Chemical Bonding
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describe metallic bonding
• giant metallic lattice structure, metal cations arranged in a lattice and surrounded by a ‘sea’ of delocalised electrons • electrostatic attraction between metal cations and the ‘sea’ of delocalised electrons form metallic bonds
factors affecting metallic bond strength
• no. of delocalised e- : due to more valence electrons • high charge density : due to higher charge and smaller ionic radius • leads to: greater electrostatic attraction between metal cation and delocalised electrons → stronger metallic bond strength. • greater electrical conductivity: more valence electrons → more mobile charge carriers
describe ionic bonding
(NaCl) has a giant ionic structure with electrostatic attraction b/w oppositely charged ions
factors affecting ionic bond strength
increase in Lattice Energy: higher charge, decrease in ionic radius, decrease in interionic distance b/w cation and anion → stronger electrostatic attractions between cation and anion, more energy to overcome
why metals conduct electricity in both solid and molten but ionic cpd only conduct in molten state? talk about MOBILE charge carriers
due to presence of charge carriers: * describe structure of metal and ionic cpd * metals have sea of delocalised e- to act as mobile charge carriers in both solid and molten * ionic cpd cation and anion are held in fixed position in solid state * able to conduct in molten state as ionic lattice is broken down to form mobile ions which act as charge carriers
describe covalent bonding
(graphite) has a giant molecular structure with c atoms held tgt by strong electrostatic attraction b/w the 2 +ve nuclei and shared pair of electrons (covalent bond)
factors affecting covalent bonding
• size of valence orbital increases • orbital overlap becomes less effective • strength of covalent bond decreases • less energy needed to overcome covalent bond
describe the bonding of H2
simple covalent structure, held together by weak id-id
Why does id-id and pd-pd arise SIGMA BOY BOY IMPORTANTE
id-id: * Electrons constantly moving at any moment * electron density of molecule unsymmetrical, * resulting in instantaneous dipole and induces a dipole in a neighbouring molecule * resulting in id-id interactions pd-pd: * Molecules are polar with permanent dipoles * pd-pd interactions arise due to electrostatic attraction between the partial +ve charge of one phosgene molecule and partial –ve charge of the other phosgene molecule.
Why is CF4 insoluble in water?
Structure→Polarity→IMF→Energy • CF 4 and water have simple molecular structure • CF 4 is non-polar with id-id interactions between the molecules • Water is polar and hydrogen bonds exist between H2O molecules. • energy evolved when CF 4 molecules form id-id interactions with water molecules insufficient to overcome the hydrogen bonds between H2O molecules and the id-id interactions between the CF4 molecules
Why CH3SCH3 has a higher boiling point than CH3OCH3?
Structure→Polarity→IMF→Energy • Both have simple molecular structure and are polar with pd-pd interactions and id-id interactions between molecules • CH3SCH3 has a larger electron cloud than CH3OCH3 and electron cloud is polarised to a larger extent → stronger id-id interactions between molecules • more energy needed to overcome stronger id-id bond, higher bp
Why N2H4 has a higher boiling point than NH3? extensiveness of h2 bond and electonegativity of NOF
extensiveness of h2 bond: * hydrazine (contains 2 N atom), able to form more extensive, 2 intermolecular hydrogen bonds per molecule * ammonia can only form 1 intermolecular hydrogen bonds per molecule * more energy is required to break the more extensive intermolecular hydrogen bonding present in hydrazine electonegativity of N,O,F in H2 Bond: * F more electronegative, dipole moment of H-F larger * stronger intermolecular h2 bond formed in H-F bond * large amt of E required to overcome presence of INTRAmolecular H2 bond: * INTRAmolecular H2 bond limit no.of sites for INTERmolecular h2 bond * INTERmolecular less extensive, less energy required to overcome hydrogen bonding based on lone pair e- on N!! NOT H bonded to N!!!
How does id-id form in CO2? USE THIS MODEL ANSWER INSTEAD!
• e- in random motion,→ uneven distribution of electrons in molecule → separation of charges and creates an instantaneous dipole in the CO2 molecule • instantaneous dipole induces formation of dipoles in the neighbouring unpolarised CO2 molecules • weak electrostatic forces of attraction (instantaneous dipole – induced dipole interactions) forms between two dipoles
How does pd-pd form in HCl?
• e- in random motion,→ uneven distribution of electrons in molecule → separation of charges and creates an instantaneous dipole in the CO2 molecule • instantaneous dipole induces formation of dipoles in the neighbouring unpolarised CO2 molecules • weak electrostatic forces of attraction (permanent dipole – permanent dipole interactions) forms between two dipoles • Molecules are polar with permanent dipoles ??
How does H2 Bond form in NH3?
• N atom more electronegative than H atom, attracts bonding e- in the N−H bond • N atom gains a partial negative charge, H atom acquires a partial positive charge • electrostatic attraction (hydrogen bond) between the e- deficient H atom of one NH3 molecule and the lone pair of e- on N atom on another NH 3 molecule
state and explain the difference in bond angles using VSEPR
• e- pairs arranged as far apart to minimise e- repulsion • lp-lp repulsion > lp-bp > bp-bp > lone e- -bp • increase in repulsion, bond angle decreases
bond strength from SP3 to SP hybridisation
bond strength increases: * s-chracter of the hybrid orbital increases * hybrid orbital is less diffused, * e- more tightly held by the nuclei —> 1) (link to basicity) e- less available for dative bonding with H+ and is a weaker base OR 2) (normal elab) e- overlap with the orbital of another atom more effective, bond is stronger and bond length is shorter
what is SP hybridisation? from sp3 to sp, decrease in p orbitals that mix with 2s orbitals
sp hybridisation results when one 2s and one 2p orbital of C mix to form two sp hybrid orbitals. There are two unhybridised p-orbitals in C. (unhybridised p orbital forms the pi bond) make sense there is 2 unhybridised as there are c triple bonds in SP
Dative Covalent Bond
one atom provides lone pair of e-, other atom has vacant low lying orbital to accept e-
factors affecting boiling point of cpds with simple molecular structure factors affecting id-id interaction
Branching: * less surface area of contact * less extensive id-id interactions b/w molecules * id-id weaker, less energy needed to overcome id-id interactions
why halogens bond energy decrease but boiling pt increase down the group?
BE decreases down the group: * size of valence orbital of atoms increases down the grp * less effective overlap b/w orbitals from Cl2 to I2 * covalent bond formed b/w the halogen atoms weaker, BE decreases Boiling pt increase down the group: * all have simple molecular structure, non-polar, with id-id interaction b/w molecule * size of e- cloud increases and becomes more polarisable * strength of id-id interactions increases, more energy need to overcome from Cl2 to I2
factors affecting boiling point of simple molecular structure
• simple molecular structure, weak id-id interactions b/w molecules • larger e- cloud, more polarisable • strength id-id interactions increases, more energy needed to overcome strong id-id