Quiz about Chemistry in Class 12

Question 1

Read the given passage and answer the questions number that follow  We know that accurate measurement of an unknown resistance can be performed on a Wheatstone bridge However for measuring the resistance of an ionic solution we face two problems. Firstly passing direct current (DC) changes the composition of the solution. Secondly a solution cannot be connected to the bridge like a metallic wire or other solid conductor. The first difficulty is resolved by using an alternating current (AC) source of power. The second problem is solved by using a specially designed vessel called conductivity cell. Basically it consists of two platinum electrodes coated with platinum black (finely divided metallic Pt is deposited on the electrodes electrochemically). These have area of cross-section equal to A and are separated by distance l. Therefore solution confined between these electrodes is a column of length l and area of cross-section A. The resistance of such a column of solution is given by the equation  R = rho 1A = 1(kA) Name the instrument for accurate measurement of an unknown resistance.

Accepted Answer

NA

Question 2

Read the given passage and answer the questions numberthat follow  Weak electrolytes like acetic acid have lower degree of dissociation at higher concentrations and hence for such electrolytes the change in Lambda_(m)with dilution is due to increase in the degree of dissociation and consequently the number of ions in total volume of solution that contains one mol of electrolyte. In such cases Lambda_(m)increases steeply on dilution especially near lower concentrations. Therefore Lambda_(m)cannot be obtained by extrapolation of Lambda_(m)to zero concentration. At infinite dilution (i.e. concentration c tozero) electrolyte dissociates completely (alpha= 1) but at such low concentration the conductivity of the solution is so low that it cannot be measured accurately. Therefore Lambda_(m)^(@)for weak electrolytes is obtained by using Kohlrausch law of independence migration of ions. Why does the change in Am take place on dilution in case of weak electrolytes ?

Accepted Answer

NA

Question 3

Read the given passage and answer the questions number that follow  We know that accurate measurement of an unknown resistance can be performed on a Wheatstone bridge However for measuring the resistance of an ionic solution we face two problems. Firstly passing direct current (DC) changes the composition of the solution. Secondly a solution cannot be connected to the bridge like a metallic wire or other solid conductor. The first difficulty is resolved by using an alternating current (AC) source of power. The second problem is solved by using a specially designed vessel called conductivity cell. Basically it consists of two platinum electrodes coated with platinum black (finely divided metallic Pt is deposited on the electrodes electrochemically). These have area of cross-section equal to A and are separated by distance l. Therefore solution confined between these electrodes is a column of length l and area of cross-section A. The resistance of such a column of solution is given by the equation  R = rho 1A = 1(kA)  If the distance between the electrodes and area of cross-section of a cell are 2.5 cm and 6.75 cm^2 respectively. Calculate the cell constant.

Accepted Answer

NA

Question 4

Read the given passage and answer the questions number that follow  We know that accurate measurement of an unknown resistance can be performed on a Wheatstone bridge However for measuring the resistance of an ionic solution we face two problems. Firstly passing direct current (DC) changes the composition of the solution. Secondly a solution cannot be connected to the bridge like a metallic wire or other solid conductor. The first difficulty is resolved by using an alternating current (AC) source of power. The second problem is solved by using a specially designed vessel called conductivity cell. Basically it consists of two platinum electrodes coated with platinum black (finely divided metallic Pt is deposited on the electrodes electrochemically). These have area of cross-section equal to A and are separated by distance l. Therefore solution confined between these electrodes is a column of length l and area of cross-section A. The resistance of such a column of solution is given by the equation  R = rho 1A = 1(kA) Where do we take the solution for the measurement of conductivity ?

Accepted Answer

NA

Question 5

Read the given passage and answer the questions number that follow  We know that accurate measurement of an unknown resistance can be performed on a Wheatstone bridge However for measuring the resistance of an ionic solution we face two problems. Firstly passing direct current (DC) changes the composition of the solution. Secondly a solution cannot be connected to the bridge like a metallic wire or other solid conductor. The first difficulty is resolved by using an alternating current (AC) source of power. The second problem is solved by using a specially designed vessel called conductivity cell. Basically it consists of two platinum electrodes coated with platinum black (finely divided metallic Pt is deposited on the electrodes electrochemically). These have area of cross-section equal to A and are separated by distance l. Therefore solution confined between these electrodes is a column of length l and area of cross-section A. The resistance of such a column of solution is given by the equation  R = rho 1A = 1(kA) The electrical resistance of a column of 0.05 mol L NaOH solution of diameter 1 cm and length 50 cm is 5.55 x 108 ohm. Calculate its resistivity.

Accepted Answer

NA

Question 6

Read the given passage and answer the questions number that follow  We know that accurate measurement of an unknown resistance can be performed on a Wheatstone bridge However for measuring the resistance of an ionic solution we face two problems. Firstly passing direct current (DC) changes the composition of the solution. Secondly a solution cannot be connected to the bridge like a metallic wire or other solid conductor. The first difficulty is resolved by using an alternating current (AC) source of power. The second problem is solved by using a specially designed vessel called conductivity cell. Basically it consists of two platinum electrodes coated with platinum black (finely divided metallic Pt is deposited on the electrodes electrochemically). These have area of cross-section equal to A and are separated by distance l. Therefore solution confined between these electrodes is a column of length l and area of cross-section A. The resistance of such a column of solution is given by the equation  R = rho 1A = 1(kA)  Why cant we use direct current for this measurement ?

Accepted Answer

NA

Question 7

Read the given passage and answer the questions number thatfollow  Molar conductivity increases with decrease in concentration. This is because the total volume V of solution containing one mole of electrolyte also increases. It has been found that decrease in K on dilution of a solution is more than compensated by increase in its volume. Physically it means that at a given concentration Am can be defined as the conductance of the electrolytic solution kept between the electrodes of a conductivity cell at unit distance but having area of cross-section large enough to accommodate sufficient volume of solution that contains one mole of the electrolyte. When concentration approaches zero the molar conductivity is known as limiting molar conductivity and is represented by the symbol Lambda_(m)The variation in Am with concentration is different for strong and weak electrolytes. How is limiting molar conductivity represented ?

Accepted Answer

NA

Question 8

Read the given passage and answer the questions number thatfollow  Molar conductivity increases with decrease in concentration. This is because the total volume V of solution containing one mole of electrolyte also increases. It has been found that decrease in K on dilution of a solution is more than compensated by increase in its volume. Physically it means that at a given concentration Am can be defined as the conductance of the electrolytic solution kept between the electrodes of a conductivity cell at unit distance but having area of cross-section large enough to accommodate sufficient volume of solution that contains one mole of the electrolyte. When concentration approaches zero the molar conductivity is known as limiting molar conductivity and is represented by the symbol Am The variation in Lambda_(m)with concentration is different for strong and weak electrolytes.  Define Lambda_(m)at a given concentration.

Accepted Answer

NA

Question 9

Read the given passage and answer the questions number thatfollow  Molar conductivity increases with decrease in concentration. This is because the total volume V of solution containing one mole of electrolyte also increases. It has been found that decrease in K on dilution of a solution is more than compensated by increase in its volume. Physically it means that at a given concentration Am can be defined as the conductance of the electrolytic solution kept between the electrodes of a conductivity cell at unit distance but having area of cross-section large enough to accommodate sufficient volume of solution that contains one mole of the electrolyte. When concentration approaches zero the molar conductivity is known as limiting molar conductivity and is represented by the symbol Lambda_(m)The variation in Am with concentration is different for strong and weak electrolytes. Is the variation in Lambda_(m)with concentration same or different for weak and strong electrolytes ?

Accepted Answer

NA

Question 10

Read the given passage and answer the questions number thatfollow  Molar conductivity increases with decrease in concentration. This is because the total volume V of solution containing one mole of electrolyte also increases. It has been found that decrease in K on dilution of a solution is more than compensated by increase in its volume. Physically it means that at a given concentration Am can be defined as the conductance of the electrolytic solution kept between the electrodes of a conductivity cell at unit distance but having area of cross-section large enough to accommodate sufficient volume of solution that contains one mole of the electrolyte. When concentration approaches zero the molar conductivity is known as limiting molar conductivity and is represented by the symbol Lambda_(m)The variation in Am with concentration is different for strong and weak electrolytes. Why is meant by limiting molar conductivity ?

Accepted Answer

NA

Question 11

Read the given passage and answer the questions number thatfollow  Molar conductivity increases with decrease in concentration. This is because the total volume V of solution containing one mole of electrolyte also increases. It has been found that decrease in K on dilution of a solution is more than compensated by increase in its volume. Physically it means that at a given concentration Am can be defined as the conductance of the electrolytic solution kept between the electrodes of a conductivity cell at unit distance but having area of cross-section large enough to accommodate sufficient volume of solution that contains one mole of the electrolyte. When concentration approaches zero the molar conductivity is known as limiting molar conductivity and is represented by the symbol Am The variation in Lambda_(m)with concentration is different for strong and weak electrolytes. Why does molar conductivity increase with decrease in concentration ?

Accepted Answer

NA

Question 12

Read the given passage and answer the questions number 1 to 5 that follow  The lowest common oxidation state of transition metals is + 2. To form the M^(2+) ions from the gaseous atoms the sum of the first and second ionisation enthalpy is required in addition to the enthsualpy of atomisation. The dominant term is the second ionisation enthalpy which shows unusually high values for Cr and Cu where M^(+) ions have the d^(5) and d^(10) configurations respectively. The value for Zn is correspondingly low as the ionisation causes the removal of 1s electron which results in the formation of stable d^(10) configuration. The trend in the third ionisation enthalpies is not complicated by the 4s orbital factor and shows the greater difficulty of removing an electron from the d^(5) (Mn^(2+)) and d^(10) (Zn^(2+)) ions. In general the third ionisation enthalpies are quite high. Also the high values for third ionisation enthalpies of copper nickel and zinc indicate why it is difficult to obtain oxidation state greater than two for these elements.  In the series Sc(Z = 21) to Zn(Z = 30) the enthalpy of atomisation of Zn is the lowest. Why?

Accepted Answer

NA

Question 13

Read the given passage and answer the questions number 1 to 5 that follow  The lowest common oxidation state of transition metals is + 2. To form the M^(2+) ions from the gaseous atoms the sum of the first and second ionisation enthalpy is required in addition to the enthsualpy of atomisation. The dominant term is the second ionisation enthalpy which shows unusually high values for Cr and Cu where M^(+) ions have the d^(5) and d^(10) configurations respectively. The value for Zn is correspondingly low as the ionisation causes the removal of 1s electron which results in the formation of stable d^(10) configuration. The trend in the third ionisation enthalpies is not complicated by the 4s orbital factor and shows the greater difficulty of removing an electron from the d^(5) (Mn^(2+)) and d^(10) (Zn^(2+)) ions. In general the third ionisation enthalpies are quite high. Also the high values for third ionisation enthalpies of copper nickel and zinc indicate why it is difficult to obtain oxidation state greater than two for these elements.  Why is the third ionisation enthalpy of the metals Mn and Zn very high?

Accepted Answer

NA

Question 14

Read the given passage and answer the questions number 1 to 5 that follow  The lowest common oxidation state of transition metals is + 2. To form the M^(2+) ions from the gaseous atoms the sum of the first and second ionisation enthalpy is required in addition to the enthsualpy of atomisation. The dominant term is the second ionisation enthalpy which shows unusually high values for Cr and Cu where M^(+) ions have the d^(5) and d^(10) configurations respectively. The value for Zn is correspondingly low as the ionisation causes the removal of 1s electron which results in the formation of stable d^(10) configuration. The trend in the third ionisation enthalpies is not complicated by the 4s orbital factor and shows the greater difficulty of removing an electron from the d^(5) (Mn^(2+)) and d^(10) (Zn^(2+)) ions. In general the third ionisation enthalpies are quite high. Also the high values for third ionisation enthalpies of copper nickel and zinc indicate why it is difficult to obtain oxidation state greater than two for these elements.  Which factors are to be considered to obtain +2 oxidation state of the metals ?

Accepted Answer

NA

Question 15

Read the given passage and answer the questions number 1 to 5 that follow  The lowest common oxidation state of transition metals is + 2. To form the M^(2+) ions from the gaseous atoms the sum of the first and second ionisation enthalpy is required in addition to the enthsualpy of atomisation. The dominant term is the second ionisation enthalpy which shows unusually high values for Cr and Cu where M^(+) ions have the d^(5) and d^(10) configurations respectively. The value for Zn is correspondingly low as the ionisation causes the removal of 1s electron which results in the formation of stable d^(10) configuration. The trend in the third ionisation enthalpies is not complicated by the 4s orbital factor and shows the greater difficulty of removing an electron from the d^(5) (Mn^(2+)) and d^(10) (Zn^(2+)) ions. In general the third ionisation enthalpies are quite high. Also the high values for third ionisation enthalpies of copper nickel and zinc indicate why it is difficult to obtain oxidation state greater than two for these elements.  Which is the lowest common oxidation state of transition metals?

Accepted Answer

NA

Question 16

Read the given passage and answer the questions number 1 to 5 that follow  The lowest common oxidation state of transition metals is + 2. To form the M^(2+) ions from the gaseous atoms the sum of the first and second ionisation enthalpy is required in addition to the enthsualpy of atomisation. The dominant term is the second ionisation enthalpy which shows unusually high values for Cr and Cu where M^(+) ions have the d^(5) and d^(10) configurations respectively. The value for Zn is correspondingly low as the ionisation causes the removal of 1s electron which results in the formation of stable d^(10) configuration. The trend in the third ionisation enthalpies is not complicated by the 4s orbital factor and shows the greater difficulty of removing an electron from the d^(5) (Mn^(2+)) and d^(10) (Zn^(2+)) ions. In general the third ionisation enthalpies are quite high. Also the high values for third ionisation enthalpies of copper nickel and zinc indicate why it is difficult to obtain oxidation state greater than two for these elements.  Why is the second ionisation enthalpy of the metals Cr and Cu very high ?

Accepted Answer

NA

Question 17

Read the given passage and answer the questions number 1 to 5 that follow  The d-block of the periodic table contains the elements of the groups 3 - 12 and are known as transition elements. In general the electronic configuration of these elements is (n - 1) d^(1-10)ns^(1-2). The d-orbitals of the penultimate energy level in their atoms receive electrons giving rise to the three rows of the transition metals i.e. 3d 4d and 5d series. However Zn Cd and Hg are not regarded as transition elements. Transition elements exhibit certain characteristic properties like variable oxidation states complex formation formation of coloured ions and alloys catalytic activity etc. Transition metals are hard (except Zn Cd and Hg) and have a high melting point.  Why is Cu^(2+) ion coloured while Zn^(2+) ion is colourless in aqueous solution ?

Accepted Answer

NA

Question 18

Read the given passage and answer the questions number 1 to 5 that follow  The d-block of the periodic table contains the elements of the groups 3 - 12 and are known as transition elements. In general the electronic configuration of these elements is (n - 1) d^(1-10)ns^(1-2). The d-orbitals of the penultimate energy level in their atoms receive electrons giving rise to the three rows of the transition metals i.e. 3d 4d and 5d series. However Zn Cd and Hg are not regarded as transition elements. Transition elements exhibit certain characteristic properties like variable oxidation states complex formation formation of coloured ions and alloys catalytic activity etc. Transition metals are hard (except Zn Cd and Hg) and have a high melting point.  Why are Zn Cd and Hg non-transition elements ?

Accepted Answer

NA

Question 19

Read the given passage and answer the questions number 1 to 5 that follow  The d-block of the periodic table contains the elements of the groups 3 - 12 and are known as transition elements. In general the electronic configuration of these elements is (n - 1) d^(1-10)ns^(1-2). The d-orbitals of the penultimate energy level in their atoms receive electrons giving rise to the three rows of the transition metals i.e. 3d 4d and 5d series. However Zn Cd and Hg are not regarded as transition elements. Transition elements exhibit certain characteristic properties like variable oxidation states complex formation formation of coloured ions and alloys catalytic activity etc. Transition metals are hard (except Zn Cd and Hg) and have a high melting point.  Why are melting points of transition metals high ?

Accepted Answer

NA

Question 20

Read the given passage and answer the questions number 1 to 5 that follow  The d-block of the periodic table contains the elements of the groups 3 - 12 and are known as transition elements. In general the electronic configuration of these elements is (n - 1) d^(1-10)ns^(1-2). The d-orbitals of the penultimate energy level in their atoms receive electrons giving rise to the three rows of the transition metals i.e. 3d 4d and 5d series. However Zn Cd and Hg are not regarded as transition elements. Transition elements exhibit certain characteristic properties like variable oxidation states complex formation formation of coloured ions and alloys catalytic activity etc. Transition metals are hard (except Zn Cd and Hg) and have a high melting point.  Why do transition metals and their compounds show catalytic activity ?

Accepted Answer

NA

Question 21

Read the given passage and answer the questions number 1 to 5 that follow  Originally the name transition metals was derived from the fact that their chemical properties were transitional between those of s and p-block elements. Now according to IUPAC transition metals are defined as metals which have incomplete d-subshell either in the neutral atom or in their ions. Zinc cadmium and mercury of Group 12 have full du configuration in their ground state as well as their common oxidation states and hence are not regarded as transition metals. However being the end members of the 3d 4d and 5d transition series respectively their chemistry is studied along with the chemistry of transition metals.  The presence of partly filled d or f-orbitals in their atoms makes transition elements different from that of the non-transition elements. Hence transition elements and their compounds are studied separately. However the usual theory of valence as applicable to the non-transition elements can be applied successfully to the transition elements also. Give the names of first four members of first transition series.

Accepted Answer

NA

Question 22

Read the given passage and answer the questions number 1 to 5 that follow  The d-block of the periodic table contains the elements of the groups 3 - 12 and are known as transition elements. In general the electronic configuration of these elements is (n - 1) d^(1-10)ns^(1-2). The d-orbitals of the penultimate energy level in their atoms receive electrons giving rise to the three rows of the transition metals i.e. 3d 4d and 5d series. However Zn Cd and Hg are not regarded as transition elements. Transition elements exhibit certain characteristic properties like variable oxidation states complex formation formation of coloured ions and alloys catalytic activity etc. Transition metals are hard (except Zn Cd and Hg) and have a high melting point.  Which transition metal of 3d series does not show variable oxidation states ?

Accepted Answer

NA

Question 23

Read the given passage and answer the questions number 1 to 5 that follow  Originally the name transition metals was derived from the fact that their chemical properties were transitional between those of s and p-block elements. Now according to IUPAC transition metals are defined as metals which have incomplete d-subshell either in the neutral atom or in their ions. Zinc cadmium and mercury of Group 12 have full du configuration in their ground state as well as their common oxidation states and hence are not regarded as transition metals. However being the end members of the 3d 4d and 5d transition series respectively their chemistry is studied along with the chemistry of transition metals.  The presence of partly filled d or f-orbitals in their atoms makes transition elements different from that of the non-transition elements. Hence transition elements and their compounds are studied separately. However the usual theory of valence as applicable to the non-transition elements can be applied successfully to the transition elements also. Is the theory of valence applicable to transition elements ?

Accepted Answer

NA

Question 24

Read the given passage and answer the questions number 1 to 5 that follow  Originally the name transition metals was derived from the fact that their chemical properties were transitional between those of s and p-block elements. Now according to IUPAC transition metals are defined as metals which have incomplete d-subshell either in the neutral atom or in their ions. Zinc cadmium and mercury of Group 12 have full du configuration in their ground state as well as their common oxidation states and hence are not regarded as transition metals. However being the end members of the 3d 4d and 5d transition series respectively their chemistry is studied along with the chemistry of transition metals.  The presence of partly filled d or f-orbitals in their atoms makes transition elements different from that of the non-transition elements. Hence transition elements and their compounds are studied separately. However the usual theory of valence as applicable to the non-transition elements can be applied successfully to the transition elements also. Do zinc cadmium and mercury come under transition elements ?

Accepted Answer

NA

Question 25

Read the given passage and answer the questions number 1 to 5 that follow  In the metallurgy of aluminium purified Al_(2)O_(3) is mixed with Na_(3)AlF_(6) or CaF_(2) which lowers the melting point of the mixture and brings conductivity. The fused matrix is electrolysed. Steel vessal with lining of carbon acts as cathode and graphite anode is used. The overall reaction may be written as  2Al_(2)O_(3)+3C to 4Al+3CO_(2)  This process of electrolysis is widely known as Hall-Heroult process. Thus electrolysis of the molten mass is carried out in an electrolysis cell using carbon electrolysis. The oxygen liberated at anode reacts with the carbon of anode producing CO and CO_(2). This way for each kg of aluminium produced about 0.5kg of carbon anode is burnt away.  In Hall-Haroults process for the extraction of Al 10kg of carbon anode was burnt. Approximately how much of Al metal was obtained?

Accepted Answer

NA

Question 26

Read the given passage and answer the questions number 1 to 5 that follow  In the metallurgy of aluminium purified Al_(2)O_(3) is mixed with Na_(3)AlF_(6) or CaF_(2) which lowers the melting point of the mixture and brings conductivity. The fused matrix is electrolysed. Steel vessal with lining of carbon acts as cathode and graphite anode is used. The overall reaction may be written as  2Al_(2)O_(3)+3C to 4Al+3CO_(2)  This process of electrolysis is widely known as Hall-Heroult process. Thus electrolysis of the molten mass is carried out in an electrolysis cell using carbon electrolysis. The oxygen liberated at anode reacts with the carbon of anode producing CO and CO_(2). This way for each kg of aluminium produced about 0.5kg of carbon anode is burnt away.  Which gases are evolved at the anode in Hall-Heroults method?

Accepted Answer

NA

Question 27

Read the given passage and answer the questions number 1 to 5 that follow  In the metallurgy of aluminium purified Al_(2)O_(3) is mixed with Na_(3)AlF_(6) or CaF_(2) which lowers the melting point of the mixture and brings conductivity. The fused matrix is electrolysed. Steel vessal with lining of carbon acts as cathode and graphite anode is used. The overall reaction may be written as  2Al_(2)O_(3)+3C to 4Al+3CO_(2)  This process of electrolysis is widely known as Hall-Heroult process. Thus electrolysis of the molten mass is carried out in an electrolysis cell using carbon electrolysis. The oxygen liberated at anode reacts with the carbon of anode producing CO and CO_(2). This way for each kg of aluminium produced about 0.5kg of carbon anode is burnt away.  Write the equation for the reduction of alumina to aluminium.

Accepted Answer

NA

Question 28

Read the given passage and answer the questions number 1 to 5 that follow  Originally the name transition metals was derived from the fact that their chemical properties were transitional between those of s and p-block elements. Now according to IUPAC transition metals are defined as metals which have incomplete d-subshell either in the neutral atom or in their ions. Zinc cadmium and mercury of Group 12 have full du configuration in their ground state as well as their common oxidation states and hence are not regarded as transition metals. However being the end members of the 3d 4d and 5d transition series respectively their chemistry is studied along with the chemistry of transition metals.  The presence of partly filled d or f-orbitals in their atoms makes transition elements different from that of the non-transition elements. Hence transition elements and their compounds are studied separately. However the usual theory of valence as applicable to the non-transition elements can be applied successfully to the transition elements also. What is the IUPAC definition of d-block elements ?

Accepted Answer

NA

Question 29

Read the given passage and answer the questions number 1 to 5 that follow  Originally the name transition metals was derived from the fact that their chemical properties were transitional between those of s and p-block elements. Now according to IUPAC transition metals are defined as metals which have incomplete d-subshell either in the neutral atom or in their ions. Zinc cadmium and mercury of Group 12 have full du configuration in their ground state as well as their common oxidation states and hence are not regarded as transition metals. However being the end members of the 3d 4d and 5d transition series respectively their chemistry is studied along with the chemistry of transition metals.  The presence of partly filled d or f-orbitals in their atoms makes transition elements different from that of the non-transition elements. Hence transition elements and their compounds are studied separately. However the usual theory of valence as applicable to the non-transition elements can be applied successfully to the transition elements also. What is the original definition of transition metals ?

Accepted Answer

NA

Question 30

Read the given passage and answer the questions number 1 to 5 that follow  In the metallurgy of aluminium purified Al_(2)O_(3) is mixed with Na_(3)AlF_(6) or CaF_(2) which lowers the melting point of the mixture and brings conductivity. The fused matrix is electrolysed. Steel vessal with lining of carbon acts as cathode and graphite anode is used. The overall reaction may be written as  2Al_(2)O_(3)+3C to 4Al+3CO_(2)  This process of electrolysis is widely known as Hall-Heroult process. Thus electrolysis of the molten mass is carried out in an electrolysis cell using carbon electrolysis. The oxygen liberated at anode reacts with the carbon of anode producing CO and CO_(2). This way for each kg of aluminium produced about 0.5kg of carbon anode is burnt away.  Describe the composition of cathode and anode in the metallurgy of aluminium.

Accepted Answer

NA

Question 31

Read the given passage and answer the questions number 1 to 5 that follow  In general ions of the same charge in a given series show progressive decrease in radius with increasing atomic number. This is because the new electron enters a d-orbitals each time the nuclear charge increases by unity. It may be recalled that the shielding effect of a d-electron is not that effective hence the net electrostatic attraction between the nuclear charge and the outermost electron increases and the ionic radius decreases. The same trend is observed in the atomic radii of a given series. However the variation within a series is quite small. An interesting point emerges when atomic sizes of one series are compared with those of the corresponding elements in the other series. We find an increase from the first (3d) to the second (d) series of the elements but the radii of the third (5d) series are virtually the same as those of the corresponding members of the second series. This phenomenon is associated with the intervention of the 4f orbitals which must be filled before the 5d series of elements begin. The filling of 4f before 5d orbitals results in a regular decrease in atomic radii called Lanthanoid contraction.  What is lanthanoid contraction ?

Accepted Answer

NA

Question 32

Read the given passage and answer the questions number 1 to 5 that follow  In the metallurgy of aluminium purified Al_(2)O_(3) is mixed with Na_(3)AlF_(6) or CaF_(2) which lowers the melting point of the mixture and brings conductivity. The fused matrix is electrolysed. Steel vessal with lining of carbon acts as cathode and graphite anode is used. The overall reaction may be written as  2Al_(2)O_(3)+3C to 4Al+3CO_(2)  This process of electrolysis is widely known as Hall-Heroult process. Thus electrolysis of the molten mass is carried out in an electrolysis cell using carbon electrolysis. The oxygen liberated at anode reacts with the carbon of anode producing CO and CO_(2). This way for each kg of aluminium produced about 0.5kg of carbon anode is burnt away.  What is the role of Na_(3)AlF_(6) or CaF_(2) in the metallurgy of aluminium Al?

Accepted Answer

NA

Question 33

Read the given passage and answer the questions number 1 to 5 that follow  In general ions of the same charge in a given series show progressive decrease in radius with increasing atomic number. This is because the new electron enters a d-orbitals each time the nuclear charge increases by unity. It may be recalled that the shielding effect of a d-electron is not that effective hence the net electrostatic attraction between the nuclear charge and the outermost electron increases and the ionic radius decreases. The same trend is observed in the atomic radii of a given series. However the variation within a series is quite small. An interesting point emerges when atomic sizes of one series are compared with those of the corresponding elements in the other series. We find an increase from the first (3d) to the second (d) series of the elements but the radii of the third (5d) series are virtually the same as those of the corresponding members of the second series. This phenomenon is associated with the intervention of the 4f orbitals which must be filled before the 5d series of elements begin. The filling of 4f before 5d orbitals results in a regular decrease in atomic radii called Lanthanoid contraction.  Name the elements with atomic number 25 26 27.

Accepted Answer

NA

Question 34

Read the given passage and answer the questions number 1 to 5 that follow  In general ions of the same charge in a given series show progressive decrease in radius with increasing atomic number. This is because the new electron enters a d-orbitals each time the nuclear charge increases by unity. It may be recalled that the shielding effect of a d-electron is not that effective hence the net electrostatic attraction between the nuclear charge and the outermost electron increases and the ionic radius decreases. The same trend is observed in the atomic radii of a given series. However the variation within a series is quite small. An interesting point emerges when atomic sizes of one series are compared with those of the corresponding elements in the other series. We find an increase from the first (3d) to the second (d) series of the elements but the radii of the third (5d) series are virtually the same as those of the corresponding members of the second series. This phenomenon is associated with the intervention of the 4f orbitals which must be filled before the 5d series of elements begin. The filling of 4f before 5d orbitals results in a regular decrease in atomic radii called Lanthanoid contraction.  The radii of 5d series are virtually the same as those of the corresponding members of the second series. Explain.

Accepted Answer

NA

Question 35

Read the given passage and answer the questions number 1 to 5 that follow  In general ions of the same charge in a given series show progressive decrease in radius with increasing atomic number. This is because the new electron enters a d-orbitals each time the nuclear charge increases by unity. It may be recalled that the shielding effect of a d-electron is not that effective hence the net electrostatic attraction between the nuclear charge and the outermost electron increases and the ionic radius decreases. The same trend is observed in the atomic radii of a given series. However the variation within a series is quite small. An interesting point emerges when atomic sizes of one series are compared with those of the corresponding elements in the other series. We find an increase from the first (3d) to the second (d) series of the elements but the radii of the third (5d) series are virtually the same as those of the corresponding members of the second series. This phenomenon is associated with the intervention of the 4f orbitals which must be filled before the 5d series of elements begin. The filling of 4f before 5d orbitals results in a regular decrease in atomic radii called Lanthanoid contraction.  Compare the shielding effect of s p d and f-orbitals.

Accepted Answer

NA

Question 36

Read the given passage and answer the questions number 1 to 5 that follow  In general ions of the same charge in a given series show progressive decrease in radius with increasing atomic number. This is because the new electron enters a d-orbitals each time the nuclear charge increases by unity. It may be recalled that the shielding effect of a d-electron is not that effective hence the net electrostatic attraction between the nuclear charge and the outermost electron increases and the ionic radius decreases. The same trend is observed in the atomic radii of a given series. However the variation within a series is quite small. An interesting point emerges when atomic sizes of one series are compared with those of the corresponding elements in the other series. We find an increase from the first (3d) to the second (d) series of the elements but the radii of the third (5d) series are virtually the same as those of the corresponding members of the second series. This phenomenon is associated with the intervention of the 4f orbitals which must be filled before the 5d series of elements begin. The filling of 4f before 5d orbitals results in a regular decrease in atomic radii called Lanthanoid contraction.  How does the radius of the same charge change as we move from manganese to nickel in first transition series ?

Accepted Answer

NA

Question 37

Read the given passage and answer the questions number 1 to 5 that follow  Froth floatation method is used for removing gangue from sulphide ores. In this process a suspension of the powdered ore is made with water. Collectors and froth stabilisers are added to it. Collectors (for examples pine oils fatty acids xanthates etc.) enhance non-wettability of the mineral particles by water and froth stabilisers (for example cresols aniline) stabilise the froth. The mineral particles become wet by oil while the gangue particle by water. A rotating paddle agitates the mixture and draws air in it. As a result froth is formed which carries the mineral particles. The froth is light and is skimmed off. It is then dried for recovery of ore particles.  Which component is present in the froth?

Accepted Answer

NA

Question 38

Read the given passage and answer the questions number 1 to 5 that follow  Froth floatation method is used for removing gangue from sulphide ores. In this process a suspension of the powdered ore is made with water. Collectors and froth stabilisers are added to it. Collectors (for examples pine oils fatty acids xanthates etc.) enhance non-wettability of the mineral particles by water and froth stabilisers (for example cresols aniline) stabilise the froth. The mineral particles become wet by oil while the gangue particle by water. A rotating paddle agitates the mixture and draws air in it. As a result froth is formed which carries the mineral particles. The froth is light and is skimmed off. It is then dried for recovery of ore particles.  Give two examples of froth stabilisers.

Accepted Answer

NA

Question 39

Read the given passage and answer the questions number 1 to 5 that follow  Froth floatation method is used for removing gangue from sulphide ores. In this process a suspension of the powdered ore is made with water. Collectors and froth stabilisers are added to it. Collectors (for examples pine oils fatty acids xanthates etc.) enhance non-wettability of the mineral particles by water and froth stabilisers (for example cresols aniline) stabilise the froth. The mineral particles become wet by oil while the gangue particle by water. A rotating paddle agitates the mixture and draws air in it. As a result froth is formed which carries the mineral particles. The froth is light and is skimmed off. It is then dried for recovery of ore particles.  Why are collectors used in froth floatation process?

Accepted Answer

NA

Question 40

Read the given passage and answer the questions number 1 to 5 that follow  Froth floatation method is used for removing gangue from sulphide ores. In this process a suspension of the powdered ore is made with water. Collectors and froth stabilisers are added to it. Collectors (for examples pine oils fatty acids xanthates etc.) enhance non-wettability of the mineral particles by water and froth stabilisers (for example cresols aniline) stabilise the froth. The mineral particles become wet by oil while the gangue particle by water. A rotating paddle agitates the mixture and draws air in it. As a result froth is formed which carries the mineral particles. The froth is light and is skimmed off. It is then dried for recovery of ore particles.  What is the principle of froth floatation method?

Accepted Answer

NA

Question 41

Read the given passage and answer the questions number 1 to 5 that follow  Colloidal particles always carry an electric charge which may be either positive or negative. For example when AgNO_3 solution is added to Kl solution a negatively charged colloidal sol is obtained. The presence of equal and similar charges on colloidal particles provide stability to the colloidal sol and if somehow charge is removed coagulation of sol occurs. Lyophobic sols are readily coagulated as compare to lyophilic sols.  Out of KI or K_(2) SO_4 which electrolyte is better in the configuration of positive sol ?

Accepted Answer

NA

Question 42

Read the given passage and answer the questions number 1 to 5 that follow  Froth floatation method is used for removing gangue from sulphide ores. In this process a suspension of the powdered ore is made with water. Collectors and froth stabilisers are added to it. Collectors (for examples pine oils fatty acids xanthates etc.) enhance non-wettability of the mineral particles by water and froth stabilisers (for example cresols aniline) stabilise the froth. The mineral particles become wet by oil while the gangue particle by water. A rotating paddle agitates the mixture and draws air in it. As a result froth is formed which carries the mineral particles. The froth is light and is skimmed off. It is then dried for recovery of ore particles.  Which type of ores are concentrated by froth floatation method?

Accepted Answer

NA

Question 43

Read the given passage and answer the questions number 1 to 5 that follow  Colloidal particles always carry an electric charge which may be either positive or negative. For example when AgNO_3 solution is added to Kl solution a negatively charged colloidal sol is obtained. The presence of equal and similar charges on colloidal particles provide stability to the colloidal sol and if somehow charge is removed coagulation of sol occurs. Lyophobic sols are readily coagulated as compare to lyophilic sols.  Name one method by which coagulation of lyophobic sol can be carried out.

Accepted Answer

NA

Question 44

Read the given passage and answer the questions number 1 to 5 that follow  Colloidal particles always carry an electric charge which may be either positive or negative. For example when AgNO_3 solution is added to Kl solution a negatively charged colloidal sol is obtained. The presence of equal and similar charges on colloidal particles provide stability to the colloidal sol and if somehow charge is removed coagulation of sol occurs. Lyophobic sols are readily coagulated as compare to lyophilic sols.  Why a negatively charged sol is obtained on adding AgNO_3 solution to KI solution ?

Accepted Answer

NA

Question 45

Read the given passage and answer the questions number 1 to 5 that follow  Colloidal particles always carry an electric charge which may be either positive or negative. For example when AgNO_3 solution is added to Kl solution a negatively charged colloidal sol is obtained. The presence of equal and similar charges on colloidal particles provide stability to the colloidal sol and if somehow charge is removed coagulation of sol occurs. Lyophobic sols are readily coagulated as compare to lyophilic sols.  Why the presence of equal and similar charges on colloidal particles provide stability ?

Accepted Answer

NA

Question 46

Read the given passage and answer the questions number 1 to 5 that follow  Colloidal particles always carry an electric charge which may be either positive or negative. For example when AgNO_3 solution is added to Kl solution a negatively charged colloidal sol is obtained. The presence of equal and similar charges on colloidal particles provide stability to the colloidal sol and if somehow charge is removed coagulation of sol occurs. Lyophobic sols are readily coagulated as compare to lyophilic sols. What is the reason for the charge on sol particles ?

Accepted Answer

NA

Question 47

Read the given passage and answer the questions number 1 to 5 that follow  Bauxite is the principal ore of aluminium. It usually contains SiO_(2) iron oxides and titanium oxide (TiO_2) as impurities. Concentration is carried out by heating the powdered ore with a concentrated solution of NaOH at 473-523 K and 35-36 bar pressure. This process is called digestion. This way Al_(2)O_(3) is extracted out as sodium aluminate. The impurities SiO_(2) too dissolves forming sodium silicate. Other impurities are left behind.  Al_(2)O_(3)(s)+2NaOH(aq)+3H_(2)O(l) to 2Na[Al(OH)_(4)](aq)  The sodium aluminate present in solution is neutralised by passing CO_2 gas and hydrated Al_(2)O_(3) is precipitated. At this stage small amount of freshly prepared sample of hydrated Al_(2)O_(3) is added to the solution. This is called seeding. It induces the precipitation.  2Na[Al(OH)_(4)](aq)+CO_(2)(g) to Al_(2)O_(3).xH_(2)O(s)+2NaHCO_(3) (aq)  Sodium silicate remains in the solution and hydrated alumina is filtered dried and heated to give back pure Al_(2)O_(3).  Al_(2)O_(3).xH_(2)O(s) overset(1470K)(to) Al_(2)O_(3)(s)+xH_(2)O(g).  What is meant by the term seeding?

Accepted Answer

NA

Question 48

Read the given passage and answer the questions number 1 to 5 that follow  Bauxite is the principal ore of aluminium. It usually contains SiO_(2) iron oxides and titanium oxide (TiO_2) as impurities. Concentration is carried out by heating the powdered ore with a concentrated solution of NaOH at 473-523 K and 35-36 bar pressure. This process is called digestion. This way Al_(2)O_(3) is extracted out as sodium aluminate. The impurities SiO_(2) too dissolves forming sodium silicate. Other impurities are left behind.  Al_(2)O_(3)(s)+2NaOH(aq)+3H_(2)O(l) to 2Na[Al(OH)_(4)](aq)  The sodium aluminate present in solution is neutralised by passing CO_2 gas and hydrated Al_(2)O_(3) is precipitated. At this stage small amount of freshly prepared sample of hydrated Al_(2)O_(3) is added to the solution. This is called seeding. It induces the precipitation.  2Na[Al(OH)_(4)](aq)+CO_(2)(g) to Al_(2)O_(3).xH_(2)O(s)+2NaHCO_(3) (aq)  Sodium silicate remains in the solution and hydrated alumina is filtered dried and heated to give back pure Al_(2)O_(3).  Al_(2)O_(3).xH_(2)O(s) overset(1470K)(to) Al_(2)O_(3)(s)+xH_(2)O(g).  Name another ore of aluminium.

Accepted Answer

NA

Question 49

Read the given passage and answer the questions number 1 to 5 that follow  Bauxite is the principal ore of aluminium. It usually contains SiO_(2) iron oxides and titanium oxide (TiO_2) as impurities. Concentration is carried out by heating the powdered ore with a concentrated solution of NaOH at 473-523 K and 35-36 bar pressure. This process is called digestion. This way Al_(2)O_(3) is extracted out as sodium aluminate. The impurities SiO_(2) too dissolves forming sodium silicate. Other impurities are left behind.  Al_(2)O_(3)(s)+2NaOH(aq)+3H_(2)O(l) to 2Na[Al(OH)_(4)](aq)  The sodium aluminate present in solution is neutralised by passing CO_2 gas and hydrated Al_(2)O_(3) is precipitated. At this stage small amount of freshly prepared sample of hydrated Al_(2)O_(3) is added to the solution. This is called seeding. It induces the precipitation.  2Na[Al(OH)_(4)](aq)+CO_(2)(g) to Al_(2)O_(3).xH_(2)O(s)+2NaHCO_(3) (aq)  Sodium silicate remains in the solution and hydrated alumina is filtered dried and heated to give back pure Al_(2)O_(3).  Al_(2)O_(3).xH_(2)O(s) overset(1470K)(to) Al_(2)O_(3)(s)+xH_(2)O(g).  Give the reaction conditions in the digestion of bauxite with sodium hydroxide.

Accepted Answer

NA

Question 50

Read the given passage and answer the questions number 1 to 5 that follow  Bauxite is the principal ore of aluminium. It usually contains SiO_(2) iron oxides and titanium oxide (TiO_2) as impurities. Concentration is carried out by heating the powdered ore with a concentrated solution of NaOH at 473-523 K and 35-36 bar pressure. This process is called digestion. This way Al_(2)O_(3) is extracted out as sodium aluminate. The impurities SiO_(2) too dissolves forming sodium silicate. Other impurities are left behind.  Al_(2)O_(3)(s)+2NaOH(aq)+3H_(2)O(l) to 2Na[Al(OH)_(4)](aq)  The sodium aluminate present in solution is neutralised by passing CO_2 gas and hydrated Al_(2)O_(3) is precipitated. At this stage small amount of freshly prepared sample of hydrated Al_(2)O_(3) is added to the solution. This is called seeding. It induces the precipitation.  2Na[Al(OH)_(4)](aq)+CO_(2)(g) to Al_(2)O_(3).xH_(2)O(s)+2NaHCO_(3) (aq)  Sodium silicate remains in the solution and hydrated alumina is filtered dried and heated to give back pure Al_(2)O_(3).  Al_(2)O_(3).xH_(2)O(s) overset(1470K)(to) Al_(2)O_(3)(s)+xH_(2)O(g).  What happens to the impurity of silica on digesting the ore with NaOH?

Accepted Answer

NA

1.	Read the given passage and answer the questions number that follow : We know that accurate measurement of an unknown resistance can be performed on a Wheatstone bridge: However, for measuring the resistance of an ionic solution we face two problems. Firstly, passing direct current (DC) changes the composition of the solution. Secondly, a solution cannot be connected to the bridge like a metallic wire or other solid conductor. The first difficulty is resolved by using an alternating current (AC) source of power. The second problem is solved by using a specially designed vessel called conductivity cell. Basically it consists of two platinum electrodes coated with platinum black (finely divided metallic Pt is deposited on the electrodes electrochemically). These have area of cross-section equal to 'A' and are separated by distance 'l'. Therefore, solution confined between these electrodes is a column of length 'l' and area of cross-section 'A'. The resistance of such a column of solution is given by the equation : R = rho 1/A = 1/(kA) Name the instrument for accurate measurement of an unknown resistance.
Answer» SOLUTION :WHEATSTONE BRIDGE.

2.	Read the given passage and answer the questions numberthat follow : Weak electrolytes like acetic acid have lower degree of dissociation at higher concentrations and hence for such electrolytes, the change in Lambda_(m)with dilution is due to increase in the degree of dissociation and consequently the number of ions in total volume of solution that contains one mol of electrolyte. In such cases Lambda_(m)increases steeply on dilution, especially near lower concentrations. Therefore, Lambda_(m)cannot be obtained by extrapolation of Lambda_(m)to zero concentration. At infinite dilution (i.e., concentration c tozero) electrolyte dissociates completely (alpha= 1), but at such low concentration the conductivity of the solution is so low that it cannot be measured accurately. Therefore, Lambda_(m)^(@)for weak electrolytes is obtained by using Kohlrausch law of independence migration of ions. Why does the change in Am take place on dilution in case of weak electrolytes ?
Answer» SOLUTION : ACETIC acid is an EXAMPLE of WEAK electrolyte.

3.	Read the given passage and answer the questions number that follow : We know that accurate measurement of an unknown resistance can be performed on a Wheatstone bridge: However, for measuring the resistance of an ionic solution we face two problems. Firstly, passing direct current (DC) changes the composition of the solution. Secondly, a solution cannot be connected to the bridge like a metallic wire or other solid conductor. The first difficulty is resolved by using an alternating current (AC) source of power. The second problem is solved by using a specially designed vessel called conductivity cell. Basically it consists of two platinum electrodes coated with platinum black (finely divided metallic Pt is deposited on the electrodes electrochemically). These have area of cross-section equal to 'A' and are separated by distance 'l'. Therefore, solution confined between these electrodes is a column of length 'l' and area of cross-section 'A'. The resistance of such a column of solution is given by the equation : R = rho 1/A = 1/(kA) If the distance between the electrodes and area of cross-section of a cell are 2.5 cm and 6.75 cm^2 respectively. Calculate the cell constant.
Answer» SOLUTION :CELL CONSTANT is 0.37

4.	Read the given passage and answer the questions number that follow : We know that accurate measurement of an unknown resistance can be performed on a Wheatstone bridge: However, for measuring the resistance of an ionic solution we face two problems. Firstly, passing direct current (DC) changes the composition of the solution. Secondly, a solution cannot be connected to the bridge like a metallic wire or other solid conductor. The first difficulty is resolved by using an alternating current (AC) source of power. The second problem is solved by using a specially designed vessel called conductivity cell. Basically it consists of two platinum electrodes coated with platinum black (finely divided metallic Pt is deposited on the electrodes electrochemically). These have area of cross-section equal to 'A' and are separated by distance 'l'. Therefore, solution confined between these electrodes is a column of length 'l' and area of cross-section 'A'. The resistance of such a column of solution is given by the equation : R = rho 1/A = 1/(kA) Where do we take the solution for the measurement of conductivity ?
Answer» SOLUTION :We TAKE it in the CONDUCTIVITY CELL.

5.	Read the given passage and answer the questions number that follow : We know that accurate measurement of an unknown resistance can be performed on a Wheatstone bridge: However, for measuring the resistance of an ionic solution we face two problems. Firstly, passing direct current (DC) changes the composition of the solution. Secondly, a solution cannot be connected to the bridge like a metallic wire or other solid conductor. The first difficulty is resolved by using an alternating current (AC) source of power. The second problem is solved by using a specially designed vessel called conductivity cell. Basically it consists of two platinum electrodes coated with platinum black (finely divided metallic Pt is deposited on the electrodes electrochemically). These have area of cross-section equal to 'A' and are separated by distance 'l'. Therefore, solution confined between these electrodes is a column of length 'l' and area of cross-section 'A'. The resistance of such a column of solution is given by the equation : R = rho 1/A = 1/(kA) The electrical resistance of a column of 0.05 mol L NaOH solution of diameter 1 cm and length 50 cm is 5.55 x 108 ohm. Calculate its resistivity.
Answer» SOLUTION :RESISTIVITY is 87.135 22 CM

6.	Read the given passage and answer the questions number that follow : We know that accurate measurement of an unknown resistance can be performed on a Wheatstone bridge: However, for measuring the resistance of an ionic solution we face two problems. Firstly, passing direct current (DC) changes the composition of the solution. Secondly, a solution cannot be connected to the bridge like a metallic wire or other solid conductor. The first difficulty is resolved by using an alternating current (AC) source of power. The second problem is solved by using a specially designed vessel called conductivity cell. Basically it consists of two platinum electrodes coated with platinum black (finely divided metallic Pt is deposited on the electrodes electrochemically). These have area of cross-section equal to 'A' and are separated by distance 'l'. Therefore, solution confined between these electrodes is a column of length 'l' and area of cross-section 'A'. The resistance of such a column of solution is given by the equation : R = rho 1/A = 1/(kA) Why can't we use direct current for this measurement ?
Answer» SOLUTION :It CHANGES the COMPOSITION of the solution.

7.	Read the given passage and answer the questions number thatfollow : Molar conductivity increases with decrease in concentration. This is because the total volume, V, of solution containing one mole of electrolyte also increases. It has been found that decrease in K on dilution of a solution is more than compensated by increase in its volume. Physically, it means that at a given concentration, Am can be defined as the conductance of the electrolytic solution kept between the electrodes of a conductivity cell at unit distance but having area of cross-section large enough to accommodate sufficient volume of solution that contains one mole of the electrolyte. When concentration approaches zero, the molar conductivity is known as limiting molar conductivity and is represented by the symbol Lambda_(m)The variation in Am with concentration is different for strong and weak electrolytes. How is limiting molar conductivity represented ?
Answer» SOLUTION :It is REPRESENTED by `Lambda_(m)^(@)`

8.	Read the given passage and answer the questions number thatfollow : Molar conductivity increases with decrease in concentration. This is because the total volume, V, of solution containing one mole of electrolyte also increases. It has been found that decrease in K on dilution of a solution is more than compensated by increase in its volume. Physically, it means that at a given concentration, Am can be defined as the conductance of the electrolytic solution kept between the electrodes of a conductivity cell at unit distance but having area of cross-section large enough to accommodate sufficient volume of solution that contains one mole of the electrolyte. When concentration approaches zero, the molar conductivity is known as limiting molar conductivity and is represented by the symbol Am The variation in Lambda_(m)with concentration is different for strong and weak electrolytes. Define Lambda_(m)at a given concentration.
Answer» Solution :It is the CONDUCTANCE of the ELECTROLYTIC solution kept between the electrodes of a conductivity cell at unit DISTANCE but having area of cross-section LARGE enough to accommodate sufficient volume of the solution that contains one mole of the electrolyte.

9.	Read the given passage and answer the questions number thatfollow : Molar conductivity increases with decrease in concentration. This is because the total volume, V, of solution containing one mole of electrolyte also increases. It has been found that decrease in K on dilution of a solution is more than compensated by increase in its volume. Physically, it means that at a given concentration, Am can be defined as the conductance of the electrolytic solution kept between the electrodes of a conductivity cell at unit distance but having area of cross-section large enough to accommodate sufficient volume of solution that contains one mole of the electrolyte. When concentration approaches zero, the molar conductivity is known as limiting molar conductivity and is represented by the symbol Lambda_(m)The variation in Am with concentration is different for strong and weak electrolytes. Is the variation in Lambda_(m)with concentration same or different for weak and strong electrolytes ?
Answer» SOLUTION : It is DIFFERENT.

10.	Read the given passage and answer the questions number thatfollow : Molar conductivity increases with decrease in concentration. This is because the total volume, V, of solution containing one mole of electrolyte also increases. It has been found that decrease in K on dilution of a solution is more than compensated by increase in its volume. Physically, it means that at a given concentration, Am can be defined as the conductance of the electrolytic solution kept between the electrodes of a conductivity cell at unit distance but having area of cross-section large enough to accommodate sufficient volume of solution that contains one mole of the electrolyte. When concentration approaches zero, the molar conductivity is known as limiting molar conductivity and is represented by the symbol Lambda_(m)The variation in Am with concentration is different for strong and weak electrolytes. Why is meant by limiting molar conductivity ?
Answer» SOLUTION :It is the MOLAR CONDUCTIVITY at ZERO CONCENTRATION.

11.	Read the given passage and answer the questions number thatfollow : Molar conductivity increases with decrease in concentration. This is because the total volume, V, of solution containing one mole of electrolyte also increases. It has been found that decrease in K on dilution of a solution is more than compensated by increase in its volume. Physically, it means that at a given concentration, Am can be defined as the conductance of the electrolytic solution kept between the electrodes of a conductivity cell at unit distance but having area of cross-section large enough to accommodate sufficient volume of solution that contains one mole of the electrolyte. When concentration approaches zero, the molar conductivity is known as limiting molar conductivity and is represented by the symbol Am The variation in Lambda_(m)with concentration is different for strong and weak electrolytes. Why does molar conductivity increase with decrease in concentration ?
Answer» Solution : TOTAL VOLUME V of the solution containing one MOLE of the electrolyte INCREASES.

12.	Read the given passage and answer the questions number 1 to 5 that follow : The lowest common oxidation state of transition metals is + 2. To form the M^(2+) ions from the gaseous atoms, the sum of the first and second ionisation enthalpy is required in addition to the enthsualpy of atomisation. The dominant term is the second ionisation enthalpy which shows unusually high values for Cr and Cu where M^(+) ions have the d^(5) and d^(10) configurations respectively. The value for Zn is correspondingly low as the ionisation causes the removal of 1s electron which results in the formation of stable d^(10) configuration. The trend in the third ionisation enthalpies is not complicated by the 4s orbital factor and shows the greater difficulty of removing an electron from the d^(5) (Mn^(2+)) and d^(10) (Zn^(2+)) ions. In general, the third ionisation enthalpies are quite high. Also, the high values for third ionisation enthalpies of copper, nickel and zinc indicate why it is difficult to obtain oxidation state greater than two for these elements. In the series Sc(Z = 21) to Zn(Z = 30), the enthalpy of atomisation of Zn is the lowest. Why?
Answer» Solution :Zinc has no unpaired electron in its atom while OTHERS have unpaired electrons which RESULTS in metal-metal bond. As Zn has no unpaired electron, metal-metal bond formation is minimised and ENTHALPY of atomisation is minimised.

13.	Read the given passage and answer the questions number 1 to 5 that follow : The lowest common oxidation state of transition metals is + 2. To form the M^(2+) ions from the gaseous atoms, the sum of the first and second ionisation enthalpy is required in addition to the enthsualpy of atomisation. The dominant term is the second ionisation enthalpy which shows unusually high values for Cr and Cu where M^(+) ions have the d^(5) and d^(10) configurations respectively. The value for Zn is correspondingly low as the ionisation causes the removal of 1s electron which results in the formation of stable d^(10) configuration. The trend in the third ionisation enthalpies is not complicated by the 4s orbital factor and shows the greater difficulty of removing an electron from the d^(5) (Mn^(2+)) and d^(10) (Zn^(2+)) ions. In general, the third ionisation enthalpies are quite high. Also, the high values for third ionisation enthalpies of copper, nickel and zinc indicate why it is difficult to obtain oxidation state greater than two for these elements. Why is the third ionisation enthalpy of the metals Mn and Zn very high?
Answer» Solution :`MN^(2+)` and `Zn^(2+)` have the electronic configuration `3d^(5)` and `3d^(10)` respectively. Both these are stable configurations. Therefore, third ionisation enthalpy has a very HIGH value.

14.	Read the given passage and answer the questions number 1 to 5 that follow : The lowest common oxidation state of transition metals is + 2. To form the M^(2+) ions from the gaseous atoms, the sum of the first and second ionisation enthalpy is required in addition to the enthsualpy of atomisation. The dominant term is the second ionisation enthalpy which shows unusually high values for Cr and Cu where M^(+) ions have the d^(5) and d^(10) configurations respectively. The value for Zn is correspondingly low as the ionisation causes the removal of 1s electron which results in the formation of stable d^(10) configuration. The trend in the third ionisation enthalpies is not complicated by the 4s orbital factor and shows the greater difficulty of removing an electron from the d^(5) (Mn^(2+)) and d^(10) (Zn^(2+)) ions. In general, the third ionisation enthalpies are quite high. Also, the high values for third ionisation enthalpies of copper, nickel and zinc indicate why it is difficult to obtain oxidation state greater than two for these elements. Which factors are to be considered to obtain +2 oxidation state of the metals ?
Answer» Solution :ENTHALPY of atomisation SUM of FIRST and SECOND ionisation enthalpies.

15.	Read the given passage and answer the questions number 1 to 5 that follow : The lowest common oxidation state of transition metals is + 2. To form the M^(2+) ions from the gaseous atoms, the sum of the first and second ionisation enthalpy is required in addition to the enthsualpy of atomisation. The dominant term is the second ionisation enthalpy which shows unusually high values for Cr and Cu where M^(+) ions have the d^(5) and d^(10) configurations respectively. The value for Zn is correspondingly low as the ionisation causes the removal of 1s electron which results in the formation of stable d^(10) configuration. The trend in the third ionisation enthalpies is not complicated by the 4s orbital factor and shows the greater difficulty of removing an electron from the d^(5) (Mn^(2+)) and d^(10) (Zn^(2+)) ions. In general, the third ionisation enthalpies are quite high. Also, the high values for third ionisation enthalpies of copper, nickel and zinc indicate why it is difficult to obtain oxidation state greater than two for these elements. Which is the lowest common oxidation state of transition metals?
Answer» Solution :`+2` is the LOWEST common OXIDATION STATE.

16.	Read the given passage and answer the questions number 1 to 5 that follow : The lowest common oxidation state of transition metals is + 2. To form the M^(2+) ions from the gaseous atoms, the sum of the first and second ionisation enthalpy is required in addition to the enthsualpy of atomisation. The dominant term is the second ionisation enthalpy which shows unusually high values for Cr and Cu where M^(+) ions have the d^(5) and d^(10) configurations respectively. The value for Zn is correspondingly low as the ionisation causes the removal of 1s electron which results in the formation of stable d^(10) configuration. The trend in the third ionisation enthalpies is not complicated by the 4s orbital factor and shows the greater difficulty of removing an electron from the d^(5) (Mn^(2+)) and d^(10) (Zn^(2+)) ions. In general, the third ionisation enthalpies are quite high. Also, the high values for third ionisation enthalpies of copper, nickel and zinc indicate why it is difficult to obtain oxidation state greater than two for these elements. Why is the second ionisation enthalpy of the metals Cr and Cu very high ?
Answer» Solution :`Cr^(+)` has the electronic CONFIGURATION `3d^(5)` and `CU^(+)` has the electronic configuration `3d^(10)`. Both these configurations are STABLE states being half-filled and completely filled. Therefore it is more difficult to take out the second electron.

17.	Read the given passage and answer the questions number 1 to 5 that follow : The d-block of the periodic table contains the elements of the groups 3 - 12 and are known as transition elements. In general, the electronic configuration of these elements is (n - 1) d^(1-10)ns^(1-2). The d-orbitals of the penultimate energy level in their atoms receive electrons giving rise to the three rows of the transition metals i.e., 3d, 4d and 5d series. However, Zn Cd and Hg are not regarded as transition elements. Transition elements exhibit certain characteristic properties like variable oxidation states, complex formation, formation of coloured ions and alloys, catalytic activity, etc. Transition metals are hard (except Zn, Cd and Hg) and have a high melting point. Why is Cu^(2+) ion coloured while Zn^(2+) ion is colourless in aqueous solution ?
Answer» Solution :In `Cu^(2+)`, there is EXCITATION of electrons from lower to higher d-orbitals. Some energy is ABSORBED and it shows COLOUR. In `Zn^(2+)`, there is no such POSSIBILITY.

18.	Read the given passage and answer the questions number 1 to 5 that follow : The d-block of the periodic table contains the elements of the groups 3 - 12 and are known as transition elements. In general, the electronic configuration of these elements is (n - 1) d^(1-10)ns^(1-2). The d-orbitals of the penultimate energy level in their atoms receive electrons giving rise to the three rows of the transition metals i.e., 3d, 4d and 5d series. However, Zn Cd and Hg are not regarded as transition elements. Transition elements exhibit certain characteristic properties like variable oxidation states, complex formation, formation of coloured ions and alloys, catalytic activity, etc. Transition metals are hard (except Zn, Cd and Hg) and have a high melting point. Why are Zn, Cd and Hg non-transition elements ?
Answer» SOLUTION :They have COMPLETED d-orbitals.

19.	Read the given passage and answer the questions number 1 to 5 that follow : The d-block of the periodic table contains the elements of the groups 3 - 12 and are known as transition elements. In general, the electronic configuration of these elements is (n - 1) d^(1-10)ns^(1-2). The d-orbitals of the penultimate energy level in their atoms receive electrons giving rise to the three rows of the transition metals i.e., 3d, 4d and 5d series. However, Zn Cd and Hg are not regarded as transition elements. Transition elements exhibit certain characteristic properties like variable oxidation states, complex formation, formation of coloured ions and alloys, catalytic activity, etc. Transition metals are hard (except Zn, Cd and Hg) and have a high melting point. Why are melting points of transition metals high ?
Answer» Solution :This is due to involvement of GREATER number of electrons from (n-1)d and NS ORBITALS in metallic BONDING.

20.	Read the given passage and answer the questions number 1 to 5 that follow : The d-block of the periodic table contains the elements of the groups 3 - 12 and are known as transition elements. In general, the electronic configuration of these elements is (n - 1) d^(1-10)ns^(1-2). The d-orbitals of the penultimate energy level in their atoms receive electrons giving rise to the three rows of the transition metals i.e., 3d, 4d and 5d series. However, Zn Cd and Hg are not regarded as transition elements. Transition elements exhibit certain characteristic properties like variable oxidation states, complex formation, formation of coloured ions and alloys, catalytic activity, etc. Transition metals are hard (except Zn, Cd and Hg) and have a high melting point. Why do transition metals and their compounds show catalytic activity ?
Answer» Solution :This is due to their ABILITY to adoptt MULTIPLE oxidation states and to FORM complexes.

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21.	Read the given passage and answer the questions number 1 to 5 that follow : Originally the name transition metals was derived from the fact that their chemical properties were transitional between those of s and p-block elements. Now according to IUPAC, transition metals are defined as metals which have incomplete d-subshell either in the neutral atom or in their ions. Zinc, cadmium and mercury of Group 12 have full du configuration in their ground state as well as their common oxidation states and hence, are not regarded as transition metals. However, being the end members of the 3d, 4d and 5d transition series, respectively their chemistry is studied along with the chemistry of transition metals. The presence of partly filled d or f-orbitals in their atoms makes transition elements different from that of the non-transition elements. Hence, transition elements and their compounds are studied separately. However, the usual theory of valence as applicable to the non-transition elements can be applied successfully to the transition elements also. Give the names of first four members of first transition series.
Answer» SOLUTION :Scandium, TITANIUM, vanadium and CHROMIUM are the first FOUR members of first transition SERIES.

22.	Read the given passage and answer the questions number 1 to 5 that follow : The d-block of the periodic table contains the elements of the groups 3 - 12 and are known as transition elements. In general, the electronic configuration of these elements is (n - 1) d^(1-10)ns^(1-2). The d-orbitals of the penultimate energy level in their atoms receive electrons giving rise to the three rows of the transition metals i.e., 3d, 4d and 5d series. However, Zn Cd and Hg are not regarded as transition elements. Transition elements exhibit certain characteristic properties like variable oxidation states, complex formation, formation of coloured ions and alloys, catalytic activity, etc. Transition metals are hard (except Zn, Cd and Hg) and have a high melting point. Which transition metal of 3d series does not show variable oxidation states ?
Answer» SOLUTION :SCANDIUM.

23.	Read the given passage and answer the questions number 1 to 5 that follow : Originally the name transition metals was derived from the fact that their chemical properties were transitional between those of s and p-block elements. Now according to IUPAC, transition metals are defined as metals which have incomplete d-subshell either in the neutral atom or in their ions. Zinc, cadmium and mercury of Group 12 have full du configuration in their ground state as well as their common oxidation states and hence, are not regarded as transition metals. However, being the end members of the 3d, 4d and 5d transition series, respectively their chemistry is studied along with the chemistry of transition metals. The presence of partly filled d or f-orbitals in their atoms makes transition elements different from that of the non-transition elements. Hence, transition elements and their compounds are studied separately. However, the usual theory of valence as applicable to the non-transition elements can be applied successfully to the transition elements also. Is the theory of valence applicable to transition elements ?
Answer» SOLUTION :Yes, THEORY of VALENCE as applicable to non-transitions metals is ALSO applicable to TRANSITION metals.

24.	Read the given passage and answer the questions number 1 to 5 that follow : Originally the name transition metals was derived from the fact that their chemical properties were transitional between those of s and p-block elements. Now according to IUPAC, transition metals are defined as metals which have incomplete d-subshell either in the neutral atom or in their ions. Zinc, cadmium and mercury of Group 12 have full du configuration in their ground state as well as their common oxidation states and hence, are not regarded as transition metals. However, being the end members of the 3d, 4d and 5d transition series, respectively their chemistry is studied along with the chemistry of transition metals. The presence of partly filled d or f-orbitals in their atoms makes transition elements different from that of the non-transition elements. Hence, transition elements and their compounds are studied separately. However, the usual theory of valence as applicable to the non-transition elements can be applied successfully to the transition elements also. Do zinc, cadmium and mercury come under transition elements ?
Answer» SOLUTION :Being the end members of 3D, 4D and 5D series, their chemistry is studied along with the chemistry of TRANSITION metals.

25.	Read the given passage and answer the questions number 1 to 5 that follow : In the metallurgy of aluminium, purified Al_(2)O_(3) is mixed with Na_(3)AlF_(6)" or "CaF_(2) which lowers the melting point of the mixture and brings conductivity. The fused matrix is electrolysed. Steel vessal with lining of carbon acts as cathode and graphite anode is used. The overall reaction may be written as : 2Al_(2)O_(3)+3C to 4Al+3CO_(2) This process of electrolysis is widely known as Hall-Heroult process. Thus electrolysis of the molten mass is carried out in an electrolysis cell using carbon electrolysis. The oxygen liberated at anode reacts with the carbon of anode producing CO and CO_(2). This way for each kg of aluminium produced, about 0.5kg of carbon anode is burnt away. In Hall-Haroult's process for the extraction of Al, 10kg of carbon anode was burnt. Approximately, how much of Al metal was obtained?
Answer» SOLUTION :20KG of ALUMINIUM was OBTAINED.

26.	Read the given passage and answer the questions number 1 to 5 that follow : In the metallurgy of aluminium, purified Al_(2)O_(3) is mixed with Na_(3)AlF_(6)" or "CaF_(2) which lowers the melting point of the mixture and brings conductivity. The fused matrix is electrolysed. Steel vessal with lining of carbon acts as cathode and graphite anode is used. The overall reaction may be written as : 2Al_(2)O_(3)+3C to 4Al+3CO_(2) This process of electrolysis is widely known as Hall-Heroult process. Thus electrolysis of the molten mass is carried out in an electrolysis cell using carbon electrolysis. The oxygen liberated at anode reacts with the carbon of anode producing CO and CO_(2). This way for each kg of aluminium produced, about 0.5kg of carbon anode is burnt away. Which gases are evolved at the anode in Hall-Heroult's method?
Answer» SOLUTION :`CO" and "CO_(2)`.

27.	Read the given passage and answer the questions number 1 to 5 that follow : In the metallurgy of aluminium, purified Al_(2)O_(3) is mixed with Na_(3)AlF_(6)" or "CaF_(2) which lowers the melting point of the mixture and brings conductivity. The fused matrix is electrolysed. Steel vessal with lining of carbon acts as cathode and graphite anode is used. The overall reaction may be written as : 2Al_(2)O_(3)+3C to 4Al+3CO_(2) This process of electrolysis is widely known as Hall-Heroult process. Thus electrolysis of the molten mass is carried out in an electrolysis cell using carbon electrolysis. The oxygen liberated at anode reacts with the carbon of anode producing CO and CO_(2). This way for each kg of aluminium produced, about 0.5kg of carbon anode is burnt away. Write the equation for the reduction of alumina to aluminium.
Answer» Solution :`2Al_(2)O_(3)+3C to 4Al+ 3CO_(2)`

28.	Read the given passage and answer the questions number 1 to 5 that follow : Originally the name transition metals was derived from the fact that their chemical properties were transitional between those of s and p-block elements. Now according to IUPAC, transition metals are defined as metals which have incomplete d-subshell either in the neutral atom or in their ions. Zinc, cadmium and mercury of Group 12 have full du configuration in their ground state as well as their common oxidation states and hence, are not regarded as transition metals. However, being the end members of the 3d, 4d and 5d transition series, respectively their chemistry is studied along with the chemistry of transition metals. The presence of partly filled d or f-orbitals in their atoms makes transition elements different from that of the non-transition elements. Hence, transition elements and their compounds are studied separately. However, the usual theory of valence as applicable to the non-transition elements can be applied successfully to the transition elements also. What is the IUPAC definition of d-block elements ?
Answer» Solution :TRANSITION metals are DEFINED as metals which have incomplete d-subshells EITHER in the neutral atom or in their ions.

29.	Read the given passage and answer the questions number 1 to 5 that follow : Originally the name transition metals was derived from the fact that their chemical properties were transitional between those of s and p-block elements. Now according to IUPAC, transition metals are defined as metals which have incomplete d-subshell either in the neutral atom or in their ions. Zinc, cadmium and mercury of Group 12 have full du configuration in their ground state as well as their common oxidation states and hence, are not regarded as transition metals. However, being the end members of the 3d, 4d and 5d transition series, respectively their chemistry is studied along with the chemistry of transition metals. The presence of partly filled d or f-orbitals in their atoms makes transition elements different from that of the non-transition elements. Hence, transition elements and their compounds are studied separately. However, the usual theory of valence as applicable to the non-transition elements can be applied successfully to the transition elements also. What is the original definition of transition metals ?
Answer» SOLUTION :The ORIGINAL INTERPRETATION of the WORDS transition metals is that there is a transition (change) in the PROPERTIES of the elements when we move from s-block elements to p-block elements.

30.	Read the given passage and answer the questions number 1 to 5 that follow : In the metallurgy of aluminium, purified Al_(2)O_(3) is mixed with Na_(3)AlF_(6)" or "CaF_(2) which lowers the melting point of the mixture and brings conductivity. The fused matrix is electrolysed. Steel vessal with lining of carbon acts as cathode and graphite anode is used. The overall reaction may be written as : 2Al_(2)O_(3)+3C to 4Al+3CO_(2) This process of electrolysis is widely known as Hall-Heroult process. Thus electrolysis of the molten mass is carried out in an electrolysis cell using carbon electrolysis. The oxygen liberated at anode reacts with the carbon of anode producing CO and CO_(2). This way for each kg of aluminium produced, about 0.5kg of carbon anode is burnt away. Describe the composition of cathode and anode in the metallurgy of aluminium.
Answer» Solution :Steel VESSAL with liningof CARBON acts as cathode. GRAPHITE acts as anode.