Chemical Bonding 2 Flashcards
Define Hybridisation.
Hybridisation is the mixing of atomic orbitals in an atom to generate a set of equivalent hybrid orbitals with the same shape and energy, each containing a single unpaired electron.
Note: The number of hybrid orbitals formed always equals the number of atomic orbitals used for mixing.
Since the theory of covalent bond formation via overlapping of atomic orbitals does not always give good agreement with observations hence hybridisation is a concept used to explain the observed shapes of molecules
Draw the displayed formula of methane. State and describe the hybridisation of carbon in methane, CH4 that gives rise to the bonding and structure in the molecule.
sp^3 hybridisation.
CH4 is tetrahedral with all the C-H bonds being of equal bond lengths and in between them, bond angles of 109.5 degrees, implying that the carbon must provide four equivalent orbitals for head-on overlap with the 1s orbital from each H atom to provide for four identical C-H sigma bonds.
In the ground state of C, there are only 2 unpaired electrons, thus it is able to form only 2 covalent bonds. When one 2s electron is promoted to the vacant 2p orbital to form C* in the excited state, it now has 4 singly occupied orbitals to form 4 covalent bonds. However, it cannot form 4 equivalent C-H bonds to give a symmetrical tetrahedral molecule. Four equivalent hybrid orbitals for carbon are generated by mixing one 2s and three 2p orbitals, resulting in the formation of four sp^3 hybrid orbitals which overlaps with the 1s orbital of the hydrogen atom to give a tetrahedral methane molecule.
Draw the displayed formula of ethene. State and describe the hybridisation of carbon in ethene, C2H4 that gives rise to the bonding and structure in the molecule.
sp^2 hybridisation.
C2H4 is planar with all the C-H bonds being of equal bond lengths and the H-C-H and C-C-H bond angles are all 120 degrees, implying that the carbon must provide three equivalent orbitals for head-on overlap with the 1s orbital from each H atom and a hybrid orbital of another C atom to provide for three identical sigma bonds, leaving one unhybridised 2pz orbital for the formation of one pi bond between the two C atoms.
In the ground state of C, there are only 2 unpaired electrons, thus it is able to form only 2 covalent bonds. When one 2s electron is promoted to the vacant 2p orbital to form C* in the excited state, it now has 4 singly occupied orbitals to form 4 covalent bonds. Three equivalent hybrid orbitals for carbon are generated by mixing one 2s and two 2p orbitals, resulting in the formation of four sp^2 hybrid orbitals, leaving one unhybridised 2p orbital.
Two of the sp2 hybrid orbitals overlap with the 1s orbital of the two H atoms while the other sp2 hybrid orbital overlaps with the sp2 hybrid orbital of another C atom and the remaining unhybridised 2pz orbitals of each C atom overlap side-on, giving rise to a pi bond. The p orbitals are perpendicular to the plane containing the unhybridised orbitals, resulting in the a trigonal planar electron pair geometry.
Draw the displayed formula of ethyne. State and describe the hybridisation of carbon in ethyne, C2H2 that gives rise to the bonding and structure in the molecule.
sp hybridisation.
C2H2 is linear with all the C-H bonds being of equal bond lengths and the H-C-H and C-C-H bond angles are all 180 degrees, implying that the carbon must provide two equivalent orbitals for head-on overlap with the 1s orbital from each H atom and a hybrid orbital of another C atom to provide for two identical sigma bonds, leaving two unhybridised 2p orbitals for the formation of two pi bond between the two C atoms.
In the ground state of C, there are only 2 unpaired electrons, thus it is able to form only 2 covalent bonds. When one 2s electron is promoted to the vacant 2p orbital to form C* in the excited state, it now has 4 singly occupied orbitals to form 4 covalent bonds. Two equivalent hybrid orbitals for carbon are generated by mixing one 2s and one 2p orbital, resulting in the formation of two sp hybrid orbitals, leaving two unhybridised 2p orbitals.
One of the sp hybrid orbitals overlaps with the 1s orbital from one H atom while the other sp hybrid orbital overlaps with the sp hybrid orbital of the other C atom. The remaining two unhybridised 2p orbitals each contain an unpaired electron and their side-on overlap give rise to two pi bonds between the C atoms.
State the (i) types of orbitals involved, (ii) total number of hybrid orbitals formed, (iii) number of unhybridised orbitals, (iv) number of regions of electron density, (v) electron pair geometry around each atom for each type of hybridisation, drawing diagrams for the hybrid orbitals. (Tip: Draw dotted lines for the geometry before drawing the orbitals along them)
Pg 6 of notes
Explain the effect of hybridisation on the bond length and bond strength of a molecule. Describe the trend in the bond length and bond strength of the C-H bond in ethane, ethene and ethyne.
The s orbital is spherical and electrons are closer to the nucleus. The higher the s character of the hybrid orbital, the less diffuse the hybrid orbital, the more tightly the shared electrons are held by the nuclei, the more effective its overlap with the orbital of the other atom and the stronger the bond and the shorter its bond length. From ethane to ethene to ethyne, the C-H bond strength increases and the C-H bond length decreases.
Explain what resonance structures are.
Resonance structures are two structures that have the same placement of atoms but a different placement of pi bonds and non-bonding electrons. The placement of the sigma bonds remain the same.
When does a molecule or ion exhibit resonance?
They exhibit resonance when there is a continuous side-on overlap of p-orbitals over at least 3 adjacent atoms, allowing for delocalisation of the pi electrons. Such molecules have two or more resonance structures.
Explain why resonance structures are not correct representations of the O3 molecule. State the name of the actual structure of the molecule and describe how the electrons are distributed in the molecule.
The resonance structures of the O3 molecule suggests that the molecule would be unsymmetrical, with the O=O bond shorter than the dative bond. However, when the oxygen-oxygen bond lengths are measured, both are found to be the same and mid-way between the O-O and O=O bond lengths in other molecules.
This suggests that the resonance structures are not correct representations of the molecule and the actual structure of the molecule is a hybrid of the resonance structures, known as the resonance hybrid.
In the resonance hybrid, the pi electrons are delocalised over the 3 adjacent oxygen atoms, via the continuous side-on overlapping of p orbitals, which are perpendicular to the plane of the molecule containing the sigma bonds. The delocalised electrons are denoted by dotted lines.
Explain why the resonance hybrid is more stable than the resonance structures.
The resonance hybrid is more stable than any one of the resonance structures because it delocalises electron density over a larger volume.
Distinguish between localised and delocalised electrons.
Electrons are localised when they are shared between 2 atoms of the bond only, for example in ethene. The electrons are delocalised when they are shared among more than two atoms through continual overlap for example in O3 or other molecules or ions that exhibit resonance.
Draw the resonance structure of benzene.
Pg 11 of notes
Explain the melting point, solubility and conductivity of diamond in terms of the structure and bonding in a diamond.
Diamond has a giant molecular structure. Each carbon atom is sp3 hybridised and is covalently bonded to four other carbon atoms arranged tetrahedrally around it. This tetrahedral arrangement is repeated throughout the molecule and gives rise to interlocking hexagonal units and an extremely strong and rigid structure with a very high melting point of 3550 degrees celsius. It does not dissolve in water or other solvents.
Diamond is an electrical insulator as it does not have delocalised electrons and is made up of neutral atoms.
Silicon has the same structure as that of a diamond. Explain why silicon has a lower melting point than diamond.
Since the valence orbitals of Si are larger and thus more diffused than that of C in diamond, there is a less effective overlap between the orbitals of Si, resulting in the Si-Si bond being weaker than the C-C bond. Less energy is required to overcome the weaker Si-Si bonds in silicon than the stronger C-C bonds in diamond, hence silicon has a lower melting point than diamond.
Explain the melting point and conductivity of graphite in terms of the structure and bonding in graphite.
Graphite has a giant molecular structure. It has a layered structure and is made up of planes of interconnected hexagonal rings of carbon atoms. Each carbon atom is sp2 hybridised and forms three sigma bonds with three other carbon atoms, leaving one unhybridised p-orbital which contains an unpaired electron. (The electron-pair geometry is trigonal planar with bond angles of 120 degrees around each carbon atom.) The p-orbital of one carbon atom overlaps collaterally with the p-orbitals of its immediate neighbours, resulting in an extended pi-electron cloud above and below the plane containing the carbon atoms. The pi electrons are delocalised over the whole layer so that graphite is an electrical conductor. (However, since the pi-electrons are delocalised within each layer, graphite can only conduct electricity in a direction parallel to the layers.) The strong covalent bonds and attraction between the different layers due to weak instantaneous dipole-induced dipol forces require lot of energy to be overcome so graphite has a high melting point of 3730 degree celsius.
