2 Marks Questions - Chemistry STD 12 Science Questions - Vidyadip

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56 questions · 5 auto-graded MCQ + 51 self-marked written.

Question 12 Marks

Copper can be extracted by hydrometallurgy but not zinc. Explain.

Answer

Hydro metallurgy method is based on the fact that more electropositive metals displace less electro positive metals from their salt solution. Copper is precipitated from copper sulphate solution by adding iron.
$\text{CuS}\text{O}_4+\text{Fe}\rightarrow\text{FeS}\text{O}_4+\text{Cu}\downarrow$
But Zinc is not precipitated from zinc sulphate solution by adding iron. Hence copper can be extracted by hydro metallurgy but not zinc.

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Question 22 Marks

Giving examples, differentiate between 'roasting' and 'calcination'.

Answer

Roasting is the process of converting sulphide ores to oxides by heating the ores in a regular supply of air at a temperature below the melting point of the metal. For example, sulphide ores of Zn, Pb, and Cu are converted to their respective oxides by this process.
$2\text{ZnS}+3\text{O}_2\xrightarrow[]{\ \ \Delta\ \ }2\text{ZnO}+2\text{SO}_2\\\text{Zinc blende}$
$2\text{Pbs}+3\text{O}_2\xrightarrow[]{\ \ \Delta\ \ }2\text{PbO}+2\text{SO}_2\\\text{Gelena}$
$2\text{Cu}_2\text{S}+3\text{O}_2\xrightarrow[]{\ \ \Delta\ \ }2\text{Cu}_2\text{O}+2\text{SO}_2\\\text{Copper glance}$
On the other hand, calcination is the process of converting hydroxide and carbonate ores to oxides by heating the ores either in the absence or in a limited supply of air at a temperature below the melting point of the metal. This process causes the escaping of volatile matter leaving behind the metal oxide. For example, hydroxide of Fe, carbonates of Zn, Ca, Mg are converted to their respective oxides by this process.
$\text{Fe}_2\text{O}_3.3\text{H}_2\text{O}\xrightarrow[]{\ \ \Delta\ \ }\text{Fe}_2\text{O}_3+3\text{H}_2\text{O}\\\ \ \text{Limonite}$
$\text{ZnCO}_{3(\text{s})}\xrightarrow[]{\ \ \Delta\ \ }\text{ZnO}_{(\text{s})}+\text{CO}_{2(\text{g})}\\\ \ \text{Calamine}$
$\text{CaMg}(\text{CO}_3)_2\xrightarrow[]{\ \ \Delta\ \ }\text{CaO}_{(\text{s})}+\text{MgO}_{(\text{s})}+2\text{CO}\\\ \ \text{Dolomite}$

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Question 32 Marks

Explain:
Column chromatography

Answer

Column Chromatography: In column chromatography an adsorbent $($e.g., $Al_2O_3)$ is packed in a glass column. The mixture to be separated or purified, taken in a suitable solvent, is applied on the top of the column. The components of the mixture get adsorbed on the column. They are then eluted out with a suitable eluent (Solvent). The weakly adsorbed component is eluted first followed by the more strongly adsorbed and so on. The method is especially suitale for such elements which are available only in minute quantities and the impurities are not very much different in their chemical behaviour from the element to be purified.

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Question 42 Marks

What is the role of graphite rod in the electrometallurgy of aluminium?

Answer

In the electrometallurgy of aluminium, oxygen gas is evolved at anode. $O_2$ reacts with graphite or carbon (graphite electrodes) to form carbon monoxide and carbon dioxide. In case if some other metal electrodes is used as anode, then oxygen will react with aluminium formed during the process to form aluminium oxide$(Al_20_3)$ which will pass into the reaction mixture resulting into wastage of $Al.$ Since graphite is cheaper than aluminium, its wastage or can be tolerated.

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Question 52 Marks

Why copper matte is put in silica lined converter?

Answer

Coppermatte contains $Cu_2$ Sand $FeS$. Copper matte is put in a silica-lined converter to remove the remaining $FeO$ and FeS present in the matte as slag $(FeSiO_3).$ Also, some silica is added to the silica-lined converter. Then, a hot air blast is blown. As a result, the remaining $FeS$ and $FeO$ are converted to iron silicate $(FeSiO_3)$ and $Cu_2S$ is converted into metallic copper.
$2FeS + 3O_2 → 2FeO + 2SO_2$
$FeO + SiO_2 → FeSiO_3$
$2Cu_2S + 3O_2 → 2Cu_2O + 2SO_2$
$2Cu_2O + Cu_2S → 6Cu + SO_2$

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Question 62 Marks

Why is zinc not extracted from zinc oxide through reduction using $CO?$

Answer

The standard Gibbs free energy of formation of $ZnO$ from $Zn$ is lower than that of $CO_2$ from $CO$. Therefore, $CO$ cannot reduce $ZnO$ to $Zn$. Hence, $Zn$ is not extracted from $ZnO$ through reduction using $CO$.

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Question 72 Marks

How can you separate alumina from silica in a bauxite ore associated with silica? Give equations, if any.

Answer

To separate alumina from silica in bauxite ore associated with silica, first the powdered ore is digested with a concentrated NaOH solution at $473-523K$ and $35-36$ bar pressure. This results in the leaching out of alumina $(Al_2O_3)$ as sodium aluminate and silica $(SiO_2)$ as sodium silicate leaving the impurities behind.
$\text{A}\text{l}_2\text{O}_{3(\text{s})}+2\text{NaOH}_{(\text{aq})}+3\text{H}_2\text{O}_{(1)}\rightarrow2\text{Na}\big[\text{Al}(\text{OH})_4\big]_{(\text{aq})}\\\text{Alumina}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{sodium aluminate}$
$\text{SiO}_2+2\text{NaOH}_{(aq)}\rightarrow\text{Na}_2\text{SiO}_{3(\text{aq})}+\text{H}_2\text{O}\\\text{Silica}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Sodium silicate}$
Then, $CO_2$ gas is passed through the resulting solution to neutralize the aluminate in the solution, which results in the precipitation of hydrated alumina. To induce precipitation, the solution is seeded with freshly prepared samples of hydrated alumina.
$2\text{Na}\big[\text{Al}(\text{OH})_4\big]_{(\text{aq})}+\text{CO}_{2(\text{g})}\rightarrow\text{Al}_2\text{O}_3.\text{xH}_2\text{O}_{(\text{s})}+2\text{NaHCO}_{3(\text{aq})}\\\text{Sodium aluminate}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Hydrate alumina}\ \ \text{Sodium hydrogen carbonate}$
During this process, sodium silicate remains in the solution. The obtained hydrated alumina is filtered, dried, and heated to get back pure alumina.
$\text{Al}_2\text{O}_3.\text{xH}_2\text{O}_{(\text{s})}\xrightarrow[]{\ \ 1470\ \ \ }\text{Al}_2\text{O}_{3(\text{s})}+\text{xH}_2\text{O}_{(\text{g})}\\\text{Hydrated alumina}\ \ \ \ \ \ \ \ \ \text{Alumina}$

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Question 82 Marks

The value of $\Delta_{\text{f}}\text{G}^\ominus$ for formation of $Cr_2O_3$ is $-540\ kJ\ mol^{-1}$ and that of $Al_2O3$ is $-827\ kJ\ mol^{-1}.$ Is the reduction of $Cr_2O_3$ possible with Al?

Answer

Chemical equation for the formation of $Cr_2O_3$ and $Al_2O_3$ are as follows:

$\frac{4}{3}\text{Cr}_{(\text{s})}+\frac{3}{2}\text{O}_{2(\text{g})}\rightarrow\frac{2}{3}\text{Cr}_2\text{O}_{3(\text{s})};\ \ \Delta\text{G}^{\circ}_{\text{f}}=-540\text{kJ mol}^{-1}$
$\frac{4}{3}\text{Al}_{(\text{s})}+\frac{3}{2}\text{O}_{2(\text{g})}\rightarrow\frac{2}{3}\text{Al}_2\text{O}_{3(\text{s})};\ \ \Delta\text{G}^{\circ}_{\text{f}}=-827\text{kJ mol}^{-1}$

Substrating equation (a) from equation (b), we get
$\frac{4}{3}\text{Al}_{(\text{s})}+\frac{2}{3}\text{Cr}_2\text{O}_{3(\text{s})}\rightarrow\frac{2}{3}\text{Al}_2\text{O}_{3(\text{s})}+\frac{4}{3}\text{Cr}_{(\text{s})}\ \ \ \Delta\text{G}^{\circ}=-287\text{kJ mol}^{-1}$
As can be seen $\Delta_{\text{r}}\text{G}^{\circ}$ is negative, thus, reduction of $Cr_2O_3$ by Al is possible.

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Question 92 Marks

Which of the ores mentioned in Table can be concentrated by magnetic separation method?

Metal	Ores	Components
Aluminium	Bauxite	$AlO_x(OH)3-2x$ $[$where $0 < x < 1]$
Iron	Kaolinite $(a$ form of clay$)$ Haematite Magnetite Siderite Iron pyrites	$[Al_2(OH)_4Si_2O_5]$ $Fe_2O_3$ $Fe_3O_4$ $FeCO_3$ $FeS_2$
Copper	Copper pyrites Malachite Cuprite Copper glance	$CuFeS_2$ $CuCO_3.Cu(OH)_2$ $Cu_2O$ $Cu_2S$
Zinc	Zinc blende or Sphalerite Calamine Zincite	$ZnS$ $ZnCO_3$ $ZnO$

Answer

If the ore or the gangue can be attracted by the magnetic field, then the ore can be concentrated by the process of magnetic separation. Among the ores mentioned in table $6.1,$ the ores of iron such as haematite $(Fe_2O_3),$ magnetite $(Fe_3O_4),$ siderite $(FeCO_3),$ and iron pyrites $(FeS_2)$ can be separated by the process of magnetic separation.

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Question 102 Marks

Predict conditions under which Al might be expected to reduce $MgO.$
$($Hint: See Intext question $6.4)$

Answer

The equations for the formation of the two oxides are
$4/3Al(s) + O_2(s) → 2/3Al_2O_3(s)$
$2Mg(s) + O_2(s) → 2MgO(s)$
If we look at the plots for the formation of the two oxides of the Ellingham diagram, we find that they intersect at certain point. The corresponding value of $\Delta\text{G}^{\circ}$ becomes zero for the reduction of $MgO$ by $Al$ metal.
$2\text{MgO}(\text{s})+4/3\text{Al(s)}\rightleftharpoons2\text{Mg}(\text{s})+2/3\text{Al}_2\text{O}_3(\text{s})$
This means that the reduction of $MgO$ by Al metal can occur below this temperature. Aluminium (Al) metal can reduce $MgO$ to $Mg$ above this temperature because $\Delta^{\circ}\text{G}$ for $Al_20_3$ is less as compared to that of $MgO.$
$3\text{MgO}(\text{s})+2\text{Al}(\text{s})\xrightarrow[]{(>1665\text{k})}\text{Al}_2\text{O}_3(\text{s})+3\text{Mg}(\text{s})$

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Question 112 Marks

Explain: Zone refining

Answer

Zone refining: The method is based on the principle that the impurities are more soluble in the melt than in the solid state of the metal. A circular mobile heater is fixed at one end of a rod of impure metal. The molten zone moves along with the heater which is moved forward. As the heater moves forward, the pure metal crystallizes out of the metal and the impurities pass on into the adjacent molten zone. The process is repeated several times and the heater is moved in the same direction. At one end impurities get con-centrated. This end is cut off. In this way, the impu-rities are swept from one end of the bar to the other. By repeating the process, ultra pure metal of silicon, germanium etc are obtained.

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Question 122 Marks

Outline the principles of refining of metals by the following methods:
Zone refining

Answer

Zone refining: This method is used for production of semiconductors and other metals of very high
purity, e.g., Ge, Si, B, Ca and In.
It is,based on the principle that the impurities are more soluble in the molten state (melt) than in the solid state of the metal.
The impure metal in the form of bar is heated at one end with a moving circular heater. As the heater is slowly moved along the length of the rod, the pure metal crystallises out of the melt whereas the impurities pass into the adjacent molten zone. This process is repeated several times till the impurities are completely driven to one end of the rod which is then cut off and discarded.

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Question 132 Marks

Outline the principles of refining of metals by the following methods:
Electrolytic refining

Answer

Electrolytic refining: Many metals, such as Cu, Ag, Au, Al, Pb, etc., are purified by this method. The irnpun metals is made the anode while a thin sheet of pure metal acts as a cathode. The electrolytic solution consists of a salt or a complex salt solution of the metal. On passing the current, the pure metal is deposited on the cathode while the impurities fall down as anode mud.

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Question 142 Marks

How is leaching carried out in case of low grade copper ores?

Answer

In case of low grade copper ores, leaching is carried out using acid or bacteria in the presence of air. In this process, copper goes into the solution as $Cu^{2+}$ ions.
$\text{Cu}_{(\text{s})}+2\text{H}^+_{(\text{aq})}+\frac{1}{2}\text{O}_{2(\text{g})}\rightarrow\text{Cu}^{2+}_{(\text{aq})}+2\text{H}_2\text{O}_{(\text{l})}$
The resulting solution is treated with scrap iron or 2 H to get metallic copper.
$\text{Cu}^{2+}_{(\text{aq})}+\text{H}_{2(\text{g})}\rightarrow\text{Cu}_{(\text{s})}+2\text{H}^+_{(\text{aq})}$

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Question 152 Marks

The choice of a reducing agent in a particular case depends on thermodynamic factor. How far do you agree with this statement? Support your opinion with two examples.

Answer

We can study the choice of a reducing agent in a particular case using Ellingham diagram. It is evident from the diagram that metals for which the standard free energy of formation oftheir oxides is more negative can reduce those metal oxides for which the standard free energy of formation of their respective oxides is less negative. It means that any metal will reduce the oxides of other metals which lie above it in the Ellingham diagram. This is because the standard free energy change $(\Delta_{\text{r}}\text{G}^{\circ})$ of the combined redox reaction will be negative by an amount equal to the difference in $\Delta_{\text{f}}\text{G}^{\circ}$ of the two metal oxides. Thus both Al and $Zn$ can reduce $FeO$ to Fe but Fe cannot reduce $Al_2O_3$ to $A1$ and $ZnO$ to $Zn$. In the same way, G can reduce $ZnO$ to Zn but not $CO.$
Note: Only that reagent will be preferred as reducing agent which will lead to decrease in free energy value $(\Delta\text{G}^{\circ})$ at a certain specific temperature.

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Question 162 Marks

Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present?

Answer

Selonium, tellurium, silver, gold and platinum are the elements present in the anode mud in electrolytic refining of copper.
These elements are less reactive than copper, hence does not participate in electrolytic refining process.
Hence, they settle down below the anode as anode mud.

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Question 172 Marks

State the role of silica in the metallurgy of copper.

Answer

During the roasting of pyrite ore, a mixture of $FeO$ and $Cu_2O$ is obtained.
$2\text{CuFe}\text{S}_2+\text{O}_2\xrightarrow[]{\Delta}\text{Cu}_2\text{S}+2\text{FeS}+2\text{S}\text{O}_2$
$2\text{Cu}_2\text{S}+3\text{O}_2\xrightarrow[]{\Delta}2\text{Cu}_2\text{O}+2\text{S}\text{O}_2$
$2\text{Fe}+3\text{O}_2\xrightarrow[]{\Delta}\text{Fe}\text{O}+2\text{S}\text{O}_2$
The role of silica in the metallurgy of copper is to remove the iron oxide obtained during the process of roasting as 'slag'. $FeS$ obtained from roasting iron sulphide ore is converted into $FeO$ which forms iron silicate with silica $(SiO_2).$ Thus $FeO$ is removed as $FeSiO_3$(slag).
$\text{FeO}\big(\text{s}\big)+\text{Si}\text{O}_2\xrightarrow[]{\Delta}\text{Fe}\text{Si}\text{O}_3\big(\text{s}\big)$

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Question 182 Marks

What is meant by the term "chromatography"?

Answer

It is the technique of separating the components of a mixture in which separation is achieved by the differential movement of individual components through a stationary phase under the influence of a mobile phase. The stationary is made of aluminum oxide or silica gel.

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Question 192 Marks

Out of $C$ and $CO$, which is a better reducing agent for $ZnO?$

Answer

The two reduction reactions are:
$ZnO(s) + C(s) → Zn(s) + CO(g) ...(i)$
$ZnO(s) + CO(g) → Zn(s) + CO_2(g) ....(ii)$
In the first case, there is increase in the magnitude of $\Delta\text{S}^{\circ}$ while in the second case, it almost remains the same. In other words, $\Delta\text{G}^{\circ}$will have more negative value in the first case when $C(s)$ is used as the reducing agent than in the second case when $CO(g)$ acts as the reducing agent. Therefore, $C(s)$ is a better reducing agent.

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Question 202 Marks

How is 'cast iron' different from 'pig iron'?

Answer

The iron obtained from blast furnaces is known as pig iron. It contains around 4% carbon and many impurities such as S, P, Si, Mn in smaller amounts. Cast iron is obtained by melting pig iron and coke using a hot air blast. It contains a lower amount of carbon (3%) than pig iron. Unlike pig iron, cast iron is extremely hard and brittle.

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Question 212 Marks

The reaction, $\text{Cr}_2\text{O}_3+2\text{Al}\rightarrow\text{Al}_2\text{O}_3+2\text{Cr}\ \ \ \ \ (\Delta\text{G}^{\ominus}=-421\text{ kJ})$ is thermodynamically feasible as is apparent from the Gibbs energy value. Why does it not take place at room temperature?

Answer

The change in Gibbs energy is related to the equilibrium constant, K as
$\Delta\text{G}=-\text{RTln}\Delta\text{G}=-\text{RT lnk}$
At room temperature, all reactants and products of the given reaction are in the solid state. As a result, equilibrium does not exist between the reactants and the products. Hence, the reaction does not take place at room temperature. Certain amount of activation energy is essential even for such reactions which are thermodynamically feasible.

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Question 222 Marks

Name the processes from which chlorine is obtained as a by-product. What will happen if an aqueous solution of $NaCl$ is subjected to electrolysis?

Answer

Down process is used for the preparation of sodium metal, where chlorine is obtained as a by- process. This -process involves the electrolysis of a fused mixture of $NaCl$ and $CaCl_2$ at $873$ K.Sodium is discharcge at the cathode while $Cl_2$ is obtained at the anode as a by-product.
$\text{NaCl}(\text{l})\xrightarrow[]{\text{Electrilysis}}\text{Na}^+(\text{melt})+\text{Cl}^{-}(\text{melt})$
$\text{At cathod}:\text{Na}^++\text{e}^-\rightarrow\text{Na}$
$\text{At anode}:\text{Cl}^-\rightarrow\frac{1}{2}\text{Cl}_2+\text{e}^-$
If, an aqueous solution of $NaCl$ is electrolysed, $H_2$ is evolved at the cathode while $Cl_2$ is obtained at the anode.

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Question 232 Marks

What criterion is followed for the selection of the stationary phase in chromatography?

Answer

The stationary phase is selected in such a way that the components of the sample have different solubility's in the phase. Hence, different components have different rates of movement through the stationary phase and as a result, can be separated from each other.

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Question 242 Marks

Differentiate between "minerals" and "ores".

Answer

Minerals are naturally occurring chemical substances containing metals. They are found in the Earth's crust and are obtained by mining.
Ores are rocks and minerals viable to be used as a source of metal.
For example, there are many minerals containing zinc, but zinc cannot be extracted profitably (conveniently and economically) from all these minerals.
Zinc can be obtained from zinc blende $(ZnS),$ calamine $(ZnCO_3),$ Zincite $(ZnO)$ etc.
Thus, these minerals are called ores of zinc.

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Question 252 Marks

Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction?

Answer

In the graph of $\triangle _rG^\circ$ Vs T for the formation of oxides, the $Cu_2O$ line is almost at the top. So it is quite easier to reduce oxide ores of copper directly to the metal by heating with coke $($both the lines of $C,\ CO$ and $C,\ CO_2$ are at much lower positions in the graph particularly after $500-600K).$
However sulphide ores are roasted/smelted to give oxides.
$2Cu_2S + 3O_2 → 2Cu_2O + 2SO_2$
The oxide can then be easily reduced to metallic copper using coke.
$Cu_2O + C → 2Cu + CO$

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Question 262 Marks

Is it true that under certain conditions, $Mg$ can reduce $SiO_2$ and Si can reduce $MgO?$ What are those conditions?

Answer

$\text{Mg}_{(\text{s})}+\frac{1}{2}\text{O}_{(\text{g})}\rightarrow\text{MgO}_{(\text{s})}\Big[\Delta\text{G}_{(\text{Mg},\ \text{MgO})}\Big]$
$\text{Si}_{(\text{s})}+\text{O}_{2(\text{g})}\rightarrow\text{SiO}_{2(\text{s})}\Big[\Delta\text{G}_{(\text{Si},\ \text{Sio}_2)}\Big]$
The temperature range in which $\Delta\text{G}_{(\text{Mg},\ \text{MgO})}$ is lesser than $\Delta\text{G}_{(\text{Si},\ \text{Sio}_2)}$ $Mg$ can also reduce to $SiO_2$ to $Si.$
$2\text{Mg}+\text{SiO}_2\Rightarrow2\text{MgO}+\text{Si};\ \ \Delta\text{G}^{\ominus}=-\text{ve}$
On the other hand, the temperatures range in which $\Delta\text{G}_{(\text{Si},\ \text{Sio}_2)}$ us less than $\Delta\text{G}_{(\text{Mg},\ \text{MgO})},$ Si can reduce $MgO$ to $Mg.$
$\text{SiO}_2+2\text{Mg}\rightarrow\text{SiO}_2+2\text{Mg};\ \ \Delta\text{G}^{\ominus}=-\text{ve}$
The temperature at which $\Delta_{\text{f}}\text{G}$ curves of these two substances intersect is $1966K$. Thus, at temperatures less than $1966K,$ $Mg$ can reduce $SiO_2$ and above $1966K$, Si can reduce $MgO.$

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Question 272 Marks

Describe the role of the following:

$SiO_2$ in the extraction of copper from copper matte.
$NaCN$ in froth floatation process.

Answer

It acts as Flux to remove iron oxide as silicate $($slag$) / FeO + SiO_2→ FeSiO_3 ($Slag$).$
$NaCN$ acts as the depressant. It selectively prevents $ZnS$ from coming to the froth but allows $PbS$ to come with the froth.

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Question 282 Marks

Explain the principle of the method of electrolytic refining of metals. Give one example.

Answer

In this method, the impure metal is made to act as anode. A strip of the same metal in pure form is used as cathode. They are put in a suitable electrolytic bath containing soluble salt of the same metal. Pure metal is deposited at the cathode and impurities remain in the solution.For example: electro refining of Cu, Ag, Au.

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Question 292 Marks

Name the method used for removing gangue from sulphide ores.
How is wrought iron different from steel?

Answer

Froth Float ation Method.
Wrought iron is the pure form of iron & Steel is an alloy of iron.

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Question 302 Marks

Which methods are usually employed for purifying the following metals:

Nickel.
Germanium.

Mention the principle behind each one of them.

Answer

Nickel:- Mond Process

Principle the metal is converted into its volatile compound which is then decomposed to give pure metal at higher temperature.

Germanium:- Zone refining
Principle that the impurities are more soluble in the melt than in the solid state of the metal.

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Question 312 Marks

Describe the underlying principle of each of the following metal refining methods:

Electrolytic refining of metals.
Vapour phase refining of metals.

Answer

Electrolytic refining: Crude metal is made as anode and pure metal as cathode. When current is passed through electrolyte of same metal ions then pure metal gets deposited at cathode and impurities settle at bottom of anode.
Vapour phase refining of metals: Crude metal is freed from impurities by first converting it into a volatile compound which decomposes into a pure metal on heating.

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Question 322 Marks

write the principle behind the following methods of refining:

Hydraulic washing.
Vapour phase refining.

Answer

Hydraulic washing: This is based on the differences in gravities of the ore and the gangue particles.
Vapour phase refining: In this, the metal forms a volatile compound which on further heating at higher temperature decomposes to pure metal.

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Question 332 Marks

Outline the principles behind the refining of metals by the following methods:

Zone refining method.
Chromatographic method.

Answer

Impurities are more soluble in melt than in solid state of the metal.
Different components of a mixture are differently adsorbed on an adsorbent.

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Question 342 Marks

What happens when:

$PCl_5$ is heated?
$H_3PO_3$ is heated?

Write the reactions involved.

Answer

$\text{PCl}_{5}\xrightarrow{heat}\text{PCl}_{3}+\text{PCl}_{2}$
$\text{4H}_{3}\text{PO}_{3}\xrightarrow{heat}\text{3H}_{3}\text{PO}_{4}+\text{PH}_{3}$

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Question 352 Marks

Describe the principle involved in each of the following processes.

Mond process for refining of Nickel.
Column chromatography for purification of rare elements.

Answer

Ability of Nickel to form volatile compound which can decompose on further heating.
Different components of a mixture are differently adsorbed on an adsorbent.

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Question 362 Marks

Write down the reactions taking place in Blast furnace related to the metallurgy of iron in the temperature range 500-800K.

Answer

The following reactions take place in blast furnace during the metallurgy of iron at the temperature range 500-800 K.
$3\text{Fe}_{2}\text{O}_{3}+\text{CO}\ \xrightarrow{\ \ \ \ \ \ \ }2\text{Fe}_{3}\text{O}_{4}+\text{CO}_{2}$
$\text{Fe}_{3}\text{O}_{3}+\text{CO}\ \xrightarrow{\ \ \ \ \ \ \ }3\text{FeO}+\text{CO}_{2}$
$\text{Fe}_{2}\text{O}_{3}+\text{CO}\ \xrightarrow{\ \ \ \ \ \ }2\text{FeO}+\text{CO}_{2}$

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Question 372 Marks

Explain the following:
Carbon and hydrogen are not used as reducing agents at high temperatures.

Answer

At the high temperature carbon and hydrogen rect with metals to from carbides and hydrides respectively hence they are not used reding agents.

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Question 382 Marks

Why are sulphide ores converted to oxide before reduction?

Answer

Sulphides are not easily reduced while oxides are easily reduced. Due to this, sulphides are first oxidised and then subjected to reduction.

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MCQ 392 Marks

Note: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Zone refining method is very useful for producing semiconductors.
Reason: Semiconductors are of high purity.

A
Both assertion and reason are true and reason is the correct explanation of assertion.
✓
Both assertion and reason are true but reason is not the correct explanation of assertion.
C
Assertion is true but reason is false.
D
Assertion is false but reason is true.

Answer

Correct option: B.

Both assertion and reason are true but reason is not the correct explanation of assertion.

Zone refining method is very useful for producing semiconductor and other metals of very high purity, e.g., germanium.

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MCQ 402 Marks

Note: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Sulphide ores are concentrated by Froth Flotation method.
Reason: Cresols stabilise the froth in Froth Flotation method.

A
Both assertion and reason are true and reason is the correct explanation of assertion.
✓
Both assertion and reason are true but reason is not the correct explanation of assertion.
C
Assertion is true but reason is false.
D
Assertion is false but reason is true.

Answer

Correct option: B.

Both assertion and reason are true but reason is not the correct explanation of assertion.

Sulphide ores are concentrated by froth floatation process. This process is used for removing gangue from sulphide ore.

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Question 412 Marks

Which method is used for refining Zr and Ti? Explain with equation.

Answer

Van Arkel method is used for refining Zr and Ti. In this method crude metal is heated with iodine to form volatile unstable compound. The compound is then decomposed to get pure metal.
$\text{Zr}(\text{s})+2\text{I}_{2}\ \xrightarrow{\ \ \ \ \ }\text{ZrI}_{4}$
$\text{ZrI}_{4}\ \xrightarrow{\ \ \ \ \ }\text{Zr}(\text{s})+2\text{I}_{2}$
$\text{Ti}+2\text{I}_{2}\ \xrightarrow{\ \ \ \ \ }\text{TiI}_{4}(\text{g})$
$\text{TiI}_{4}\ \xrightarrow{\ \ \ \ \ \ \ }\text{Ti}+2\text{I}_{2}$

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Question 422 Marks

Explain the following:
$CO_2$ is a better reducing agent below $710K$ whereas $CO$ is a better reducing agent above $710K.$

Answer

According to Ellingham diagram, at temperature below $710K,$
$\triangle\text{G}^{0}_{(\text{C, CO}_{2})} < \triangle\text{G}^{0}_{(\text{C, CO})}$
Hence $CO_2$ is better reducing agent. At above $710K,$ $\triangle\text{G}^{0}_{(\text{C, CO}_{2})} < \triangle\text{G}^{0}_{(\text{C, CO})}$ Hence, $CO$ is brtter agent.

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Question 432 Marks

Why is sulphide ore of copper heated in a furnace after mixing with silica?

Answer

Iron oxide present as impurity in sulphide ore of copperforms slag which is iron silicate and copper is produced in the form of copper matte.
$\text{FeO}+\text{SiO}_{2}\ \xrightarrow{\ \ \ \ \ \ \ }\text{FeSiO}_{3}$

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Question 442 Marks

Explain the following:
Vapour phase refining method is used for the purification of Ti.

Answer

Ti reacts with iodine to from TiI$_4$ which is volatile and decomposes tp give Ti at high to give extra titanium.
$\text{Ti}+2\text{I}_{2}\ \xrightarrow{530\text{K}}\text{TiI}_{4}\ \xrightarrow{{1800\text{K}}}\text{Ti}+2\text{I}_{2}$

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Question 452 Marks

Write two basic requirements for refining of a metal by Mond process and by Van Arkel Method.

Answer

Two basic requirements of both processes are:

The metal should form a volatile compound with an available reagent.
The volatile compound should be easily decomposable, so that the recovery of metal is easy.

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MCQ 462 Marks

Note: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Zirconium can be purificed by Van Arkel method.
Reason: $ZrI_4$ is volatile and decomposes at $1800K.$

✓
Both assertion and reason are true and reason is the correct explanation of assertion.
B
Both assertion and reason are true but reason is not the correct explanation of assertion.
C
Assertion is true but reason is false.
D
Assertion is false but reason is true.

Answer

Correct option: A.

Both assertion and reason are true and reason is the correct explanation of assertion.

Explanation: van Arkel method is very useful for removing all the oxygen and nitrogen present in the form of impurity in certain metals like $Zr$ and $Ti.$ The crude metal is heated in an evacuated vessel with iodine. The metal iodide being more covalent, volatilises and then decomposes at $1800K$ to given pure $Zr.$

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MCQ 472 Marks

Note: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Hydrometallurgy involves dissolving the ore in a suitable reagent followed by precipitation by a more electropositive metal.
Reason: Copper is extracted by hydrometallurgy.

A
Both assertion and reason are true and reason is the correct explanation of assertion.
✓
Both assertion and reason are true but reason is not the correct explanation of assertion.
C
Assertion is true but reason is false.
D
Assertion is false but reason is true.

Answer

Correct option: B.

Both assertion and reason are true but reason is not the correct explanation of assertion.

Copper is extracted by hydrometallurgical process. In this process, salts of metal are dissolved in suitable solvent like water and then reduced by more electropositive element.

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Question 482 Marks

What should be the considerations during the extraction of metals by electrochemical method?

Answer

Generally two things are considered so that proper precautions can be taken.

Reactivity of metal produced.
Suitability of electrodes.

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MCQ 492 Marks

Note: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Nickel can be purified by Mond process.
Reason: Ni $(CO)_4$ is a volatile compound which decomposes at $460K$ to give pure $Ni.$

✓
Both assertion and reason are true and reason is the correct explanation of assertion.
B
Both assertion and reason are true but reason is not the correct explanation of assertion.
C
Assertion is true but reason is false.
D
Assertion is false but reason is true.

Answer

Correct option: A.

Both assertion and reason are true and reason is the correct explanation of assertion.

Nickel treated with carbon monoxide forms nickel tetracarbonyl $Ni\ (CO)_4$ while impurities are left behind. When the vapour of $Ni\ (CO)_4$ is heated at $460K,$ it is decomposed to give pure nickel while carbon monoxide is removed as gas.

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Question 502 Marks

Explain the following:
Generally sulphide ores are converted into oxides before reduction.

Answer

Sulphide ores cannot be reduced easily but oxide orde can be easily reduced, hrnce sulphild ores are grnerally converted into oxides brfore reduction.

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Question 512 Marks

How do we separate two sulphide ores by Froth Floatation Method? Explain with an example.

Answer

Two sulphide ores can be separated by adjusting proportion of oil to water or by using depressants. For example, in the case of an ore containing ZnS and PbS, the depressant NaCN is used. It forms complex with ZnS and prevents it from coming with froth but PbS remains with froth.

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Question 522 Marks

How are metals used as semiconductors refined? What is the principle of the method used?

Answer

Semiconducting metal is produced by zone refining method which is basedon the principle that the impurities are more soluble in melt than in the solid state of metals.

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Question 532 Marks

Give two requirements for vapour phase refining.

Answer

The two requirements for vapour phase refining are:

The metal should form a volatile compound with an available reagent.
The volatile compound should be easily decomposable so that the recovery is easy.

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Question 542 Marks

Explain the following:
Silica is added to the sulphide ore of copper in the reverberatory furnace.

Answer

Copper pyrites contain iron sulphide in addition to copper sulphide. in the reverberatory furnace copper ore is to given oxides. FeO is removed by adding silica from the matte containing $Cu_2S$ and FeS.
$2\text{FeS}+3\text{O}_{2}\ \xrightarrow{\ \ \ \ \ \ }2\text{FeO}+2\text{SO}_{2}$
$\text{FeO}+\text{SiO}_{2}\ \xrightarrow{\ \ \ \ \ \ \ \ }\text{FeSiO}_{3}$

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Question 552 Marks

What is a slag?

Answer

A slag is an easily fusible material which is formed when gangue still present in the roasted or the calcined ore combines with the flux. For example, in the metallurgy of iron, CaO (flux) combines with silica gangue to form easily fusible calcium silicate $(CaSiO_3)$ slag.
$CaCO_3 → CaO + CO_2$
$CaO + SiO_2 → CaSiO_3$ (slag)

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Question 562 Marks

How is copper extracted from low grade copper ores?

Answer

Copper is extracted by hydrometallurgy from low grade ores. It is leached out using acid or bacteria. The solution containing copper ions $(Cu^{+2})$ is treated with scrap iron, zinc or $H_2$ as:
$\text{Cu}^{2+}(\text{aq})+\text{H}_{2}(\text{g})\ \xrightarrow{\ \ \ \ \ \ \ }\text{Cu}(\text{s})+2\text{H}^{+}(\text{aq})$
$\text{Cu}^{2+}(\text{aq})+\text{Fe}(\text{s})\ \xrightarrow{\ \ \ \ \ \ \ }\text{Cu}(\text{s})+\text{Fe}^{+2}(\text{aq})$
In this way, copper, is obtained.

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