4.1 Atomic structure and the periodic table — coverage pack

Answer

• Element.
• It contains only one type of atom.

Method: Identify the number of atom types present. Every particle is an argon atom, so only one type of atom is present; the substance is therefore an element.

Answer

• magnesium + oxygen → magnesium oxide
• 2Mg + O₂ → 2MgO

Method: Write the named reactant and product first. In the symbol equation oxygen enters as O₂, so place 2 before MgO to give two oxygen atoms on the right, then place 2 before Mg to balance magnesium.

Answer

• C₂H₆O + 3O₂ → 2CO₂ + 3H₂O
• Atoms have been rearranged and new chemical bonds have formed.
• The products have different properties from the reactants.

Method: Balance carbon first by placing 2 before CO₂, then hydrogen by placing 3 before H₂O. The products now contain seven oxygen atoms in total; one comes from ethanol, so six must come from 3O₂. A chemical reaction rearranges atoms into substances with new bonding and properties, unlike physical mixing.

Answer

• Filtration.
• Water passes through as the filtrate.

Method: Chalk is an insoluble solid with particles too large to pass through filter paper, so filtration traps it as the residue while the liquid water becomes the filtrate.

Answer

• Heat gently to evaporate some water and make a concentrated solution.
• Leave the solution to cool so crystals form.
• Filter the crystals from the remaining solution.
• Dry the crystals, for example between sheets of filter paper.

Method: Concentrate rather than boil the solution dry. Cooling reduces the amount of solute that remains dissolved, so crystals form. Separate them by filtration and remove surface solution by drying.

Answer

• Use fractional distillation and collect propanone near 56 °C.
• The fractionating column allows repeated condensation and evaporation, improving separation of the liquids.
• Then use simple distillation to collect most of the water near 100 °C, stopping before the flask is dry.
• The non-volatile solid remains in the concentrated solution in the flask.
• Cool the concentrated solution, then filter and dry the solid crystals.

Method: Exploit the two different boiling points first. Fractional distillation separates the more volatile propanone from water; subsequent simple distillation collects most of the water. Because the blue solid does not vaporise, stop before dryness and cool the concentrated residue so it crystallises.

Answer

• plum pudding model → nuclear model → Bohr's shells → Chadwick's neutron evidence

Method: The electron prompted the plum pudding model. Scattering evidence then produced the nuclear model, Bohr refined electron positions, and the neutron was identified last.

Answer

• Most of an atom is empty space because nearly all particles passed through.
• Positive charge is concentrated in a very small nucleus because some positive alpha particles were strongly repelled.
• Most of the atom's mass is concentrated in the nucleus.

Method: Match each observation to a structural inference. Easy passage requires mostly empty space. Rare, very large deflections require a small concentration of charge and mass capable of exerting a strong repulsive force.

Answer

• The plum pudding model had diffuse positive charge with embedded electrons and no nucleus.
• The nuclear model concentrated positive charge and most mass in a tiny central nucleus, with electrons outside it.
• Large alpha deflections contradicted diffuse positive charge and supported a concentrated nucleus.
• Bohr proposed electrons at specific distances or energy levels.
• Positive nuclear charge was divided into protons.
• Chadwick supplied evidence for neutrons in the nucleus.

Method: Begin with one direct contrast between the models. Link the rare large deflections to the replacement of spread-out positive charge by a nucleus. Then add the later refinements in sequence: fixed electron levels, protons and finally neutrons.

Answer

• Proton: +1.
• Neutron: 0.
• Electron: −1.

Method: Recall the relative charge table: the proton is positive, the neutron is neutral and the electron has an equal-magnitude negative charge.

Answer

• Overall charge: +3.
• It is a positive ion.

Method: Neutrons contribute no charge. Add the charges from protons and electrons: $13(+1)+10(-1)=+3$. Because proton and electron numbers are unequal, the particle is an ion rather than a neutral atom.

Answer

• Atomic number: 26.
• Mass number: 56.
• Charge: 3+.
• X is Fe³⁺, an iron ion.

Method: Atomic number equals protons, so it is 26. Mass number is protons plus neutrons: $26+30=56$. There are three fewer electrons than protons, so the charge is 3+. Element 26 in the periodic table is iron, Fe.

Answer

• $1.2\times10^{-10}\,\text{m}$

Method: The prefix nano means $10^{-9}$, so $0.12\,\text{nm}=0.12\times10^{-9}\,\text{m}=1.2\times10^{-10}\,\text{m}$.

Answer

• 8 protons.
• 10 neutrons.
• 8 electrons.
• Almost all the mass is in the nucleus.

Method: Atomic number gives 8 protons. Neutrons equal $18-8=10$. A neutral atom has the same number of electrons as protons, so it has 8 electrons. Protons and neutrons carry almost all the mass and both are in the nucleus.

Answer

• The atomic radius is $1.2\times10^4$ times the nuclear radius.
• The ion has mass number 56 and charge 2+.
• $\,^{56}_{26}\mathrm{Fe}^{2+}$

Method: Divide the radii: $(9.0\times10^{-11})/(7.5\times10^{-15})=1.2\times10^4$. The mass number is $26+30=56$. Two electrons are missing relative to the proton count, so the ion is 2+. Atomic number 26 identifies Fe, giving $\,^{56}_{26}\mathrm{Fe}^{2+}$.

Answer

• A_r = 63.62

Method: Weight each isotope by its percentage abundance: $A_r=(63\times69+65\times31)/100=(4347+2015)/100=63.62$.

Answer

• A_r = 24.32

Method: Use all three weighted contributions: $A_r=(24\times79+25\times10+26\times11)/100=(1896+250+286)/100=24.32$.

Answer

• Mass-79 abundance: 55%

Method: Let the mass-79 abundance be x%, so mass-81 has abundance (100 − x)%. Then $[79x+81(100-x)]/100=79.90$. Expanding gives $79x+8100-81x=7990$, so $-2x=-110$ and $x=55$. Therefore the mass-79 isotope is 55% abundant.

Answer

• 2,8,3

Method: A neutral aluminium atom has 13 electrons. Fill 2 into the first shell and 8 into the second, leaving 3 for the third shell: 2,8,3.

Answer

• Calcium atom: 2,8,8,2.
• Ca²⁺ ion: 2,8,8.

Method: Distribute the atom's 20 electrons as 2,8,8,2. A 2+ ion has lost two electrons from its outer shell, leaving 18 electrons arranged 2,8,8.

Answer

• X is chlorine, Cl.
• Atomic number: 17.
• Atom: 2,8,7.
• After gaining one electron: 2,8,8.

Method: Three occupied shells place X in period 3, and seven outer electrons place it in Group 7. The period-3 Group-7 element is chlorine. Counting $2+8+7$ gives atomic number 17. Adding one electron completes the outer shell, producing 2,8,8.

Answer

• Atomic number: 12.
• Period 3.
• Group 2.

Method: The electron total is $2+8+2=12$, so a neutral atom has atomic number 12. Three occupied shells mean period 3, and two outer electrons mean Group 2.

Answer

• Both have one electron in the outer shell.
• Both are in Group 1 and tend to react in similar ways by losing that electron.
• Q has the larger atomic number: 11 compared with 3 for P.

Method: Compare the last number in each structure: both atoms have one outer electron, so they occupy the same group and react similarly. Add all electrons to obtain atomic numbers: P has $2+1=3$ and Q has $2+8+1=11$.

Answer

• R: 2,8,6; period 3, Group 6.
• S: 2,8,8,1; period 4, Group 1.
• S is more likely to form a positive ion because it can lose one outer electron to obtain a stable arrangement.

Method: Distribute 16 electrons as 2,8,6 and 19 electrons as 2,8,8,1. The number of occupied shells gives periods 3 and 4; the outer electron counts give Groups 6 and 1. Losing S's single outer electron is the simpler route to a stable shell, so S is the likely positive-ion former.

Answer

• He left gaps for elements that had not yet been discovered.
• He changed the strict atomic-weight order where properties showed an element fitted a different group.

Method: Recall the two deliberate departures from simply listing known elements by weight: preserve a gap when evidence suggested a missing element, and prioritise matching properties over strict weight order.

Answer

• The prediction was made before the element was discovered.
• The observed compound matched the predicted property.
• This agreement supplied evidence that the gap and group placement were valid.

Method: Treat the prediction as a test of the model. An independently discovered element displaying the forecast property is unlikely to be explained by chance alone and supports both the missing-element prediction and its position.

Answer

• Strict atomic-weight order would place 39.1 before 40.0.
• That order could put elements into groups with the wrong chemical properties.
• Mendeleev could reverse the order to preserve property patterns.
• The modern table uses atomic number, not relative atomic mass.
• Isotopes explain why relative atomic masses do not always rise in the same order as atomic numbers.

Method: State the conflict between numerical weight order and repeating properties. Mendeleev prioritised the property pattern even without knowing why. Proton discovery supplied atomic number as the correct ordering variable, while isotope mixtures explained the apparently reversed average masses.

Answer

• T is a metal.
• It forms T²⁺ ions.

Method: Loss of electrons produces a positive ion, which is the defining chemical behaviour of a metal. Losing two negatively charged electrons leaves charge 2+.

Answer

• Magnesium is on the left of the table and has two outer electrons.
• It can lose those electrons to form Mg²⁺, so it is a metal.
• Sulfur is towards the right and has six outer electrons.
• Sulfur does not form positive ions in its typical reactions, so it is a non-metal.

Method: Link location to outer-shell behaviour. Magnesium reaches a stable arrangement by losing its two outer electrons and therefore forms a positive ion. Sulfur lies in the non-metal region and does not show positive-ion formation.

Answer

• A is a metal: it is lustrous, malleable and conducts electricity.
• Formation of A³⁺ confirms positive-ion behaviour.
• A is expected on the left or towards the lower part of the table.
• B is a non-metal: it is brittle and non-conducting.
• B is expected towards the upper-right part of the table.

Method: Use several independent observations. A shows the physical pattern of a metal and the decisive chemical evidence of positive-ion formation. B lacks the characteristic metallic properties. Map the classifications to the broad metal and non-metal regions of the table.

Answer

• Neon has a full outer shell of eight electrons.
• This is a stable electron arrangement, so neon has little tendency to react.

Method: Inspect the outer shell in 2,8. Because it is complete, neon does not need to gain, lose or share electrons to obtain a stable arrangement.

Answer

• −108 °C.
• Xenon is below krypton in Group 0 and has a greater relative atomic mass.
• Boiling point increases down Group 0.

Method: Continue the stated group trend. Xenon must have a boiling point higher than krypton's −153 °C; only −108 °C satisfies that direction.

Answer

• The gas is argon.
• Its 18 electrons give the structure 2,8,8.
• Its full outer shell is stable, making it unreactive and unlikely to form molecules.
• Argon is below neon and has greater relative atomic mass, so its boiling point is higher.

Method: Count the electrons to obtain atomic number $2+8+8=18$, which identifies argon. Connect the full outer shell to unreactive monatomic particles, then apply the increasing boiling-point trend down Group 0.

Answer

• Sodium hydroxide and hydrogen.
• Suitable observation: fizzing, movement on the surface, melting into a ball, or eventual disappearance.

Method: Apply the general alkali-metal reaction: metal + water → metal hydroxide + hydrogen. Hydrogen production causes effervescence; sodium also moves and may melt because the reaction releases heat.

Answer

• 2Na + Cl₂ → 2NaCl
• Both sodium and potassium have one outer electron.
• Each loses one electron to form a 1+ ion, while chlorine forms a 1− ion, giving a one-to-one ratio.

Method: Chlorine is diatomic, so make two NaCl units and then balance sodium with a coefficient of 2. The common single outer electron explains why either Group 1 metal forms a 1+ ion and hence a 1:1 chloride formula.

Answer

• Lithium < sodium < potassium in vigour.
• The outer electron is farther from the nucleus down the group.
• There are more inner shells and greater shielding.
• The nuclear attraction to the outer electron is weaker, so it is lost more easily.
• Rubidium would react even more vigorously than potassium.

Method: Use the known downward increase in Group 1 reactivity. Explain it through distance and shielding rather than electron number: both reduce the attraction holding the single outer electron. Extrapolate the same pattern to rubidium below potassium.

Answer

• Their atoms each have seven electrons in the outer shell.
• The elements form diatomic molecules made from pairs of atoms.

Method: Use one atomic feature and one molecular feature: seven outer electrons fixes the group, while the elemental halogens exist as X₂ molecules.

Answer

• Potassium chloride and bromine form.
• Cl₂ + 2KBr → 2KCl + Br₂
• Chlorine is above bromine in Group 7 and is more reactive.
• Chlorine therefore displaces bromine from bromide ions.

Method: Compare group positions to select the more reactive halogen. Chlorine displaces bromine, so swap the halogens in the salt. Balance the two atoms in each diatomic molecule by using two formula units of each potassium halide.

Answer

• No reaction occurs with sodium chloride because bromine is less reactive than chlorine.
• Bromine displaces iodine from sodium iodide.
• Br₂ + 2NaI → 2NaBr + I₂
• Reactivity decreases down Group 7, so bromine is more reactive than iodine but less reactive than chlorine.
• Down the group, increased distance and shielding make attraction for an incoming electron weaker.

Method: Place the halogens in reactivity order chlorine > bromine > iodine. Bromine cannot displace chloride but can displace iodide. Form sodium bromide and iodine, then balance the diatomic halogens and sodium salts. Relate the trend to the increasing difficulty of gaining an electron down the group.

Answer

• Iron is denser.
• Iron is stronger or harder.
• Iron has a higher melting point.
• Iron is less reactive with water, oxygen or halogens.

Method: Treat iron as a named transition metal and sodium as a Group 1 metal. Any three correct contrasts from density, strength, hardness, melting point or reactivity earn the marks.

Answer

• Nickel is a transition metal whereas potassium is Group 1.
• Nickel has a much higher melting point.
• Nickel is stronger and harder, so it resists deformation.
• Nickel is much less reactive with water.

Method: Translate each design condition into a property, then apply the transition-metal versus Group 1 comparison. Nickel meets the high-temperature, mechanical-strength and low-reactivity requirements; potassium does not.

Answer

• U is a Group 1 metal.
• Its low melting point, low density, softness and vigorous water reaction support this.
• V is a transition metal.
• Its high melting point, high density, hardness and lower reactivity support this.
• A valid example of V is chromium, manganese, iron, cobalt, nickel or copper.

Method: Compare the whole pattern rather than relying on one value. U matches the soft, low-density, low-melting and highly reactive Group 1 pattern. V shows the opposing transition-metal pattern. Finish with one of the named exemplars required by AQA.

Answer

• They can form ions with different charges.
• They form coloured compounds.
• They or their compounds can act as catalysts.

Method: Recall the three specification headings for typical transition chemistry: variable ion charge, coloured compounds and catalytic activity.

Answer

• Iron is 2+ in FeCl₂.
• Iron is 3+ in FeCl₃.
• Iron forms ions with different charges.
• Its different compounds can have different colours.

Method: A neutral formula must have total charge zero. Two chloride ions contribute 2−, so Fe is 2+ in FeCl₂; three contribute 3−, so Fe is 3+ in FeCl₃. The formulas and observations show both variable charge and coloured compounds.

Answer

• The copper compound is the more effective catalyst.
• Its trial is 4 times faster than the uncatalysed trial.
• A catalyst increases reaction rate.
• A catalyst is not used up overall or remains chemically unchanged at the end.
• Transition elements and their compounds commonly show catalytic activity.

Method: For equal reaction amounts, rate is proportional to $1/t$. The copper trial has the shortest time, so it is fastest. The rate factor is $(1/60)/(1/240)=240/60=4$. Link the result to the typical catalytic property and state that a catalyst speeds the reaction without being consumed overall.