Chemical Bonding And Molecular Structure

Atoms, except those of noble gases, do not have free existence. They readily combine with one another to form molecules. According to the classical view, the phenomenon of combination, called chemical bonding, “involves the union of two or more atoms through redistribution of electrons in their outermost shells involving either transference of electrons from one atom to another or sharing of electron amongst themselves so that all the atoms acquire the stable noble gas configurations of minimum energy”.

1. If the union takes place by the transference of electrons from one atom to another, the bonding is said to be ionic.

2. If it takes place by the mutual sharing of electrons between the atoms concerned, the bonding is said to be covalent.

Note: There is yet another type of bonding called coordinate covalent bonding in which both the electrons shared between that atoms are contributed by one atom only.

Steps to draw Lewis symbols:

⇒Step 1: Count the number of valence electrons (electrons in the outermost shell)

⇒Step 2: Write the symbol of element in the centre.

⇒Step 3: Draw dots around the symbol according to the number of valence electrons.

a) H₂S:

The atomic number of S is 16 and the no. of valence electrons is 6. The atomic number of H is 1 and no. of valence electron is 1. Hence, the total number of valence electrons is:

Two hydrogens (2× 1 valence electron) + sulphur (6 valence electrons) = Total 8 valence electrons

[Each line represents a pair of electrons]

The central atom S shares its two valence electrons out of six valence electrons to complete its octet and each hydrogen atom shares its one valence electron to complete the octet. The remaining four electrons act as a lone pair (electrons which do not participate in bonding) which is placed on the sulphur atom.

b) SiCl₄

The atomic number of Si is 14 and the no. of valence electrons is 4. The atomic number of Cl is 17 and no. of valence electron is 7. Hence, the total number of valence electrons is:

Four chlorines (4× 7 valence electrons) + silicon (4 valence electrons) = Total 32 valence electrons

The central atom Si shares it’s all valence electrons and each chlorine atom shares its one valence electron out of seven valence electrons to complete the octet.

c) BeF₂:

The atomic number of Be is 4 and the no. of valence electrons is 2. The atomic number of F is 9 and no. of valence electron is 7. Hence, the total number of valence electrons is:

Four fluorine (4× 7 valence electrons) + Beryllium (2 valence electrons) = Total 30 valence electrons

The central atom Be shares its all valence electrons and each fluorine atom shares its one valence electron out of seven valence electrons to complete the octet.

d) CO₃^2-

The atomic number of C is 6 and the no. of valence electrons is 4. The atomic number of O is 8 and no. of valence electron is 6. Hence, the total number of valence electrons is:

Three oxygen (3× 6 valence electrons) + Carbon (4 valence electrons) + negative charge (2 extra electrons) = Total 24 valence electrons

The central atom C shares its all valence electrons with each oxygen atom. Carbon requires only two electrons to complete its octet hence, only one of the oxygen atom shares its two valence electrons out of six valence electrons which are represented by a double bond.

e) HCOOH:

The atomic number of C is 6 and the no. of valence electrons is 4. The atomic number of O is 8 and no. of valence electron is 6.

The atomic number of hydrogen is 1 and no. of valence electron is 1. Hence, the total number of valence electrons is:

two oxygen (2× 6 valence electrons) + Carbon (4 valence electrons) + two hydrogen (2 × 1 valence electron) = Total 18 valence electrons

The central carbon atom shares it’s all 6 valence electrons with one hydrogen and two oxygen atoms to complete its octet. One of the oxygen atoms shares its two valence electrons with carbon which is represented by a double bond.

Octet rule states that “Atoms can combine either by the transfer of valence electrons from one atom to another by losing, sharing or gaining to acquire noble gas configuration (to complete octet in the outermost shells)”

Significance:

⇒With the help of octet rule, we can find out the nature of the bond.

⇒Octet rule helps out in providing a picture of bonding in molecules.

⇒It is useful for understanding the structure of most of the organic compounds.

Limitations:

⇒The incomplete octet of the central atom

It has been observed that in some compounds, the number of electrons around the central atom is less than 8. For example, in BCl₃

The number of electrons surrounding the central atom boron is less than eight. Some other such compounds are AlCl₃, BF₃, LiCl, BeH₂ etc.

⇒Odd-electron molecules

In compounds like NO and NO₂^-, which have odd number of electrons around the central atom N,

The favourable factors for the formation of the ionic bond are given below:

⇒Ionization Enthalpy: It is defined as the amount of energy required to remove the most loosely bound electron from an isolated gaseous atom of an element. The lesser the ionization energy, the greater is the ease of the formation of a cation. Alkali metals and alkaline earth metals have low ionization energies and, therefore, they form metals cations very easily.

⇒Electron Affinity: It is defined as the amount of energy released when an electron is added to an isolated gaseous atom of an element. The formation of an anion occurs with the addition of one or more electrons to an atom. The greater the energy released during this process, the easier will be the formation of an anion. Thus, high electron affinity of a non-metals favours the formation of an anion.

⇒Lattice Energy: It is defined as the amount of energy released when cations and anions are brought from infinity to their respective equilibrium sites in the crystal lattice to form one mole of the ionic compound. The higher the magnitude of the lattice energy, the greater is the tendency of the formation of an ionic bond.

Thus, the lower ionization energy of an atom (generally a metal atom), the high electron affinity of another atom (generally a non-metal atom) and a high lattice energy of the compound formed to facilitate the formation of an ionic bond.

Note: Each line in a Lewis symbol represent the bond pairs. The lone pairs on the central atom should be counted.

Remember these geometries:

In NH₃, there is only one lone pair on nitrogen atom to repel the bond pairs whereas in H₂O there are two lone pairs on O-atom to repel the bond pairs as shown below:

Hence, the repulsion on bond pairs are greater in H₂O than in NH₃ and hence the bond angle is less.

Note More the no. of lone pairs on an atom, greater will be the repulsion and as a result, less will be the bond angle.

1. Energy requires breaking a chemical bond. For example, in the breaking of one mole of hydrogen gas into atoms, 458 kJ of energy is required. The bond strength, in this case, is said to be 458 kJ per mole, i.e., per Avogadro’s number of bonds.

2. Bond strength or bond energy of a particular type of bond is defined as the energy required to break one mole of bonds of that type in a substance in a gaseous state.

3. Bond order is given by the number of bonds between the two atoms in a molecule. For example, in H_2, there is only one single shared electron pair, hence the bond order is 1.

4. The bond strength in terms of bond order is that higher will be the bond strength, it means more strong will be the bond, and as a result, greater will be the bond order.

5. It means greater is the bond order, higher the bond energy is required to break the strong bond.

The bond length is defined as the average distance between the nuclei of two bonded atoms in a molecule.

Example:

Let us consider a diatomic molecule. The atoms in this molecule are always vibrating with respect to each other. The question of any fixed distance between them, therefore, does not arise. We can, however, think of an average distance between the nuclei of the two atoms bonded to each other. This is called bond length or bond distance.

Thus,

The bond length in a covalent molecule AB.

R= r₁ + r₂

Where R is the bond length

R₁ and r₂ are the covalent radii of atoms A and B respectively

Note: The bond length decreases with a multiplicity of the bond formed between the two atoms. Thus, the C Ξ C bond is shorter than C=C bond which in turn, is shorter than C—C bond.

It has been observed experimentally that the carbon and oxygen bonds in CO₃^2- are equivalent. As we know that Lewis structure represents only unequal bonds. Hence, it is not enough to represent the molecule by Lewis structure accurately as it represents unequal bonds.

Hence it can be best explained by making its canonical structures and its resonance hybrid.

The structure I, II and III represent the three canonical forms (resonating structures) of CO₃^2-

The structure IV represents the structure of CO₃^2-more accurately. This is called resonance hybrid.

No, the structures 1 and 2 cannot be taken as canonical forms because the position of atoms has been changed.

While making resonating structures, following conditions must be followed:

⇒We cannot change the position of any atom. Atoms must be in their own position.

⇒We can only shift the bonds from one atom to another.

⇒We can only change the position of electrons from one atom to another.

The single Lewis structure based on the presence of two single bonds and one double bond is not enough to represent the molecule precisely as it presents unequal bonds. Hence, the concept of resonance was introduced to deal with this difficulty.

a) There are three possible resonance structures for SO₃ which are given below:

The structure I, II and III represent the three canonical forms (resonating structures) of SO₃.

The structure IV represents the structure of SO₃ more accurately. This is also called resonance hybrid.

b) There are two possible resonance structures for SO₃ which are given below:

The structure I and II represent the two canonical forms (resonating structures) of NO₂

The structure III represents the structure of NO₂ more accurately. This is also called resonance hybrid.

c) There are three possible resonance structures for NO₃^- which are given below:

The structure I,II and III represent the three canonical forms (resonating structures) of NO₃^-

The structure IV represents the structure of NO₃^- more accurately. This is also called resonance hybrid.

A diatomic molecule is polar if the bond formed between the atoms is polar. In such a molecule, the dipole moment of the bond gives the dipole moment of the molecule. For example, the dipole moment of HCl molecule is the same as the dipole moment of HCl bond.

However, in the case of molecules containing more than two atoms, the dipole moment not only depends upon the individual dipole moments but also on the spatial arrangement of various bonds in the molecule.

Carbon dioxide (CO₂) and water (H₂O) molecules are both triatomic molecules. It has been experimentally observed that dipole moment of CO₂ is zero and dipole moment of H₂O is 1.85D

If the dipole moment is zero, it is only possible when the molecule is linear so that the dipole moment of C=O bond on one side of the molecule is cancelled by that on the other side of the molecule giving net zero dipole moment.

μ = 0, shape is linear

If the dipole moment of H₂O is 1.85D, which shows that the molecule is non- linear. The H₂O molecule is bent, which is only possible when the dipole moment of the two O—H bonds inclined at an angle of 104.5^° and the contribution of electrons of the two lone pairs of electrons on the oxygen atom.

μ = 1.85D, shape is bent

When a covalent bond is formed between two dissimilar atoms or we can say, heteronuclear molecules like HF, the electron pair shared between two atoms displaced towards the more electronegative atom(fluorine) since electronegativity of fluorine is far greater than that of hydrogen. The resultant covalent bond is a polar covalent bond. This is called polarisation.

As a result of polarization, the molecule acquires the dipole moment which can be defined as the product of the magnitude of the charge and the distance between the centers of positive and negative charge.

Dipole moment (u) = Charge (Q) × distance of separation (r)

Significances/Applications:

⇒The SI unit of Dipole moment is Debye. [ 1Debye = 3.35 × 10^-30m]

⇒Dipole moment is a vector quantity. It has a magnitude and a direction.

⇒It is represented by an arrow with tail on a negative charge and head pointing towards the positive charge. []

⇒It is measure of the polarity of a bond. For example, in HF molecule, the dipole moment for the H⁺F^- when an electron is completely transferred from hydrogen atom to fluorine atom is given by 4.48D.

⇒The dipole moment also used to calculate the per cent ionic character.

Per cent ionic character = × 100

⇒On the basis of dipole moment, we can also find out the structure of molecule

Electronegativity is defined as the ability to attract a shared pair of electrons to itself is called electronegativity. The electronegativity of any given element is not constant, it changes from element to element to which it is bound. Though it is not a measurable quantity, it does provide a means predicting the nature of the force that holds a pair of electrons.

On the other hand, Electron gain enthalpy is defined as the amount of energy released when an electron is added to a gaseous isolated atom or ion. It a measurable quantity. The electron gain enthalpy of an atom measures the tightness with which it binds an additional electron to itself. The electron gain enthalpy is expressed in eV per atom or kJmol^-1

X(g) + e^-→ X^- (g) + Energy (Electron gain enthalpy)

Atom Anion

As we know that, if the bond is formed by the mutual sharing of electrons between the atoms, then it is called the covalent bond.

A covalent bond formed between two dissimilar atoms, the bond formed is said to be a polar covalent bond. A polar covalent bond is more stable than a pure covalent bond

The bond is formed between two dissimilar atoms means both having different electronegativities and which means the bond pair of electrons is not shared equally. The atom which is having high electronegativity, its bond pair shift towards the nucleus. As a result, the atom with higher electronegativity generally has a greater tendency to attract the electron towards itself.

This unsymmetrical distribution of electrons leads to partial charge separation. The electronegative atom acquires partial negative charged and the other atom acquires a slightly positive charge.

For example, HF, which contains a polar covalent bond (having different electronegativities). As we know that fluorine is more electronegative atom than the hydrogen. The electron pair shared between hydrogen and fluorine atoms, is, therefore, remains closer to the fluorine atom. As a result, fluorine acquires a partial negative charge and hydrogen acquires a partial positive charge.

To denote the partial charge on an atom, we use the symbol “δ”.

Electron pair is shifting towards

more electronegative atom, i.e., F

One more example:

Transfer of electron pair towards more electronegative chlorine.

N₂ < SO₂ <CIF₃ < K₂O < LiF

Explanation: First, we have to find out the ionic character in each molecule. As we know that ionic character in a molecule depends upon the electronegativity difference between the atoms. The higher the electronegativity difference, higher will be the ionic character.

LiF = The electronegativity of Li is 0.98 and electronegativity of F are 3.98. Hence the electronegativity difference between Li—F atoms is 3

K₂O = The electronegativity of K is 0.82 and the electronegativity of oxygen is 3.4. Hence the electronegativity difference between

K₂—O atoms is 2.58

N₂ = The electronegativity of N is 3.04. Hence the electronegativity difference between N—N atoms is 0

SO₂ = The electronegativity of S is 2.58 and the electronegativity of oxygen is 3.4. Hence the electronegativity difference between

S—O₂ atoms is 0.82

CIF₃ = The electronegativity of F, I and C is 3.9, 2.6 and 2.5 respectively. Hence the electronegativity difference between

C—I—F₃ atoms is 1.2

As 0 < 0.9 < 1.2 < 2.58 < 2.92

The ionic character increases with increase in the electronegativity difference.

∴ N₂ < SO₂ <CIF₃ < K₂O < LiF

The correct structure of CH₃COOH is:

Explanation:

The atomic number of C is 6 and no. of valence electrons is 4.

The atomic number of O is 8 and no. of valence electrons is 6.

The atomic number of H is 1 and no. of valence electron is 1.

Hence, the total number of valence electrons in the element is:

Two carbons (2× 4 valence electrons) + two oxygen (2× 6 valence electrons) + 4 hydrogen (4× 1 valence electron) = 24 total valence electrons. Hence, the skeletal structure of CH₃COOH is:

Here consider the electronic configuration of carbon atom:

₆C: 1s² 2s² 2p²

Whereas in the excited stage the orbital diagram can be,

Therefore, the carbon atom undergoes sp³ hybridisation in CH₄ and takes the tetrahedral shape.

But for a square planar shape, the hybridization of the central atom should be dsp² Since the atom of carbon does not have any d-orbitals to undergo hybridisation, the structure of CH₄ cannot be square planar. Also, with a bond angle of 90° in square planar, the stability of CH₄ will be very less because of the repulsion between the bond pairs. Hence, VSEPR theory also favours tetrahedral structure for CH₄.

The Lewis structure for BeH₂ is as follows:

H:Be:H

Here there is no lone pair at the central atom (Be) and there are two bond pairs on either side. Hence, BeH₂ is of the type AB₂ having linear structure.

Also, dipole moments of each H–Be bond are equal and are in opposite directions. Therefore, they nullify each other and hence, BeH₂ molecule has zero dipole moment.

Here, in both molecules i.e., NH₃ and NF₃, the central atom (N) has a lone pair electron and there are three bond pairs in total. Therefore, both the molecules have a pyramidal shape. Since fluorine is more electronegative than hydrogen, it is expected that the net dipole moment of NF₃ is greater than NH₃. But in real, the net dipole moment of NH₃ (1.46 D) is greater than that of NF₃ (0.24 D).

This phenomenon is explained on the basis of the direction of dipole moments extended by each individual bonds of the molecules

. These directions can be shown as:

From this figure, the net dipole moment of N-H bonds can be found greater than that of counterpart N-F bonds, where there is partial cancelling rather than adding up of moments. Hence, the net dipole moment of NF₃ is less than that of NH₃.

Hybridization can be defined as an intermixing of a set of atomic orbitals of slightly different energies, thereby forming a new set of orbitals having equivalent energies and shapes.

For example, one 2s-orbital hybridizes with two 2p-orbitals of carbon to form three new sp² hybrid orbitals.

These hybrid orbitals have minimum repulsion between their electron pairs and thus, are more stable. Hybridization helps indicate the geometry of the molecule.

Shape of sp hybrid orbitals:

Shape of sp hybrid orbitals:

sp hybrid orbitals have a linear shape. They are formed by the intermixing of s and p orbitals.

Shape of sp² hybrid orbitals:

sp² hybrid orbitals are formed because of the intermixing of one s-orbital and two p orbitals.

The shape of sp³ hybrid orbitals:

Four sp³ hybrid orbitals are formed by intermixing one s-orbital with three p-orbitals.

The valence orbital picture of aluminum in the ground state can be shown as:

The orbital picture of aluminum in the excited state can be shown as:

Hence, it undergoes sp² hybridization to give a trigonal planar arrangement (in AlCl₃). To form AlCl₄^–, the empty 3pz orbital also gets involved and the hybridization changes from sp² to sp³. As a result, the shape becomes tetrahedral.

Boron atom in BF3 is sp² hybridized. The excited state of boron can be represented by:

While the nitrogen atom in NH3 is sp³ hybridized. The excited state can be shown as:

On completion of this reaction, an adduct F₃B.NH₃ is formed as the hybridization of Boron changes to sp³, while the hybridization of nitrogen remains intact.

Case I:

C₂H_4: The electronic configuration of C-atom in the excited state is:

6C=1s²2s¹2px¹2py¹2pz¹

For the formation of the ethane molecule, one sp² hybrid orbital of carbon overlaps another sp² hybridised carbon orbital, forming a C-C sigma bond and the remaining two sp² orbitals of each carbon forms a sp²s sigma bond with the hydrogen atoms.

The unhybridized orbital of one carbon atom undergoes sidewise overlap with the orbital of a similar kind present on another carbon atom to form a weak π-bond.

CASE II:

C₂H₂ : In the formation ethane molecule, each of the C atoms is sp hybridized with two 2-p orbitals in the unhybridized state.

One sp orbital of each carbon atom overlaps with the other along the internuclear axis forming a C–C sigma bond. The second sp orbital of each C–atom overlaps a half-filled 1s-orbital to form a σ bond.

The two unhybridized 2p-orbitals of the first carbon undergo sidewise overlap with the 2p orbital of another carbon atom, thereby forming two pi bonds between carbon atoms. Hence, the triple bond between two carbon atoms is made up of one sigma and two π-bonds.

There are three sigma’s and two pi-bonds in C₂H₂.

Explanation: A single bond is a result of the axial overlap of bonding orbitals. Hence, it contributes a sigma bond. A multiple bond (double or triple bond) is always formed as a result of the sidewise overlap of orbitals. A pi-bond is always present in it. A triple bond is a combination of two pi-bonds and one sigma bond.

B.

There are five sigma bonds and one pi-bond in C₂H_4.

Explanation: A single bond is a result of the axial overlap of bonding orbitals. Hence, it contributes a sigma bond. A multiple bond (double or triple bond) is always formed as a result of the sidewise overlap of orbitals. A pi-bond is always present in it. A triple bond is a combination of two pi-bonds and one sigma bond.

The axial overlapping results in the formation of sigma bond. and sideways overlapping results in the formation of pi bond .
sigma bond involves overlapping of s - s , s - p and p - p atomic orbitals .


hence, option c is wrong because 2p_y and 2p_y orbitals will not a form a sigma bond. Taking x-axis as the internuclear axis, 2p_y and 2p_y orbitals will undergo lateral overlapping, thereby forming a pi bond.

A.

According to structure, C1 & C2 are making 4 sigma bonds(single bond) each with the help of ones hybrid orbital & 3 p hybrid orbital, hence C1& C2 are sp³ hybridized

B.

Here C1 is making 4 sigma bond, therefore, its sp³ hybridized, while C₂ and C₃ are making a double bond( 1 sigma + 1 pie bond ) therefore they both are sp² hybridized.

C.

Both the carbons C1 & C2 are making single bond(sigma bond), therefore they are sp³hybridized.

D.

From the structure it is clear that C1 is making sigma bonds only, therefore it is sp³hybridised.C2 is making a double bond therefore it is sp²hybridised.

C₁issp³hybridized and C₂issp²hybridized.

E.

Here C1 is making a sigma bond, therefore, it is in sp³hybridization state, while C2 is making a double bond, it is in the sp² hybridized state.

Lone pair means those pair of electrons which are not forming any bond with any other atom. For example, say a compound NH3, here in the central atom i.e. Nitrogen if you see you will observe that it is forming bond with three hydrogen atoms by sharing three electrons, but we also know that nitrogen contains five electrons at its outermost shell, that means nitrogen could have form five bond by sharing five electrons but here in this case only three electrons are only used to form bond with hydrogen and still two electrons i.e. a pair of electron is left as such. This pair of electron which is not taking part in bond formation will be called as lone pair and those which are forming bonds will be called as bond pair.

For Example:

In H₂O, there are two bond pairs and two lone pairs on the central atom (oxygen).

The differences of each type are listed below:

Sigma Bond:

1. The covalent bond formed by the overlap of atomic orbitals along the internuclear axis is called sigma bond

2. The overlapping orbitals are oriented along the internuclear axis.

3. s as well as p orbitals can form this type of bonds.

4. It is stronger than pi bond

Pi Bond:

1. The covalent bond formed by the lateral overlap of two p orbitals which are mutually parallel but oriented perpendicular to the internuclear axis is called the pi bond.

2. The overlapping orbitals are oriented perpendicular to the inter nuclear axis.

3. The bond is not rotationally symmetrical around the internuclear axis.

4. Only p orbitals can form this type of bond.

Assume that two hydrogen atoms (A and B) with nuclei (NA and NB) and electrons (e_A and e_B) are taken to undergo a reaction to form a hydrogen molecule.

When A and B are at a large distance, there is no interaction between them. As they begin to approach each other, the attractive and repulsive forces start operating.

Attractive force arises between:

(a) Nucleus of one atom and its own electron i.e., NA–e_A, and NB–e_B.

(b) The nucleus of one atom and electron of another atom i.e., NA–e_B and NB–e_A.

Repulsive force arises between:

(a) Electrons of two atoms i.e., e_A – e_B.

(b) Nuclei of two atoms i.e., NA – NB.

The force of attraction brings the two atoms together, whereas the force of repulsion tends to push them apart.

The magnitude of the attractive forces is more than that of the repulsive forces. Hence, the two atoms approach each other. As a result, the potential energy decreases. Finally, a state is reached when the attractive forces balance the repulsive forces and the system acquires minimum energy. This leads to the formation of a dihydrogen molecule.

The given conditions should be satisfied by atomic orbitals to form molecular orbitals:

(a) The combining atomic orbitals must have the same or nearly the same energy. This means that in a homonuclear molecule, the 1s-atomic orbital of an atom can combine with the 1s-atomic orbital of another atom, and not with the 2s-orbital.

(b) The combining atomic orbitals must have proper orientations to ensure that the overlap is maximum.

(c) The extent of overlapping should be large.

The electronic configuration of Beryllium is 1S² 2S^2.

The molecular orbital electronic configuration for Be2 molecule can be written as:

Where,

N_b = Number of electrons in bonding orbitals = 4 [ which are in (σ_1s) and (σ_2s) ].

N_a = Number of electrons in anti-bonding orbitals = 4 [ which are in (σ^∗_1s) and (σ^∗_2s) ].

Bond order of Be₂ = N_b - N<sub>a

= 4 - 4

= 0</sub>

Bond order of Be₂ is 0 _. Hence, Be₂molecule does not exist.

The electronic configuration of oxygen molecule can be written as:

Here number of bonding electrons is 8 (N_a) and number of antibonding electrons is 4 (N_b)

Therefore, bond order can be found out by,

=1/2(8-4)

=2

The electronic configuration of O₂⁺ can be written as:

N_b = 8

N_a = 3

Bond order is 1/2(8-3)

=2.5

The electronic configuration of O₂^- will be:

N_b = 8

N_a = 5

Bond order is 1/2 (8-5)

=1.5

The electronic configuration of O₂^2-will be:

N_b = 8

N_a = 6

Bond order is 1/2 (8-6)

=1

Bond dissociation energy is directly proportional to bond order. Thus, the higher the bond order, the greater will be the stability. On this basis, the order of stability is:

O₂⁺ > O₂> O₂^- > O₂^2-

Molecular orbitals are represented by wave functions. A plus sign in an orbital indicates a positive wave function while a minus sign in an orbital represents a negative wave function.

The ground state and excited state outer electronic configurations of phosphorus (Z = 15) are:

Ground state:

Excited state:

Phosphorus atom is sp³d hybridized in the excited state. These orbitals are filled by the electron pairs donated by five Cl atoms as: PCl_5.

The five sp³d hybrid orbitals are directed towards the five corners of the trigonal bipyramidal. Hence, the geometry of PCl₅ can be represented as:

There are five P–Cl sigma bonds in PCl₅. Three P–Cl bonds lie in one plane and make an angle of 120° with each other. These bonds are called equatorial bonds.

The remaining two P–Cl bonds lie above and below the equatorial plane and make an angle of 90° with the plane. These bonds are called axial bonds.

As the axial bond pairs suffer more repulsion from the equatorial bond pairs, axial bonds are slightly longer than equatorial bonds.

A hydrogen bond is defined as an attractive force acting between the hydrogen attached to an electronegative atom of one molecule and an electronegative atom of a different molecule. Due to a difference between electronegativities, the bond pair between hydrogen and the electronegative atom gets drifted far away from the hydrogen atom. As a result, a hydrogen atom becomes electropositive with respect to the other atom and acquires a positive charge.

There are two types of H-bonds:

(i) Intermolecular H-bond e.g., HF, H₂O etc.

(ii) Intramolecular H-bond e.g., o-nitrophenol

Hydrogen bonds are stronger than Van der Walls forces since hydrogen bonds are regarded as an extreme form of dipole-dipole interaction.

Bond order is defined as one half of the difference between the number of electrons present in the bonding and anti-bonding orbitals of a molecule.

If N_a is equal to the number of electrons in an anti-bonding orbital, then N_b is equal to the number of electrons in a bonding orbital.

Bond Order:

If N_b > N_a, then the molecule is said be stable. However, if N_b≤ N_a, then the molecule is considered to be unstable.

Bond order of N₂ :

Number of bonding electrons, N_b = 10

Number of anti-bonding electrons, N_a = 4

Bond order of nitrogen molecule= 1/2 (10-4)=3

Bond order of O₂:

Number of bonding electrons, N_b = 8

Number of anti-bonding electrons, N_a = 4.

Bond order of Oxygen molecule = 1/2 (8-4)=2

Bond order of O₂⁺ :

Number of bonding electrons, N_b = 8

Number of anti-bonding electrons, N_a = 3

Bond order of O₂^- = 1/2 (8-3) = 2.5

Bond order of O₂^-:

Number of bonding electrons, N_b = 8

Number of anti-bonding electrons, N_a = 5

Bond order of O₂^- = 1/2 (8-5) = 1.5