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Chapter 4 Chemical Bonding And Molecular Structure
Matter is composed of elements, and except for noble gases, atoms of elements typically do not exist independently in nature. Instead, they form groups of atoms called molecules or are organized into larger structures like ionic lattices. This suggests that there are attractive forces holding these constituent atoms or ions together. This attractive force is known as a chemical bond.
Why Do Atoms Combine?
Atoms combine to form chemical bonds because bonding generally leads to a lowering of energy for the system, resulting in increased stability. Different theories have been developed over time to explain why atoms combine, which combinations are possible, why some atoms combine while others do not, and why molecules have specific shapes. These theories include the Kössel-Lewis approach, Valence Shell Electron Pair Repulsion (VSEPR) Theory, Valence Bond (VB) Theory, and Molecular Orbital (MO) Theory. The understanding of chemical bonding has evolved alongside advancements in atomic structure, electronic configurations, and the periodic table.
Kössel-Lewis Approach To Chemical Bonding
The initial understanding of chemical bonding focused on the role of electrons. In 1916, G.N. Lewis and W. Kössel independently proposed explanations based on the observed inertness of noble gases.
Lewis's Model:
- Lewis visualized an atom as having a positively charged 'Kernel' (nucleus plus inner electrons) and an outer shell containing the valence electrons.
- He proposed that the outer shell could hold a maximum of eight electrons, arranged at the corners of a cube surrounding the kernel. This stable arrangement of eight electrons was called an octet.
- Lewis postulated that atoms form chemical bonds to achieve this stable octet configuration in their valence shell, resembling the electronic configuration of noble gases.
- Bonding could occur either by the transfer of electrons from one atom to another (forming ions) or by the sharing of electrons between atoms (forming covalent bonds).
Lewis Symbols: Lewis introduced a simple notation to represent valence electrons. The Lewis symbol of an element consists of the element's symbol surrounded by dots or crosses representing its valence electrons. For instance, the Lewis symbols for the second-period elements show the number of valence electrons increasing from one to eight.
The number of dots in a Lewis symbol indicates the number of valence electrons, which helps predict an element's common valence (typically equal to the number of valence electrons or 8 minus that number).
Kössel's Model (Ionic Bonding):
- Kössel focused on the formation of ionic bonds through electron transfer.
- He noted that highly electropositive alkali metals and highly electronegative halogens are located near noble gases in the periodic table.
- Alkali metals readily lose their single valence electron to form positive ions (cations) with a noble gas configuration.
- Halogens readily gain one electron to form negative ions (anions) with a noble gas configuration (specifically, an $ns^2np^6$ octet, except for He's duplet $1s^2$).
- The resulting positive and negative ions are held together by strong electrostatic attraction, forming an electrovalent bond (or ionic bond).
- The electrovalence of an element is equal to the magnitude of the charge on its ion.
Example of NaCl formation (Kössel's approach):
Na ($\text{[Ne]}3s^1$) $\rightarrow$ Na⁺ ($\text{[Ne]}$) + e⁻
Cl ($\text{[Ne]}3s^23p^5$) + e⁻ $\rightarrow$ Cl⁻ ($\text{[Ne]}3s^23p^6$ or $\text{[Ar]}$)
Na⁺ + Cl⁻ $\rightarrow$ Na⁺Cl⁻ (NaCl)
Similarly for CaF₂:
Ca ($\text{[Ar]}4s^2$) $\rightarrow$ Ca²⁺ ($\text{[Ar]}$) + 2e⁻
F ($\text{[He]}2s^22p^5$) + e⁻ $\rightarrow$ F⁻ ($\text{[He]}2s^22p^6$ or $\text{[Ne]}$)
Ca²⁺ + 2F⁻ $\rightarrow$ Ca²⁺(F⁻)₂ (CaF₂)
Kössel's ideas provided a valuable framework for understanding ionic compound formation and the stability gained by achieving noble gas configurations. However, it was recognized that many compounds could not be explained by simple electron transfer.
Octet Rule
The electronic theory of chemical bonding, developed by Kössel and Lewis, is centered around the octet rule: Atoms combine by transferring or sharing valence electrons to achieve eight electrons in their valence shell, thereby attaining a stable configuration similar to that of noble gases. (Hydrogen and some other elements near Helium tend to achieve a duplet, $1s^2$ configuration).
Covalent Bond
Irving Langmuir (1919) built upon Lewis's ideas by introducing the concept of the covalent bond, formed by the mutual sharing of electron pairs between atoms. He discarded Lewis's rigid cubical electron arrangement.
Formation of Cl₂ (Langmuir's model): A chlorine atom ($\text{[Ne]}3s^23p^5$) needs one electron to complete its octet. When two Cl atoms approach, they can share one pair of electrons, with each atom contributing one electron to the shared pair. This allows both atoms to effectively have eight electrons in their valence shell (counting the shared pair), achieving the stable argon configuration.
This is represented by Lewis dot structures. A line is often used to denote a shared electron pair (single bond).
Multiple Covalent Bonds: Atoms can share more than one pair of electrons to satisfy the octet rule:
- Double Bond: Two atoms share two pairs of electrons. Example: Carbon dioxide (CO₂), Ethene (C₂H₄).
- Triple Bond: Two atoms share three pairs of electrons. Example: Nitrogen (N₂), Ethyne (C₂H₂).
Conditions for covalent bond formation in Lewis-Langmuir theory:
- Each bond involves sharing of an electron pair.
- Each participating atom contributes at least one electron to the shared pair.
- Combining atoms achieve a stable outer shell configuration (octet or duplet).
Lewis Representation Of Simple Molecules (The Lewis Structures)
Drawing Lewis structures is a systematic way to visualize bonding and non-bonding electron pairs (lone pairs) in molecules and polyatomic ions based on the octet rule. While simplified, they provide a basic understanding of molecular structure.
Steps to Write Lewis Structures:
- Count Total Valence Electrons: Sum the valence electrons of all atoms in the molecule or ion.
Example: CH₄: C (4) + 4 $\times$ H (1) = 8 valence electrons.
- Adjust for Charge (for ions): Add electrons for each negative charge (anions) or subtract electrons for each positive charge (cations).
Example: CO₃²⁻: C (4) + 3 $\times$ O (6) + 2 (for -2 charge) = 4 + 18 + 2 = 24 valence electrons.
Example: NH₄⁺: N (5) + 4 $\times$ H (1) - 1 (for +1 charge) = 5 + 4 - 1 = 8 valence electrons.
- Determine Skeletal Structure: Arrange atoms with the least electronegative atom usually occupying the central position. Hydrogen and Fluorine are almost always terminal atoms.
- Draw Single Bonds: Place one shared electron pair (a single bond) between the central atom and each surrounding atom. Subtract these electrons from the total.
- Complete Octets (and Duplets): Distribute the remaining electrons as lone pairs around the terminal atoms first to satisfy their octets (Hydrogen needs only 2 electrons). Then, place any leftover electrons on the central atom.
- Form Multiple Bonds (if necessary): If the central atom does not have an octet after distributing all electrons as lone pairs, move lone pairs from terminal atoms to form double or triple bonds until the central atom achieves an octet.
| Molecule/Ion | Total Valence Electrons | Lewis Structure |
|---|---|---|
| H₂O | O(6) + 2H(1) = 8 | |
| CO₂ | C(4) + 2O(6) = 16 | |
| CO₃²⁻ | C(4) + 3O(6) + 2 = 24 | |
| NH₃ | N(5) + 3H(1) = 8 |
Example 1. Write the Lewis dot structure of CO molecule.
Answer:
Step 1: Count valence electrons: Carbon (Group 14) has 4 valence electrons, Oxygen (Group 16) has 6. Total = 4 + 6 = 10 electrons.
Step 2: Skeletal structure: C O (for diatomic molecule).
Step 3: Draw a single bond: C—O (uses 2 electrons). Remaining electrons = 10 - 2 = 8.
Step 4: Complete octets (terminal first): Place 6 electrons around O to complete its octet (uses 6 electrons). Remaining electrons = 8 - 6 = 2. Place the last 2 electrons on C as a lone pair. Structure looks like C : — O : : .
Current state: C has 2 (from lone pair) + 2 (from bond) = 4 electrons (not octet). O has 6 (from lone pairs) + 2 (from bond) = 8 electrons (octet).
Step 5: Form multiple bonds: C needs 4 more electrons. Move two lone pairs from O to form two more bonds with C (a triple bond). Uses 4 electrons from O's lone pairs. O now has 2 lone pairs instead of 3. Structure looks like : C $\equiv$ O : .
Check octets: C has 2 (lone pair) + 6 (triple bond) = 8 electrons. O has 2 (lone pair) + 6 (triple bond) = 8 electrons. Both have octets.
The Lewis structure of CO is : C $\equiv$ O : .
Example 2. Write the Lewis structure of the nitrite ion, NO₂⁻.
Answer:
Step 1: Count valence electrons: Nitrogen (Group 15) has 5, Oxygen (Group 16) has 6 each. Add 1 for the -1 charge. Total = 5 + (2 $\times$ 6) + 1 = 5 + 12 + 1 = 18 electrons.
Step 2: Skeletal structure: O N O (N is less electronegative than O, so N is central).
Step 3: Draw single bonds: O—N—O (uses 4 electrons). Remaining electrons = 18 - 4 = 14.
Step 4: Complete octets (terminal first): Place 6 electrons around each O (uses 12 electrons). Remaining electrons = 14 - 12 = 2. Place the last 2 electrons on N as a lone pair. Structure looks like : : O : — N — : O : : with a lone pair on N.
Current state: Each O has 6 + 2 = 8 electrons (octet). N has 2 (lone pair) + 2 + 2 (from bonds) = 6 electrons (not octet).
Step 5: Form multiple bonds: N needs 2 more electrons. Move a lone pair from one of the O atoms to form a double bond with N. Uses 2 electrons from one O's lone pair. That O now has 4 lone pair electrons + 4 bond electrons = 8 (octet). N now has 2 (lone pair) + 4 (double bond) + 2 (single bond) = 8 (octet). The other O still has 6 + 2 = 8 (octet).
There are two possible options for where the double bond goes, leading to two resonance structures:
The Lewis structure of NO₂⁻ is represented by these two resonance forms.
Formal Charge
In polyatomic molecules or ions, the overall charge is distributed across the entire species. Formal charge (F.C.) is a conceptual charge assigned to each atom in a specific Lewis structure to help assess the distribution of electrons and compare different possible Lewis structures for the same species. It is calculated based on the assumption that bonding electrons are shared equally between atoms.
The formula for formal charge on an atom is:
$$ \text{F.C.} = \left( \begin{array}{c} \text{Total number of} \\ \text{valence electrons} \\ \text{in the free atom} \end{array} \right) - \left( \begin{array}{c} \text{Total number of} \\ \text{non-bonding} \\ \text{(lone pair) electrons} \end{array} \right) - \frac{1}{2} \left( \begin{array}{c} \text{Total number of} \\ \text{bonding} \\ \text{(shared) electrons} \end{array} \right) $$Example: Ozone molecule (O₃), with Lewis structure O=O-O and a lone pair on the central O, two lone pairs on the double-bonded O, and three lone pairs on the single-bonded O.
Formal charge on:
- Central O (atom 1, bonded to two Os, one lone pair): Valence e⁻ = 6. Non-bonding e⁻ = 2. Bonding e⁻ = 6 (double bond + single bond). F.C. = $6 - 2 - \frac{1}{2}(6) = 6 - 2 - 3 = +1$.
- Double-bonded O (atom 2, bonded to central O, two lone pairs): Valence e⁻ = 6. Non-bonding e⁻ = 4. Bonding e⁻ = 4 (double bond). F.C. = $6 - 4 - \frac{1}{2}(4) = 6 - 4 - 2 = 0$.
- Single-bonded O (atom 3, bonded to central O, three lone pairs): Valence e⁻ = 6. Non-bonding e⁻ = 6. Bonding e⁻ = 2 (single bond). F.C. = $6 - 6 - \frac{1}{2}(2) = 6 - 6 - 1 = -1$.
The Lewis structure with formal charges is:
The sum of formal charges on all atoms in a molecule must equal zero, and in an ion, it must equal the charge of the ion. Formal charges help in choosing the most stable (lowest energy) Lewis structure among several possibilities; the preferred structure is generally the one with the smallest formal charges (closest to zero) and negative formal charges located on the more electronegative atoms.
Limitations Of The Octet Rule
Despite its usefulness, especially for organic molecules and second-period elements, the octet rule is not universally applicable. Key limitations and exceptions include:
- Incomplete Octet: The central atom has fewer than eight valence electrons. This occurs in compounds of elements with less than four valence electrons, like Group 1, 2, and some Group 13 elements. Examples: LiCl (Li has 2 e⁻), BeH₂ (Be has 4 e⁻), BCl₃ (B has 6 e⁻), AlCl₃, BF₃.
- Odd-electron Molecules: Molecules with a total odd number of valence electrons cannot satisfy the octet rule for all atoms. At least one atom will have an odd number of electrons (a single unpaired electron), preventing a complete octet. Examples: Nitric oxide (NO, 11 valence e⁻), Nitrogen dioxide (NO₂, 17 valence e⁻).
- Expanded Octet: The central atom has more than eight valence electrons. This is possible for elements in the third period and beyond because they have vacant d-orbitals available in their valence shell that can accommodate additional electron pairs. Examples: PF₅ (P has 10 e⁻), SF₆ (S has 12 e⁻), H₂SO₄ (S has 12 e⁻). Phosphorus and Sulfur can also form compounds obeying the octet rule (e.g., PCl₃, SCl₂), but they can expand their octet in other compounds.
Other drawbacks of the octet theory:
- It assumes the inertness of noble gases as the basis for stability, but some noble gases (Xe, Kr) are known to form compounds (e.g., XeF₂, KrF₂).
- It does not provide information about the actual shapes of molecules.
- It does not explain the relative stability or energy of different molecules.
Ionic Or Electrovalent Bond
An ionic bond is formed by the electrostatic attraction between a positive ion (cation) and a negative ion (anion), which are created by the complete transfer of one or more electrons from one atom to another. This concept was primarily advanced by Kössel.
The formation of an ionic compound involves:
- Formation of a gaseous cation from a neutral gaseous atom (requires energy - Ionization Enthalpy, $\Delta_i H$).
M(g) $\rightarrow$ M⁺(g) + e⁻ - Formation of a gaseous anion from a neutral gaseous atom (energy change is Electron Gain Enthalpy, $\Delta_{eg} H$).
X(g) + e⁻ $\rightarrow$ X⁻(g) - Packing of gaseous ions into a crystalline solid lattice (releases energy - Lattice Enthalpy).
M⁺(g) + X⁻(g) $\rightarrow$ MX(s)
Ionic bonds are most likely to form between elements with low ionization enthalpies (metals, easy to lose electrons) and elements with highly negative electron gain enthalpies (non-metals, easy to gain electrons). Typically, cations are derived from metals and anions from non-metals, though polyatomic ions like NH₄⁺ can also participate in ionic bonding.
Ionic compounds in the solid state exist as a three-dimensional network of ions (a crystal lattice, e.g., the rock salt structure of NaCl) held together by strong coulombic forces. The stability of an ionic compound is primarily determined by the large amount of energy released during the formation of this crystal lattice (lattice enthalpy).
Even if the sum of ionization enthalpy and electron gain enthalpy for the formation of gaseous ions is positive (endothermic), the very large, negative value of lattice enthalpy typically compensates for this, making the overall process of forming the solid ionic compound exothermic and stable.
Lattice Enthalpy
The Lattice Enthalpy of an ionic solid is defined as the energy required to completely separate one mole of a solid ionic compound into its constituent gaseous ions. For example, the lattice enthalpy of NaCl is +788 kJ mol⁻¹, meaning 788 kJ of energy is needed to convert 1 mole of NaCl(s) into 1 mole of Na⁺(g) and 1 mole of Cl⁻(g).
A higher (more positive) lattice enthalpy indicates stronger forces between the ions and greater stability of the ionic crystal. Lattice enthalpy depends on the charges of the ions (higher charges lead to stronger attraction and higher lattice enthalpy) and the distance between the ions (smaller ions lead to closer packing, stronger attraction, and higher lattice enthalpy), as well as the crystal structure.
Bond Parameters
Chemical bonds, whether primarily ionic or covalent, have characteristics that can be quantified. These quantitative aspects are called bond parameters.
Bond Length
Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule. It is usually measured in picometers (pm) or Ångstroms (Å) and determined using experimental techniques like spectroscopy, X-ray diffraction, and electron diffraction.
Each atom in a bonded pair contributes to the bond length. For covalent bonds, this contribution is approximated by the covalent radius, which is roughly half the distance between the nuclei of two identical atoms joined by a single covalent bond in the same molecule.
For comparison, the van der Waals radius is defined for non-bonded atoms and is half the distance between the nuclei of two identical atoms in separate molecules in a solid. Van der Waals radii are generally larger than covalent radii as they represent the overall size of the non-interacting atom's electron cloud.
Bond lengths are influenced by the type of bond (single, double, or triple), the size of the atoms, and the atoms involved. Generally, multiple bonds are shorter than single bonds between the same two atoms.
| Bond Type | Covalent Bond Length (pm) | Bond Type | Covalent Bond Length (pm) |
|---|---|---|---|
| O–H | 96 | C=O | 121 |
| C–H | 107 | N=O | 122 |
| N–O | 136 | C=C | 133 |
| C–O (single) | 143 | C=N | 138 |
| C–N (single) | 143 | C$\equiv$N | 116 |
| C–C (single) | 154 | C$\equiv$C | 120 |
Bond Angle
The bond angle is defined as the angle between the orbitals containing bonding electron pairs around the central atom in a molecule or complex ion. It is expressed in degrees and can be determined experimentally using spectroscopic methods.
Bond angles provide information about the spatial arrangement of bonds around the central atom and are crucial in determining the molecule's overall shape or geometry. For example, in a water molecule (H₂O), the bond angle is the angle between the two O-H bonds, which is approximately 104.5°.
Bond Enthalpy
Bond enthalpy (or bond dissociation enthalpy) is the amount of energy required to break one mole of a specific type of bond between two atoms in the gaseous state. It is a measure of bond strength; a higher bond enthalpy indicates a stronger bond. The unit is typically kJ mol⁻¹.
For diatomic molecules like H₂, O₂, or N₂, the bond enthalpy is the energy required to break the bond in one mole of molecules:
H₂(g) $\rightarrow$ 2H(g); $\Delta_a H^\ominus = +435.8$ kJ mol⁻¹ (H-H bond enthalpy)
O₂(g) $\rightarrow$ 2O(g); $\Delta_a H^\ominus = +498$ kJ mol⁻¹ (O=O bond enthalpy)
N₂(g) $\rightarrow$ 2N(g); $\Delta_a H^\ominus = +946$ kJ mol⁻¹ (N≡N bond enthalpy)
For polyatomic molecules, breaking different bonds, even if of the same type, may require different amounts of energy due to changes in the chemical environment. For example, in water (H₂O), breaking the first O-H bond requires 502 kJ/mol, while breaking the second O-H bond requires 427 kJ/mol.
In such cases, the concept of mean or average bond enthalpy is used, calculated as the total bond dissociation enthalpy divided by the number of bonds broken. For H₂O, the average O-H bond enthalpy is $(502 + 427)/2 = 464.5$ kJ mol⁻¹.
Bond Order
In the Lewis description of covalent bonding, the bond order is defined as the number of covalent bonds (shared electron pairs) between two atoms in a molecule.
- H₂ (H-H): 1 shared pair, Bond Order = 1 (Single bond)
- O₂ (O=O): 2 shared pairs, Bond Order = 2 (Double bond)
- N₂ (N≡N): 3 shared pairs, Bond Order = 3 (Triple bond)
- CO (C≡O): 3 shared pairs, Bond Order = 3 (Triple bond)
Species with the same number of electrons (isoelectronic species) often have the same bond order (e.g., N₂, CO, and NO⁺ all have bond order 3). There is a general correlation between bond order, bond enthalpy, and bond length: As bond order increases, bond enthalpy increases, and bond length decreases.
Resonance Structures
Sometimes, a single Lewis structure cannot accurately represent a molecule or ion based on experimental data (like bond lengths). For example, in the ozone molecule (O₃), experiments show both oxygen-oxygen bonds have the same length (128 pm), intermediate between a typical single bond (148 pm) and a double bond (121 pm). A single Lewis structure with one single and one double bond is inadequate.
The concept of resonance is used to describe such species. When a single Lewis structure is insufficient, the molecule or ion is considered to be a resonance hybrid of two or more contributing structures called canonical forms or resonance structures.
Canonical forms are hypothetical structures that:
- Have the same atomic nuclei arrangement.
- Have the same number of electron pairs (bonding and non-bonding).
- Differ only in the distribution of electron pairs (positions of double/triple bonds and lone pairs).
The actual structure of the molecule is the resonance hybrid, which is an average of the canonical forms. Resonance is indicated by a double-headed arrow ($\leftrightarrow$) between the canonical forms.
Examples of species exhibiting resonance include the carbonate ion (CO₃²⁻) and the carbon dioxide molecule (CO₂).
Example 3. Explain the structure of CO₃²⁻ ion in terms of resonance.
Answer:
A single Lewis structure for CO₃²⁻ based on octets typically involves one C=O double bond and two C-O single bonds. This would imply one shorter double bond and two longer single bonds.
However, experimental evidence shows that all three C-O bonds in the CO₃²⁻ ion are identical in length and strength. To accurately represent this, we use resonance. The double bond can be placed between the carbon atom and any of the three oxygen atoms, leading to three equivalent canonical forms:
The actual structure of the carbonate ion is a resonance hybrid of these three forms. The hybrid has a partial double bond character distributed equally among the three C-O bonds. The bond order for each C-O bond is $(1+1+2)/3 = 4/3 \approx 1.33$, consistent with bonds intermediate between single and double.
Example 4. Explain the structure of CO₂ molecule.
Answer:
The typical Lewis structure for CO₂ shows two C=O double bonds.
The normal C=O double bond length is about 121 pm. However, the experimentally determined bond length for the C-O bonds in CO₂ is 115 pm, which is shorter than a typical double bond but longer than a typical triple bond (C≡O is about 110 pm).
This indicates that the bonds have some triple bond character. Resonance structures involving triple bonds can be drawn:
The actual structure of CO₂ is a resonance hybrid of these three canonical forms. The hybrid has a bond order for each C-O bond of $(2+3+1)/3 \approx 2$, but considering the structures contributing to the hybrid, the average character is slightly more than a double bond ($115 \text{ pm} < 121 \text{ pm}$). More detailed MO theory is needed for a complete picture, but resonance helps explain the observed bond length being shorter than a pure double bond.
Key points about resonance:
- Resonance stabilizes the molecule; the resonance hybrid is more stable (lower energy) than any single canonical form.
- Resonance leads to average bond characteristics (lengths, strengths).
- Canonical forms are hypothetical; the molecule exists only as the resonance hybrid. There is no rapid interconversion between canonical forms like in tautomerism.
Polarity Of Bonds
Pure ionic and pure covalent bonds are idealized concepts. In reality, most chemical bonds have some degree of both ionic and covalent character.
- Nonpolar Covalent Bond: Formed between identical atoms (e.g., H₂, O₂, Cl₂). The shared electron pair is equally attracted by both nuclei and is located symmetrically between them.
- Polar Covalent Bond: Formed between different atoms with different electronegativities (e.g., HF, H₂O). The shared electron pair is pulled closer to the more electronegative atom, creating a partial negative charge ($\delta^-$) on that atom and a partial positive charge ($\delta^+$) on the less electronegative atom.
The degree of polarity in a bond is measured by its dipole moment ($\mu$). For a diatomic molecule, the dipole moment is the product of the magnitude of the partial charge (Q) and the distance (r) between the centers of positive and negative charge ($\mu = Q \times r$).
Dipole moment is a vector quantity. By convention, it is represented by an arrow pointing from the positive end to the negative end ($\longrightarrow$). In chemistry diagrams, a crossed arrow ($\longrightarrow$) is often used on the Lewis structure, with the cross indicating the positive end and the arrowhead pointing towards the negative end (direction of electron density shift).
The unit of dipole moment is typically Debye (D), where $1 \text{ D} = 3.33564 \times 10^{-30}$ C m.
Dipole Moment in Polyatomic Molecules: For molecules with more than two atoms, the overall dipole moment is the vector sum of the individual bond dipoles. Molecular geometry is crucial here. Even if a molecule contains polar bonds, its overall dipole moment can be zero if the bond dipoles are oriented symmetrically and cancel each other out.
- H₂O: Bent shape (bond angle 104.5°), polar O-H bonds. The bond dipoles do not cancel and vectorially add up, resulting in a net dipole moment (1.85 D). H₂O is a polar molecule.
- BeF₂: Linear shape (bond angle 180°), polar Be-F bonds. The bond dipoles are equal in magnitude and opposite in direction, canceling out. Net dipole moment is zero. BeF₂ is a nonpolar molecule.
- BF₃: Trigonal planar shape (bond angle 120°), polar B-F bonds. The three bond dipoles are oriented symmetrically in a plane and vectorially add to zero. Net dipole moment is zero. BF₃ is a nonpolar molecule.
Comparison of NH₃ and NF₃: Both molecules have a trigonal pyramidal shape with a lone pair on the central N atom. The bond dipoles point towards the more electronegative atom (towards N in N-H bonds, towards F in N-F bonds). The lone pair on N also contributes an "orbital dipole moment" (pointing away from the nucleus). The resulting molecular dipole moment is the vector sum of bond dipoles and the lone pair dipole.
- NH₃: Bond dipoles (towards N) and the lone pair dipole are roughly in the same direction (upwards, away from the H atoms). They reinforce each other, leading to a large net dipole moment (1.47 D).
- NF₃: Bond dipoles (towards F, away from N) and the lone pair dipole (away from N) are in opposite directions. The lone pair dipole partially cancels the resultant of the N-F bond dipoles, leading to a small net dipole moment (0.23 D).
Partial Covalent Character in Ionic Bonds (Fajans' Rules): Just as covalent bonds have some ionic character, ionic bonds also have partial covalent character. Fajans' rules summarize the factors that increase the covalent character in an ionic bond:
- Cation Size/Charge: Smaller cation and larger charge on the cation lead to greater polarising power (ability to distort the anion's electron cloud).
- Anion Size/Polarisability: Larger anion and higher charge on the anion lead to greater polarisability (ease of being distorted by the cation).
- Cation Electronic Configuration: Cations with a $(n-1)d^{10}ns^0$ configuration (like transition metal ions) are generally more polarising than cations with a noble gas $ns^2np^6$ configuration (like alkali/alkaline earth metal ions) of similar size and charge.
The distortion of the anion's electron cloud by the cation increases the electron density in the region between the nuclei, which is characteristic of a covalent bond.
The Valence Shell Electron Pair Repulsion (VSEPR) Theory
The VSEPR theory provides a simple model for predicting the geometrical shapes of covalent molecules and polyatomic ions. It is based on the idea that electron pairs in the valence shell of the central atom repel each other and arrange themselves in space to minimize these repulsions.
Postulates of VSEPR Theory:
- The shape of a molecule is determined by the total number of valence shell electron pairs (both bonding pairs and lone pairs) around the central atom.
- These electron pairs repel each other due to their negative charge.
- Electron pairs arrange themselves as far apart as possible in space to minimize repulsion and maximize stability.
- The valence shell can be considered a sphere, with electron pairs localized on its surface at maximum distance from each other.
- A multiple bond (double or triple bond) is treated as a single "super" electron pair or "domain" when determining the electron pair geometry. However, it exerts greater repulsion on adjacent electron pairs compared to a single bond.
- If resonance structures are possible, the VSEPR model can be applied to any of them.
Repulsion Order: Repulsive interactions between electron pairs decrease in the order:
Lone pair - Lone pair (lp-lp) > Lone pair - Bond pair (lp-bp) > Bond pair - Bond pair (bp-bp)
Lone pairs are localized on the central atom and occupy more space than bonding pairs, which are shared between two atoms. This difference in space occupancy leads to varying degrees of repulsion, which can cause deviations from ideal geometries and affect bond angles.
Predicting Shapes: The VSEPR model predicts the shape based on the total number of electron pairs (bonding pairs + lone pairs) around the central atom and the specific number of lone pairs.
- Molecules with No Lone Pairs on the Central Atom (AB$_x$ type): The electron pairs are all bonding pairs, and they arrange themselves symmetrically to minimize bp-bp repulsion. The electron pair geometry is the same as the molecular geometry.
- Molecules with One or More Lone Pairs on the Central Atom (AB$_x$E$_y$ type): The lone pairs influence the spatial arrangement and repel bonding pairs, distorting the geometry from the ideal electron pair arrangement. The molecular geometry considers only the positions of the atoms (bond pairs). The electron pair geometry is still based on the total pairs, but the molecular shape is described differently due to the invisible lone pairs.
| Number of Electron Pairs | Arrangement of Electron Pairs | Molecular Geometry | Examples |
|---|---|---|---|
| 2 | Linear | Linear | BeCl₂, CO₂ |
| 3 | Trigonal Planar | Trigonal Planar | BF₃, SO₃ |
| 4 | Tetrahedral | Tetrahedral | CH₄, SiCl₄ |
| 5 | Trigonal Bipyramidal | Trigonal Bipyramidal | PCl₅, AsF₅ |
| 6 | Octahedral | Octahedral | SF₆, IF₅⁺ |
| Molecule Type | No. of Bonding Pairs | No. of Lone Pairs | Arrangement of Electrons | Molecular Shape | Reason for the shape acquired / Bond angle deviation |
|---|---|---|---|---|---|
| AB₂E | 2 | 1 | Trigonal Planar (total 3 pairs) | Bent (V-shape) | lp-bp repulsion > bp-bp repulsion. Angle reduced from 120° (ideal) e.g., NO₂⁻ angle $\approx$ 119.5°. |
| AB₃E | 3 | 1 | Tetrahedral (total 4 pairs) | Trigonal Pyramidal | lp-bp repulsion > bp-bp repulsion. Angle reduced from 109.5° (ideal) e.g., NH₃ angle $\approx$ 107°. |
| AB₂E₂ | 2 | 2 | Tetrahedral (total 4 pairs) | Bent (Angular) | lp-lp repulsion > lp-bp repulsion > bp-bp repulsion. Angle reduced from 109.5° e.g., H₂O angle $\approx$ 104.5°. |
| AB₄E | 4 | 1 | Trigonal Bipyramidal (total 5 pairs) | See-Saw | Lone pair occupies an equatorial position in trigonal bipyramid to minimize lp-bp repulsions. |
| AB₃E₂ | 3 | 2 | Trigonal Bipyramidal (total 5 pairs) | T-Shaped | Two lone pairs occupy equatorial positions to minimize lp-lp and lp-bp repulsions. |
| AB₂E₃ | 2 | 3 | Trigonal Bipyramidal (total 5 pairs) | Linear | Three lone pairs occupy equatorial positions. |
| AB₅E | 5 | 1 | Octahedral (total 6 pairs) | Square Pyramidal | Lone pair occupies one vertex of octahedron. |
| AB₄E₂ | 4 | 2 | Octahedral (total 6 pairs) | Square Planar | Two lone pairs occupy opposite vertices of octahedron to minimize lp-lp repulsion. |
The VSEPR theory successfully predicts the shapes of many molecules and ions, especially those of p-block elements, although the theoretical basis for the electron pair repulsions is not fully explained within the theory itself.
Valence Bond Theory
The Valence Bond (VB) theory, developed by Heitler and London, and later by Pauling, provides a quantum mechanical description of covalent bond formation. Unlike Lewis or VSEPR theories, VB theory addresses the energetics of bond formation and explains bond strengths and directional properties.
VB theory describes covalent bond formation primarily through the overlap of atomic orbitals.
Consider the formation of a hydrogen molecule (H₂) from two hydrogen atoms (A and B). Each atom has one electron in a 1s atomic orbital.
When the two atoms are far apart, there's no interaction. As they approach, new forces arise:
- Attractive forces: Between the nucleus of one atom and the electron of the other (N$_A$ - e$_B$, N$_B$ - e$_A$).
- Repulsive forces: Between the electrons (e$_A$ - e$_B$) and between the nuclei (N$_A$ - N$_B$).
At intermediate distances, the attractive forces are stronger than the repulsive forces, leading to a net attraction. As the atoms get closer, the potential energy decreases. An equilibrium distance is reached where the net attraction balances the repulsion, and the system's energy is minimum. This corresponds to the formation of a stable covalent bond (H-H bond length is 74 pm). The energy released at this minimum is the bond enthalpy.
Orbital Overlap Concept
According to VB theory, the minimum energy state (bond formation) occurs when the atomic orbitals of the two approaching atoms undergo partial interpenetration or overlap. This overlap of orbitals containing unpaired electrons with opposite spins leads to the formation of a covalent bond by pairing of electrons.
The strength of the covalent bond is proportional to the extent of overlap; greater overlap leads to a stronger bond.
Directional Properties Of Bonds
VB theory explains the specific geometries of polyatomic molecules (like tetrahedral CH₄, pyramidal NH₃, bent H₂O) by considering the overlap of specific atomic orbitals and, more importantly, the concept of hybridisation (discussed below).
Simple overlap of pure atomic orbitals alone cannot explain the observed bond angles. For instance, the three 2p orbitals in carbon are mutually perpendicular (90° angles), suggesting bond angles of 90° in methane if only p orbitals overlapped with H 1s. However, the actual HCH angle in CH₄ is 109.5° (tetrahedral), indicating pure orbital overlap is not the complete picture for polyatomic molecules.
Overlapping Of Atomic Orbitals
When atomic orbitals overlap, the overlap can be constructive (positive overlap), destructive (negative overlap), or zero (no effective interaction). Constructive overlap, where the overlapping regions of the orbital wave functions have the same sign (phase), leads to bond formation. Destructive overlap (opposite signs) leads to the formation of antibonding interactions (discussed in MO theory).
Positive overlap requires orbitals to have both the same phase and appropriate orientation in space.
Types Of Overlapping And Nature Of Covalent Bonds
Based on the pattern of orbital overlap, covalent bonds are classified into two main types:
- Sigma ($\sigma$) Bond: Formed by the end-to-end (head-on) overlap of atomic orbitals along the internuclear axis. The electron density is concentrated symmetrically around the internuclear axis. Sigma bonds can form from:
- s-s overlap (e.g., H-H in H₂)
- s-p overlap (e.g., H-Cl in HCl, overlap of H 1s and Cl 3p$_z$ if z is internuclear axis)
- p-p axial overlap (e.g., F-F in F₂, overlap of F 2p$_z$ and F 2p$_z$)
- Pi ($\pi$) Bond: Formed by the sidewise (lateral) overlap of atomic orbitals whose axes are parallel to each other and perpendicular to the internuclear axis (typically p-p overlap). The electron density is concentrated in two lobes, one above and one below the internuclear axis. A $\pi$ bond is formed in addition to a sigma bond.
Strength Of Sigma And Pi Bonds
The strength of a bond is related to the extent of overlap. Sigma bonds are generally stronger than pi bonds because the end-to-end overlap in $\sigma$ bonds is usually more extensive than the sidewise overlap in $\pi$ bonds.
Multiple bonds consist of one $\sigma$ bond and one or two $\pi$ bonds:
- Double bond: 1 $\sigma$ bond + 1 $\pi$ bond
- Triple bond: 1 $\sigma$ bond + 2 $\pi$ bonds
In a multiple bond, the $\sigma$ bond is formed first by axial overlap, and then the $\pi$ bond(s) are formed by sidewise overlap of remaining parallel p orbitals.
Hybridisation
To explain the observed geometries and equivalent bonds in polyatomic molecules (like CH₄ having four identical C-H bonds oriented tetrahedrally, not three at 90° and one in an arbitrary direction), Pauling introduced the concept of hybridisation.
Hybridisation is the process of intermixing atomic orbitals of slightly different energies to form a new set of equivalent orbitals called hybrid orbitals. These hybrid orbitals have the same energy and shape and are more effective in forming stable bonds than pure atomic orbitals.
Hybrid orbitals are directed in specific orientations in space to minimize repulsion between electron pairs, which determines the molecule's geometry.
Salient Features of Hybridisation:
- The number of hybrid orbitals formed is equal to the number of atomic orbitals that hybridize.
- Hybrid orbitals are always equivalent in energy and shape.
- Hybrid orbitals form stronger bonds than pure atomic orbitals due to better overlap.
- Hybrid orbitals point in specific directions to minimize electron pair repulsion, thus determining the molecular geometry.
Important Conditions for Hybridisation:
- Only orbitals in the valence shell participate.
- Orbitals undergoing hybridisation must have almost equal energy.
- Electron promotion to higher energy subshells before hybridisation is not essential but can occur.
- Orbitals involved can be half-filled, fully filled, or even empty.
Types Of Hybridisation
Different combinations of s, p, and d orbitals lead to various types of hybridisation:
- sp Hybridisation: Mixing of one s and one p atomic orbital to form two equivalent sp hybrid orbitals. These hybrid orbitals are oriented 180° apart, resulting in a linear geometry. Each sp hybrid has 50% s character and 50% p character. Example: BeCl₂.
- Be (ground state): 1s² 2s². Be (excited state): 1s² 2s¹ 2p¹.
- 2s and one 2p orbital hybridize to form two sp hybrid orbitals.
- Each sp hybrid orbital of Be overlaps axially with a 2p orbital of Cl to form two Be-Cl $\sigma$ bonds.
- Molecule is linear, bond angle 180°.
- sp² Hybridisation: Mixing of one s and two p atomic orbitals to form three equivalent sp² hybrid orbitals. These hybrid orbitals lie in a plane and are oriented 120° apart, resulting in a trigonal planar geometry. Example: BCl₃.
- B (ground state): 1s² 2s² 2p¹. B (excited state): 1s² 2s¹ 2p².
- 2s and two 2p orbitals hybridize to form three sp² hybrid orbitals.
- Each sp² hybrid orbital of B overlaps axially with a 2p orbital of Cl to form three B-Cl $\sigma$ bonds.
- Molecule is trigonal planar, bond angle 120°.
- sp³ Hybridisation: Mixing of one s and three p atomic orbitals to form four equivalent sp³ hybrid orbitals. These hybrid orbitals are directed towards the corners of a tetrahedron, resulting in a tetrahedral geometry. Bond angle is 109.5°. Each sp³ hybrid has 25% s character and 75% p character. Example: CH₄.
- C (ground state): 1s² 2s² 2p². C (excited state): 1s² 2s¹ 2p³.
- 2s and three 2p orbitals hybridize to form four sp³ hybrid orbitals.
- Each sp³ hybrid orbital of C overlaps axially with the 1s orbital of H to form four C-H $\sigma$ bonds.
- Molecule is tetrahedral, bond angle 109.5°.
sp³ hybridisation can also explain the shapes of molecules with lone pairs, like NH₃ and H₂O, leading to distorted tetrahedral electron pair geometries and pyramidal or bent molecular shapes due to lone pair repulsions (as explained by VSEPR theory). In these cases, the central atom (N or O) is still considered sp³ hybridized, but some hybrid orbitals contain lone pairs instead of forming bonds.
- NH₃: N (ground state): [He] 2s² 2p³. 4 sp³ hybrid orbitals. One contains a lone pair, three contain unpaired electrons forming N-H $\sigma$ bonds. Electron pair geometry: tetrahedral. Molecular geometry: pyramidal (bond angle $\approx$ 107° due to lp-bp repulsion).
- H₂O: O (ground state): [He] 2s² 2p⁴. 4 sp³ hybrid orbitals. Two contain lone pairs, two contain unpaired electrons forming O-H $\sigma$ bonds. Electron pair geometry: tetrahedral. Molecular geometry: bent (V-shape) (bond angle $\approx$ 104.5° due to lp-lp and lp-bp repulsions).
Other Examples Of Sp3, Sp2 And Sp Hybridisation
Hybridisation explains bonding in molecules with multiple bonds as well:
- sp³ Hybridisation in C₂H₆ (Ethane): Each carbon atom is sp³ hybridized. One sp³ orbital from each carbon overlaps axially to form a C-C $\sigma$ bond. The remaining three sp³ orbitals on each carbon form C-H $\sigma$ bonds with hydrogen atoms. All bond angles around each carbon are approximately 109.5°.
- sp² Hybridisation in C₂H₄ (Ethene): Each carbon atom is sp² hybridized. One sp² orbital from each carbon overlaps axially to form a C-C $\sigma$ bond. The remaining two sp² orbitals on each carbon form C-H $\sigma$ bonds. The unhybridized p orbital on each carbon (perpendicular to the sp² plane) overlaps sidewise to form a C-C $\pi$ bond. The C=C bond is $1\sigma + 1\pi$. The molecule is planar with bond angles near 120°.
- sp Hybridisation in C₂H₂ (Ethyne): Each carbon atom is sp hybridized. One sp orbital from each carbon overlaps axially to form a C-C $\sigma$ bond. The other sp orbital on each carbon forms a C-H $\sigma$ bond. The two unhybridized p orbitals on each carbon (mutually perpendicular and perpendicular to the sp axis) overlap sidewise to form two C-C $\pi$ bonds. The C$\equiv$C bond is $1\sigma + 2\pi$. The molecule is linear with bond angles of 180°.
Hybridisation Of Elements Involving D Orbitals
Elements from the third period onwards have vacant d orbitals available for hybridisation, allowing them to form more than four bonds (expanded octet).
Some hybridisation types involving d orbitals:
| Hybridisation Type | Atomic Orbitals Mixed | Number of Hybrid Orbitals | Shape of Molecules/Ions | Examples |
|---|---|---|---|---|
| sp³d | 1 s + 3 p + 1 d | 5 | Trigonal Bipyramidal | PCl₅, AsF₅ |
| sp³d² (or d²sp³) | 1 s + 3 p + 2 d | 6 | Octahedral | SF₆, [CrF₆]³⁻, [Co(NH₃)₆]³⁺ |
| dsp² | 1 d + 1 s + 2 p | 4 | Square Planar | [Ni(CN)₄]²⁻, [PtCl₄]²⁻ |
| sp³d³ | 1 s + 3 p + 3 d | 7 | Pentagonal Bipyramidal | IF₇ |
Example: Formation of PCl₅ (sp³d hybridisation):
- P (ground state): [Ne] 3s² 3p³. P (excited state): [Ne] 3s¹ 3p³ 3d¹.
- One 3s, three 3p, and one 3d orbital hybridize to form five sp³d hybrid orbitals.
- These five hybrid orbitals are arranged in a trigonal bipyramidal geometry.
- Each hybrid orbital overlaps with a Cl atom's p orbital to form five P-Cl $\sigma$ bonds.
In PCl₅, there are three equatorial bonds in a plane (120° apart) and two axial bonds perpendicular to the plane (90° to equatorial bonds). Axial bonds are slightly longer and weaker than equatorial bonds because they experience more repulsion from the equatorial bond pairs.
Example: Formation of SF₆ (sp³d² hybridisation):
- S (ground state): [Ne] 3s² 3p⁴. S (excited state): [Ne] 3s¹ 3p³ 3d².
- One 3s, three 3p, and two 3d orbitals hybridize to form six sp³d² hybrid orbitals.
- These six hybrid orbitals are arranged in an octahedral geometry.
- Each hybrid orbital overlaps with a F atom's p orbital to form six S-F $\sigma$ bonds.
- Molecule is octahedral, all bond angles are 90° or 180°.
Molecular Orbital Theory
Molecular Orbital (MO) theory provides a more advanced, quantum mechanical description of bonding, viewing electrons as occupying orbitals that are spread over the entire molecule, rather than localized between two atoms or on a single atom.
Salient Features of MO Theory:
- Electrons in a molecule reside in molecular orbitals (MOs), similar to how electrons in an atom occupy atomic orbitals (AOs).
- MOs are formed by the combination of comparable energy and proper symmetry.
- While AOs are monocentric (influenced by one nucleus), MOs are polycentric (influenced by two or more nuclei).
- The number of MOs formed is equal to the number of combining AOs. When AOs combine, they form two types of MOs: bonding molecular orbitals and antibonding molecular orbitals.
- Bonding MOs have lower energy and are more stable than the original AOs, while antibonding MOs have higher energy and are less stable.
- MOs represent the probability distribution of electrons around the group of nuclei in a molecule.
- MOs are filled with electrons following the same rules as AOs: Aufbau principle (lowest energy first), Pauli exclusion principle (max 2 electrons per MO with opposite spins), and Hund's rule (degenerate MOs are singly occupied before pairing).
Formation Of Molecular Orbitals Linear Combination Of Atomic Orbitals (LCAO)
The Linear Combination of Atomic Orbitals (LCAO) is an approximate method used to form molecular orbitals from atomic orbitals. It treats atomic orbitals as wave functions ($\psi$) and assumes that molecular orbitals can be described as linear combinations (sums or differences) of these atomic wave functions.
For two atoms A and B with atomic orbitals $\psi_A$ and $\psi_B$, two molecular orbitals are formed:
- Bonding MO: Formed by the addition of atomic wave functions ($\psi_{\text{bonding}} = \psi_A + \psi_B$). This leads to constructive interference and increased electron density in the region between the nuclei, strengthening the bond. This MO is lower in energy.
- Antibonding MO: Formed by the subtraction of atomic wave functions ($\psi_{\text{antibonding}} = \psi_A - \psi_B$). This leads to destructive interference, decreased electron density between the nuclei, and the formation of a nodal plane (zero electron density) between the nuclei. This MO is higher in energy and destabilizes the molecule.
The energy gained by forming the bonding MO is slightly less than the energy lost by forming the antibonding MO, resulting in a net stabilization when electrons occupy bonding MOs.
Conditions For The Combination Of Atomic Orbitals
Atomic orbitals can combine to form molecular orbitals only if they meet certain criteria:
- They must have the same or very similar energies. Orbitals from the same shell or adjacent shells can combine (e.g., 1s with 1s, 2s with 2s, 2p with 2p).
- They must have the same symmetry about the molecular axis. (e.g., a $2p_z$ orbital can combine with another $2p_z$ orbital if the z-axis is the molecular axis, but not with $2p_x$ or $2p_y$).
- They must overlap significantly. Greater overlap leads to stronger bonding and antibonding interactions.
Types Of Molecular Orbitals
Molecular orbitals are classified based on their symmetry around the internuclear axis:
- Sigma ($\sigma$) MOs: Symmetrical around the bond axis. Formed by axial overlap of s-s, s-p$_z$, or p$_z$-p$_z$ AOs. Include bonding ($\sigma$) and antibonding ($\sigma^*$) types.
- Pi ($\pi$) MOs: Not symmetrical around the bond axis. Formed by sidewise overlap of p$_x$-p$_x$ or p$_y$-p$_y$ AOs. Electron density is above and below the plane containing the axis. Include bonding ($\pi$) and antibonding ($\pi^*$) types.
Energy Level Diagram For Molecular Orbitals
The relative energies of MOs are determined experimentally. For homonuclear diatomic molecules of Period 2 elements, the order of increasing energy is generally:
- For B₂, C₂, N₂: $\sigma 1s < \sigma^*1s < \sigma 2s < \sigma^*2s < (\pi 2p_x = \pi 2p_y) < \sigma 2p_z < (\pi^*2p_x = \pi^*2p_y) < \sigma^*2p_z$. (Note the $\pi 2p$ MOs are lower in energy than the $\sigma 2p_z$ MO).
- For O₂, F₂, Ne₂: $\sigma 1s < \sigma^*1s < \sigma 2s < \sigma^*2s < \sigma 2p_z < (\pi 2p_x = \pi 2p_y) < (\pi^*2p_x = \pi^*2p_y) < \sigma^*2p_z$. (Note the $\sigma 2p_z$ MO is lower in energy than the $\pi 2p$ MOs).
The K shell (1s orbitals) AOs form $\sigma 1s$ and $\sigma^*1s$ MOs. These are often grouped as KK and don't significantly contribute to bonding in molecules of heavier elements.
Electronic Configuration And Molecular Behaviour
The electronic configuration of a molecule is written by filling electrons into the molecular orbitals according to increasing energy, following Pauli's exclusion principle and Hund's rule.
From the electronic configuration, we can deduce molecular properties:
- Stability: A molecule is stable if the number of electrons in bonding MOs ($N_b$) is greater than the number of electrons in antibonding MOs ($N_a$) ($N_b > N_a$). A molecule is unstable if $N_b \le N_a$.
- Bond Order (b.o.): Defined as half the difference between the number of electrons in bonding and antibonding orbitals.
$$ \text{b.o.} = \frac{1}{2} (N_b - N_a) $$ A positive bond order indicates stability. Bond order values often correspond to single (b.o.=1), double (b.o.=2), or triple (b.o.=3) bonds. - Bond Length: Inversely related to bond order; higher bond order means shorter bond length.
- Magnetic Nature: Molecules with unpaired electrons in their MOs are paramagnetic (attracted by a magnetic field). Molecules with all electrons paired are diamagnetic (repelled by a magnetic field).
Bonding In Some Homonuclear Diatomic Molecules
Let's apply MO theory to some simple diatomic molecules:
- Hydrogen (H₂): Total 2 electrons (1 from each H). Electronic configuration: $(\sigma_{1s})^2$. $N_b=2, N_a=0$. Bond order = ½(2-0) = 1 (single bond). Stable. Diamagnetic (no unpaired electrons). Bond length 74 pm, Bond energy 438 kJ/mol.
- Helium (He₂): Total 4 electrons (2 from each He). Electronic configuration: $(\sigma_{1s})^2 (\sigma^*_{1s})^2$. $N_b=2, N_a=2$. Bond order = ½(2-2) = 0. Unstable, does not exist.
- Lithium (Li₂): Total 6 electrons (3 from each Li). Electronic configuration: $(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2$ or KK$(\sigma_{2s})^2$. $N_b=4, N_a=2$. Bond order = ½(4-2) = 1 (single bond). Stable. Diamagnetic. Exists in vapor phase.
- Carbon (C₂): Total 12 electrons (6 from each C). Using the MO order for lighter elements: $(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2$. $N_b=8, N_a=4$. Bond order = ½(8-4) = 2 (double bond). Stable. Diamagnetic. Note that according to this MO diagram, the double bond in C₂ consists of two pi bonds.
- Oxygen (O₂): Total 16 electrons (8 from each O). Using the MO order for heavier elements: $(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi^*_{2p_x})^1 (\pi^*_{2p_y})^1$. $N_b=10, N_a=6$. Bond order = ½(10-6) = 2 (double bond). Stable. Contains two unpaired electrons in the $\pi^*$ antibonding orbitals $\rightarrow$ Paramagnetic. This paramagnetism is a key experimental observation that MO theory successfully explains, unlike Lewis or VB theories.
MO diagrams for other homonuclear diatomic molecules from Period 2 (B₂ through Ne₂) illustrate how filling the MOs predicts bond order, stability, and magnetic properties. For example, B₂ (10 electrons) is predicted to have unpaired electrons and be paramagnetic (b.o.=1), while N₂ (14 electrons) is diamagnetic (b.o.=3).
Hydrogen Bonding
Hydrogen bonding is a special type of weak electrostatic attraction that occurs when a hydrogen atom is bonded to a highly electronegative atom (typically F, O, or N). The covalent bond becomes polar, with hydrogen acquiring a partial positive charge ($\text{H}^{\delta+}$) and the electronegative atom acquiring a partial negative charge ($\text{X}^{\delta-}$).
This partially positive hydrogen atom is then attracted to a lone pair of electrons on another electronegative atom (Y, which can be F, O, or N) in the same or a different molecule. The hydrogen bond is represented by a dotted line ($\cdots$).
Example: In hydrogen fluoride (HF), the hydrogen bond forms between the $\text{H}^{\delta+}$ of one HF molecule and the $\text{F}^{\delta-}$ of an adjacent HF molecule.
$\text{H}^{\delta+} - \text{F}^{\delta-} \cdots \text{H}^{\delta+} - \text{F}^{\delta-} \cdots \text{H}^{\delta+} - \text{F}^{\delta-}$
Hydrogen bonds are weaker than covalent or ionic bonds but are stronger than typical van der Waals forces. They act like a bridge between two electronegative atoms via a hydrogen atom.
Cause Of Formation Of Hydrogen Bond
The cause of hydrogen bonding is the polarity of the covalent bond between hydrogen and a strongly electronegative atom (X). The electronegative atom pulls electron density away from H, leaving H with a significant partial positive charge ($\delta^+$). This exposed positive charge on the small hydrogen nucleus is strongly attracted to the lone pair (electron density) on another nearby electronegative atom (Y).
The strength of hydrogen bonding depends on the electronegativity of X and Y and the distance between H and Y. It is also influenced by the physical state, being strongest in solids and weakest in gases.
Types Of H-Bonds
Hydrogen bonds are classified based on where they occur:
- Intermolecular Hydrogen Bond: Formed between two different molecules. The molecules can be the same (e.g., water molecules forming H-bonds with each other in liquid water or ice) or different (e.g., water and alcohol forming H-bonds). This leads to association of molecules, affecting properties like boiling point and solubility.
- Intramolecular Hydrogen Bond: Formed within the same molecule, where a hydrogen atom is bonded to an electronegative atom and is also attracted to another electronegative atom elsewhere in the same molecule. This typically occurs when a molecule contains two polar groups appropriately positioned to allow for a ring-like structure involving the hydrogen bond. Example: In o-nitrophenol, the hydrogen on the -OH group forms a hydrogen bond with an oxygen atom of the nearby -NO₂ group within the same molecule.
Hydrogen bonding plays a significant role in determining the structure and properties of many substances, including water, proteins, and DNA.
Exercises
Question 4.1. Explain the formation of a chemical bond.
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Question 4.2. Write Lewis dot symbols for atoms of the following elements : Mg, Na, B, O, N, Br.
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Question 4.3. Write Lewis symbols for the following atoms and ions:
S and $S^{2-}$; Al and $Al^{3+}$; H and $H^-$
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Question 4.4. Draw the Lewis structures for the following molecules and ions :
$H_2S$, $SiCl_4$, $BeF_2$, $CO_3^{2-}$, HCOOH
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Question 4.5. Define octet rule. Write its significance and limitations.
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Question 4.6. Write the favourable factors for the formation of ionic bond.
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Question 4.7. Discuss the shape of the following molecules using the VSEPR model:
$BeCl_2$, $BCl_3$, $SiCl_4$, $AsF_5$, $H_2S$, $PH_3$
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Question 4.8. Although geometries of $NH_3$ and $H_2O$ molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.
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Question 4.9. How do you express the bond strength in terms of bond order ?
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Question 4.10. Define the bond length.
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Question 4.11. Explain the important aspects of resonance with reference to the $CO_3^{2-}$ ion.
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Question 4.12. $H_3PO_3$ can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing $H_3PO_3$ ? If not, give reasons for the same.
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Question 4.13. Write the resonance structures for $SO_3$, $NO_2$ and $NO_3^-$.
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Question 4.14. Use Lewis symbols to show electron transfer between the following atoms to form cations and anions : (a) K and S (b) Ca and O (c) Al and N.
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Question 4.15. Although both $CO_2$ and $H_2O$ are triatomic molecules, the shape of $H_2O$ molecule is bent while that of $CO_2$ is linear. Explain this on the basis of dipole moment.
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Question 4.16. Write the significance/applications of dipole moment.
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Question 4.17. Define electronegativity. How does it differ from electron gain enthalpy ?
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Question 4.18. Explain with the help of suitable example polar covalent bond.
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Question 4.19. Arrange the bonds in order of increasing ionic character in the molecules: LiF, $K_2O$, $N_2$, $SO_2$ and $ClF_3$.
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Question 4.20. The skeletal structure of $CH_3COOH$ as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid.
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Question 4.21. Apart from tetrahedral geometry, another possible geometry for $CH_4$ is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why $CH_4$ is not square planar ?
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Question 4.22. Explain why $BeH_2$ molecule has a zero dipole moment although the Be–H bonds are polar.
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Question 4.23. Which out of $NH_3$ and $NF_3$ has higher dipole moment and why ?
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Question 4.24. What is meant by hybridisation of atomic orbitals? Describe the shapes of $sp$, $sp^2$, $sp^3$ hybrid orbitals.
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Question 4.25. Describe the change in hybridisation (if any) of the Al atom in the following reaction.
$AlCl_3 + Cl^- \rightarrow AlCl_4^-$
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Question 4.26. Is there any change in the hybridisation of B and N atoms as a result of the following reaction?
$BF_3 + NH_3 \rightarrow F_3B.NH_3$
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Question 4.27. Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in $C_2H_4$ and $C_2H_2$ molecules.
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Question 4.28. What is the total number of sigma and pi bonds in the following molecules?
(a) $C_2H_2$
(b) $C_2H_4$
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Question 4.29. Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why? (a) 1s and 1s (b) 1s and $2p_x$; (c) $2p_y$ and $2p_y$ (d) 1s and 2s.
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Question 4.30. Which hybrid orbitals are used by carbon atoms in the following molecules?
(a) $CH_3–CH_3$
(b) $CH_3–CH=CH_2$
(c) $CH_3-CH_2-OH$
(d) $CH_3-CHO$
(e) $CH_3COOH$
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Question 4.31. What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.
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Question 4.32. Distinguish between a sigma and a pi bond.
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Question 4.33. Explain the formation of $H_2$ molecule on the basis of valence bond theory.
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Question 4.34. Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
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Question 4.35. Use molecular orbital theory to explain why the $Be_2$ molecule does not exist.
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Question 4.36. Compare the relative stability of the following species and indicate their magnetic properties;
$O_2$, $O_2^+$, $O_2^-$ (superoxide), $O_2^{2-}$ (peroxide)
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Question 4.37. Write the significance of a plus and a minus sign shown in representing the orbitals.
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Question 4.38. Describe the hybridisation in case of $PCl_5$. Why are the axial bonds longer as compared to equatorial bonds?
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Question 4.39. Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?
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Question 4.40. What is meant by the term bond order? Calculate the bond order of : $N_2$, $O_2$, $O_2^+$, and $O_2^-$.
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