number of lone pair in addition to bond pair greater would be the decrease in the bond angle.

VSEPR theory is applied to predict the shape of simple molecules or ions in which the atoms are bonded by single bond. This theory is based upon the fact that, the direction of bonds around the central atom in polyatomic molecule depends upon the interaction of electron pairs of the central atom will arrange themselves in space in such a way that their minimum repulsion between them so that the molecule will have definite geometry with minimum energy and maximum stability.

 

 

 

 

b.

Hybridization is the process of mixing of dissimilar atomic orbitals of same atoms giving rise to equal number of a new set of orbitals having same energy contents.

Features of hybridization:

(i) Hybridization is the process of mixing of energies not the electron.

(ii) The hybrid orbitals have shape and same energy.

(iii) The number of Hybrid orbital formed is equal to the number of atomic orbitals mixing together.

(iv) The hybrid orbitals are situated in fire position in the space giving a definite geometrical shape.

(v) Only the bigger lobe takes part in overlapping.