Chem bonding Flashcards
Electronegativity
Electronegativity of an atom is a measure of its ability to attract the electrons in a covalent bond to itself
Structure of metallic bonding
Giant metallic lattice structure which is a lattice of metal cations and the sea of delocalised electrons held together by metallic bonding
Strength of metallic bonding
- Number of valence electrons contributed per metal atom.
- Charge and radius of the metal cation
The higher the charge and the smaller the radius of the metal cation, the higher its charge density and the stronger the metallic bonding
Boiling points and melting points of metals
Metals have high melting and boiling points as a large amount of energy is needed to overcome the strong attraction between the metal cations and the delocalised electrons.
Why is the boiling point of Mg higher than that of Na, quoting relevant data from the data booklet.
Each Mg atom has two valence electrons while each Na atom has one valence electron. Hence, magnesium has a larger number of delocalised electrons.
Mg2+ has a larger charge and smaller ionic radius than Na+. Mg has a higher charge density than Na+.
More energy is needed to overcome the stronger metallic bonding in magnesium than that in sodiu,.
Explain electrical conductivity and thermal conductivity of metals
Metals are good electrical conductors because the delocalised electrons act as mobile charge carriers.
Thermal energy causes the electrons to move more quickly, allowing heat energy is transferred to the other parts of the metal by the mobile electrons.
Why are metals malleable and ductile?
When a large stress is applied to a piece of solid metal, the layers of ions will slide over one another into new positions. The overall shape of the metal changes, but it does not break apart because the sea of delocalised electrons prevents repulsion between the cations as they move past one another, and metallic bonding remains intact.
Lattice energy
The heat evolved when 1 mole of a pure ionic solid is formed from its constituent gaseous ions.
Explain melting point boiling point of ionic compounds
Ionic solids have high melting and boiling points as a large amount of energy is needed to overcome the strong electrostatic attractions between oppositely charged ions.
Explain why magnesium oxide has a melting point than sodium chloride, quoting relevant data from the Data Booklet
Mg2+ has a smaller radius than Na+ while O2- has a smaller radius than Cl-.
Mg2+ and O2- have higher charges than Na+ and Cl- respectively
(Insert LE formula here )
Lattice energy of MgO is larger in magnitude. More energy is needed to overcome the stronger ionic bonding in MgO.
Explain electrical conductivity of ionic compounds.
In the molten or aqueous state, ionic compounds are good electrical conductors because the ions can act as mobile charge carriers.
The electrical conductivity of an aqueous solution of an ionic compound increases as its concentration increases, because there are more ions to carry the charge.
Explain hardness of ionic compounds
Ionic compounds are hard and rigid but brittle.
In an ionic lattice, oppositely charged ions are held in fixed positions throughout the crystal lattice by strong ionic bonding. Moving the ions out of position requires large amounts of energy to overcome these bonds. Ionic lattices are therefore quite hard and rigid.
If a strong enough force is applied, it will force ions of like charges to move next to each other. Repulsion between ions of like charges will cause the lattice to shatter. Ionic lattices are therefore brittle
Bond energy
The average amount of energy required to break 1 mole of a covalent bond in the gaseous state to form gaseous atoms.
Factors affecting bond energy
- Bond length
- smaller atoms form shorter and stronger bonds, as the overlap between orbitals is more effective - Bond order
- multiple bonds are shorter and stronger than single bonds. When more electrons are being shared, the attraction between the two positively charged nuclei and the shared electrons is stronger.
Why can suffer form SF6 but oxygen can only form up to OF2
Both oxygen and sulfur have six valence electrons so they use two valence electrons to form two bonds to attain octet.
Unlike oxygen in period 2, sulfur in Period 3 is able to use its energetically accessible 3d orbitals to accommodate more than 8 electrons. Hence, sulfur can use four to six valence electrons in bonding with fluorine to form SF4 and SF6 respectively.
VSEPR Theory
- The electron pairs around the central atom of a molecule arrange themselves as far apart as possible
- The repulsion between lone pair and lone pair > lone pair and bond pair > bond pair bond pair
Bond angle difference
(Compound) has (no.) lone pairs, which exert a stronger force of repulsion than the bond pairs. The bond pair of (compound) are pushed closer together resulting in slightly smaller bond angles
Hybridisation
(Geometry of carbon atom), (type of hybrid orbital)
Tetrahedral, sp3
Trigonal planar, sp2
Linear, sp
Hybridisation and strength of covalent bond
- A p orbital is more elongated as compared to a spherical s orbital. Hence, a hybrids orbital that has more p character tends to form a longer and weaker bond.
- The bond formed by an sp3 hybrid orbital is hence actually longer and weaker than a bond formed by an sp hybrid orbital as it has a higher percentage of the characteristic of a p orbital
Factors affecting degree of covalent character
- Polarising power of the cation
Smaller and highly charged cations have high charge densities and high polarising power. These cations have a higher tendency to distort the anion’s electron cloud, resulting in greater covalent character - Polarisability of the cation
Anions that are relatively large have high polarisability . Their valence electrons are further from and less strongly attracted by the nucleus so the electron cloud is easily distorted by a cation, resulting in greater covalent character in the ionic bonding
Why does this compound have a zero net dipole moment? (CCl4)
If the compound has polar bonds: Though (C-Cl) bond is polar, (compound (CCl4)) has a (shape of compound (tetrahedral)) shape and thus all dipole moments cancel each other and there is a overall zero net dipole moment.
Why is the net dipole moment of this compound (CH3CL) larger than this other compound (CH3I)?
Down group (7), the electronegativity of the elements decreases. As (chlorine) is more electronegative than iodine, and (C-Cl) bond is more polar than (C-I) bond, the net dipole moment produced would be larger.
Factors affecting strength of dispersion forces
- The number of electrons in the molecule
The larger the number of electrons, the more polarisable the electron cloud, the stronger the dispersion forces
E.g. F2 and Cl2 are gases, Br2 is a liquid and I2 is a solid
Group 17 elements are non-polar simple covalent molecules. As the number of electrons increases, the instantaneous dipole - induced dipole attractions between the molecules become stronger. More energy is needed to overcome the stronger intermolecular dispersion forces so volatility of the elements decreases.
- The surface area of contact between adjacent molecules (ORGANIC CHEM MOSTLY)
- The larger the surface area of contact, the more easily induced dipoles are formed, the stronger the dispersion forces. The surface area of contact depends on the overall shape of the molecules.
E.g. between pentane and 2.2 dimethylpropane
- These isomers have the same number of electrons. Pentane has an elongated shape and a larger surface area than 2,2 dimethulpropane which has a spherical shape. Hence, the dispersion forces between pentane molecules are stronger
Strength and extent of hydrogen bonding
Electronegativity of the atom that H is bonded to
F is more electronegative than N. The δ+ on H in HF is larger than that in NH3, hence the hydrogen bonding between the HF molecules is stronger. More energy is needed to overcome the stronger hydrogen bonding between HF molecules than that between NH3 molecules.
Number of H atoms bonded to F, O or N and the number of lone pairs available
The hydrogen bonding in H2O is more extensive than that in HF or NH3. H2O has 2 Lone pairs of electrons and two electron deficient hydrogen atoms. H2O can form on average 2 hydrogen bonds per molecule. On the other hand, NH3 has only one lone pair while HF has only one electron deficient hydrogen atom. NH3 and HF can only form an average of one hydrogen bond per molecule.
Why are the boiling points of NH3, H2O and HF higher than CH4?
CH4, NH3, H2O and HF have the same number of electrons. However, the boiling points of NH3, H2O and HF are considerably higher than that of CH4. The hydrogen bonding in NH3, H2O and HF are stronger than the dispersion forces in CH4.
Intermolecular and intramolecular hydrogen bonding (compare boiling points of 2-nitrophenol and 4-nitrophenol)
2-nitrophenol is able to form intramolecular hydrogen bonding so the intermolecular hydrogen bonding is less extensive. It’s isomer 4-nitrophenol is only able to form intermolecular hydrogen bonding. Less energy is needed to break the less extensive intermolecular hydrogen bonding in 2-nitrophenol hence it has a lower boiling point
Dimerisation of carboxylic acids in non-polar solvents (ethanoic acid)
In water, ethanoic acid forms hydrogen bonds with H2O molecules. However, in non-polar solvents such as benzene, the carboxylic acid molecules form dimers via hydrogen bonding.
Ice is less dense than liquid water
In ice, each H2O molecule is hydrogen bonded to four other molecules in a roughly tetrahedral arrangement. This produces an open lattice, with empty space between the H2O molecules. The more random arrangement of hydrogen bonding in liquid water takes up less space as the H2O molecules are closer together.
For the same mass, the lattice structure of ice occupies a larger volume, hence the lower density
Liquid water has high surface tension
Hydrogen bonded H2O molecules form an array across the surface of water, allowing objects that are expected to sink in water to float on water
Argon having a lower bp than hydrogen chloride
Argon has a simple atomic structure while hydrogen chloride has a simple molecular structure Stronger permanent dipole - permanent dipole attractions exist between polar HCl molecules
Less energy is needed to overcome the weaker dispersion forces between argon atoms so it has a lower boiling point
Hydrogen iodide has a higher boiling point than hydrogen bromide
Both Hydrogen iodide and hydrogen bromide are polar simple covalent molecules
Permanent dipole- permanent dipole attractions and dispersion force exist between the molecules
HI molecules have more electrons so more energy is needed to overcome the stronger dispersion forces between the molecules
Explain electrical conductivity across period 3
- Na Mg and Al have giant metallic structures with strong metallic bonding between cations and the sea of delocalised valence electrons. Metals are good electrical conductors because the delocalised electrons are able to act as mobile charge carriers
- From Na to Al, the number of valence electrons contributed per atom increases. Therefore, electrical conductivity increases
- Si has a giant molecular structure with covalent bonds between atoms. As a metalloid, it is a semiconductor
- P4, S8 and Cl2 have simple molecular structures. Ar has simple atomic structure. In P4, S8 and Cl2, the valence electrons are held strongly in covalent bonds. There are no mobile charge carriers in these molecules as well as in the argon atoms so they do not conduct electricity.
Explain physical periodicity across Period 3
- Na, Mg and Al have giant metallic structures and relatively high melting points due to the strong metallic bonding between cations and the sea of delocalised valence electrons
- From Na to Al, melting point increases as more energy is needed to overcome the stronger metallic bonding due to
- The increase in the number of valence electrons contributed per atom
- The increase in the charge density (directly proportional to charge/radius)
- Si has a giant molecular structure and a very high melting point as a lot of energy is needed to overcome the many strong covalent bonds between the Si atoms during melting
- Non-polar P4, S8 and Cl2 have simple molecular structures. Ar has simple atomic structure. Little energy is needed to overcome the weak dispersion forces between the molecules or between argon atoms hence the low melting points.
- The number of electrons and hence strength of dispersion forces decreases from S8 to P4 to Cl2 to Ar. Hence, energy needed to overcome the dispersion forces during melting and the resulting melting point decreases in the same order.
Explainations for solubility of substances in solvents
- (solute) is (type of compound, polar/non polar if covalent) and (solvent) is (type of compound, polar/non polar if covalent).
(If soluble) - The energy released when forming (type of bonds (ion dipole > hydrogen bonding > pd-pd > pd-id> id-id) between (solute molecules/cation/anion) is comparable to the energy required to overcome (type of bond in solvent) between (solvent molecules) and (type of bond in solute) between (solute molecules/ ions)
(If insoluble) - The energy released when forming (type of bonds (ion dipole > hydrogen bonding > pd-pd > pd-id> id-id) between (solute molecules/cation/anion) is insufficient to overcome (add ‘stronger’ here if type of bond is stronger than solute and solvent) (type of bond in solvent) between (solvent molecules) and (add ‘stronger’ here if type of bond is stronger than solute and solvent) (type of bond in solute) between (solute molecules/ ions)
