(A) Study Material for Supportive Learners -:
1.
General characteristics of solid state:
Definite mass, volume and shape
Short intermolecular distances
Strong intermolecular forces
Fixed lattice positions of the constituent particles
Incompressibility and rigidity
Classification of the solid state:
Crystalline
Amorphous (sometimes called pseudo solids or super-cooled liquids)
Differences between the crystalline and amorphous solids
Crystalline
Amorphous
Have definite characteristic
geometrical shape
Have irregular shape
Melt at a sharp and
characteristic temperature
Gradually soften over a range
of temperature
When cut with a sharp edged
tool, the newly generated
pieces are plain and smooth
When cut with a sharp edged
tool, the newly generated
pieces are with irregular
surfaces
Have definite and
characteristic heat of fusion
Do not have definite heat of
fusion
Are true solids
Are pseudo solids
Anisotropic in nature
Isotropic in nature
Have long-range order
Have only short-range order
Classification of crystalline solids (On the basis of the nature of intermolecular forces)
Molecular solids
Non-polar molecular solids →These consist either atoms or the molecules formed
by non-polar covalent bonds. Example: H2, Cl2, I2
Polar molecular solids → The molecules ion these types of solids are held together by
strong dipole-dipole interactions. Example: Solid SO2, solid NH3
Hydrogen-bonded molecular solids → The molecules of such solids contain polar
covalent bonds between H and f, O or N atoms. Example: Ice (H2O)
Ionic solids
Ions are the constituent particles; e.g., NaCl, KNO3
Metallic solids
Each metal atom is surrounded by electrons; e.g., Fe, Cu
Covalent or network solids
Formed by covalent bonds; e.g., diamond, silicon carbide
Crystal lattice: A regular three-dimensional arrangement of points in space.
There are 14 Bravais lattices
Unit cells: There are two categories of unit cells –
Primitive unit cells: There are seven types of primitive unit cells –
Cubic (a = b = c, α =β = γ = 90°); e.g., NaCl
Tetragonal (a = b ≠ c, α =β = γ = 90°); e.g., CaSO4
Orthorhombic (a ≠ b ≠ c, α =β = γ = 90°); e.g., KNO3
Hexagonal (a = b ≠ c, α =β = 90°, γ = 120°); e.g., ZnO
Rhombohedral or Trigonal (a = b = c, α =β = γ ≠ 90°); e.g., CaCO 3
Monoclinic (a ≠ b ≠ c, α = γ = 90°, β ≠ 120°)
Triclinic (a ≠ b ≠ c, α ≠ β ≠ γ ≠ 90°); e.g., Na2SO4.10H2O
Centred unit cells: There are three types of centred unit cells –
Body-centred unit cells: Contain one constituent particle at its body centre along
with the ones present at corners
Face -centred unit cells: Contain one constituent particle at the centre of each face
along with the ones present at corners
End-centred unit cells: Contain one constituent particle at the centre of any two faces
along with the ones present at corners
Number of atoms in a unit cell:
Total number of atoms in one primitive cubic unit cell = 1
Total number of atoms in one body-centred cubic unit cell = 2
Total number of atoms in one face-centred cubic unit cell = 4
Close-packed structures:
Coordination number: The number of the nearest neighbours of a particle
Packing efficiency: The percentage of total space filled by the particles
Close-Packing in One dimension
Only one way of arrangement, i.e., the particles are arranged in a row, touching
each other
Close-Packing in Two Dimensions
Square close-packing in two dimensions
Close-Packing in Three Dimensions
Three-dimensional close-packing is obtained by stacking two -dimensional layers (square
close-packed or hexagonal close-packed) one above the other.
There are two highly efficient lattices of close-packing –
Hexagonal close-packed (hcp)
Cubic close-packed (ccp) [also called face-centred cubic (fcc) lattice]
In hcp and ccp, 74% space is filled, i.e., packing efficiency = 74%. The
remaining space is present in the form of voids.
There are two types of voids –
Octahedral voids
Tetrahedral voids
The packings other than hcp and ccp are not close packings as they have less packing
efficiency.
Packing efficiency in bcc = 68%
Packing efficiency in simple cubic lattice = 52.4%
2.
Adsorption: Phenomenon of accumulation of molecules at the surface of a solid, which
results in a higher concentration on the surface than in the bulk
Adsorbate: The substance which gets adsorbed onto a surface
Adsorbent: The substance on which adsorption takes place
Desorption: The phenomenon of removing an adsorbate from an adsorbent
Types of adsorption –
1. Physical adsorption or physisorption
2. Chemical adsorption or chemisorption
Physisorption
2.
3.
Arises because of van der
Waals forces
Not specific in nature
Reversible
4.
Enthalpy of adsorption is low
1.
5.
6.
No appreciable activation
energy is required
Results in multi-molecular
layers
Chemisorption
1.
2.
3.
4.
5.
6.
Arises because of chemical
forces
Highly specific in nature
Irreversible
Enthalpy of adsorption is
high
Sometimes, high activation
energy is required
Results in uni-molecular
layer
Gas is adsorbed in almost all solids.
Factors on which the extent of adsorption of a gas on a solid depends upon:
Nature of the gas
Nature of the solid
Surface area of the solid
Pressure of the gas
Temperature of the gas
Adsorption isotherm:
The relation between the extent of adsorption (x/m) and the pressure of a gas at constant
temperature
1. Freundlich adsorption isotherm –
Applications of Adsorption
Production of high vacuum.
Gas mask.
Silica and aluminium gels are used as adsorbents for controlling
humidity.
A number of drugs are used for killing germs; the drugs get adsorbed on the
germs.
Used in heterogeneous catalysis Used
in chromatographic analysis.
The substance which alters the rate of a reaction, but itself remains chemically and
quantitatively unaltered after the reaction is called a catalyst.
The phenomenon of using a catalyst is called catalysis.
The substances which enhance the activity of a catalyst are called promoters.
The substances which decrease the activity of a catalyst are called poisons.
Broadly divided into two groups:
Homogeneous catalysis – The reactants and the catalyst are in the same phase
Heterogeneous catalysis – The reactants and the catalyst are in different phases
Shape selective catalysis:
Catalysis that depends upon the pore structure of the catalyst, and the size of the reactant
and the product
Enzyme catalysis or biochemical catalysis: An enzyme acts as the catalyst
Highly specific in nature
Highly active under optimum temperature and pH
Influenced by inhibitors and poisons
Industrial catalytic processes:
Haber’s process
Ostwald’s process
Contact process
Colloids:
Heterogeneous system
Intermediate between true solutions and suspensions
Consist of two phases – Dispersed phase and dispersion medium
Classification of colloids: Colloids are classified depending upon the following
three criteria.
Physical state of the dispersed phase and the dispersion medium
Dispersed
phase
Solid
Dispersion
medium
Solid
Type of
colloid
Solid sol
Solid
Solid
Liquid
Liquid
Liquid
Gas
Gas
Liquid
Gas
Solid
Liquid
Gas
Solid
Liquid
Sol
Aerosol
Gel
Emulsion
Aerosol
Solid sol
Foam
Examples
Gem stones, some
coloured glasses
Paints, cell fluids
Smoke, dust
Cheese, butter, jellies
Milk, hair cream
Fog, mist, cloud
Stone, foam rubber
Froth, soap lather
Nature of interaction between the dispersed phase and the dispersion medium
Lyophilic colloids (solvent attracting):
Also called reversible sols
Quite stable and cannot be easily coagulated
Lyophobic colloids (solvent repelling):
Also called irreversible sols
Can be easily coagulated
Type of particles of the dispersed phase
Multi-molecular colloids
Macromolecular colloids
Associated colloids (Micelles)
The temperature above which the formation of micelles takes place is called
Kraft temperature (Tk).
The concentration above which the formation of micelles takes place is called
critical micelle concentration (CMC).
Preparation of colloids:
Chemical method
Electrical disintegration or Bredig’s Arc method
Peptisation – Process of converting a precipitate into colloidal sol by shaking it with
the dispersion medium in the presence of a small amount of electrolyte
Purification of colloidal solutions:
Reduction of the amount of impurities to a requisite minimum
3.
Group 15 elements:
2 3
The valence shell electronic configuration is ns np .
Nitrogen differs in chemical properties from other elements of the group due to its
small size, high electronegativity, high ionisation enthalpy and nonavailability of d-orbitals.
They exhibit two oxidation states, +3 and +5. Heavier elements exhibit mainly +3
oxidation state due to inert pair effect.
The main use of nitrogen is in the manufacture of ammonia
Dinitrogen (N2)
Colourless, odourless, tasteless, and non-toxic gas
Very low solubility in wate, low freezing point and low boiling point. Inert at room
temperature due to High bond enthalpy of N≡ N bond. Commercially produced by
the liquefaction and frictional distillation of air
In the laboratory − Prepared by treating an aqueous solution of NH 4Cl with NaNO2
Uses
3. In the manufacture of NH3.
4. In the manufacture of industrial chemicals containing nitrogen.
5. As a refrigerant to preserve biological materials and food items
6. I n cryosurgery.
7. To generate inert atmosphere.
Ammonia:
Colourless gas with a pungent odour
Freezing point = 198.4 K
Boiling point = 239.7 K
Haber’s process
It acts as Lewis base.
Nitric acid (HNO3)
1. Preparation: Ostwald’s process
Detection of the presence of nitrate:
(Brown ring test)
Phosphorus exists as P4 in elemental form.
Allotropic forms of phosphorus:
1. White phosphorus
2. Red phosphorus
3. Black phosphorus (a-block phosphorus and β-block phosphorus
Phosphorus forms two types of halides, PX3(X = F, Cl, Br, I) and PX5(X = F, Cl, Br).
The structure of PCl5 is trigonal bipyramidal
Phosphorus forms a number of oxoacids such as ortho-phosphoric acid
(H3PO4), ortho-phosphorus acid (H3PO3), hypo-phosphorus acid (H3PO2).
The oxoacids containing P – H bond are strong reducing agents.
Group 16 elements: (known as chalcogens)
Oxygen
Sulphur
Selenium
Tellurium
Polonium
2 4
The valence shell electronic configuration is ns np .
Like nitrogen, oxygen differs from other elements of the group due to its small size and high
electronegativity
1. Preparation:
1. Three stable isotopes –
16
O,
17
O,
18
O
Uses
In normal respiration and combustion
As an oxidant (in liquid state) for propelling rockets In
oxyacetylene welding
In the manufacture of many metals (particularly steel)
Oxygen cylinders are used in hospitals, high altitude flying and
mountaineering.
1.
Acidic oxides − Combine with water to give an acid
Example − SO2, Cl2O7, CO2, N2O5
Basic oxides − Combine with water to give bases
Examples − Na
2O, CaO, BaO
Amphoteric oxides − Show the characteristics of both acidic as well as basic oxides
React with both acids and alkalies
4.
The d-block elements are known as transition metals.
The f-block elements are known as inner transition metals.
4f – metals → lanthanoids
5f – metals → actinoids
Electronic Configuration
General outer electronic configuration is (n −1)d
1−10
ns
1−2
.
Zn, Cd, and Hg are not regarded as transition elements.
Reason − The orbitals of these elements are completely-filled. [Electronic
10 2
configuration is (n − 1) d ns ]
The d-block elements (transition metals):
The d-block elements (group 3 – 12) occupy the large, middle section of the periodic
table.
1–10 1–2
In general, their valence shell electronic configuration is (n – 1)d
ns
.
There are three series of transition metals, corresponding to the filling of 3d, 4d and 5d
orbitals.
All the transition metals exhibit typical metallic properties such as high tensile
strength, ductility, malleability, high thermal and electrical conductivity, metallic
lustre, etc.
The melting and boiling points of transition metals are high due to the
involvement of (n – 1)d electrons in interatomic bonding.
The maxima of these properties occur at about the middle of each series, which
indicates that one unpaired electron per d-orbital is particularly favourable for
strong interatomic interaction.
Variation of atomic radii:
Lanthanoid contraction – Regular decrease in atomic radii due to the filling of the 4f orbital
before the 5d orbital. As a result, with increase in atomic number, the expected increase in
size is not observed.
Ionisation Enthalpies
In each of the three transition series, the first ionisation enthalpy increases from left to
right. However, there are some exceptions.
The first ionisation enthalpies of the third transition series are higher than those of
the first and second transition series.
Reason: Poor shielding effect of 4f electrons in the third transition series
Exhibit variable oxidation states
Exhibit paramagnetic behaviour. Their magnetic moment values can be
calculated using the expression:
where, μ is the magnetic moment
and n is the number of unpaired electrons.
Have the tendency to form coloured ions
Form interstitial and complex compounds
Act as catalysts
Form alloys
Many of the transition metals are electropositive, and can dissolve in mineral acids
while a few of them are noble.
Transition metals form oxides and oxoanions of metals.
Magnetic Properties
Diamagnetic substance − Repelled by the applied field
Paramagnetic substance − Attracted by the applied field
Ferromagnetic substance − Attracted very strongly
Paramagnetism arises due to the presence of unpaired electrons.
Magnetic moment can be calculated by using ‘spin-only’ formula, i.e.,
Where,
n = Number of unpaired electrons
μ = Magnetic moment in Bohr magneton (BM)
Formation of Coloured Ions
An electron from a lower energy d-orbital is excited to a higher energy d-orbital when
the energy of excitation corresponds to the frequency of light absorbed.
This frequency of light generally lies in the visible region.
The colour observed is the complementary colour of the light absorbed.
Formation of Complex Compounds
Transition metals form a large number of complex compounds.
Reason: Comparatively smaller size of metal ions, high ionic charges and
availability of d-orbitals for bond formation
Catalytic Properties
Transition metals and their compounds are known for their catalytic activity.
Formation of Interstitial Compounds
Interstitialcompounds formed when small atoms like H, C, N are trapped inside
the crystal lattices of metals
Usually non-stoichiometric
Neither typically ionic nor covalent
Example: TiC, Mn4N, Fe3H, etc.
Alloy Formation
Alloys are readily formed by these metals.
Reason: Because of similar radii and other characteristics of transition metals
Oxides and Oxoanions of Metals
Oxides are generally formed by the reaction of metals with oxygen at higher
temperatures.
Potassium Dichromate (
)
Preparation:
Potassium dichromate being less soluble than sodium dichromate can be
obtained in the form of orange coloured crystals by treating sodium dichromate
with potassium chloride and can be removed by filtration.
The dichromate ion
exists in equilibrium with chromate
ion at pH =
4. However, by changing the pH, they can be inter-converted.
5.
Coordination compounds:
These are the complex compounds in which transition metal atoms are bound to a number
of anions or neutral molecules.
Postulates of Werner’s theory of coordination compounds:
In coordination compounds, there are two types of linkages (valences) – primary
and secondary.
The primary valences are ionisable and are satisfied by negative ions.
The secondary valences are non-ionisable and are satisfied by negative ions or neutral
molecules. The secondary valence is equal to the coordination number of a metal and
remains fixed for a metal.
Different coordination numbers have characteristic spatial arrangement of the ions or
groups bound by the secondary linkages.
Difference between a double salt and a complex:
In water, a double salt dissociates completely to give simpler ions. Examples of double
salt: carnallite (KCl. MgCl2. 6H2O), Mohr’s salt [FeSO4. (NH4)2 SO4. 6H2O]
Complex ions do not dissociate further to give simpler ions. For example,
4–
3–
[Fe(CN)6] , [Fe(C2O4)3]
Ligands:
Ions or molecules bound to the central metal atom or ion in the coordination entity
Didentate –
Polydentate
Ambidentate –
(Can bind through two different atoms)
Coordination number: Number of ligand donor atoms bonded directly to the metal
Coordination polyhedral:
Homoleptic and heteroleptic complexes:
Homoleptic complexes: In these complexes, the metal is bound to only one kind of
3+
donor group, e.g., [Co(NH3)6]
Heteroleptic complexes: In these complexes, the metal is bound to more than one kind
+
of donor groups, e.g., [Co(NH3)4Cl2]
Naming of mononuclear coordination compounds:
The cation is named first in both positively and negatively charged
coordination entities.
The ligands are named in alphabetical order, before the name of the central atom/ion.
Names of the anionic ligands end in –o.
[Exceptions: aqua (H2O), ammine (NH3), carbonyl (CO), nitrosyl (NO)]
To indicate the number of the individual ligands, the prefixes mono–, di–, tri–, etc.,
are used. If these prefixes are present in the names of ligands, then the terms bis–,
tris–, tetrakis–, etc., are used.
Oxidation state of the metal is indicated in Roman numerals, in parentheses.
If the complex ion is cation, then the metal is named as the element.
If the complex ion is anion, then the metal is named with ‘–ate’ ending.
The neutral complex molecule is named as the complex cation.
Isomerism in coordination compounds:
Stereoisomerism –
Geometrical isomerism:
This type of isomerism arises in heteroleptic complexes due to different geometric
arrangements of the ligands.
Optical isomerism:
Optical isomers (enantiomers) are mirror images which cannot be superimposed on one
another.
The molecules or ions which cannot be superimposed on their mirror images are
called chiral or optically active.
6.
Carbohydrates: Most carbohydrates have the general formula Cx(H2O)y.
Classification of carbohydrates: Three groups (on the basis of their behaviour on
hydrolysis)
(I) Monosaccharides: Cannot be hydrolysed further to yield simpler units. Example:
glucose, fructose
(II) Oligosaccharides: On hydrolysis, give two to ten units of monosaccharides
Disaccharides (give two units). Example: sucrose (gives glucose and fructose)
maltose (gives two molecules of glucose).
Trisaccharides (give three units)
Tetrasaccharides (give four units), etc.
(III) Polysaccharides: On hydrolysis, give a large number of units of monosaccharides.
They are also called non-sugars as they are not sweet to taste The carbohydrates which
are sweet to taste are also called sugars. All monosaccharides are reducing sugars.
Classification of monosaccharides
Carbon atomsGeneral termAldehyde
Ketone
3
Triose
Aldotriose
Ketotriose
4
Tetrose
Aldotetrose Ketotetrose
5
Pentose
AldopentoseKetopentose
6
Hexose
Aldohexose Ketohexose
7
Heptose
AldoheptoseKetoheptose
Glucose:
Evidences for the given structure
8. Molecular formula, C6H12O6
9. The six C-atoms are linked in a straight chain.
(iii) Presence of a carbonyl group
4.
Presence of carboxyl groups as an aldehydic group
5. Presence of five –OH groups attached to different carbon atoms.
6.
Presence of a primary alcoholic (–OH) group
Cyclic structure of glucose: Reactions that are not explained by the open-chain structure.
(i) Does not give 2, 4–DNP test, Schiff’s test, and does not form hydrogensulphite addition
product with NaHSO3
(ii) There is no reaction between pentaacetate of glucose and hydroxylamine (i.e., free –
CHO group is absent)
(iii) There are two different crystalline forms of glucose i.e. α–form and β-form. They are
called anomers.
(iv) Haworth structure
Fructose:
(i) Molecular formula: C6H12O6
(ii) Structure:
(iii) Cyclic structure
(iv) Haworth structure:
Disaccharides: Glycosidic linkage Linkage between two monosaccharide units through
oxygen atom.
7.
Polymers are large molecules having high molecular mass (103 – 107 u). They are also
called macromolecules.
Polymerisation is the process of formation of polymers from respective monomers.
Classification of polymers:
(I) On the basis of source
10. Natural polymers: Found in plants and animals, e.g., proteins, cellulose, starch.
11. Semi-synthetic polymers: Cellulose acetate (rayon), cellulose nitrate
12. Synthetic polymers: Synthetic fibre → nylon 6, 6 Synthetic rubber → Buna-S
(II) On the basis of structure
7. Linear polymers: High-density polythene, Polyvinyl chloride.
8.
Branched-chain polymers: Low-density polythene.
9. Cross-linked or network polymers: They contain strong covalent bonds between
various linear polymer chains, e.g., bakelite, melamine
(III) On the basis of mode of polymerisation
(iv)Addition polymer: Polymers formed by repeated addition of monomers containing double
or triple bonds Addition polymers formed from single monomeric species are called
homopolymers. Addition polymers formed from two different monomeric species are called
copolymers.
(v) Condensation polymers: Polymers formed by repeated condensation reaction between
two different bi-functional or trifunctional monomers Condensation polymerisation involves
elimination of small molecules such as water, alcohol, hydrogen chloride, etc.
(IV) On the basis of molecular forces
(iii)Elastomers: Polymer chains are held together by the weakest intermolecular forces,
e.g., buna N, buna S, neoprene
(iv) Fibres: Possess strong intermolecular forces like H-bonding, e.g., nylon-6, 6, terylene
(v) Thermoplastic polymers: Possess intermolecular forces of attraction
intermediate between elastomers and fibres,e.g polythene, polystyrene, PVC
(vi) Thermosetting polymers: Cross-linked or highly branched molecules, e.g., bakelite,
urea-formaldehyde resins.
(V) On the basis of growth polymerisation (Depending upon the type of
polymerisation mechanism)
(i) Chain-growth polymers
(ii) Step-growth polymers
Types of polymerisation reactions:
(I) Addition polymerisation or chain-growth polymerisation: Molecules of the same
monomer or different monomers add together on a large scale to form a polymer.
(a) Polytetrafluoroethene (Teflon): Chemically inert and resistant to attack by corrosive
reagents. It is used for making oil seals and gaskets, and for non-stick-surface-coated
utensils.
(b) Polyacrylonitrile: It is used as a substitute for wool in making commercial fibres as
orlon or acrilan.
(II) Condensation polymerisation or step- growth polymerisation: Involves a repetitive
condensation reaction between two bi-functional monomers. This results in the loss of some
simple molecules as water, alcohol, etc., and leads to the formation of high molecular mass
condensation polymers.
(a) Polyamides
Nylon, 6, 6 : Used in making sheets, bristles for brushes and in textile industry.
Nylon 6: Used for the manufacture of tyre cords, fabrics and ropes
(b) Polyesters
Dacron: Used in blending with cotton and wool fibres and as glass-reinforcing materials
in safety helmets.
(c) Phenol-formaldehyde polymer: E.g., Bakelite Novolac, when heated with formaldehyde,
undergoes cross linking to form bakelite, which is an infusible solid mass.
Novolac, obtained on heating with formaldehyde, undergoes cross- linking to form an
infusible solid mass called bakelite.
(d) Melamine-formaldehyde polymer: Used in the manufacture of unbreakable crockery.
8.
Drugs: Chemicals of low molecular masses (~100–500 u) used to produce biological
response by interacting with target macromolecules. Drugs used for therapeutic effect
are called medicines. Chemotherapy refers to the use of chemicals for therapeutic
effect.
Classification of drugs
(i) Based on pharmacological effect
Useful for doctors as it provides the whole range of drugs available to cure a particular
type of problem.
(ii) Based on drug action
On the basis of action of a drug on a particular biochemical process.
13. Based on chemical structure
14. Based on molecular targets
Useful for medicinal chemists
(ii) Target molecules or molecular targets are the biomolecules with which drugs interact.
For example, carbohydrates, lipids, proteins and nucleic acids.
Drug-Target interaction:
(A) Enzymes as drug targets
(1) Catalytic action of enzymes
Two major functions of enzymes are:
(i) Substrate molecules are held in suitable positions by active sites of enzymes through
ionic bonding, H-bonding, van der Waals interaction or dipole–dipole interaction.
(ii) To attack the substrate molecules, functional groups are provided by enzymes.
(2) Drug-enzyme interaction
Drugs inhibit the catalytic action of enzyme by blocking the binding site of enzyme. Such
drugs are called enzyme inhibitors.
The two ways of drug enzyme interaction are:
(i) Drugs compete with substrate molecules to attack on the active sites of enzymes. Such
drugs are called competitive inhibitors.
(ii) Some drugs change the shape of the active site of enzyme by binding to a
different site not to the active site of enzyme so that substrate cannot recognize it. This site
is called allosteric site.
(B) Receptors as drug targets
(i) Receptors are proteins that are crucial to communication process of body.
(ii) Chemical messenger are chemicals through which message between two neurons
and that between neurons to muscles is communicated.
(iii) Antagonists are drugs which inhibit natural function of receptor by binding to its site.
(iv)Agonists are drugs that mimic the natural messenger by switching on the receptor.
Therapeutic action of different classes of drugs:
Antacids
(i) Used for the treatment of acidity.
(ii) Metal hydroxides are generally used as antacids.
Antihistamines
Drugs
(i) Cimetidine and ranitidine prevent the interaction of histamine with the receptors present in
the stomach wall and as a result lesser amount of acid is released.
(ii) Antihistamine drugs – brompheniramine (Dimetapp), terfenadine (Seldane). They
interfere with the natural action of histamine by competing with histamine for finding sites of
receptors.
(iii) Antihistamines do not affect the secretion of acid in stomach.
Neurologically active drugs
They affect the mechanism of message transfer from nerve to receptor.
(a) Tranquilizers
(i) Used for the treatment stress and mental diseases
(ii) Antidepressant drugs are used when a person is suffering from
depression. E.g. iproniazid, phenelzine.
(iii) Barbiturates constitute an important class of tranquilizers.
(b) Analgesics– Used to reduce pain.
Two types –
(i) Non-narcotic (non-addictive) analgesics: Example: Asprin
(ii) Narcotic drugs: Example: Morphine and many of its homologues
Antimicrobials: Inhibit the pathogenic action of microbes such as bacteria, fungi, virus or
other parasites. Examples – Antibiotics, antiseptics, disinfectants.
(a) Antibiotics
(i) Used to treat infections
(ii) Penicillin is an antibacterial drug
(iii) Antibiotics which kill or inhibit a wide range of Gram-positive and Gramnegative bacteria are called broad spectrum
antibiotics. Ampicillin and amoxicillin which are synthetic modifications of penicillin
are broad spectrum antibiotics.
(iv) Chloramphenicol is a broad spectrum antibiotic which is used orally for
treatment of typhoid, dysentery, acute fever
pneumonia. Vancomycin and ofloxacin are broad spectrum antibiotics.
(v) Antibiotics which are effective mainly against Gram-positive or Gram-negative
bacteria are called narrow spectrum
antibiotics. E.g. Penicillin G.
(vi) Antibiotics which are effective against a single organism or disease are called limited
spectrum antibiotics.
(b) Antiseptics and disinfectants: Chemicals which kill or prevent the growth of
microorganisms.
(i) Antiseptic are applied to living tissues such as wounds, cuts, ulcers, diseased skin
surfaces while disinfectants are applied to
inanimate objects such as floors, drains, instruments etc.
(ii) Examples of antiseptics: Furacine, soframicine, dettol (mixture of chloroxylenol and
terpineol), tincture of iodine (2-3%) solution of iodine in alcohol-water mixture), iodoform,
basic acid in dilute aqueous solution (weak antiseptic for eyes).
(ii) Examples of disinfectants: 0.2 to 0.4 ppm of chlorine in aqueous solution, sulphur
dioxide in low concentration. 0.2% solution of phenol is used as antiseptic while 1%
solution of phenol is used as disinfectant.
Antifertility drugs: Used in the direction of family planning. E.g. norethindrone,
ethynylestradiol (novestrol).
Chemical in food:
Artificial sweetening agents: They are required to control intake of calories.
Practice Questions
1.
Q1)
Which type of unit cell does cinnabar (HgS) correspond to?
Answer - Cinnabar (HgS) adopts a hexagonal unit cell.
Q2) Which type of molecular solid can CCl4 be classified into?
Answer CCl4 can be classified into a covalent molecular solid.
Q3) What is the formula of a compound having a cubic structure formed by elements A and B in which the atoms of A are
present at the corner and the atoms of B are present at the face centres of the cube?
Answer -If the atoms of Aare present at the 8 corners of the cube, then the number of atoms present in the unit
cell will be If the atoms of B are present at the 6 face centres of the cube, then the number of atoms present in the
unit cell will be Thus, the ratio of the atoms of A to the atoms of B is 1:3.
Hence, the formula of the compound is AB3.
Q4) What is meant by the term 'coordination number'?
Answer-The number of closest (or nearest) neighbours of any constituent particle in the crystal lattice is called
the coordination number.
Q5) In corundum, the oxide ions are arranged in hexagonal close-packing, while aluminium ions occupy two-thirds of the
octahedral voids. What is the formula of corundum?
Answer-Let the number of octahedral voids be N. Accordingly, Number of octahedral voids occupied by Al3+ ions
In H.C.P., there is one octahedral void corresponding to each atom that constitutes the close-packing.Ratio of Al3+: Hence, the
formula of corundum is Al2O3.
Q6) Why does urea exhibit a definite heat of fusion, while glass exhibits an indefinite heat of fusion?
Answer
Urea is a crystalline solid, whereas glass is an amorphous solid. Crystalline solids possess sharp melting points, while
amorphous solids gradually soften over a temperature range. It is for this reason that urea has a definite heat of fusion, whereas
glass does not.
Q6)-Why does urea exhibit a definite heat of fusion, while glass exhibits an indefinite heat of fusion?
Answer Urea is a crystalline solid, whereas glass is an amorphous solid. Crystalline solids possess sharp
melting points, while amorphous solids gradually soften over a temperature range. It is for this reason that urea
has a definite heat of fusion, whereas glass does not.
Q7) Calculate the number of atoms present in a Face-Centred Cubic Lattice (FCC).
Answer A Face-Centred Cubic Lattice contains 8 atoms at the corner and 6 atoms at the centre of faces.
Contribution by atoms present at the corner Contribution by atoms present at the faces
Thus, the total number of atoms in one unit cell can be calculated as:
8 corner atoms
atom per unit cell + 6 face-centred atoms
atom per unit cell
Number of atoms present in the unit cell = 1 + 3 = 4
Hence, the number of atoms present in a Face-Centred Cubic Lattice (FCC) is 4.
Q8) The density of KBr is 2.75 gcm−3. The length of the edge of the unit cell is 654 pm. Predict the type of cubic lattice to
which the unit cell of KBr belongs. (N A = 6.023 × 1023 mol−1; atomic mass of K = 39; atomic mass of Br = 80)
Answer-For cubic crystals, the density of unit cells can be calculated using the following formula:
Hence, KBr crystallizes in a face-centred cubic structure.
Q9)
a. What is the total number of voids present in 1 mole of a compound forming hexagonal close-packed structure?
b. Write the coordination number of each ion in CaF2.
Answer
a. The number of atoms in the close-packing is 1 mole i.e., 1 × 6.022 × 10 23 = 6.022 × 1023 atoms
In closed-packing:
Number of octahedral voids = Number of atoms = 6.022 × 10 23
Number of tetrahedral voids = 2 × Number of atoms in the close-packing
= 2 × 6.022 × 1023
= 12.044 × 1023
Thus, total number of voids =6.022 × 10 23 + 12.044 × 1023 = 18.066 × 1023
Hence, the total number of voids in 1 mole of a compound forming hexagonal close-packed structure is 18.066 × 1023.
b. In a molecule of CaF2, the coordination number of Ca2+is 8 and that of F−is 4.
Q10) a. If the atoms of an element in a cubic unit cell have radius r, then calculate the length of the
i. Face diagonal
ii. Body diagonal
b. When is Schottky defect observed in a crystal?
Answer
a.
In the given figure, AC represents the face diagonal and CD represents the body diagonal of a cubic unit cell.
Now,
AB = BC = AD = 2r
ΔABC is right-angled at B.
Now, ΔACD right-angled at A
b. Schottky defect is observed in a crystal when an equal number of cations and anions are missing from their
lattice sites. This is done in order to maintain electrical neutrality.
Q11)
a. AgI crystallizes in the cubic close-packed ZnS structure. What fraction of tetrahedral sites is occupied by Ag+ ions?
b. The compound AB2 possesses CaF2-type crystal structure. What is the coordination number of A2+ and B− in its crystals?
c. CsCl forms a body-centred cubic lattice. Caesium and chloride ions are in contact along the body diagonal of a cell. The
length of the side of the unit cell is 412 pm and the Cl − ion has a radius of 181 pm. Calculate the radius of the Cs + ion.
Answer
a. In a face-centred unit cell, there are eight tetrahedral voids. Out of these, half of the voids are occupied by silver cations.
b. In compound AB2, the coordination number of A is 8, whereas the coordination number of B is 4.
c.
Body diagonal, CD =
Hence, the radius of the Cs+ ion is 175.8 pm.
2.
Q1) Why is the heat of adsorption greater for chemisorption than physisorption?
Answer-Physisorption involves weak Van der Waals forces of attraction between the adsorbate and the
adsorbent. So, the heat evolved is less. On the other hand, chemisorption involves relatively stronger forces of
attraction between the adsorbate and the adsorbent. Thus, the heat evolved is much higher.
Q2)-Define the process that involves breaking of precipitates into small particles of the size of colloid.
Answer--Peptization is the process of converting a precipitate into a colloidal sol by shaking it with a dispersion
medium in the presence of small amount of electrolytes (peptizing agenst). It involves
the breaking of precipitates into small particles of the size of colloid.
Q3) What are emulsions?
Answer-Emulsions are colloidal dispersions in which both the dispersed phase and the dispersion medium are
liquids, which are either immiscible or partially miscible with each other.
Q4) Why are lyophilic colloidal sols more stable than lyophobic colloidal sols?
Answer-Lyophilic colloidal sols are more stable than lyophobic colloidal sols because the former are highly
solvated in solutions.
Q5) What happens when a beam of light is passed through As2S3 sol?
Answer-Metallic sulphide such as As2S3 acts as a negatively charged sol. When a beam of light is passed
through As2S3 sol, the path of light becomes visible. This is because, colloidal particles scatter light in all directions
in space. This effect is known as the Tyndall effect.
Q6) Define the colloids formed above Kraft’s temperature.
Answer-Some substances at low concentrations behave as normal electrolytes. But at higher concentrations,
they exhibit colloidal behaviour due to the formation of aggregates. These aggregated particles are
called associated colloids or micelles. They are formed above a certain temperature known as Kraft’s temperature.
Q7) Why is the process of adsorption always exothermic in nature?
Answer
During the process of adsorption, the residual forces on the surface of the adsorbent decreases. Consequently, the
surface energy decreases. This energy disappears in the form of heat of adsorption. Hence, adsorption is always
exothermic in nature.
Q8) Give three differences between physisorption and chemisorption.
Answer
Physisorption and chemisorption can be differenciated as follows:
Physisorption
Chemisorption
Physisortion arises because of Van der Waals forces of
This type of adsorption arises because of strong chemical
attraction.
forces.
It is not specific in nature.
It is highly specific in nature.
It is reversible in nature.
It is irreversible in nature.
It also depends upon the nature of gas.
It depends upon the nature of gas.
Gases which can react with adsorbents show
More easily liquefiable gases are adsorbed readily.
chemisorption.
The enthalpy of adsorption for physisorption is low.
The enthalpy of adsorption for chemisorption is high.
Sometimes, high activation energy is required for
No appreciable activation energy is required for physisorption.
chemisorption.
Physisorption results in multi-molecular layers.
Chemisorption results in uni-molecular layers.
Low temperature is favourable for this type of adsorption. It
High temperature is favourable for chemisorption. It
decreases with an increase in temperature.
increases with an increase of temperature.
Physisorption increases with an increase in the surface area.
Chemisorption increases with an increase in the surface
area.
Note- Any three of the above differences can fulfill the purpose
Q9) Discuss the steps involved in the adsorption theory of heterogeneous catalysis?
Answer
The adsorption theory of heterogenous catalysis involves the following steps:
i.
Diffusion of reactants to the surface of the catalyst
ii.
Adsorption of reactant molecules on the surface of the catalyst
iii.
Occurrence of chemical reactions on the catalyst’s surface through the formation of an intermediate
iv.
Desorption of reaction products from the catalyst surface (making the surface available for more reactions to occur)
v.
Diffusion of reaction products away from the surface of the catalyst
Q10)
a. Why is the sol of As2S3 prepared by the action of H2S on As2O3 negatively charged?
b. The colloidal solution of gold is red in colour. Why?
c. Name the process by which an adsorbate is removed from the adsorbent.
Answer
a. The sol of As2S3 is negatively charged because of the adsorption of S2− ions on the surface of the colloidal particles. On the
other hand, H+ ions are adsorbed in the diffused layer.
b. The colour of the colloidal solution depends on the wavelength of light scattered by its particles. The wavelength of light
further depends upon the size of particles. The finest gold sol has red colour. As the size of the particle increases, it first
becomes purple, then blue, and finally yellow.
c. The process of removing the adsorbate from the adsorbent is called desorption.
Q11)
a. Discuss the process for purification of colloidal solutions, which involves the use of a colloidion.
b. Name an experiment to confirm the existence of charge on colloidal particles.
c. Which of the given electrolytes is the most effective in the coagulation of As 2S3 sol?
AlCl3, BaCl2, NaCl
Answer
a. Ultra-filtration is the process of separating the colloidal particles from the solvent and the soluble solutes (present in the
colloidal solutions) by specially prepared filters. These ultra-filter papers are prepared by soaking the filter paper in a
colloidion solution. The filters are permeable to all substances, except the colloidal particles. It is a slow process and to speed
it up, pressure or suction is applied.
b. The existence of charge on colloidal particles is confirmed by electrophoresis experiment. In this experiment, when
electric potential is applied across two platinum electrodes, dipping in a colloidal solution, the colloidal particles move
towards one or another electrode. As soon as the colloidal particles reach the oppositely charged electrode, they get neutralized
and coagulated.
c. As2S3 is a negatively charged sol. According to Hardy-Schulze rule, the greater is the charge on the oppositely charged ion
of the electrolyte added, the more effective will be the coagulation. Hence, AlCl 3 containing the tri-positive Al3+ ions is the
most effective in the coagulation of As2S3 sol.
3.
Q1) Why does NCl3 get readily hydrolysed but NF3 does not?
Answer
In NCl3, Cl has vacant d−orbitals to accept the lone pair of electrons donated by the O−atom of H 2O molecule. On
the other hand, the F in NF3 does not have d−orbitals. Thus, NCl3 gets readily hydrolysed, whereas NF3 does not.
Q2) Draw the structure of PCl5.
Answer
PCl5 adopts the given structure.
Q3) Why is PCl5 ionic in solid state?
Answer
PCl5 is ionic in solid state because it exists as
octahedral.
in which the cation is tetrahedral and the anion is
Q4) Why is H2S acidic, whereas H2O is neutral?
Answer
The S−H bond in H2S is weaker than the O−H bond in water. This is because of the fact that the size of sulphur
atom is bigger than that of oxygen atom. Thus, H2S can dissociate easily to give H+ ions in aqueous solution.
Hence, H2S is acidic, whereas H2O is neutral.
Q5) Why is neon generally used in warning signals?
Answer
Neon is generally used in warning signals as neon lights are visible from long distances even in mist and fog.
Q6)
Explain the order of acidic strength of various oxoacids of chlorine.
Answer
The acidic strength the oxoacids of chlorine increases with increase in the oxidation number of the halogen.
Acidic strength:
This can be explained on the basis of the stabilities of the oxoanions.
Stability:
As the stability of oxoanion increases, its tendency to decompose and give O 2 decreases. Hence, the oxidising
power of oxoanion decreases.
Q7) What is the order of ionic character of metal halides?
Answer
The ionic character of metal halides is as follows:
For the same metal, metal fluoride is more ionic than metal chloride, which in turn is more ionic than metal bromide
and metal iodide. This happens because as the electronegativity of the halogen decreases from F to I, the ionic
character also decreases.
Q8) Why is it that NCl3 gets hydrolysed to form NH3 and HCl but PCl3 gives H3PO3 and HCl on hydrolysis?
Answer
N does not have d−orbitals to accommodate the electrons donated by the O of H2O. Hence, the attack of H2O takes place on
the Cl atom, which has d−orbitals, to accommodate the extra electrons denoted by H 2O. Thus, Cl−O bond is formed, which
leads to the formation of products HOCl and NH3.
On the other hand, both P and Cl have d−orbitals to accommodate electrons donated by H2O. P−O bond is comparatively
much stronger than Cl−O bond. Thus, the attack of H2O molecules takes place on P of PCl3 to form the products H3 PO3 and
HCl.
Q9)
a. Why does the addition of Cl2 to KI solution give it a brown colour but excess of Cl2 turns it colourless?
b. ClF3 exists but FCl3 does not. Why?
Answer
a. Cl2 is a stronger oxidising agent than I2. Hence, it first oxidises KI to give I2, which imparts brown colour to the solution.
If Cl2 is passed in excess, then the I2 thus formed gets further oxidized to iodic acid (KIO3), which is colourless.
b. Cl has vacant d−orbitals. Hence, it can show an oxidation state of +3. On other hand, F has no d−orbitals. Also, owing to
bigger size, Cl can accommodate three small F atoms around it. F being smaller in size cannot accommodate three large-sized
Cl atoms around it. Hence,ClF3 exists but FCl3 does not.
Q10)
a. Give reasons for the following observations.
(i) HF is liquid, while HBr, HCl and HI are gases.
(ii) The order of thermal stability of hydrogen halides varies as follows:
b. Explain the oxidising action of chlorine in aqueous solutions.
[[SLO,1.12]]
Answer
a.
(i) HF is liquid because of the presence of intermolecular hydrogen bonding.
On the other hand, hydrogen bonding is absent in HCl and HI. These compounds are held together by covalent bonds.
(ii) The thermal stability of a compound is directly proportional to the bond dissociation energy. Among the hydrogen halides,
HF has the highest bond dissociation energy, whereas that of HI is the least. This is because of the similar size of H and F. As
the size difference between hydrogen and halogen increases, the bond dissociation energy also increases. Hence, HF is the
most stable halogen acid, while HI is the least stable.
b. Chlorine, when dissolved in water, forms chlorine water. Chlorine water, on standing, loses its yellow colour because of the
formation of HCl and HClO. Hypochlorous acid in turn decomposes to form hydrochloric acid and nascent oxygen. The latter
is responsible for the oxidising as well as the bleaching properties of chlorine.
Thus, in the presence of moisture or in aqueous solution, Cl2 acts as a powerful oxidising agent.
Q11)
a. Write the chemical equation to show the hydrolysis of xenon hexafluoride.
b. Why are all bonds in PCl5 molecule not equivalent?
c. Complete the following equations.
(i)
(ii)
Answer
a. On hydrolysis, xenon hexafluoride forms xenon trioxide.
b. In gaseous and liquid state, PCl5 has a trigonal bipyramidal structure. In this structure, the two axial P−Cl bonds are longer
and less stable than the three equatorial P−Cl bonds. This is because of the greater bond pair−bond pair repulsion in the axial
bonds. Hence, all the bonds in PCl5 are not equivalent.
c.
(i)
(ii)
4.
Q1) Why do all transition elements (except mercury) form metallic bonds?
Answer
Transition elements have relatively low ionization energies and have one or two electrons in their outermost energy
level (ns1 or ns2). As a result, they form metallic bonds.
Q2) Why is the d1 configuration unstable in ions?
Answer
The d1 configuration is unstable in ions as it has a greater tendency to acquire more stable d0 configuration by
losing one d-electron.
Q3) Name the transition metals used in the following:
(i) Haber’s process
(ii) Catalytic hydrogenation
Answer
(i) Finely divided iron is used in Haber’s process.
(ii) Nickel is used in catalytic hydrogenation.
Q4) Calculate the 'spin only' magnetic moment of Fe2+.
Answer
The outer electronic configuration of Fe2+ is 3d6. There are four unpaired electrons.
So, its magnetic moment is calculated as:
Q5) Name the species obtained when V2O5 reacts with alkalies and acids.
Answer
V2O5 reacts with alkalies as well as acids to give
and
respectively.
Q6) Explain the variations in the melting points of the first row transition series elements.
Answer
The melting points of the first row transition series elements rises to a maximum and then falls as the atomic number
increases. The graphical variation in the melting points of these elements is as below:
The melting point increases upto Cr due to an increase in the number of unpaired electrons. After Cr, the number of unpaired
electrons decreases which results in a decrease in the melting point. The dip in the melting point of Mn is observed due to the
stability of exactly half-filled d-orbitals.
Q7)
Why is the separation of lanthanide elements difficult?
Answer
Due to lanthanide contraction, the change in the atomic or ionic radii of transition elements is very small. Thus, their chemical
properties are similar. This makes their separation difficult.
Q8)
Account for the following statements:
a. The difference in the ionization energies between any two successive d-block elements is less than the difference in case of
successives-block elements.
b. The order of the second ionization energy for the elements Ni, Cu, and Zn is
.
Answer
a. The addition of d-electrons in penultimate shell with an increase in the atomic number provides a screening effect and thus,
shields the outer s-electrons from the inward nuclear pull. The effects of increased nuclear charge and addition of d-electrons
tend to oppose each other. This is due to their counter effects that the ionization energies show a little variation on moving
along a period of d-block.
b. The second ionization energy falls from Cu to Zn because after the removal of one electron, Cu acquires a stable
configuration (3d10) and the removal of the second electron is difficult.
Q9)
a. If X is the last element in the series of lanthanoids, then what is the electronic configuration of X3+ ?
b. What is the composition of alloy mischmetal?
Answer
a. The last element in the series of lanthanoids is lutetium (Lu). Its atomic number is 71 and its electronic configuration is
. Thus, the electronic configuration of
.
b. The alloy mischmetal consists of a lanthanoid metal (∼95%) and iron (∼5%). It contains traces of S, C, Ca, and Al.
Q10)
a. Write the steps involved in the preparation of potassium dichromate from chromite ore.
b. Write the ionic equations involved in the reaction of potassium dichromate with:
(i) Potassium iodide
(ii) Acidified ferrous sulphate
c. Draw the structures of chromate and dichromate ions.
Answer
a. The steps involved in the preparation of potassium dichromate from chromite ore (FeCr 2O4) are:
The yellow solution of dichromate is filtered and acidified with sulphuric acid.
Sodium dichromate formed is then treated with potassium chloride.
The orange crystals of potassium dichromate crystallize out.
b.
(i)
(ii)
c.
Q11)
a. How does the atomic size vary along the first transition series elements?
b. Why do transition elements exhibit higher enthalpies of atomization?
c. Comment on the magnetic properties of the following elements:
Zn2+, Mn2+, Cr2+
Answer
a. The atomic size decreases in the beginning i.e., from Sc to Cr due to an increase in the nuclear charge. However, the
increased nuclear charge is counter balanced by the screening effect and the atomic radii become almost constant (in Mn and
Fe). The atomic radii increase towards the end due to the electron-electron repulsion, arising because of the pairing of
electrons.
b.Transition elements have very strong interatomic bonds due to their small size and also due to the presence of large number
of unpaired electrons in their atoms. As a result, they have high enthalpies of atomization.
c. The outer shell electronic configuration of Zn2+ is 3d10. In this shell, all the d-electrons are paired. Hence, it is diamagnetic.
In the case of Mn2+, the outer shell electronic configuration is 3d5. The electrons are unpaired. Hence, it is paramagnetic.
In the case of Cr2+, the outer shell electronic configuration is 3d4.. Hence, it is paramagnetic due to the presence of unpaired
electrons.
5.
Q1) Classify the following ligands as unidentate, bidentate, or hexadentate ligands.
H2O, H2NCH2 CH2NH2, EDTA4−, NH3
Answer
The given ligands can be classified as follows:
Unidentate ligands: H2O, NH3
Bidentate ligands: H2NCH2 CH2 NH2
Hexadentate ligands: EDTA4−
Q2) What is the oxidation number and coordination number of Fe in K3[Fe(C2O4)3].
Answer
The oxidation number and coordination number of Fe is + 3 and is 6 respectively.
Q3)
Write the formula for the following coordination compounds:
i. Potassium tetracyanocuperate (II)
ii. Diamminesilver (I) dicyanoargentate(I)
Answer
i.
K2[Cu(CN)4]
ii.
[Ag(NH3)2][Ag(CN)2]
Q4) What type of isomerism is exhibited by the given pair of complexes?
[Cr (SCN) (H2O)5]2+ and [Cr (NCS) (H2O)5]2+
Answer
The complexes given above contain ambident ligands. Hence, they exhibit linkage isomerism.
Q5) Predict the number of ions present per mole in the following complexes:
i.
ii.
Answer
i.
Here, 3 moles of ions are present.
ii.
Here, 2 moles of ions are present.
Q6)
A coordination compound has the formula PtCl4 . 3NH3. It does not liberate ammonia, but precipitates chloride ion as silver
chloride.
i. Write the IUPAC name and the structural formula of the compound.
ii. What is the secondary valency of the compound in part (i)?
Answer
i. The structural formula of the compound is [PtCl3 (NH3)3] Cl.
Its IUPAC name is triamminetrichloridoplatinum(IV)chloride.
ii. This compound has a secondary valency of 6.
Q5)
Predict the number of ions present per mole in the following complexes:
i.
ii.
Answer
i.
Here, 3 moles of ions are present.
ii.
Here, 2 moles of ions are present.
Q7) Write down the IUPAC nomenclature of the following compounds:
i. [CoCl2 (en)2] SO4
ii.[(NH3)5 Cr − OH − Cr (NH3)5] Cl5
Answer
The IUPAC nomenclatures of the given compounds are as follows:
i. Dichlorobis (ethylenediamine) cobalt (IV) sulphate
ii. Pentaamine chromium (III) − μ − hydroxo pentaamine chromium (III) chloride
Q8)
For the complex [Fe (CN)6]4−, predict the following:
i Type of complex and Hybridization
ii Magnetic behavior of the complex on the basis of valence bond theory
Answer
i. The atomic number of Fe is 26. Its valence shell configuration is 3d6 4s2.
Now, Fe is in +2 state in the complex [Fe (CN)6]4−.
Fe atom (Z = 26) is ground state
Fe2+ ion
Hence, the complex [Fe (CN)6]4− forms a low spin complex and is i d2 sp3 hybridized.
ii. Diamagnetic, as it does not contain any unpaired electrons.
Q9)
Account for the following statements:
a. [Sc (H2O)6]3+ is colourless while [Ti (H2O)6]3+ is coloured.
b. In octahedral crystal field splitting, the eg set of orbitals are higher in energy than the t2g orbitals.
Answer
a. The outer electronic configuration of Sc is 3d1 4s2.
In [Sc (H2O)6]3+, Sc is in + 3 oxidation state. The outer electronic configuration of Sc3+ is 3d0. Hence, there are no delectrons available for the d-d transition and the complex [Sc (H2O)6]3+ is colourless.
b. In crystal field splitting of the octahedral complexes, the eg set has two orbitals i.e., dx2-y2 and dz2 that point towards the
axes along the direction of the ligand and experience more repulsion. As a result, they are raised higher in energy than t2g set
(which has the orbitalsdxy, dyz, and dxz ), and are directed between the axes.
Q10)
a. Explain the bonding in metal−carbonyl complexes.
b. Draw the structure of the following complexes:
i. Fe (CO)5
ii. [Co2 (CO)8]
Answer
a. The metal−carbon bond (in metal carbonyl) possesses both s and p character. The metal − carbon bond is formed by the
donation of lone pair of electrons on the carbonyl carbon into a vacant orbital of the metal. The metal−carbon π bond is
formed by the donation of a pair of electrons from a filled d orbital of metal into vacant anti-bonding π* orbital of carbonyl .
The metal to ligand bonding creates a synergic effect, which strengthens the bond between CO and the metal.
b.
Q11)
a. Compare and contrast double salts and complex compounds with the help pf examples.
b. How does the magnitude of Δ0 and P decide the actual configuration of d orbital in a d6 ion?
c. Give one limitation of the crystal field theory.
Answer
a. Both double salts and complexes are formed by the combination of two or more stable compounds in stiochiometric ratio.
They differ in the facr that double salts such as Mohr’s salt, FeSO4. (NH4)2. SO4. 6H2O dissociate into simpler substances or
ions completely when dissolved in water as:
On the other hand, in complexes, the ions do not dissociate. For example,
K4 [Fe (CN)6] does not dissociate into Fe2+ and CN−.
b. For a d6 ion, if Δ0 < P, then the first three electrons enter the t2g orbital. The fourth and fifth electrons enter the eg orbital
and the sixth enters the t2g orbital, giving configuration
.
On the other hand, if Δ0 < P, then all the six electrons enter the t2g orbital, giving configuration
.
c. Limitation of the crystal field theory:
i. The crystal field theory assumes the ligands as point charges. It follows that anionic ligands should exert the greatest
splitting effect. But, these were found at the low end of the spectrochemical series.
6.
Q1) Write the structure of D − (+) − Glyceraldehyde.
Answer
The structure of D − (+) − Glyceraldehyde can be represented as:
Q2) Give a reaction to confirm the presence of carbonyl group as aldehydic group in a glucose molecule.
Answer
The carbonyl group in a glucose molecule undergoes oxidation in the presence of Br 2 water to form gluconic acid.
Q3) Which of the following amino acids are essential amino acids?
Lysine, Tyrosine, Valine, Aspartic acid, Arginine
Answer
The essential amino acids are Lysine, Valine, and Arg
Q4) How many peptide linkages are present in Gly-Ala-Val?
Answer
Two peptide linkages are present in Gly-Ala-Val.
Q5) Show the linkage of β-D-Glucose in cellulose?
Answer
The following diagram illustrates the linkage of β-D-Glucose in cellulose.
.
Q6) How many −OH groups are present in a glucose molecule. Give the reaction to confirm the number of −OH groups
present?
Answer
There are five −OH groups present in a glucose molecule.
Glucose, on acetylation with acetic anhydride, gives glucose pentacetate which confirms the presence of five −OH
groups.
Q7) Why is sucrose also known as 'invert sugar'?
Answer
Sucrose is dextrorotatory, but after hydrolysis, it gives dextrorotatory glucose and laevorotatory fructose. The
resultant mixture is laevorotatory since the laevorotation of fructose (−92.4°) is more than dextrorotation of glucose
(+52.5). So, hydrolysis of sucrose brings about a change in the sign of rotation and therefore, it is also known as'
invert sugar'.
Q8)
Give the products formed when glucose reacts with the following:
a. Hydroxylamine
b. Hydrogen cyanide
c. Hydroiodic acid
Answer
a.
b.
c.
Q9)
a. What are anomers?
b. Show the formation of ring forms of glucose through the open chain form.
Answer
a. A pair of stereoisomers which differ in configuration around C-1 are called anomers.
b.
Q10)
a. A base ‘B’ is only found in RNA and not in DNA. When this base in attached to 1′ position of sugar moiety present in
RNA, then a unit ‘C’ is formed. On linking phosphoric acid at 5′ position of ‘C’, a unit ‘D’ is obtained?
Identify B, C, and D?
b. What happens when two units of ‘D’ are joined between 5′ and 3′ carbon atoms?
Answer
a. The base uracil is only found in RNA.
Hence, B is uracil.
The sugar moiety present in RNA is β-D-2-deoxy ribose. When the base uracil is attached to 1′ position of the sugar moiety, it
forms a nucleoside. Hence, C is a nucleoside.
When nucleoside is linked to phosphoric acid at 5′ position of the sugar moiety, a nucleotide is formed. Hence, D is
a nucleotide.
b. Nucleotides are joined together by phosphodiester linkage between 5′ and 3′ carbon atoms, forming dinucleotide.
Q11)
a. Briefly discuss the primary, secondary, and tertiary structures of proteins.
b. Explain denaturation of proteins.
Answer
a.
i. Primary structure of proteins: Each polypeptide chain in a protein has amino acids linked with each other in a specific
sequence. This sequence of amino acids is said to be the primary structure of proteins.
ii. Secondary structure of proteins: The secondary structure of proteins refers to the shape in which a long polypeptide can
exist.
The two different secondary structures possible are α-helix structure and β-pleated sheet structure.
α-Helical structure: In α-Helix structure, a polypeptide chain forms all possible hydrogen bonds by twisting into a helix with
−NH group of each amino acid residue and hydrogen bonded to >C=O of an adjacent turn of helix.
β-pleated structure: In a β-pleated structure, all peptide chains are stretched out to nearly maximum extensions and then laid
side by side which are held together by intermolecular hydrogen bonds.
Tertiary structure of proteins: The tertiary structure of proteins represents an overall folding of polypeptide chains. The
main forces which stabilize the secondary and tertiary structures are hydrogen bonding, disulphide linkage, Van der Waals,
and electrostatic forces of attraction.
b. When a protein, in its native form, is subjected to physical changes like change in temperature or chemical change like
change in pH, then the hydrogen bonds are disturbed. Due to this, globules unfold, helix get uncoiled, and protein looses its
biological activity. This is called denaturation of proteins.
7.
Q1) How is teflon obtained through polymerization reaction?
Answer
Teflon is obtained by polymerizing tetrafluoroethene.
Q2) Which polymer is used in the manufacturing of unbreakable crockery?
Answer
Melamine-formaldehyde polymer is used in the manufacturing of unbreakable crockery.
Q3) Between rayon and nylon, which polymer is a semi-synthetic polymer? Give reasons for your answer.
Answer
Rayon is a semi-synthetic polymer as it is derived from naturally occurring polymers by chemical modification,
while nylon is a synthetic polymer.
Q4) Name the catalyst used in the formation of high density polyethene.
Answer
Ziegler-Natta catalyst is used in the formation of high density polyethene.
Q5) Which of the following polymers has the maximum elastic property?
Polyethene, Nylon 6,6, Buna − N
Answer
From the given polymers, Buna − N has the maximum elasticity.
Q6)
a. What are the monomer units of Buna − N?
b. Give one property and one use of Buna − N.
Answer
a. The monomer units of Buna − N are H2C = CH − CH = CH2 (1, 3-Butadiene) and
b. Property of Buna − N:
It is resistant to the action of petrol, lubricating oil, and organic solvents.
Use of Buna − N:
It is used in making oil seals.
Q7)
What are polyesters? Explain the formation of any one polyester.
Answer
Polyesters are polycondensation products of dicarboxylic acids and diols.
For example: TeryleneThey are manufactured by heating ethylene glycol and terephthalic acid.
.
Q8)
a. Write the mechanism involved in the polymerization of ethene molecule.
b. Name two more polymers which can be prepared by this mechanism.
Answer
a. The following steps are involved in the polymerization of ethene:
i.
Chain initiation step:- In this step, the benzoyl peroxide initiator generates a free radical which adds to the ethene double
bond, generating new and larger free radicals.
ii.
Chain propogation step:
The free radical formed in step (i) reacts with another molecule of ethene forming bigger- sized radicals.
iii.
Chain terminating step:
b. Polyacrylonitrile and polystyrene can be prepared by this method.
Q9)
a. What is meant by condensation polymerization?
b. Name the polymer used in the making of bristles for brushes.
c. Give the reaction involved in the formation of the polymer in part (b) from its monomer unit?
Answer
a. Condensation polymerization involves the reaction between two different bi-functional or tri-functional monomeric units
with the elimination of small molecules such as water, alcohol, etc. For example: Terylene
b. Nylon 6,6 is used in the making of bristles for brushes.
c. Nylon 6,6 is prepared by condensation polymerization of hexamethlylenediamine with adipic acid.
Q10)
a. Give the structure of natural rubber.
b. Write two physical properties of natural rubber.
c. Explain the process carried out to improve the physical property of natural rubber.
d. Explain the preparation and the use of a synthetic rubber formed by free radical polymerization reaction.
Answer
a.
b. Physical properties of natural rubber are:
i.
It becomes soft at high temperature (> 335 K) and brittle at low temperature (< 283 K).
ii.
It is soluble in non-polar solvents and is non-resistant to attack by oxidizing agents.
c. A process of vulcanization is carried out to improve upon the physical properties of natural rubber. In this process, raw
rubber is heated with sulphuric acid at a temperature range between 373 K to 415 K. On vulcanization, sulphur forms cross
links at reactive sites of double bonds and thus, the rubber gets stiffened. The probable structure of vulcanized rubber is:
d. Neoprene is the synthetic rubber formed by the free radical polymerization of chloroprene.
Uses of Neoprene:- It is used in the manufacturing of conveyor belts, gaskets, and hoses.
Q11)
a. What was the need for developing biodegradable polymers?
b. What does PHBV stand for? Give its preparation and uses.
c. Write the monomer units of Nylon 2-Nylon 6?
Answer
a. A large number of polymers are quite resistant to the environmental degradation process and are thus, responsible for the
accumulation of polymeric solid waste materials. These solid wastes cause acute environmental problems and remain
undergraded for quite a long time. In view of these problems, biodegradable synthetic polymers have been developed.
b. PHBV stands for poly β-hydroxybutyrate-co-β-hydroxyvalerate.
It is obtained by the co-polymerization of 3-hydroxybutanoic acid and 3-hydroxypentanoic acid.
It is used in speciality packaging, orthopaedic devices, and in controlled release of drugs.
c. Monomer units of Nylon 2-Nylon 6 are glycine (H2N − CH2 − COOH) and amino caproic acid [H2N − (CH2)5 − COOH].
8.
Q1) Define the term antagonists.
Answer
Drugs that bind to the receptor site and inhibit its natural function are called antagonists.
Q2) What is the name of the chemical that communicates messages between two neurons?
Answer
Chemical messengers communicate messages between two neurons.
Q3) Name a drug which is used in controlling depression and hypertension.
Answer
The drug which is used in controlling depression and hypertension is equanil.
Q4) Draw the structure of drug veronal.
Answer
The Structure of drug veronal can be represented as:
Q5) Name the medicine developed by Paul Ehrlich for the treatment of syphilis.
Answer
Salvarsan is the medicine developed by Paul Ehrlich for the treatment of syphilis.
Q6) Mention two functions performed by enzymes in their catalytic activity?
Answer
The following functions are performed by enzymes in their catalytic activity:
i. To hold the substrate for a chemical reaction: Active sites of enzymes hold the substrate molecules in a suitable
position so that it can be attacked by reagents effectively.
ii. To provide functional groups that will attack the substrate and carry out chemical reactions.
Q7)
a. Name two antihistamines.
b. How do antihistamines function?
Answer
a. Brompheniramine and terfenadine are two antihistamines.
b. Antihistamines interfere with the natural action of histamine. They compete with histamine for binding sites of
receptors, where the latter exerts its effect.
Q8) Discuss the basis on which drugs are classified.
Answer
Drugs are classified in the following ways:
i. On the basis of pharmacological effect:- This classification is based on the pharmacological effect of the drugs. It is useful
for doctors because it provides them the whole range of drugs available for treatment for a particular type of problem.
ii. On the basis of drug action:- It is based on the action of a drug on a particular biochemical process.
iii. On the basis of chemical structure:- It is based on the chemical structure of the drug. Drugs classified according to this
category have common structural features and often exhibit similar pharmacological activities.
iv. On the basis of molecular targets:- Drugs usually interact with target molecules. Drugs possessing some common
structural features may have the same mechanism of action on targets.
Q9)
a. Write the name and structure of the drug which was designed to prevent the interaction of histamine with stomach wall
receptors.
b. What are the two major functions of histamine?
Answer
a. Drug cimetidine was designed to prevent the interaction of histamine with stomach wall receptors.
Structure of cimetidine:
i.
b. Histamine has the following functions:
It contracts the smooth muscles in the bronchi and gut, and relaxes other muscles.
ii.
It is responsible for the nasal congestion associated with common cold and allergic responses to pollen.
Q10)
a. Write the chemical equation for the preparation of sodium stearate.
b. Why cant soaps form lather easily in hard water?
c. Give two differences between soaps and synthetic detergents.
Answer
a.
b. Hard water contains calcium and magnesium ions.
These ions form insoluble calcium and magnesium soaps (respectively) when sodium or potassium soaps are dissolved in hard
water.
These insoluble soaps separate as scum in water and makes washing difficult.
c.
Soap
Synthetic detergent
i.
Soaps are sodium salts of higher fatty acids.
Synthetic detergents are sodium alkyl sulphates or
sodium alkyl benzene sulphonates with alkyl groups
having more than ten carbon atoms.
ii.
Soaps form insoluble salts with calcium and
magnesium in hard water. Hence, they
cannot be used in hard water.
Calcium and magnesium salts of detergents are
soluble in water. Hence, they can be used even in
hard water.
Q11)
a. Discuss two ways by which drugs inhibit the attachment of substrate on the active site of enzymes.
b. Diagrammatically, represent how chemical messengers give messages to cells.
Answer
a. Drugs inhibit the attachment of substrate on the active site of enzymes in the following two ways.
i.
Drugs compete with the natural substrate for their attachment on the active sites of enzymes. Such drugs are competitive
inhibitors.
ii.
Some drugs do not bind to the enzymes active site. They bind to a different site of enzymes, called allosteric site. This
binding of the inhibitor at allosteric site changes the shape of the active site in such a way that the substrate cannot recognize
it.
b. Chemical messengers communicate the message between two neurons or between neurons and muscles. To accommodate a
messenger, the shape of the receptor site changes. This brings about the transfer of messages to the cells.