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Published on: 20/10/2025
Download Tamil Nadu 12th Standard Chemistry question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
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1.
Arrange the following in order of increasing molar conductivity
(i) Mg[Cr(NH3)(Cl)5]
(ii) Cr(NH3)5Cl]3[CoF6]2
(iii) [Cr(NH3)3Cl3]
2.
What is crystal field stabilization energy (CFSE)?
3.
How are vitamins classified?
4.
Write the structure of α-D (+) glucophyranose
5.
Write the structural formula of aspirin.
6.
Define anode and cathode
7.
Complete the following reaction

8.
Why does bleeding stop by rubbing moist alum
9.
Comment on the statement: Colloid is not a substance but it is a state of substance.
10.
Write the postulates of Werner’s theory.
11.
Give the differences between primary and secondary structure of proteins.
12.
Write a short note on peptide bond.
13.
Explain the mechanism of cleansing action of soaps and detergents.
14.
What are bio degradable polymers? Give examples.
15.
How will you convert nitrobenzene into
i. 1,3,5 - trinitrobenzene
ii. o and p- nitrophenol
iii. m – nitro aniline
iv. azoxybenzene
v. hydrozobenzene
vi. N – phenylhydroxylamine
vii. aniline
16.
Write short notes on the following
i. Hofmann’s bromide reaction
ii. Ammonolysis
iii. Gabriel phthalimide synthesis
iv. Schotten – Baumann reaction
v. Carbylamine reaction
vi. Mustard oil reaction
vii. Coupling reaction
viii. Diazotisation
ix. Gomberg reaction
17.
Differentiate physisorption and chemisorption.
18.
What is the difference between homogenous and hetrogenous catalysis?
19.
20.
Which one of the following will give a pair of enantiomorphs?
[Cr(NH3)6][Co(CN)6]
[Co(en)2Cl2]Cl
[Pt(NH3)4][PtCl4]
[Co(NH3)4Cl2]NO2
21.
Which one of the following pairs represents linkage isomers?
[Cu(NH3)4][PtCl4] and [Pt(NH3)4][CuCl4]
[Co(NH3)5(NO3)]SO4 and [Co(NH3)5(ONO)]
[Co(NH3)4(NCS)2]Cl and [Co(NH3)4(SCN)2]Cl
both (b) and (c)
22.
The molar conductivity of a 0.5 mol dm-3 solution of AgNO3 with electrolytic conductivity of 5.76 ×10−3 S cm−1at 298 K is______.
2.88 S cm2mol-1
11.52 S cm2mol-1
0.086 S cm2mol-1
28.8 S cm2mol -1
23.
Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because _______.
Zinc is lighter than iron
Zinc has lower melting point than iron
Zinc has lower negative electrode potential than iron
Zinc has higher negative electrode potential than iron
24.
A gas X at 1 atm is bubbled through a solution containing a mixture of 1MY- and 1MZ- at 25oC. If the reduction potential of Z>Y>X, then_____.
Y will oxidize X and not Z
Y will oxidize Z and not X
Y will oxidize both X and Z
Y will reduce both X and Z
25.
Which of the following is incorrect for physisorption?
reversible
increases with increase in temperature
low heat of adsorption
increases with increase in surface area
26.
Fog is colloidal solution of _______.
solid in gas
gas in gas
liquid in gas
gas in liquid
27.
28.
The order of basic strength for methyl substituted amines in aqueous solution is _________.
N(CH3)3> N(CH3)2H> N(CH3)H2> NH3
N(CH3)H2>N(CH3)2H > N(CH3)3>NH3
NH3> N(CH3)H2> N(CH3)2 H>N(CH3)3
N(CH3)2H>N(CH3)H2> N(CH3)3> NH3
29.
P Product ‘P’ in the above reaction is _______.
30.
Which one given below is a non-reducing sugar?
Glucose
Sucrose
maltose
Lactose
31.
Which of the following vitamins is water soluble?
Vitamin E
Vitamin K
Vitamin A
Vitamin B
32.
Insulin, a hormone chemically is ______.
Fat
Steroid
Protein
Carbohydrates
33.
Antiseptics and disinfectants either kill or prevent growth of microorganisms. Identify which of the following statement is not true.
dilute solutions of boric acid and hydrogen peroxide are strong antiseptics.
Disinfectants harm the living tissues
A 0.2% solution of phenol is an antiseptic while 1% solution acts as a disinfectant
Chlorine and iodine are used as strong disinfectants
34.
The polymer used in making blankets (artificial wool) is _______.
polystyrene
PAN
polyester
polythene
35.
What are the limitations of VBT?
36.
What are reducing and non – reducing sugars?
37.
38.
How is terylene prepared?
39.
State Faraday’s Laws of electrolysis
40.
Addition of Alum purifies water. Why?
41.
How is Nylon - 6,6 prepared? Give it use.
42.
Write short notes on the following
iv. Schotten – Baumann reaction
43.
Write short notes on the following
v. Carbylamine reaction
1.
(i) \(\mathrm{Mg}\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right) \mathrm{Cl}_{5}\right]^{2-} \rightleftharpoons \mathrm{Mg}^{2+}+\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right) \mathrm{Cl}_{5}\right]^{2-} (2 ions)\)
(ii) \(\begin{aligned}
{\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}_3\left[\mathrm{CoF}_6\right]_2 \rightleftharpoons 3\right.} & {\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right]^{2+} } \\
& (5 \text { ions })
\end{aligned}\)\(+2\left[\mathrm{CoF}_6\right]^{3-}\)
(iii) \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\right]= \text{ No ions}\)
If no of ions increases, molar conductivity increases molar conductivity of the complex also INCREASES.
\(\therefore\) The order of the given compound is
[Cr(NH3)3Cl3]<Mg[Cr(NH3)3Cl3] < [Cr(NH3)5Cl3] [CoF6]2
(No ion) (2 ions) (5 ions)
2.
The CFSE is defined as the energy of the electronic configuration in the ligand field minus the energy of the electronic configuration in the isotropic field.
CFSE (\(\Delta\)Eo) = {ELf}-{Eiso}
={[nt2g(-0.4) + neg(0.6)]\(\Delta\)o + npP} - {n'pP}
Here, nt2g is the number of electrons in t2g orbitals;
neg is number of electrons in eg orbitals;
np is number of electron pairs in the ligand field; &
n'p is the number of electron pairs in the isotropic field (barycenter).
P - pairing energy
3.
Vitamins are classified into two groups based on their solubility in water or in fat.
(a) Fat soluble vitamins:
These vitamins absorbed best when taken with fatty food and are stored in fatty tissues and livers. These vitamins do not dissolve in water. Hence they are called fat soluble vitamins. Vitamin A, D, E & K are fatsoluble vitamins.
(b) Water soluble vitamins:
Vitamins B (B1, B2, B3, B5, B6, B7, B9 & B12) and C are readily soluble in water. On the contrary to fat soluble vitamins, these can't be stored. The excess vitamins present will be excreted through urine and are not stored in our body. Hence, these two vitamins should be supplied regularly to our body.
4.
α-D (+) glucophyranose
5.
Aspirin is o-acetyl salicylic acid
6.
(i) Anode: The electrode at which the oxidation occurs is called anode. (loss of electrons)
(ii) Cathode: The electrode at which the reduction occurs is called cathode. (gain of electrons)
7.
8.
(a) Blood is a colloidal sol. When we rub the injured part with moist alum, coagulation of blood takes place.
(b) Coagulation stops bleeding. Moist alum is an electrolyte.
9.
(i) A Colloid depends on the size of the particle. A Colloid is formed when the size of the particle lies between 1 nm and 100 nm. For example soap dissolves in water to form colloidal soap solution whereas it dissolves in alcohol to form a true solution. Thus change of state takes place. A colloidal state maybe an intermediate between a true solution and a suspension.
(ii) Also some crystalloids under certain conditions can be colloids. NaCl is a crystalloid in aqueous medium; but when mixed with benzene it acts as a colloid.
10.
Most of the elements exhibit, two types of valence namely primary valence and secondary valence and each element tend to satisfy both the valences.
The primary valence is referred the oxidation state of the metal atom.
The secondary valence as the coordination number. For example, according to Werner, the primary and secondary valences of cobalt are 3 and 6 respectively.
The primary valence of a metal ions ae always satisfied by negative ions.
For example in the complex CoCI3.6NH3. The primary valence of Co is +3 and is satisfied by 3CI- ions.
The secondary valence is satisfied by negative ions, neutral molecules, positive ions or the combination of these.
For example, in CoCl3.6NH3 complex primary valence of cobalt +3 and it is satisfied by 3 CI-.
The secondary valence of cobalt is 6 and is satisfied by six neutral ammonia molecules. where as in CoCI6.NH3.
Secondary valence of Co = 5{It is satisfied five neutral molecules and a Cl- ion}
According to Werner, there are two spheres of attraction around a metal atom/ion in a complex.
The inner /coordination sphere:
The groups present in this sphere are firmly attached to the metal.
The outer sphere / ionisation sphere:
The groups present in this sphere are loosely bound to the central metal ion and hence can be separated into ions upon dissolving the complex in a suitable solvent.
The primary valencies are non-directional. while the secondary valencies are directional.
The geometry of the complex is determined by the special arrangement of the groups which satisfy the secondary valence.
| Secondary valence | Geometry |
| 4 | Tetrahedral / Square planar |
| 6 | Octahedral |
11.
Primary structure of Proteins:
Proteins are polypeptide chains, made up of amino acids are connected through peptide bonds. The relative arrangement of the amino acids in the polypeptide chain is called the primary structure of the protein. Knowledge of this is essential as even small changes have potential to alter the overall structure and function of a protein.
\(\mathrm{H}_{2} \mathrm{~N}-\mathrm{Gly}-\mathrm{Met}-\mathrm{Phe}-\mathrm{Cys}-\mathrm{Arg}-\mathrm{Asp}-\mathrm{COOH}\)
α - Helix:
In the α-helix sub-structure, the amino acids are arranged in a right handed helical (spiral) structure and are stabilised by the hydrogen bond between the carbonyl oxygen of one amino acid (n residue) with amino hydrogen of the fifth residue (n + 4th residue). The side chains of the residues protrude outside of the helix. Each turn of an α-helix contains about 3.6 residues and is about 5.4 Å long. The amino acid proline produces a kink in the helical structure and often called as a helix breaker due to its rigid cyclic structure.
1. Linear sequence of aminoacids
2. Linear
3. Composed of peptide bonds formed between amino acids.
Secondary structure of Proteins:
The amino acids in the polypeptide chain forms highly regular shapes (sub-structures) through the hydrogen bond between the carbonyl oxygen (-C=O) and the neighbouring amine hydrogen (-NH) of the main chain. α-Helix and β-strands or sheets are two most common substructures formed by proteins.
β-Strand:
β-Strands are extended peptide chain rather than coiled. The hydrogen bonds occur between main chain carbonyl group one such strand and the amino group of the adjacent strand resulting in the formation of a sheet like structure. This arrangement is called β-sheets.
12.
(i) The amino acids are linked covalently by peptide bonds. The carboxyl group of the first amino acid react with the amino group of the second amino acid to give an amide linkage between these amino acids. This amide linkage is called peptide bond. The resulting compound is called a dipeptide. Addition an another amino acid to this dipeptide a second peptide bond results in tripeptide.
(ii) Thus we can generate tetra peptide, penta peptide etc... When you have more number of amino acids linked this way you get a polypeptide. If the number of amino acids are less it is called as a polypeptide, if it has large number of amino acids (and preferably has a function) then it is called a protein.
(iii) The amino end of the peptide is known as N - terminal or amino terminal while the carboxy end is called C-terminal or carboxy terminal. In general protein sequences are written from N-Terminal to C-Terminal.The atoms other than the side chains (R-groups) are called main chain or the back bone of the polypeptide.
13.
(i) To understand how a soap works as a cleansing agent, let us consider sodium palmitate an example of a soap. The cleansing action of soap is directly related to the structure of carboxylate ions (palmitate ion) present in soap. The structure of palmitate exhibit dual polarity. The hydrocarbon portion is non polar and the carboxyl portion is polar.
(ii) The nonpolar portion is hydrophobic while the polar end is hydrophilic. The hydrophobic hydro carbon portion is soluble in oils and greases, but not in water. The hydrophilic carboxylate group is soluble in water.
(iii) The dirt in the cloth is due to the presence of dust particles intact or grease which stick. When the soap is added to an oily or greasy part of the cloth, the hydrocarbon part of the soap dissolve in the grease, leaving the negatively charged carboxylate end exposed on the grease surface.
(iv) At the same time the negatively charged carboxylate groups are strongly attracted by water, thus leading to the formation of small droplets called micelles and grease is floated away from the solid object. When the water is rinsed away, the grease goes with it. As a result, the cloth gets free from dirt and the droplets are washed away with water. The micelles do not combine into large drops because their surfaces are all negatively charged and repel each other. The cleansing ability of a soap depends upon its tendency to act as a emulsifying agent between water and water insoluble greases.
The cleansing action of detergents are similar to cleansing action of soap. Eg: the structure of a cationic detergents is:
14.
1. The materials that are readily decomposed by microorganisms in the environment are called biodegradable.
2. Natural polymers degrade on their own after certain period of time but the synthetic polymers do not.
3. It leads to serious environmental pollution. One of the solution to this problem is to produce biodegradable polymers which can be broken down by soil micro organism.
Examples:
(i) Polyhydroxy butyrate (PHB)
(ii) Polyhydroxy butyrate-co-A- hydroxyl valerate (PHBV)
(iii) Polyglycolic acid (PGA), Polylactic acid (PLA)
(iv) Poly ( E caprolactone) (PCL)
(v) Biodegradable polymers are used in medical field such as surgical sutures, plasma substitute etc...
4. these polymers are decomposed by enzyme action and are either metabolized or excreted from the body.
15.
Nitrobenzene into 1,3,5 - trinitro benzene
Nitrobenzene 1,3,5 - trinitro benzene
ii. Nitrobenzene into o - and p - nitro phenols
iii. Nitrobenzene into m - nitro aniline
+ 6NH3 + 2H2O + 3S
iv. Nitrobenzene into azoxybenzene
+3H2O
v. Nitrobenzene into hydrazobenzene
vi. Nitrobenzene into N - phenyl hydrazobenzene
vii. Nitrobenzene into aniline
16.
Hoffmann's bromide reaction:
When Amides are treated with bromine in the presence of aqueous or ethanolic solution of KOH, primary amines with one carbon atom less than the parent amides are obtained.
\(\underset { \quad \quad \quad amide\\ R=Alkyl(or)Aryl }{ R-\overset { \underset { || }{ O } }{ C } -{ NH }_{ 2 } } \overset { { Br }_{ 2 }/KOH }{ \longrightarrow } \underset { Primary\quad amine }{ R-{ NH }_{ 2 }+{ K }_{ 2 }{ CO }_{ 3 } } +KBr+{ H }_{ 2 }O\)
(ii) Hoffmann's ammonolysis:
When Alkyl halides (or) benzylhalides are heated with alcoholic ammonia in a sealed tube, mixtures of 1°, 2° and 3° amines and quaternary ammonium salts are obtained.
\( { CH }_{ 3 }-Br\overset { \ddot { N } { H }_{ 3 } }{ \underset { \Delta }{ \longrightarrow } } \underset { { 1 }^{ 0 }-amine }{ { CH }_{ 3 }-\ddot { N } { H }_{ 2 } } \overset { { CH }_{ 3 }-Br }{ \longrightarrow } \underset { { 3 }^{ o }-amine }{ \left( { CH }_{ 3 } \right) _{ 2 }\ddot { N } H } \overset { { CH }_{ 3 }Br }{ \longrightarrow } \underset { 3^{ o }-amine }{ \left( { { CH }_{ 3 } } \right) _{ 3 }\ddot { N } } \overset { CH_{ 3 }Br }{ \longrightarrow } \underset { Quartenary \ ammonium\ bromide }{ \left( { CH }_{ 3 } \right) _{ 4 }\overset { + }{ N } { Br }^{ - } } \)
This is a nucleophilic substitution, the halide ion of alkyl halide is substituted by the -NH2 group. The product primary amine so formed can also has a tendency to act as a nucleophile and hence if excess alkyl halide is taken, further nucleophilic substitution takes place leading to the formation of quarternary ammonium salt. However, if the process is carried out with excess ammonia, primary amine is obtained as the major product. The order of reactivity of alkylhalides with amines
RI > RBr > RCl
(iii) Gabriel phthalimide synthesis:
Gabriel synthesis is used for the preparation of Aliphatic primary amines. Phthalimide on treatment with ethanolic KOH forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis gives primary amine. Aniline cannot be prepared by this method because the arylhalides do not undergo nucleophilic substitution with the anion formed by phthalimide.
(iv) Schotten - Baumann reaction :
Aniline reacts with benzoylchloride (C6H5COCI) in the presence of NaOH to give N - phenyl benzamide. This reaction is known as Schotten - Baumann reaction. The acylation and benzoylation are nucleophilic substitutions.
\(\underset { Aniline }{ { C }_{ 6 }{ H }_{ 5 }-{ NH }_{ 2 } } +\underset { Benzoyl\quad chloride }{ { C }_{ 6 }{ H }_{ 5 }-\overset { \underset { || }{ O } }{ C } -Cl } \overset { Pyridine }{ \longrightarrow } \underset { N-phenyl\quad benzamide }{ { C }_{ 6 }{ H }_{ 5 }-NH-\overset { \underset { || }{ O } }{ C } -{ C }_{ 6 }{ H }_{ 5 } } +HCl\)
(v) Carbylamine reaction :
Aliphatic (or) aromatic primary amines react with: chloroform and alcoholic KOH to give isocyanides (carbylamines), which has an unpleasant smell. This reaction is known as carbylamines test. This test used to identify the primary amines.
\(\underset { Ethylamine }{ { C }_{ 2 }{ H }_{ 5 }-{ NH }_{ 2 } } +\underset { Chloroform }{ { CHCl }_{ 3 }+3KOH } \longrightarrow \underset { Ethylisocyanide }{ { C }_{ 2 }{ H }_{ 5 }-NC } +3KCl+3{ H }_{ 2 }O\)
(vi) Mustard oil reaction:
(a) When primary amines are treated with carbon disulphide (CS2), N - alkyldithio carbonic acid is formed which on subsequent treatment with HgCI2, gives an alkyl isothiocyanate.
(b) When aniline is treated with carbon disulphide, or heated together, S-diphenylthio urea is formed, which on boiling with strong HCI, phenyl isothiocyanate (phenyl mustard oil), is formed.
These reactions are known as Hofmann - Mustard oil reaction. This test is used to identify the primary amines.
(vii) Coupling reactions (or) p-hydroxyazobenzene, p-aminoazobenzene, 2-phenyl azo-4-methyl phenol:
Benzene diazonium chloride reacts with electron rich aromatic compounds like phenol, aniline to form brightly coloured azo compounds. Coupling generally occurs at the para position. If para position is occupied then coupling occurs at the ortho position. Coupling tendency is enhanced if an electron donating group is present at the para - position to -N2CI- group. This is an electrophilic substitution.
Aryl fluorides and iodides cannot be prepared by direct halogenation and the cyano group cannot be introduced by nucleophilic substitution of chlorine in chlorobenzene. For introducing such a halide group. cyano group -OH, NO2, etc.. benzenediazonium chloride is a very good intermediate Diazo compounds obtained from the coupling reactions of diazonium salts are coloured and are used as dyes.
(viii) Diazotisation :
Aniline reacts with nitrous acid at low temperature (273 - 278 K) to give benzene. Diazonium chloride which is stable for a short time and slowly decomposes even at low temperatures.This reaction is known as diazotization
(ix) Gomberg reaction :
Benzene diazonium chloride reacts with benzene in the presence of sodium hydroxide to give biphenyl. This reaction in known as the Gomberg reaction
17.
| S.No | Physical adsorption or vander waals adsorption or Physisorption | Chemical adsorption or Chemisorption or Activated adsorption |
|---|---|---|
| 1. | It is instantaneous | It is very slow |
| 2. | It is non-specific | It is very specific depends on nature of adsorbent and adsorbate |
| 3. | In Physisorption, when pressure increases the amount of adsorption increases | Chemical adsorption is fast with increase pressure, it can not alter the amount. |
| 4. | Physisorption decreases with increase in temerature | When temperature is raised chemisorption first increases and then decreases |
| 5. | No transfer of electrons | Chemisorption involves transfer of electrons between the adsorbent and adsorbate |
| 6. | Heat of adsorption is low in the order of 40kJ/ mole. | Heat of adsorption is high i.e., from 40- 400kJ/mole. |
| 7. | Multilayer of the adsorbate is formed on the adsorbent. | Monolayer of the adsorbate is formed |
| 8. | It occurs on all sides | Adsorption occurs at fixed sites called active centres. It depends on surface area |
| 9 | Activation energy is insignificant. | Chemisorption involves the formation of activated complex with appreciable activation energy. |
18.
| Homogenous catalysis | Heterogeneous catalysis |
|---|---|
| 1. In a catalysed reaction, the reactants, products and catalyst are present in the same phase. Ex: \( 2\mathrm{SO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}+[\mathrm{NO}]_{(\mathrm{g})} \rightarrow 2 \mathrm{SO}_{3(\mathrm{~g})}+[\mathrm{NO}]_{(\mathrm{g})} \) [NO] - catalyst; gaseous state SO2, O2 & SO3 are gases. |
1. In a catalysed reaction, the catalyst is present in a different phase (ie) it is not present in the same phase as that of the reactants or products. Ex: \(\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \stackrel{\mathrm{Fe}_{(\mathrm{s})}}{\longrightarrow} 2 \mathrm{NH}_{3(\mathrm{~g})}\) Fe - catalyst; solid N2, H2 & NH3 are gases. |
| 2. It is not a contact catalysis. | 2. It is a contact catalysis and the mental catalyst will be in finely divided metal or as gauze. |
| 3. It is explained by intermediate compound formation theory. |
3. It is explained by adsorption theory. |
19.
20.
Complexes given in other options (a), (c) and (d) have symmetry elements and hence they are optically inactive.
21.
(a) coordination isomers
(b) no isomerism ( different molecular formula)
(c)⬅NCS; ⬅SCN coordinating atom differs : linkage isomers
22.
Λ = k/W x 10−3 mol−1 dm3
\(=\frac{ 5.76 \times 10^{−3} S cm^{−1} \times 10^{−3}}{0.5} = mol^{-1} dm^{3}\)
\(=\frac{ 5.76 \times 10^{−3} S cm^{−1} \times 10^{−3}}{0.5}\) S cm-1 mol−1 dm3
= 11.52 S cm2mol-1
23.
EoZn2+[Zn] = 0.76V and EoFe2+[Fe] = -0.44V
Zinc has higher negative electrode potential than iron, iron cannot be coated on zinc
24.
Z is tie strongest oxidising agent (High SRP)
X is the strongest reducing agent (Low SRP)
25.
The incorrect statement is option (n)
Physisorption is an exothermic process. Hence increases in temperature decreases the Physisorption
26.
dispersion medium-gas
dispersed phase-liquid
27.
(d)
28.
(d)
N(CH3)2H>N(CH3)H2> N(CH3)3> NH3
29.
(b)
30.
(b)
Sucrose
31.
(d)
Vitamin B
32.
(c)
Protein
33.
(a)
dilute solutions of boric acid and hydrogen peroxide are strong antiseptics.
34.
(b)
PAN
35.
Even though VBT explains many of the observed properties of complexes, it still has following limitations.
(i) It does not explain the colour of the complex.
(ii) It considers only the spin only magnetic moments and does not consider the other components of magnetic moments.
(iii) It does not provide a quantitative explanation as to why certain complexes are inner orbital complexes and the: others are outer orbital complexes for the same metal. For example, [Fe(CN)6]4- is diamagnetic (low spin) whereas [FeF6]4-is paramagnetic (high spin).
36.
i) Reducing sugars:
1. Sugars which reduce Tollen's reagent or Fehling's solution or Benedict's solution are called reducing sugars.
2. These contain either α - hydroxyl ketone or cyclical hemi acetal or hemi ketal or structures in equilibrium with open chain forms having a free- CHO or C=O group.
3. E.g. a) All monosaccharide's like D - glucose, D - fructose (aldoses and ketoses)
b) Sugars like Lactose and maltose except sucrose.
ii) Non - reducing sugars:
1. Sugars which do not reduce either Tollen's reagent, Fehling's solution or Benedict's solution are called non-reducing sugars.
2. They contain a stable acetal or ketal structures which cannot be opened into a free carbonyl group.
E.g. Sucrose, starch, cellulose, glycogen, dextrin etc.
37.
38.
(i) Monomers :Ethylene glycol and terepathalic acid (or) dimethyl terephthalate.
(ii) Catalyst : Zinc acetate and antimony trioxide.
(iii) Temperature : 500 K
(iv) Product : Terylene
(v) Uses : blending with cotton or wool fibres and as glass reinforcing materials in safety helmets
39.
First law:
The mass of the substance (m) liberated at an electrode during electrolysis is directly proportional to the quantity of charge (Q) passed through the cell.
m α Q \(\left[\because \mathrm{I}=\frac{\mathrm{Q}}{\mathrm{t}} \Rightarrow \mathrm{Q}=\mathrm{It}\right]\)
m α It
m = ZIt
Where Z = electro chemical equivalent of the substance
I = current
t = time of passage of current
Second law:
When the same quantity of charge is passed through the solutions of different electrolytes, the amount of substances liberated at the respective electrodes are directly proportional to their electrochemical equivalents.
m α Z
\(\frac{m_{1}}{Z_{1}}=\frac{m_{2}}{Z_{2}}\)
m = mass of the metal deposited
Z = electro chemical equivalent
40.
(i) Purification of drinking water is activated by coagulation of suspended impurities in water by using alums containing \(\mathrm{Al}^{3+}\left(\mathrm{K}_{2} \mathrm{SO}_{4} \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot 24 \mathrm{H}_{2} \mathrm{O}\right)\) Alum has a negative charge and tends to disperse in water very fast.
(ii) The increased size as well as the lack of repelling charges cause the alum particles to settle down at the bottom or rise up and float in water. After the particles are neutralized, they clump together because of the London dispersive force which are part of vander Waal's forces. The weak inter molecular force arising from quantum induced instantaneous polarisation multi poles in molecules causes even non polar particles to attract each other due to the corelated movements of the electrons in interacting molecules. Then they settle down.
41.
Nylon - 6,6 can be prepared by mixing equimolar adipic acid and hexamethylene - diamine to form a nylon salt which on heating eliminate a water molecule to form amide bonds.
It is used in textiles, manufacture of cards etc ...
42.
Schotten - Baumann reaction :
Aniline reacts with benzoylchloride (C6H5COCI) in the presence of NaOH to give N - phenyl benzamide. This reaction is known as Schotten - Baumann reaction. The acylation and benzoylation are nucleophilic substitutions.
\(\underset { Aniline }{ { C }_{ 6 }{ H }_{ 5 }-{ NH }_{ 2 } } +\underset { Benzoyl\quad chloride }{ { C }_{ 6 }{ H }_{ 5 }-\overset { \underset { || }{ O } }{ C } -Cl } \overset { Pyridine }{ \longrightarrow } \underset { N-phenyl\quad benzamide }{ { C }_{ 6 }{ H }_{ 5 }-NH-\overset { \underset { || }{ O } }{ C } -{ C }_{ 6 }{ H }_{ 5 } } +HCl\)
43.
Carbylamine reaction :
Aliphatic (or) aromatic primary amines react with: chloroform and alcoholic KOH to give isocyanides (carbylamines), which has an unpleasant smell. This reaction is known as carbylamines test. This test used to identify the primary amines.
\(\underset { Ethylamine }{ { C }_{ 2 }{ H }_{ 5 }-{ NH }_{ 2 } } +\underset { Chloroform }{ { CHCl }_{ 3 }+3KOH } \longrightarrow \underset { Ethylisocyanide }{ { C }_{ 2 }{ H }_{ 5 }-NC } +3KCl+3{ H }_{ 2 }O\)
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