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Atomic Structure
C
hem
F
actsheet
September 2000 Number 01
Fundamental particles
An atom is the smallest particle of a chemical element. Atoms themselves consist of protons, neutrons (in the nucleus) and electrons (see Fig 1). The mass of an atom is concentrated in the nucleus. The nucleus is very small and massive, so therefore has an incredibly high density.
To succeed in this topic you need to:
- be able to use the periodic table to find atomic and mass numbers.
After working through this Factsheet you will be able to:
- describe and understand the accepted model of an atom
- interpret simple mass spectra
- understand the importance of ionisation energies of different elements, and use them as evidence in describing atomic structure
- assign electronic configurations to atoms
- understand the impact of electronic configuration on chemical properties
- define electron affinity
Fig 1. Fundamental particles of an atom
electrons
nucleus (contains protons and neutrons
Table 1 Shows the relative charge , relative mass and position of atomic particles. (For most purposes, the mass of the electron is taken as 0).
Particle
Proton
Neutron
Electron
Relative mass
Relative charge
Table 1. Atomic particles data
Position within atom
nucleus
nucleus
shells
The particles within a nucleus are drawn together by extremely powerful forces capable of overcoming the repulsion of the protons (+ve to +ve) However, these forces act over a short distance as they do not pull the electrons in.
The electrons are in constant motion, orbiting the nucleus.
Atomic number, atomic mass number and isotopes
Be
9
4
mass number
atomic number
Atomic number = 4 ∴ number of protons = 4 No. protons = no. electrons ∴ number of electrons = 4 Mass number = no. p + no. n ∴ number of neutrons = 9 – 4 = 5
All atoms of the same element have the same atomic number - e.g atomic number of magnesium = 12, therefore all magnesium atoms contain 12 protons.
Mass numbers of atoms of the same element may, however, vary – due to different atoms containing different numbers of neutrons. These atoms are called isotopes.
Exam Hint: A thorough understanding of atomic structure is essential for success throughout AS Chemistry. Questions often require candidates to
- work out the electronic configuration of an atom
- deduce chemical properties from electronic configuration
- describe and explain trends in ionisation energy
- interpret mass spectra
Atomic number (Z): the number of protons in an atom Mass number (A): the number of protons + neutrons in an atom.
Remember also that the number of protons (+ve charges) will be equal to the number of electrons (-ve charges) in a neutral atom.
Given the information on the periodic table, it is possible to calculate the number of protons, neutrons and electrons present in an atom of any given element. Fig 2.
Fig 2.
Remember - Isotopes of one particular element are atoms which have the same atomic number and so the same number of protons (therefore are atoms of the same element) but different mass numbers, because they have different numbers of neutrons.
One example is hydrogen - there are three isotopes, all with atomic number 1, but with mass numbers 1, 2 and 3.
Elements with isotopes do pose a problem when wanting to assign a mass number on for the element. An average atomic mass is calculated - called the relative atomic mass, A (^) r. The calculation of this average mass needs to take account of the relative abundance of each isotope; the method for doing this is illustrated in the following example:
Atomic Structure (^) Chem Factsheet
For example, naturally occuring chlorine consists of 2 isotopes. 75% of the atoms have a mass of 35 (Cl-35) 25% of the atoms have a mass of 37 (Cl-37)
To work out A (^) r, we do
The actual definition of relative atomic mass involves carbon - 12 - this must be learnt.
Ionisation Energies
If an atom is supplied with enough energy, it will lose an electron, and additional supplies of energy may cause the loss of a second electron, then a third and so on. If a neutral atom loses an electron, it becomes a positively
charged ion (cation).
Relative Atomic Mass (Ar ) = mass of one atom of an element 1/12 mass of one atom of carbon–
75 100 ×^ 35 +^ ×^ 37 =^ 35.
25 100
Carbon – 12 is used to because carbon is a common element and as a solid is easy to store and transport.
The mass spectrometer
The mass spectometer is a machine which provides chemists with a way to measure and compare masses of atoms and molecules (Fig 3.)
Fig 3. Mass Spectrometer
The way in which the mass spectrometer works can be broken down into 5 stages.
1. Vaporisation - The sample being tested has to be turned into a gas, so individual atoms/molecules are separated. 2. Ionisation - A heated filament gives out electrons into the ionisation chamber. As the sample enters the ionisation chamber, its atoms/ molecules are bombarded by these electrons. The collisions cause electrons to be removed from the atoms/molecules of the sample, so positive ions are formed. (This is where fragmentation can occur – molecules may break into pieces.) 3. Acceleration - An electric field is applied, which will accelerate the positive ions (as they are charges particles). 4. Deflections - A magnetic field deflects the beam of ions. Ions with a high mass/charge ratio (eg. Heavy, 1+ charge) will be deflected less than ions with a low mass/charge (e.g. light 13+ charge) 5. Detection - Those ions which have the correct mass/charge ratio will be detected. If the magnetic field is kept constant whilst the electric field causing acceleration is continuously varied, one species after another will be detected, so a complete spectrum, or trace, is obtained.
From the mass spectra shown, it is clear that the sodium sample tested consisted solely of sodium –23.
The sample of iron was a mixture of 4 isotopes, iron –54, iron –56, iron –57 and iron –58. The percentage abundancies are given (i.e. 91.68% iron –56, showing it to be the most common isotope) so now the relative atomic mass of iron can be calculated.
Ar = avge mass =
Note:many ions have a charge of +1, so the mass/charge (m/e) ratio is equal to the mass (m) of the ion.
Fig 4. Mass spectrometer traces
100
100
100
× 54 + 91.68× 56 + × 57 + × 58= 55.
100
Examining successive ionisation energies for an element can give us further insight into atomic structure – specifically the arrangement of electrons (Fig 6 overleaf)).
First Ionisation Energy - the energy required to remove 1 mole of electrons from 1 mole of gaseous atoms. e.g. Na(g) → Na+^ (g) + e- ∆ HIE = +494 KJ mol -
Second Ionisation Energy - the energy required to remove 1 mole of electrons from 1 mole of gaseous 1+ charged cations e.g. Na+(g) → Na 2+(g) + e- ∆ H (^) IE2 = +4564 KJ mol-
5
12345678 (^1234567812345678) 12345678 (^1234567812345678) (^1234567812345678)
(^12345678901234561234567890123456) (^12345678901234561234567890123456) (^12345678901234561234567890123456) (^12345678901234561234567890123456) (^12345678901234561234567890123456) 1234567890123456 (^12345678901234561234567890123456) (^12345678901234561234567890123456)
○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○
- vaporisation
- ionisation
- accleration (by electric field)
- deflection (by magnetic field)
- detection
T o vacuum pump
% abundace
0 1 0 2 0^ 3 0^ 4 0 7 0 8 0
100
0
Mass / Charge Ratio
2 0
4 0
6 0
8 0
sodium (Na)
2 0 3 0 4 0 5 0 6 0 7 0 (^) 8 0
100
5 0
0
Relative Abundance / %
Mass / Charge Ratio
(91.68%)
Sodium
Iron
(2.17%)
(0.31%)
Chem Factsheet Atomic Structure
Order of filling of sub-shells
We already know that the electrons in an atom fill up the first shell before going into the second, and fill that before going into the third shell. We now need to look at the order in which the sub-shells fill up. This is described by Fig 9.
Fig 9. Order of filling of sub-shells
Using the diagram, the order in which the subshells fill is:
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p .... etc
We are now able to write out electronic configurations in more detail, using the following notation:
or using arrows in boxes
Using the principles covered so far, we can now find the electronic configurations of the first 36 elements (Table2)
Evidence for sub-shells from ionisation energies
Whilst there is a general increase in IE across the period, this is not a smooth trend, which can be explained by the existence of sub-shells.
2p
shell number subshell
number of electrons in subshell
Fig 10. First ionisation energies across period 2
1s
2s
3s
4s
5s
6s
2p
3p
4p
5p
6p
3d
4d
5d
6d
4f
5f
atomic number
first ionisation energyLi
Be
B
C
N
O
F
Ne
2p sub-shell half-fills
pairing of 2p electrons
filling of 2p sub-shell
filling of 2s sub-shell
Element Atomic 1s 2s 2p 3s 3p 3d 4s 4p number
Hydrogen 1 1 Helium 2 2 !s level full
Lithium 3 2 1 Beryllium 4 2 2 2s level full
Boron 5 2 2 1 Carbon 6 2 2 2 Nitrogen 7 2 2 3 Oxygen 8 2 2 4 Fluorine 9 2 2 5 Neon 10 2 2 6 2p level full
Sodium 11 2 2 6 1 Magnesium 12 2 2 6 2 3s level full
Aluminium 13 2 2 6 2 1 Silicon 14 2 2 6 2 2 Phosphorus 15 2 2 6 2 3 Sulphur 16 2 2 6 2 4 Chlorine 17 2 2 6 2 5 Argon 18 2 2 6 2 6 3p level full
Potassium 19 2 2 6 2 6 1 Calcium 20 2 2 6 2 6 2 4s level full
Scandium 21 2 2 6 2 6 1 2 Titanium 22 2 2 6 2 6 2 2 Vanadium 23 2 2 6 2 6 3 2 Chromium 24 2 2 6 2 6 5 1 ** Manganese 25 2 2 6 2 6 5 2 Iron 26 2 2 6 2 6 6 2 Cobalt 27 2 2 6 2 6 7 2 Nickel 28 2 2 6 2 6 8 2 Copper 29 2 2 6 2 6 10 1 ** Zinc 30 2 2 6 2 6 10 2 3d level full
Gallium 31 2 2 6 2 6 10 2 1 Germanium 32 2 2 6 2 6 10 2 2 Arsenic 33 2 2 6 2 6 10 2 3 Selenium 34 2 2 6 2 6 10 2 4 Bromine 35 2 2 6 2 6 10 2 5 Krypton 36 2 2 6 2 6 10 2 6 4p level full
Table 2. Electronic arrangement of the elements from hydrogen
to krypton
Note: a) how the 4s level fills up before the 3d level, b) for chromium and copper (labelled **) the sequence is out of step; you will meet this point again when studying the transition elements. You will be required to write the electronic arrangement (or configuration ) for elements and there are two accepted ways for doing this. For example, the electronic arrangement for iron could be written: 1s^2 , 2s 2 , 2p^6 , 3s^2 , 3p^6 , 3d^6 , 4s 2 or [Ar] 3d^6 , 4s 2 where [Ar] represents the electronic arrangement of the noble gas argon.
2p
Chem Factsheet Atomic Structure
You will notice from Fig 10 that the first ionisation energy decreases from Be to B and from N to O. A similar phenomenon occurs between groups 2 to 3 and 5 to 6 in the other periods. This can be explained by examining the electron configurations of the elements:
Comparing Be and B
Beryllium has full-subshell stability , as the highest occupied subshell is complete. Boron has one electron in a higher (2p) sub-shell, which is easier to remove, hence its 1st^ IE is lower than that of Be
Comparing N and O
Nitrogen has half-shell stability.
Oxygen has one 2p orbital which has a pair of electrons and paired electrons repel, so one of these electrons is easier to remove, hence it has a lower 1 st IE than that of nitrogen.
Electronic Structure and Chemical Properties
Chemical reactions involve the making and/or breaking of bonds. Bond involve the movement of electrons.
e.g. Covalent bond - sharing of electrons Ionic bond - transfer of electrons
It makes sense therefore that the electronic configuration of an atom has an impact on its chemical properties.
Any group on the periodic table can be considered a ‘family of elements’ as the elements in that group will exhibit similar chemical properties. This is due to each member of the group having the same number of electrons in its outer shell.
Even within groups, trends in reactivity can be explained by electronic configuration:
e.g.1 For Group 1 – the alkali metals – reactivity increases down the group.
If we consider lithium and sodium, their electron configurations are as follows: Li: 1s 2 2s^1 Na: 1s^2 2s^2 2p 6 3s 1
Both Li and Na require to lose 1 electron to gain a stable noble gas electronic configuration. It is difficult to remove the outer electron from lithium as it is close to the nucleus and experiencing little shielding. These factors mean lithium:
- has a higher 1 st^ IE than sodium
- is less reactive than sodium This trend continues down the group.
e.g.2 For Group 7 – the halogens – reactivity decreases down the group.
If we consider fluorine and chlorine, their electron configurations are as follows: F: 1s^2 2s^2 2p^5 Cl: 1s 2 2s 2 2p 6 3s 2 3p^5
Both F and Cl need to gain 1 electron to acquire a stable noble gas electronic configuration. An electron would be more strongly attracted to F rather than the larger Cl, as it could join a lower energy level closer to the nucleus with less shielding from it. Therefore, the smaller the halogen atom, the more reactive it is. So reactivity decreases down the group.
Electron Affinities
As shown with the halogens, it is possible to add electrons to an atom, forming a negative ion (anion). The energy required to do this is the electron affinity.
First Electron Affinity - the energy required to add1 mole of electrons to 1 mole of gaseous atoms. e.g. O (g) + e- → O^ −^ (g) ∆ HEA1 = -141 kJ mol-
Second Electron Affinity - the energy required to add 1 mole of electrons to 1 mole of gaseous 1- charged anions e.g. O^ −^ (g) + e- → O^2^ −^ (g) ∆ HEA2 = +798 kJ mol -
The 1 st^ EA is always exothermic (energy is released) because the electron goes into a vacancy in the outer energy level. This is ‘bond-making’ so energy is released.
However, this creates a 1- ion so to put the second electron into the vacant site needs energy to be put in to overcome the repulsion (–ve to –ve) between the ion and the electron - so the 2 nd^ EA is always endothermic (energy is absorbed).
Questions
- a) Define the terms first ionisation energy.
b) The graph shows a plot of lg (ionisation energy) vs number of the electron removed for aluminium.
Explain the form of this graph in terms of the electron structure of aluminium.
- a) Define the terms (i) atomic number (ii) mass number (iii) relative atomic mass (iv) isotope
b) Describe in detail the five stages in the operation of a mass spectrometer.
c) The mass spectrometer analysis of neon shows it exists of two isotopes of different relative abundances ie. 20 (90%) and 22 (10%). Calculate the relative atomic mass of neon.
number of electrons removed
lg(ionisation energy)
Be
1s 2s 2p B
1s 2s 2p
N
1s 2s 2p O
1s 2s 2p
Moles and Formulae
C
hem
F
actsheet
September 2000 Number 02
To succeed with this topic you need to:
- be able to find Ar (atomic mass number) values from the Periodic Table
- use a calculator to do basic arithmetic
After working through this Factsheet you will be able to:
- calculate M (^) r (molecular mass number) from Ar values
- calculate percentage composition by mass for a compound
- calculate moles from grams and grams from moles of a substance
- calculate empirical formulae using a variety of different methods
- convert empirical formulae into molecular formulae
Examination guide
The calculations covered by this Factsheet can appear in nearly every module of the AS and A2 specification. The concepts and methods introduced are the basis of all quantitative chemistry and it is vital you can handle them.
1. Finding Mr (relative molecular mass or formula mass)
The M (^) r of a compound is found by adding up the relative atomic masses (Ar) of the elements in the compound's formula.
Remember: The A (^) r value is found from the Periodic Table.
11
23 Na
A (^) r value
Atomic number Hint: Ar is always the larger of the two number in the boxes
Example 1: What is the M (^) r of C 2 H 6?
C 2 H 6 includes 2 C atoms and 6 H atoms. So M (^) r (C 2 H 6 )= 2 × Ar(C) + 6 × Ar(H) (Ar = 12 for C and 1 for H) ∴ Mr = (2 × 12) + (6 × 1) = 30
Example 2: What is the Mr of Ca(NO 3 ) 2?
NB: The small 2 outside the brackets multiplies everything inside the bracket - just like in maths.
So Ca(NO 3 ) 2 includes 1 Ca atom, 2 N atoms and 6 O atoms. So M (^) r(Ca(NO 3 ) 2 )= Ar(Ca) + 2 × A (^) r(N) + 6 × Ar(O) (Ar = 40 for Ca, 14 for N and 16 for O) ∴ Mr = 40 + (2 × 14) + (6 × 16) = 164
2. The percentage composition of a compound
You may be asked to find (for example) the percentage of sodium nitrate that is nitrogen. N.B this is a commonly asked examination question!
Example: What is the percentage by mass of each of the elements present in C 2 H 5 Br?
M (^) r (C 2 H 5 Br) = (12 × 2) + (1 × 5) + (80 × 1) = 109
% C =
×
× 100 = 22.02 %
% H =
× 5
× 100 = 4.59 %
% Br =
× 1
× 100 = 73.39 %
3. Moles
Definition: One mole is the amount of substance which contains the same number of particles (atoms, molecules or ions) as there are atoms in 12.00g of 12 C
Why that particular number? Avagadro's number is chosen to "fiddle" it so that one mole of a substance has mass (in grams) equal to the Aror M (^) r of that substance. So as hydrogen atoms have A (^) r = 1, one mole of hydrogen atoms will have mass one gram. Similarly, as C 2 H 6 has M (^) r = 30, one mole of C 2 H 6 will have mass 30 grams. This makes moles easier to work with!
Definition of a mole The definition of a mole given below probably seems a bit odd, but it is the one that must be given in an exam!
The important thing is to be able to use the mole in calculations. The basic equation is:
Number of mass (g) moles =^ Ar or M (^) r
or rearranged
Mass (g) = moles × Ar or Mr
Method Step 1:- find M (^) r (by totalling A (^) r values)
Step 2:- find % of an element using:
no. of atoms of element A M
r r
×
× 100%
Calculating with Moles
Exam Hint: - Be careful what it's a mole of! A mole of hydrogen atoms (H) is not the same as a mole of hydrogen molecules (H 2 ). A mole of hydrogen molecules contains 2 moles of hydrogen atoms! If a question refers to a mole of an element, it means a mole of molecules of that element.
Check! these should add up to 100%
What is a mole?
A mole of something is just 6.023 × 1023 of it. A mole of hydrogen atoms
is 6.023 × 10 23 hydrogen atoms, a mole of water molecules is 6.023 × 1023
water molecules - you could even imagine a mole of people or a mole of
cars! The number 6.023 × 1023 is called the Avagadro Number - you do
NOT have to learn it!
A (^) r or Mr= mass (g) moles
Moles and Formulae (^) Chem Factsheet
Example 2: How many moles are there in 36g of H 2 O?
Since Mr is in the formula, we must calculate this first
Mr (H 2 O) = (1 × 2) + (16 × 1) = 18
moles =
mass(g) M (^) r^ =^
= 2 moles
Example 3: What is the mass of 0.5 moles of H 2 S?
Mr (H 2 S) = (1 × 2) + (32 × 1) = 34
mass(g) = moles × M (^) r = 0.5 × 34 = 17 g
4. Empirical and Molecular Formulas For calculation purposes there are 2 types of formulae you need to know:
- The empirical formula (ef) shows the ratio of the atoms present in their lowest terms i.e. cancelled down to smallest whole numbers.
- The molecular formula (mf) shows the actual number of each type of atom present in one molecule.
Finding Empirical Formulae
There are several ways to calculate empirical formulae, and these are shown below in order of increasing difficulty:-
1. Calculating EF from Moles What is the EF of the compound formed when 6 moles of potassium atoms react with 3 moles of oxygen atoms?
K:O Moles 6 : Simplest ratio 2 :1 (divided by 3) EF = K 2 O
2. Calculating EF from Mass What is the EF of the compound formed when 6g of carbon reacts with 32g of sulphur?
First find moles:
moles C =
= 0.5 Moles S =
C : S
moles 0.5 : 1 (now divide by 0.5 - the smaller number) Simplest ratio 1 : 2 EF =CS (^2)
3. Calculating EF from Percentage Composition
NB. This is the most commonly examined method of finding EF. The approach is exactly the same as calculating from mass; you treat the percentages as if they are masses. One method of approaching these is using a table, as shown below - but you must use whichever style of presentation you are most comfortable with.
What is the empirical formula for the compound which contains the following elements by percentage composition of mass? C = 66.67%, H = 11.11%, O = 22.22%
*To find the ratio column, take the smallest of the %÷A (^) r values (which is 1.39 here) and divide all the %÷Ar values by it.
EF = C 4 H 8 O
4. Calculating EF from Combustion Data This method is one step up in difficulty from the last example because you have to calculate the masses of the elements first. The combustion products are always oxides. CO 2 and H 2 O are the commonest and the following example uses these, although the method can be adapted for others.
Example 4: 0.2 moles of a metal have a mass of 4.6g. i) Calculate the element's atomic mass ii) Suggest an identity for the metal.
i) A (^) r =
mass(g) moles
=
ii) Looking at the Periodic Table, we see that sodium has an atomic mass of 23, and it is a metal.
Challenge Question : Does the metal HAVE to be sodium?
Answer: No. It is the most likely answer, but isotopes of other
metals could have atomic mass 23.
Some students find the 'triangle method' useful in remembering and rearranging equations.
Mass
Moles × Mr
You cover up the thing in the triangle you want to find. Then, what you can see tells you the calculation to do. For example, if you want to find moles, cover it up and you are left with mass/M (^) r
This method can be used for ANY equation that has a fraction on one side and just one thing on the other side. Whatever is on the top of the fraction goes in the top of the triangle. Here are some examples:
Method Step1:- find mass of the carbon and hydrogen using
mass of oxide ×
no. of atoms of element × Ar M (^) r for oxide Find the mass of any other element in the compound by subtraction.
Step 2:- convert mass to moles for each element by dividing by A (^) r
Step 3:- find the simplest ratio by dividing all the values from step 2 by the smallest of them.
Example 1: How many moles are there in 6g of C?
moles =
mass(g) A (^) r
(A (^) r = 12 for C)
∴ (^) moles = 6 12
= 0.5 moles
Element % Ar % ÷Ar Ratio * C 66.67 12 5.56 5.56 ÷1.39 = 4 H 11.11 1 11.11 11.11 ÷1.39 = 8 O 22.22 16 1.39 1.39 ÷1.39 = 1
Chem Factsheet
- Calculating Mr values from mass and moles What is the M (^) r value for each of the following compounds? (a) 1.0g of compound A contains 0.0208 moles (b) 1.5 moles of compound B has a mass of 105g (c) 14.8g of compound C contains 0.117 moles (d) 7.0g of compound D contains 0.219 moles (e) 0.24 moles of compound E has a mass of 13.92g
- Find the percentage composition by mass of elements in a compound What is the percentage composition by mass of each element in the following compounds? (a) SiCl 4 (b) C 2 H 6 (c) Na 2 CO 3 (d) CaBr 2 (e) CuSO 4 .5H 2 O
- Calculating empirical formula from moles What is the empirical formula of compounds with the following composition? (a) 2 moles Na with 2 moles I (b) 0.1 moles K with 0.05 moles O (c) 0.5 moles N with 1.5 moles H (d) 0.2 moles Mg with 0.4 moles Cl (e) 1.2 moles of a carbon oxide contains 0.4 moles of carbon
- Calculating empirical formula from mass What is the empirical formula of compounds with the following composition by mass? (a) 12g C with 16g O (b) 6g Mg with 4g O (c) 46g Na with 80g Br (d) 14g N reacting with H to form 17g of compound (e) 22g Sr reacting with O to form 26g of compound
- Calculating empirical formula from percentage composition What is the empirical formula of each of the following compounds? (a) 80% C, 20%H (b) 52.2% C, 13.1% H, 34.7% O (c) 40.4% C, 7.9% H, 15.7% N, 36.0% O (d) 38.7% C, 9.7% H, 51.6% S (e) 40.2% K, 26.9% Cr, 32.9% O (f) 85.25% BaCl 2 , 14.75% water of crystallisation
- Calculating empirical formula from combustion data What is the empirical formula of each of the following compounds? (a) Complete combustion of 1.0g of a compound produced 2.99g CO 2 and 1.64g H 2 O (b) 1.0g of a compound underwent complete combustion and produced 3.035g CO 2 and 1.55g H 2 O (c) 2.0g of a compound produced 5.86g CO 2 and 3.6g H 2 O on complete combustion (d) A compound made of carbon, hydrogen and oxygen produced 2.2g CO 2 and 1.2g H 2 O when 1.0g of it underwent complete combustion
- Finding molecular formula from empirical formula and M (^) r What is the molecular formula of the following? (a) E.F. = CH Mr = 72 (b) E.F. = C 2 H 2 O Mr = 42 (c) E.F. = C 2 H 3 Br Mr = 214 (d) E.F. = CH 2 O Mr = 120 (e) E.F. = NaO Mr = 78
Moles and Formulae
Answers
- (a) 2 (b) 0.5 (c) 0. (d) 3 (e) 2 (f) 0. (g) 2 (h) 0.25 (i) 0. (j) 2 (k) 0.078 (l) 0. (m) 2.11 (n) 0.048 (o) 0.
- (a) 160g (b) 20g (c) 18g (d) 10.5g (e) 112g (f) 80g (g) 1778g (h) 80 (i) 193.5g (j) 3g (k) 3.24g (l) 70g (m) 11.02g (n) 2587.5g (o) 89.7g
- (a) 207 (b) 32 (c) 32 (d) 40 (e) 24
- (a) 44 (b) 18 (c) 98 (d) 80 (e) 16 (f) 58 (g) 46 (h) 84 (i) 188. (j) 170 (k) 286 (l) 249. (m) 123 (n) 248 (o) 183
- (a) 0.5 (b) 3 (c) 2 (d) 0.5 (e) 0.5 (f) 0. (g) 1 (h) 0.071 (i) 0. (j) 1.74 (k) 0.44 (l) 0. (m) 0.11 (n) 0.063 (o) 0.
- (a) 112g (b) 9.24g (c) 280g (d) 42g (e) 6g (f) 12.4g (g) 50g (h) 98.55g (i) 40.95g (j) 1096g (k) 450.8g (l) 0.36g (m) 398g (n) 42.75g (o) 53.2g
- (a) 48 (b) 70 (c) 126. (d) 32 (e) 58
- (a) 16.57% Si, 83.43% Cl (b) 80 % C, 20 % H (c) 43.40%Na, 11.32%C, 45.28% O (d) 20 % Ca, 80 % Br (e) 25.45% Cu, 12.83% S, 4.00 % H, 57.72% O
- (a) Na I (b) K 2 O (c) NH (^3) (d) MgCl 2 (e) CO (^2)
- (a) CO (b) MgO (c) Na 2 Br (d) NH 3 (e) Sr O
- (a) CH 2 (b) C 2 H 6 O (c) C 3 H 7 NO 2 (d) K 2 CrO 4 (e) BaCl 2. 2H 2 O
- (a) C 3 H 8 (b) C 2 H 5 (c) CH (^3) (d) C 3 H 8 O
- (a) C 6 H 6 (b) C 2 H 2 O (c) C 4 H 6 Br (^2) (d) C 4 H 8 O 4 (e) Na 2 O (^2)
Acknowledgements: This Factsheet was researched and written by Sam Goodman. Curriculum Press, Unit 305B, The Big Peg, 120 Vyse Street, Birmingham, B18 6NF ChemistryFactsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 1351-
Moles and Equations
C
hem
F
actsheet
September 2000 Number 03
Writing chemical formulae
Chemical formulae fall into three main categories:
a) Formulae which must be learnt :
Water H 2 O Sulphuric acid H 2 SO (^4) Oxygen O 2 Hydrochloric acid HCl Nitrogen N 2 Nitric acid HNO (^3) Ammonia NH 3 Phosphoric acid H 3 PO (^4) Ozone O 3 Chlorine Cl (^2) Argon Ar Bromine Br (^2) Neon Ne Iodine I (^2) Hydrogen H (^2)
N.B. Use the Periodic Table whenever you can for elements eg. Mg, Fe, Na, etc., but notice those elements in the list above are diatomic (two atoms in a molecule) - O2, H (^) 2, and the Halogens (Cl (^) 2, Br2, I 2 ). Noble gases (Ne, Ar, etc.) are monatomic.
b) Formulae that can be worked out from their names alone:
The list of terms used is shown below mono=1 penta=5 octa= di=2 hexa=6 nona= tri=3 septa=7 deca= tetra=
Carbon dioxide CO 2 Carbon monoxide CO Sulphur trioxide SO 3 Phosphorus pentachloride PCl (^5) Dinitrogen trichloride N 2 Cl 3 Sulphur dioxide SO (^2)
NB: Hydrocarbons have a different system of naming eg methane,CH 4 ethane, C 2 H 6 This is covered in the Factsheet No. 13.
To succeed with this topic, you need to:
- ensure you understand the work on ‘moles’ from Factsheet No. 2
- learn thoroughly the valencies of the commonest cations and anions
- learn thoroughly those common formulae you are expected to know
- practice writing formulae and balancing equations because unless these are correct your calculations will always give the wrong answers!
After working through this Factsheet, you will understand:
- how to write chemical formulae
- putting formulae together in an equation to describe a chemical reaction
- balancing chemical equations
- calculating quantities from balanced equations
- using molar volumes of gas in equations
- writing ionic equations
Exam Hint: Writing formulae and balanced chemical equations is central to all questions at AS level. The time spent working on these will repay you in terms of marks and grades.
c) Formulae that can be worked out from the charge on their ions :
These compounds usually contain metals (the cations) and non-metals (the anions). You are expected to know these cations and anions; some have to be learnt, but you can use the Periodic Table to help you for the ions of elemental atoms - the box below reminds you how to do this.
Using the Periodic Table to help you find the charges on ions
- Group 1 form ions with charge +
- Group 2 form ions with charge + 2 (but beryllium compounds may not be ionic)
- Group 6 form ions with charge -
- Group 7 form ions with charge -
The table below contains the commonest ions at AS level, but more are used as you progress through the course. You must learn these - and you will find later work much easier if you do it now, rather than waiting until the exam.
Table 1. Cations and Anions for AS-level
ANIONS
Ions that can be worked out from Periodic Table rules above Name Formula chloride Cl – bromide Br – iodide I – oxide O 2- sulphide S 2-
Other Anions Name Formula hydroxide OH – nitrate (V) NO 3 – nitrate (III) NO 2 – cyanide CN – hydrogencarbonate HCO 3 – hydrogensulphate HSO 4 – carbonate CO 3 2 – sulphate (IV) SO 3 2 – sulphate (VI) SO 4 2 – phosphate PO 4 3 – manganate (VII) MnO 4 –
CATIONS
Ions that can be worked out from Periodic Table rules above Name Formula lithium Li + sodium Na+ potassium K+ magnesium Mg2+ calcium Ca2+ strontium Sr2+ barium Ba2+
Other Cations Name Formula hydrogen H+ zinc Zn 2+ aluminium Al3+ silver Ag+ cobalt Co 2+
- copper Cu+/Cu2+
- iron Fe 2+/Fe3+
- lead Pb 2+^ /Pb4+
- manganese Mn 2+^ /Mn 4+ ammonium NH 4 +
- = elements with more than one valency For these, Roman Numerals are used to show which valency is being used - eg copper (II) hydroxide contains Cu 2+^ , whilst copper (I) oxide contains Cu+ For non-metals, this refers to the oxidation state (see Factsheet 11) of the non-metal involved - eg sulphate (VI) contains sulphur in the +6 oxidation state
Exam Hint: If a question uses compound names with roman numerals
- eg sodium sulphate (IV) - make sure you take note of them! Many students lose marks through assuming sodium sulphate (IV) is Na 2 SO 4 , rather than Na 2 SO (^3)
Moles and Equations (^) Chem Factsheet
(c) Using the description to find the word equation
The process is:
description word balanced chemical of reaction equation equation
Sometimes the description gives you all the reactants and products, and in other cases you have to apply your knowledge of chemical reactions to find the products. Both types are shown in the following examples:
eg 1. Calcium hydroxide is the only product when calcium oxide dissolves in water.
Calcium oxide + water ‘ calcium hydroxide CaO + H 2 O ‘ Ca(OH) (^2)
eg 2. Sulphuric acid reacts with potassium hydroxide
Acid + Alkali ‘ salt + water
Sulphuric + potassium ‘ potassium + water acid hydroxide sulphate
H 2 SO 4 + KOH ‘ K 2 SO 4 + H 2 O H 2 SO 4 + 2KOH ‘ K 2 SO 4 + 2H 2 O
Question 4 at the end provides further practise with writing equations.
Calculation work based on equations
All calculation work must be based on balanced chemical equations - ones that have the same number of atoms of each type on each side of the arrow. Any calculation based on an unbalanced equation will automatically be wrong!
Calculation work also requires you to use moles. You need to ensure you are happy converting between masses and moles (see Factsheet 2 - Moles and Formulae) before you go any further.
e.g. 2Mg + O 2 ‘ 2 MgO
The numbers in front of the formulae tell us the mole ratio in the reaction
- for every 2 moles of magnesium, we need 1 mole of oxygen, and we will produce 2 moles of magnesium oxide.
So if you wanted to react 4 moles of magnesium, you'd need 2 moles of oxygen and you would get 4 moles of magnesium oxide. Similarly, if you reacted 1 mole of magnesium, you'd need 0.5 moles of oxygen and you'd get 1 mole of magnesium oxide.
In all mass calculations based on equations, you must always follow these steps: 1 Write a balanced equation 2 Work out the mass ratio 3 Use the mass ratio, together with the information given in the question, to find the unknown masses.
The following examples illustrate how to do this:
eg 1. How much magnesium oxide will be made by burning 12g of magnesium?
2Mg + O 2 ‘ 2MgO Mass ratio 48 32 80 Actual mass 12?
To work out the required mass of magnesium oxide, we use the fact that 48:80 and 12:? are in the same ratio. One easy way of dealing with ratios is using the "cross method", shown in the box below
Tip: The equation can never tell you how much of a substance is actually reacting - that depends on how much of the chemicals you decide to use! It only tells you the ratios
We can use these mole ratios to find out mass ratios - which are what we need to work with to find masses. To find these, we need to use the equation mass = moles ××××× Mr. So for the equation above, we have:
2Mg + O 2 ‘ 2 MgO Mole ratio 2 1 2 Multiply by Mr : 2 × 24 1 × 32 2 × 40 Mass ratio 48 32 80
Note that the masses balance - the total is 80 on both sides of the equation.
So mass of MgO = 20g
eg 2. What mass of calcium carbonate is needed to make 0.12g of calcium oxide? (Mr values: Ca = 40, C = 12, O = 16)
CaCO 3 ‘ CaO + CO (^2) 1 mole ‘ 1 mole + 1 mole Mass ratios 100 ‘ 56 + 44 Actual mass? 0.
So our ratios are 100 : 56 and? : 0.
So? = 100 × 0.12 ÷ 48 = 0.25g
Gas Molar Volumes
Up to this point state symbols have not been used in any of the equations: (s) = solid (l) = liquid (g) = gas (aq) = solution (i.e. dissolved in water ≡ ‘aqueous’) They are important and in effect complete any balanced chemical equation. They are used in examination questions, and are useful because you may need to specify that something is precipitated (so it will have an (s) not an (aq)), and, most importantly in this section, because some formulae - and hence some methods - only work for gases! The key fact is:
Working with ratios using the "cross method" This method involves 3 easy steps: 1 Write down the two ratios underneath each other, putting in a? for the number you don't know. 2 Draw a cross 3 Multiply the two joined numbers and divide by the other one.
Using the example above, our two ratios are:
So we do 12 × 80 ÷ 48 = 20 If you are not happy working with ratios, see Factsheet 14: Maths for Chemists 1
1 mole of any gas has a volume of 24 dm^3 (24000cm 3 ) at room temperature and pressure (rtp) - which is 1 atmosphere and 25o^ C
At standard temperature and pressure (stp) - which is 1 atmosphere and 0 oC - the volume of 1 mole of any gas is 22.4dm 3 (22400cm^3 )
Moles and Equations (^) Chem Factsheet
There are two types of questions that use this fact:
- Equations involving only gas volumes
The method here relies on the fact that mole ratio = gas volume ratio So the steps are:
1 Write a balanced equation 2 Write down the mole ratio 3 Use the mole ratio, together with the gas volume information given in the question, to find the unknown volume.
eg 1. What volume of SO 3 would be made from 200cm^3 O 2 reacting with SO 2?
2SO 2 (g) + O 2 (g) ‘ 2SO 3 (g) mole ratio 2 1 2 volumes 200?
So 1: 2 and 200 :? are in the same ratio.
So? = 200 × 2 ÷ 1 = 400cm^3
- Equations involving gas volumes and masses. Here, we cannot rely on simple ratio methods. We need to convert between masses/volumes and moles, then work with moles and the mole ratio. NB: This approach will also work with all the previous types of calculation, so if you'd rather remember just one method, use this one! The procedure is:
1 Write a balanced equation 2 Write down the mole ratio 3 For the substance you have information about, work out how many moles of it there are 4 Use the mole ratio to find out how many moles there are of the substance you're asked about 5 Find out the mass or volume you're asked for using the correct moles equation
You may find the "triangles" below helpful:
eg 1. What volume of CO 2 would be made by heating 20g of CaCO 3?
So we have 0.2 moles of CO 2.
Now we must find the volume of CO 2.
Volume CO 2 = moles × 24000 cm 3 = 4800cm^3
eg 2. Iron reacts with oxygen to make iron (III) oxide. Calculate the mass of iron and the volume of oxygen required to produce 3 grammes of iron (III) oxide. (A (^) r values are Fe: 56 O:16)
4Fe + 3O 2 ‘ 2Fe 2 O (^3) Mole ratio 4 3 2
We're told about iron (III) oxide, so find moles of it: M (^) r (Fe 2 O 3 ) = 112 + 48 = 160
moles of Fe 2 O 3 = 3 ÷ 160 = 0.
NB: We do not round at this stage - it leads to loss of accuracy.
We need to find the moles of iron and the moles of oxygen:
Moles of iron = 0.01875 × 4 ÷ 2 = 0.
Moles of oxygen = 0.01875 × 3 ÷ 2 = 0.
Now we need to find the mass of iron and volume of oxygen:
Mass Fe = moles × 56 = 2.1g
Volume O 2 = moles × 24000 cm 3 = 675cm 3
Triangles for moles formulae
Mass
Moles × Mr
Cover up the thing in the triangle you want to find. Then, what you can see tells you the calculation to do. For example, if you want to find moles, cover it up and you are left with mass/M (^) r
NB: Convert litres (= dm^3 ) to cm^3 by multiplying by 1000. Convert cm^3 to litres by dividing by 1000
Volume (cm 3 )
Moles × 24000
CaCO 3 (s) ‘ CaO(s) + CO 2 (g)
Mole ratio 1 ‘ 1 + 1
We know we have 20g of CaCO 3. So we work out how many moles this is: M (^) r (CaCO 3 ) = 40 + 12 + 48 = 100
moles of CaCO 3 = mass÷M r
= 20÷ 100 = 0.2 moles
We're asked about CO 2. The mole ratio CaCO 3 : CO 2 is 1:1.
Tip: If you are at all unsure what to multiply by and what to divide by when you are using ratios:
_- Use the "cross method"
- Double check that the substances with the larger number in the_ mole ratio has the higher the number of moles.
Writing ionic equations for precipitation reactions
In precipitation reactions, the reactants are solutions, but one of the products is a solid. So state symbols are very important when writing equations for these reactions.
The key idea used here is that when an ionic substance is in solution, the ions seperate - so we can consider sodium chloride solution, for example, to consist of Na+^ (aq) and Cl−^ (aq).
The worked example below shows how it works:
eg. Write the following equation in its ionic form.
FeSO 4 (aq) + 2NaOH(aq) ‘ Fe (OH) 2 (s) + Na 2 SO 4 (aq)
Three of the substances are in solution (we know this from the (aq)), so we split them into ions:
Fe2+(aq) + SO 4 2-^ (aq) + 2Na+(aq) +2OH-(aq)‘ Fe(OH) 2 (s) + 2Na +^ (aq) + SO 4 2-^ (aq)
Note that some ions appear on both sides of the equation - they started off in solution and they stay in solution. These are called spectator ions. We must "cancel them out" to give the final ionic equation:
Fe2+(aq) + SO 4 2-^ (aq) + 2Na+(aq) +2OH-(aq)‘ Fe(OH) 2 (s) + 2Na +^ (aq) + SO 4 2-^ (aq)
Fe^2 (aq) + 2 OH -^ (aq) ‘ Fe (OH) 2 (s)
Shapes of Molecules and Ions
C
hem
F
actsheet
September 2000 Number 04
To succeed with this topic you need to:
- be able to find an atom/element in the Periodic Table and decide which group it is in
- know that the number of the group an element is in tells you how many electrons it has in its outer shell
- understand how to draw dot-and-cross diagrams to show covalent bonding
- understand that positive ions have lost electrons, and negative ions have gained them, and that the size of the charge on the ion tells
you how many electrons are involved
After working through this Factsheet you will:
- have met the commonest examples of the shapes used in ‘AS’ level examination questions
- be able to work out the shape of any molecule from its dot-and-cross diagram
Examination guide
Success in this topic relies on you understanding how shapes of molecules are determined and learning the names of, and diagrams for, the basic shapes. It is very important to name the shapes correctly and present the shapes clearly as 3D structures.
What determines the shape of a molecule?
In a molecule there are covalent bonds that hold the group of atoms together. A single covalent bond is a shared pair of electrons (called a bond pair ) between two nuclei (the central parts of the two atoms involved). It is negative in charge because the electrons have a negative charge themselves.
Not all electrons around an atom are in a bond. We can see this by looking at a dot and cross diagram, like the one below for water (H 2 O). A pair of electrons not in a bond is called a lone pair.
If you have several electron pairs (bond pairs or lone pairs) around an atom then they will repel one another (two negatives repel -electrostatic repulsion) and because of this the electron pairs become as far apart as possible. This is the basic principle on which this topic is based. This theory is called 'valence shell electron pair repulsion (vsepr) ' and you may be asked to explain it.
Bond pairs, lone pairs and electron clouds The electrons in a bond pair are shared between two atoms. The positive nuclei of both atoms are attracting the negative electrons, so the electron cloud for a bond pair is pulled between the nuclei.
In a lone pair, only one nucleus is attracting the electrons, so the electron cloud is close to that nucleus. This means that the electron cloud for a lone pair looks "short and fat", and for a bond pair "long and thin" (Fig 1.)
Fig 1. Lone pair and bond pair electron clouds
nucleus
lone pair
H
H
O
bond pairs
lone pairs
Electron clouds and orbitals Although at GCSE we always thought of electrons as simple particles, and we show them like that in dot and cross diagrams, it's not really as simple as that! We never know exactly where an electron is in an atom - just that it's somewhere in a region of space called an orbital (see Factsheet 1 - Atomic Structure). Each pair of electrons has its own orbital. Because we can't know where in the orbital the electrons are, the most helpful way of thinking about an electron pair is as a kind of cloud of negative charge - the shape of the orbital tells you the shape of the cloud. Electron clouds are three dimensional - it can help to think of them as like balloons!
VSEPR theory states that each electron pair tries to separate itself as much as possible from other electron pairs (due to electrostatic repulsion.)
The shapes of molecules and ions are thus determined by the number of electron pairs not by the atoms.
bond pair
nuclei
Repulsion, bond pairs and lone pairs The shape of the electron cloud for bond pairs and lone pairs matters because it affects the amount of repulsion between them. Both types of electron cloud have the same total charge (-2 from the two electrons), but with the lone pair, the charge is concentrated close to the nucleus. This means the lone pair has a higher charge density - so it repels more strongly than a bond pair does. (You can imagine this with balloons as well - each lone pair is a short fat balloon, and each bond pair a long thin balloon. To model the water molecule shown in the first column, (which has two LP and two BP around the oxygen atom) we'd tie two short fat balloons and two long thin balloons to one point. The short fat balloons will push the other balloons away more effectively!) Fig 2 shows how repulsion varies for lone pairs and bond pairs.
Two lone pairs
Lone pair and bond pair
Two bond pairs
Fig 2. Differences in repulsion with lone pairs and bond pairs
Increasing Repulsion
N.B. The difference between lone and bond pairs does not alter the basic shape of a molecule BUT it does distort it and so alter bond angles (i.e. the angle between two adjacent bond pairs).
Shapes of molecules and ions (^) Chem Factsheet
Table 1. Shapes of molecules and ions
Example molecule
electron pairs around central atom
dot - cross diagram Name of shape Bond angles
BeCl 2 linear Cl Be^ Cl
180 o
BF 3
trigonal planar B
F F
F
120 o
H
H
H
(^4) CH (^4) C H tetrahedral
H H
C
H
H
109.5o
5 PCl (^5) trigonal bipyramidal
6 SF 6
octahedral S
F
F F
F
F
F
Finding the shape To find the shape of any molecule, go through the following procedure:
- Draw a dot and cross diagram
- Count the number of electron pairs (both types) around the central atom. 3 Decide the shape adopted by the electron pairs (see table below) 4 Look at the number of lone pairs, and decide the shape adopted by the atoms (see table on page 3)
- Draw the shape, including bond angles
Note that in these first two, the central atom does not have 8 electrons
- they are electron deficient. You'll learn more about this later!
Note in these last two, the central atom has more than 8 electrons. That's because it's using its d-orbitals, so there's extra space.
Exam Hint: - Although you have to work out the shape adopted by the electron pairs first, your answer must always be the shape adopted by the atoms
Cl
Cl
P
Cl
Cl
Cl
Cl Be Cl
F
F
B
F
S
F
F
F
F
F
F
Drawing 3D shapes No one finds drawing 3D shapes very easy! One way of showing things more clearly is to use different sorts of line to show bonds coming out of the page towards you and bonds going into the page away from you. Other bonds - shown with a normal line - are in the plane of the page, going along it. You can also add dotted lines to diagrams to make the shape clearer.
all angles 90o
Bond coming out of the page
Bond going into the page
Exam Hint: - Learn the names, examples and diagrams (with angles!) in tables 1 and 2 - they cover all the common cases you can be asked in the exam.
120 o^ P
Cl
Cl
Cl
Cl
Cl
90 o
Chem Factsheet Shapes of molecules and ions
Practice Questions
Examination questions on this topic usually form one part of a complete question and ask for the dot and cross diagram of a molecule/ion along with its shape and bond angles. The molecules below have all been asked in AS and A2 specimen questions.
- For each of the molecules/ions below (i) draw its dot-cross diagram (ii) sketch its shape (iii) show its bond angles (iv) give the name of the shape
(a) H 2 S (b) CHCl (^3) (c) F 2 O (d) PF (^5) (e) PCl 6 - (f) SiCl 4 (g) HCN (h) POCl (^3) (i) BeF (^2) (j) PCl (^3) (3 marks each)
- Explain, using a dot-cross diagram, why XeF 4 has a square planar structure. (4 marks)
- What factors affect the shape of a molecule? (3 marks)
Shapes of ions
Working out shapes of ions is very similar to molecules. You draw a dot and cross diagram as before, but you must remember:
- If it's a positive ion (cation), remove the same number of electrons
as the charge on the ion.
- If it's a negative ion (anion), add the same number of electrons as the charge on the ion This will give the central atom a full outer shell. 1. Cations You only need to know two of these - NH 4 +^ and H 3 O+ 2. Anions You need to know NO 3 - , SO 4 2-^ and CO 3 2-
All of these ions, from the dot and cross diagrams, appear to contain a mixture of normal single bonds, dative single bonds and double bonds. Although the dative bonds wouldn't affect the shape, the double bonds normally would. But they don't in these anions! All the bond angles are the same!
This is because of something called delocalisation , which you will study properly in organic chemistry. What it basically means is that in, for example, the nitrate ion, instead of having one normal single bond, one dative bond and one double bond, the "spare" electrons in the double bond are "shared" between all three bonds, so that all the bonds are the same. You don't have to worry about the details of this yet!
H
H
H
N H
H H
N
H
H
109.5o
tetrahedral
NH 4 +
(4 bond pairs, so the same shape as CH 4 )
H 3 O +
H
H
H
O
H
H H
O
(3 bond pairs + 1 lone pair, so the same shape as NH 3 )
trigonal pyramidal
Fig 4. Shapes of cations
e
O O
O
120 o
N
O
O
N
O
NO 3 −
O O
O
120 o
C
- O
S
O
O
109.5o
O
Exam Hint:- Learn the dot and cross diagrams for the ions - they may not be that easy to work out in an exam!
Fig 4. Shapes of anions - Remember all the bond angles are equal!
Useful Websites There are many useful resources on the internet for this topic, including animated diagrams and free software for drawing and viewing molecules. Here are three!
http://wunmv.wustl.edu/EduDev/Vsepr http://www.spusd.k12.ca.us/chemmybear/shapes.html http://www.chem.ufl.edu/~myers/chm2045/shapes.htm
e
e
CO 32 −
O
O
C
O
S
O O
O
O
e
e
SO 42 −
NB: the "extra" electrons in these ions are shown as "e"
Answers
marks:- (1) dot-cross (1) shape (1) bond angles
- (a)
(b)
(c)
(d)
(e)
(f)
(g)
Chem Factsheet Shapes of molecules and ions
octahedral
P
Cl
Cl Cl
Cl
Cl
Cl
P
Cl
Cl Cl
Cl
Cl
Cl
90 o e
Cl
Cl
Cl
Si Cl
tetrahedral
Cl Cl
Si
Cl
Cl
109.5o
H C N
180 o
linear
H C N
Cl
O
Cl
P Cl
tetrahedral
Cl Cl
P
Cl
Cl
109.5 o
Be F F Be (^) F
180 o
F linear
Cl
Cl
P Cl
Cl
Cl Cl
P
F
F
F
F
Xe
Xe
F F
F F
Acknowledgements: This Factsheet was researched and written by Kieron Heath Curriculum Press, Unit 305B, The Big Peg, 120 Vyse Street, Birmingham, B18 6NF ChemistryFactsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 1351-
S
H H
S
105 o^ H
H
bent
H
Cl
Cl
C Cl
tetrahedral
Cl Cl
C
Cl
H
109.5o
O
F F
O
F 105 o F
bent
P
F
F
Cl
F
F
120 o
F P
F
F
F
90 o trigonal bipyramidal
F
(h)
(i)
(j)
Around the central Xe atom there are 6 electron pairs - 2 lone pairs and 4 bond pairs (1) 6 pairs give an octahedral shape.(1) Lone-pairs repel more than bond- pairs, so the lone-pairs position themselves away form each other on opposite sides
- The factors are (a) the number of electron pairs in total (1) (b) how many are bond-pairs and how many are lone-pairs (1) (c) lone-pairs repel more than bond-pairs. (1)
