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1. 1. Name the radiation with their corresponding frequency or wavelength range
associated with the given spectroscopic methods .
i. FT-IR
ii. NMR
iii. XRD
iv. UV-Visible
v. Raman spectroscopy
2. The best spectroscopic method for identification of hydroxyl group (OH
group) in a compound is
i. UV-Visible
ii. SEM
iii. EDS
iv. IR
3. A chiral molecule can be identified from achiral molecule by
i. IR
ii. Optical rotation
iii. Mass spectroscopy
iv. Lewis structure
4. Burning of wood is a
i. Chemical reaction
ii. Nuclear reaction
iii. A physical change
5. Flash point of kerosene is
i. Higher than petrol
ii. Lower than petrol
iii. Same as that of petrol
6. Identify the synthetic polymer
i. Rubber
ii. Cellulose
iii. Teflon

2. iv. Asphalt
7. Asphalt (also known as slime, pitch, bitumen, asphaltum, naphtha) is used
on a surface for waterproof paints (slime – very fluid), while pitch (thicker) would
be packed as mortar into cracks to stop leaks. The ancient Mesopotamians used it
to waterproof temple baths, baskets and water tanks. The hardness of asphalt
depend on the amount of trapped volatile gases. This black or dark brown form of
petroleum which melts and flows readily with heat Is an example of
i. Hydrocarbon composition
ii. Polymer material
iii. Inorganic material
iv. A molecule with long chain alcohols
8. paper glue or carpenter’s glue is poly(vinyl acetate) and easily hydrolysed by
base to give poly(vinyl alcohol), also called
i. poly(ethenyl alcohols) - answer
ii. poly(vinyl alcohol-co-vinyl acetate) - copolymer of poly(vinyl alcohol) and
poly(vinyl acetate)
iii. poly propyl alcohol
iv. vinyl acetate
9. Cyanoacrylates are used as super glue`. In presence of moisture methyl
cyanoacrylate form close-fitting chains in between the two surfaces in contact .
The mechanism of the adhesive property is
i. Polymerization
ii. Condensation
iii. Elimination
iv. addition
10. Usually esters are reduced to alcohols using LiAlH4, In a Chinese factory,
more cheaper complex of AlCl3/NaBH4/DME is used as alternative in the step .

3. During the first industrial batch (150 kg batch size) the reactor blow up, causing
fire.
This is an example of
i. Reduction
ii. Oxidation
iii. Disproportionation
iv. Addition
The cause for explosion is
i. NaBH4 form dangerously explosive hydrogen gas on contact with alkali,
water and other protic solvents
ii. Al(BH4)3 formed ignite spontaneously when exposed to air
iii. Na metal reacts with air violently
iv. NaCl reacted with moisture in exothermic manner
11. The scanning electron microscope ( SEM ) generates images of the
sample surface by scanning it with a high -energy electron beam in a raster scan
pattern . The electrons interact with the sample atoms producing secondary
electrons, backscattered electrons, x-Rays hence probing the sample's surface
topography, composition and other properties such as electrical conductivity.
Improved modification of SEM is Energy Dispersive X-ray Spectroscopy (EDS),
it studies a sample through interactions between electromagnetic radiation and
matter, analyzing x -rays emitted by the matter in response to being hit with
charged particles. Each element has a unique atomic structure permitting x -rays
that are distinctive of an element's atomic structure to be identified distinctively
from each other. Why does each element has a its own unique atomic structure ?
i. Due to unique number of protons and any neutrons in the nucleus.
ii. Due to unique atomic number
iii. Due to unique quarks present
iv. Due to different electronegativities of elements

4. The EDS system makes the spectrum correlating the energy of the x-Rays and
records the number of signals for specific energy.
Higher atomic number cause more interactions of electrons with the atom. So the
number of x-rays coming from the specimen increases with the atomic number of
the atom.
wavelength dispersive X-ray spectrometry (WDS) enables quantitative analysis of
sample below the minimum detection limits of SEM-EDS.
Beer Lambert’s law connects
a) Reflected radiation and concentration
b) Scattered radiation and concentration
c) Energy absorption and concentration
d) Energy absorption and reflected radiation
12. Give the hybridization of the heteroatoms found in the molecules
i. Saturated Amine
ii. R-N=R,
iii.
iv. R-O-R
v.
13. The term chromatography generally stands for a technique which separates
mixtures on the basis of different dynamic distribution of their components
between two distinct physico-chemical environments called mobile and stationary
phase The mobile phase is passed or forced over a stationary phase which is fixed
in a column or on a solid surface. The components of the sample distribute
themselves in the mobile and stationary phase to a different extents. Thus, the
components that are not strongly held by the stationary phase move faster down the
column than those which are retained by it. This difference in migration rates
through the column results in discrete bands for sample components .
i. Which of the following cannot be used as adsorbent in Column adsorption
chromatography?
a) Magnesium oxide
b) Silica

5. c) Alumina
d) Mohr salt
ii. In Thin layer chromatography, the stationary phase is made of _________
and the mobile phase is made of _________
a) Solid, liquid
b) Liquid, liquid
c) Liquid, gas
d) Solid, gas
iii. Amino-acid analysis can be done efficiently using
a) Gas chromatography
b) Ion exchange chromatography
c) Paper electrophoresis
d) affinity chromatography
14. Why is infrared analyzers not used commonly in the analysis of inorganic
compound
i. no absorption found
ii. Less accurate results occur
iii. Strong absorption of IR radiation by water
iv. Unimportant range
15. In H2O2, extra stabilization occurs with overlap of filled lone pair of O to
vacant AMBO σ* of O-H. this occurs in anti-position.
Account for the following observation

6. i. In axial form overlap of vacant σ* C-N with filled lone pair of N (with axial
t-Bu) occur.
ii. Lone pair obtain less repulsion in equatorial position
iii. T-butyl group in axial position is sterically less demanding due to 1,3 diaxial
interaction.
16. Account for In the IR spectrum of C-H bond is much weaker in aldehyde
i. Overlap between lone pair of O and σ* C-H , (which are much closer than
in σ* C-H overlap of σ C-C in alkene.)
ii. Electron withdrawing effect by alkyl group in aldehyde
iii. O form triple bond with C , hence weakening C-H bond in aldehyde
17. A molecule has following degrees of freedom. Which degree give no
spectroscopic technique ?
i. Translational
ii. Rotational
iii. Vibrational
iv. Electronic energy
v. Nuclear energy
18. Why is …….. degree of freedom, not used in spectroscopic technique
i. Due to low intensity
ii. Due to lack of interacting energy sources
iii. Energy spacing associated is very small quantity compared to thermal
energy 10-22
J

7. 19.In March 2007, Mexico, the USA's primary source for D-methamphetamine,
closed a chemical company accused of illegally importing more than 60 tons
of pseudoephedrine, a precursor of the drug.
Suggest suitable reagents for both reactions. Please avoid radical
halogenations.
(R-OH + HBr  R-Br , NaBr + H2SO4 , PX3 , red P/X2 also possible
R-Br + LiAlH4  R-H, )
20.Sherlock Holmes investigated a case involving D-methamphetamine, where
Jim Moriarty had flasks of 1-chloropropan-2-one, benzene, anhydrous
aluminium chloride, methylamine and hydrogen gas. Help him solve the
reaction involving reagents, mention the reactions used and give structures
of all chemical species involved.
Answer
21.Pure water is a
Weak electrolyte
Strong electrolyte
Not an electrolyte
22. Indigo is produced from indican , which occurs in Indigofera species, . the
indigo plant Indigofera tinctoria L.

8. Identify reactions a, b,c from the choices given
i. Hydrolysis
ii. Tautomerism
iii. Oxidation
iv. Reduction
v. Elimination
23.Identify the UV spectrum of indigo
i.

9. ii.
iii.
24.Atracurium is a neuromuscular blocking agent with nondepolarizing action.
It degrades non-enzymatically into innocuous metabolites under
physiological conditions.
This is an example of

10. i. 1,2 elimination
ii. 1,1 elimination
iii. 1,3 elimination
25.Coca-Cola European Partners use a very small quantity of phosphoric acid in
some of the Coca-Cola system’s soft drinks for their tangy flavor. The vast
amount of sugar acts to mask and balance the acidity. Give reaction
occurring on adding coco-cola to rusted iron
Fe₂O₃ + 2H₃PO₄ yields 2FePO₄ + 3H₂O
26. Black Ferric Phosphate- is used as a ‘Rust converter’.( naval jelly.).
If coating is left on, it will keep providing further corrosion resistance.
Account for the reason
27.Arrange according to increasing acidity
i. Sulphuric acid
ii. Acetic acid
iii. Phosphoric acid
iv. Hydrochloric acid
28.Arrange given molecules in increasing order of acidity (taken in 1 mM
solution )
i. aqueous solutions of aspirin (acetyl salicylic acid) pH 2.62.
ii. citric acid ((2-hydroxy-1,2,3-propane tricarboxylic acid) pH 3.2
iii. lactic acid (2-hydroxypropanoic acid) pH 3.51
iv. acetic acid pH 3.91
v. ascorbic acid pH 3.59 (Vitamin C)
vi. cinnamic acid pH 3.76 (from oil of cinnamon)
vii.viii.
i.ii.
29.The sugary food in mouth is degraded by bacteria , creating an acidic
medium. When pH decreases than 5.5, acidic medium can corrode enamel
(demineralize). Saliva (alkaline pH – 7.4) neutralizes some acid. But some
acid remains. Arrange given molecules in increasing basicity, so as to rinse
the mouth with mild base.
i. Aluminum hydroxide (insoluble, so donot trigger secretion of excess
acid by the stomach) pH 7.7
ii. Colgate with fluoride ion pH 7.62

11. iii. Neem stick (best otion to clean teeth) pH 6.9
iv. fallen neem leaves pH 8.2
v. Peroxides (hydrogen peroxide or carbamide peroxide) active
ingredient in most tooth-whitening agents pH 5.0-4.0
vi. sodium hydrogen carbonate (baking soda) 8.27
30.The difference in the electronegativity of two atoms determines their bond
type. Based on electronegativity differences, mention the type of bonds
formed.
i. Hydrogen (2.0) oxygen ( 3.5)
ii. Sodium (0.9) chlorine (3.0)
iii. Hydrogen (2.0) fluorine (4.0)
iv. Cesium (0.8) fluorine (4.0)
v. Carbon (2.55) Hydrogen (2.0)
vi. Cesium (0.8) iodide (2.66)
31.The molecule which donot show hydrogen bonding is
i.
ii.
iii.
J Nat Prod. 2012 Aug 24;75(8):1441-50. doi: 10.1021/np300341z.
32.Gold is the least reactive of coinage metals, with only one electron in its
outermost shell. [Xe] 4f14 5d10 6s1
Being heavier metal, it’s relativistic electrons gain mass and as a result, their
orbits contract. Unlike d and f orbitals, s orbitals, and to a smaller degree p

12. orbitals, have an significant electron density near the nucleus. 6s orbital with
one electron is contracted, this electron is more tightly bound to the nucleus,
hence have high ionization energy 890 kJ mol.
Why is gold yellowish coloured, while other metals in silvery colour.?
i. Gold has inert pair effect
ii. Gold has s and p orbitals near nucleus , hence emit yellow colour in
energy region
iii. Fairly low energy difference between 6s and 5d, so reflects light in
low energy region
33.A blackbody is a bounded body that absorbs all incident radiation upon
it.and at equilibrium, it will emit radiation at the equal rate as it absorbs
energy from the surrounding medium.
Which one is better black body.
i. Star
ii. Hydrogen Atom
iii. Sodium atom
iv. Soot from flames (emissivity of soot does depend on temperature-
Soot, carbon black, platinum black, and carborundum come closest to
having blackbody properties)
34.Why is copper sulfate is a blue colour, but zinc sulfate is a white compound
i. Zn d orbitals are completely filled , electron to make a d-d transition
as they are all filled up
ii. Copper sulphate has water of hydration, hence blue colour, but zinc
sulphate donot.
iii. Copper sulphate has vacant electron in them, emiiting blue colour
When the metal starts bonding with other ligands,
degenerate partially filled (n-1)d orbitals split apart and become non-degenerate
(have different energy levels), due to different symmetries of the d orbitals and the
inductive effects of the ligands on the d electrons

13. 35. purple colour KMnO4 and reddish yellow colour of K2Cr2O7 is due to
i. charge transfer spectra (both metal ions have no d electrons)
ii. d-d transition
iii. free electron present
iv. water of hydration
If the ligand molecular orbitals are full, charge transfer may occur from the ligand
molecular orbitals to the empty or partially filled metal d-orbitals. The absorptions
cause ligand-to-metal charge-transfer bands (LMCT) in KMnO4
Similarly a low oxidation state (electron rich) metals and the ligand possesses
low-lying empty orbitals with π* (e.g., CO or CN−) form a metal-to-ligand charge
transfer (MLCT) transition .
36. Organic molecules contain bonding orbitals (σ & π) and non-bonding
(lone pair) orbitals (usually designated the letter n and have anti-bonding
molecular orbitals (σ*
& π*
).
the colour corresponds to the wavelength of visible light emitted by the
compound, rest is absorbed for electron excitation in relevant energy levels.
Identify the coloured molecules, mentioning possible electronic transitions
possible
a.

14. b.
ii.
iii.

15. 1.
i. FT-IR – uses infrared region (750 nm – 1mm or 102-
104
cm-1
or 3x1011
- 3.9x1014
Hz), to study vibration change in bond length and bond angle keeping the center
on mass constant) and change in dipole moment of molecule. Eg. hydrocarbons
ii. NMR – uses radiowaves ( above 0.3 m or below 3000 cm-1
– 1 cm-1
or below
1GHz, 109
Hz), to study nuclear energy levels (based on spin angular momentum
of spin in nuclei) eg. ketones
iii. XRD - uses X-rays , to relative molecular structure of a crystal (3.4x1016
- 5x1019
Hz or 0.1 -10 nm) and arrangement of atoms based on diffraction pattern of
incident X-rays. Eg. Tartaric acid crystals
iv. UV-visible – uses UV (7.9x1014
- 3.4x1016
Hz or 10-400 nm) and visible region
(3.9x1014
- 7.9x1014
Hz or 400-750 nm) to study electronic energy levels and their
excitations. Eg. porphyrins
v. Raman spectroscopy – uses laser radiation in the visible (3.9x1014
- 7.9x1014
Hz
or 400-750 nm) or near-infrared (2.1x1014
- 4x1014
Hz or 1100-780 nm) studies
molecular vibration causing a change in polarizability (distortion of the electron
cloud). Eg. alcohols
2. IR gives O-H stretch of non H bonded or free hydroxyl groups – 3700-3584 (sharp), O–
H stretch, hydrogen bonded 3550-3200 (broad), O-H wag (out of plane ) 600-700
(confirms presence of alcohol), O-H bend in plane 1420-1330
3. Polarimeter
i. glucose is converted to carbondioxide and water (gas), Na and K to their respective
carbonates and oxides ( Emmanuel A. Okunade , 2008. The Effect of Wood Ash and

16. Sawdust Admixtures on the Engineering Properties of a Burnt Laterite-Clay Brick.
Journal of Applied Sciences, 8: 1042-1048. )
4. Flash point of kerosene is near 65 °C (so used in aviation), of petrol is −43 °C (relatively
low flash point , so used in piston engines of vehicles )
5. Teflon
Rubber – has isoprene (2-methyl-1,3-butadiene) units
Cellulose – has Beta D glucose monomer
6. Asphalt is of Hydrocarbon composition
7. Acetate (conjugate base of acetic acid)
8. Polymerization
9. Reduction
i. The cause for explosion is Al(BH4)3 formed ignite spontaneously when exposed to air
10. Due to unique atomic number
Beer Lambert’s law connects Energy absorption and concentration
.
11. .
a. Sp3 N
b. Sp2 N
c. Sp2 N
d. Sp3 O
e. Sp2 O
12. .
i. Mohr salt
ii. Solid, liquid

17. iii. Ion exchange chromatography
iv. Thin layer chromatography
v. Gas chromatography
vi. affinity chromatography
13. no absorption found
14. Overlap between lone pair of O and σ* C-H , (which are much closer than in σ* C-H
overlap of σ C-C in alkene.)
15. Translational
16. Energy spacing associated is very small quantity compared to thermal energy 10-22
J
17. Weak electrolyte
18. Reaction a- ester hydrolysis
Reaction b- keto-enol tautomerism
Reaction c – oxidation- removal of H2
19.
Shows no absorption in 400-500nm (blue- violet) and is emitted
20. Fe₂O₃ + 2H₃PO₄ yields 2FePO₄ + 3H₂O
21. FePO₄ coating on metal surface protects it from atmospheric oxygen, by masking it.

18. 22. Sulphuric acid > Hydrochloric acid> Phosphoric acid> Acetic acid
23. acetic acid pH 3.91< cinnamic acid pH 3.76 (from oil of cinnamon)< ascorbic acid pH
3.59 (Vitamin C)< lactic acid (2-hydroxypropanoic acid) pH 3.51 < citric acid ((2-
hydroxy-1,2,3-propane tricarboxylic acid) pH 3.2< aqueous solutions of aspirin (acetyl
salicylic acid) pH 2.62.
24. Peroxides (hydrogen peroxide or carbamide peroxide pH 5.0-4.0 < Neem stick pH 6.9 <
Colgate with fluoride ion pH 7.62 < Aluminum hydroxide pH 7.7 < fallen neem leaves
pH 8.2 < sodium hydrogen carbonate (baking soda) 8.27
25.
i. Hydrogen (2.0) oxygen ( 3.5) – polar covalent bond
ii. Sodium (0.9) chlorine (3.0) – ionic bond
iii. Hydrogen (2.0) fluorine (4.0) – polar covalent bond (H-F bond strong ,so donot
form ions in aqueous medium, H-F bond so not ionic bond)
iv. Cesium (0.8) fluorine (4.0)- ionic bond
v. Carbon (2.55) Hydrogen (2.0) – nonpolar covalent
vi. Cesium (0.8) iodide (2.66) – ionic bond
26. .
27. Fairly low energy difference between 6s and 5d, so reflects light in low energy region
28. Star
29. Zn d orbitals are completely filled , hence no electron to make a d-d transition as they are
all filled up
30. charge transfer spectra (both metal ions have no d electrons)
31.