chem_bonding Flashcards
Define valence electrons.
Valence electrons are those electrons furthest from the nucleus (i.e. in the outer shell) - which are typically involved in covalent bonding.
Describe a reliable method to determine the number of valence electrons for an element.
The number of valence electrons can be determined by refering to which main-group column the element is found in. For example nitrogen is found in main-group 5 and therefore nitrogen has 5 valence electrons. These are the electrons described by the 2s22p3 electron configuration.
List the steps in drawing a Lewis structure.
The steps are
- Count the total number of valence electrons (add 1 for a negative charge, subtract 1 for a postive charge).
- Draw a molecular skeleton generally with the least electronegative element in the center and joining each atom with a dashed line.
- Place enough electrons to give each element an octet (hydrogen only needs a duplet).
- If Step 3 uses too many electrons- take away the excess electrons and make the appropriate double bond or triple bonds. If Step 3 uses too few electrons, add the required number of electrons to the central atom.
- Assign formal charges to all atoms.
- Draw all necessary resonance forms.
Define formal charge.
Formal charge is an apparent charge which results if an atom “owns” more or less electrons than its normal count of valence electrons. All lone pairs plus 1/2 of the shared pairs are “owned”. Formal charge is therefore calculated by subtracting the number of electrons “owned” from the normal count of valence electrons.
Define resonance form.
Resonance forms are two or more Lewis structures drawn (seperated by double headed arrows) to indicate the existance of delocalized electrons. Delocalized electrons are the result of pi electrons being shared between three or more atoms (instead of just two atoms).
A negative formal charge is generally more stable on a more electronegative atom:
T or F?
This statement is true and is important when comparing two or more possible resonance forms.
The resonance form which contributes the most to the actual structure is the one with fewer non-zero formal charges.
If there are formal charges, it is better for a negative charge to be on a more electronegative atom. Likewise it is better for a positive charge to be on an atom with a smaller electronegativity.
Under what conditions will carbon have a non-zero formal charge?
Carbon will have a non-zero formal charge when it is not making four bonds (as per usual). The only other option for carbon is to make three bonds.
If carbon is only making three bonds, it is a carbocation and has a positive charge (see diagram).
If carbon is making three bonds and has a lone pair of electrons, it is a carbanion and has a negative charge (see diagram).
Under what conditions will nitrogen have a non-zero formal charge?
Nitrogen will have a non-zero formal charge when it is not making three bonds (as per usual).
If nitrogen is making four bonds, it is a cation and has a positive charge (see diagram).
If nitrogen is making two bonds and has a lone pair of electrons, it is a anion and has a negative charge (see diagram).
Under what conditions will oxygen have a non-zero formal charge?
Oxygen will have a non-zero formal charge when it is not making two bonds (as per usual).
If oxygen is making three bonds, it is a cation and has a positive charge (see diagram).
If oxygen is only making one bond, it is an anion and has a negative charge (see diagram).
How are bond length and bond order related?
The greater the bond order, the shorter the bond length.
Bond order indicates the number of bonds between two atoms. The more bonds “holding” two atoms, the shorter will be the internuclear distance.
One sigma bond = bond order = 1. One sigma bond + one pi bond = bond order = 2 . One sigma bond + two pi bonds = bond order = 3 .
Describe atomic orbital hybridization.
Atomic orbital hybridization is the result of the original atomic orbitals “fusing” to produce unique hybrid orbitals. These hybrid orbitals have different energies and “shapes” as compared to the original atomic orbitals.
For example, carbon’s valence electrons sit in the 2s and 2p atomic orbitals according to the electon configuration (2s22p2). When carbon forms four single bonds, these atomic orbitals hybridize into four unique sp3 orbitals. Collectively the four sp3 orbtitals produce a tetrahedral orientation. All four sp3 hybrid orbitals are degenerate and their energies are distinct from the energies of the 2s and 2p orbitals.
Compare and contrast sigma versus pi bonds.
A sigma bond is the result of a head to head overlap of orbitals. This results in what we normally label as a single bond.
A pi bond is the result of a side by side overlap of p orbitals. This side by side overlap occurs both above and below the plane of the sigma bond.
A sigma bond and a pi bond between two atoms would constitute a double bond. A sigma bond and two pi bonds would constitute a triple bond.
Define steric number (SN).
Steric number (SN) is a parameter about a central atom which accounts for the number of regions of electron density. It is calculated by: number of sigma bonds + number of lone pairs.
Describe the relationship of steric number (SN) and hybridization.
The steric number (which is determined by considering the number of sigma bonds and the number of localized lone pair of electrons about an atom) is a good predictor of the atom’s hybridization. SN = 2 = sp hybridization; SN = 3 = sp2 hybridization; SN = 4 = sp3.
If there are lone pair of electrons, the key is to distinguish between localized and delocalized lone pairs. Delocalized lone pairs are those that are involved in pi bonds and are identified through consideration of possible resonance forms.
Explain VSEPR theory.
Valence Shell Electron Pair Repulsion (VSEPR) theory predicts that electron pairs about a central atom will arrange themselves so as to minimize repulsion. Electron pairs include both bonding and lone pairs.
