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2. Chemistry A-level > 1.1 + 1.2 Formulae and equations/Basic Ideas about Atoms > Flashcards

1.1 + 1.2 Formulae and equations/Basic Ideas about Atoms Flashcards

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1
Q

name of:
positive ions
negative ions

A

cations

anions

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2
Q

order of balancing

A

CARBON
HYDROGEN
OXYGEN

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3
Q

relative isotopic mass

A

weighted average of several iostopes relative to 1/12th mass of a carbon 12 atom

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4
Q

relative atomic mass

A

weighted average of an atom compared to 1/12 mass of one carbon 12 atom

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5
Q

diatomic

A

N O F F CL BR I

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6
Q

ammonia

A

NH3

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7
Q

methane

A

CH4

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8
Q

hydrogen sulfide

A

H2S

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9
Q

giant covalent elements

A

Diamond
graphite
graphine
silicon

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10
Q

giant covalent compounds

A

silicon dioxide (SiO2

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11
Q

ionic compounds

A

HCL - hydrochloric acid
H2SO4 - sulfuric acid
HNO3 - nitric acid
H3PO4 - phosphoric acid

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12
Q

hydrogen sulfide

A

H2S

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13
Q

Positive ions 1+

A

Li+
Na+
K+
Ag+
NH4+ (ammonium)
H+a

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14
Q

Positive ions 2+

A

Mg2+
Ca2+
Ba2+
Zn2+

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15
Q

Positive ions 3+

A

Al3+

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16
Q

Negative 1- ions

A

F-
Cl-
Br-
I-
NO3- (nitrite)
HCO3 - (hydrogencarbonate)
OH-
MnO4- (Magnate VII)

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17
Q

Negative 2- ions

A

O2-
S2-
CO3 2-
NO 2-
SO3 2- (sulfate)
Cr2O7 2- (dicromate VI)

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18
Q

Negative 3- ions

A

PO4 3- (phosphate)
N 3- (nitride)

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19
Q

mass = Mm (Mr) x mol

A
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20
Q

calculate empirical formula

A

find out mol using mass/percentage

divide by smallest mol value

give mol ratio

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21
Q

molecular formula from empirical

A

relative mass of empirical

relative molecular mass (given) divided by empirical formula mass

number of empirical formula units

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22
Q

find formula of hydrated salt

A

determine mass of water in hydrated salt

determine mole ratio by g divided by Mr

divide by smallest calue

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23
Q

mole

A

amount of substance that has the same number of particles as there are number of atoms in C12/Avogadros number

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24
Q

empirical formula

A

the simplest whole number ratio of atoms present in a compound

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25
Q

cm3 to dm3

A

divide by 1000

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26
Q

dm3 to cm3

A

multiply by 1000

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27
Q

oxidation number/state

A

number of electrons that nned to be added or taken away to make an element neutral

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28
Q

oxidation rule 1

A

All uncombined elements have an oxidation number of zero.
E.g. in O2 gas the oxidation number = 0

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29
Q

oxidation rule 2

A

The sum of oxidation numbers in a compound is zero.
E.g. in MgO, Mg = +2, O = –2, +2 – 2 = 0

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30
Q

oxidation rule 3

A

In an ion, the sum equals the overall charge.
E.g. in NO3– the sum of oxidation numbers = –1, Nitrogen = +5, Oxygen = -2, +5 + (–2 x3) = –1

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31
Q

oxidation rule 4

A

Group 1 metals have an oxidation number of +1; Group 2 metals have an oxidation number of +2.

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32
Q

oxidation rule 5

A

Group 6 elements usually have an oxidation number of –2; Group 7 elements usually have an oxidation number of –1.

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33
Q

oxidation rule 6

A

The oxidation number of oxygen is –2, except in peroxides (in H2O2 it is –1) or when combined with fluorine.

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34
Q

oxidation rule 7

A

The oxidation number of hydrogen is +1, except in metal hydrides. For example, in NaH, hydrogen is –1 because Group 1 metals like Na always have an oxidation number of +1.

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35
Q

balance equations with oxidising and reducing agents

A

Oxidising agents - O

Reducing agents - H

then balance

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36
Q

mol = conc x dm3

A

conc (mol/dm3)

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37
Q

atoms = mol x avo

A
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38
Q

vol = mol x 24

A
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39
Q

radioactive emission

A

occurs when unstable nucleus becomes more stable

via giving out energy

eg electrons

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40
Q

alpha particles

A

two protons, two neutrons

helium neclei

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41
Q

beta particles

A

electrons

42
Q

deflection of radioactive particles in an electric field

A

alpha - positive, slowmoving, weakly attracted to negative plate of electric field

beta - light, fastmoving, deviation towards positive plate of electric field

gamma - short wavelenth, therefore unaffected by an electric field

positron - more attracted to negative plate than alpha, as lower mass

43
Q

penetrating power of radioactive particles

A

alpha - least penetrating, paper

beta - thin layer of foil

gamma - stopped thick plate of lead

44
Q

ionising power of radioactive particles

A

alpha - most ionising, remove electrons from atoms, high positive charge, strongly attracts electrons

beta - less ionising, collide with electrons in atoms, knock them out, ionise the atom

gamma - least ionising, ionises atoms if electron absorbs energy of gamma unlikely due to short wavelength

45
Q

positrons

A

anti electrons

same mass but positive charge

46
Q

positron + electron

A

when positrons and electron come in contact, they annihilate, creating gamma radiation

47
Q

gamma radiation

A

high energy EM radiation

short wavelength

high frequency

emited after electron capture (proton rich nucleus absorbs inner electron, which combines with proton to form a neutron)

48
Q

alpha equation

A
49
Q

beta equation

A
50
Q

positron equation

A
51
Q

electron capture equation

A
52
Q

atomic number

atomic mass

A

atomic mass - bigger, top

atomic number - smaller, bottom

53
Q

half life

A

time takes for the mass of a radioactive substance to fall to half its original value

54
Q

ionising radiation

A

radiation absorbed by a neutral atom

electron is removed

atom is ionised

55
Q

Radiation affect on living cells

A

Late Effect of Ionising Radiation:

DNA of a cell is damaged

becomes cancerous

divides uncontrollably

tumors

Acute Effect:
Large doses of radiation can cause cell death

Kill cancer cells, harmful bacteria, and other microorganisms

56
Q

Effects on the body INSIDE

A

Inside the body:
alpha - most ionising, most damage to living cells
beta and gamma - less ionising, more likely to pass through

57
Q

Effects on the body OUTSIDE

A

Outside the body:
alpha - least penetrating, can’t reach living cells under dead cells
beta and gamma - more dangerous, can penetrate dead cells on surface and damage living cells beneath

58
Q

uses of radiation:
tracers

A

radioactive materials injected, radiation detectors, beta or gamma used, short half-life, easily pass through body, not easily absorbed as alpha by cells

59
Q

uses of radiation:
radiotherapy

A

beam of gamma focused on tumor, kills cancer cells, radiation can damage/kill healthy cells, rotating gamma source focus on tumor, minimise exposure to healthy cells

60
Q

uses of radiation:
radio dating

A

eg carbon 14, long half-life
carbon 14 decays
by counting remaining carbon atoms

61
Q

uses of radiation:
thickness of metal eg foil

A

beta source and detector, rollers move depending on level of radiation recieved

62
Q

ionisation energy

A

measure of energy required to remove one or more electron from an atom

63
Q

first ionisation energy

A

energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions

64
Q

Second ionisation energy

A

the energy required to remove one mole of electrons from one mole of gaseous 1+ ions to form one mole of gaseous 2+ ions.

65
Q

large increase in successive ionisation energies

A

change in principle energy level

electron removed from an energy level closer to the nucleus (less shielding)

so more energy required

eg if large jumpe between group 2 and 3
Element is in group 2

66
Q

factors affecting ionisation energy

A
  1. atomic radius - greater radius, e- easier to remove, outer e- feels reduced nuclear charge
  2. number of protons - increase protons, increase force of attraction to nucleus
  3. shielding - repulsion by electrons in shells between electron and nucleus, outershell electrons are held less tightly

SO
weaker force of electrostatic attraction between nucleus and outer shell electron

67
Q

nuclear charge

A

Nuclear charge refers to the total positive charge found in the nucleus of an atom

the higher the nuclear charge, the higher the ionisation energy

the greater the positive charge of the nucleus, the stronger the attraction for the outer electrons.

68
Q

IE
trend down groups

A

first IE decreases down groups

increasing atomic radius, increasing shielding, reduces electrostatic attraction

69
Q

IE
trend across groups

A

generally increases, number of protons in nucleus increase, increases nuclear charge
Each electron in an energy level is not identical

70
Q

shape of S orbital

A
71
Q

shape of P orbital

A
72
Q

shell and subshell table

A
73
Q

How to write electron configuration

A
74
Q

Order of orbitals increasing energy

and exception

A

The 4s is filled before the 3d because it is lower in energy.

75
Q

watch out for chromium and copper

A

the 3d orbital becomes lower in energy than the 4s orbital when filled

4s electrons are removed before the 3d electrons when transition metals are ionised

76
Q

Eg of orbitals

A

1s2
2s2
2p6
3s2
3p6
4s2
3d10
4p6

77
Q
A
78
Q

explain increase in IE across period

A

increased nuclear attraction

distance to nucleus is constant, shielding is constant, number of protons steadily incrasing

79
Q

chromium

A

expected configuration (based on the Aufbau principle) would be:

Ar 3d4 4s2

However, the actual configuration is:

Ar 3d5 4s1

This is because having a half-filled 3d subshell provides extra stability due to reduced electron repulsion

80
Q

noble gas electron configuration

A

configuration
2p6
meaning it is a noble gas with a completely filled outer shell, which is the most stable electron configuration in nature.

2p1
This lone electron is less stable because it is far from the nucleus and not part of a stable full or half-full configuration.

C has a greater effective nuclear charge (ENC) than A, which means its electrons are pulled closer to the nucleus and are harder to remove.
ENC increases as the number of protons increases and the outermost electrons are shielded less by inner electrons.

In C, the electrons are evenly distributed in filled subshells, minimizing electron-electron repulsion.
In A, the single electron in the
2
p
2p orbital experiences less stability because it is less shielded and subject to greater interactions.

81
Q

EM spectrum

A

radio waves (at the lowest frequency and longest wavelength) to gamma rays (at the highest frequency and shortest wavelength).

82
Q

E = hf

A

E=J h=Js f=Hz

h is Planck’s constant and has a value of 6.63 × 10-34 Js. This constant is given on the AS examination data sheet.
Since h is a constant, then E α f. Therefore, if the frequency increases, the energy increases.

83
Q

c = fλ

A

c=ms-1 f=Hz wvleng=m
c is the speed of light and is a constant, with a value of 3 × 10^8 ms-1.

nm to m 1x10^-9

As c is a constant, when frequency increases, wavelength decreases.

84
Q

To calculate energy in kJ mol-1, you must:

A

calculate the frequency using the given wavelength

use the wavelength to calculate the energy in J

multiply the energy in joules by Avogadro’s number (NA, 6.02 × 1023 mol-1) to get the energy in J mol-1

divide the energy value by 1000 to convert from J mol-1 to kJ mol-1.

85
Q

absorption spectrum

A

certain specific energy has been absorbed, then only a specific frequency of light is absorbed from the electromagnetic spectrum - this appears as a “missing” line or frequency in the spectrum.

86
Q

emission spectrum

A

An emission spectrum is a series of lines which correspond to particular frequencies of energy emitted by an excited species.

87
Q

why do atoms absorb or emit certain frequencies of light

A

the gap between energy levels is fixed

88
Q

where is the majority of ligh absorbed or emitted by atoms and molecules found

A

in ultra violet and visable regions of the EM spectrum

89
Q

The Balmer Series

A

If the excited electron falls back into the n=2 energy level (second shell), then the energy of light emitted is in the visible region of the electromagnetic spectrum and appears as coloured lines.

You need to remember the origin of the first four lines in the visible emission spectrum.

90
Q

The Lyman Series

A

If the excited electron falls back into the n=1 energy level (first shell), then the energy of light emitted is in the ultraviolet part of the electromagnetic spectrum and thus can only be detected electronically, such as with a UV-Vis spectrometer.

91
Q

Examination of the Lyman series gives us information about two key features of the hydrogen atom:]

Convergence Limit

A

The Convergence Limit
-spectral lines become closer and closer as frequency of radiation increases until they converge to a limit

-Within the Lyman series, the frequency of this convergence limit corresponds to the energy required to remove the electron (i.e. the ionisation energy).

-The lines get closer together toward the convergence limit because the energy levels get closer together the further away they are from the nucleus.

92
Q

Examination of the Lyman series gives us information about two key features of the hydrogen atom:]

Ionisation Energy

A

Transition between n=1 level and the convergence limit, where the energy levels are so far from the nucleus that they are effectively the same energy. At this point, the electron is effectively removed from the atom and therefore corresponds to ionisation.

Therefore, the transition from n=1 to n = ∞ corresponds to the atom losing the electron completely. The ionisation energy of the hydrogen atom corresponds exactly to the highest frequency line in the Lyman series of the hydrogen atom spectrum since they both refer to exactly the same process.

Measuring the convergent frequency allows the ionisation energy to be calculated from E = hf.

93
Q

When the energy levels in hydrogen get closer in energy as they get further from the nucleus, it is known as the

A

convergence limit

94
Q

The frequency associated with the convergence limit in the Lyman series can be measured and used to calculate the

A

ionisation energy

95
Q

The series of lines produced when the electron in hydrogen drops down to the n = 1 energy level is known as the

A

Lyman Series

96
Q

The series of lines produced when the electron in hydrogen drops down to the n = 2 energy level is known as the

A

Balmer Series

97
Q

effect of bond energy (KJ mol -1) on absorbtion wavelentgh (nm)

A

lower bond energy, bond can absorb lower energy photons, correspond to longer wavelentghs

98
Q

absorbtion range relation to colour

A

if absorbtion within visible range

will absorb specific wavelengths of visible light

unabsorbed wavelengths of visible light will be transmitted or reflected

99
Q

Explain how the Lyman series can be used to calculate the ionisation energy of
hydrogen

A

The ionization energy of hydrogen is the energy required to completely remove an electron from the hydrogen atom, i.e., to transition the electron from the ground state (n=1) to n=∞, where the electron is no longer bound to the atom.

100
Q

hydrogen emission spectrum frequency and wavelength

A

2. Chemistry A-level (7 decks)

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