Mix Examples - Metals and Non-metals Questions in English

(A) Platinum is a noble metal,which means it is highly resistant to corrosion and oxidation. Due to its lustrous appearance,durability,and malleability,it is widely used in the manufacturing of jewelry. Aluminum,lead,and magnesium are not typically used for jewelry because they are either too reactive,too soft,or lack the aesthetic appeal required for fine jewelry.

(B) Most metals are solids at room temperature. However, $\text{Mercury}$ $(Hg)$ is a unique metal that exists in a liquid state at room temperature. Sodium is a soft solid, Magnesium is a solid, and Lead is a solid metal.

(B) The Earth's crust is the largest source of metals. Many metals are found in the Earth's crust in the form of minerals.
Most of the metals are found in the form of oxides or sulphides because these are the most stable forms of metals in the presence of oxygen and sulphur in the Earth's crust.
Therefore,the correct answer is $B$.

(C) Metals that are at the bottom of the reactivity series are the least reactive.
These metals are often found in a free or native state in nature because they do not react easily with oxygen,moisture,or other substances in the environment.
Gold $(Au)$,silver $(Ag)$,and platinum $(Pt)$ are examples of such metals.
Iron $(Fe)$,Aluminum $(Al)$,and Calcium $(Ca)$ are more reactive and are typically found in the form of oxides,sulfides,or carbonates (ores).

(B) Copper pyrites,also known as chalcopyrite,is the most important ore of copper.
Its chemical formula is $CuFeS_{2}$.
- $Cu_{2}S$ is Chalcocite.
- $Cu_{2}O$ is Cuprite.
- $Cu(OH)_{2} \cdot CuCO_{3}$ is Malachite.

(D) The primary ore of aluminium is $Bauxite$ $(Al_2O_3 \cdot 2H_2O)$.
$Siderite$ is an ore of iron $(FeCO_3)$.
$Horn \text{ } silver$ is an ore of silver $(AgCl)$.
$Cuprite$ is an ore of copper $(Cu_2O)$.
Therefore, the correct option is $D$.

(A) Bauxite is the primary ore of aluminum.
Its chemical composition is primarily hydrated aluminum oxide.
The general chemical formula for bauxite is $Al_{2}O_{3} \cdot 2H_{2}O$ (or sometimes represented as $Al_{2}O_{3} \cdot nH_{2}O$ where $n$ is between $1$ and $3$).
Therefore,the correct option is $A$.

(B) Malachite is a green-colored carbonate mineral with the chemical formula $Cu_2CO_3(OH)_2$.
It is a significant ore of the metal Copper $(Cu)$.
Therefore,the correct option is $B$.

(C) Dolomite is a sedimentary carbonate rock that contains a high proportion of the mineral dolomite,$CaMg(CO_3)_2$.
It is primarily considered an important source of $Calcium$ and $Magnesium$.
Among the given options,$Calcium$ is the correct metal associated with dolomite.

(D) The chemical formulas for the given ores are:
$1$. Dolomite: $CaCO_3 \cdot MgCO_3$ (Carbonate ore)
$2$. Siderite: $FeCO_3$ (Carbonate ore)
$3$. Horn silver: $AgCl$ (Halide ore)
$4$. Bauxite: $Al_2O_3 \cdot 2H_2O$ (Oxide ore)
Since Bauxite contains aluminum oxide, it is classified as an oxide ore. Therefore, the correct option is $D$.

(A) The chemical formula for Copper glance is $Cu_2S$,which is a sulphide ore.
Cuprite is $Cu_2O$ (oxide ore).
Haematite is $Fe_2O_3$ (oxide ore).
Malachite is $Cu(OH)_2 \cdot CuCO_3$ (carbonate/hydroxide ore).
Therefore,the correct option is $A$.

(B) The chemical formula for Siderite is $FeCO_3$, which is an iron carbonate ore.
Bauxite is an oxide ore of aluminum $(Al_2O_3 \cdot 2H_2O)$.
Copper pyrites is a sulfide ore of copper $(CuFeS_2)$.
Cuprite is an oxide ore of copper $(Cu_2O)$.
Therefore, Siderite is the correct answer as it is a carbonate ore.

(B) The ores of iron are minerals from which iron can be extracted profitably. Common iron ores include Hematite $(Fe_2O_3)$ and Magnetite $(Fe_3O_4)$.
- $A$: Cuprite is an ore of copper $(Cu_2O)$.
- $B$: Magnetite is an ore of iron $(Fe_3O_4)$.
- $C$: Malachite is an ore of copper $(Cu_2CO_3(OH)_2)$.
- $D$: Bauxite is an ore of aluminium $(Al_2O_3 \cdot 2H_2O)$.
Therefore, the correct option is $B$.

(A) Noble metals are metals that are resistant to corrosion and oxidation in moist air.
They are found in their native state in the Earth's crust because they are chemically inert or very less reactive.
Examples of noble metals include Gold $(Au)$,Silver $(Ag)$,and Platinum $(Pt)$.
Iron $(Fe)$,Calcium $(Ca)$,and Magnesium $(Mg)$ are reactive metals that readily react with oxygen and moisture to form oxides or other compounds.
Therefore,Platinum is the correct answer.

(C) The process of extracting a metal from its ore and refining it to obtain it in a pure form is known as $Metallurgy$.
$Calcination$ and $Roasting$ are specific chemical processes used during the extraction of metals from ores,while $Smelting$ is a form of extractive metallurgy used to produce a metal from its ore. $Metallurgy$ is the comprehensive term that encompasses all these processes.

(C) The concentration of sulphide ores is typically carried out using the $Froth$ $Floatation$ method.
In this process,the finely powdered ore is mixed with water and pine oil in a tank.
Compressed air is blown through the mixture,creating froth.
Since sulphide ores are preferentially wetted by oil,they stick to the froth and rise to the surface,while the gangue particles (impurities) are wetted by water and settle at the bottom.
The froth containing the concentrated ore is then skimmed off.

(B) The froth flotation process is primarily used for the concentration of sulphide ores.
Copper pyrites $(CuFeS_2)$ is a sulphide ore of copper.
Hematite $(Fe_2O_3)$ is an oxide ore,Cuprite $(Cu_2O)$ is an oxide ore,and Siderite $(FeCO_3)$ is a carbonate ore.
Therefore,Copper pyrites is the correct answer as it is a sulphide ore.

(B) The process of heating carbonate ores in the absence of air or in a limited supply of air to convert them into their respective metal oxides is called $Calcination$.
For example, $ZnCO_3 \xrightarrow{\Delta} ZnO + CO_2$.
In contrast, $Roasting$ involves heating sulfide ores in the presence of excess air.

(B) The process of heating sulphide ores strongly in the presence of excess air is known as $Roasting$.
In this process, the metal sulphide reacts with oxygen to form metal oxide and sulphur dioxide gas.
For example: $2ZnS + 3O_2 \rightarrow 2ZnO + 2SO_2$.
$Calcination$ is used for carbonate ores in limited air, while $Roasting$ is specific to sulphide ores.

(A) The process of removing gangue (impurities like sand,soil,etc.) from the ore is known as concentration of ore or ore dressing.
Common methods for concentration include hydraulic washing,magnetic separation,froth flotation,and leaching.
$A$. Centrifugation is a physical method used for separation based on density,which is often employed in the concentration of ores.
$B$. Calcination is a process of heating ore in the absence of air to remove volatile impurities and moisture.
$C$. Roasting is a process of heating sulfide ores in the presence of excess air.
$D$. Chemical reduction is used to extract metal from the concentrated ore.
Therefore,centrifugation is a method used for the concentration of ore.

(D) The process of removing gangue from an ore is known as concentration or dressing of ore.
Common methods for concentration include Froth flotation,Magnetic separation,and Centrifugation (or Gravity separation).
Calcination is a process used in the extraction of metals from concentrated ore,where the ore is heated strongly in the absence of air to convert carbonates into oxides.
Therefore,Calcination is a metallurgical process used for refining/extraction,not for the concentration of ore.

(D) The froth flotation process is primarily used for the concentration of sulphide ores.
In this process,the ore is finely powdered and mixed with water.
Collectors like pine oil or turpentine are added,which make the sulphide ore particles water-repellent (hydrophobic).
These sulphide particles get wetted by the oil and rise to the surface along with the froth,while the gangue particles (like sand or clay) are wetted by water and settle at the bottom.

(B) The method described involves placing finely powdered ore on a rotating grooved table.
As the table rotates,the heavier ore particles settle in the grooves,while the lighter gangue (impurities) particles are washed away by a stream of water.
This process is known as Gravity Separation or the Wilfley table method,which relies on the difference in densities between the ore and the gangue.
Therefore,the correct option is $B$.

(A) The magnetic separation method is used for the concentration of ores where either the ore or the impurities (gangue) are magnetic in nature.
Iron ores,such as magnetite $(Fe_3O_4)$ or hematite $(Fe_2O_3)$,are magnetic.
In this process,the powdered ore is passed over a conveyor belt moving over magnetic rollers.
The magnetic ore particles are attracted to the magnetic roller and fall closer to it,while the non-magnetic impurities fall further away,allowing for effective separation.
Therefore,iron ore is concentrated using this method.

(A) Roasting is a metallurgical process used for sulphide ores.
In this process, the concentrated sulphide ore is heated strongly in the presence of excess air.
As a result, the sulphide ore is converted into its corresponding metal oxide and sulphur dioxide gas is released.
For example: $2ZnS + 3O_2 \rightarrow 2ZnO + 2SO_2$.

(C) The reactivity series of metals lists metals in decreasing order of their reactivity. The order is: $K > Na > Ca > Mg > Al > Zn > Fe > Pb > H > Cu > Hg > Ag > Au$.
Comparing the given options:
$1$. Sodium $(Na)$ and Potassium $(K)$ are highly reactive.
$2$. Aluminum $(Al)$ is moderately reactive.
$3$. Copper $(Cu)$ is placed below hydrogen in the reactivity series,making it the least reactive among the given options.

(D) The reactivity series of metals lists metals in decreasing order of their reactivity.
According to the reactivity series: $K > Na > Ca > Mg > Al > Zn > Fe > Pb > H > Cu > Hg > Ag > Au$.
Among the given options, $Sodium$ $(Na)$ is placed at the top of the reactivity series compared to $Iron$ $(Fe)$, $Zinc$ $(Zn)$, and $Copper$ $(Cu)$.
Therefore, $Sodium$ is the most reactive metal among the choices provided.

(D) reducing agent is a substance that donates electrons or increases the oxidation state of another substance while being oxidized itself.
In the context of metallurgy and chemical reactions, metals like $Al$ (Aluminum) and $Fe$ (Iron) act as strong reducing agents.
However, $Al$ is a much stronger reducing agent than $Fe$, commonly used in the thermite reaction to reduce metal oxides (e.g., $Fe_{2}O_{3} + 2Al \rightarrow 2Fe + Al_{2}O_{3} + \text{Heat}$).
Therefore, $Al$ is the most appropriate answer among the given choices as it is widely recognized as a powerful reducing agent in chemical processes.

(C) The process of extracting metals from their oxides using aluminium as a reducing agent is known as the thermite reaction or aluminothermic process.
In this process, highly reactive metals like aluminium are used to reduce the oxides of less reactive metals such as chromium $(Cr_2O_3)$ and manganese $(MnO_2)$.
The reaction is highly exothermic, releasing a large amount of heat, which melts the metal produced.
For example: $Cr_2O_3 + 2Al \rightarrow 2Cr + Al_2O_3 + \text{Heat}$.

(D) Aluminum is a highly reactive metal,and its oxide,alumina $(Al_2O_3)$,is very stable.
It cannot be reduced by common reducing agents like carbon or carbon monoxide because the affinity of aluminum for oxygen is much higher than that of carbon.
Therefore,aluminum is extracted from alumina by the process of electrolytic reduction (electrolysis of molten alumina dissolved in cryolite).

(A) Metals that are high in the reactivity series,such as $Al$,$Mg$,$Ca$,$Na$,and $K$,have a higher affinity for oxygen than carbon does.
Therefore,their oxides cannot be reduced by carbon.
Among the given options,$Al$ (Aluminium) is the most reactive metal and its oxide $(Al_2O_3)$ cannot be reduced by carbon.
Other metals like $Fe$,$Zn$,and $Pb$ are less reactive and their oxides can be reduced by carbon.

(C) The most common and effective method for the refining of copper metal is $Electrolytic$ $refining$.
In this process, an impure copper block is used as the anode, and a thin strip of pure copper is used as the cathode.
An aqueous solution of copper sulfate $(CuSO_4)$ mixed with dilute sulfuric acid $(H_2SO_4)$ is used as the electrolyte.
When an electric current is passed through the solution, pure copper from the anode dissolves into the electrolyte and deposits onto the cathode, while impurities settle at the bottom of the anode as 'anode mud'.

(D) Refining is the process of purifying an impure metal. Common methods of refining include Liquation,Electrolytic refining,and Zone refining.
Calcination is a process used in the extraction of metals,specifically for converting carbonate ores into oxides by heating them in a limited supply of air. It is a step in the extraction process,not a method of refining.

(B) In the electrolytic refining of copper,an electrolytic cell is set up using an aqueous solution of copper sulfate $(CuSO_4)$ acidified with sulfuric acid as the electrolyte.
An impure block of copper is used as the anode,while a thin strip of pure copper is used as the cathode.
When an electric current is passed through the electrolyte,pure copper from the anode dissolves into the solution,and an equivalent amount of pure copper from the electrolyte gets deposited on the cathode.
Therefore,a pure copper rod is used as the cathode.

(A) In the electrolytic refining of copper,an impure copper rod is used as the anode and a thin strip of pure copper is used as the cathode.
When an electric current is passed through the electrolyte (acidified copper sulfate solution),copper ions $(Cu^{2+})$ from the solution move towards the cathode.
At the cathode,these $Cu^{2+}$ ions gain two electrons and get reduced to form pure copper metal,which deposits on the cathode.
Therefore,pure copper deposits at the cathode.

(C) In the electrolytic refining of copper,an impure copper rod is used as the anode and a thin strip of pure copper is used as the cathode.
When an electric current is passed through the electrolyte (acidified copper sulfate solution),the impure copper from the anode dissolves into the solution as $Cu^{2+}$ ions.
These $Cu^{2+}$ ions then migrate to the cathode,where they gain electrons and deposit as pure copper metal.
As the impure copper anode continuously loses copper atoms to the solution,its weight decreases over time.
Conversely,the weight of the pure copper cathode increases as more copper is deposited on it.

(C) Liquation is a metallurgical process used for the purification of metals that have a low melting point compared to their impurities.
In this process,the crude metal is placed on the sloping hearth of a furnace and heated.
The metal melts and flows down the slope,leaving behind the non-fusible impurities.
Tin $(Sn)$ is a classic example of a metal purified by this method because it has a relatively low melting point $(231.9 \ ^\circ C)$ compared to the impurities associated with it.

(B) Liquation is a metallurgical process used for the purification of metals that have low melting points,such as tin $(Sn)$,lead $(Pb)$,and bismuth $(Bi)$.
In this process,the impure metal is placed on the sloping hearth of a reverberatory furnace and heated gently.
The metal melts and flows down the slope,leaving behind the infusible impurities on the hearth.
Thus,the correct option is $B$.

(A) Fractional distillation is a process used for the separation of a mixture into its components or fractions.
It works on the principle of the difference in the boiling points of the components present in the mixture.
When the mixture is heated,the component with the lower boiling point vaporizes first,followed by the component with the higher boiling point.
This method is commonly used to separate components of crude oil or to separate gases from the air.

(B) In the process of electrolytic refining, an impure metal rod is used as the anode and a thin strip of pure metal is used as the cathode.
When an electric current is passed through the electrolyte, the pure metal from the anode dissolves into the electrolyte, and an equivalent amount of pure metal from the electrolyte deposits on the cathode.
The insoluble impurities present in the impure metal, which do not dissolve in the electrolyte, settle down at the bottom of the electrolytic cell below the anode.
This collection of insoluble impurities is known as $Anode \text{ } mud$.

(C) semiconductor is a material that has electrical conductivity between that of a conductor (like copper) and an insulator (like glass).
Germanium $(Ge)$ is a well-known metalloid that acts as a semiconductor.
Aluminium,Magnesium,and Lead are metals that act as good conductors of electricity.
Therefore,the correct option is $C$.

(B) Metalloids are elements that possess properties intermediate between those of metals and non-metals.
Common metalloids include Boron $(B)$,Silicon $(Si)$,Germanium $(Ge)$,Arsenic $(As)$,Antimony $(Sb)$,and Tellurium $(Te)$.
Tin $(Sn)$ is a chemical element with atomic number $50$ and is classified as a post-transition metal,not a metalloid.
Therefore,Tin is the correct answer.

(C) Zone refining is a method used to purify semiconductors like $Si$,$Ge$,and $Ga$.
This process is based on the principle that impurities are more soluble in the molten state of the metal than in the solid state.
$A$ circular mobile heater is fixed at one end of a rod of the impure metal.
As the heater moves,the pure metal crystallizes out of the melt,and the impurities pass into the adjacent molten zone.
This process is repeated several times until the impurities are concentrated at one end,which is then cut off.

(A) In an electrolytic cell,the anode is the positive electrode. During electrolysis,negatively charged ions (anions) migrate towards the anode. At the anode,these anions lose electrons to form neutral atoms or molecules. The loss of electrons is defined as oxidation. Therefore,oxidation always occurs at the anode.

(D) In an electrolytic cell,the cathode is the negatively charged electrode.
During electrolysis,positively charged ions (cations) move towards the cathode.
At the cathode,these cations gain electrons to become neutral atoms or molecules.
The process of gaining electrons is defined as reduction.
Therefore,reduction always occurs at the cathode.

(A) The process of extracting pure alumina $(Al_2O_3)$ from bauxite ore is known as the Bayer's process.
In this process,bauxite is treated with a concentrated solution of sodium hydroxide $(NaOH)$ at high pressure and temperature.
This dissolves the alumina to form sodium aluminate,while impurities like silica,iron oxide,and titanium oxide remain as insoluble residues (red mud).
The solution is then filtered,and the pure alumina is precipitated out.

(C) The Bayer process is the principal industrial means of refining bauxite to produce alumina $(Al_2O_3)$.
In this process,bauxite ore is treated with a hot concentrated solution of sodium hydroxide $(NaOH)$.
The concentration of the sodium hydroxide solution used is typically around $45\%$ by weight.
This treatment dissolves the aluminum-bearing minerals to form sodium aluminate,while impurities like iron oxide remain as a solid residue (red mud).

(B) The $Bayer$ process is the principal industrial means of refining bauxite to produce alumina $(Al_2O_3)$.
In this process,finely ground bauxite ore is digested with a hot concentrated solution of $NaOH$ under pressure.
The $Al_2O_3$ present in the bauxite reacts with $NaOH$ to form soluble sodium aluminate ($NaAlO_2$ or $Na[Al(OH)_4]$).
The chemical reaction is: $Al_2O_3(s) + 2NaOH(aq) + 3H_2O(l) \rightarrow 2Na[Al(OH)_4](aq)$.

(B) Sodium aluminate is an inorganic chemical compound that acts as an effective source of aluminum hydroxide for many industrial and technical applications.
It is formed by the reaction of aluminum oxide $(Al_{2}O_{3})$ with sodium hydroxide $(NaOH)$.
The chemical reaction is: $Al_{2}O_{3} + 2NaOH \rightarrow 2NaAlO_{2} + H_{2}O$.
Therefore,the molecular formula of sodium aluminate is $NaAlO_{2}$.

(C) Aluminum oxide $(Al_{2}O_{3})$ is an amphoteric oxide,meaning it reacts with both acids and bases.
When $Al_{2}O_{3}$ reacts with a strong base like sodium hydroxide $(NaOH)$,it forms sodium aluminate $(NaAlO_{2})$ and water.
The balanced chemical equation for this reaction is:
$Al_{2}O_{3(s)} + 2NaOH_{(aq)} \xrightarrow{\Delta} 2NaAlO_{2(aq)} + H_{2}O_{(l)}$
Therefore,the missing product is $2NaAlO_{2}$.