ALCHEMY--THE CHEMISTRY TEACHER BLOG: I COMPLEMENTARY CHEMISTRY

MODULE—I BASIC CHEMICAL CONCEPTS

Periodic Table

Section—A

1. Modern periodic law states that properties of elements are a periodic function of their --------------------------

2. Lanthanides and actinides are also called ----- block elements or ------------------------elements.

3. Group 17 elements are also called -------------------

4. Lithium has diagonal relationship with ----------------

5. Colour of transition elements is due to ---------------------------------------

6. Give the names of any 4 noble gases --------------- ---------------- --------------- ----------

7. Magnetic property of transition elements is ----------------------- because of the presence of --------------------

8. Decrease in the force of attraction of the nucleus for an electron caused by the presence of electrons in the inner shell is called ------------------- effect

9. Half the distance between the centres of the nuclei of two like atoms bonded together by a single covalent bond is called -----------------------------

10. Which of the following has smallest covalent radius : Be, N, Li, O --------------------

11. Atomic radius ----------------- down a group

12. Na⁺ is ----------------- than Na and Cl^¯is ---------------------- than Cl

13. Three factors affecting ionization enthalpy are ----------------------------, ---------------------------- and ----------

14. Which group of elements in the periodic table has lowest ionization enthalpy? ----------------------

15. Ionization enthalpy of N is more than that of O because ------------------------------------------------------------------------------------------------------------

16. Why is electron affinity of Be, N and Ne positive?

17. How does ionization enthalpy change along a period and down a group?

18. Group of elements having the smallest atomic radius in any period is --------------------

19. Which of the following has highest electron affinity: F,Cl, Br, I --------------------

20. Number of elements in the 6^th period is --------------

Answer Key

1. Atomic numbers 2. F-block, inner transition 3. Halogens 4. Magnesium 5. d-d transition 6. Helium, Neon, Argon, Krypton, Xenon, Radon (any three) 7. Paramagnetic, unpaired electrons 8. Shielding/ Screening 9. Covalent Bond 10. O (covalent radius decreases from left to right along a period) 11. Increases 12. Smaller, Bigger 13. Atomic size, Nuclear charge, Screening effect of inner electrons 14. Alkali metals 15. N—2s² sp³— has exactly half filled p-subshell which is more stable. So removing one electron is difficult, and it has higher ionization enthalpy. Whereas O has outer shell electronic configuration 2s² 2p⁴, and removing one electron gives stable configuration 2s² 2p³. Therefore easy to remove electron and so low ionization enthalpy. 16. Outer shell electronic configuration of Be—2s² 2p⁰, N—2s² 2p³, Ne—2s² 2p⁶. All are stable electronic configurations (completely filled s or p subshell and exactly half filled p-subshell). Therefore difficult to remove electron and so higher ionization enthalpy than expected value. 17. Along a period—increases, Down a group—decreases 18. Halogens 19. Cl 20. 32

Section—B

1. Explain why elements with similar chemical properties are repeated at regular intervals?

Chemical properties of elements are based on their outermost shell electronic configuration. It is found that similar electronic configurations are repeated at regular intervals of 2, 8, 18. 18. 32 … Therefore elements with similar chemical properties also get repeated at these intervals.

Element	At. no	Difference in at. No.	Elec. Confg.
H	1		1s¹
Li	3	2	2s¹
Na	11	8	3s¹
K	19	8	4s¹
Rb	37	18	5s¹
Cs	55	18	6s¹
Fr	87	32	7s¹

2. What are inner-transition elements? Why are they so named/

The two series of f-block elements—Lanthanides and Actinides are together called inner transition elements. This is because these elements resemble the transition elements in many of their properties, but the last electron or the differentiating electron is filled in a shell inner to the transition elements. In transition elements the last electron is filled in (n—1)d subshell, whereas in inner transition elements the last electron enters the (n-2)f subshell.

3. Why are noble gases inert?

Noble gases have all their subshells completely filled. The octet is complete, and they have no tendency to form bonds for completeing their octet. Hence they are inert.

4. How does the arrangement in the periodic table reflect the electronic configuration of elements?

In the periodic table the elements are so arranged that similar electronic configurations are repeated at regular intervals. All the elements belonging to a particular group have the same outermost shell electronic configuration. For example the alkaline earth metals have configuration—2s², nitrogen group—2p³, halogens—2p⁵, copper group—d¹⁰ s¹ and zinc group—d¹⁰ s².

5. Explain the term diagonal relationship?

Diagonal relationship is exhibited between elements in the 1^st and 2^nd periods in the periodic table. An element in the 1^st period resembles the element in the 2^nd period placed diagonal to it in position. For example Li resembles Mg, Be resembles Al, B resembles Si.

6. What is meant by the term screening effect of inner electrons?

The electrons in the inner shells exert a repulsive force on the electrons in the outer shells. So the attraction of the nucleus on the outer shell electrons decreases, and the outer electrons feel as though the nucleus has less number of electrons than the actual number. i.e the effective nuclear charge felt by outer electrons is less. This effect is called shielding or screening effect.

7. Explain the variation of covalent radius along a period.

8. Why is a cation smaller than the neutral atom from which it is formed?

A cation is formed by the loss of one or more electrons from the neutral atom and it is positively charged. Hence the attraction of the nucleus on the electrons is more. This pulls the electrons closer and leads to decrease in size of the cation.

9. Which is larger Cl or Cl¯? Why?

Cl¯ is formed by the addition of one electron to the neutral atom. There are more electrons than protons. Hence the attraction of nucleus on electrons is less, leading to increase in size.

10. Why is second ionization enthalpy more than the first?

First ionization enthalpy is the energy required to remove the electron from outermost shell of a neutral gaseous atom. After the first electron is removed the atom becomes positively charged. Removing an electron from this positive ion is more difficult because the effective nuclear charge is more. Thus second ionization enthalpy is more than the first.

11. Explain how the magnitude of screening effect of inner electrons affects ionization enthalpy.

Due to screening effect of inner electrons the effective nuclear charge felt by the outermost electron decreases. Therefore it is easier to remove the outermost shell electron. As the screening effect increases the effective nuclear charge decreases, it becomes easier to remove the most loosely bound outermost shell electron and the ionization enthalpy decreases.

12. . Which has higher first ionization enthalpy-N or O? explain.

N has higher first ionization enthalpy. N has outermost shell electronic configuration—2s² 2p³. The p-subshell is exactly half filled and is hence more stable. Removing one electron destroys this stability. Hence I ionization enthalpy is more. O has the outermost shell configuration—2s² 2p⁴. Removing one electron gives the more stable configuration—2s² 2p³. Hence it easily loses one electron and the first ionization enthalpy is less.

13. Arrange the following in the increasing order of I ionization enthalpy—Li, Be, B. Explain your answer.

The order is Li < B < Be. As we move along a period the atomic size decreases and therefore the ionization enthalpy is expected to increase. Thus Li has smallest value. We expect Be to come next and B to have the highest value. But Be with electronic configuration 2s² has fully filled shells, is more stable and so has higher I.E than expected. On the other hand B with configuration 2s² 2p1 can attain stable fully filled configuration by losing one electron and thus has less I.E than expected. Thus B has less I.E than Be, and the order is as given above.

14. Among the atoms represented by the following electronic configurations which would have lowest I.E-1s² 2s²2p⁶; 1s² 2s² 2p⁵; 1s² 2s² 2p⁶ 3s¹? Why?

The configuration 1s² 2s² 2p⁶ 3s¹ would have the lowest I.E. This is i because in this configuration the outermost electron is in the 3^rd shell and in the others it is in the 2^nd shell. Addition of new shell increases atomic radius, the electron is father away, it has less attractive force from the nucleus, and so it is easily removed. Hence low I.E. Next higher value of I.E is for the configuration 1s² 2s² 2p⁵, and the highest value of I.E is for 1s² 2s²2p⁶, because the latter has completely filled electronic configuration, it has special stability, removal of electron from it is difficult, and so I.E is high.

15. Explain why electron affinity of Be is negative and its electron gain enthalpy is positive.

Electron affinity of an element is define as the energy emitted when an electron if added to the outermost shell of a neutral gaseous atom. Be with electronic configuration 1s²2s² has all the shells completely filled and is hence very stable. If an electron is added the stability is lost, and so energy has to be supplied for adding one electron. Therefore, since it is energy added, the electron affinity is negative. Electron gain enthalpy is the enthalpy change (ΔH ) of the process. Since it is an endothermic process ΔH is negative.

16. Explain why F has less electron affinity than Cl and iths electron gain enthalpy is less negative than Cl?

F is the first element in the Halogen group with outermost shell configuration 2s² 2p⁵. Addition of 1 electron to the very small 2psubshell leads to high inter-electronic repulsion. Whereas in Cl the electron is added in the bigger 3p subshel and the repulsion is less. Thus F has less tendency t ad electron than Cl and its electron affinity is less. In both cases energy is evolved when one electron is added and so ΔH is negative. But magnitude of energy evolved is less in F than in Cl and so ΔH is less negative for fluorine than chlorine ( F = —328, Cl = –349).

17. Distinguish between electronegativity and electron affinity.

Electronegativity is the tendency of an electron in covalent bond to attract the shared pair of electrons toward itself. It is a purely qualitative term and does not involve any energy change. The electronegativity if an element is a relative term and is a measure of its tendency to attract electrons in bond compared to F taken as =4.

Electron affinity is a quantitative term and is the energy evolved when an electron is added to the outermost shell of a neutral gaseous atom in the ground sate. It is an energy term and each element has a definite absolute value, irrespective of whether it is in the free or combined state.

Section—C

1. Discuss the general characteristics of s-block elements

The s-block elements are those in which the last or distinguishing electron enters the s-subshell. They have the following general characteristics

· They are metals and like all metals are good conductors of heat and electricity and show metallic luster.

· They have low I.E and hence easily loose electrons. Therefore they are good reducing agents, since the electron can be used for reduction.

· They are highly electropositive and form compounds with highly electronegative elements like halogens—eg NaCl.

· They impart characteristics colour to Bunsen flame. The electron in the outermost shell can be easily excited to higher level. When they are deexcited back to ground sate they emit the excess energy in the form of radiation. In this case the energy of the emitted radiation falls in the visible region.

· They form mostly ionic compounds, except for the compounds with Be which are covalent. They form ionic hydrides.

· They are highly reactive.

· The I.E decreases down the group, and the electropositivity increases down the group. Cs is the most electropositive element in the periodic table

2. What are transition elements? Discuss their general characteristics

The elements in which the last or distinguishing electron is added in the (n-1) d-subshell are called transition or d-block elements. They are called transition elements because their properties show a gradual transition from the highly electropositive metals of the groups 1 & 2 to the electronegative non-metals of the p-block. Their characteristics are as follows

· They are very good metals and show all the metallic properties like malleability, ductility, good conduction of heat and electricity, metallic luster. In fact all the metals used for making jewelery are transition metals like gold, platinum & silver, prized for their shine as well as the ease with which they can be drawn into very thin wires and sheets.

· They exhibit variable oxidation state. Electrons from the inner (n-1)d subshell also can take part in bonding. Thus they can exhibit more than one oxidation state. For example Fe has the outermost shell electronic configuration 3d⁶ 4s². The two electrons of the outermost 4s subshell can ionize to give the +2 oxidation state in ferrous compounds. When one more electron from the inner 3d subshell also ionizes then we get the +3 oxidation state in ferric compounds.

· They have very high melting and boiling points. They have strong metallic bond and, in addition to this, they can also form strong covalent bonds using the unpaired d-orbitals. Because of these two strong bonds they have higher melting and boiling points

· They can form alloys with each other

· Their compounds are coloured. The energy absorbed for transition between lower energy d orbital and higher energy d orbital is in the visible region. For example cupric compounds have blue colour and nickel compounds have green colour.

· Most of these metals have unpaired electrons and so they are paramagnetic

· They can easily form coordination complexes. This is because, the metal ions are small in size and have high effective nuclear charge and they also have empty d-orbitals for accepting electron pairs from ligands.

· The metal sand their compounds act as catalysts. For example Fe is the catalyst in Haber process for Ammonia , Vanadium pentoxide is catalyst in contact process for manufacture of H₂SO₄.

· They form interstitial compounds where small atoms like H, B, C N are filled in the tetrahedral and octahedral voids in their crystal lattice.

3. Discuss the position of Zinc group elements in the periodic table.

The Zinc group elements have the outermost shell electronic configuration (n-1) d¹⁰ ns². They have completely filled (n-1)d and ns shells. Therefore they cannot actually be classified under transition or d-block elements. The definition of transition elements is that they have incomplete (n-1)d shells. But they show more similarities with transition elements than with main group elements. They are

· They show variable valency to a certain extent. Even though zinc and cadmium show only +2 oxidaion state, mercury can show +1 oxidation state in Hg₂²⁺ compounds, and may exhibit +4 oxidation state in [HgF₄]

· Zinc forms many complexes with the same stoichiometry as complexes of copper(II),

· Zinc forms many alloys with Cu, Sn etc. just like transition metals.

At the same time they also show great dissimilarities with transition metals. They are

· They have comparatively low melting and boiling points. Mercury is a liquid at room temperature. This is because of weak metallic bond.

· Their compounds are colourless whereas transition metal compounds are coloured

· They and their compounds are diamagnetic

Since they have ns² configuration with all the inner shells fully filled, they also resemble alkaline earth metals

· For instance, zinc and cadmium are similar to beryllium and magnesium in their atomic radii, ionic radii, electronegativities,

· They form hydrides and amines like Be and Mg

Thus zinc, cadmium and mercury of group 12 of the periodic table can be considered to be d-block elements but not transition elements. They are sometimes also called pseudo-transition elements.

4. Discuss the variation of covalent radius along a period and down a group.

Along a period the covalent radius decreases. This is because, in successive elements along a period the atomic number is increasing by one unit, i.e the number of protons and hence the positive charge is increasing. But electrons are added in the same shell at the same distance from the nucleus. Thus the attraction of the nucleus on the electron increases and the nuclear charge increases, the electrons are pulled inward and the covalent radius decreases. Thus in any period the Alkali metals have the largest covalent radius and the Halogens have the lowest. Inert gases form no bond between themselves in the gaseous state and so exist as atoms only. Hence their atomic radii are expressed only in terms of their van der Waal’s radii.

Down a group the covalent radius increases. As we go down a group, always a new shell is being formed. This increases the distance of the outermost electron from the nucleus. There is also the effect due to increased screening effect of inner electrons. These two effects lead to increase in covalent radius. But at the same time the nuclear charge is increasing by a large amount. But this effect is cancelled by the other two effects, leading to an overall increase in covalent radius.

5. Define ionization enthalpy and electron affinity. State how they vary down a group of the periodic table.

Ionization enthalpy is defined as the enthalpy change associated with the removal of the most loosely bound electron from an isolated neutral gaseous atom in the ground state.

X(g) → X⁺(g) ΔH = Δ_iH = ionization enthalpy

Down a group the ionization enthalpy decreases. This is because, as we go down a group the atomic radius increases and the effective nuclear charge decreases. Thus the attractive force of the nucleus on the outermost shell electron decreases and it requires less energy to remove the most loosely bound electron.

Electron affinity is the energy released when one electron is added to the outermost shell of an isolated gaseous atom.

X(g) → X^¯(g) + EA ---Energy Evolved

X(g) → X^¯(g) – EA ---Energy Supplied

Electron affinity decreases down a group. As we move down the group, the atomic radius increases, screening effect of inner electrons increases and so the effective nuclear charge decreases, even though there is an increase in the number of protons in the nucleus. Thus, the added electron in the outermost shell feels lesser and lesser attractive force. Attraction always leads to decrease in energy, and the excess energy emitted is measured as electron affinity. As the attraction decreases, energy evolved becomes less, and the electron affinity decreases.

6. Explain the irregularities observed in the variation of electron affinity along the second period.

The second period consists of elements starting from Li to Ne. As we go along a period the electron affinity is expected to increase. This is found to be so, since the Li has EA= + 60 and F has EA= + 328. The positive value indicates that energy is evolved when electron is added. But it is found that the increase is not smooth and continuous. There are some exceptions. They are

· The EA of Be is –66—this is because Be has the outermost shell electronic configuration 2s². The 2s subshell is fully filled, and is stable. When electron is added the stability is lost. Thus instead of energy released, energy has to be supplied in order to add one electron to Be, and the EA becomes negative.

· The E.A of N is –31—N has the outermost shell electronic configuration 2s²2p³. The p-subshell is exactly half filled and is stable. Adding one electron destroys the stability. Thus energy has to be supplied for adding one electron and the E.A becomes negative.

· The EA of Ne is –116—Ne has outermost shell configuration 2s²2p⁶. The p-subshell is completely filed and is exceptionally stable. Adding one electron destroys the stability. Thus energy has to be supplied for adding one electron and the E.A becomes negative.

7. Discuss the general trend in the variation of electronnegativity along a period and down a group of the periodic table.

Electronegativity is defined as the tendency of an atom in a covalent bond to attract the shared pair of electrons towards itself. It is purely qualitative and not an energy term. Fluorine is found to have the highest value of electronegativity. The values for the other elements are measured compared to the lectronegativity of fluorine taken as = 4.

Along a period the electronegativity increases. This is because along a period the atonic radius decreases and the effective nuclear charge increases. Thus its nucleus can more effectively attract the electron of the outermost shell of the other atom with which it is forming bond, and its electronegativity increases.

Down a group electronegativity decreases. This is because as we move down a group the atomic radius increases and the effective nuclear charge decreases. Thus attraction for the shared pair of electrons decreases for successive elements down a group, and the electronegativity decreases.

Section—D

1. A) Define Ionization Enthalpy and discuss the factors that determine I.E of an element

B) Explain variation of I.E along a perod and down a group

A) Ionization enthalpy is defined as the enthalpy change associated with the removal of the most loosely bound electron from an isolated neutral gaseous atom in the ground state.

X(g) → X⁺(g) ΔH = Δ_iH = ionization enthalpy

Factors Affecting ionization enthalpy

a) Atomic size—as the atomic size increases, the distance of nucleus to outermost electron increases, the nuclear attraction decreases, and, therefore, the I.E decreases.

b) Nuclear charge—as nuclear charge increases the attractive force on the outermost electron increases. Thus more energy would be needed to remove the most loosely bound electron and the ionization enthalpy increases

c) Screening effect of inner electrons—as the number of inner shells or the number of electrons in the inner shell increases, they screen the outermost electron more effectively from feeling the attraction of the nucleus. Thus the effective nuclear charge decreases, and the attractive force on the electron decreass, leading to decrease in I.E

d) Type of electron being removed—the electron in the s-subshell being spherically symmetrical is closer to the nucleus. An electron in ssubshell would feel greater attraction from nucleus and would be most difficult to remove. P-subshell has more diffuse shape and this increases to d and f. thus the s-electron is most tightly bound and most difficult toremove and has slightly higher I.E than p, which is higher than d which is higher than f-electron. Thus I.E increases in the orde

f < d < p < s

e) Stable electronic configuration—a completely filled and exactly half-filled subshell has very less tendency to remove electrons. This is because it will become less stable. So more energy has to be supplied to remove the electron and the element would have exceptionally high value of I.E. on the other hand if the configuration of an element is such that it becomes completely filled or exactly half filled on loosing an electron, it will easily loose the electron to attain stability. It will therefore have exceptionally low value of I.E.

B) Down a group the ionization enthalpy decreases. This is because, as we go down a group the atomic radius increases and the effective nuclear charge decreases. Thus the attractive force of the nucleus on the outermost shell electron decreases and it requires less energy to remove the most loosely bound electron.

Along a period the I.E increases. This is because, as we move along successive elements in a period, the atomic radius decreases and the effective nuclear charge increases, and it becomes more and more difficult to remove the most loosely bound electron, and the energy required increases.

But there are many exceptions to this general trend. Be, N and Ne, having outermost shell electronic configurations 2s²; 2s²2p³;2s²2p⁶ respectively have fully filled or exactly half filled shells and are exceptionally stable. Thus they have exceptionally high value of I.E. On the other hand B and O, which are 2s²2p¹and 2s²2p⁴, will become completely filled (similar to Be) and exactly half filled (similar to N) respectively when they lose one electron. Hence they have greater tendency to lose electrons and become stable, and have lower I.E than expected.

Mole Concept

Section—A

1. Exactly 1/12^th the mass of one atom of Carbon-12 is called ---------------

2. Gram atomic mass is atomic mass expressed in -----------------

3. Avogadro number = 1mol = ------------------------

4. 64 grams of oxygen contains ------------------ moles of oxygen atoms

5. Number of atoms in 28 grams of nitrogen = --------------

6. Volume occupied by 1 mol of a substance at a given temperature and pressure is called its ---------------

7. Volume occupied by 44 grams of carbon dioxide at STP = ---------

8. Equal volumes of all gases at the same temperature and pressure contain equal number of molecules is the ------------------- law

9. Valency of aluminium in Al₂(SO₄)₃ is -----------

10. Valency of phosphate ion in H₃(PO₄) is ------------

11. Fe exhibits variable valency of ----- and -------

12. Basicity of acids is ----------------------

13. Acidity of bases is -------------------

14. Number of gram moles of a substance that is present in 1000 grams of solvent is called ------------

15. Mass of 5 moles of nitrogen atoms is = -----------

Answer Key

1. amu 2.grams 3. 6.02x 10²³ 4.8 moles 5. 2 mols=2x6.02x 10²³ atoms 6. Molar volume 7. 22.414L 8.Avogadro’s 9. 3 10. 3 11. 2 and 3 12. Number of replaceable hydrogen atoms 13. Number of replaceable OH groups 14. molality 15. 5x14 = 70g

Section—B

1. What is meant by variable valency? Give one example.

Some elements can exhibit different valencies in different compoundsd. i.e they can exhibit than one valency or, we can say, they show variable valency. For example Fe shows valency =2 in FeCl₂ and FeSO₄ and valency =3in FeCl₃ and Fe₂(SO₄)₃.

2. What is the difference between valency and oxidation state ⁄ oxidation number?

Valency is just a number (neither positive nor negative)and is the number of hydrogen atoms with which one atom of the element can combine. Whereas the oxidation state or oxidation number of an element is negative or positive and is the formal charge on an element in a compound when the electrons are counted according to certain rules.

3. What is the basicity of the following acids? H₂(SO₄), HNO₃, HCl, CH₃COOH, H₃(PO₄), H₂C₂O₄(oxalic acid), HNO₂, H₂(SO₃) ?

Basicity of an acid is the number of replaceable H atoms in 1 molecule. Therefore basicity of H₂(SO₄) = 2, HNO₃= 1, HCl = 1, CH₃COOH = 1( only the H atom of COOH is replaceable.), H₃(PO₄) = 3, H₂C₂O₄(oxalic acid) = 2 (since it is COOH—COOH the H atoms on both COOH are replaceable, HNO₂ = 1, H₂(SO₃) =1

4. What is the valency of a) Al in AlCl₃ b) SO₄^2– in H₂(SO₄) c) PO₄^3– in Mg₃(PO₄)₂ d) Mg in Mg₃N₂?

a) Valency of Al is 3, since it is forming bonds with 3 monovalent Cl¯ ions

b) Valency of SO₄^2– is 2 since it bonds with 2 H atoms

c) Valency of PO₄^3– is 3 since the charge on the ion is 3

d) Valency of Mg is 2

5. Why is the molarity of HCl equal to that of its normality?

Equivalent mass of an acid is = Molecular Mass

Basicity

Basicity of HCl is 1 since it has only 1 replaceable H atom. Therefore its Equivalent mass is same as its Molecular mass, and the number of gram moles will be equal to the number of gram equivalents. Since molarity is the number of gram moles per litre of solution and normality is the number of gram equivalents per litre of solution, they would be the same since the number of gram moles is equal to the number of gram equivalents of HCl.

S ec

ASSIGNMENT PROBLEMS

1. Calculate the number of molecules at 273K and 760mm pressure present in 200mL of CO₂ and 0.025L of ammonia

a) Molecular mass of CO₂ is 44

44g of CO₂ contains 6.02x10²³ molecules of CO₂ and has a volume of 22.414L

i.e22.414L or 22414mL contains 6.02x10²³ molecules

Therefore 200 mL contains 6.02x10²³ x 200

22414

= 5.37 x 10²¹ molecules

b) Molecular mass of NH₃ is 17

17g of CO₂ contains 6.02x10²³ molecules of CO₂ and has a volume of 22.414L

i.e22.414L or 22414mL contains 6.02x10²³ molecules

Therefore 0.025L contains 6.02x10²³ x 0.025

22.414

= 6.72 x 10²⁰molecules

2. Calculate the number of atoms present in 10.6g of Na₂CO₃

Molecular mass of Na₂CO₃ is 106

106g of CO₂ contains 6.02x10²³ molecules of Na₂CO₃

Therefore 10.6g contains 6.02x10²³ x 10.6

106

= 6.02x10²² molecules

Each molecule contains 3atoms of oxygen. Therefore number of atoms in 6.02x10²²molecules is 3 x 6.02x10²² atoms.= 18.02 atoms

3. What is the total mass of the mixture of 3.011x10²⁴ dinitrogen molecules and 1.2044x10²⁴ dioxygen molecules?

Molecular mass of dinitrogen is 14g and molecular mass of dioxygen is 32g.

6.02x10²³molecules of dinitrogen has a mass of 28g. Therefore mass of 3.011x10²⁴molecules is

28 x 3.011x10²⁴

6.02x10²³

= 140.04

6.02x10²³molecules of dioxygen has a mass of 32g. Therefore mass of 1.2044x10²⁴ molecules is

32 x 1.2044x10²⁴

6.02x10²³

= 64.02g

Total mass = 140.04 + 64.02 = 204.06

4. A mixture of Helium and dioxygen in the 1:1 molar ratio occupies a volume of 11.207 dm³ at 273k and 760mm. What is the total mass of mixture?

1 mol of He = 4g = 22.414L

1 mo of dioxygen = 32g = 22.414L

1 mol of He + 1 mol of dioxygen = 4 + 32 =36g = 22.414 + 22.414 = 44.818L

Therefore 11.207L has a mass of -- 36x 11.207

44.818

= 9g

5. 3.75g of a gas occupies a volume of 2.8L at 273K and 1.01325 bar. Calculate the molecular mass of gas.

22.141L of the gas is 1 mol. Therefore 2.8L is 1 x 2.8

22.414

= 0.1265 moles

0.1265 moles of the gas has a mass of 3.75g

Therefore mass of 1 mol = 3.75

0.1265

= 30g = molecular mass of the gas

6. Calculate the number of moles of water in 488g of BaCl₂. 2H₂O.

Gram-molecular mass of BaCl₂. 2H₂O = 244g=1 mol

Therefore number of moles in 488g = 1 x 488

244

= 2 moles

1 mol of BaCl₂. 2H₂O contains 2 moles of water (from formula)

Therefore 2 moles of BaCl₂. 2H₂O contains 4 moles of H₂O

7. How many gram moles will be present in 0.5L of a 4M solution?

M = n / V. \Number of moles = n = M x V = 4 x 0.5 = 2 moles

8. 11.7g of NaCl dissolved in water to give 500mL of the solution. Calculate molality and molarity of solution. Density of solution is 1.1g/mL.

Molarity of solution = w₂ x 1000 = 11.7 x 1000 = 0.4003M

M₂ x V58.45 x 500

Molality of solution = w₂ x1000

M₂ x w₁

Mass of solution = density x volume == 1.1 x 500 = 550

Mass of solute ( NaCl) = 11.7g

Mass of water = 550—11.7 = 539 = w₁

\ Molality = 11.7 x 1000 = 0.3714m

58.45 x 539

9. How many mole of non-volatile solute of molar mass 60 will be present in 2.240kg of a 2 molal solution?

A 2 molal solution contains 2 moles = 2 x 60 = 120g of solute present in 1000g of water.

Total mass of water + solute = 1000 + 120 =1120g of solution = 1.120kg of solution

\ Mass of solute present in 2.240kg of solution = 120 x 2.240 = 240g =240/60 moles

1.120 = 4 moles

10. Calculate mass of Na₂CO₃ that is to be dissolved to prepare 500mL of 0.1M solution.

W = MEV/1000 = 0.1 x 126 x 500 /1000 = 5.3g

11. Calculate the molarity and molality of an aqueous solution of HCl that contains 37% w/w of HCl. Density of solution is 1.18g/mL.

A 37% w/w solution of HCl contains 37g in 100g of solution

Mass of water = 100—37 = 63g

Volume of 100g of solution = Mass /density = 100 / 1.18 =84.75mL

Molality = w₂ x1000 = 37 x 1000 = 16.11m

M₂ x w₁ 36.45 x 63

Molarity = w₂ x1000 = 37 x 1000 = 11.97M

M₂ x V 36.45 x84.75

12. You are given 0.64g of oxygen and 0.42g of nitrogen gas in two separate jars. Calculate for each the volume at STP, the number of moles and the number of molecules.

Oxygen gas = O₂. Molar mass = 32g.

Volume of 32g (1 gram mol)at STP = 22.414L

Therefore volume of 0.64g = 22.414 x 0.64 = 0 .44828L

32g = 1 mol therefore 0.64g =0.64 / 32 mols = 0.02 mols

1 mol contains Avogadro number of molecules = 6.02 x10²³ molecules

\Number of molecules in 0.02 mols = 0.02 x 6.02 x10²³ = 1.204 x 10²²molecules

Nitrogen gas = N₂. Molar mass = 28g.

Volume of28g (1 gram mol)at STP = 22.414L

Therefore volume of 0.42g = 22.414 x 0.42 = 0 .33621L

32g = 1 mol therefore 0.64g =0.42 / 28 mols = 0.015 mols

1 mol contains Avogadro number of molecules = 6.02 x10²³ molecules

\Number of molecules in 0.015 mols = 0.015 x 6.02 x10²³ = 9.03 x 10²¹ molecules

13. If there are 6x10⁸⁰ atoms in the known universe, how many moles do they represent?

1 mol = 6.02 x10²³. Therefore 6x10⁸⁰ atoms = 6x10⁸⁰ / 6.02 x10²³ = 0.9967 x 10⁵⁷ atoms

14. Which has more mass—(a) 50g of Fe (b) 5g atom of N₂ (c) 0.1g atom of Ag (d)1x10²³ atoms of C

Mass of Fe = 50g

Mass of 5 gram atom of nitrogen = 14 x 5 = 70g

Mass of 0.1 gram atom of silver = 0.1 x 170 = 17g

Mass of 1 x 10²³ atoms of carbon = 12 x 1 x 10²³ / 6.02 x10²³ = 1.99g

\ 5 gram atom of nitrogen has more mass

15. Mass of 5600mL of a gas at STP is 12g. Calculate molecular mass

22414 ml is the volume of 1 gram mol

Therefore 5600 ml = 5600 / 22414 = 0.25 mols

Mass of 0.25 mols = 12g. Therefore mass of 1 mol = 12 / 0.25 = 48= molecular mass

16. 4.86L of CO₂ at 12ºC and 770mm pressure has mass = 9.27g. Calculate molecular mass

First convert the volume to STP.

P₀V₀ = P₁V₁ V₀= P₁V₁ x T₀ = 770 x 4.86 x273 = 4.72L

T₀T₁T₁x P₀285x 760

22.414 L is the volume of 1 gram mol

Therefore 4.72L = 4.72 / 22.414 = 0.21mols

Mass of 0.21 mols = 9.27g. Therefore mass of 1 mol = 9.27 / 0.21 = 44.14

17. 3.5g of Cu gave 4.5g of its oxide. Calculate equivalent mass

Mass of Cu =3.5g

Mass of oxide = 4.5g

Therefore mass of oxygen combined with 3.5g Cu = 4.5 – 3.5 = 1g

Therefore mass of Cu that will combine with 8g of Oxygen = 3.5 x 8 = 28= equivalent mass

18. 0.33g of a metal reacted with acid to give 113mL of H₂ gas at STP. Calculate equivalent mass.

Mass of metal reacted == 0.33g

Mass of hydrogen liberated at STP == 113mL

Mass of metal that will liberate 11212 mL of hydrogen at STP == 0.33 X 11212 / 113

= 32.71=EQ. MASS

Oxidation and Reduction—Redox Reactions

Section— A

1. Oxidation is ----- and reduction is ----------

2. In the reaction between sodium and chlorine the oxidizing agent is -------

3. Oxidation number of Cr in Cr₂O₇^2—is ----------

4. Oxidation number of O in OF₂ is ---------

5. Reduction involves ----- in oxidation state and oxidation involves ---- in oxidation state

6. An oxidizing agent undergoes ---------

7. KMnO₄is an example for -------agent.

8. ------- and ----- are two oxidizing agents

9. Oxalic acid -------- permanganate to Mn²⁺

10. In the reaction between nitric acid and hydrogen sulphide, H₂S ------ HNO₃

Answer Key

1. De-electronation, electronation 2. Chlorine 3. +6 4.+2

5. decrease, increase 6. reduction 7. oxidising 8. KMnO₄, K₂Cr₂O₇ 9. reduces 10.reduces

Section—B

1. Define oxidation and reduction in terms of electronic concept

According to electronic concept oxidation is de-electronation and reduction is electronation. Or in other words oxidation is loss of electrons and reduction is gain of electrons

2. Define oxidation number

Oxidation number of an element is defined as the formal charge the element apers to have when the electrons are counted according to the following rules: a) electrons shared between like atoms are shared equally b) electrons shared between unlike atoms are counted with more electronegative atom

3. What is meant by redox reactions? Give one example.

Reactions in which oxidation and reduction take place side by side are called redox reactions. For example in the reaction—3H₂S +HNO₃ → 2NO + 3S + 4H₂O, H₂S is oxidised to S and HNO₃is reduced to NO. therefore it is a redox reaction

Section—C

4. Find oxidation number of S in the following: H₂SO₄, Na₂S₂O₃, Na₂S₄O₆

H₂SO₄

Oxidation number of H = +1

Oxidation Number of O = —2

(+1x2 ) + (--2 x 4) + (oxidation number of S) = 0

+2 –8 + X = 0. \ X = +6

Na₂S₂O₃

Oxidation number of Na = +1

(+1 x 2) + (2X) + (3x –2 ) = 0

X= 4/2 = 2

Na₂S₄O₆

(2 x +2) + (4X) + (A—2 x 6) = 0

X = 10/4 = 2.5

5. Identify the oxidising and reducing agents in the following reactions

2FeSO₄ + H₂SO₄ + Cl₂ → Fe₂(SO₄)₃ + HCl

Oxidation number of Fe increases from +2 to +3. FeSO₄ is getting oxidized. \ FeSO₄ is reducing agent

Oxidation number of Cl decreases from 0 to –1 . Cl is getting reduced. \ Cl is oxidizing agent

Cu +4HNO₃ → Cu(NO₃)₂ +2NO +H₂O

Oxidation number of Cu increases from 0 to +2. Cu is getting oxidized. \ Cu is reducing agent

Oxidation number of N in HNO₃decreases from +6 to +2. HNO₃ is getting reduced. \ HNO₃ is oxidizing agent

2KMnO₄ + 8H₂SO₄ +10FeSO₄ → K₂SO₄ + 2MnSO₄ + 8H₂O + 5Fe₂(SO₄)₃

Oxidation number of Mn in KMnO₄ decreases from +7 to +2. KMnO₄ is getting oxidized. \ KMnO₄ is oxidising agent

Oxidation number of Fe increases from +2 to +3. FeSO₄ is getting oxidized. \ FeSO₄ is reducing agent

3I₂ + 6NaOH → NaOI + 5NaI + 3 H₂O

Oxidation number of I decreases from 0 to –1 in NaI. I is getting reduced. \ I is oxidizing agent

Oxidation number of I increases from 0 to +1 in NaOI. I is getting oxidized. \ I is reducing agent. I acts both as oxidizing and reducing agent

K₂Cr₂O₇ + 14HCl → 2KCl + 2CrCl₃ + 3Cl₂ + 7 H₂O

Oxidation number of Cr in K₂Cr₂O₇ decreases from +6 to +3. K₂Cr₂O₇ is getting reduced. \ K₂Cr₂O₇ is oxidizing agent.

Oxidation number of Cl in HCl increases from –1 to 0. Cl is getting oxidized. \ Cl is reducing agent

2MnO₄^— + H₂S → 2MnO₂+3S +2 H₂O + 2OH^—

MnO₄^—1is getting reduced to MnO₂. So MnO₄^—1 is oxidizing agent. H₂S is getting oxidized to S.So H₂S is reducing agent

Cr₂O₇^2— + SO₂ + 2H⁺→ 2Cr3⁺ + 3SO₄^2— + H₂O

Cr₂O₇^2— is getting reduced to Cr³⁺. So it is oxidizing agent. SO₂ is getting oxidized to SO₄^2—. So it is reducing agent.

6. Explain the rules for determining oxidation number of an element in a compound

The rules are as follows:-

· Oxidation number of elements in uncombined state is 0

· Oxidation number of F in all compounds is –1. For other halogens it is –1 when bonded to less electronegative atom.

· Alkali and alkaline earth metals have +1 and +2 Oxidation number respectively,

· Oxidation number of H is +1 in all compounds. But in ionic hydrides it is –1

· Oxidation number of O is –2. But in peroxides it is –1 and in superoxides it is –1/2 and in F₂O & F₂O₂ it is +2 and +1 respectively

· Sum of Oxidation number of all elements in neutral compounds is 0

· Sum of Oxidation number of all elements in ions is equal to charge on the ion

Theories of Acids and Bases

Section—A

1 . According to Arrhenius concept acid is a ------containing compound and base is a ----containing compound

2 . In the Lowry-Bronsted theory acid is a------ and base is a ----------

3 . A pair of acid and base that can be converted into each other by gain or loss of proton is called ------------

4 . ---------- is a substance that can act as both acid and base

5 . Example of amphiprotic substance is ------ and -----

6 . According to Lewis concept of acids and bases an electron pair donor is a ----- and electron pair acceptor is a --------

7 . Example for Lewis acid is ----- and Lewis base is -----

Answer key

1. hydrogen, hydroxyl 2. Proton donor, proton acceptor 3. Conjugate acid base pair 4. Amphiprotic substance 5. H₂O, CH₃COOH 6. base, acid 7. BF₃, NH₃

Section—B

1. What is the Arrhenius concept of acids and bases?

According to the Arrhenius theory of acids and bases, an acid is a substance containing hydrogen that ionizes in water to release a proton and a base a compound containing hydroxyl group that ionizes to produce an OH^— ion.

2. Define Bronsted acid and Bronsted base

According to Lowry –Bronsted theory an acid is a substance that can donate a proton to any other substance and a base is any substance that can accept a proton from any other substance.

3. Define conjugate acid base pair. Give 2 examples

A pair of acids and bases that can be converted into each other by gain or loss of proton is called a conjugate acid base pair. Examples—

+H⁺ -H⁺

HCl → Cl^— and H₂O → H₃O⁺

-H⁺ +H⁺

4. What are amphiprotic substances?

Substances that can act both as acids and bases are called amphoprotic substances or amphoteric substances. They can act as proton donors or acceptors. Example HCl + H₂O → H₃O⁺ + Cl^—

NH₃ + H₂O → NH₄⁺ + OH^—

We can see that H₂O acts as acid when it reacts with HCl. It accepts proton to form H₃O⁺. When it reacts with NH₃it acts as an acid and donates proton to NH₃and itself gets converted to OH^—. Therefore H₂O is an amphiprotic substance.

HCO₃^— + NH3 → CO₃^2— + NH₄⁺

HCO₃^— + HCl → H₂CO₃ + Cl^—

We can see that HCO₃^— can act as both acid and base. It can donate electrons to NH₃ and accept electrons from HCl. Thus it is an amphiprotic substance.

5. Define Lewis acid and base

A substance that can donate a pair of electrons to any other substance is called a Lewis base and a substance that can accept a pair of electrons from any other substance is called a Lewis acid. Thus a Lewis acid is an electron pair acceptor and a Lewis base is an electron pair donor

6. What is the conjugate acid of the following—H₂O, NH_3,Cl^— ,HCO₃^—, NH₂^—, OH^—

Conjugate acid of H₂O is H₃O⁺

NH₃is NH₄⁺

Cl^—is HCl

HCO₃^— is H₂CO₃

NH₂^— is NH₃

OH^— is H₂O

7. What is the conjugate base of -- H₂O, NH₄⁺ , HCl, H₂CO₃, NH₃, H₃O⁺

Conjugate base of H₂O is OH^—, NH₄⁺is NH₃, HCl is Cl^—, H₃CO₃ is HCO₃^—, NH₃ is NH₂^—, H₃O⁺ is H₂O

Section—C

1. A) Discuss Lowry-Bronsted theory of acids and bases

The answer should include the following points—1) definition 2) equation showing which is acid and which is base 3) conjugate acid base pair 4) amphiprotic substances-definition and examples with equations

B) Explain conjugate acid base pair. Which are the conjugate acid base pairs in the following?

HCl + CH₃COOH → CH₃COOH⁺ + Cl^—

NH₃ + HCl → NH₄⁺ + Cl^—

A pair of substances which can be converted to each other by gain or loss of protons is called a conjugate acid-base pair.

In the first equation HCl → Cl^— make a conjugate acid-base pair because HCl loses a proton to become Cl^— and Cl^— gains a proton and gets converted to HCl. Another conjugate acid-base pair is CH₃COOH → CH₃COOH⁺. CH₃COOH gains proton and gets converted to CH₃COOH⁺.and CH₃COOH⁺.loses a proton and gets converted to CH₃COOH.

In the second equation the conjugate acid-base pairs are NH₃→NH₄⁺ and HCl → Cl^—. When proton is added to NH₃ it gets converted to NH₄⁺ and when NH₄⁺ loses a proton it gets converted to NH₃. Similarly HCl loses proton to give Cl^— and Cl^— gains a proton and becomes HCl

2. Discuss Lewis theory of acids and bases. Indicate Lewis acid and base in the following

The answer should include definition of Lewis acid and base, examples with explanation, which substances can be classified as Lewis acid, limitations of the Lewis acid-base concept

Cu²⁺ + 4NH₃_→ [Cu(NH₃)₄]²⁺ Lewis acid Cu²⁺, base NH₃ because NH₃ donates electron pair to Cu²⁺.

Na⁺F^— + BF₃→ Na⁺BF₄^—Lewis acid is BF₃because it accepts pair of electrons from F^—. F^—is a Lewis base because it donated pair of electrons to BF₃

MODULE—II ANALYTICAL CHEMISTRY

Section—A

1. A solid compound of sufficient purity from which a standard solution can be prepared by direct weighing is called----------

2. A solution whose concentration is known is called a --------- solution

3. The stage in the titration at which the reactants are used up in the exact stoichiometric proportions is called -------------

4. The point of completion of reaction determined using an indicator is called-------

5. A solution which can be used as a standard solution if its concentration can be determined using a standard solution is called ----

6. Titrations using standard acid solution is called------

7. Titration using standard alkali solution is called -------

8. ------ and ------ are two indicators that can be used in acid-base titrations

9. ------- is the indicator for the titration of weak base

10. ------ is the indicator for titration of weak acid

11. Permanganometry and dichrometry are examples of -------- titrations

12. N-phenyl anthranilic acid is a ---- indicator

13. ------ and ----- are oxidizing agents used in redox titrations

14. Examples for redox indicators are – and ------

15. Redox indicator undergoes a colour change due to the sudden change in ------ in the vicinity of the equivalence point

16. Direct titration of standard iodine against a reducing agent is called ------

17. Titration of iodine liberated during a reaction with a suitable reducing agent is called -------

18. Indicator used in iodimetry and iodometry is ----

19. Iodine is soluble in --- due to the formation of ----

20. Iodometry titration can be used to estimate ----- and ------ and ------

21. Reducing agent used in iodimetry- iodometry titrations is -----

22. EDTA is --------

23. EDTA is a -------- ligand

24. Example for complexometric / metallochromic indicators are ---- and ------

25. Colour change for Eriochrome Black—T indicator is ----- to -----

26. Complexometric titrations can be used for estimation of ------- and --------

27. Hardness of water can be determined using --- titrations

28. Product of concentration of ions in a solution is called ------

29. Product of concentration of ions in a saturated solution is -------

30. Precipitation occurs when --------- exceeds -----------

31. Suppression of dissociation of weak electrolyte by the addition of strong electrolyte containing a common ion is called--------

32. HCl is added in Group II of qualitative analysis to suppress the dissociation of ----

33. Dissociation of NH₄OH is suppressed in Group—III by adding -------

34. Solubility product of Group—IV metal ion sulphides is ----- than that of Group—II sulphides

35. The --- of a determination is the concordance of observed value and the true value or most probable value

36. ----- is defined as the concordance of a series of measurements of the same quantity.

37. Ratio of the error to the true or most probable value is called ------ error

38. Relative error can be expressed as ---- or -----

39. In inorganic qualitative analysis --------Group cations are precipitated as carbonates

40. Group – II cations are precipitated as -------

41. Addition of sodium acetate suppresses the dissociation of acetic acid. This is an example for ---- effect.

Answer key

1. Primary standard 2. Standard solution 3. Equivalence point. 4. End point 5. Secondary standard. 6. Acidimetry. 7. Alkalimetry. 8. Phenolphthalein and Methyl Orange. 9. Phenolphthalein. 10. Methyl Orange. 11. Redox. 12. Redox. 13. Potassium permanganate and potassium dichromate. 14. N—phenyl anthranilic acid and diphenyl amine. 15. Potential. 16. Iodimetry. 17. Iodometry. 18. Starch solution. 19. KI, KI₃^—. 20. KMnO₄, K₂Cr₂O₇, and Cu²⁺ ion. 21. Sodium thiosulphate. 22. Ethylene Diamine Tetraacetic Acid. 23. Chelating or Hexadentate (both are correct). 24. Eriochrome Black-T, Murexide. 25. Wine red, sky blue. 26. Mg²⁺, Zn²⁺. 27. Complexometric titrations. 28. Ionic product. 29. Solubility product. 30. Ionic product exceeds solubility product. 31. Common ion effect. 32. H₂S, 33. NH₄Cl. 34. More / greater. 35. Accuracy. 36. Precision. 37. Relative. 38. Percentage, parts per thousand. 39. V. 40. Sulphides. 41. Common ion effect.

Section—B

1. Which is the indicator used in titration of weak acid vs strong base, strong acid vs weak base and strong acis vs strong base?

Weak acid vs strong base – phenolphthalein

Strong acis vs weak base –methyl orange

Strong acid vs strong base –both phenolphthalein and methyl orange.

2. What is meant by permanganometry titrations?

Redox titrations involving potassium permanganate as oxidizing agent are called permanganometry titrations. Example—titration of KMnO₄ vs oxalic acid

3. What is meant by dichrometry titrations?

Redox Titrations involving potassium dichromate as oxidizing agent are called dichrometry titrations. Example—K₂Cr₂O₇vs Fe²⁺

Direct titration of standard iodine solution against reducing agent is called iodimetry titrations whereas titration of iodine liberated from a chemical reaction with a reducing agent is called iodometry titrations.

5. How does N-phenyl anthranilic acid act as indicator in dichrometry titrations?

N-phenyl anthanilic acid indicator is in the reduced form and is colourless. At the end point it gets oxidized to the violet coloured oxidized form. Thus it shows a colour change from colourless to violet at the end point of the titration

6. What are metallochromic indicators? Give one example. Which titration is it used?

Indicators used in complexometric titrations to indicate the presence of metal ions are called metallichromic indicators or complexometric indicators.They are responsive to metal ion concentration and change colour with the presence or absence of metal ions.

7. Iodine is prepared by dissolving in KI solution. Why?

Iodine has very little solubility in water. Iodine is highly volatile and the concentration of aqueous iodine solutions will decrease on keeping. But iodine is readily soluble in KI due to the formation of the complex KI₃. The solution is fairly stable and the iodine is not lost from it. Therefore iodine solution is prepared in KI.

8. What is meant by solubility product? Write equation for solubility product of calcium phosphate

Product of concentration of ions in a saturated solution is called solubility product.

Solubility product of calcium phosphate –Ca₃(PO₄)₂can be written as

K_sp = [Ca²⁺]³[(PO₄)^3—]²

9. Differentiate between ionic product and solubility product

Product of concentration of ions in a solution having any concentration is called ionic product. Solubility product is a special case of ionic product and refers to the ionic product of a saturated solution.

10. What happens when sodium acetate is added to acetic acid?

Acetic acid—CH₃COOH and sodium acetate—CH₃COONa both have the common ion CH₃COO^–. Acetic acid is a weak acid and the addition of CH₃COONa suppresses the dissociation of CH₃COOH.

11. Why do we add HCl before passing H₂S in Group—II?

H₂S is a weak acid. Addition of HCl which contains the common ion H⁺ suppresses the dissociation of H₂S, thus decreases concentration of S^2—. Group II sulphides having very low solubility product can be selectively precipitated. Other suphides having higher solubility product cannot be precipitated

12. CuS is precipitated by H₂S in presence of HCl but CoS is precipitated in presence of NH₄OH. Why?

HCl suppresses the dissociation of H₂S and leads to decrease in concentration of S^2—. (H₂S D H⁺+ S^2--). CuS having low solubility product is easily precipitated. But CoS has higher solubility product and is precipitated only in presence of NH₄OH. NH₄OH is a base and so it reacts with H⁺ions produced by H₂S leading to increase in concentration of S^2--. So CoS is easily precipitated in presence of NH₄OH.

Section—C

1. What are the characteristics of a primary standard?

A substance should possess the following properties if it is to be used as a primary standard

· It should be easily available in the pure form

· It should not absorb water, i.e, it should not be hygroscopic or deliquescent

· Its composition should not change on keeping. For example ferrous sulphate changes to ferric on keeping. Therefore it cannot be used. Instead ferrous ammonium sulphate or Mohr’s salt is used, which is not easily degraded

· It should have high molecular mass so that mass required to prepare standard solution is more, which eliminates weighing errors

· Must be readily soluble in water

2. Discuss the Ostwald’s theory of acid base indicators

According to Ostwald’s theory, an acid base indicator is either a weak acid or a weak base. If it is a weak acid we can write the following equilibrium—

HIn + H₂O D In^— + H⁺

The HIn has one color, say acid colour. The indicator ion In^— has a different colour, say base colour. When indicator is added to acid solution, concentration of H⁺ ion increases. Then the equilibrium shifts to the left. This leads to increase in concentration of HIn, and the colour of solution will be the color of HIn, i.e, the acid colour. If it is added to an alkali, the OH^— ions of alkali react with H⁺ ions to form unionized H₂O. Concentration of H⁺ ions decreases and the equilibrium shifts to the right. Then concentraton of In^—increases and the colour will be that of In^—, i.e, base colour.For example phenolphthalein is a weak acid. HPh + H₂O D Ph^— +H⁺ The undissociated form HPh exists in acid medium and is colourless. In alkaline medium it has the colour of Ph^—, i.e base colour, which is pink.

If it is a weak base we can write the following equilibrium In + H₂O D InH⁺ + OH^—. In acid medium H⁺ in the acid reacts with OH^— to form unionized H₂O. This decreases concentration of OH^— and the equilibrium shifts to the right. Then concentration of InH⁺ increases and the solution gets the colour of InH⁺. In alkaline medium, concentration of OH^— will be greater. Then equilibrium shifts to the left, and concentration of In will be more. The colour of the solution will then be the colour of In. for example we can consider methul orange as a weak base MeOH. MeOH + H₂O D Me⁺ + OH^—. In acid medium it will have colour of Me⁺ i.e pink, and in alkaline medium colour will be that of MeOH, i.e, yellow.

3. Discuss the Quinonoid theory of acid base indicators

According to quinonoid theory, acid—base indicators exist as equilibrium mixture of two structures—benzenoid form and quinonoid form. The quinoid form usually has a brighter colour, and the benzenoid form has lighter colour. The benzenoid form is usually present in acid medium and quinonoid form is present in alkaline medium. Therefore at the end point the colour change occurs due to the conversion of one form to the other. For example phenolphthalein exists in the quinonoid form in alkaline medium and has pink colour. When this solution is titrated against acid in the burette, the solution is slightly acidic at the end point. Then it changes to the benzenoid form which is colourless. So we get the colour change from pink to colourless.

4. Briefly discuss about permanganometry titrations

Titrations involving potassium permanganate as oxidizing agent are called permanganometry titration. Potassium permanganate acts as oxidizing agent in acid medium.

2KMnO₄ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 3H₂O + 5[O]

It can oxidize ferrous iron to ferric iron and oxalic acid to CO₂ & H₂O.

10FeSO₄ + 5H₂SO₄ + 5[O] → Fe₂(SO₄)₃+ 5H₂O

5H₂C₂O₄ + 5[O] → 5H₂O + 10CO₂

FeSO₄ and oxalic acid solutions have to be acidified before titration. Only dilute sulphuric acid can be used for acidification, conc. H₂SO₄ and conc. and dilute HNO₃ cannot be used as these acids are also oxidizing agents and it will lead to unwanted side reactions. If HCl is used KMnO₄ can oxidize it to chlorine. Here again we have unwanted reactions taking place which will not give correct value for the estimation.

Potassium permanganate cannot be used as a primary standard since it cannot be obtained with purity. It always contains some manganese dioxide as impurity. Therefore KMnO₄ solution has to be first standardized.

Oxalic acid solution has to be heated to 70⁰ before titration. Otherwise reaction will be slow and we are likely to get a brown precipitate of manganese dioxide. This is because in the oxidation of oxalic acid covalent bonds have to be broken. This requires higher energy and also it will not take place instantaneously. But once the reaction has started, the Mn²⁺ions formed will act as auto catalyst for the reaction. This will speed up the reaction, so that even if the solution cools down the reaction will easily take place.

5. Briefly discuss about dichrometry titrations

Titrations involving potassium dichromate as oxidizing agent are called dichrometry titrations. K₂Cr₂O₇ acts as oxidizing agent in acid medium

K₂Cr₂O₇ + 4H₂SO₄→ K₂SO₄ + Cr₂(SO₄)₃ + 4H₂O + 3[O]

It can oxidize Fe²⁺ to Fe³⁺.

6FeSO₄ + 3H₂SO₄ + [3O] → 3Fe₂(SO₄)₃ + 3H₂O

The FeSO₄solution has to be acidified with dil. H₂SO₄ before titration since K₂Cr₂O₇ acts as oxidizing agent only in acid medium.

Potassium dichromate can be obtained in pure form; therefore it can be used as a primary standard

The indicator used in the titration can be external indicator or internal indicator. Potassium ferricyanide is used as external indicator, which gives dark blue colour with Fe²⁺ ion. At the end point when all the Fe²⁺ has completely reacted, the test drop does not give blue colour with ferricyanide. The internal indicators used in this kind of titration are called redox indicators. These are themselves redox systems, existing in both oxidized form and reduced form. These two forms have different colours. At the end point there is sudden change of potential and the indicator shows sharp colour change. Examples of redox indicators are a) diphenyl amine b) N-phenyl anthranilic acid. The reduced form of both indicators is colourless and the oxidized form is violet colour. Thus the colour change at the end point is from colourless to violet.

6. Discuss the theory of redox indicators

Redox indicators are substances which can show sharp colour change in response to sudden change of potential at the end point of a redox titration. They are themselves redox systems, and can exist in both oxidized form and reduced form. These two forms have different colours. The equilibrium can be represented as follows

In_ox + ne^— → In_red

Colour-A Colour-B

The EMF of the system is given by the Nernst Equation

E_ind = E⁰_ind — 0.0591 log [In_red] ⁄ [I_ox]

It is found that colour A predominates when [In_red] ⁄ [I_ox]ratio is 1/10, i.e, when the potential is = E_ind= E⁰_ind—( 0.0591 / n )log1/10 = E⁰_ind+ 0.0591/ n and the colour-B predominates when the [In_red] ⁄ [I_ox] ratio is 1/10. i.e, when E_ind = E⁰_ind— 0.0591

The change of colour from Colour-A to Colour-B occurs only when there is change of potential = E_ind = E⁰_ind± 0.0591 / n.

Consider the titration of Fe²⁺ against K₂Cr₂O₇. At first there will be only Fe²⁺. But when small amount of dichromate is added, small amount of Fe³⁺ is produced. Then the Fe²⁺/ Fe³⁺ electrode is set up, and the potential of the solution becomes equal to that of this electrode. As more of dichromate is added, the potential of this electrode increases according to the Nernst Equation : E = E⁰ + 0.0591 log [Fe³⁺ / Fe²⁺]

At the equivalence point the whole of Fe²⁺ has completely reacted. When excess dichromate is added, the Cr₂O₇^2— / Cr³⁺ electrode is set up and the solution has the potential of this electrode. Thus there is sudden increase in potential. Potential of indicator system for colour change is in between these two potentials. Thus after all ferrous has been oxidized; the dichromate oxidizes the reduced form of the indicator. The colour changes from Colour—B to Colour—A

7. What are the conditions necessary for a redox indicator

For a substance to be acceptable as indicator it must satisfy the following conditions

· The colour change of the indicator at the end point should be rapid and reversible

· It should not undergo any side reaction with the reactants

· It should change colour only just after the equivalence point. And the colour change should be complete before too much excess of titrant has been added. Thus the standard potential of the indicator should be in between the standard potentials of the two systems being titrated. In the titration where dichromate is added to ferrous solution, the indicator potential should be more than that of the ferrous-ferric system and less than that of dichromate-chromate system. If it is less than that of Fe²⁺/ Fe³⁺ system it will undergo colour change before the end point and if it is more than Cr₂O₇^2— / Cr³⁺ it will not change colour at all.

8. What are the characteristics of a metal ion indicator

The metallochromic indicator or complexometric indicator should hve the following properties:

· It should have a characteristic colour at the pH at which titration is carried out

· It should form a chelating complex with the metal ion. The colour of the complex should be different from that of the free indicator.

· It metal ion-indicator complex should be less strong than the complex formed by metal ion with the titrating chelating ligand (eg. EDTA). At the equivalence point when the entire free metal ion has reacted, the indicator – metal ion complex should release the metal ion so that it can react with the titrant, thus leading to complete reaction.

· Colour of free indicator should be very much different from that of the metal ion – indicator complex, so that we can easily identify the end point

· It should be very much sensitive to metal ion concentration so that colour change occurs as near to equivalence point as possible

9. Explain double burette method of titration

In the double burette method of titration no pipette is used. Two burettes are set up one with the titrant and the other with the solution to be titrated. Consider the titration of HCl against NaOH. Two burettes are set up—one containing HCl and the other containing NaOH. A measured amount say 16 mL of NaOH is added into a conical flask from the burette, a drop of indicator is added and titrated against HCl taken in the other burette. Titre value is noted, which is the titre value for 16 mL.Then a known amount say 2mL of NaOH is again added into the solution in the conical flask and titration is carried out and titre value is noted as before—which is the value for 18 mL. This is repeated until we get 5 or 6 titre values say for 16, 18, 20, 22, 24 mL. The normality of the solution can be determined either as an average of all these trials or by graphical method.

The advantages of this method are A) It avoids mouth pipetting, which is dangerous for poisonous solutions. B) It saves a lot of chemicals, because, there is no repetition for concordant values and indicator and other reagents like acids etc. has to be added only once. C) It saves time since repletion for concordant values is avoided and heating of solution before titration etc. has to be done only once. D) More accurate results can be obtained, since the calculation can be done by graphcal method.

10. What is microanalysis? Give 3 examples of microanalysis techniques

Microanalysis is the quantitative or qualitative analysis of very tiny amounts of the sample in the microgram microlitre or micromolar range or even less. These are usually instrumental analysis using modern automated systems using microscopes or microprobes. Analysis of chemical, isotopic and molecular constituents can be carried out using these techniques. The instrument usually consists of an input beam—which may be an electromagnetic radiation like X-rays. UV etc., or beam of particles like electrons, protons or ions. This input beam interacts with the atoms and molecules in the sample resulting in an output beam—which again may be electromagnetic radiations or stream of particles. This output beam is analyzed by a suitable detector and the results are usually displayed on a computer screen, either as numerical data or as graphical plots.

A few examples of microanalysis techniques are:

· Photons of electromagnetic radiation are use as probes in molecular structure determination in

o FTIR Spectroscopy—Infrared radiations used as probe

o RAMAN spectroscopy—near infrared, UV laser etc.

o X-ray Diffraction (XRD)—Xrays are used as input—used for analysis of crystal structure and molecular structure

· Electrons are used as probe for surface topography, elemental analysis and elemental distribution

o Scanning Electron Microscopy (SEM)

o Electron Micro Probe Analysis (EPMA)

o Scanning Auger Microscopy (SAM)

· Ions are used as probes for surface structure and composition analysis

o Secondary Ion Mass Spectroscopy

Advantages of Microanalysis techniques are

· Requires less time and very small quantity of sample, solvent etc.

· Enables in situ analysis of the sample—for example, reactants or products can be monitored during the course of the reaction, without having to isolate it from the reaction mixture.

· Sample size as small as parts permilion and even parts per billion can be analysed.

· Elemental analysis including isotopic analysis of all elements can be carried out

· Takes only very little time to complete the analysis

11. Discuss briefly about the principles underlying the separation of cations into groups in qualitative analysis

The cations have been divided into six analytical Groups, based on their mode of separation from the mixture by selective precipitation

GROUP	IONS
I	Pb²⁺, Hg₂²⁺, Ag⁺
II	Pb²⁺ Hg²⁺ Bi³⁺ Cu²⁺ As³⁺ Sb³⁺ Sn²⁺ Sn⁴⁺
III	Fe²⁺ Fe³⁺ Cr³⁺ Al³⁺
IV	Co²⁺ Ni²⁺ Mn²⁺ Zn²⁺
V	Ba²⁺ Sr²⁺ Ca²⁺
VI	Mg²⁺ Na+ K+ NH₄

⁺

The basic principles underlying the separation of cations into groups are the solubility product principle and the common ion effect. The two principles are used to selectively precipitate out the ions of that particular group. Once this separation is done the two precipitates are separately analyzed to identify the particular cation present in that group.

Solubility product is the ionic product of a saturated solution. If the ionic product exceeds the solubility product, the solution becomes supersaturated and precipitation takes place. This is called solubility product principle. If a substance has low solubility product, the ionic product easily exceeds the solubility product and it is easily precipitated.

Common ion effect is the suppression of ionization of weak electrolyte by the addition of strong electrolyte containing a common ion. This is done in order to reduce the concentration of certain ions so that compounds having low value of solubility product are selectively precipitated out.

The Group I cations are precipitated as chlorides by adding dil.HCl. This is because these chlorides have lowest value of solubility product than the chlorides of other group cations. Thus only these chlorides are selectively precipitated. Chlorides of other group catons will not be precipitated.

The Group-II cations are precipitated as sulphides by passing H₂S gas in presence of dil.HCl. HCl, which can supply the common ion H₃O⁺, suppresses the dissociation of the weak electrolyte H₂S by common ion effect and so the concentration of S^2— ions is less.

H₂S + H₂O D H₃O⁺ + S^2—

The sulphides of Group II cations are least soluble, i.e they have low solubility product. Thus, they are selectively precipitated even with less concentration of S^2— ions.

The Group III cations are precipitated as hydroxides by adding NH₄OH in presence of NH₄Cl. NH₄Cl suppresses the dissociation of weak electrolyte NH₄OH, since both contain the common ion NH₄⁺, and so concentration of OH^— ions will be very less.

NH₄OH D NH₄⁺ + OH^—

The solubility product of hydroxides of Group III cations are very less compared to hydroxides of cations of other groups. Thus they are selectively precipitated.

The Group IV cations are precipitated as sulphides by passing H₂S in presence of NH₄OH. OH^— ions of NH₄OH reacts with the H⁺ ions of H₂S to give unionized H₂O. This leads to increase in concentration of S^2— ions. The Group IV sulphides have high solubility product. In presence of this high concentration of S^2— ions, the ionic product of Group IV sulphides exceed the solubility product and are selectively precipitated.

Group V cations are precipitated as carbonates by the addition of (NH₄)₂CO₃ in presence of NH₄Cl. NH₄Cl suppresses the dissociation of (NH₄)₂CO₃, which decreases the concentration of carbonate ions. Carbonates of Group-V cations, which have low solubility product, are selectively precipitated out even with this low concentration of carbonate ions.

In Group VI magnesium phosphate has lower solubility than ammonium, potassium or sodium phosphate and can be selectively precipitated by disodium hydrogen phosphate.

12. A) Briefly explain the use of the principles of solubility product and common ion effect in the separation of cations in qualitative analysis

B) A solution contains Cu²⁺and Ba²⁺ ions. How will you separate the ions and identify them?

A) Refer previous question

B) The solution containing Cu²⁺ and Ba²⁺ ions is acidified with HCl and H₂S is passed through it till the copper ions are completely precipitated as black CuS. The precipitate is filtered out. The filtrate is then boiled to expel H₂S and NH₄OH is added followed by NH₄OH and (NH₄)₂CO₃. The precipitated barium carbonate is filtered out. The two precipitates can then be analyzed to identify the ions

ALCHEMY--THE CHEMISTRY TEACHER BLOG

I COMPLEMENTARY CHEMISTRY

No comments:

Blog Archive