Bonding and molecular structure

Question 1

Schrödinger’s equation

Answer 1

Used to show that electrons behave in a wave-like manner.

Used to show the probability of finding an electron at a point in space.

Only applicable to Hydrogen.

Question 2

Principal quantum number (n)

Answer 2

Represents the energy and size of the atomic orbital.

Orbitals within the same “n” are in the same energy level.

Only integer values are used.

Question 3

The angular momentum quantum number (l)

Answer 3

Represents different shapes with different orbits.

Restricted by “n” as l = 0 to (n-1)

Question 4

“n” values and their effect on the l value and the corresponding orbit:
n = 1
n = 2
n = 3
n = 4

Answer 4

⠀
⠀
l = 0 = s (sharp) orbital
l = 0,1 = s and p (principal) orbitals
l = 0,1,2 = a,p,d (diffuse)
l = 0,1,2,3 = s,p,d,f (fundamental)

Question 5

The magnetic quantum number (mₗ)

Answer 5

Represents the orientation of orbitals in space.

-l to +l

if n = 2, l = 1, and mₗ = -1, 0, 1

Question 6

The spin quantum number (mₛ)

Answer 6

Each electron has an opposite spin.

Values are -1/2 and 1/2.

Question 7

s orbital

Answer 7

Spherical, the larger the value of n, the bigger the orbital.

Question 8

p orbital

Answer 8

Bi-lobed / dumbbell-shaped, they have 3 different orientations (called degenerate orbitals).

py, px, pz

Question 9

Energy conversion

Answer 9

As the value for n gets higher, the energy levels converge.

Question 10

Periodicity as you go across the PT:
Atomic radii
Electronegativity

Answer 10

⠀
Decreases (stronger nucleus charge)
Increases (stronger nucleus charge)

Question 11

Ionic bonding

Answer 11

Positively and negatively charged ions held together in a lattice.

Involves a transfer of electrons and then they stay attracted due to strong, opposite charges.

Question 12

Lewis structures

Answer 12

Draw 3 random ones out.

// go revise topic 14.1

Question 13

Formal charge

Answer 13

Number of valence electrons - 1/2 x bonding electrons - number of non-bonding electrons

The most likely structure is the one that is most correct (ie has the correct charge on the molecule and also has the - or + charge on the most or least electronegative atom)

Question 14

Polarity

Answer 14

A bond is polar if its centres of positive and negative charge do not coincide.

Question 15

Hybridisation

Answer 15

The overlap of different orbitals.

It causes the s orbital to raise up to the p orbital causing sp¹/²/³ orbitals to formed

Question 16

Sigma bonds (σ)

Answer 16

Caused by the axial overlap of orbitals.

Can be done with two s orbitals, one s orbital and one p orbital, or two p orbitals

Single bonds

Question 17

Pi bonds (π)

Answer 17

Caused by the side-to-side overlap of orbitals.

Occurs only with two p orbitals.

Double bonds, triple bonds.

Question 18

NH₃ hybridisation

Answer 18

Has sp³ hybridisation because the lone pair can form a dative covalent bond so it behaves

Question 19

Amine

Answer 19

A nitron bonded to a carbon

Question 20

Amide

Answer 20

A nitrogen bonded to the same carbon that an oxygen is also double bonded to.

R-C=O
⠀⠀⠀\N

Question 21

Imine

Answer 21

Nitrogen has a double bond with the carbon(s)

R=N-R

Question 22

Why hydrogen bonding occurs so easily and strongly

Answer 22

It is the only atom present in molecules for which the only electrons are valence electrons;
Thus:
* Hydorgen’s electrons are located between the bonded atoms, so the positive charge is relatively exposed, attracting a negative charge.

Question 23

The relative length of hybrid “s” orbitals.

Answer 23

As the “s” character of a hybrid orbital increases, the electrons are held closer to the nucleus: sp hybrid orbitals hold electrons closer to the nucleus than sp³ hybrid orbitals

Question 24

Strength of acids formed by hybridised carbons

Answer 24

Acidic character is judged by the stability of the anion formed after releasing a proton.

As the “s” character increases, the easier it is to remove a proton, the carbanion is stronger and more stable, and it is more acidic (it holds tighter to the lone pairs)

Question 25

Imines:
⠀
* hybridisation
* Geometry
* σ bond overlap
* π bond overlap
* Lone pair location
* Basic in comparison to amines

Answer 25

Imines:
⠀
* Both C & N - sp2 hybridised
* Both planar geometry
* σ from sp2 orbital overlap
* π from remaining p orbital overlap
* Lone pair is in an sp2 orbital;
* Imine is less basic than amines (electron pair less available)

Question 26

Nitriles:
⠀
* hybridisation
* Geometry
* σ bond overlap
* π bond overlap
* Lone pair location
* Basic in comparison to amines

Answer 26

Nitriles:
⠀
* Both C & N - sp hybridised
* Both linear geometry
* σ from sp orbital overlap
* 2π from two 2p orbital overlaps
* Lone pair is in an sp orbital;
* Nitrile is less basic than amines and nitriles (electron pair less available)

Question 27

Imines:
⠀
* hybridisation
* Geometry
* σ bond overlap
* π bond overlap
* Lone pair location
* Basic in comparison to amines

Answer 27

Imines:
⠀
* Both C & N - sp2 hybridised
* C - planar geometry, N - bent
* σ from sp2 orbital overlap
* π from remaining p orbital overlap
* Lone pair is in an sp2 orbital;
* Imine is less basic than amines (electron pair less available)

Question 28

Alcohol:

• Geometry:
• Lone pair availability:
• Bond polarity

Answer 28

Alcohol:

• C, tetrahedral; O, bent
• in sp³ orbitals, OH can be a base
• O is more electronegative than C/H

Question 29

Ketones and aldehydes:

• Geometry:
• hybridisation
• Lone pair availability
• Location of lone pairs
• Formation of σ bond
• Formation of π bond

Answer 29

Alcohol:
* C = trigonal planar
* Both have sp² hybridisation
* sp² orbitals, less available.
* Lone pairs are in 2 sp² orbitals on O
* sp² orbitals overlap
* p orbitals on both C and O overlap

Question 30

Macromolecule

Answer 30

A molecule comprised of many molecules

Question 31

Intramolecular interactions

Answer 31

Influence the shape and stability of a macromolecule:

Ionic bonding, covalent bonding

Question 32

Intermolecular interactions

Answer 32

Affect physiochemical properties and are the basis for selective molecular recognition between molecules:

Dispersion, dipole-dipole, hydrogen

Question 33

Dispersion forces

Answer 33

Caused when one molecule is near another molecule.

Since electrons are always moving, at any given point, the electron density will be closer to one atom out of the two atoms at any specific point.

This will cause a temporary dipole to be induced in a nearby molecule and cause a very weak, basic attractive force to exist

Question 34

Dipole-dipole forces

Answer 34

Occurs between polar molecules.

Molecules line up to satisfy intermolecular attraction: atoms that are δ+ in one molecule will line up next to atoms that are δ⁻ in another molecule.

This will form a relatively strong attraction that is weaker than hydrogen bonding but stronger than dispersion forces

Question 35

Constructive interference

Answer 35

Ideal for bonding: the + and - waves are aligned and there are electrons in the molecular orbital (a sigma bond (σ))

Question 36

Destructive interference

Answer 36

Not ideal for bonding: the + and - waves are aligned and there are no electrons in the molecular orbital (an antibonding sigma bond (σ*))

Question 37

Electrons in σ orbitals: what does it do to the species?

Answer 37

Stabiilise a species

Question 38

Electrons in σ* orbitals: stabilises or destabilise a species?

Answer 38

Destabilise a species

Question 39

When can a species not exist

Answer 39

If both σ and σ* bonds are fully occupied and there is no net stabilisation

Question 40

When can a species exist

Answer 40

If the σ bond contains more electrons than the σ* bond, and there is net stabilisation

Question 41

Key parts to an energy diagram regarding bonding and antibonding orbitals

Answer 41

1) π/π* 2px & π/π* 2py are degenerate (same);

2) The π/ π∗ energy gap is less than the σ/σ* gap – more efficient overlap than pz orbitals;

3) electrons are filled from the bottom up, filling the bonding orbitals first.

Question 42

Paramagnetism

Answer 42

Occurs as a species is attracted to a magnetic field due to having unpaired electrons

Question 43

Diamagnetism

Answer 43

Occurs as a species is weakly repelled by a magnetic field due to having only paired electrons