Welcome to Edugrown – Your Learning Partner EduGrown
🕑 Aaj ke free content khatam — ye page khula rahega, naya content kal se. Login karein →
← Back to UG & PG Content
Exam Papers

B.Tech 1st Year Engineering Chemistry (BT-101) Solved Question Paper June 2025 – RGPV

RGPV University (RGPV) · MP · Engineering (B.Tech / B.E.) · Automobile · 2025 · Semester 1 · 9 views

Advertisement

RGPV · BT-101 (GS) · B.Tech I & II Semester

Engineering Chemistry · June 2025

Solved paper · Time: 3 hours · Maximum marks: 70 · Attempt any five questions; all carry equal marks

1 (a)

Write short note on: (i) Sludge and Scale (ii) Caustic Embrittlement (iii) Boiler Corrosion

(i) Sludge and scale

In a boiler, water evaporates continuously and the concentration of dissolved salts keeps rising. When the solution becomes saturated, the salts separate out as a precipitate on the inner walls. If the precipitate is loose and slimy it is called sludge; if it forms a hard, adherent coating it is called scale.

Sludge

Sludge is a soft, loose, slimy precipitate formed in the comparatively colder parts of the boiler and where the flow is slow (bends, bottom of the boiler). It is formed by salts whose solubility is greater in hot water than in cold, such as MgCO3, MgCl2, MgSO4 and CaCl2.

Disadvantages: sludge is a poor conductor of heat, so fuel is wasted; excess sludge settles in pipes and chokes them, disturbing circulation; when formed along with scale it gets entrapped in the scale and both are deposited together.

Prevention: use well-softened water and carry out frequent blow-down.

Scale

Scale is a hard deposit firmly sticking to the inner surface of the boiler, very difficult to remove even with a hammer and chisel. Causes of scale formation:

  1. Decomposition of calcium bicarbonate (low-pressure boilers): Ca(HCO3)2 → CaCO3↓ + H2O + CO2. CaCO3 scale is soft and is the main scale in low-pressure boilers.
  2. Deposition of calcium sulphate (high-pressure boilers): solubility of CaSO4 decreases with rise in temperature, so it separates out as a hard, adherent scale on the hot portions.
  3. Hydrolysis of magnesium salts: MgCl2 + 2H2O → Mg(OH)2↓ + 2HCl. Mg(OH)2 forms a soft scale.
  4. Presence of silica: SiO2 forms calcium and magnesium silicates (CaSiO3, MgSiO3), which give a very hard, glassy scale that is very difficult to remove.

Bad effects of scale

  • Wastage of fuel: scale is a poor conductor of heat, so the metal must be heated more to transfer the same heat. Approximate fuel wastage: 0.325 mm scale → 10%, 0.625 mm → 15%, 1.25 mm → 50%, 2.5 mm → 80%, 12 mm → 150%.
  • Lowering of boiler safety: the metal is overheated, becomes soft and weak, and may fail under high steam pressure.
  • Decrease in efficiency: scale deposits in valves and condensers and partially chokes them.
  • Danger of explosion: scale expands differently from the metal; when it cracks, water suddenly comes in contact with the overheated metal, a large amount of steam forms at once and high pressure may burst the boiler.

Difference between sludge and scale

SludgeScale
Soft, loose and slimy deposit.Hard, adherent coating.
Forms in colder parts and where flow is slow.Forms on the heated surfaces of the boiler.
Due to MgCO3, MgCl2, MgSO4, CaCl2.Due to CaSO4, CaCO3, Mg(OH)2, silicates.
Easily removed by scraping or blow-down.Difficult to remove; needs chemical or thermal methods.
Less harmful; poor heat conductor.More harmful; may cause explosion.

Prevention and removal of scale

  • External treatment: soften feed water by lime-soda, zeolite or ion-exchange process before use.
  • Internal treatment (conditioning):
    • Colloidal conditioning (low-pressure boilers): kerosene, tannin or agar-agar coat the particles so they stay as loose sludge.
    • Phosphate conditioning (high-pressure boilers): 3CaCl2 + 2Na3PO4 → Ca3(PO4)2↓ + 6NaCl (soft sludge).
    • Carbonate conditioning (low-pressure boilers): CaSO4 + Na2CO3 → CaCO3↓ + Na2SO4.
    • Calgon conditioning: sodium hexametaphosphate forms a soluble complex: Na2[Na4(PO3)6] + 2CaSO4 → Na2[Ca2(PO3)6] + 2Na2SO4.
    • Sodium aluminate: gives gelatinous Al(OH)3 and Mg(OH)2 that entrap fine particles as sludge.
  • Removal: scraping with wire brush (soft scale), thermal shocks (heating and sudden cooling), chemical washing (5–10% HCl for CaCO3; EDTA for CaSO4), and frequent blow-down.

(ii) Caustic embrittlement

Definition: Caustic embrittlement is a type of boiler corrosion in which the boiler metal becomes brittle and develops intergranular (inter-crystalline) cracks at highly stressed parts such as bends, joints and rivet holes, due to a high concentration of caustic soda (NaOH) in the boiler water.

Cause and mechanism

Water softened by the lime-soda process contains some residual Na2CO3. In high-pressure boilers it hydrolyses to NaOH:

Na2CO3 + H2O → 2NaOH + CO2

This alkaline water enters the minute hair-cracks in the boiler walls by capillary action. Water evaporates inside the cracks and NaOH concentration builds up there (it can exceed 10%). Concentrated NaOH attacks iron and dissolves it as sodium ferroate:

Fe + 2NaOH → Na2FeO2 + H2
3Na2FeO2 + 4H2O → 6NaOH + Fe3O4 + H2

The NaOH is regenerated, so the attack continues and the cracks spread. The process can also be explained by a concentration cell, in which iron in contact with concentrated NaOH (inside cracks, under stress) acts as anode and iron in contact with dilute NaOH acts as cathode:

Iron at bends (anode) | Conc. NaOH ‖ Dilute NaOH | Iron at plane surface (cathode)

Prevention

  • Use sodium phosphate instead of sodium carbonate as the softening reagent.
  • Add tannin or lignin to boiler water; they block the hair-cracks and pits and prevent infiltration of NaOH.
  • Add sodium sulphate, which also blocks the cracks. Recommended Na2SO4 : NaOH ratio is 1:1, 2:1 and 3:1 for boilers working up to 10, 20 and above 20 atm respectively.
  • Keep the pH of boiler water around 8–9 and avoid excess alkalinity.
  • Reduce stresses by proper design and fabrication: use welding instead of riveting and avoid sharp bends.

(iii) Boiler corrosion

Boiler corrosion is the decay or disintegration of boiler material by chemical or electrochemical attack of its environment (the water and gases inside). Its main causes are:

1. Dissolved oxygen

Water usually contains about 8 mL of dissolved O2 per litre at room temperature. At high temperature it attacks the boiler metal:

2Fe + 2H2O + O2 → 2Fe(OH)2
4Fe(OH)2 + O2 → 2[Fe2O3·2H2O] (rust)

Removal: (a) chemically, by adding sodium sulphite or hydrazine:

2Na2SO3 + O2 → 2Na2SO4
N2H4 + O2 → N2 + 2H2O (best: no salt is added to the water)

(b) mechanically, by de-aeration: water is sprayed over perforated plates in a tower that is heated and connected to a vacuum pump, so dissolved gases escape.

2. Dissolved carbon dioxide

CO2 comes from dissolved air and from decomposition of bicarbonates (Mg(HCO3)2 → MgCO3 + H2O + CO2). It forms carbonic acid, which slowly attacks the metal: CO2 + H2O → H2CO3.

Removal: by adding a calculated amount of ammonia (2NH4OH + CO2 → (NH4)2CO3 + H2O), or by mechanical de-aeration along with O2.

3. Acids from hydrolysis of salts

MgCl2 present in water hydrolyses at high temperature to give HCl, which attacks iron in a chain-like reaction:

MgCl2 + 2H2O → Mg(OH)2↓ + 2HCl
Fe + 2HCl → FeCl2 + H2
FeCl2 + 2H2O → Fe(OH)2 + 2HCl (HCl regenerated)

So even a small amount of MgCl2 causes extensive corrosion. Prevention: soften the water properly and neutralise any acid with a calculated amount of alkali.

1 (b)

100 ml water sample required 20 ml of N/50 H₂SO₄ for neutralisation to Phenolphthalein end point. After Methyl orange was added to this and further acid required was 2.5 ml. Calculate the alkalinity type and extent in terms of CaCO₃ equivalent.

Alkalinity (ppm as CaCO3) = Vacid × Nacid × 50 × 1000 / Vsample (50 = equivalent weight of CaCO3).

Phenolphthalein alkalinity (P)

P = 20 × (1/50) × 50 × 1000 / 100 = 200 ppm

Methyl orange (total) alkalinity (M)

Total acid used = 20 + 2.5 = 22.5 mL
M = 22.5 × (1/50) × 50 × 1000 / 100 = 225 ppm

Type of alkalinity

Here P = 200 ppm and M/2 = 112.5 ppm, so P > M/2. Hence the alkalinity is due to OH⁻ and CO3²⁻ ions (HCO3⁻ is absent).

OH⁻ alkalinity = 2P − M = 2 × 200 − 225 = 175 ppm
CO₃²⁻ alkalinity = 2(M − P) = 2 × (225 − 200) = 50 ppm
HCO₃⁻ alkalinity = 0
Hydroxide alkalinity = 175 ppm · Carbonate alkalinity = 50 ppm · Bicarbonate = nil (total 225 ppm as CaCO₃)
2 (a)

An oil sample under test has a Saybolt Universal Viscosity same as that of standard Gulf oil (low viscosity standard) and Pennsylvanian oil (high viscosity standard) at 210°F. Their Saybolt Universal Viscosities at 100°F are 61, 758 and 420 seconds respectively. Calculate viscosity index of the sample.

The three viscosities at 100 °F are given in the order: test oil, Gulf oil, Pennsylvanian oil. All three have the same viscosity at 210 °F.

U (test oil) = 61 s, L (Gulf oil, low-VI standard) = 758 s, H (Pennsylvanian oil, high-VI standard) = 420 s
VI = (L − U) / (L − H) × 100
= (758 − 61) / (758 − 420) × 100
= 697 / 338 × 100 = 206.2
Viscosity index of the sample ≈ 206

A VI above 100 means the test oil's viscosity changes with temperature even less than that of the Pennsylvanian standard (its viscosity at 100 °F, 61 s, is lower than H = 420 s), as is typical of synthetic oils or oils containing VI improvers.

2 (b)

Describe ion exchange process used for softening of water with neat diagram. A zeolite softener was 90% exhausted by removing the hardness completely when 10,000 L of hard water sample passed through it. The exhausted zeolite bed required 200 L of 3% NaCl solution for its complete regeneration. Calculate the hardness of water sample.

The ion-exchange (demineralisation) process removes all cations and anions from water using ion-exchange resins. These are insoluble, cross-linked long-chain organic polymers (for example styrene–divinylbenzene copolymer) with a microporous structure, carrying functional groups that can exchange ions.

  • Cation exchange resins (RH2): contain acidic groups –SO3H or –COOH; they exchange H⁺ for cations.
  • Anion exchange resins (R′(OH)2): contain basic groups such as quaternary ammonium –NR3⁺OH⁻ or amino groups; they exchange OH⁻ for anions.

Process

Hard water is passed first through the cation-exchange column, which removes Ca²⁺, Mg²⁺, Na⁺ etc. and releases an equivalent amount of H⁺:

RH2 + Ca²⁺ → RCa + 2H⁺
RH2 + CaCl2 → RCa + 2HCl
RH2 + MgSO4 → RMg + H2SO4

The acidic water then passes through the anion-exchange column, which removes Cl⁻, SO4²⁻, HCO3⁻ and releases OH⁻:

R′(OH)2 + 2HCl → R′Cl2 + 2H2O
R′(OH)2 + H2SO4 → R′SO4 + 2H2O
H⁺ + OH⁻ → H2O

The water coming out is free from all ions and is called deionised or demineralised water. Dissolved CO2 may be removed in a degasifier.

Cation resinRH₂Anion resinR′(OH)₂Hard wateracidicwaterDeionised waterHCl (regenerant)Washings(CaCl₂, MgCl₂)NaOH (regenerant)Washings (NaCl)Cation exchangerAnion exchanger
Ion exchange (demineralisation) plant: cation column followed by anion column

Regeneration

When the resins are exhausted, the cation exchanger is regenerated with dilute HCl (or H2SO4) and the anion exchanger with dilute NaOH. The washings are drained off.

RCa + 2HCl → RH2 + CaCl2
R′SO4 + 2NaOH → R′(OH)2 + Na2SO4

Advantages

  • Produces water of very low hardness (about 0–2 ppm) — suitable for high-pressure boilers.
  • Removes all ions, including silica (with strongly basic resin), so total dissolved solids become very low.
  • Works for highly acidic or alkaline water.

Disadvantages

  • Equipment and resins are costly; regenerating chemicals (acid, alkali) are expensive.
  • Turbid water reduces efficiency, since suspended matter clogs the resin; water must be filtered first.

Numerical

The NaCl used for regeneration is exactly equivalent to the hardness removed by the exhausted (90%) part of the zeolite bed, which removed all the hardness of 10,000 L of water.

3% NaCl = 3 g per 100 mL = 30 g/L
NaCl in 200 L = 200 × 30 = 6000 g = 6 × 10⁶ mg
CaCO₃ equivalent = 6 × 10⁶ × 50 / 58.5 = 5.128 × 10⁶ mg
(2NaCl ≡ CaCO₃, i.e., 117 g NaCl ≡ 100 g CaCO₃)
This hardness was present in 10,000 L of water
Hardness = 5.128 × 10⁶ / 10,000 = 512.8 mg/L
Hardness of the water sample ≈ 512.8 ppm
3 (a)

Explain the following properties (any two): (i) Aniline Point (ii) Cloud and Pour Point (iii) Flash and Fire Point

All three are answered below; write any two in the exam.

(i) Aniline point

Aniline point is the minimum equilibrium solution temperature at which equal volumes of aniline and the oil are completely miscible.

Determination: equal volumes of aniline and oil are taken in a test tube and heated with stirring until they mix to form a homogeneous solution. The mixture is then cooled slowly; the temperature at which the two phases separate (turbidity appears) is the aniline point.

Significance: aniline dissolves aromatic hydrocarbons easily, so an oil with high aromatic content has a low aniline point. Aromatic hydrocarbons dissolve and swell natural and many synthetic rubbers. Therefore a lubricant that comes in contact with rubber seals, gaskets, packings and hoses should have a high aniline point. A high aniline point indicates a mainly paraffinic oil with low aromatic content.

(ii) Cloud and pour point

Cloud point: when a lubricating oil is cooled slowly, the temperature at which it becomes cloudy or hazy (because wax or other solids begin to separate) is called the cloud point.

Pour point: the lowest temperature at which the oil just ceases to flow (or, the lowest temperature at which it can still be poured) is called the pour point.

Determination: the oil is taken in a flat-bottomed glass tube fitted with a thermometer and placed in a cooling bath. The tube is taken out at every 3 °C fall in temperature and examined; the temperature at which haziness first appears is the cloud point. The temperature at which the oil does not flow when the tube is held horizontal for 5 seconds is the pour point.

Significance: these indicate the suitability of a lubricant for use in cold conditions. Oils for refrigeration machinery, aircraft at high altitude and vehicles in cold regions must have low cloud and pour points, otherwise they solidify and choke the oil lines and bearings. Pour point is lowered by removing wax (dewaxing) or adding pour-point depressants such as polymethacrylates.

(iii) Flash and fire point

Flash point is the lowest temperature at which an oil gives off enough vapour to ignite for a moment (give a flash) when a small flame is brought near it.

Fire point is the lowest temperature at which the vapour of the oil burns continuously for at least 5 seconds when a small flame is brought near it. The fire point is usually 5–40 °C higher than the flash point.

Determination (Pensky-Martens closed cup)

The oil cup is filled with oil up to the mark and covered with the lid, which carries a stirrer, a thermometer and a shutter with a test flame. The oil is heated slowly in an air bath at about 5–6 °C per minute with continuous stirring. At every 1 °C rise, the shutter is opened and the test flame is dipped into the vapour space. The temperature at which a distinct flash appears inside the cup is the flash point. For the fire point, heating is continued (usually in an open cup) until the vapour catches fire and burns for at least 5 seconds. Abel's apparatus is used for oils with flash point below 50 °C, and Cleveland open cup for open-cup values.

StirrerThermometerTest flame + shutterOil cup (filled to mark)Air bathHeater
Pensky-Martens closed-cup apparatus (simplified)

Significance

  • A good lubricant should have a flash point well above its working temperature, so that it does not catch fire during use; this ensures safety against fire hazards.
  • Indicates the safe temperature for storage, transport and handling of the oil.
  • A low flash point shows the presence of volatile components (or adulteration with lighter oils); these evaporate, thinning the lubricant film.
  • Helps in identifying and comparing oils.
3 (b)

Distinguish between thermoplastic and thermosetting resins. Give preparation and application of Urea-Formaldehyde and PVC.

Thermoplastic versus thermosetting resins

Thermoplastic resinsThermosetting resins
Soften on heating and harden on cooling; can be remoulded many times.Become permanently hard on heating; cannot be remoulded.
Linear or slightly branched chains with weak van der Waals forces.Three-dimensional cross-linked network with strong covalent bonds.
Formed generally by addition polymerisation.Formed generally by condensation polymerisation.
Soft, weak and less brittle; soluble in some organic solvents.Hard, strong and brittle; insoluble in solvents.
Scrap can be reused (recyclable).Scrap cannot be reused.
Examples: polythene, PVC, polystyrene, PMMA.Examples: Bakelite, urea-formaldehyde, epoxy resins.

Urea-formaldehyde resin

Urea-formaldehyde (UF) resin is a thermosetting amino resin prepared by condensation of urea and formaldehyde.

Preparation

Urea is treated with formaldehyde in neutral or slightly alkaline medium to give monomethylol urea and dimethylol urea:

NH2CONH2 + HCHO → NH2CONHCH2OH (monomethylol urea)
NH2CONH2 + 2HCHO → HOCH2NHCONHCH2OH (dimethylol urea)

On heating in the presence of an acid catalyst, these condense with elimination of water to form linear chains with –NH–CH2–NH– and –CH2–O–CH2– links, which then cross-link through the remaining –NH groups to form a hard, three-dimensional thermoset.

Properties

  • Clear, colourless and can be given light, bright colours; hard and scratch-resistant.
  • Good electrical insulator; resistant to oils, greases and weak acids.
  • Less resistant to boiling water and heat than Bakelite.

Applications

  • Electrical fittings — switches, plugs, lamp holders.
  • Crockery, buttons, bottle caps, cosmetic containers.
  • Adhesive for plywood, particle board and laminates.
  • Textile finishing (crease-resistant and shrink-proof fabrics), paper treatment for wet strength.

PVC

Polyvinyl chloride (PVC) is a thermoplastic made by free-radical addition polymerisation of vinyl chloride.

Preparation

Vinyl chloride is obtained from acetylene and HCl, or from ethylene via ethylene dichloride. It is polymerised in emulsion or suspension at 40–60 °C using initiators such as benzoyl peroxide or H2O2:

HC≡CH + HCl → CH2=CHCl
n CH2=CHCl —peroxide→ [–CH2–CHCl–]n

Properties

  • Colourless, odourless, rigid and chemically inert to acids, alkalis and many solvents.
  • Poor heat stability; decomposes above about 140 °C liberating HCl, so stabilisers are added.
  • Rigid PVC becomes soft and flexible when plasticisers such as dioctyl phthalate are added (plasticised PVC).

Applications

  • Rigid PVC: pipes and fittings, sheets, safety helmets, bottles, window frames.
  • Plasticised PVC: insulation of electric cables, raincoats, hand-bags, upholstery, floor tiles, toys, curtains.
4 (a)

Differentiate chain growth and step growth polymerization. Explain Free radical and Ionic polymerization with the help of suitable example.

Chain growth versus step growth

Chain-growth (addition) polymerisationStep-growth (condensation) polymerisation
Monomers are unsaturated (contain C=C).Monomers have two or more functional groups (–OH, –COOH, –NH2).
No by-product; polymer has the same empirical formula as monomer.Small molecules (H2O, HCl, NH3, CH3OH) are eliminated.
Needs an initiator; monomers add one at a time to an active centre (radical or ion).Any two molecules (monomer, dimer, oligomer) can react with each other.
High molecular weight is formed quickly; monomer is present until late.Monomer disappears early; high molecular weight only at very high conversion.
Three distinct steps: initiation, propagation, termination.No distinct steps; reaction proceeds stepwise.
Generally gives thermoplastics. Examples: PE, PVC, PS, PMMA.Gives thermoplastics or thermosets. Examples: Nylon-6,6, terylene, Bakelite.

Free radical polymerization

Free-radical polymerisation is a chain-growth polymerisation of vinyl monomers (CH2=CHX) initiated by free radicals produced from initiators such as benzoyl peroxide, azobisisobutyronitrile (AIBN) or H2O2, by heat or light. It takes place in three steps.

1. Chain initiation

(C6H5COO)2 —heat→ 2C6H5COO• → 2C6H5• (R•) + 2CO2
R• + CH2=CHX → R–CH2–ĊHX

The primary radical adds to the double bond of the monomer and forms a new, larger free radical.

2. Chain propagation

R–CH2–ĊHX + CH2=CHX → R–CH2–CHX–CH2–ĊHX
… + n CH2=CHX → R–(CH2–CHX)n–CH2–ĊHX

The growing radical adds monomer units one after another very rapidly, and the chain grows.

3. Chain termination

(a) By combination (coupling): two growing radicals join to form one dead polymer molecule.

~CH2–ĊHX + ĊHX–CH2~ → ~CH2–CHX–CHX–CH2~

(b) By disproportionation: one radical abstracts a hydrogen atom from the other, giving one saturated and one unsaturated chain end.

~CH2–ĊHX + ĊHX–CH2~ → ~CH2–CH2X + XCH=CH~

Chain transfer to monomer, solvent or polymer may also stop growth of one chain while starting another (this causes branching, as in LDPE).

Example — polyethylene: n CH2=CH2 → [–CH2–CH2–]n (at high pressure with traces of O2 or peroxide). Other examples: PVC (X = Cl), polystyrene (X = C6H5), PMMA.

Ionic polymerization

In ionic polymerisation the active centre of the growing chain is an ion (carbocation or carbanion) instead of a free radical.

Cationic polymerisation

Initiated by Lewis acids such as BF3, AlCl3 or SnCl4 with a co-catalyst (traces of water). It is suitable for monomers with electron-donating groups, such as isobutylene, styrene and vinyl ethers.

Initiation: BF3 + H2O → H⁺[BF3OH]⁻; H⁺ + CH2=C(CH3)2 → CH3–C⁺(CH3)2
Propagation: CH3–C⁺(CH3)2 + n CH2=C(CH3)2 → ~CH2–C⁺(CH3)2
Termination: loss of a proton (~CH=C(CH3)2 + H⁺) or combination with the counter-ion

Example: polyisobutylene and butyl rubber.

Anionic polymerisation

Initiated by strong bases or organometallic compounds such as n-butyl lithium, NaNH2 or KNH2 in liquid ammonia. It is suitable for monomers with electron-withdrawing groups, such as acrylonitrile, methyl methacrylate and styrene.

Initiation: C4H9⁻Li⁺ + CH2=CHX → C4H9–CH2–CH⁻X Li⁺
Propagation: C4H9–CH2–CH⁻X + n CH2=CHX → C4H9–(CH2–CHX)n–CH2–CH⁻X
Termination: by adding a proton donor (water, alcohol, acid) or CO2

In the absence of impurities, anionic chains have no natural termination step and remain active ("living polymers"); this allows block copolymers and polymers of narrow molecular-weight distribution to be made.

4 (b)

Define Corrosion. Write types and prevention of corrosion. Explain theory of mechanism of corrosion. How much rust (Fe₂O₃·3H₂O) will be formed when 100 kg of iron have completely rusted away? (Mol. Wt. of rust = 214 g)

Corrosion is the gradual destruction or deterioration of a metal by chemical or electrochemical reaction with its environment. Examples: rusting of iron (formation of reddish-brown hydrated ferric oxide, Fe2O3·xH2O), green coating on copper (basic copper carbonate, CuCO3·Cu(OH)2) and tarnishing of silver (Ag2S). Most metals occur in nature as oxides, carbonates or sulphides; corrosion is the reverse of extraction (metallurgy), in which the metal returns to a more stable, combined state.

Types of corrosion

  1. Dry (chemical) corrosion: oxidation corrosion, corrosion by other gases, liquid-metal corrosion.
  2. Wet (electrochemical) corrosion: galvanic (bimetallic) corrosion; concentration-cell corrosion (differential aeration, water-line, pitting, crevice corrosion); stress corrosion and caustic embrittlement; intergranular corrosion; soil and microbiological corrosion; erosion corrosion.

Prevention of corrosion

  • Proper design: avoid contact of dissimilar metals, crevices, sharp corners and stagnant areas; allow free drainage; keep anode area large compared with cathode area.
  • Proper choice of material: use pure metals, noble metals or corrosion-resistant alloys (stainless steel).
  • Cathodic protection: sacrificial anode method and impressed current method.
  • Modifying the environment: de-aeration, dehumidification, removal of acidic gases, and addition of inhibitors (anodic inhibitors such as chromates, phosphates; cathodic inhibitors such as amines, thiourea, salts of As and Sb).
  • Protective coatings: metallic coatings (galvanising with Zn, tinning, electroplating, metal cladding), inorganic coatings (anodising, phosphating, chromate coating, enamels) and organic coatings (paints, varnishes, lacquers).

Theory of mechanism of corrosion

(a) Dry corrosion occurs by direct reaction of the metal with gases, forming a surface film; its progress depends on the nature of the film (Pilling–Bedworth rule): M → Mn+ + ne⁻ and ½O2 + 2e⁻ → O²⁻, giving the oxide.

(b) Wet (electrochemical) corrosion occurs through anodic and cathodic areas:

Electrochemical (wet) corrosion occurs when (a) two dissimilar metals are in contact with each other in a conducting liquid (electrolyte), or (b) a metal is exposed to different concentrations of oxygen or electrolyte. Separate anodic and cathodic areas are formed and a current flows between them, as in a galvanic cell. Corrosion always takes place at the anode.

Mechanism

At anode (oxidation): Fe → Fe²⁺ + 2e⁻

At cathode (reduction) — two possible reactions:

  1. Hydrogen evolution (acidic medium, no oxygen; e.g., iron in an industrial acid waste): 2H⁺ + 2e⁻ → H2↑. Overall: Fe + 2H⁺ → Fe²⁺ + H2. Usually the anode area is large and cathode small.
  2. Oxygen absorption (neutral or basic aqueous medium with dissolved oxygen, the most common case): ½O2 + H2O + 2e⁻ → 2OH⁻.

The Fe²⁺ ions (from anode) and OH⁻ ions (from cathode) diffuse towards each other and form ferrous hydroxide, which is further oxidised to rust:

Fe²⁺ + 2OH⁻ → Fe(OH)2↓
4Fe(OH)2 + O2 + 2H2O → 4Fe(OH)3 → Fe2O3·xH2O (yellowish-brown rust)

With limited oxygen, black magnetite Fe3O4·6H2O is formed. Note that the metal is lost at the anode, while rust is deposited near the cathode or between anode and cathode.

Numerical: rust formed from 100 kg iron

4Fe + 3O2 + 6H2O → 2(Fe2O3·3H2O)
4 × 56 = 224 g Fe → 2 × 214 = 428 g rust
100 kg Fe → 100 × 428 / 224 = 191.07 kg rust
Rust formed ≈ 191.07 kg
5 (a)

Define and explain the terms involved in the phase rule. Draw a neat labelled phase diagram of water system and explain the areas and curves in it. What is the significance of the triple point and meta stable curve in this system?

Phase rule (J. W. Gibbs): for a heterogeneous system in equilibrium, influenced only by temperature, pressure and concentration,

F = C − P + 2

where F = degrees of freedom, C = number of components and P = number of phases.

Phase (P)

A phase is any homogeneous, physically distinct and mechanically separable part of a system, separated from other parts by definite boundary surfaces.

  • A gas or a mixture of gases is always one phase (gases mix completely).
  • Two completely miscible liquids form one phase; two immiscible liquids (benzene–water) form two phases.
  • Each solid is a separate phase: CaCO3(s) ⇌ CaO(s) + CO2(g) has 3 phases.
  • Ice ⇌ water ⇌ vapour has 3 phases; a sugar solution in water is 1 phase.

Component (C)

The number of components is the smallest number of independently variable chemical constituents by means of which the composition of every phase can be expressed, directly or in the form of a chemical equation.

  • Water system (ice, water, vapour): every phase is H2O, so C = 1.
  • CaCO3 ⇌ CaO + CO2: any two species fix the third (CaO = CaCO3 − CO2), so C = 2.
  • NH4Cl(s) ⇌ NH3(g) + HCl(g): C = 1 if NH3 and HCl are in equal amounts; C = 2 if one is in excess.
  • Sulphur system (rhombic, monoclinic, liquid, vapour): C = 1. Salt solution: C = 2.

Degree of freedom (F)

The degree of freedom (variance) is the minimum number of independent variable factors — temperature, pressure and concentration — that must be specified to define the system completely.

  • Ice ⇌ water ⇌ vapour at the triple point: F = 1 − 3 + 2 = 0 (non-variant).
  • Water ⇌ vapour: F = 1 − 2 + 2 = 1 (univariant); fixing either T or P fixes the other.
  • Water vapour alone: F = 1 − 1 + 2 = 2 (bivariant); both T and P must be stated.
  • A mixture of N2 and H2 gases: C = 2, P = 1, F = 3 (T, P and composition).

Phase diagram of water system

The water system is a one-component system (C = 1). Phase rule gives F = 1 − P + 2 = 3 − P. Since F cannot be negative, at most 3 phases (ice, water, vapour) can coexist, and the maximum F is 2, so the system is represented on a pressure–temperature diagram.

Temperature →Pressure →Acritical point 374 °C, 218 atmBCA′OTriple point: 0.0098 °C, 4.58 mm HgICEWATERVAPOUROA vaporisationOB sublimationOC fusion
Phase diagram of the water system (not to scale). OA′ (dashed) is the metastable curve of supercooled water.

Curves (two phases in equilibrium, P = 2, F = 1, univariant)

  • OA — vaporisation curve: water ⇌ vapour. It shows the vapour pressure of liquid water at different temperatures. It ends at A, the critical point (374 °C, 218 atm), beyond which liquid and vapour become indistinguishable.
  • OB — sublimation curve: ice ⇌ vapour. It shows the vapour pressure of ice at different temperatures and extends down to absolute zero.
  • OC — fusion (melting) curve: ice ⇌ water. It slopes towards the pressure axis (negative slope), showing that the melting point of ice decreases with increase of pressure. This is because ice occupies more volume than water; by Le Chatelier's principle, pressure favours melting.
  • OA′ — metastable curve (dashed): water can be cooled below its freezing point without forming ice (supercooled water). OA′ is the vapour-pressure curve of supercooled water ⇌ vapour, a continuation of OA. It lies above OB, i.e., supercooled water has a higher vapour pressure than ice at the same temperature, so it is metastable; slight disturbance or adding an ice crystal converts it to ice.

Areas (one phase, P = 1, F = 2, bivariant)

Area AOC — water; area BOC — ice; area below AOB — vapour. Within an area, both T and P can be changed independently without changing the number of phases.

Triple point O (P = 3, F = 0, non-variant)

The three curves meet at O, where ice, water and vapour coexist in equilibrium, at 0.0098 °C (≈ 0.01 °C) and 4.58 mm Hg. If temperature or pressure is changed even slightly, one of the phases disappears.

Significance of triple point and metastable curve

  • Triple point: it is the only condition (0.0098 °C, 4.58 mm) where ice, water and vapour exist together; it is non-variant (F = 0), exactly reproducible and is therefore used as the fixed point for defining the Kelvin scale (273.16 K). It is also the lowest pressure at which liquid water can exist; below it, ice sublimes directly (basis of freeze-drying).
  • Metastable curve OA′: it shows that water can be supercooled below its freezing point and still exist as liquid. Because its vapour pressure is higher than that of ice at the same temperature, supercooled water is unstable and changes to ice on slight disturbance or seeding. It explains phenomena such as freezing rain and the need for nucleation in ice formation.
5 (b)

What are the important properties that an efficient lubricant should possess? 1.5 gm of a lubricant oil dissolved in alcohol require 24 ml of 0.0125 N alcoholic KOH for neutralization. What is neutralization value of the lubricant? If an oil to be used in machinery should not have neutralization value greater than 8.0, is it safe to use the test sample as a lubricant in the machinery?

An efficient lubricant should have the following properties:

  • Suitable viscosity for the load and speed, and a high viscosity index, so viscosity changes little with temperature.
  • High flash and fire point, above the working temperature, for safety against fire.
  • Low cloud and pour point, so it flows in cold conditions.
  • High aniline point, so it does not attack rubber seals.
  • Low neutralisation (acid) value, so it does not corrode metal parts; a good oil has acid value below about 0.1.
  • High oiliness, i.e., good ability to adhere to the metal surface under high load.
  • Low volatility and good oxidation stability; low carbon residue.
  • Good demulsifying property (low steam emulsion number), so it separates quickly from water.
  • Chemically stable, non-corrosive and free from moisture and suspended impurities.

Numerical: neutralisation value

Neutralisation (acid) value = mg of KOH required to neutralise the free acids in 1 g of oil.

KOH used = 24 mL × 0.0125 N × 56 mg/meq = 16.8 mg
Neutralisation value = 16.8 / 1.5 = 11.2 mg KOH per g of oil
Neutralisation value = 11.2 mg KOH/g. Since 11.2 > 8.0, the oil is not safe to use as a lubricant in this machinery.

A high neutralisation value shows that the oil contains excess free acids (from oxidation or poor refining). Such acids corrode machine parts, and the oil also tends to form sludge and gum.

6 (a)

How can a combination of UV and IR spectroscopy be used to determine the structure of an unknown organic compound? Illustrate with an example.

No single technique gives the complete structure of a compound. UV and IR give complementary information:

  • UV spectroscopy tells whether the compound contains chromophores and conjugation, and the extent of conjugation (from λmax and εmax). Strong absorption above 200 nm indicates conjugated double bonds or an aromatic ring; a weak band at 270–300 nm indicates n→π* of a carbonyl group; no absorption above 200 nm indicates a saturated compound.
  • IR spectroscopy identifies the functional groups present (O–H, N–H, C=O, C≡N, C=C, aromatic ring) and, through the fingerprint region, confirms the identity of the compound by comparison with a reference.

Combining these with the molecular formula (and degree of unsaturation) narrows the possibilities to one structure.

Example

An unknown compound has molecular formula C8H8O.

Degree of unsaturation = (2C + 2 − H)/2 = (16 + 2 − 8)/2 = 5
⇒ a benzene ring (4) + one more double bond (1)

UV data: strong band at λmax ≈ 246 nm (ε ≈ 13,000) and a weak band near 280 nm. The strong band is a π→π* transition of a carbonyl group conjugated with a benzene ring (benzoyl chromophore); the weak band is n→π* of C=O. So C=O is conjugated with the ring.

IR data:

Band (cm⁻¹)Inference
~1685 (strong)C=O of a ketone, lowered from 1715 by conjugation with the ring
~3060Aromatic C–H stretching
~2925Aliphatic C–H (CH3)
1600, 1450C=C stretching of the aromatic ring
~760, 690Monosubstituted benzene
no band at 3200–3600No O–H, so not a phenol or alcohol
no band at 2720–2820No aldehyde C–H, so not an aldehyde

Conclusion: a monosubstituted benzene ring bearing a conjugated ketone group with a methyl group: C6H5–CO–CH3, acetophenone. UV established conjugation of C=O with the ring; IR established that the carbonyl is a ketone and that there is no –OH or –CHO group.

6 (b)

What is Electronic spectroscopy? An unknown substance absorbs 40% of the light of wavelength 43.5 nm in a cell of 2 cm dimension when it has a concentration of 20 gL⁻¹. Under the similar conditions, what percentage of light will be absorbed by a solution containing 40 gL⁻¹ of the same unknown substance?

Electronic spectroscopy (UV-visible spectroscopy) deals with the absorption of ultraviolet and visible radiation (about 200–800 nm), which causes transitions of electrons between electronic energy levels (σ→σ*, n→σ*, π→π*, n→π*). Each electronic transition is accompanied by vibrational and rotational changes, so broad absorption bands are observed. Quantitative work is based on the Beer–Lambert law, A = log(I0/I) = εcl.

Numerical

Case 1: 40% absorbed ⇒ 60% transmitted, T₁ = 0.60
A₁ = −log T₁ = −log 0.60 = 0.2218 = ε × 20 × 2
Case 2: same ε and l, concentration doubled (40 g/L)
A₂ = A₁ × (40/20) = 2 × 0.2218 = 0.4437
T₂ = 10^(−0.4437) = 0.36 (equivalently T₂ = T₁² = 0.6² = 0.36)
Light absorbed = 1 − 0.36 = 0.64
The 40 g L⁻¹ solution absorbs 64% of the light.
7 (a)

What is Lime Soda process? Calculate the amount of lime and soda required for softening of 20,000 liters of water whose analysis is: Ca(HCO₃)₂ = 40.0 ppm; Mg(HCO₃)₂ = 36.5 ppm; CaSO₄ = 34.0 ppm; MgSO₄ = 30.0 ppm; CaCl₂ = 27.75 ppm.

Principle: In the lime-soda process, calculated amounts of lime [Ca(OH)2] and soda [Na2CO3] are added to hard water. They convert the soluble hardness-causing salts into insoluble CaCO3 and Mg(OH)2, which settle out and are removed by filtration.

Reactions of lime

Ca(HCO3)2 + Ca(OH)2 → 2CaCO3↓ + 2H2O
Mg(HCO3)2 + 2Ca(OH)2 → 2CaCO3↓ + Mg(OH)2↓ + 2H2O
MgCl2 + Ca(OH)2 → Mg(OH)2↓ + CaCl2
MgSO4 + Ca(OH)2 → Mg(OH)2↓ + CaSO4
CO2 + Ca(OH)2 → CaCO3↓ + H2O
2HCl + Ca(OH)2 → CaCl2 + 2H2O
FeSO4 + Ca(OH)2 → Fe(OH)2↓ + CaSO4; Al2(SO4)3 + 3Ca(OH)2 → 2Al(OH)3↓ + 3CaSO4
2NaHCO3 + Ca(OH)2 → CaCO3↓ + Na2CO3 + 2H2O

Reactions of soda

CaCl2 + Na2CO3 → CaCO3↓ + 2NaCl
CaSO4 + Na2CO3 → CaCO3↓ + Na2SO4
Ca(NO3)2 + Na2CO3 → CaCO3↓ + 2NaNO3

Note that permanent Mg hardness (MgCl2, MgSO4) needs both lime and soda: lime converts it to an equivalent Ca salt, which is then removed by soda.

Formulae for calculation

Lime required = 74/100 × [Temp. Ca + 2 × Temp. Mg + Perm. (Mg + Fe + Al) + CO₂ + H⁺ (HCl/H₂SO₄) + HCO₃⁻ (as NaHCO₃)]
(all as CaCO₃ equivalent) × volume of water
Soda required = 106/100 × [Perm. (Ca + Mg + Fe + Al) + H⁺ − HCO₃⁻] × volume of water

74 and 106 are the molecular weights of Ca(OH)2 and Na2CO3; 100 is that of CaCO3. NaCl, KCl, Na2SO4, SiO2 do not consume lime or soda.

Numerical

ImpurityppmMol. wt.CaCO₃ eq. (ppm)Needs
Ca(HCO3)240.016240 × 100/162 = 24.69L
Mg(HCO3)236.514636.5 × 100/146 = 25.002L
CaSO434.013634 × 100/136 = 25.00S
MgSO430.012030 × 100/120 = 25.00L + S
CaCl227.7511127.75 × 100/111 = 25.00S
Lime = 74/100 × [Ca(HCO₃)₂ + 2 Mg(HCO₃)₂ + MgSO₄] × V
= 0.74 × [24.69 + 50 + 25] × 20,000 mg = 0.74 × 99.69 × 20,000 mg
= 1,475,432 mg ≈ 1.475 kg
Soda = 106/100 × [CaSO₄ + MgSO₄ + CaCl₂] × V
= 1.06 × [25 + 25 + 25] × 20,000 mg = 1,590,000 mg = 1.59 kg
Lime required ≈ 1.475 kg · Soda required = 1.59 kg (for 20,000 L, pure chemicals)
7 (b)

Write brief note on (Any two): (i) Electron affinity (ii) Effective Nuclear Charge (iii) Aufbau rule (iv) Atomic and Ionic size

All four are answered below; write any two in the exam.

(i) Electron affinity

Electron affinity is the amount of energy released when an electron is added to a neutral, isolated gaseous atom to form a gaseous anion: X(g) + e⁻ → X⁻(g) + energy. (In terms of electron gain enthalpy, ΔHeg is negative when energy is released.) It is expressed in kJ mol⁻¹ or eV.

Across a period electron affinity generally increases from left to right, because atomic size decreases and effective nuclear charge increases, so the incoming electron is attracted more strongly. Halogens have the highest values. Exceptions: Be and Mg (stable ns² configuration), N (stable half-filled 2p³) have very low or nearly zero values; noble gases (complete octet) have zero or positive values.

Down a group electron affinity generally decreases, because atomic size increases. But F (−328 kJ/mol) has a lower value than Cl (−349 kJ/mol), and O less than S, because the very small 2p subshell of F and O is compact and the added electron feels strong repulsion from the electrons already present. Order for halogens: Cl > F > Br > I.

Second electron affinity is always endothermic (energy must be supplied) because the incoming electron is repelled by the negative ion: O⁻(g) + e⁻ → O²⁻(g), ΔH = +744 kJ/mol.

Factors: atomic size, effective nuclear charge and the stability of the electronic configuration.

(ii) Effective nuclear charge

In a multi-electron atom, an outer electron does not feel the full nuclear charge Z, because inner electrons shield (screen) it from the nucleus. The net positive charge actually felt by the electron is called the effective nuclear charge:

Z_eff = Z − σ (σ = screening constant)

Slater's rules for σ

  • Group orbitals as (1s)(2s,2p)(3s,3p)(3d)(4s,4p)(4d)(4f)(5s,5p)…; electrons to the right of the chosen group contribute 0.
  • For an ns or np electron: each other electron in the same group contributes 0.35 (0.30 for 1s); each electron in shell (n−1) contributes 0.85; each electron in shell (n−2) or lower contributes 1.00.
  • For an nd or nf electron: same group 0.35; all electrons to the left contribute 1.00.
Na (1s² 2s² 2p⁶ 3s¹), 3s electron: σ = 8 × 0.85 + 2 × 1.00 = 8.8; Z_eff = 11 − 8.8 = 2.2
N (1s² 2s² 2p³), 2p electron: σ = 4 × 0.35 + 2 × 0.85 = 3.1; Z_eff = 7 − 3.1 = 3.9

Trends: Zeff increases across a period (by about 0.65 per element, since same-shell electrons shield poorly) and changes only slightly down a group. Higher Zeff means smaller atomic size, higher ionisation energy, higher electron affinity and higher electronegativity.

(iii) Aufbau rule

Aufbau principle ("building up" in German): in the ground state of an atom, electrons enter orbitals in order of increasing energy; an orbital of higher energy is filled only after lower-energy orbitals are filled.

The energy order is given by the (n + l) rule (Bohr–Bury / Madelung rule): the orbital with the lower value of (n + l) has lower energy; if two orbitals have the same (n + l), the one with lower n is filled first. For example, 4s (n + l = 4) is filled before 3d (n + l = 5); 3d and 4p both have n + l = 5, so 3d (n = 3) comes first.

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p

The principle is applied together with the Pauli exclusion principle (an orbital holds at most two electrons of opposite spin) and Hund's rule (orbitals of a subshell are singly occupied first with parallel spins).

Examples: K (19): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹; Fe (26): [Ar] 3d⁶ 4s².

Exceptions: Cr (24) is [Ar] 3d⁵ 4s¹ and Cu (29) is [Ar] 3d¹⁰ 4s¹ instead of 3d⁴ 4s² and 3d⁹ 4s², because half-filled and completely filled d-subshells are extra stable (symmetrical distribution and greater exchange energy).

(iv) Atomic and ionic size

Atomic radius is the distance from the centre of the nucleus to the outermost shell of electrons. Since an atom has no sharp boundary, it is measured as covalent radius (half the internuclear distance between two bonded identical atoms), metallic radius, or van der Waals radius.

Across a period (left to right) atomic radius decreases: Li 152, Be 111, B 88, C 77, N 74, O 66, F 64 pm. Reason: electrons are added to the same shell, while nuclear charge increases by one unit each step; electrons in the same shell shield each other poorly, so the effective nuclear charge increases and the electron cloud is pulled closer to the nucleus.

Down a group atomic radius increases: Li 152, Na 186, K 231, Rb 244, Cs 262 pm. Reason: a new shell is added at each step and the shielding effect of inner electrons increases, which outweighs the increase in nuclear charge.

Ionic radius is the effective distance from the nucleus of an ion up to which it exerts its influence on the electron cloud.

  • Cations are smaller than their parent atoms (Na 186 pm → Na⁺ 95 pm): the outermost shell is usually lost, and the same nuclear charge now acts on fewer electrons, so the cloud contracts.
  • Anions are larger than their parent atoms (Cl 99 pm → Cl⁻ 181 pm): the added electron increases inter-electronic repulsion and decreases the effective nuclear charge per electron, so the cloud expands.
  • Isoelectronic species (same number of electrons): size decreases as nuclear charge increases: N³⁻ (171) > O²⁻ (140) > F⁻ (133) > Na⁺ (95) > Mg²⁺ (65) > Al³⁺ (50) pm.
  • For ions of the same charge, ionic radius increases down a group (Li⁺ < Na⁺ < K⁺ < Rb⁺ < Cs⁺; F⁻ < Cl⁻ < Br⁻ < I⁻) and decreases across a period for cations of increasing charge.
8 (a)

Write brief note on (Any two): (i) Electro negativity (ii) Polarizability (iii) Oxidation states

All three are answered below; write any two in the exam.

(i) Electronegativity

Electronegativity is the tendency of an atom in a molecule to attract the shared pair of electrons towards itself. It is a property of a bonded atom and has no units. Common scales: Pauling scale (based on bond energies; F = 4.0, highest; Cs = 0.7, lowest), Mulliken scale (EN = (IE + EA)/2) and Allred–Rochow scale (EN = 0.359 Zeff/r² + 0.744).

Variation across a period: electronegativity increases from left to right — Li 1.0, Be 1.5, B 2.0, C 2.5, N 3.0, O 3.5, F 4.0. Reason: atomic size decreases and effective nuclear charge increases, so the nucleus attracts the bonding electrons more strongly.

Variation down a group: electronegativity decreases — F 4.0, Cl 3.0, Br 2.8, I 2.5; Li 1.0, Na 0.9, K 0.8, Rb 0.8, Cs 0.7. Reason: atomic size increases and the bonding electrons are farther from the nucleus and more shielded by inner electrons.

Other factors: higher oxidation state gives higher EN (Fe³⁺ > Fe²⁺); more s-character in hybrid orbitals gives higher EN (sp > sp² > sp³).

Applications: predicting bond polarity and ionic character (ΔEN > 1.7 → mainly ionic), metallic and non-metallic character, acidic/basic nature of oxides, and bond lengths.

(ii) Polarizability

Polarisability is the ease with which the electron cloud of an atom, ion or molecule can be distorted by an electric field, such as that of a neighbouring ion. The induced dipole moment is proportional to the field: µind = αE, where α is the polarisability. The ability of a cation to distort an anion is called its polarising power.

Polarisability increases with size and number of electrons, because outer electrons of large atoms are loosely held (F⁻ < Cl⁻ < Br⁻ < I⁻). It increases down a group and generally decreases across a period.

Fajans' rules

When a cation polarises an anion, electron density is drawn into the region between them and the ionic bond acquires partial covalent character. Covalent character increases when:

  1. the cation is small (high polarising power): LiCl is more covalent than NaCl;
  2. the anion is large (high polarisability): LiI is more covalent than LiF;
  3. the charge on either ion is high: AlCl3 > MgCl2 > NaCl in covalent character;
  4. the cation has a pseudo-noble-gas (18-electron) configuration, e.g., Cu⁺, Ag⁺, Zn²⁺, which shield the nucleus poorly: CuCl is more covalent than NaCl though Cu⁺ and Na⁺ are of similar size.

Consequences: greater covalent character lowers melting point (AlCl3 sublimes at 180 °C, NaCl melts at 801 °C), lowers solubility in water, and can cause colour (AgI is yellow). Higher polarisability also increases London dispersion forces, which is why boiling points of halogens and noble gases increase down the group.

(iii) Oxidation states

The oxidation state (oxidation number) of an atom is the charge it would carry if all its bonds were completely ionic, i.e., the apparent charge assigned by a set of rules.

Rules

  • An element in the free state has oxidation state 0 (Na, H2, O2, S8).
  • For a monatomic ion, it equals the charge on the ion.
  • Oxygen is −2 (peroxides −1, superoxides −½, OF2 +2); hydrogen is +1 (−1 in metal hydrides); fluorine is always −1.
  • The sum of oxidation states is zero for a neutral molecule and equals the charge for a polyatomic ion.
K₂Cr₂O₇: 2(+1) + 2x + 7(−2) = 0 ⇒ x = +6
KMnO₄: +1 + x + 4(−2) = 0 ⇒ x = +7
Fe₃O₄: 3x = +8 ⇒ x = +8/3 (average of Fe²⁺ and Fe³⁺)

Trends

  • s-block: group 1 elements show +1 and group 2 show +2 only (equal to group number).
  • p-block: the maximum oxidation state equals the number of valence electrons (group number − 10); non-metals also show negative states (group number − 18), e.g., N from −3 to +5, Cl from −1 to +7.
  • Across a period, the highest positive oxidation state increases: Na +1, Mg +2, Al +3, Si +4, P +5, S +6, Cl +7.
  • Inert pair effect: down groups 13–15, the lower oxidation state (two less than the group valency) becomes more stable — Tl⁺, Pb²⁺, Bi³⁺ — because the ns² electrons are reluctant to take part in bonding due to poor shielding by d and f electrons.
  • d-block: elements show variable oxidation states because (n−1)d and ns electrons have similar energies (Mn shows +2 to +7).
8 (b)

Calculate temporary, permanent and total hardness of water sample containing Mg(HCO₃)₂ = 7.3 mg/L, Ca(HCO₃)₂ = 16.4 mg/L, MgCl₂ = 9.5 mg/L and CaSO₄ = 13.6 mg/L.

Saltmg/LMol. wt.CaCO₃ eq. (mg/L)Type
Mg(HCO3)27.31467.3 × 100/146 = 5.00Temporary
Ca(HCO3)216.416216.4 × 100/162 = 10.12Temporary
MgCl29.5959.5 × 100/95 = 10.00Permanent
CaSO413.613613.6 × 100/136 = 10.00Permanent
Temporary hardness = 5.00 + 10.12 = 15.12 ppm
Permanent hardness = 10.00 + 10.00 = 20.00 ppm
Total hardness = 15.12 + 20.00 = 35.12 ppm
Temporary = 15.12 ppm · Permanent = 20 ppm · Total = 35.12 ppm
Advertisement

💬 Comments & Doubts

💭

Abhi tak koi comment nahi

Sabse pehle comment karne wale bano!

Apna comment likhein

Doubt ho ya suggestion — kuch bhi poochh sakte hain. Login zaroori nahi hai.

7 × 8 =

Ye sirf ye confirm karne ke liye hai ki aap robot nahi hain.

Comment publish hone se pehle admin check karta hai.

🔒 Content copy nahi kar sakte