The Covalent Model

விளையாடிக் கற்றுக்கொள்ளுங்கள்

ஆற்றல் சம்பாதிக்க இந்த கேள்விகளுக்குப் பதிலளியுங்கள், பின்னர் மீன் பிடித்து ஆராயுங்கள். கணக்கு தேவையில்லை.

கல்வியாளர்களுக்கு: The Covalent Model (Chemistry, HL)-க்கான தயாரான பாட ஸ்லைடுகள், திருப்புதல் குறிப்புகள் — உங்கள் பாடத்தில் அவற்றைப் பயன்படுத்தவும், அல்லது கற்பவர்கள் நேரலை விளையாட்டாக விளையாடும் ஊடாடும் வகுப்பு செயல்பாடாக தலைப்பை இயக்கவும்.

பாட குறிப்புகள்

Covalent Bonds

Covalent bonding occurs between two non-metals and is the electrostatic attraction between a shared pair of electrons and the positively charged nuclei of two atoms.
No electrons are transferred; electrons are shared between atoms, and each atom typically contributes one electron to the shared pair.
When a covalent bond forms, two atomic orbitals overlap to form a molecular orbital.
Covalent bonding occurs because electrons are more stable when shared between two nuclei than when held by one atom alone.
By sharing electrons, both atoms achieve a noble gas configuration, making them more stable.
A single covalent bond is represented by a short straight line, e.g. H–H, Cl–Cl, H–Br.
Covalent bonds are not fixed lines; the shared electrons are constantly moving and are best represented as charge clouds.
The octet rule refers to the tendency of atoms to gain a valence shell with a total of 8 electrons; exceptions include electron-deficient and expanded octet species.

A single covalent bond

Lewis Formulas

Lewis formulas (electron dot diagrams) show all valence electrons in a covalently bonded species, including bonding pairs and lone pairs.
Electron pairs can be represented by dots, crosses, a combination of dots and crosses, or a line.
Steps for drawing Lewis formulas: count total valence electrons, draw skeletal positions with the central atom (usually least electronegative) in the middle, add bonding pairs then lone pairs, and check the structure.
When adding electron pairs, first add bonding pairs between bonded atoms, then add lone pairs to complete octets (or duets for hydrogen), starting with outer atoms.
For methane, CH₄: total valence electrons = (1 × 4) + (4 × 1) = 8; carbon has a full octet and each hydrogen has a duet.
For ammonia, NH₃: total valence electrons = (1 × 5) + (3 × 1) = 8; nitrogen has a full octet and one lone pair.
For tetrachloromethane, CCl₄: total valence electrons = (1 × 4) + (4 × 7) = 32; carbon has a full octet and each chlorine has three lone pairs.
Exceptions to the octet rule include hydrogen (stable with 2 electrons), beryllium (4 valence electrons in BeCl₂), and boron (6 valence electrons in BF₃).

Multiple Bonds

Non-metal atoms can share more than one pair of electrons to form single, double, or triple covalent bonds.
A single bond (C–C) has 2 shared electrons, a double bond (C=C) has 4 shared electrons, and a triple bond (C≡C) has 6 shared electrons.
Quadruple covalent bonds are not possible because the repulsion between 8 electrons between the two nuclei is too great.
Bond energy is the amount of energy required to break one mole of a covalent bond in the gaseous state, with units kJ mol⁻¹.
The higher the bond energy, the stronger the bond.
Bond length is the distance between the nuclei of two covalently bonded atoms.
As more electrons are shared, electron density between the atoms increases, increasing electrostatic attraction and pulling atoms closer together.
Triple bonds are the shortest and strongest covalent bonds; single bonds are the longest and weakest.

A double covalent bond

Coordinate Bonds

In a regular covalent bond, each atom contributes one electron to the shared pair.
In a coordinate bond, both electrons in the bond come from the same atom.
Coordinate bonding typically occurs when one atom has a lone pair of electrons and the other atom is electron-deficient (has an unfilled outer shell).
Example: the ammonium ion (NH₄⁺) forms when the nitrogen atom in ammonia (NH₃) donates a lone pair to an electron-deficient hydrogen ion (H⁺).
Coordinate bonding also occurs in transition metal complexes, such as the aluminium(III) hexaaqua ion ([Al(H₂O)₆]³⁺).
In [Al(H₂O)₆]³⁺, each oxygen atom in H₂O donates a lone pair to the Al³⁺ ion, forming six coordinate bonds.

Shapes of Molecules

VSEPR theory states that bonding and non-bonding electron pairs around a central atom repel each other and arrange themselves as far apart as possible in three-dimensional space.
VSEPR rules: all electron pairs spread out as far as possible; lone pairs repel more strongly than bonding pairs; multiple bonds behave like a single bond when determining shape.
Each region of electron density around the central atom is called an electron domain; a domain may contain one, two, or three pairs of electrons.
Two electron domains give a linear shape with a bond angle of 180° (e.g. BeCl₂, CO₂, HC≡CH).
Three electron domains give a trigonal planar electron domain geometry with a bond angle of 120°; with one lone pair the molecular shape is bent with a bond angle of approximately 118° (e.g. SO₂).
Four electron domains give a tetrahedral electron domain geometry with an ideal bond angle of approximately 109.5°.
With one lone pair, the molecular shape is trigonal pyramidal with a bond angle of around 107° (e.g. NH₃); with two lone pairs, the shape is bent with a bond angle of approximately 104.5° (e.g. H₂O).
The order of electron pair repulsion is: lone pair–lone pair > lone pair–bonding pair > bonding pair–bonding pair.

Bond and Molecular Polarity

Electronegativity is the ability of an atom to attract the shared pair of electrons in a covalent bond; values are shown on the Pauling scale (Data Booklet Section 9).
In diatomic molecules with identical atoms (e.g. H₂, O₂, Cl₂), electron density is shared equally, resulting in a nonpolar covalent bond.
When two atoms have different electronegativities, the bonding electrons are drawn closer to the more electronegative atom, creating a polar covalent bond with δ⁺ and δ⁻ charges.
The greater the difference in electronegativity, the more polar the bond and the greater the dipole moment.
A dipole forms when there is a separation of charge across a covalent bond; the dipole arrow points toward the more electronegative atom (δ⁻).
To determine molecular polarity, consider both the bond polarity and the arrangement of bonds in space (molecular geometry).
Molecules with polar bonds can be nonpolar overall if the bond dipoles are arranged symmetrically and cancel out (e.g. CO₂, CCl₄).
Molecules with polar bonds are polar overall if the bond dipoles are arranged asymmetrically and do not cancel (e.g. H₂O, NH₃, CH₃Cl).

Giant Covalent Structures

Giant covalent structures (covalent network structures) are large networks of atoms joined by strong covalent bonds extending throughout the entire lattice.
Examples include silicon, silicon dioxide, diamond, graphite, buckminsterfullerene, and graphene.
In diamond, each carbon atom is covalently bonded to four others in a tetrahedral geometry with bond angles of approximately 109.5°; it is extremely hard and does not conduct electricity.
In graphite, each carbon atom is bonded to three others in hexagonal rings arranged in flat layers with bond angles of 120°; the fourth electron is delocalised, so graphite conducts electricity and layers can slide (soft and slippery).
Graphene is a single layer of carbon atoms in a hexagonal lattice with trigonal planar geometry and bond angles of 120°; it conducts electricity due to delocalised electrons and is strong and flexible.
Buckminsterfullerene (C₆₀) is a molecular form of carbon made of 60 atoms in a spherical structure of interlocking hexagons and pentagons; it is a semiconductor.
Giant covalent lattices have very high melting and boiling points because they contain a large number of strong covalent bonds requiring large amounts of energy to break.
Most covalent network structures are insoluble in water; graphite and graphene conduct electricity due to delocalised electrons.

Intermolecular Forces

Covalent molecular substances are held together by intermolecular forces, which are much weaker than covalent or ionic bonds.
There are four types of intermolecular force: London (dispersion) forces, dipole–dipole attractions, dipole–induced dipole attractions, and hydrogen bonding.
Van der Waals forces is a collective term that includes London (dispersion) forces, dipole–dipole attractions, and dipole–induced dipole attractions.
London (dispersion) forces arise from temporary dipoles and induced dipoles; they are present between all atoms and molecules but are usually very weak (1–50 kJ mol⁻¹).
The strength of London forces depends on the number of electrons and the surface area available for contact; more electrons and larger surface area increase the strength.
Dipole–dipole attractions occur between the δ⁺ end of one polar molecule and the δ⁻ end of a neighbouring polar molecule, in addition to London forces.
Dipole–induced dipole attractions occur when a polar molecule distorts the electron cloud of a nonpolar molecule, creating a temporary dipole.
Hydrogen bonding is the strongest type of intermolecular force and requires a hydrogen atom covalently bonded to O, N, or F, and a lone pair on an O, N, or F atom in a neighbouring molecule.

Simple molecular structure of water: strong covalent bonds within each molecule, weak intermolecular forces between molecules.

Physical Properties of Covalent Substances

The physical properties of molecular covalent compounds are largely influenced by their intermolecular forces.
When molecular covalent substances change state, the intermolecular forces between their molecules are overcome, not the covalent bonds.
The stronger the intermolecular forces, the higher the melting and boiling point; substances with low boiling points are described as volatile.
The strength of intermolecular forces depends on the size (molar mass) of the molecule, the polarity of the molecule, and the type of intermolecular force present.
The order of intermolecular force strength is: London < dipole–dipole < hydrogen bonding.
The general rule for solubility is “like dissolves like”: nonpolar substances dissolve in nonpolar solvents, and polar substances dissolve in polar solvents.
Most covalent network structures are insoluble in water and do not conduct electricity, except graphite and graphene which conduct due to delocalised electrons.

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இலவச முன்னோட்டம் — 62-இல் 8 கேள்விகள். அனைத்தையும் பார்க்க பதிவு செய்யவும்.

1.Which statement best defines a covalent bond?
Easy
- AThe electrostatic attraction between a shared pair of electrons and the positively charged nuclei of two atoms
- BThe transfer of electrons from a metal to a non-metal
- CThe attraction between oppositely charged ions in a lattice
- DA weak force of attraction between molecules
2.In a typical covalent bond, each atom contributes one electron to the shared pair.
Easy
True or false?
3.Which of the following molecules contains a coordinate bond?
Easy
- ANH4+
- BCH4
- CH2O
- DCO2
4.Which of the following statements about sigma (σ) and π (π) bonds are correct? (select all that apply)
Medium
- AA sigma bond is formed by the end-on overlap of atomic orbitals.
- BA π bond is formed by the sideways overlap of p-orbitals.
- CA sigma bond is weaker than a π bond.
- DA π bond allows free rotation about the bond axis.
- EA sigma bond has electron density concentrated along the internuclear axis.
5.Match each molecule with its molecular geometry.
Medium
- CH4
- NH3
- H2O
- CO2
- Tetrahedral
- Trigonal pyramidal
- Bent
- Linear
6.Arrange the following carbon-carbon bonds in order of increasing bond length.
Medium
- C≡C (triple bond)
- C=C (double bond)
- C–C (single bond)
7.Which of the following molecules is polar overall?
Medium
- ACCl4
- BCO2
- CCH3Cl
- DBF3
8.Graphite conducts electricity because each carbon atom forms four covalent bonds, leaving no free electrons.
Medium
True or false?

மேலும் கேள்விகள்

9.Which of the following statements about London (dispersion) forces are correct? (select all that apply)
Medium
- AThey arise from temporary dipoles caused by the constant motion of electrons
- BThey are present between all atoms and molecules
- CThey are the only intermolecular forces in nonpolar substances
- DThey are stronger than covalent bonds
- EThey are classified as a type of van der Waals force
10.Which of the following molecules contain at least one lone pair on the central atom? (select all that apply)
Medium
- ANH3
- BH2O
- CCH4
- DCO2
- ESO2
11.In a coordinate bond, both electrons in the shared pair come from the same atom.
Easy
True or false?
12.Hydrogen bonding is the strongest type of intermolecular force.
Medium
True or false?
13.Match each molecule with its strongest type of intermolecular force.
Medium
- CH4
- CH3Cl
- CH3OH
- HCl and Cl2
- London (dispersion) forces only
- Dipole-dipole attractions
- Hydrogen bonding
- Dipole-induced dipole attractions
14.Arrange the following molecules in order of increasing boiling point.
Medium
- CH3CH2CH2CH3
- CH3COCH3
- CH3CH2CH2OH
15.Which of the following statements about the polarity of a molecule with polar bonds is correct?
Medium
- AIt is always polar.
- BIt is always nonpolar.
- CIt is polar if the bond dipoles are arranged asymmetrically.
- DIt is nonpolar if the bond dipoles are arranged asymmetrically.
16.Which of the following correctly describes the formation of a sigma (σ) bond?
Medium
- ASideways overlap of p orbitals
- BHead-on overlap of atomic orbitals
- CTransfer of electrons from one atom to another
- DSharing of two pairs of electrons between two atoms
17.Which of the following statements about the bond angles in methane (CH4), ammonia (NH3), and water (H2O) is correct?
Medium
- ACH4 > NH3 > H2O
- BH2O > NH3 > CH4
- CNH3 > CH4 > H2O
- DCH4 > H2O > NH3
18.Which of the following is the correct order of increasing strength of intermolecular forces?
Medium
- ALondon < dipole-dipole < hydrogen bonding
- BDipole-dipole < London < hydrogen bonding
- CHydrogen bonding < dipole-dipole < London
- DLondon < hydrogen bonding < dipole-dipole
19.An unknown molecule has 4 bonding pairs and 2 lone pairs around its central atom. What is its molecular geometry?
Medium
- Atetrahedral
- Btrigonal pyramidal
- Cbent
- Dlinear
20.Which of the following molecules has a net dipole moment?
Medium
- ACO2
- BCCl4
- CCH3Cl
- DBF3
21.Which of the following statements about sigma (σ) and π (π) bonds is correct?
Medium
- AA σ bond is formed by sideways overlap of p orbitals, while a π bond is formed by end-on overlap.
- BA σ bond allows rotation about the bond axis, while a π bond restricts rotation.
- CA π bond is stronger than a σ bond because it has more electron density between the nuclei.
- DA triple bond consists of three π bonds.
22.In the ethanoate ion, CH3COO⁻, the two carbon–oxygen bonds are found to be identical in length. Which statement best explains this?
Hard
- AThe ion rapidly interconverts between two resonance structures, so the bonds appear identical on average.
- BThe double bond is localised on one oxygen, but the other oxygen also has a double bond due to formal charge.
- CThe lone pair on the negatively charged oxygen is delocalised over both oxygen atoms, giving each C–O bond a bond order of 1.5.
- DThe carbon atom expands its octet to form two double bonds, making both C–O bonds equal.
23.The enthalpy change of hydrogenation of benzene is −208 kJ mol⁻¹, whereas the expected value based on the Kekulé structure (three isolated C=C bonds) is −360 kJ mol⁻¹. What is the best explanation for this difference?
Hard
- ABenzene has three C=C bonds that are weaker than expected due to ring strain.
- BThe delocalisation of π electrons in benzene makes the molecule more stable than the Kekulé structure, so less energy is released on hydrogenation.
- CBenzene undergoes addition reactions more readily than cyclohexene, so the enthalpy change is smaller.
- DThe hydrogenation of benzene produces cyclohexane, which has a higher enthalpy of formation than expected.
24.Which of the following statements about formal charge are correct? (select all that apply)
Hard
- AFormal charge = (valence electrons) − ½(bonding electrons) − (non-bonding electrons).
- BThe most stable resonance structure has the largest possible formal charges.
- CIn XeF4, the formal charge on xenon is 0.
- DIn the preferred Lewis structure of CO2, the formal charge on each oxygen is 0.
- EFormal charge is always equal to the oxidation state of the atom.
25.In a molecule of sulfur hexafluoride, SF6, the sulfur atom has an expanded octet with 12 electrons in its valence shell.
Easy
True or false?
26.Arrange the following carbon–carbon bonds in order of increasing bond length.
Medium
- C–C (single)
- C=C (double)
- C≡C (triple)
27.The electronegativity values of some elements are: C = 2.6, N = 3.0, O = 3.4, F = 4.0. Which of the following bonds is the most polar?
Medium
- AC–O
- BC–N
- CN–O
- DC–F
28.Which of the following statements about the delocalisation of electrons in benzene are correct? (select all that apply)
Hard
- AAll six carbon–carbon bonds in benzene are of equal length.
- BThe delocalised π electrons are located above and below the plane of the ring.
- CBenzene readily undergoes addition reactions because it has three localised double bonds.
- DThe delocalisation of electrons makes benzene more stable than the hypothetical Kekulé structure.
- EBenzene does not conduct electricity because the delocalised electrons are localised.
29.In graphite, each carbon atom forms three covalent bonds and the fourth valence electron is delocalised, allowing graphite to conduct electricity.
Medium
True or false?
30.In which of the following molecules does the central atom have an expanded octet?
Hard
- ABF3
- BBeCl2
- CSF6
- DCH4
31.Which of the following statements about hybridisation in ethyne, C2H2, are correct? (select all that apply)
Hard
- AEach carbon atom is sp hybridised.
- BThe molecule contains two sigma (σ) bonds and two π (π) bonds.
- CThe H–C–C bond angle is 180°.
- DEach carbon atom forms four sigma bonds.
- EThe triple bond consists of three π bonds.
32.Which of the following correctly states the F–P–F bond angles in PF5?
Hard
- A90° and 120°
- B109.5° only
- C90° only
- D120° only
33.Which of the following statements about the formal charge on the atoms in carbon dioxide, CO2, is correct?
Hard
- AThe formal charge on carbon is 0 and on each oxygen is 0.
- BThe formal charge on carbon is +1 and on each oxygen is −1.
- CThe formal charge on carbon is −1 and on each oxygen is +1.
- DThe formal charge on carbon is +2 and on each oxygen is −1.
34.Which of the following statements about the physical properties of giant covalent structures are correct? (select all that apply)
Hard
- ADiamond does not conduct electricity because all four valence electrons are used in bonding.
- BGraphite conducts electricity because it has delocalised electrons between its layers.
- CBuckminsterfullerene is a good conductor of electricity like graphite.
- DGraphene is a single layer of carbon atoms with delocalised electrons and conducts electricity.
- ESilicon dioxide has a low melting point because it consists of discrete molecules.
35.Match each molecule with its correct number of sigma (σ) and π (π) bonds.
Medium
- CH4
- C2H4
- C2H2
- CO2
- 4 σ, 0 π
- 5 σ, 1 π
- 3 σ, 2 π
- 2 σ, 2 π
36.A student sets up a paper chromatography experiment to separate a mixture of amino acids. After the solvent front has moved up the paper, the chromatogram is dried and sprayed with ninhydrin. Which statement best explains why ninhydrin is used?
Hard
- ANinhydrin reacts with amino acids to form coloured spots, locating them on the paper
- BNinhydrin dissolves the amino acids so they move faster up the paper
- CNinhydrin increases the Rf value of each amino acid
- DNinhydrin acts as the mobile phase
37.In a chromatography experiment, a dye travels 4.8 cm while the solvent front travels 12.0 cm. A second dye in the same mixture travels 7.2 cm. What is the Rf value of the second dye?
Hard
- A0.40
- B0.60
- C1.67
- D0.72
38.Which of the following statements about the chromatography of amino acids are correct? (select all that apply)
Hard
- AThe Rf value of a given amino acid depends on the solvent used
- BAmino acids are colourless and must be located with a locating agent such as ninhydrin
- CThe solvent front must be marked before the solvent reaches the top of the paper
- DAmino acids with the same Rf value in one solvent must be identical
- EThe stationary phase in paper chromatography is the water held in the paper fibres
39.Which technique would best separate a mixture of two coloured dyes that are both soluble in ethanol?
Medium
- AFiltration
- BPaper chromatography
- CDistillation
- DRecrystallisation
40.Two isomers, pentane and 2,2-dimethylpropane, have the same molecular formula C5H12. Pentane boils at 36 °C while 2,2-dimethylpropane boils at 9.5 °C. Which statement best explains this difference?
Medium
- APentane has hydrogen bonding while 2,2-dimethylpropane does not
- BPentane has more electrons than 2,2-dimethylpropane
- CPentane has a more extended shape and greater surface area, so stronger London forces
- D2,2-dimethylpropane has a permanent dipole
41.How many valence electrons are in a molecule of carbon dioxide, CO2?
Medium
42.Complete the sentence about benzene.
Easy
The bond angle in the planar regular hexagon structure of benzene is ____ degrees.
43.State the term used to describe the delocalisation of electrons in a molecule such as benzene.
Easy
44.How many sigma (σ) bonds are present in a molecule of ethyne, C2H2?
Medium
45.How many π (π) bonds are present in a molecule of benzene, C6H6?
Medium
46.Complete the sentence about the VSEPR theory.
Medium
According to VSEPR theory, electron pairs around a central atom arrange themselves to be as ____ apart as possible.
47.Complete the sentence about bond polarity.
Medium
In a polar covalent bond, the more electronegative atom acquires a partial ____ charge.
48.What is the term for the type of bond formed when both electrons in the shared pair come from the same atom?
Medium
49.What is the molecular geometry of a molecule with four bonding pairs and no lone pairs around the central atom?
Medium
50.Calculate the total number of valence electrons in a molecule of carbon dioxide, CO2.
Medium
51.Complete the statement about VSEPR theory.
Medium
According to VSEPR theory, lone pairs repel more strongly than bonding pairs, so they ____ the bond angles.
52.Calculate the formal charge on the xenon atom in xenon tetrafluoride, XeF4, using the formula: FC = (valence electrons) − ½(bonding electrons) − (non-bonding electrons).
Medium
53.Explain why the enthalpy change of hydrogenation of benzene is less exothermic than expected from its Kekulé structure.
Hard
54.Explain why the melting point of diamond is much higher than that of buckminsterfullerene.
Hard
55.Explain how the delocalisation of electrons in benzene accounts for its resistance to addition reactions.
Hard
56.A student obtains a chromatogram on which a single spot appears for an unknown amino acid. Explain why this does not prove the sample is a pure single amino acid.
Hard
57.Benzene reacts with bromine only in the presence of a halogen carrier such as FeBr3, whereas cyclohexene decolourises bromine water readily at room temperature. Explain this difference.
Challenge
58.How many valence electrons are in a molecule of carbon dioxide, CO2?
Medium
- A15 electrons
- B18 electrons
- C16 electrons
- D17 electrons
59.How many sigma (σ) bonds are present in a molecule of ethyne, C2H2?
Medium
- A3
- B5
- C4
- D2
60.How many π (π) bonds are present in a molecule of benzene, C6H6?
Medium
- A3
- B5
- C4
- D2
61.Calculate the total number of valence electrons in a molecule of carbon dioxide, CO2.
Medium
- A17 electrons
- B18 electrons
- C15 electrons
- D16 electrons
62.Calculate the formal charge on the xenon atom in xenon tetrafluoride, XeF4, using the formula: FC = (valence electrons) − ½(bonding electrons) − (non-bonding electrons).
Medium
- A1
- B2
- C-1
- D0

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