Chapter 3: Metals and Non-metals

Step 1 – Recall the everyday observation
When a black-smith hammers a piece of hot iron, it spreads out and takes the shape of a thin plate. In another workshop the same iron can be pulled through a narrow hole to get a long wire. These two observations correspond to two different mechanical properties of solids.

Step 2 – Define the first property: malleability
A solid is said to be malleable if it can be hammered, rolled or pressed into very thin sheets without breaking or cracking.

• This property is quantified in industries by the minimum thickness (in µm) that a sheet can attain before it fractures.
• Metals such as $$\mathrm{Au}$$ (gold), $$\mathrm{Ag}$$ (silver), $$\mathrm{Cu}$$ (copper) and $$\mathrm{Al}$$ (aluminium) show very high malleability; for example, 1 g of gold can be beaten into a sheet of area about 1 m².
• The underlying reason is the metallic bond: positive metal ions are embedded in a ‘sea’ of mobile electrons, so rows of ions can slide over one another when hammered without disrupting the bond network.

Step 3 – Define the second property: ductility
A solid is said to be ductile if it can be drawn (stretched) into long, thin wires without snapping.

• The property is measured by the extent (in metres) to which a material can be elongated per unit cross-sectional area before fracture.
• Gold and silver are again the best examples; copper and aluminium follow close behind and are widely used for electrical wiring because they combine high ductility with good conductivity.
• The same metallic bonding that permits atoms to slide over one another under tensile stress allows the metal to neck down smoothly into a wire.

Step 4 – Distinguish clearly

• Malleability deals with compressive stress (hammering); the result is a sheet.
• Ductility deals with tensile stress (pulling); the result is a wire.

Conclusion
Therefore, malleable means “capable of being shaped into thin sheets by hammering or rolling,” while ductile means “capable of being drawn into wires by stretching.” Both properties are characteristic of most metals because of their unique metallic bonding.

Why special storage is needed. Sodium belongs to Group 1 (alkali metals). It has a single valence electron, so it loses that electron very readily and is therefore extremely reactive.

Step 1 – Reaction with the oxygen of air.
Whenever a fresh surface of sodium is exposed, it combines almost instantly with oxygen:
$$4\,\mathrm{Na}+\mathrm{O_2}\;\rightarrow\;2\,\mathrm{Na_2O}$$ (or, in excess oxygen, $$2\,\mathrm{Na_2O_2}$$).
The reaction is highly exothermic; the heat produced can ignite the metal itself.

Step 2 – Reaction with moisture (water vapour).
Water vapour present in air also attacks sodium vigorously:
$$2\,\mathrm{Na}+2\,\mathrm{H_2O}\;\rightarrow\;2\,\mathrm{NaOH}+\mathrm{H_2}$$.
The hydrogen liberated may catch fire because even this reaction releases a great deal of heat.

Resulting hazard. Because of the above two rapid, exothermic reactions, sodium left in the open can catch fire or even explode.

Step 3 – Why kerosene is chosen.
Kerosene is a mixture of saturated hydrocarbons that does not react with sodium at ordinary temperatures. When sodium pieces are completely submerged in kerosene, air (oxygen) and moisture cannot reach the metal surface, so no reaction takes place.

Conclusion. Sodium is stored under kerosene oil so that it remains isolated from both air and water and hence is prevented from undergoing dangerous, spontaneous reactions.

Step 1 : Identify the gas evolved when a reactive metal reacts with dilute $$\mathrm{HCl}$$

Metals that lie above hydrogen in the reactivity series displace hydrogen from dilute acids. The general reaction with dilute hydrochloric acid is :

$$\mathrm{Metal + 2HCl \;\longrightarrow\; MetalCl_2 + H_2\uparrow}$$

The gas released is therefore molecular hydrogen, $$\mathrm{H_2}$$.

Step 2 : Write and balance the equation for iron with dilute $$\mathrm{H_2SO_4}$$

Iron is above hydrogen in the reactivity series, so it displaces hydrogen from dilute sulphuric acid:

$$\mathrm{Fe + H_2SO_4 \;\longrightarrow\; FeSO_4 + H_2\uparrow}$$

• Number of Fe atoms : 1 on each side.
• Number of S atoms : 1 on each side.
• Number of O atoms : 4 on each side.
• Number of H atoms : 2 on each side.

Thus the equation is already balanced.

Step 3 : State the observations

• Effervescence is seen due to the evolution of colourless and odourless $$\mathrm{H_2}$$ gas.
• If a burning splint is brought near the gas, it burns with a characteristic ‘pop’ sound, confirming hydrogen.

Step 1 – Compare reactivities
According to the reactivity series of metals
$$\mathrm{K \gt Ca \gt Na \gt Mg \gt Al \gt Zn \gt Fe \gt Pb \gt Cu}$$
Zinc is placed above iron, therefore zinc is more reactive than iron.

Step 2 – Predict the type of reaction
When a more reactive metal is brought into contact with a salt of a less reactive metal, the more reactive metal displaces the less reactive one. This is called a displacement reaction.

Step 3 – Write the word and chemical equations
Word equation: Zinc + Iron(II) sulphate → Zinc sulphate + Iron
Chemical equation: $$\mathrm{Zn(s) + FeSO_4(aq) \rightarrow ZnSO_4(aq) + Fe(s)}$$

Step 4 – Observations in the test-tube

• The original light-green colour of $$\mathrm{FeSO_4}$$ solution gradually fades and finally becomes colourless, because $$\mathrm{ZnSO_4}$$ formed in solution is colourless.
• A grey deposit of metallic iron appears on the surface of the zinc granule/pellet and may settle at the bottom.
• Occasionally, if the moist iron layer gets oxidised by air it may look slightly brown, but the fresh deposit is grey.

Hence the experimental evidence (colour change and metal deposit) confirms that zinc has displaced iron from iron(II) sulphate.

Step 1 · Nature of an ionic compound
An ionic compound (for example, $$\mathrm{NaCl}$$) is made up of positive ions (cations) and negative ions (anions) that are arranged in a three-dimensional crystal lattice.
Each $$\mathrm{Na^+}$$ is surrounded by six $$\mathrm{Cl^-}$$ ions and vice-versa. No single ‘molecule’ exists; the entire crystal is one giant aggregation of alternate charges.

Step 2 · Origin of the forces holding the lattice together
The binding force between any pair of opposite charges is the electrostatic (Coulombic) attraction given by
$$F = \dfrac{1}{4\pi\varepsilon_0}\,\dfrac{q_1 q_2}{r^2}$$
where $$q_1$$ and $$q_2$$ are integral multiples of the electronic charge $$e$$ and $$r$$ is the distance between the ion centres. Because $$|q_1| = |q_2| = e$$ (or multiples of it) and $$r$$ is very small (≈ $$2\text{–}3\,\text{Å}$$), $$F$$ is numerically very large. Thus each pair of neighbouring ions experiences a strong attraction, and every ion is simultaneously attracted by many oppositely charged neighbours.

Step 3 · Concept of lattice enthalpy
The total energy that holds one mole of such a crystal together is expressed as its lattice enthalpy (or lattice energy). Typical values are hundreds of kilojoules per mole; e.g. $$\Delta H_{\text{lattice}}(\mathrm{NaCl})\approx 787\,\text{kJ mol}^{-1}$$. Such large positive values indicate that a very large amount of energy must be supplied to separate the ions.

Step 4 · Why a high melting point follows
Melting requires the crystal lattice to be disrupted so that the ions can move past one another. To achieve this, enough thermal energy must be supplied to overcome the very strong electrostatic attractions described in Step 2. Because these attractions are so strong (large $$F$$ and large lattice enthalpy), the required thermal energy corresponds to a very high temperature. Consequently ionic compounds possess high melting points—often well above $$800\,{}^{\circ}\!\mathrm{C}$$ for common salts like $$\mathrm{NaCl}$$ or $$\mathrm{MgO}$$.

Step 5 · Summary
The high melting points of ionic compounds arise from the strong Coulombic (electrostatic) forces between the oppositely charged ions in their rigid lattice; breaking these forces needs large amounts of energy, which is only available at high temperatures.

Metals are arranged in a reactivity series.

• Highly reactive metals (e.g. $$\mathrm{Na}$$, $$\mathrm{K}$$, $$\mathrm{Ca}$$, $$\mathrm{Al}$$) readily combine with other elements and therefore occur in nature as compounds, usually oxides, carbonates, sulphides or chlorides.

• Moderately reactive metals (e.g. $$\mathrm{Zn}$$, $$\mathrm{Fe}$$, $$\mathrm{Pb}$$) also occur mostly as compounds, though sometimes they can be obtained in the native form after simple roasting or reduction.

• Very low-reactivity or noble metals do not combine easily with air, water or other chemicals. Because of this chemical inertness, they can remain unchanged for long geological periods and are often found in the native (free) state in the earth’s crust.

The most common examples of such chemically inert (noble) metals are:

1. Gold  $$\mathrm{Au}$$
2. Platinum  $$\mathrm{Pt}$$

Hence, gold and platinum occur naturally in the free, metallic state.

To extract a free metal from its oxide we must remove the oxygen that is chemically combined with the metal. In chemical language, removal of oxygen from a compound is called reduction.

Therefore the overall process used is reduction of the metal oxide. Depending on the reactivity of the particular metal, different reducing agents or methods are chosen:

• Reduction by carbon or carbon monoxide (applicable to moderately reactive metals such as Fe, Zn, Pb, Cu). Example:  $$\mathrm{2Fe_2O_3 + 3C \rightarrow 4Fe + 3CO_2}$$

• Displacement (thermite) reduction with a more reactive metal like Al, e.g. $$\mathrm{Fe_2O_3 + 2Al \rightarrow 2Fe + Al_2O_3}$$

• Electrolytic reduction for very highly reactive metals (Na, Mg, Al, etc.), where the oxide (or another suitable compound) is reduced at the cathode: $$\mathrm{Al_2O_3 \xrightarrow[\text{electrolysis}][] 2Al + \tfrac32 O_2}$$

In every case, the essential step is the reduction of the metal oxide to the free metal.

Basic principle
When a metal that is more reactive is heated with the oxide of a less reactive metal, it removes (displaces) the oxygen from that oxide:

$$\text{More–reactive metal + Metal oxide} \longrightarrow \text{Oxide of more–reactive metal + Free metal}$$

From the reactivity series that you memorise in Class 10:

$$\mathrm{Mg > Zn > Cu}$$

Hence $$\mathrm{Mg}$$ is the most reactive of the three, followed by $$\mathrm{Zn}$$, while $$\mathrm{Cu}$$ is the least reactive.

Step-by-step testing of every combination

1. Heating zinc oxide (ZnO) with the metals

• With zinc: no reaction (same metal).
• With magnesium: $$\mathrm{Mg + ZnO \rightarrow MgO + Zn}$$  ✓ displacement.
• With copper: $$\mathrm{Cu}$$ is less reactive than $$\mathrm{Zn}$$ ⇒ no reaction.

2. Heating magnesium oxide (MgO) with the metals

• With zinc: $$\mathrm{Zn}$$ is less reactive than $$\mathrm{Mg}$$ ⇒ no reaction.
• With magnesium: same metal ⇒ no reaction.
• With copper: $$\mathrm{Cu}$$ is still less reactive ⇒ no reaction.

3. Heating copper oxide (CuO) with the metals

• With zinc: $$\mathrm{Zn + CuO \rightarrow ZnO + Cu}$$  ✓ displacement.
• With magnesium: $$\mathrm{Mg + CuO \rightarrow MgO + Cu}$$  ✓ displacement.
• With copper: no reaction (same metal).

Summary table

Therefore, displacement reactions are observed only in these three cases.

Step 1  — Recall the idea of corrosion
Corrosion is the slow oxidation of a metal in the presence of air, moisture or other chemicals, leading to the formation of compounds such as oxides, carbonates, sulphides, etc. A metal that is very unreactive toward oxygen, water and other common corrosive agents will therefore resist corrosion.

Step 2  — Connect corrosion resistance with a metal's position in the reactivity series
Metals placed low in the reactivity series are the least reactive because their tendency to lose electrons and form positive ions is very small. Owing to this low reactivity, they do not combine readily with $$\mathrm{O_2}$$, $$\mathrm{H_2O}$$, $$\mathrm{CO_2}$$, acids or bases, and hence they do not corrode.

Step 3  — Identify such metals
The metals lying at the bottom of the series are mainly noble metals. Two common examples are:

• Gold $$\left(\mathrm{Au}\right)$$
• Platinum $$\left(\mathrm{Pt}\right)$$

Step 4  — State the result
Because gold and platinum are extremely low in the reactivity series (they are often called noble metals), they show negligible chemical interaction with the atmosphere and therefore do not corrode easily.

Step 1 – Recall the definition of an alloy

An alloy is a homogeneous mixture (solid–solution) that is obtained when two or more elements, at least one of which is a metal, are melted together and then allowed to solidify.

Step 2 – Essential features

• The constituents may be metal + metal (e.g. brass — copper and zinc) or metal + non-metal (e.g. steel — iron and carbon).
• The mixture is uniform throughout; therefore it is called a homogeneous solid mixture.
• Alloys usually possess properties superior to those of the pure metals, such as greater strength, resistance to corrosion, lower melting point, etc.

Step 3 – Conclusion

Thus, alloys are homogeneous mixtures of two or more elements (at least one being a metal) formed by fusing the components together and then solidifying the melt.

Key principle — Displacement reaction (Class 10 rule)

A metal placed higher in the reactivity series can displace a metal that is lower from its salt solution. If the free metal is lower than the metal present in the compound, no reaction occurs.

Simplified reactivity series needed here

$$\mathrm{K \;>\; Na \;>\; Ca \;>\; Mg \;>\; Al \;>\; Zn \;>\; Fe \;>\; Pb \;>\; (H) \;>\; Cu \;>\; Hg \;>\; Ag \;>\; Au}$$

We now test each pair.

1. (a) $$\mathrm{NaCl}$$ solution + copper metal

   • Position comparison: $$\mathrm{Cu}$$ is below $$\mathrm{Na}$$ in the series.
   • Requirement: free metal must be higher → not satisfied.
   No displacement.

2. (b) $$\mathrm{MgCl_2}$$ solution + aluminium metal

   • Positions: $$\mathrm{Al}$$ lies below $$\mathrm{Mg}$$.
   • Rule not satisfied.
   No displacement.

3. (c) $$\mathrm{FeSO_4}$$ solution + silver metal

   • Positions: $$\mathrm{Ag}$$ is well below $$\mathrm{Fe}$$.
   • Rule not satisfied.
   No displacement.

4. (d) $$\mathrm{AgNO_3}$$ solution + copper metal

   • Positions: $$\mathrm{Cu}$$ lies above $$\mathrm{Ag}$$ in the reactivity series.
   • Rule satisfied → copper can replace silver.
   Balanced equation:
   $$\mathrm{Cu + 2\,AgNO_3 \;\longrightarrow\; Cu(NO_3)_2 + 2\,Ag}$$
   Displacement occurs.

Conclusion: only pair (d) undergoes a displacement reaction.

To minimise rusting (formation of hydrated iron(III) oxide) the surface of iron must be kept away from both moist air and oxygen:

$$\mathrm{4Fe + 3O_2 + 2H_2O \rightarrow 2Fe_2O_3\cdot 2H_2O}$$

1. Applying grease − the oily layer forms a barrier that repels water and air, so the iron is no longer exposed.
2. Applying paint − paint dries to an impermeable film, again blocking contact with air and moisture.
3. Coating with zinc (galvanisation) − the zinc layer not only isolates iron from the environment but also offers sacrificial protection; zinc oxidises preferentially if the coating is damaged.

All three techniques therefore satisfy the requirement of cutting the iron off from oxygen and water, so any of them is suitable for an iron frying pan.

Hence the correct choice is option (d).

Step 1 – Interpret the data given in the question
The oxide described is said to have

• a very high melting point, and
• is soluble in water.

Step 2 – Analyse each option

Step 3 – Select the correct element
Only calcium produces an oxide that is both highly refractory and capable of dissolving/reacting with water. Hence, the element must be calcium.

First, recall why a metal coating is applied inside food-storage cans:

• The coating metal must prevent corrosion of the iron/steel body of the can.
• It must not react with the food contents; otherwise the food will be spoiled or become toxic.

The tendency of a metal to react with common substances (air, water, acids present in food, etc.) is governed by its chemical reactivity.

Now compare the two possible coating metals:

1. Zinc is placed above tin in the reactivity series. Hence zinc readily reacts with dilute acids and even with weak organic acids that are naturally present in many foods (e.g. citric acid, lactic acid). Therefore, if the inner surface of a food can were coated with zinc, the zinc layer would gradually dissolve, contaminating the food and also exposing the underlying iron to rusting.
2. Tin is located much lower in the reactivity series. It is considerably less reactive, so it remains unaffected by the weak acids found in food. Thus tin provides an inert, protective layer and keeps the food safe for long periods.

Hence the decisive factor is the difference in reactivity. Cost or melting point does not determine the suitability here.

Therefore the correct statement is:

(c) zinc is more reactive than tin.

Definition of Amphoteric Oxide

An oxide that exhibits dual behaviour — it reacts with both acids and bases to yield salt and water — is termed an amphoteric oxide.

Chemical evidence

1. Behaviour as a base (reaction with an acid):
   $$\mathrm{Al_2O_3 + 6HCl \;\longrightarrow\; 2AlCl_3 + 3H_2O}$$
   $$\mathrm{ZnO + 2HCl \;\longrightarrow\; ZnCl_2 + H_2O}$$

2. Behaviour as an acid (reaction with a base):
   $$\mathrm{Al_2O_3 + 2NaOH \;\longrightarrow\; 2NaAlO_2 + H_2O}$$
   $$\mathrm{ZnO + 2NaOH \;\longrightarrow\; Na_2ZnO_2 + H_2O}$$

Because they can react in both ways, $$\mathrm{Al_2O_3}$$ and $$\mathrm{ZnO}$$ are classified as amphoteric oxides.

Step 1 · Position of hydrogen in the reactivity series
In the reactivity series, hydrogen is placed below metals such as $$\mathrm{K,\,Na,\,Ca,\,Mg,\,Al,\,Zn,\,Fe,\,Pb}$$ and above metals such as $$\mathrm{Cu,\,Hg,\,Ag,\,Au,\,Pt}$$.

Step 2 · Criterion for displacement of hydrogen from dilute acids
A metal that lies above hydrogen in the series can donate electrons to the hydrogen ions of a dilute acid, thereby reducing them to molecular hydrogen $$\mathrm{H_2}$$ gas. Metals placed below hydrogen cannot do so under ordinary conditions.

Step 3 · Two metals that will displace hydrogen

• Magnesium $$\mathrm{(Mg)}$$
• Zinc $$\mathrm{(Zn)}$$

Illustrative equations:

• $$\mathrm{Mg + 2HCl \rightarrow MgCl_2 + H_2\uparrow}$$
• $$\mathrm{Zn + 2HCl \rightarrow ZnCl_2 + H_2\uparrow}$$

Step 4 · Two metals that will not displace hydrogen

• Copper $$\mathrm{(Cu)}$$
• Silver $$\mathrm{(Ag)}$$

Because both of these metals lie below hydrogen, reactions such as

• $$\mathrm{Cu + 2HCl \nrightarrow}$$
• $$\mathrm{Ag + H_2SO_4(dil.) \nrightarrow}$$

do not occur under normal laboratory conditions. Hence, copper and silver fail to liberate hydrogen from dilute acids, whereas magnesium and zinc do so readily.

Step 1 — Principle of electrolytic refining.
During refining, the impure metal must dissolve from the anode into the electrolyte, while pure metal must be deposited from the electrolyte onto the cathode. This requires:

• an anode made of the impure sample of the metal,
• a cathode made of a thin sheet of pure metal, and
• an electrolyte that already contains a soluble salt of the same metal, so that ions of M are present in solution.

Step 2 — Apply the principle to metal M.

• Anode: a thick rod or block of impure M.
• Cathode: a thin strip of pure M.
• Electrolyte: an aqueous solution of a soluble salt of M. The exact formula of the salt depends on the usual valency of M, and must follow ordinary valency rules. Common choices are listed below.

For example, in the commercial electrorefining of copper (a divalent metal) the electrolyte is acidified $$\mathrm{CuSO_4}$$ — exactly of the divalent form $$\mathrm{MSO_4}$$.

Step 3 — Electrode reactions (illustrated for a divalent M).

• At the anode (oxidation): $$\mathrm{M\,(s) \;\longrightarrow\; M^{2+}\,(aq) + 2e^-}$$
• At the cathode (reduction): $$\mathrm{M^{2+}\,(aq) + 2e^- \;\longrightarrow\; M\,(s)}$$

Impure metal therefore dissolves from the anode as $$\mathrm{M^{n+}}$$ ions and deposits as pure metal on the cathode. Insoluble impurities sink below the anode as anode mud, while any soluble impurities stay dissolved in the electrolyte.

Key idea : Rusting is an oxidation of iron to hydrated iron(III) oxide, represented as $$\mathrm{Fe_2O_3\;\cdot xH_2O}$$. The reaction needs both oxygen and water, so any method that keeps at least one of these away from the metal, or makes iron chemically less reactive, will stop rust from forming.

1. Barrier (coating) methods
   
   • Painting, oiling, greasing or plastic coating puts an impermeable film over the surface. The film blocks contact with moist air, so neither $$\mathrm{O_2}$$ nor $$\mathrm{H_2O}$$ can reach the iron and the rusting reaction cannot occur.
   • Galvanisation is a special barrier method: the iron object is coated with a thin layer of molten zinc. Besides acting as a physical shield, zinc is more reactive than iron, so it also offers sacrificial protection if the coat is scratched.
2. Changing the composition of the metal (alloying)
   

   Making stainless steel by mixing iron with at least $$\mathrm{12\%}$$ chromium (often with $$\mathrm{Ni}$$ and $$\mathrm{C}$$ as well) produces a hard, passive surface layer of $$\mathrm{Cr_2O_3}$$. This compact oxide firmly adheres to the alloy and blocks further entry of air or moisture, so the bulk iron beneath never rusts.

Any two of the above techniques, clearly stated, earn full marks because each removes a necessary condition for the rusting reaction.

When a non-metal reacts with oxygen, an oxide of the non-metal is produced:

$$\mathrm{Non\!\!-\!metal + O_2 \rightarrow Non\!\!-\!metal\ oxide}$$

To classify the oxide we examine how it behaves with water and with indicators.

1. Acidic oxides
   Most non-metal oxides dissolve in water (or react with the alkali present in moist litmus paper) to give an acid; they turn blue litmus red.
   Example:

   $$\mathrm{SO_2 + H_2O \rightarrow H_2SO_3}$$

   The solution contains $$\mathrm{H_2SO_3}$$ (sulphurous acid), so $$\mathrm{SO_2}$$ is an acidic oxide. Other common acidic oxides are $$\mathrm{CO_2, SO_3, P_2O_5}$$.