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Engineering Chemistry Notes

Dr. Bathini Srinivas

Ph.D (JNTUH)

Unit I: Molecular structure and Theories of Bonding

Topic1: Molecular orbitals

Q1) Define degenerate orbitals? Give the examples.

Ans. Orbital’s which is having same energy is called degenerate orbital’s. Examples Px, Py, Pz orbital having same energy and dxy, dyz, dzx, dx2-y2, dz2 orbital’s having same energy ,these orbital’s are called degenerate orbital’s.

Q2) Write the difference between the molecular orbitals and atomic orbitals?

Differences between Molecular Orbital and Atomic Orbital

Molecular Orbital

Atomic Orbital

1. An electron Molecular orbital is under the influence of two or more nuclei depending upon the number of atoms present in the molecule.

2. Molecular orbitals are formed by combination of atomic orbitals

3. They have complex shapes.

1. An electron in atomic orbital is under the influence of only one positive nucleus of the atom.

2. Atomic orbitals are inherent property of an atom.

3. They have simple shapes.

Q 3) Draw the shapes and structures of S, P,d orbital’s ?

S Orbital shape spherical, P orbital shape dumbel

Figure: d orbital shape double dumbel

Explain about molecular orbital diagram of N2 Molecule?

Nitrogen:

This molecule has ten electrons. The atomic orbitals combine to produce the following molecular orbital diagram:

σ 1s, σ *1s, σ 2s, σ *2s, [π 2px = π 2py], σ 2pz [π *2px= π *2py], σ*2pz

Here the 2pg orbital is occupied by two electrons to give a total bond order of three. This corresponds well with the Lewis structure (), although the orbital approach tells us that there is one s and two p.

Q4. Explain about the molecular orbital diagram of O2 molecule?

Oxygen: This molecule has twelve electrons, two more than nitrogen - and these extra two are placed in a pair of degenerate g orbitals. The atomic orbitals combine to produce the following molecular orbital diagram:

For O2 and higher molecules →

σ1s, σ *1s, σ 2s, σ *2s, σ 2pz, [π2px = π2py], [π*2px= π*2py], σ *2pz

FigOrder of Energy for O2 and Higher molecules

Comparison of the above energy level diagram wit hthat for nitrogen - you can see that the 2sg level lies lower than pu. Here, we are starting to fill the anti-bonding orbitals originating from the p orbital interactions and so the bond order decreases from three to two.The lowest energy arrangement (Hund's rule) - has a single electron, each with parallel spins, in each of the pgx and pgy orbitals. This produces a paramagnetic molecule, with a double bond and has two unpaired electrons.

Q5) Explain about molecular orbital diagram of N2 Molecule?

Nitrogen:




Topic2: Complexo metric compounds

Q6) Why do transition metal compounds forms complexes?

The transition metals are almost unique in their tendency to form coordination complexes. The tendency of cations of transition elements to form complexes is due to two factors;

1. These ions are very small in size and, therefore, have high positive charge density. This facilitates acceptance of lone pair of electrons from other molecules.

2 .They have vacant orbitals and these orbitals have the appropriate type of energy to accept lone pair of electrons.

Q7) Write the silent features of CFT

Important Features of Crystal Field theory are

1. Transition metal ion is surrounded by ligands with lone pair of electrons and the complex is a combination of central ion surrounded by other ions or molecules or diploes i.e ligand

2. All types of ligands are regarded as point charges.

3. The interaction between the metal ion and the negative ends of anion (or ion dipoles) are purely electrostatic i.e bond between the metal and ligand is considered 100 percent ionic.

4.The ligands surrounding the metal ion produce electric field influences the energies of the orbitals of central metal ion particularly d-orbitals.

5. In the case of free metal ion all the five d-orbitals have the same energy. Such orbital having the same energies are called degenerate orbital’s.

Q8) Write a note on spectro chemical series?

The tool used often in calculations or problems regarding spin is called the spectro chemical series. The spectro chemical series is a list that orders ligands on the basis of their field strength. Ligands that have a low field strength, and thus high spin, are listed first and are followed by ligands of higher field strength, and thus low spin. This trend also corresponds to the ligands abilities to split d orbital energy levels. The ones at the beginning, such as I−, produce weak splitting (small Δ) and are thus weak field ligands. The ligands toward the end of the series, such as CN−, will produce strong splitting (large Δ) and thus are strong field ligands. A picture of the spectrochemical series is provided below.

(weak) I− < Br− < S2− < SCN− < Cl− < NO3− < N3− < F− < OH− < C2O42− ≈ H2O <

NCS− < CH3CN < py < NH3 < en < bipy < phen < NO2− < PPh3 < CN− ≈ CO (strong)

Q 9) Explain about d orbital splitting in octahedral complexes?

In an octahedral complex, there are six ligands attached to the central transition metal. The d orbital splits into two different levels. The bottom three energy levels are named dxy, dxz, and dyz (collectively referred to as t2g). The two upper energy levels are named d(x2−y2), and dz2 (collectively referred to as eg).

The reason they split is because of the electrostatic interactions between the electrons of the ligand and the lobes of the d-orbital. In an octahedral, the electrons are attracted to the axes. Any orbital that has a lobe on the axes moves to a higher energy level. This means that in an octahedral, the energy levels of eg are higher (0.6∆o) while t2g is lower (0.4∆o).

Figure 4: Splitting of the degenerate d-orbitals (without a ligand field) due to an octahedral ligand field shown in Figure 3.

The distance that the electrons have to move from t2g from eg and it dictates the energy that the complex will absorb from white light, which will determine the color. Whether the complex is paramagnetic or diamagnetic will be determined by the spin state. If there are unpaired electrons, the complex is paramagnetic; if all electrons are paired, the complex is diamagnetic.

Q 10. Why do transition metal compounds forms complexes?

Sol. The transition metals are almost unique in their tendency to form coordination complexes. The tendency of cations of transition elements to form complexes is due to two factors;



Unit-II: Water and Its Treatment

Topic1. Hardness of water

Q1. Define hardness of water? What are the causes of hardness?

Hardness of water:

When soap comes in contact with hard water, sodium stearate will react with dissolved Ca and Mg salts and produce Ca-stearate or Mg- stearate which is white precipitate.

C17H35COONa+ CaCl2 → (C17H35COO)2 Ca↓ + 2NaCl (soluble)

Hardness precipitation

Substance (insoluble)

Soft water:

Soap is sodium or potassium salts of higher fatty acids like stearic, oleic and palmetic acids. When soap soap is mixed with soft water lather is produced due to stearic acid and sodium stearate.

Ex: 2C17H35COONa + H2O →2C17H35COOH + NaOH

Sodium stearate (soap)Stearic acid

Stearic acid + Sodium stearate → Formation of lather

Q2. What is the difference between temporary and permanent hardness of water?

Types of Hardness: Hardness in water is of two types.

(1) Temporary hardness and (2) permanent hardness

1. Temporary hardness: Temporary hardness is due to presence of dissolved bicarbonates of calcium and magnesium salts present in water.

By boiling Temporary hardness can be removed [bicarbonates converts into carbonates as precipitate].

Ex: Ca(HCO3)2 CaCO3↓+ CO2 +H2O

Mg(HCO3)2Mg(OH)2↓+ 2CO2

Permanent Hardness: Permanent hardness is due to presence of chlorides [Cl-], sulphates [SO42-] and Nitrates [NO3-] of calcium, magnesium and other heavy metals.

This hardness cannot be removed easily by boiling. Hence, it is called permanent hardness.

Total hardness of water = Temporary hardness + Permanent Hardness

Q3. Write the relationship between units of hardness of water.

The following are common units for hardness of water.

1. Parts per million [ppm]

2. Milligram per litre [mg/lt]

3. Degree Clark [0cl]

4. DegreeFrench [0Fr]

5. Milli equivalents per litre [m eq/lt]

1. Parts Per Million [PPM]:-The number of parts by weight of CaCO3 equivalents hardness causing salt present in one million parts of water.

[One million = 10 lakhs [106]].

1ppm = 1 part CaCO3 eq. hardness in 106 parts of water.

1. Milligrams per Litre [mg/Lt]:-The number of milligrams of CaCO3 equivalent hardness causing salt present in one litre of water. i.e 1mg/lit

As density of water is unity i.e. 1lit of water = 1Kg = 1000X1000 mgs (at 4oC)

= 106 parts

Hence 1 mg/lt = 1 ppm

1. Drgree Clark [0Cl]:- The number of parts of CaCO3 equivalent hardness causing salt present in 70,000 parts of water.

10 Cl = 1 parts of CaCO3 eq .hardness per 70,000 parts of water.

1 ppm = 0.070Cl

1. Degree French [0Fr]:-The number of parts by weight of CaCO3 equivalent hardness causing salt per 105 parts by weight of water.

10Fr = 1 parts of CaCO3 eq. hardness per 105 arts of water 1 ppm = 0.1 0Fr

1. Milli equivalent per Litre [Mg/Lt]:-The number of milli equivalent of CaCO3 eq. hardness causing salt per one litre of water.

1meq/lt = 50 mg/lt CaCO3eq. =1ppm = 0.02 meq/lt

Interconversion:- 1ppm =1mg/lt =0.070Cl = 0.10Fr =0.02 meq/lt.

Topic 2: Treatment of Water

Q4. Describe the internal treatment of water?

Internal treatments are followed by blow down operation, so that accumulated sludge is removed. Important internal conditioning/treatment methods are:

1. Colloidal conditioning: In low pressure boiler scale formation can be avoided by adding organic substances like kerosene, tarnnin, agar-agar, etc. which get coated over the scale forming precipitates, there by yielding non-sticky and loose deposits, which can be removed by blow down operation.

1. Calgon conditioning :It involves in adding calgon [sodium hexa meta phosphate] to boiler water .it prevents the scale and sludge formation by forming soluble complex compound with CaSO4.

Na2[Na4(PO3)3] → 2Na+ + [Na4(PO3)3]

(Calgon)

2CaSO4+ [Na4(PO3)6]2-[Ca2(PO3)6]2- +2Na2SO4 .

1. Phosphate conditioning: The scale formation can be avoided in high pressure boilers by adding tri sodium phosphate or other types of phosphates according to the pH of boiler water.

3CaSO4+2Na3PO4Ca3 (PO4) 2+3Na2SO4

The different phosphates used are:

2. Na3PO4 (trisodium phosphate) - used to acidic water.

2. Na2HPO4 (disodium phosphate) - used to weak alkaline.

2. NaH2PO4 (monosodium phosphate) - used to alkaline water.

Q5. Explain the process of reverse osmosis?

Reverse Osmosis:

When two solutions of different concentrations are separated by a semi-permeable membrane, flow of solvent from low concentration to high concentration takes place due to difference in concentration, this is said to be “osmosis”.

If a hydrostatic pressure in excess of osmotic pressure is applied on the concentrated side the flow of solvent reverses as it is forced to move from high concentration to low concentration through the membrane known as “reverse osmosis”.

In this process, semi-permeable membranes are made by thin film of cellulose acetate (or) polyamide polymers (or) polymethylene are used. A pressure of 15 -40 kg/cm2 is applied for separating the water from its contaminants. This process also known as super or hyper filtration.

Advantages:

1. It is simple and reliable process. Low maintenance cost and also pollution free.

1. The life of semi permeable membranes is about 2 years , and it can be easily replaced within minutes, and there by nearly un interrupted water supply can be provided.

Q6. Which one is most widely used chlorination process?

Chlorination:

Chlorination is the process of purifying the drinking water by producing a powerful Germicide like hypochlorous acid. When this chlorine is mixed with water it produces Hypochlorous acid which kills the Germs present in water.

H2O+Cl2→ HOCl + HCl

Chlorine is basic (means PH value is more than 7) disinfectant and is much effective over the germs. Hence chlorine is widely used all over the world as a powerful disinfectant. Chlorinator is an apparatus, which is used to purity the water by chlorination process.

Q7. Why is Ion exchange process preferred over Zeolite-process for the softening?

Ion exchange process (or) deionization or demineralization:

Ion exchanges are of two types they are anionic and cationic. These are co-polymers of styrene & divinyl benzene i.e., long chain organic polymers with a micro porous structure.

Cation exchange resins: The resins containing acidic functional groups such as -COOH, -SO3H etc. are capable of exchanging their H+ ions with other cations are cation exchange resins , represented as RH+.

Anion exchange resins: The resins containing amino or quaternary ammonium or quaternary phosphonium (or) Tertiary sulphonium groups, treated with “NaOH solution becomes capable of exchanging their OH- ions with other anions. These are called as Anion exchanging resins represented as ROH-

Process: The hard water is passed first through cation exchange column. It removes all the cation (ca2+ & Mg2+) and equivalent amount of H+ icons are released from this column.

2RH+ + Ca2+ → R2Ca2+ + 2H+

After this the hard water is passed through anion exchange column, which removes all the anions like SO42-, Cl-, CO32- etc and release equal amount of OH- from this column.

R1OH + Cl- → R1Cl + OH-

2R1OH +SO42- → R21SO4 +2OH-

The output water is also called as de-ionized water after this the ion exchanges get exhausted. The cation exchanges are activated by mineral acid (HCl) and anion exchanges are activated by dil NaOH solution.

R2Ca + 2H+ → 2RH + Ca+2

R21SO4 + 2OH- → 2R1OH + SO42-

Advantages:

1. The process can be used to soften highly acidic or alkaline water.

(2) It produces water of very low hardness. So it is very good for treating for use in high pressure boilers.

Disadvantages:-

(1) The equipment is costly and common expensive chemicals required.

(2) It water contains turbidity, and then output of this process is reduced. The turbidity must below 10 ppm.

Topic 3: Boiler Troubles

Q8. Define priming and foaming?

a). Priming: - The carrying out of water droplets with steam in called “priming” Because of rapid and high velocities of steam, the water droplets moves out with steam from the boiler. This process of wet steam generation is caused by

(i) The presence of large amount of dissolved solids.

(ii)High stream velocities (iii) sudden boiling (iv) improper designing of boilers (v) sudden increase in stream production rate and (vi) The high levels of water in boilers.

Prevention of priming: - The priming is avoided by

(i)Fitting mechanical steam purifiers

(ii)Avoiding rapid change in steaming rate

(iii) Maintaining low water levels in boilers and

(iv)Efficient softening and filtration of boiler feed water.

b). Foaming: - Formation of stable bubbles at the surface of water in the boiler is calling foaming. More foaming will cause more priming. It results with the formation of wet steam that harms the boiler cylinder and turbine blades. Foaming is due to the presence of oil drops, grease and some suspended solids.

Prevention of Foaming: Foaming can be avoided by

(1)Adding antifoaming chemicals like castor oil. The excess of castor oil addition can cause foaming.

(2) Oil can be removed by adding sodium aluminates or alum.

(3) Replacing the water concentrated with impurities with fresh water.

Q 9. Explain the reasons of caustic embrittlement?

Caustic embrittlement:

Caustic embrittlement is a term used for the appearance for cracks inside the boiler, particularly at those places which are under stress such as riverted joints due to the high concentration of alkali leading to the failure of the boiler. The cracks have appearance of brittle fracture. Hence, the failure is called 'caustic embrittlement'.

Reasons for the formation of caustic embrittlement:

During the softening process by lime soda process, free Na2CO3 is usually present in small portion in soft water which decomposes to give sodium hydroxide and CO2 at high pressure of the boilers.

Na2CO3 + H2O → 2NaOH + CO2

The precipitation of NaOH makes the boiler water 'caustic'. The NaOH containing water flows into small pits and minute hair cracks present on the boiler. As the vapour evaporates, the concentration of caustic soda increases progressively creating a concentration cell as given below, thus dissolves in the iron of boiler as sodium ferrate.

(-) Iron at bends, rivets and joints

/

Concentrated NaOH solution

//

Dilute NaOH solution

/

Iron at plane (+) surfaces

The iron at plane surfaces surrounded by dilute NaOH becomes cathodic while the iron at bends, rivets, joints are surrounded by highly concentrated NaOH becomes anionic which consequently decayed or corroded. The cracking of the boiler occurs particularly at stress parts like bends, joints, rivets etc., causing the failure at bolier.

Thus, the cracks present at such places are intercrystalline, irregular running from one rivet to another without joining each other. These cracks have the appearance of brittle fracture, hence, known as caustic embrittlement.

Preventions of caustic embrittlement:

1. By using sodium phosphate as softening reagent instead of sodium carbonate, disodium hydrogen phosphate is the best softening reagent because it not only forms complex with Ca+2 and Mg+2 resulting the softening of water but also maintains pH of water 9-10. The Phosphates used are trisodium phosphate, sodium dihydrogen phosphate etc.

1. By adding tanning or legnin to boiler water, which blocks the hair cracks and pits that are present on the surface of the boiler plate, preventing the infiltration of the caustic soda solution

1. By adding sodium sulphate to boiler water, which also blocks the hair cracks and pits that are present on the surface of the boiler plate, preventing the infiltration of the caustic soda solution. The amount of sodium sulphate added to the boiler be in the ratio [Na2SO4 conc. / NaOH conc.] kept as 1:2 , 2:1 and 3:1 in boilers working as pressures upto 10, 20 and above 30 atmospheres respectively.

Disadvantages of caustic embrittlement:

The cracking or weakening of the boiler metal causes the failure of boiler.

Q10. Discuss the sludge and scale formation in boilers?

In boilers more water is removed in form of steam during boiling, hence boiler water gets concentrated with dissolved salts and reaches saturation point. At this point the dissolved salts precipitated and slowly settle on the inner walls of the boiler plate. The precipitation takes place in two ways

a). Sludge formation:

The precipitation in the form of soft loose and slimy deposits formed comparatively in the colder portions of boiler is called sludge.

Reasons for the formation of sludge:

The dissolved salts whose solubility is more in hot water and less in cold water produce sludges.

Ex: MgCO3, MgCl2, CaCl2, and MgSO4. The sludges were formed at comparatively colder portions of the boiler and get collected where rate of flow of water is low.

Disadvantages of sludges:

i) Sludges are bad conductors of heat and results in wastage of heat and fuel.

ii) Sometimes sludges were entrapped in the scales and are deposited as scale, which causes more loss of efficiency of the boiler.

iii) Excessive sludge formation leads to the settling of sludge in slow circulation areas such as pipe connections, plug openings leading to the chocking of pipes.

Prevention of sludge formation:

i) By using soft water which is free from dissolved salts like MgCO3, MgCl2, CaCl2, and MgSO4 can prevent sludge formation.

b) Scale formation:

The precipitation in the form of hard deposits, stick very firmly on the inner walls of the boiler is called scales.


i) Decomposition of Ca(HCO3)2:

Due to the high temperature and pressure present in the bolers, the Ca(HCO3)2 salt decomposes to CaCO3 ↓, an insoluble salt, forms scale in low pressure boilers.

Ca (HCO3)2 → CaCO3 + H2O + CO2

ii) Decomposition of CaSO4

CaSO4 is more soluble in cold water hence its solubility decreases and precipitates out to produce hard scale on the surface of the boiler. The solubility of CaSO4 is 3200ppm at 150C, reduces to 27ppm at 3200C and completely insoluble in super heated water. CaSO4 scale is very hard, highly adherent and difficult to remove.

iii) Hydrolysis of magnesium salts:

Dissolved magnesium salts undergo hydrolysis at high temperature prevailing inside the boiler forming magnesium hydroxide precipitate, which form salt type of scale.

MgCl2 + 2H2O → Mg (OH)2 ↓ + 2HCl

iv) Presence of silica:

Silica present in small quantities deposits as calcium silicate (CaSiO3) or Magnesium silicate (MgSiO3). The deposits form hard scale very difficult to remove.

Disadvantages of scale formation:

i) wastage of fuel: Scales are bad conductors of heat due to which the flow of heat to inside water is decreased hence excessive heating is required which increases fuel consumption causing wastage of fuel. The wasage of fuel increases with increases of the thichness of the scale.

ii) Lowering the boiler safety: Due to scale formation overheating of the boiler done to maintain the constant supply of steam. due to over heating the boilermaterial become softer and weaker, which causes distortion of boiler.

iii) Decrease in efficiency: Scales deposited in the valves and condensers of the boiler cause chocking which results in decrease in efficiency of the boiler.

iv) Danger of explosion: Because of the formation of the scales, the boiler plate faces higher temperature outside and lesser temperature inside due to uneven heat transfer resulting cracks in the layer of scales. Water passes through the crack and comes in contact with boiler plate having high temperature. This causes formation of large amount of steam suddenly developing sudden high pressure. This causes the explosion of the boiler.

Removal scales:

i) If the scale formed is soft it can be removed by a scrapper, wire brush etc.

ii) By giving thermal shocks done by heating the boiler to high temperature and suddenly cooling with cold water, if the scale is brittle in nature.

iii) If the scale is very adherent and hard, chemical treatment must be given. Ex: CaCO3 scale is removed by washing the boiler plate with EDTA solution.

iv) Frequent blowdown operation can remove the scales, which are loosely adhering.

Prevention of Scale formation:

Scale formation can be prevented by softening water by following methods.

Unit-III

ELECTRO CHEMISTRY AND CORROSION

Topic 1: Galvanic Cells

Q 1.What is galvanic cell? Explain the construction and reactions of Daniel cell?

Galvanic cell:

Galvanic cell is a device in which chemical energy is converted into electrical energy. These cells are called Electrochemical cells or voltaic cells. Daniel cell is an example for galvanic cell.

This cell is made up of two half cells. One is oxidation or anodic half cell. The other is reduction or catholic half cell. The first half cell consists of ‘Zn’ electrode dipped in ZnSO4solution and second half cell consists of ‘Cu” electrode dipped in Cuso4 solution. Both the half cells are connected externally by metallic conductor. And internally by ‘salt bridge’ salt bridge is a U- tube containing concentrated solution of Kcl or NH4 NO3 in agar-agar gel contained porous pot. It provides electrical contact between two solutions.

The following reactions take place in the cell.

At cathode:

Zn → Zn+2 +2e- (oxidation or de-elecronation)

At cathode:

Cu+2 +2e- → Cu (Reduction or electronatioin)

The movement of electrons from Zn to cu produces a current in the circuit.

The overall cell reaction is: Zn +Cu+2 → Zn+2 +Cu

The galvanic cell can be represented by

Zn / ZnSO4 // CuSO4/ Cu

The passage of electrons from one electrode to other causes the potential difference between them which is called E.M.F.

Q2. Derive the Nernst equation? Explain the application of this equation?

Ans: We have considered only standard reduction potentials, which refer to solution concentrations of 1M. It is common in the laboratory to work with solutions of lower concentrations, and reduction potentials depend on the concentration of the solutions in the electrochemical cells. The dependence is given by the Nernst equation. Temperature is another variable in the equation, although normally experiments will be carried out at a specified temperature.

Consider a reaction,

(K)

𝑎𝐴+𝑏𝐵 ⇔ 𝑐𝐶+𝑑𝐷

−Δ𝐺= −Δ𝐺0− 𝑅𝑇𝑙𝑛𝐾

Δ𝐺= Δ𝐺0+ 𝑅𝑇𝑙𝑛𝐾

𝑠𝑖𝑛𝑐𝑒 Δ𝐺 = −𝑛𝐹𝐸𝑐𝑒𝑙𝑙 𝑎𝑛𝑑Δ𝐺0= −𝑛𝐹𝐸0𝑐𝑒𝑙𝑙

𝑛𝐹𝐸𝑐𝑒𝑙𝑙=𝑛𝐹𝐸0𝑐𝑒𝑙𝑙− 𝑅𝑇𝑙𝑛𝐾

𝐸𝑐𝑒𝑙𝑙= 𝐸0𝑐𝑒𝑙𝑙− 𝑙𝑛𝐾

𝐸𝑐𝑒𝑙𝑙= 𝐸𝑐𝑒𝑙𝑙0− 𝑙𝑜𝑔𝐾

R- Gas Constant = 8.314J/sec

T- Absolute temperature = 298K

F = Faraday = 96500c

= 0.0591

n- No.of electrons involved in the reaction

1. Reduction:

Mn+ + ne- → M(s)

𝐸𝑐𝑒𝑙𝑙= 𝐸0𝑐𝑒𝑙𝑙− 2.303𝑙𝑜𝑔



2. Oxidation:

M(s) → Mn+ + ne-


𝐸𝑐𝑒𝑙𝑙= 𝐸0𝑐𝑒𝑙𝑙− 2.303[Mn+]

Applications of Nernst Equation:

1. One of the major applications of Nernst equation is in determining ion concentration

2. It is also used to calculate the potential of an ion of charge “z” across a membrane.

3. It is used in oxygen and the aquatic environment.

4. It is also used in solubility products and potentio-metric titrations.

5. It is also used in pH measurements.

TOPIC 2: TYPES OF ELECTRODES

Q3. Explain the functioning of SCE?

Ans: It consists of mercury at the bottom over which a paste of mercury- mercurous chloride is placed. A solution of potassium chloride is then placed over the paste. A platinum wire sealed in a glass tube helps in making the electrical contact. The electrode is connected with the help of the side tube on the left through a salt bridge with the other electrode to make a complete cell. The electrode is represented as

Pt, Hg/ Hg2Cl2, Cl-(aq)

The potential of the calomel electrode depends upon the concentration of the potassium chloride solution.

If potassium chloride solution is saturated, the electrode is known as saturated calomel electrode (SCE) and if the potassium chloride solution is 1 N, the electrode is known as normal calomel electrode (NCE) while for 0.1 N potassium chloride solution, the electrode is referred to as decinormal calomel electrode (DNCE).

The reduction potentials of the calomel electrodes on hydrogen scale at 298K are as follows:

Saturated KCl = 0.2415 V

1.0N KCl= 0.2800 V

0.1N KCl= 0.3338 V

Calomel electrode

Calomel electrode acts as either anode or cathode w.r.to the other electrode connected to it. If it acts as anode, it involves oxidation:

2Hg → Hg22+ + 2e─

Hg2+ + 2 Cl─ → Hg2Cl2

--------------------------------------------------

2Hg + 2Cl─ → Hg2Cl2 + 2e─

--------------------------------------------------

Oxidation half reaction, which results in fall of concentration of Cl- ions

If it acts as cathode, it involves reduction

Hg2Cl2 → Hg22+ + 2Cl−

Hg22+ + 2e− → 2Hg

--------------------------------

Hg2Cl2 + 2e−→ 2Cl−+ 2Hg

--------------------------------

Reduction half reaction, which results increase in concentration of Cl- ions. Thus, Calomel electrode is reversible to Cl- ions. The reduction potential of calomel electrode is given by

Since [Hg] = [Hg2Cl2] = 1 and [Cl-] ≈ 4M, then ESCE = 0.242 V

The electrode potential of any other electrode on hydrogen scale can be measured when it is combined with calomel electrode. The emf of such a cell is measured. From the value of electrode potential of calomel electrode, the electrode potential of the other electrode can be evaluated.

Advantages:

0. Its construction is very easy

0. Results of cell potential measurements are reproducible.

Disadvantages:

Since Hg2Cl2 breaks at 500C, it can’t be used above this temperature.

Q4. Differentiate metallic and electrolytic conductors?

Ans:

Metallic conductors

Electrolytic conductors

1. Conductance is due to the flow of electrons.

2. It does not result any chemical change.

3. Metallic conduction decreases with increase in temperature.

4. It does not involve any transfer of matter.

1. Conductance is due to the movement of ions in a solution.

2. Chemical reactions take place at the electrodes.

3. Electrolytic conduction increases with increase in temperature.

4. It involves transfer of matter.

Q5. Give the differences between Primary and Secondary cells?

Differences between Primary

Secondary cells

1. These are non-rechargeable and meant for a single use and to be discarded after use.

2. Cell reaction is not reversible.

3. Cannot be rechargeable.

4. Less expensive.

5. Can be used as long as the materials are active in their composition.

Eg: Leclanche cell, ‘Li’ Cells.

1. These are rechargeable and meant for multi cycle use.

2. Cell reaction can be reversed.

3. Can be rechargeable.

4. Expensive.

5. Can be used again and again by recharging the cell.

Eg; Lead- acid cell, Ni-cd cells.

Q6. Define the terms Equivalent conductance? With units.

Ans: It is defined as the conductance of all ions produced by the dissociation of Igm equivalent of an electrolyte dissolved in certain volume ‘V’ of the solvent at const temperature

Units = = Ohm-1 cm2 eq

Q7. What is EMF of cell? How the emf of cell is calculated? Remember b E.M.F:-

Ans: The difference of potential which causes flow of electrons from an electrode of higher potential to an electrode of lower potential is called Electro motive force (EMF) of the cell.

The E.M.F of galvanic cell is calculated by the reduction half – cell potentials using to following ex. Ecell = E (right) - E(left)

Ecell EMF of the cell.

Eright reduction potential of right hand side electrode.

Eleft reduction potential of left hand side electrode.

Applications of EMF measurement:-

1. Potentiometric titrations can be carried out.

2. Transport number of ions can be determined.

3. PH can be measured.

4. Hydrolysis const, can be determined.

5. Solubility of sparingly soluble salts can be found.

Q8. The resistance of 0.1 N solution of an electrolyte is 40 ohms. If the distance between the electrodes is 1.2cm and the area of cross-section is 2.4cm. Calculate the equivalent conductivity.

Distance between electrodes l = 1.2cm

Area of cross-section a = 2.4cm2

Cell const. = 0.5cm-1

Normality of given solution = 0.1 N.

Resistance R = 40 ohms.

Specific conductance K = 0.0125

Equivalent Conductivity = 125 ohm-1cm2 eq-1

Q 9: Calculate the emf for the cell,

Ans: Zn/Zn+ // Ag+ /Ag given E0Zn+/Zn+2 / Zn = 0.762v and E0Ag+/Ag = 0.8 v

Given cell is Zn /Zn+2 //Ag+/Ag.

E0 = Zn+2/Zn = 0.762 v

E0 = Ag+/Ag =0.8 v

E0cell = E0right – E0left

= 0.8 – (-0.762) = 1.562 v.

Q10 : Calculate the cell Constant of a cell having a solution of concentration N/30 gm eq /li of an electrolyte which showed the equivalent conductance of 120 Mhos cm2 eq- 1, resistance 40 ohms.

Ans: Resistance R = 40 ohms.

Equivalent conductance of solution (A) = 120 mho cm2eq-1

Concentration of sol. N= gm eq/li

=0.033N.

Cell const =?

Equivalent conductance =

Specific cond. (K) = 0.00396

Cell constant = s p. conductance x Resistance

= 0.00396 x 40

= 0.1584 cm-1

Q 11. Write brief about working and construction of Glass Electrode?

Ans: Glass Electrode:

Credit for the first glass sensing pH electrode is given to Cremer, who first described it in his paper published in 1906. Later several groups contributed for development of different ion selective electrodes An Ionselective electrode (ISE) (also known as a specific ion electrode, or SIE) is a sensor which converts the activity of a specific ion dissolved in a solution into an electrical potential which can be measured by some potentio-metric devise like a voltmeter or pH meter. As we know, the emf is theoretically dependent on the logarithm of the ionic activity (concentration), in accordance with the Nernst equation. Basically a concentration cell is developed with respect to the ion under observation. The sensing part of the electrode is usually

Made as an ion-specific membrane which is coupled with a reference electrode. So we need to have different ISE s for different ions.

Glass Electrode: Most often used pH electrodes are called glass electrodes and belong to the family of ISEs. They are sensitive only to H+ ions. Typical glass electrode is made of glass tube engaged with small glass bubble sensitive to protons. Inside of the electrode is usually filled with buffered solution of chlorides in which silver wire covered with silver chloride is immersed. pH of internal solution varies- for example it can be 1.0(0.1M HCl) or 7.0 Active part of the electrode is the glass bubble. While tube has strong and thick walls, bubble is made to be as thin as possible.

Surface of the glass is protonated by both internal and external solution till equilibrium is achieved. Both sides of the glass are charged by the adsorbed protons, this charge is responsible for potential difference. This potential in turn is described by the Nernst equation and is directly proportional to the pH difference between solutions on both sides of the glass.

The majority of pH electrodes available now a day are combination electrodes that have both glass H+ ion sensitive electrode and reference electrode compartments, conveniently placed in one housing.

Range of a pH glass electrode

The pH range at a constant concentration can be divided into 3 parts

Useful Working Range: Dependence of potential on pH has linear behavior and within which such electrode really works as ion-selective electrode for pH and obeys Nernst equation.

Alkali error range: At very low concentration of hydrogen-ions (high values of pH) metal ions interfere. In this situation dependence of the potential on pH become non-linear.

Acidic error range: At very high concentration of hydrogen-ions (low values of pH) the anions plays a big role as interfering ions, in addition to destruction of the glass matrix used for making glass bulb. Almost all commonly used glass electrodes have a working.

pH range from pH = 1 till

pH = 12. So specially designed electrodes should be used only for working in aggressive conditions.

Unit IV: Stereochemistry, Reaction Mechanism and synthesis of drug molecules

Topic 1 Stereochemistry

Q1. Define Enantiomers and Diastereomers?

A). Diastereomers (sometimes called diastereoisomers) are a type of a stereoisomer. Diastereomerism occurs when two or more stereoisomer’s of a compound have different configurations at one or more (but not all) of the equivalent (related) stereo centers and are not mirror images of each other.

Enantiomers are chiral molecules that are mirror images of one another. Furthermore, the molecules are non-super imposable on one another. ... It is sometimes difficult to determine whether or not two molecules are Enantiomers. The enantiomers differ only in their spatial arrangements at the stereo center.

Consider 2-bromo-3-chlorobutane, which has stereo centers at C2 and C3. In general, a molecule with n stereo centers has 2n stereo isomers, so there are a total of four possibilities for 2-bromo-3-chlorobutane:

Q2. Indicate, with a suitable diagram, the potential energy changes during rotation about C (2) -C (3) bond of n-butane.

Q3. Name the elements of symmetry. Discuss, with the help of an example, an optically active compound without chirality.

The compound that has a plane of symmetry will show optical activity. The compound has to be non-planar. There are some compounds, which do not have a chiral carbon, that show optical activity. The best example is biphenyls. Take the example of the one above (the picture ). It should have been a planar compound ( obviously, each carbon on the benzene ring is sp2 hybridized) but, because of the repulsion between the two NO2 groups attached ( it is a big group and their electron clouds repel), one of the NO2 moves out of the plane, thus making the compound optically active. This is how a compound without chiral carbon becomes optically active.

Each of the four stereo isomers of 2-bromo-3-chlorobutane is chiral. There are two pairs of enantiomers. Any given molecule has its enantiomer; the two other molecules are its diastereomers.

Q4. Define Relative Configuration & Absolute Configuration?

An absolute configuration refers to the spatial arrangement of the atoms of a chiral molecular entity (or group) and its stereo chemical description e.g. R or S, referring to Rectus, or Sinister, respectively.

The precise arrangement of substituent’s at a stereogenic center is known as the absolute configuration of the molecule.

The arrangement of atoms in an optically active molecule, based on chemical interconversion from or to a known compound, is a relative configuration. Relative configurations of optically active compounds by chemically interconnecting them.

Topic 2: Reaction Mechanism and synthesis of drug molecules

Q5. Explain Reduction of carbonyl compound using LiAlH4

LiAlH4 is one of the good reducing agent , it reduce the formaldehyde to methyl alcohol, aldehyde to primary alcohol , ketones to secondary alchol

Reaction type: Nucleophilic Addition

Q6. Write about Electrophilic Addition Reactions?

A. Electrophillic addition :In organic chemistry, an electrophilic addition reaction is an addition reaction where, in a chemical compound, a π bond is broken and two new σ bonds are formed. The substrate of an electrophilic addition reaction must have a double bond or triple bond.

The driving force for this reaction is the formation of an electrophile X+ that forms a covalent bond with an electron-rich unsaturated C=C bond. The positive charge on X is transferred to the carbon-carbon bond, forming a carbo cation during the formation of the C-X bond.

In step 2 of an electrophilic addition, the positively charged intermediate combines with (Y) that is electron-rich and usually an anion to form the second covalent bond.

Step 2 is the same nucleophilic attack process found in an SN1 reaction. The exact nature of the electrophile and the nature of the positively charged intermediate are not always clear and depend on reactants and reaction conditions.

In all asymmetric addition reactions to carbon, regioselectivity is important and often determined by Markovnikov's rule. Organoborane compounds give anti-Markovnikov additions. Electrophilic attack to an aromatic system results in electrophilic aromatic substitution rather than an addition reaction.

The driving force for this reaction is the formation of an electrophile X+ that forms a covalent bond with an electron-rich unsaturated C=C bond. The positive charge on X is transferred to the carbon-carbon bond, forming a carbo cation during the formation of the C-X bond.

Q7. What is Aspirine? Write any two pharmaceutical applications.

Ans. Aspirin is a non-steroidal anti-inflammatory drug .Aspirin, or acetylsalicylic acid (ASA), is commonly used as a pain reliever for minor aches and pains and to reduce fever. It is also an anti-inflammatory drug and can be used as a blood thinner.People with a high risk of blood clots, stroke, and heart attack can use aspirin long-term in low doses.

Q8. What is the Markownikoff rule?

Ans. Markovnikov’s Rule: This rule states that an addition reaction of an asymmetric alkene by an asymmetric reagent the negative part of the reagent attacks the carbon containing the less number of hydrogen across the double bond.

Markovnikov’s rule is mostly carried out using HBr as it is a good reagent for this process, neither highly exothermic nor highly endothermic.

Propene reacts with HBr:

CH3−CH=CH2+HBr−−>CH3−CH(Br)−CH3+CH3−CH2−CH2(Br)CH3−CH=CH2+HBr−−>CH3−CH(Br)−CH3+CH3−CH2−CH2(Br)

We obtain two products 2-Bromopropane and 1-Bromopropane.

according to Markovnikov’s rule the major product the major product will be decide by carbo cation stability, we see that between 2-Bromopropane and 1-Bromopropane if we remove the bromine atom, 2-Bromopropane will have a 2°2°carbocation and 1-Bromopropane will have 1°1° carbo cation. Hence the major product will be 2-Bromopropane.

2. Show, how SN2 reaction give rise to inverted product

SN2- Nucleophilic substitution Bi-molecular reaction?

In this rate of reaction depends on the concentration of substrate as well as concentration reagent. This is single step mechanism , Nucleophilic addition and elimination takes place simultaneously ,and inversion of the molecule takes place , this inversion is called Walden inversion .

Q9. Write the structure and synthesis of Paracetamol?

The original method for production involves the nitration of phenol with sodium nitrate gives a mixture of two isomers, from which the wanted 4-nitrophenol (bp 279 °C) can easily be separated by steam distillation. In this electrophilic aromatic substitution reaction, phenol's oxygen is strongly activating, thus the reaction requires only mild conditions as compared to nitration of benzene itself. The nitro group is then reduced to an amine, giving 4-aminophenol. Finally, the amine is acetylated with acetic anhydride.

10. Explain Reduction of carbonyl compound using LiAlH4

LiAlH4 is one of the good reducing agent , it reduce the formaldehyde to methyl alcohol, aldehyde to primary alcohol , ketones to secondary alchol

Reaction type: Nucleophilic Addition

Step 1: The nucleophilic H in the hydride reagent adds to the electrophilic C in the polar carbonyl group in the aldehyde, electrons from the C=O move to the O creating an intermediate metal alkoxide complex. (note that all 4 of the H atoms can react)

Step 2: This is the work-up step, a simple acid/base reaction. Protonation of the alkoxide oxygen creates the primary alcohol product from the intermediate complex.

Unit V: Spectroscopic Techniques and Applications

Topic 1: IR spectroscopy

Q1. IR spectra is often characterized as molecular finger-prints. Comment on it.

IR spectrum showing fingerprint region

The region to the right-hand side of the diagram (from about 1500 to 500 cm-1) usually contains a very complicated series of absorptions. These are mainly due to all manner of bending vibrations within the molecule. This is called the fingerprint region.

Q2. What is λmax in UV-visible spectrum?

The λmax in UV-visible spectrum is the wave length where the molecule has maximum absorption of light (maximum absorption coefficient)

UV-visible spectrum

Q3: A solution of X of concentration 0.010 mol dm–3 gives an absorbance of 0.5. What concentration is a solution of X which gives an absorbance reading of 0.25? Assume that the same optical cell is used for both readings.

Answer: Solution X concentration is (C1) = 10-2 mol dm-3

Absorbance (A1) = 0.5

For the same molecule if Absorbance (A2) is = 0.25

Than the concentration C2 is =?

From Beer Lamberts A = εCl

We can write A1/A2 = C1/C2

By substituting above values

0.5/0.25 = 10-2/C2

i.e C2 = 0.5x10-2 mol. dm-3

Q4. The four central lines in the high resolution υ =1←υ = 0 infrared spectrum of HCl37 occur at 2837.6, 2858.8, 2899.2 and 2918.6 cm-1. Deduce as much as possible about the molecule. Would the corresponding lines in HCl35 lie at the same spectral positions?

Answer: The band centre is at the average of the two central lines, i.e. 2879.0 cm-1.

This is equal to the fundamental frequency, ω0.

The 4B separation at the centre is 40.4 cm-1, giving a value of 10.1 cm-1 for B.

The moment of inertia is (I) = h/8π2Bc = 2.771x10-47 kg m2

The reduced mass is given by

Therefore,

Since the positions of the lines depend on the reduced mass, the lines for HCl35 will be at a different position. The reduced mass of HCl35 is smaller and hence its moment of inertia is smaller and rotational constant larger, so the lines will have larger separation. The fundamental vibration frequency will also be higher since it is inversely proportional to the square root of the reduced mass.

Q5. Explain batho-, hipso-, hyper- and hypochromic shifts with neat diagramme

Changes in chemical structure or the environment lead to changes in the absorption spectrum of molecules and materials. There are several terms that are commonly used to describe these shifts that you will see in the literature, and with which you should be familiar.

Bathochromic: a shift of a band to lower energy or longer wavelength (often called a red shift).

Hypsochromic: a shift of a band to higher energy or shorter wavelength (often called a blue shift).

Hyperchromic: an increase in the molar absorptivity.

Hypochromic: an decrease in the molar absorptivity.

UV-visible spectrum showing red, blue, hyper and hypo chromic shifts

Q6. Deduce the Beer-Lambert law for absorptivity and concentration.

Answer: According to this equation the Absorbance (A) is directly proportional to concentration (C) of the solution

i.e A α C ------------------------(i)

and the Absorbance is also directly proportional to path length of the light travelled (l)

i.e A α l ------------------------(ii)

From equation (i) and (ii)

A α Cl ------------------------(iii)

The equation can be written as

A = εCl

(where ‘ε’ is Absorptivity constant )

Topic2: Rotational and NMR spectroscopy

Q7. How many different types of H-atom environments are present in methyl alcohol? Also mention the ratio of peak areas due to –CH3 group and –OH group in NMR spectrum

Methanol NMR shows two types of peaks one belongs to –CH3 and another belongs to –OH The peak area ratios (number of hydrogens present on functional group -CH3: Number of hydrogens present on functional group –OH) i.e 3: 1

Q 8: What is Magnetic resonance imaging (MRI)? Describe the applications of MRI.

Magnetic resonance imaging is a scan that produces detailed pictures of organs and other internal body structures while a CT scan forms images inside of the body. CT scans use radiation, which may be harmful to the body, while MRIs do not. MRI’s cost more than CT scans

Q 9: What type of information is obtained by studying the UV, IR, H1-NMR.

Nuclear Magnetic Resonance (NMR) Spectroscopy is one of the most useful analytical techniques for determining the structure of an organic compound

UV/Vis spectroscopy is an absorption spectroscopy technique that utilizes electromagnetic radiation in the 10 nm to 700 nm range. The energy associated with light between these wavelengths can be absorbed by both non-bonding n-electrons and π-electrons residing within a molecular orbital.

The wavenumbers of the absorbed IR radiation are characteristic of many bonds, so IR spectroscopy can determine which functional groups are contained in the sample.

Q 10.What is the principle of Nuclear Magnetic Resonance (NMR) spectroscopy?

Answer: Nuclear magnetic resonance is defined as a condition when the frequency of the rotating magnetic field becomes equal to the frequency of the processing nucleus.

Principle of NMR:

The principle of nuclear magnetic resonance is based on the spins of atomic nuclei. The magnetic measurements depend upon the spin of unpaired electron whereas nuclear magnetic resonance measures magnetic effect caused by the spin of protons and neutrons. Both these nucleons have intrinsic angular momenta or spins and hence act as elementary magnet.

The existence of nuclear magnetism was revealed in the hyper fine structure of spectral lines. If the nucleus with a certain magnetic moment is placed in the magnetic field, we can observe the phenomenon of space quantization and for each allowed direction there will be a slightly different energy level.

Tutorial topics

Q1) What is the difference between temporary and permanent hardness of water?

Types of Hardness: Hardness in water is of two types.

(1) Temporary hardness and (2) permanent hardness

1. Temporary hardness: Temporary hardness is due to presence of dissolved bicarbonates of calcium and magnesium salts present in water.


Ex: Ca(HCO3)2 CaCO3↓+ CO2 +H2O


Permanent Hardness: Permanent hardness is due to presence of chlorides [Cl-], sulphates [SO42-] and Nitrates [NO3-] of calcium, magnesium and other heavy metals.



Q 2. Write the relationship between units of hardness of water.







1. Parts Per Million [PPM]:-The number of parts by weight of CaCO3 equivalents hardness causing salt present in one million parts of water.



1. Milligrams per Litre [mg/Lt]:-The number of milligrams of CaCO3 equivalent hardness causing salt present in one litre of water. i.e 1mg/lit


= 106 parts


1. Drgree Clark [0Cl]:- The number of parts of CaCO3 equivalent hardness causing salt present in 70,000 parts of water.


1 ppm = 0.070Cl

1. Degree French [0Fr]:-The number of parts by weight of CaCO3 equivalent hardness causing salt per 105 parts by weight of water.


1. Milli equivalent per Litre [Mg/Lt]:-The number of milli equivalent of CaCO3 eq. hardness causing salt per one litre of water.


Interconversion:- 1ppm =1mg/lt =0.070Cl = 0.10Fr =0.02 meq/lt.

Q 3. Which one is most widely used chlorination process?

Ans:

Chlorination:

Chlorination is the process of purifying the drinking water by producing a powerful Germicide like hypochlorous acid. When this chlorine is mixed with water it produces Hypochlorous acid which kills the Germs present in water.

H2O+Cl2→ HOCl + HCl

Chlorine is basic (means PH value is more than 7) disinfectant and is much effective over the germs. Hence chlorine is widely used all over the world as a powerful disinfectant. Chlorinator is an apparatus, which is used to purity the water by chlorination process.

Q 4. Why is Ion exchange process preferred over Zeolite-process for the softening?

Ion exchange process (or) deionization or demineralization:

Ion exchanges are of two types they are anionic and cationic. These are co-polymers of styrene & divinyl benzene i.e., long chain organic polymers with a micro porous structure.

Cation exchange resins: The resins containing acidic functional groups such as -COOH, -SO3H etc. are capable of exchanging their H+ ions with other cations are cation exchange resins , represented as RH+.

Anion exchange resins: The resins containing amino or quaternary ammonium or quaternary phosphonium (or) Tertiary sulphonium groups, treated with “NaOH solution becomes capable of exchanging their OH- ions with other anions. These are called as Anion exchanging resins represented as ROH-

Process: The hard water is passed first through cation exchange column. It removes all the cation (ca2+ & Mg2+) and equivalent amount of H+ icons are released from this column.

2RH+ + Ca2+ → R2Ca2+ + 2H+

After this the hard water is passed through anion exchange column, which removes all the anions like SO42-, Cl-, CO32- etc and release equal amount of OH- from this column.

R1OH + Cl- → R1Cl + OH-

2R1OH +SO42- → R21SO4 +2OH-

The output water is also called as de-ionized water after this the ion exchanges get exhausted. The cation exchanges are activated by mineral acid (HCl) and anion exchanges are activated by dil NaOH solution.

R2Ca + 2H+ → 2RH + Ca+2

R21SO4 + 2OH- → 2R1OH + SO42-

Advantages:

(1) The process can be used to soften highly acidic or alkaline water.

(2) It produces water of very low hardness. So it is very good for treating for use in high pressure boilers.

Disadvantages:-

(1) The equipment is costly and common expensive chemicals required.

(2) It water contains turbidity, and then output of this process is reduced. The turbidity must below 10 ppm.

Q5) Discuss the sludge and scale formation in boilers?

In boilers more water is removed in form of steam during boiling, hence boiler water gets concentrated with dissolved salts and reaches saturation point. At this point the dissolved salts precipitated and slowly settle on the inner walls of the boiler plate. The precipitation takes place in two ways










Prevention of sludge formation:

i) By using soft water which is free from dissolved salts like MgCO3, MgCl2, CaCl2, and MgSO4 can prevent sludge formation.

b) Scale formation:

The precipitation in the form of hard deposits, stick very firmly on the inner walls of the boiler is called scales.


i) Decomposition of Ca(HCO3)2:

Due to the high temperature and pressure present in the bolers, the Ca(HCO3)2 salt decomposes to CaCO3 ↓, an insoluble salt, forms scale in low pressure boilers.

Ca (HCO3)2 → CaCO3 + H2O + CO2

ii) Decomposition of CaSO4

CaSO4 is more soluble in cold water hence its solubility decreases and precipitates out to produce hard scale on the surface of the boiler. The solubility of CaSO4 is 3200ppm at 150C, reduces to 27ppm at 3200C and completely insoluble in super heated water. CaSO4 scale is very hard, highly adherent and difficult to remove.

iii) Hydrolysis of magnesium salts:

Dissolved magnesium salts undergo hydrolysis at high temperature prevailing inside the boiler forming magnesium hydroxide precipitate, which form salt type of scale.

MgCl2 + 2H2O → Mg (OH)2 ↓ + 2HCl

iv) Presence of silica:

Silica present in small quantities deposits as calcium silicate (CaSiO3) or Magnesium silicate (MgSiO3). The deposits form hard scale very difficult to remove.

Disadvantages of scale formation:

i) wastage of fuel: Scales are bad conductors of heat due to which the flow of heat to inside water is decreased hence excessive heating is required which increases fuel consumption causing wastage of fuel. The wasage of fuel increases with increases of the thichness of the scale.

ii) Lowering the boiler safety: Due to scale formation overheating of the boiler done to maintain the constant supply of steam. due to over heating the boilermaterial become softer and weaker, which causes distortion of boiler.

iii) Decrease in efficiency: Scales deposited in the valves and condensers of the boiler cause chocking which results in decrease in efficiency of the boiler.

iv) Danger of explosion: Because of the formation of the scales, the boiler plate faces higher temperature outside and lesser temperature inside due to uneven heat transfer resulting cracks in the layer of scales. Water passes through the crack and comes in contact with boiler plate having high temperature. This causes formation of large amount of steam suddenly developing sudden high pressure. This causes the explosion of the boiler.

Removal scales:

i) If the scale formed is soft it can be removed by a scrapper, wire brush etc.

ii) By giving thermal shocks done by heating the boiler to high temperature and suddenly cooling with cold water, if the scale is brittle in nature.

iii) If the scale is very adherent and hard, chemical treatment must be given. Ex: CaCO3 scale is removed by washing the boiler plate with EDTA solution.

iv) Frequent blowdown operation can remove the scales, which are loosely adhering.

Prevention of Scale formation:

Scale formation can be prevented by softening water by following methods.

Q6. Write the structure and synthesis of Paracetamol?

The original method for production involves the nitration of phenol with sodium nitrate gives a mixture of two isomers, from which the wanted 4-nitrophenol (bp 279 °C) can easily be separated by steam distillation. In this electrophilic aromatic substitution reaction, phenol's oxygen is strongly activating, thus the reaction requires only mild conditions as compared to nitration of benzene itself. The nitro group is then reduced to an amine, giving 4-aminophenol. Finally, the amine is acetylated with acetic anhydride.

Q5. Explain batho-, hipso-, hyper- and hypochromic shifts with neat diagramme

Changes in chemical structure or the environment lead to changes in the absorption spectrum of molecules and materials. There are several terms that are commonly used to describe these shifts that you will see in the literature, and with which you should be familiar.

Bathochromic: a shift of a band to lower energy or longer wavelength (often called a red shift).

Hypsochromic: a shift of a band to higher energy or shorter wavelength (often called a blue shift).

Hyperchromic: an increase in the molar absorptivity.

Hypochromic: an decrease in the molar absorptivity.

UV-visible spectrum showing red, blue, hyper and hypo chromic shifts

Unit wise Question Bank

Unit I: Molecular structure and Theories of Bonding

Two marks questions

Q1. Define degenerate orbitals? Give the examples.

Ans. Orbital’s which is having same energy is called degenerate orbital’s. Examples Px, Py, Pz orbital having same energy and dxy, dyz, dzx, dx2-y2, dz2 orbital’s having same energy ,these orbital’s are called degenerate orbital’s.

Q2. What is nodal plane?

Ans. Plane which is having probability of finding the electron is equal to zero is called nodal plane, number of orbitals is equal to n, and number of nodal planes are (n-1)

Q3. What are the shapes and structures of S, P,d orbital’s ?

S Orbital shape spherical, P orbital shape dumbel

Figure: d orbital shape double dumbel

Q4. What is a Coordinate Bond?

Sol. When both the electrons are being shared between the atoms are contributed by one atom only, then this type of bond is called Coordination Bond. It is also called Dative Bond.

A coordinate bond established between two atoms; one of which has a complete octet with at least one pair of unshared electrons while the other is short of two electrons. The Coordinate Bond is shown by (a) sign.

Example: Formation of hydronium ion from water molecules. In this, oxygen atom in water molecule is the donor and hydrogen ion is the acceptor.

Q5. Why do transition metal compounds forms complexes?

Sol. The transition metals are almost unique in their tendency to form coordination complexes. The tendency of cations of transition elements to form complexes is due to two factors;



Three marks questions

Q1. Explain about hydrogen molecules?

The simplest molecule is hydrogen, which can be considered to be made up of two separate protons and electrons. There are two molecular orbital’s for hydrogen, the lower energy orbital has its greater electron density between the two nuclei. This is the bonding molecular orbital - and is of lower energy than the two 1s atomic orbitals of hydrogen atoms making this orbital more stable than two separated atomic hydrogen orbital’s. The upper molecular orbital has a node in the electronic wave function and the electron density is low between the two positively charged nuclei. The energy of the upper orbital is greater than that of the 1s atomic orbital, and such an orbital is called an anti bonding molecular orbital. Normally, the two electrons in hydrogen occupy the bonding molecular orbital, with anti-parallel spins. If molecular hydrogen is irradiated by ultra-violet (UV) light, the molecule may absorb the energy, and promote one electron into its anti bonding orbital (*), and the atoms will separate. The energy levels in a hydrogen molecule can be represented in a diagram - showing how the two 1s atomic orbitals combine to form two molecular orbitals, one bonding () and one anti bonding (*). This is shown graphically below

Q2. Write the difference between the molecular orbitals and atomic orbitals?

Differences between Molecular Orbital and Atomic Orbital

Molecular Orbital

Atomic Orbital

4. An electron Molecular orbital is under the influence of two or more nuclei depending upon the number of atoms present in the molecule.

5. Molecular orbitals are formed by combination of atomic orbitals

6. They have complex shapes.

4. An electron in atomic orbital is under the influence of only one positive nucleus of the atom.

5. Atomic orbitals are inherent property of an atom.

6. They have simple shapes.

Q3. Write the silent features of CFT

Important Features of Crystal Field theory are

1. Transition metal ion is surrounded by ligands with lone pair of electrons and the complex is a combination of central ion surrounded by other ions or molecules or diploes i.e ligand

2. All types of ligands are regarded as point charges.

3. The interaction between the metal ion and the negative ends of anion (or ion dipoles) are purely electrostatic i.e bond between the metal and ligand is considered 100 percent ionic.

4.The ligands surrounding the metal ion produce electric field influences the energies of the orbitals of central metal ion particularly d-orbitals.

5. In the case of free metal ion all the five d-orbitals have the same energy. Such orbital having the same energies are called degenerate orbital’s.

Q4. Write the applications of co-ordination compounds?

Coordination compounds are widely use now days. Some of their applications are listed below:

· Extraction process of gold and silver

· Used as a catalyst in many industrial processes.

Example: Nickel, Copper can be extracted by using hydrometallurgical process involving coordination compounds.

1. Used in hardness of water.

Example: EDTA (Ethylenediaminetetraacetate) is used in the estimation of Ca+2 and Mg+2 in hard water.

1. Cyanide complexes are used in electroplating.

Q5. Write a note on spectro chemical series?

The tool used often in calculations or problems regarding spin is called the spectro chemical series. The spectro chemical series is a list that orders ligands on the basis of their field strength. Ligands that have a low field strength, and thus high spin, are listed first and are followed by ligands of higher field strength, and thus low spin. This trend also corresponds to the ligands abilities to split d orbital energy levels. The ones at the beginning, such as I−, produce weak splitting (small Δ) and are thus weak field ligands. The ligands toward the end of the series, such as CN−, will produce strong splitting (large Δ) and thus are strong field ligands. A picture of the spectrochemical series is provided below.

(weak) I− < Br− < S2− < SCN− < Cl− < NO3− < N3− < F− < OH− < C2O42− ≈ H2O <

NCS− < CH3CN < py < NH3 < en < bipy < phen < NO2− < PPh3 < CN− ≈ CO (strong)

Five marks questions

Q1. Explain about linear combination of atomic orbitals?

According to this method the formation of orbital’s is because of Linear Combination (addition or subtraction) of atomic orbital’s which combine to form molecule. Consider two atoms A and B which have atomic orbital’s described by the wave functions ΨA and ΨB .If electron cloud of these two atoms overlap, then the wave function for the molecule can be obtained by a linear combination of the atomic orbital’s ΨA and ΨB i.e. by subtraction or addition of wave functions of atomic orbital’s ΨMO= ΨA + ΨB

The above equation forms two molecular orbital’s

Bonding Molecular Orbitals

When addition of wave function takes place, the type of molecular orbitals formed are called Bonding Molecular orbital’s and is represented by ΨMO = ΨA + ΨB.

They have lower energy than atomic orbital’s involved. It is similar to constructive interference occurring in phase because of which electron probability density increases resulting in formation of bonding orbital. Molecular orbital formed by addition of overlapping of two s orbitals shown in figure no. 2. It is represented by s.

Anti-Bonding Molecular Orbital’s

When molecular orbital is formed by subtraction of wave function, the type of molecular orbitals formed are called Anti-bonding Molecular Orbitals and is represented by ΨMO = ΨA - ΨB.

They have higher energy than atomic orbitals. It is similar to destructive interference occurring out of phase resulting in formation of anti-bonding orbitals. Molecular Orbital formed by subtraction of overlapping of two s orbitals are shown in figure no. 2. It is represented by s* (* is used to represent anti-bonding molecular orbital) called Sigma Anti-bonding.

Fig. Formation of Bonding and Anti-Bonding Orbital

Therefore, Combination of two atomic orbitals results in formation of two molecular orbitals, bonding molecular orbital (BMO) whereas other is anti-bonding molecular orbital (ABMO).

Q2. Explain about molecular orbital diagram of N2 Molecule?

Nitrogen:




Q3. Explain about the molecular orbital diagram of O2 molecule?

Oxygen:This molecule has twelve electrons, two more than nitrogen - and these extra two are placed in a pair of degenerate g orbitals. The atomic orbitals combine to produce the following molecular orbital diagram:

For O2 and higher molecules →

σ1s, σ *1s, σ 2s, σ *2s, σ 2pz, [π2px = π2py], [π*2px= π*2py], σ *2pz

Fig. no. 6 Order of Energy for O2 and Higher molecules

Comparison of the above energy level diagram wit hthat for nitrogen - you can see that the 2sg level lies lower than pu. Here, we are starting to fill the anti-bonding orbitals originating from the p orbital interactions and so the bond order decreases from three to two.The lowest energy arrangement (Hund's rule) - has a single electron, each with parallel spins, in each of the pgx and pgy orbitals. This produces a paramagnetic molecule, with a double bond and has two unpaired electrons.

Q4. Explain about d orbital splitting in octahedral complexes?

In an octahedral complex, there are six ligands attached to the central transition metal. The d orbital splits into two different levels. The bottom three energy levels are named dxy, dxz, and dyz (collectively referred to as t2g). The two upper energy levels are named d(x2−y2), and dz2 (collectively referred to as eg).

The reason they split is because of the electrostatic interactions between the electrons of the ligand and the lobes of the d-orbital. In an octahedral, the electrons are attracted to the axes. Any orbital that has a lobe on the axes moves to a higher energy level. This means that in an octahedral, the energy levels of eg are higher (0.6∆o) while t2g is lower (0.4∆o). The distance

Figure 4: Splitting of the degenerate d-orbitals (without a ligand field) due to an octahedral ligand field shown in Figure 3.

that the electrons have to move from t2g from eg and it dictates the energy that the complex will absorb from white light, which will determine the color. Whether the complex is paramagnetic or diamagnetic will be determined by the spin state. If there are unpaired electrons, the complex is paramagnetic; if all electrons are paired, the complex is diamagnetic.

Q5. Discuss about the band structure and effect of doping in solid materials?

A useful way to visualize the difference between conductors, insulatorsand semiconductors is to plot the available energies for electrons in the materials. Instead of having discrete energies as in the case of free atoms, the available energy states form bands. Crucial to the conduction process is whether or not there are electrons in the conduction band. In insulators the electrons in the valence band are separated by a large gap from the conduction band, in conductors like metals the valence band overlaps the conduction band, and in semiconductors there is a small enough gap between the valence and conduction bands that thermal or other excitations can bridge the gap. With such a small gap, the presence of a small percentage of a dopingmaterial can increase conductivity dramatically.

An important parameter in the band theory is the Fermi level, the top of the available electron energy levels at low temperatures. The position of the Fermi level with the relation to the conduction band is a crucial factor in determining electrical properties.

Figure: Semiconductor Energy Bands

For intrinsic semiconductors like silicon and germanium, the Fermi level is essentially halfway between the valence and conduction bands. Although no conduction occurs at 0 K, at higher temperatures a finite number of electrons can reach the conduction band and provide some current. In doped semiconductors, extra energy levels are added.

The increase in conductivity with temperature can be modeled in terms of the Fermi function, which allows one to calculate the population of the conduction

UNIT-II: Water and its treatment

Two marks questions with answers

Q1. Distinguish between hard water and soft water.

The water, which does not produce lather with soap solution is called “Hard water”. But soft water readily produce lather with soap.

The salts responsible for hardness are bicarbonates, chlorides, sulphates and nitrates of bivalent metal ions like calcium, magnesium and other heavy metals.

Soap is sodium or potassium salts of higher fatty acids like stearic, oleic and palmetic acids. When soap soap is mixed with soft water lather is produced due to stearic acid and sodium stearate.

Ex: 2C17H35COONa + H2O →2C17H35COOH + NaOH

Sodium stearate (soap)Stearic acid

Stearic acid + Sodium stearate → Formation of lather

When soap comes in contact with hard water, sodium stearate will react with dissolved Ca and Mg salts and produce Ca-stearate or Mg- stearate which is white precipitate.

C17H35COONa+ CaCl2 → (C17H35COO)2 Ca↓ + 2NaCl (soluble)

Hardness precipitation

substance (insoluble)

Q2. Differentiate between temporary and permanent hardness of water. Discuss in detail the various units of hardness

Temporary [Or] Carbonate Hardness: - Temporary hardness is due to presence of dissolved bicarbonates of calcium and magnesium salts present in water.


Ex: - Ca(HCO3)2 CaCO3↓+ CO2 +H2O


Permanent [or] Non-Carbonate Hardness:- Permanent hardness is due to presence of chlorides [Cl-], sulphates [SO42-] and Nitrates [NO3-] of calcium, magnesium and other heavy metals.



Q3. Describe the causes and harmful effects of sludge formation in boilers.










Q4. Discuss break point of chlorination in treatment of potable water.

Break-point chlorination:

Calculated amount of chlorine must be added to water because chlorine after reacting with bacteria and organic impurity or ammonia remains in water as residual chlorine, which gives bad taste, odor and is toxic to human beings. The exact amount of chlorine required to kill bacteria and to remove organic matter is called break-point chlorination.

Water sample is treated with chlorine and estimated for the residual chlorine in water and plotted a graph as shown below which gives the break-point chlorination:

From the above graph, it is clear that A grams of chlorine added oxidizes reducing impurities of water. 'B' grams of chlorine added forms chloramine and other chloro compounds. 'C' grams of chlorine added forms destruction of bacteria and destruction of chloramines. Further amount of chlorine is residual chlorine. ‘C’ grams are the breakpoint for addition of chlorine to water.

Advantages of break-point chlorination:

It removes taste, color and oxidizes completely organic compound like ammonia and other reducing impurities. It destroys completely all disease causing bacteria.

Q5. Write about calgon conditioning in internal treatment of boiler feed water.

Calgon conditioning :It involves in adding calgon [sodium hexa meta phosphate] to boiler water .it prevents the scale and sludge formation by forming soluble complex compound with CaSO4.

Na2[Na4(PO3)3] → 2Na+ + [Na4(PO3)3]

(Calgon)

2CaSO4+ [Na4(PO3)6]2- [Ca2(PO3)6]2- +2Na2SO4 .

Phosphate conditioning: The scale formation can be avoided in high pressure boilers by adding tri sodium phosphate or other types of phosphates according to the pH of boiler water.

3CaSO4+2Na3PO4Ca3 (PO4) 2+3Na2SO4

The different phosphates used are:

· Na3PO4 (trisodium phosphate) - used to acidic water.

· Na2HPO4 (disodium phosphate) - used to weak alkaline.

· NaH2PO4 (monosodium phosphate) - used to alkaline water.

Three marks questions with answers

Q1. Express the units of hardness of water

EXPRESSION OF HARDNESS:-

Hardness of water is expressed in terms of “calcium carbonates (CaCO3) equivalents”. The weights of different hardness causing salts are converted into weight equivalents to that of calcium carbonate.

If sample contains two or more than two salts, their quantities are converted in equivalent to CaCO3 and then sum will give the total hardness.

Thus 120 parts by weight (1gr mol wt) of MgSO4 would react with the same amount of soap as 100 parts by wight of CaCO3 (1gr mol wt) hardness. Hence weight in terms of CaCO3 would be equal to weight of MgSO4 in water multiplied by 100/120. Therefore 100 parts by weight of CaCO3 hardness must be equal to

1. 162 parts by weight of Ca (HCO3)2 hardness

2. 146 parts by weight of Mg (HCO3)2 hardness

3. 136 parts by weight of CaSO4 hardness

4. 111 parts by weight of CaCl2 hardness

The choice of CaCO3 is due to its molecular weight being 100 and also it is the most insoluble salt that can be precipitated in water treatment.

Equivalents of CaCO3

Amount of hardness causing salt X molecular weight of CaCO3 [100]

=

M.w of ℎardeness causing salt

UNITS OF HARDNESS AND INTERCONVERSION:-







1 Parts Per Million [PPM]:-The number of parts by weight of CaCO3 equivalents hardness causing salt present in one million parts of water.



2 Milligrams per Litre [mg/Lt]:-The number of milligrams of CaCO3 equivalent hardness causing salt present in one litre of water. i.e 1mg/lit


= 106 parts


3 Drgree Clark [0Cl]:- The number of parts of CaCO3 equivalent hardness causing salt present in 70,000 parts of water.


1 ppm = 0.070Cl

4 Degree French [0Fr]:-The number of parts by weight of CaCO3 equivalent hardness causing salt per 105 parts by weight of water.


5 Milli equivalent per Litre [Mg/Lt]:-The number of milli equivalent of CaCO3 eq. hardness causing salt per one litre of water.


Q2. What is reverse osmosis? Describe reverse osmosis method of desalination of brackish water.

Ans: Reverse Osmosis:

When two solutions of different concentrations are separated by a semi-permeable membrane, flow of solvent from low concentration to high concentration takes place due to difference in concentration, this is said to be “osmosis”.

If a hydrostatic pressure in excess of osmotic pressure is applied on the concentrated side the flow of solvent reverses as it is forced to move from high concentration to low concentration through the membrane known as “reverse osmosis”.

In this process, semi-permeable membranes are made by thin film of cellulose acetate (or) polyamide polymers (or) polymethylene are used. A pressure of 15 -40 kg/cm2 is applied for separating the water from its contaminants. This process also known as super or hyper filtration.

Advantages:

· It is simple and reliable process. Low maintenance cost and also pollution free.

· The life of semi permeable membranes is about 2 years , and it can be easily replaced within minutes, and there by nearly un interrupted water supply can be provided.

Q3. Calculate the temporary [carbonate] and permanent [Non-carbonate] hardness of water sample having following values:- Ca(HCO3)2 = 3.24 mg/lt , mg(HCO3)2 = 14.6 mg/lt, CaCl2 = 22.2 mg/lt and MgSO4 = 60 mg/lt

Ans:

S.No

Constituent

Amount mg/l

Mol.wt

1

Mg(HCO3)2

14.6

146

2

Ca(HCO3)2

32.4

162

3

MgSO4

60

120

4

CaCl2

22.2

111

Hardness of Ca (HCO3)2 in terms of CaCO3 equivalent = 3.24×100/162

= 20

Hardness of Mg (HCO3)2 in terms of CaCO3 equivalent = 14.6×100/146

= 10

Hardness of MgSO4 in terms of CaCO3 equivalent = 60×100/120

= 50

Hardness of CaCl2 in terms of CaCO3 equivalent = 22.2×100/111

= 20

Temporary hardness= Ca (HCO3)2 Hardness + Mg (HCO3)2 Hardness

=20+10 =30 ppm (or) 30Fr

Permanent hardness=CaCl2 hardness + MgSO4 hardness

=20 + 50= 70 ppm or 70Fr

Total Hardness = Temporary hardness + Permanent Hardness

= 30 + 70 = 100 ppm or 100Fr

Q4. One litre of water from an underground reservoir in tirupathi town in Andhra Pradesh showed the following analysis for its contents. Mg (HCO3)2= 42 mg, Ca (HCO3)2= 146 mg, CaCl2= 71 mg, NaOH= 40 mg, MgSO4=48 mg, organic impurities=100 mg, Calculate temporary, permanent and total hardness?

Ans:

Hardness causing salt (H.C.S)

Quantity (H.C.S)

Mol.Wt.of (H.C.S)

CaCl2

71

111

MgSO4

48

120

Ca(HCO3)2

146

162

Mg(HCO3)2

42

146

NaOH

40

-

Temporary Hardness = Mg(HCO3)2 + Ca(HCO3)2

= 28.7 + 90.1 = 118.8ppm

Permanent Hardness =CaCl2 + MgSO4

= 64 + 40 = 104ppm

Total Hardness = Temporary Hardness + Permanent Hardness

= 118.8 + 104

= 222.8ppm.

NaOH (compounds of alkali metals) does not give hardness to water.

Q5. Mention the steps involved in the treatment of potable water

Treatment of water for municipal supply:

Ans: The treatment of water for drinking purpose mainly includes colloidal impurities and harmful pathogenic bacteria. The following is the flow diagram of water treatment for domestic purpose and various stages involved in the purification, given as:

(a) Screening: In this, the water is passed through screens having number of holes in it to remove floating impurities.

(b) Sedimentation with coagulation:

In this, the suspended and colloidal impurities are allowed to settle under gravitation. The basic principle of this treatment is to allow water to flow at a very slow velocity so that the heavier particles coagulate like alums, sodium alluminate and salts of iron are added which produces gelatinous precipitates called floc. Floc attracts and helps the accumulation of the colloidal particles, resulting in setting of the colloidal particles.

(c) Sterilization and disinfection:

Destruction of harmful pathogenic bacteria from the drinking water is carried out of sterilization and disinfection.

The following are the methods adopted for domestic purpose.

(i) Boiling:

By boiling water for 15-20 minutes, harmful bacteria are killed. This is not possible for municipal supply of water. This method of sterilization is adapted for domestic purpose.

(ii) Passing ozone:

Ozone is an unstable isotope of oxygen, produces Nascent oxygen which is powerful disinfectant.

O3 → O2 + O

(iii) By ultraviolet light:

UV light is used as disinfectant for swimming pool water as no chemicals are used. It is safe for skin. In this process water is exposed to UV rays which are generated foe an electric mercury vapor lamp.

Five marks questions with answers

Q1. Write the experimental procedure for the determination of total hardness by EDTA method.

Ans: Estimation of Hardness of Water by EDTA Method [Complexometric Method]:-

In this complexometric method, disodium salt of Ethylene diamine tetra acetic acid is used as complexometric agent.

NaOOC-H2C CH2COOH

HOOC-H2C N-CH2-CH2-N CH2COONa

Basic principle:

When hard water comes in contact with EDTA at pH 9-10 (pH maintained by buffer solution prepared by using NH4Cl + NH4OH) the Ca2+ and Mg2+ forms colourless stable complex with EDTA.

Erichrome Black – T [EBT] indicator forms an unstable, wine red coloured complex with Ca2+ and Mg2+.

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