Section A 1 Flashcards

1
Q

So we define a mole (mol) of a substance as containing …. molecules.
This value is …

A

6.022 x 10^23

Avogadros number, NA (or sometimes L)

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2
Q

M(Mg) = 24.31 x 1 g mol-1 = 24.31 g mol-1

This value is called the….. and it … units, in contrast to the relative ….., which ….

A

molar mass, M
has

molar (atomic) mass, Mr
is unitless

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3
Q

Using ….. the mass of the individual isotopes can be determined.

A

a mass spectrometer

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4
Q

Relative atomic mass is used to calculate ….

And the molar mass of a substance, M, is obtained by …..

A

relative molecular mass, Mr:

e.g. Mr(H2O) = (1.008 x 2) + (15.999 x 1) = 18.015

multiplying Mr by 1 g mol-1

e.g. M (H2O) = 18.015 x 1 g mol-1 = 18.015 g mol-1

This means that if we want to use 1 mol of water in the laboratory we would need 18.015 g.

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5
Q

No of mols (mol) =

A

mass (g) / molar mass (g mol-1)

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6
Q

An implication of describing the electron as a wave is the uncertainty principle –

A

its position and momentum (energy) cannot not simultaneously be determined.

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7
Q

Each electron in an atom is labelled with four quantum numbers. The first is….

It tells you about ….

A

the Principal quantum number:
n (the shell):
n = 1, 2, 3, …

how big the orbital is, or how far (on average) an electron
occupying it is from the nucleus.

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8
Q

The second quantum number is ….. :

This tells you about….:

A

the angular momentum quantum number: l
l = 0, … n-1

the shape of the orbital

l = 0 is an s orbital,
l = 1 is a p orbital,
l = 2 is a d orbital, etc.
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9
Q

The third quantum number is …. :

These are … orbitals, and tell you about ….

A

the magnetic quantum number: ml :

ml = -l, … +l
e.g. if l = 1, ml = -1, 0, +1.

the px, py, and pz

the orientation of the orbital

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10
Q

Wavefunctions can be used to calculate the radial distribution function of the orbital i.e. …

A

how the probability of finding the electron varies with distance from the nucleus.

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11
Q

The position where the function equals zero is called a ….and there is zero probability of the electron being found there.

A

node

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12
Q

the Pauli exclusion principle -

A

A maximum of two electrons can share each orbital, with the same values of n, l, ml but different spin (ms = +1/2 and -1/2).

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13
Q

Hunds Rule of maximum multiplicity -

A

If two or more orbitals have the same energy, then electrons will spread out to occupy the maximum possible number of these, maximising the number of parallel spins.

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14
Q

If the electrons occupy separate orbitals then they occupy different regions of space so there is …. electrostatic repulsion. This is a … energy arrangement.

A

less

lower

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15
Q

Why is cobalt essential for vitamin B12 function?

A

Vitamin B12 has an unusual Co-C bond which
can be broken three ways:
1. To form Co(III) and CH3-
(anion)

  1. To form Co(II) and CH3.
    (rodnik)
  2. To form Co(I) and CH3+
    (kation)

Essential to its function that Co is stable in three oxidation states and they are relatively close in energy to allow fast conversion between them.

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16
Q

Przypomnienie:

Compounds which require more than an octet of electrons in order to draw a Lewis structure are called …..

A

Ans. hypervalent compounds.

Although period 2 elements obey the octet rule very well,

Period 3 elements begin to show some deviations.
e.g. PCl5 can only be explained if P has 10 electrons in its outer shell
In SF6 S must have 12 electrons in its outer shell

17
Q

Znając: Number of electron pairs around central atom: 2

Podaj:

Shape:
Bond Angles:
Example:

A

Linear
180
BeF2

18
Q

Znając: Number of electron pairs around central atom: 3

Podaj:

Shape:
Bond Angles:
Example:

A

Trigonal planar
120
BF3

19
Q

Znając: Number of electron pairs around central atom: 4

Podaj:

Shape:
Bond Angles:
Example:

A

Tetrahedral
109.5
CF4

20
Q

Znając: Number of electron pairs around central atom: 5

Podaj:

Shape:
Bond Angles:
Example:

A

Trigonal bipyramidal
90, 120
PF5

21
Q

Znając: Number of electron pairs around central atom: 6

Podaj:

Shape:
Bond Angles:
Example:

A

Octahedral
90
SF6

22
Q

Znając: Number of electron pairs around central atom: 7

Podaj:

Shape:
Bond Angles:
Example:

A

Pentagonal bipyramidal
90, 72
IF7

23
Q

Structures based on tetrahedral geometry:

  1. No lone pairs
  2. One lone pair
  3. Two lone pairs
A
  1. Tetrahedral
  2. Trigonal pyramidal
  3. Bent
24
Q

Structures based on trigonal bipyramidal geometry

  1. No lone pairs
  2. One lone pair
  3. Two lone pairs
A
  1. Trigonal bipyramidal
  2. Disphenoidal
  3. T-shaped
25
Q

Structures based on octahedral geometry

  1. No lone pairs
  2. One lone pair
  3. Two lone pairs
A
  1. Octahedral
  2. Square pyramidal
  3. Square planar
26
Q

Bond angles:

H20
NH3
CH4

HCN
CO2
Acetone

SF4

A

H2O:
e = 104.5 ‘bent’

NH3:
e = 107.3 ‘trigonal pyramidal’

CH4:
e = 109.5 ‘tetrahedral’

bond angle gets smaller due to lone pair repulsion (od dołu do góry)

HCN:
e = 180 ‘linear’

CO2:
e = 180 ‘linear’

Acetone
‘Trigonal planar’

SF4:
The F-S-F angle involving the axial F atoms is 173

27
Q

Limitations of the Lewis model of bonding

A

It also does not explain some other observations, for example that the oxygen molecule is paramagnetic (it has unpaired electrons), which is not predicted by the Lewis model.

28
Q

ATP is made up from three sections:

A

A purine base – Adenine
A pentose sugar – Ribose
A phosphate section - Triphosphate

Adenine + Ribose = Adenosine

29
Q

These bonds are called pi or π-bonds. They have two …., above and below the line between the atoms.

A

lobes

30
Q

Combinations of waves can either be constructive or destructive.
We use this argument when combining the s orbitals to from a bond:

If they combine constructively then electron density between the atoms is …… This is called ….
If the s orbitals combine destructively then electron density between the atoms is and this is called ….

A

increased
a bonding molecular orbital.

decreased
an anti-bonding molecular orbital.

31
Q

Bond Order =

A

[(number of bonding electrons) – (number of anti-bonding electrons) ] /2

32
Q

Bond order of:
1. He2

  1. H2
  2. He2 +
A
  1. So for He2: Bond Order = (2 – 2) / 2 = 0
    A bond order of zero means there is no bond, so the He2 molecule does not exist.
  2. H2 = (2 – 0) / 2 = 1
    A bond order of 1 is equivalent to a single bond using the Lewis model, and is consistent with observations.
  3. He2 + = (2 – 1) / 2 = ½
    A bond order of ½ is something that is not encountered in
    the Lewis electron pair model, but something that MO
    theory can predict with ease.

What this means is that a bond of about half the strength
of a single bond can form between the He atoms of He2+

This means that this molecule can exist, but will not be
very stable. Its existence is likely to be transient.

33
Q

Rozwiń skróty:
LUMO
HOMO

A

Lowest unoccupied MO (Molecular Orbital)

Highest occupied MO