Bonding Flashcards

1
Q

Orbitals

A
  • Orbitals are representations of the area
    of space occupied by an electron (the
    likelihood of finding an electron in a
    given space)
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2
Q

Pauli Exclusion Principle:

A
  • A maximum of two electrons can
    occupy any one orbital (must have opposite charge)
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3
Q

Carbon

A
  • Atomic number 6
  • Got 4 valence electron
  • Every stable covalent bond must have 2 electrons
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4
Q

(Hund’s Rule)

A
  • For orbitals of the same energy a single electron is added to each before pairing begins.
    –This is to avoid e to e replusion
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5
Q

Valence-Shell Electron-Pair Repulsion

A
  • In a molecule all groups that are attached to a single atom should be as far away from each other in space as possible to minimize electron repulsions.
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6
Q

Methane (CH2)

A

C has 4 valence electron
2 will bond in 1S and 2S
2 will go to 2px, 2py and H will bond here with their electron

Two of carbon’s valence electrons are in a filled 2s orbital and can’t participate in bonding. We have the 2px and 2py orbitals, each with 1 electron, which can covalently
bond with two hydrogens

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7
Q

Hybridization

A
  • The atomic orbital of carbon mix to form a set of four hybrid orbital that we call sp3 orbitals
    They are equal in energy and are directed to the corner of a regular tetrahedron (109.5 degree)
  • 2s, 2px,2py, 2pz will turn to 4 sp3 when hybridized (when you combine 4 orbitals you will end up with 4 new orbitals)
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8
Q

Bonds

A
  • When orbitals overlap, electrons will be shared by both atoms. This will form MO (molecular orbitals)
  • The number of AOs or hybrids is equal to the number of MOs
  • There will be an MO lower in energy to a constituent orbital (bonding MO) and the other higher in energy (anitibonding MO)
  • A covalent bond is formed whenever two electrons occupy a molecular orbital
  • There need to be two orbitals overlapping to be called sigma bond is formed)
  • Anti-bonding orbital has no electron in a stable compounds but it still exist and IS able to accept e under certain circumstances
  • Having e in anti-bonding orbitals make compounds unstable.
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9
Q

Nitrogen

A
  • Got 5 valence electrons
  • In sp3 hybrid orbital the electron is placed in each orbital until they all have one then the remaining electron is paired up in one of the orbital
  • The filed orbital is called lone pair (push down on bonds)
    – NH3 got 110.6 degree because of lone pair
  • Lone pair can interact with vacant orbital (since the e both originally belonged to nitrogen but are now shared this mean that the N is now electron deficient and DEVELOPS a Positive charge)
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10
Q

Oxygen

A
  • Got 6 valence electrons
  • It got two lone pairs (filled sp3) and two singly occupied orbitals that can overlap
  • H2O got 104.5 degree
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11
Q

Fluorine

A
  • Got 7 valence electrons
  • 3 lone pairs and F will only be able to form one covalent bond
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12
Q

Ethylene

A
  • 3 atoms bonded to carbon: needs to stay as far as possible from each other (120degree)
  • A trigonal planar does that and keep them at 120 degree
  • An ethylene is two carbons bound together and have only 2H on each
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13
Q

Hybridization

A
  • sp3 will not give us a trigonal planar arrangement so an alternative hybridization is required
  • We need 3 hybrid orbital so we mix 3 atomic orbital (2s+2px+2py) and we get 3 sp2 hybrid orbital (gonna get an equilateral triangle 120 degree)
  • Each sp2 will allow for forming three single bonds)
  • The remaining orbital is carried unchanged (2pz) and it will remain perpendicular to the plane formed by the sp2
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14
Q

Hybridization (2pz)

A
  • Orbital that exist in perpendicular planes cannot interact with one another (stay independent of the hybrid orbital)
  • Form a double bond (pi bond) with another C in C2H4 as they are next to each other and perpendicular to the rest)
  • Double bonds are more reactive than single bonds
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15
Q

Double bonding

A
  • Form from the mixing of two Pz orbitals and they form two set of new MO
  • The lower energy MO is called bonding pi orbital
  • The higher energy (vacant) is called an anti bonding pi-orbital
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16
Q

Oxygen and Nitrogen Bonds

A
  • Oxygen: got two lone pairs with one electron in the remaining hybrid orbital (for forming single bond) and the other is in the Pz orbital (used to form double bonds)
  • Nitrogen: Got one lone pair while the remaining two will be occupied with one electron (will form two single bonds) and the remaining electron is in the Pz orbital (used to form double bond)
17
Q

Acetylene (C2H2)

A
  • Carbon is attached to only two other atoms
  • You will mix two 2s and a 2px orbital which will form two sp hybrid orbital needed to allow the bonding of the two atoms
  • There will be 2py and 2pz left which will be used to form a triple bonds (perpendicular)
  • Each orbital will have one electron and they will stay at 180 degree to stay away from each other (2py and 2pz will be perpendicular to one another)
    When they bond the Hs will bond with C using a sp orbitals and C will bond with C also using sp. Then the two p-orbitals will overlap and form two bonds that between C:C which mean we will end up with a triple bond
18
Q

Hybridization Summary

A

The number of groups attached to an atom determine the hybridization of the atom
- 4 groups : sp3 (tetrahedral geometry, 109.5)
- 3 groups : sp2 (trigonal, planar geometry, 120)
- 2 groups : sp (linear geometry, 180)
Bent (with two lone pairs):104.5
Trigonal bipyramidal (1 lone pair):110.6

19
Q

Stereochemistry

A
  • In an sp3 hybridization (central atom and four surrounding groups) only THREE of the groups (atoms or lone pair) will occupy the same plane
  • The remaining two groups will be out of the plane
    – Central C and two Hs in a plane
    – One H in front of plane and one H behind plane
20
Q

Benzene

A
  • C6H6: Each carbon is bonded with two other carbon and one hydrogen
    – This mean the hybridization is sp2 with each carbon having one p-orbital that is perpendicular
  • Adjacent p-orbital overlap to form pi-bonds (double) so you have alternating single and double bonds (conjugated system)
  • Equally likely that the double form in either direction so you get resonance structure direction (lower the E of the system)
21
Q

Aromatic (benzene is aromatic)

A
  • Cyclic
  • Flat (allow max overlap of p-orbitals)
  • Every atom in the ring possess a p-orbital
  • 4n+2 rules: a compound is aromatic if the total number of electron in the pi-system is equal to 4n+2)
    – As long as N is a whole number it’s an aromatic
    – You add the electron (each pi-bond is 2 electron)
22
Q

Antiaromatic (doesn’t like to exist)

A
  • Flat
  • Cyclic
  • Each atom has a p-orbital
  • total number of electron in pi system (double bond) is equal to 4n
23
Q

Carbon with a positive charge (+)

A
  • Happen when the 4th group left with pair of electron
  • This mean that the carbon only has 3 groups left and is sp2-hybridized
    • charge is counted as a vacant p-orbital (doesn’t do anything for the 4n+2 formula only (-) charged contribute 2 e)
24
Q

Pyridine

A
  • Heterocycles (Cyclic compounds with atoms other than C in the ring)
    – Can be aromatic (pyridine is cyclic, flat, and each C and N have p-orbital (6e- in the pi-system)
  • The N (third group is lone pair) and each carbon all have sp2hybridized
    – The lone pair MUST be in the same plane to be contained within sp2
    – There is no interaction between lone pair and pi-system as pi-system are perpendicular
25
Q

Pyrrole

A
  • 5-membered ring nitrogen heterocyclic
  • Got 2-double bonds and an N-H bond
  • The N is attached to 2 carbons, an H, and has a lone pair (4 groups)-This might make it seem like an sp3 hybridized (meaning no p-orbital) but pyrrole is flat and the H is in the plane of the ring
    – This mean N is sp2 hybridized
  • The lone pair is located in the p-orbital (as the other three hybrid orbital of nitrogen are used to form bond bonds with the two carbon and one hydrogen)
    – Pyrrole is aromatic
    The E increase that result from brining the 4th group (lone pair) closer to other e- is offset by the lowering in E by making the rink aromatic
    – A compound with an sp2 hybridized N is lower in E than the same compound with an sp3
26
Q

Other Aromatic system

A
  • Pyridine is aromatic and it’s trrogen act as a base
    – The lone pair was in the plane of the ring and therefore not part of the pi-system
  • For pyrrole the lone pair IS PART of the pi-system and this mean the pyrrole is NOT BASIC because it just ensure the ring is aromatic
27
Q

Other Aromatic Systems II

A
  • Furan
  • Imidazole (the double bond N is basic and the other N is not as the lone pair is incorporated into the pi-system to make the ring aromatic)
  • Thiazole (The double bond N is basic)
  • Isoxazole (The double bond N is basic)
28
Q

Cyclooctatraene

A

Not aromatic as it doesn’t have 4n+2