Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2014 Outside Delhi to familiarize themselves with the exam format and marking scheme.
CBSE Class 12 Chemistry Question Paper 2014 Outside Delhi with Solutions
General Instructions
- All questions are compulsory.
- Questions number 1 to 8 are very short answer questions and carry 1 mark each.
- Questions number 9 to 18 are short answer questions and carry 2 marks each.
- Questions number 19 to 27 are also short answer questions and carry 3 marks each.
- Questions number 28 to 30 are long answer questions and carry 5 marks each.
-
Use log tables if necessary. Use of calculators is not allowed.
† Deleted from Syllabus.
Set I
Question 1.
What is the effect of temperature on chemisorption? [1]
Answer:
Chemisorption increases with increase of temperature.
Question 2.
What is the role of zinc metal in the extraction of silver? [1]
Answer:
Question 3.
What is the basicity of H
3
PO
3
? [1]
Answer:
Basicity of H
3
PO
3
= 2
Because basicity is the number of replaceable H
+
ions in an acid and H
3
PO
3
is a Dibasic acid.
Question 4.
Identify the chiral molecule in the following pair: [1]
Answer:
Question 5.
Which of the following is a natural polymer? [1]
Buna-S, Proteins, PVC Economics type
Answer:
Protein is a natural polymer
Question 6.
The conversion of primary aromatic amines into diazonium salts is known as ____ [1]
Answer:
Diazotization.
Question 7.
What are the products of hydrolysis of sucrose? [1]
Answer:
Question 8.
Write the structure of p-methylbenzaldehyde.
Answer:
p-methylbenzaldehyde
Question 9.
An element with density 2.8 g cm
-3
forms an f.c.c. unit cell with edge length 4 × 10
-8
cm. Calculate the molar mass of the element. (Given N
A
= 6.022 × 10
23
mol
-1
) [2]
Answer:
Question 10.
(i) What type of non-stoichiometric point defect is responsible for the pink colour of LiCl?
(ii) What type of stoichiometric defect is shown by NaCl? [2]
OR
How will you distinguish between the following pairs of terms?
(i) Tetrahedral and octahedral voids
(ii) Crystal lattice and unit cell
Answer:
(i) This is due to metal excess defect due to anionic vacancies in which the anionic sites are occupied by unpaired electrons (F-centres).
(ii) Schottky defect is shown by NaCl.
OR
(i)
| Tetrahedral voids | Octahedral voids |
| 1. It is much smaller than the size of spheres in the packing. | Size is much larger than tetrahedral voids |
| 2. Each tetrahedral void is surrounded by 4 spheres. Hence, co-ordination no. is 4. | Each octahedral void is surrounded by 6 spheres. Hence, its co-ordination no. is 6 |
(ii) A regular arrangement of the constituent particles of a crystal in a three dimensional space is called crystal lattice.
The smallest three dimensional portion of a complete crystal lattice, which when repeated over and again in different directions produces the complete crystal lattice is called the unit cell.
Question 11.
State the Kohlrausch law of independent migration of ions. Why does the conductivity of a solution decrease with dilution? [2]
Answer:
Kohlrausch law of independent migration of ions : The limiting molar conductivity of an electrolyte (i.e. molar conductivity at infinite dilution) is the sum of the limiting ionic conductivities of the cation and the anion each multiplied with the number of ions present in one formula unit of the electrolyte
Λ°
m
for A
x
B
y
= \(x \lambda_{+}^0+y \lambda_{-}^0\)
For acetic acid Λ° (CH
3
COOH) = \(\lambda_{\mathrm{CH}_3 \mathrm{COO}^{-}}^{-}+\lambda_{\mathrm{H}^{+}}^{\circ}\)
Λ° (CH
3
COOH) = Λ° (CH
3
COOK) + Λ° (HCl) – Λ° (KCl)
Question 12.
For a chemical reaction R → P, the variation in the concentration (R) Vs. time (t) plot is given as [2] (i)
(i) Predict the order of the reaction.
(ii) What is the slope of the curve?
Answer:
(i) It is zero order reaction.
(ii) Slope of the curve = -K
Question 13.
Explain the principle of the method of electrolytic refining of metals. Give one example. [2]
Answer:
Electrolytic refining: Here the impure metal is made to act as anode and a strip of the same metal in pure form is used as cathode. When they both are put in suitable electrolyte containing soluble salt of same metal, the more basic metal remains in the solution and the less basic ones go to the anode mud.
Example: In refining of Cu
At anode: (oxidation)
Cu → Cu
2+
+ 2e
–
At cathode : (reduction)
Cu
2+
+ 2e
–
→ Cu
Question 14.
Complete the following equations: [2]
(i) P
4
+ H
2
O →
(ii) XeF
4
+ O
2
F
2
→
Answer:
(i) P
4
+ 6H
2
O → 2PH
3
+ 2H
3
PO
3
(ii)
Question 15.
Draw the structures of the following:
(i) XeF
2
(ii) BrF
3
[2]
Answer:
(i) XeF
2
:
(ii) BrF
3
:
Question 16.
Write the equations involved in the following reactions: [2]
(i) Reimer-Tiemann reaction
(ii) Williamson synthesis
Answer:
(i) Reimer – Tiemann reaction :
(ii) Williamson synthesis : Alkyl Halide reacts with Alkoxide
Question 17.
Write the mechanism of the following reaction:
Answer:
Question 18.
Write the name of monomers used for getting the following polymers: [2]
(i) Bakelite
(ii) Neoprene
Answer:
(i) Bakelite : Phenol and formaldehyde → Condensation.
(ii) Neoprene :
Question 19.
(a) Calculate Δ
r
G
0
for the reaction
Mg(s) + Cu
2+
(aq) → Mg
2+
(aq) + Cu(s)
Given: E
0
cell
= 2.71 V, 1F = 96500 C mol
-1
(b) Name the type of cell that was used in Apollo space program for providing electrical power. [3]
Answer:
(a) ∆
r
G° = – nFE°
= -2 × 96500 × 2.71 (∵ n = 2)
= -523,030 J mol
-1
= -523.03 KJ mol
-1
(b) Fuel cell was used in Apollo space programme for providing electrical power.
Question 20.
The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume:
SO
2
Cl
2
(g) → SO
2
(g) + Cl
2
(g)
Calculate the rate constant. (Given: log 4 = 0.6021, log 2 = 0.3010) [3]
Answer:
Question 21.
What are emulsions? What are their different types? Give one example of each type. [3]
Answer:
Emulsion : The colloidal solution in which both dispersed phase and medium are in liquid state is called emulsion. This is liquid-liquid colloidal system.
Different types of emulsions are :
Emulsions of oil in water in which oil is the dispersed phase and water is the dispersion medium.
Example : Milk is an emulsion of liquid fat dispersed in water.
Emulsions of water in oil in which water is the dispersed phase and oil is the dispersion medium.
e.g. Cod liver oil is an emulsion of oil i.e. water is the dispersed phase and oil is the dispersion medium.
Two applications of emulsions are :
- The digestion of fats in the intestines takes place by the process of emulsification.
- Several oily drugs are prepared in the form of emulsion.
Question 22.
Given the reasons for the following:
\(\dagger\)(i) (CH
3
)
3
P = O exists but (CH
3
)
3
N = O does not.
(ii) Oxygen has less electron gain enthalpy with a negative sign than sulphur.
(iii) H
3
PO
2
is a stronger reducing agent than H
3
PO
3
. [3]
Answer:
(i) (CH
3
)
3
P = 0 exists due to presence of empty d-orbitals and thus can expand its covalency upto 6 but (CH
3
)
3
N = O cann’t expand its covalency due to absence of d-orbitals.
(ii) The least negative electron gains enthalpy of oxygen is due to small size and more interelectronic repulsion with coming electron.
(iii) H
3
PO
2
contains two P-H bonds while H
3
PO
3
contains only one P-H bond therefore H
3
PO
2
is stronger reducing agent.
Question 23.
(i) Write the IUPAC name of the complex [Cr(NH
3
)
4
Cl
2
]Cl.
(ii) What type of isomerism is exhibited by the complex [Co(en)
3
]
3+
? (en = ethane-1,2-diamine)
(iii) Why is [NiCl
4
]
2-
paramagnetic but [Ni(CO)
4
] is diamagnetic [3]
(At. nos.: Cr = 24, Co = 27, Ni = 28)
Answer:
(i) IUPAC name : Tetraammine dichlorido chromium (III) chloride
(ii) Optical isomerism is exhibited by the complex [Co(en)
3
]
3+
(iii) In [NiCl
4
]
2-
, Ni
2+
has 3d
8
4s
0
configuration and due to weak ligand i.e. Cl
–
, electrons cannot pair up hence show paramagnetism while in [Ni(CO)
4
]. Ni is in zero oxidation state with 3d
8
4s
2
configuration and the 4s electrons are used up in pairing of 3d electrons as carbonyl ligand is strong hence diamagnetic.
Question 24.
(a) Draw the structure of major moon halo products in each of the following reactions:
(b) which halogen compound in each of the following pairs will react faster in SN2 reaction:
(i) CH
3
Br or CH
3
I
(ii) (CH
3
)
3
C-Cl or CH
3
-Cl
Answer:
(a)
(b) (i) CH
3
Br will react faster in S
N
2 reaction.
(ii) CH
3
— Cl will react faster in S
N
2 reaction.
Question 25.
Account for the following:
(i) Primary amines (R-NH
2
) have a higher boiling point than tertiary amines (R
3
N).
(ii) Aniline does not undergo Friedel-Crafts reaction.
(iii) (CH
3
)
2
NH is more basic than (CH
3
)
3
N in an aqueous solution. [3]
OR
Give the structures of A, B and C in the following reactions:
Answer:
(i) Due to presence of two H-atoms on N-atom of primary amines, they undergo extensive intermolecular H-bonding while tertiary amines due to the absence of a H-atom on the N-atom, do not undergo H-bonding. As a result, primary amines have higher boiling points than 3° amines.
(ii) Aniline being a Lewis base reacts with Lewis acid AlCl
3
to form a salt.
As a result, N of aniline acquires positive charge and hence it acts as a strong deactivating group for electrophilic substitution reaction. Consequently, aniline does not undergo Freidel Craft reaction.
(iii) Due to more steric hindrance in (CH
3
)
3
N it is less basic than (CH
3
)
2
NH.
OR
Question 26.
Define the following terms related to proteins: [3]
(i) Peptide linkage
(ii) Primary structure
(iii) Denaturation
Answer:
(i) Peptide linkage : It is an amide linkage formed between – COOH group of one a-amino acid and NH
2
group of the other α-amino acid by loss of a molecule of water. – CO – NH – bond is called Peptide linkage.
(ii) Primary structure : Proteins may have one or more polypeptide chains. Each polypeptide in a protein has amino acids linked with each other in a specific sequence which is known as primary structure of protein.
(iii) Denaturation of protein : Due to coagulation of globular protein under the influence of change in temperature, change in pH etc., the native shape of the protein is destroyed and biological activity is lost and the protein so formed is called denaturated proteins and the phenomenon is denaturation.
Question 27.
On the occasion of World Health Day.
Dr Satpal organized a ‘health camp’ for the poor fanners living in a nearby village. After checkup, he was shocked to see that most of the farmers suffered from cancer due to regular exposure to pesticides and many were diabetic. They distributed free medicines to them. Dr Satpal immediately reported the matter to the National Human Rights Commission (NHRC). On the suggestions of NHRC, the government decided to provide medical care, financial assistance, setting up of super-speciality hospitals for treatment and prevention of the deadly disease in the affected villages all over India.
(i) What type of analgesics are chiefly used for the relief of pains of terminal cancer?
(ii) Give an example of artificial sweetener that could have been recommended to diabetic patients. [3]
Answer:
(i) Morphine, Codeine are used for the relief of pains of terminal cancer.
(ii) Saccharin
Question 28.
(a) Define the following terms:
(i) Molarity
(ii) Molal elevation constant (k
b
)
(b) A solution containing 15 g urea (molar mass = 60 g mol
-1
) per litre of a solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass = 190 g mol
-1
) in water. Calculate the mass of glucose present in one litre of its solution. [2, 3]
OR
(a) What type of deviation is shown by a mixture of ethanol and acetone? Give reason.
(b) A solution of glucose (molar mass = 180 g mol
-1
) in water is labelled as 10% (by mass). What would be the molality and molarity of the solution? (Density of solution = 1.2 g mL
-1
)
Answer:
(a) (i) Molarity is the number of moles of solute dissolved in one litre of solution.
(ii) Molal elevation constant may be defined as the elevation in boiling point when the molality of solution is unity i.e. 1 mole of solute is dissolved in 1 kg of the solvent.
(b) For urea, concentration = \(\frac{15}{60}\) moles/lt.
For glucose, concentration = \(\frac{w}{180}\)moles/lt.
∵ Solutions are isotonic
∴ \(\frac{w}{180}\) = \(\frac{15}{60}\) ∴ w = \(\frac{15 \times 180}{60}\) = 45 g
Or
(a) It shows positive deviation due to weaker interactions between acetone and ethanol- ethanol interactions.
(b) 10% glucose means 10 g in 100 g solution or, 90 g of water = 0.090 kg of water
∴ Molality = \(\frac{10}{180 \times 0.090}\) = 0.617 m
100 g of solution = \(\frac{100}{1.2}\) mL = 83.33 mL = 0.08333 L
∴ Molality = \(\frac{10}{180 \times 0.0833}\) = 0.67 M
Question 29.
(a) Complete the following equations:
(i) Cr
2
O
7
2-
+ 2OH
–
→
(ii) MnO
4
–
+ 4H
+
+ 3e
–
→
(b) Account for the following:
(i) Zn is not considered as a transition element.
(ii) Transition metals form a larger number of complexes.
(iii) The E value for the Mn
3+
/Mn
2+
couple is much more positive than that for C
3+
/Cr
2+
couple. [2, 3]
OR
(i) With reference to structural variability and chemical reactivity, write the difference between lanthanoids and actinoids.
(ii) Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state.
(iii) Complete the following equation :
MnO
4
–
+ 8H
+
+ 5e
–
→
(iv) Out of Mn
3+
and Cr
3+
, which is more paramagnetic and why? (atomic nos : Mn = 25, Cr = 24) [5]
Answer:
(a) (i) Cr
2
\(\mathrm{O}_7^{2-}\) + 20H 2CrO42 + H20
(ii) Mn\(\mathrm{O}_4^{-}\) + 2H
2
O + 3e
–
→ MnO
2
+ 4OH
–
(Q is wrong)
(b) (i) Zinc in its common oxidation state of +2 has completely filled d-orbitals. Hence considered as non-transition elements.
(ii) Because of smaller size of their ions, high ionic charge and availability of vacant d-orbitals.
(iii) The large positive E° value for Mn
3+
/ Mn
2+
shows that Mn
2+
is much more stable than Mn
3+
because Mn
2+
has half filled 3d
5
stable configuration, therefore the 3
rd
ionization energy of Mn will be very high.
Or
(i) Difference between lanthanoids and actinoids :
- The general electronic configuration of lanthanoids is [Xe] 54 4f 1-14 5d 0-1 6s 2 whereas that of actinoids is [Rn] 86 5f 1-14 6d 0-1 7s 2 . Thus lanthanoids belong to 4f series whereas actinoids belong to 5f series.
- The first few members of the lanthanoids series are quite reactive. Therefore they combine with H 2 on gentle heating while the actinoids are highly reactive especially in the finely divided state, therefore they combine with most of the non-metals at moderate temperature.
(ii) Cerium
(iii) Mn\(\mathrm{O}_4^{-}\) + 8H
+
+ 5e
–
→ Mn
2+
+ 4H
2
O
(iv) Mn
3+
is more paramagnetic due to presence of 4 unpaired electrons than Cr
3+
having 3 unpaired electrons.
Question 30.
(a) Write the products formed when CH
3
CHO reacts with the following reagents:
(i) HCN
(ii) H
2
N-OH
(iii) CH
3
CHO in the presence of dilute NaOH
(b) Give simple chemical tests to distinguish between the following pairs of compounds.
(i) Benzoic acid and Phenol
(ii) Propanal and Propanone. [3, 2]
OR
(a) Account for the following: (2, 2, 1)
(i) Cl
–
CH
2
COOH is a stronger acid than CH
3
COOH.
(ii) Carboxylic acids do not give reactions of the carbonyl group.
(b) Write the chemical equations to illustrate the following name reactions:
(i) Rosenmund reduction
(ii) Cannizzaro’s reaction
(c) Out of CH
3
CH
2
-CO-CH
3
and CH
3
CH
2
-CH
2
-CO-CH
3
, which gives iodoform test?
Answer:
(a)
(b) (i) Benzoic acid and Phenol : On addition of NaHCO
3
to both solutions carbondioxide
gas is evolved with benzoic acid while phenol does not form CO
2
.
(ii) Propanal and Propane : Propanal gives positive test with Fehling solution in which a red ppt. of cuprous oxide is obtained while propanone does not respond to test.
OR
(a) (i) Because — I effect of Cl decreases the electron density in the O-H bond thereby making the release of a proton easier.
(ii) Carboxylic acids do not give the reactions of carbonyl group as there is no carbonyl group present due to resonance.
(b) (i) Rosenmund reduction
(ii) Cannizzaro’s reaction : Aldehydes, which do not have an α-hydrogen atom undergo self oxidation and reduction on treatment with conc. alkali and produce alcohol and carboxylic acid salt.
(c) CH 3 CH 2 — CH 2 — CO — CH 3 gives iodoform test.
Set II
Note: Except for the following questions, all the remaining questions have been asked in the previous set.
Question 1.
Why is adsorption always exothermic? [1]
Answer:
Adsorption is accompanied by decrease of randomness. For the process to be spontaneous, ∆G must be negative.
Hence, according to equation ∆G = ∆H — T∆S, ∆G can be -ve only if ∆H is negative.
Question 2.
Name the method used for refining of Nickel. [1]
Answer:
Mond’s process is used for refining of nickel.
Question 3.
Why does NO
2
dimerise? [1]
Answer:
NO
2
contains 7 + 2 × 8 i.e. 23 odd electrons. In the valence shell N has seven electrons and hence less stable. To acquire stability it dimeizes to form N
2
O
4
Question 4.
Based on molecular forces, what type of polymer is neoprene? [1]
Answer:
Elastomers
Question 5.
What are the products of hydrolysis of maltose? [1]
Answer:
Question 6.
Write the structure of 4-chloropentan-2 one. [1]
Answer:
4-Chloropentan-2-one:
Question 9.
Write the name of monomers used for getting the following polymers: [2]
(i) Terelyne
(ii) Nylon-6,6
Answer:
(i) Monomers of terylene are : ethylene glycol and terephthalic acid.
(ii) Monomers of nylon-6, 6 are : adipic acid and hexarnethylene diamine.
Question 10.
Describe the role of the following: [2]
(i) SiO
2
in the extraction of copper from copper matte.
(ii) NaCN in froth floatation process.
Answer:
(i) During roasting, copper pyrites are converted into a mixture of FeO and Cu
2
O.
To remove FeO (basic), an acidic flux silica is added during smelting. FeO then combines with SiO
2
to form ferrous silicate (FeSiO
3
) slag which floats over molten matte and hence can be easily removed
The silica is used to remove ferrous oxide and ferrous sulphide as slag.
(ii) NaCN is used as a depressant to separate PbS ore from ZnS in froth flotation process.
Question 11.
Complete the following equations: [2]
(i) Ag + PCl
5
(ii) CaF
2
+ H
2
SO
4
Answer:
(i) 2.Ag + PCl
5
→ 2AgCl + PCl
3
(ii) CaF
2
+ H
2
SO
4
→ CaSO
4
+ 2Hf
Question 12.
Draw the structures of the following: [2]
(i) XeF
4
(ii) HClO
4
Answer:
(i) XeF
4
SP3d2 hybridization
Shape – Square planar
(ii) HClO
4
(Perchioric acid)
Question 13.
(i) Write the type of magnetism observed when the magnetic moment is oppositely aligned and cancel out each other.
(ii) Which stoichiometric defect does not change the density of the crystal? [2]
Answer:
(i) Diamagnetism is observed when the magnetic moments are oppositively aligned and cancel out each other.
(ii) Frenkel defect does not change the density of the crystal.
Question 14.
Define the following terms: [2]
(i) Fuel cell
(ii) Limiting molar conductivity (∧
m
°)
Answer:
(i) Fuel cell : These cells are the deviœs which convert the energy produced during combustion of fuels like H
2
, CH
4
, etc. directly into electrical energy.
(ii) The molar conductivity of a solution at infinite dilution is called limiting molar conductivity and is represented by the symbol \(\Lambda_{\mathrm{m}}^{\circ}\).
Question 19.
Define the following terms: [3]
(i) Glycosidic linkage
(ii) Invert sugar
(iii) Oligo saccharides
Answer:
(i) Glycosidic linkage : The two monosaccharide units are joined together through an etheral or oxide linkage formed by loss of a molecule of water. Such a linkage between two monosaccharide units through oxygen atom is called glycosidic linkage
(ii) Invert sugar : An equimolar mixture of glucose and fructose obtained by hydrolysis of sucrose in presence of an acid such as dil. HCl or the enzyme invertase or sucrase is called invert sugar.
(iii) Oligosaccharides: Those carbohydrates which on hydrolysis give 2-10 molecules of monosaccharides are called oligosaccharides. Example: sucrose, maltose.
Set III
Note: Except for the following questions, all the remaining questions have been asked in previous sets.
Question 1.
What are the dispersed phase and dispersion medium in milk? [1]
Answer:
Milk : Dispersed phasë → Fat (Liquid);
Dispersion medium → Liquid
Question 2.
Name the method used for refining of copper metal. [1]
Answer:
Electrolytic refining is used for refining of copper metal.
\(\dagger\)Question 3.
Why does NH
3
act as a Lewis base? [1]
Answer:
Due to presence of Ione pair on nitrogen NH
3
acts as a Lewis base.
Question 5.
Which of the following is a fibre? [1]
Nylon, Neoprene, PVC
Answer:
Nylon is a fibre.
Question 6.
Write the products of hydrolysis of lactose. [1]
Answer:
Lactose on hydrolysis with dilute acids gives an equimoar mixture of D-glucose and D-galactose.
Question 8.
Write the structure of 2-hydroxybenzoic acid. [1]
Answer:
2-hydroxybenzoic acid:
Question 9.
Complete the following equations: [2]
(i) Cu + 2H
2
SO
4
(Conc.) →
(ii) XeF
2
+ H
2
O →
Answer:
(i) C + 2H
2
SO
4
(conc.) → CO
2
+ 2SO
2
+ 2H
2
O
(ii)
Question 10.
Draw the structure of the following: [2]
(i) XeO
3
(ii) H
2
SO
4
Answer:
(i) XeO
3
:
(ii) H
2
SO
4
:
Question 11.
Write the name of monomers used for getting the following polymers: [2]
(i) Teflon
(ii) Buna-N
Answer:
(i) Teflon :
(ii) Buna-N : Monomers of Buna-N are 1, 3-Butadiene and aczylonitrile.
\(\dagger\)Question 13.
(i) Write the type of magnetism observed when the magnetic moment is aligned in parallel and anti-parallel directions in unequal numbers. [2]
(ii) Which stoichiometric defect decreases the density of the crystal?
Answer:
(i) Ferrimagnetism is observed.
(ii) Schottky defect decreases the density of the crystal.
Question 14.
Define the following terms:
(i) Molar conductivity (∧
m
)
(ii) Secondary batteries [2]
Answer:
(i) Molar conductivity: Molar conductivity of a solution at a given concentration is the conductance of the volume ‘V’ of a solution containing one mole of electrolyte kept between two electrodes with area of cross section ‘A’ and distance of unit length. It is represented by Λ
m
(lamda).
Λ
m
= \(\frac{\mathrm{KA}}{l}\) ∴ l = 1 and A = V
∴ Λ
m
= KV Unit = S cm
2
mol
-1
(ii) Secondary batteries : Those batteries which can be recharged by passing an electric current through them and can be used again and again are called secondary batteries.
Question 17.
Write the principle behind the froth floatation process. What is the role of collectors in this process? [2]
Answer:
Principle:
Froth floatation process : This method is used for removing gangue from suiphide ores. In this powdered ore is mixed with collectors (e.g. pine oils, fatty acids etc.) and froth stabilisers (e.g. cresols, aniline) which enhance non-wettability of the mineral particles and froth stabilisation respectively. As a result of which ore comes with froth and gangue remain in the solution.
Role : Collectors e.g. pine oils, fatty acids etc. enhance non-wettability of the mineral particles.
Question 23.
Define the following terms : [3]
(i) Nucleotide
(ii) Anomers
(iii) Essential amino acids
Answer:
(i) Nucleotide : A nucleotide contains all the three basic components of nucleic acids, i.e. a phosphoric acid group, a pentose sugar and a nitrogenous base. These are formed by estenfication of C
5
’—OH of the sugar of the nucleoside with phosphoric acid.
(ii) Anomers: A pair of stereoisomers which differ in configuration only around C
1
are called anomers. Two isomers are said to be anomers if the isomerisation in the molecule is at first carbon.
(iii) Essential amino acids: Amino acids which the body cannot synthesize are called essential amino acids.
Examples : Valine, leucine etc. Therefore they must be supplied in diet.