ENGINEERING CHEMISTRY DEPARTMENT OF S&H

UNIT-5

SPECTROSCOPIC TECHNIQUES AND APPLICATIONS

ANALYTICAL CHEMISTRY

· Analytical chemistry is concerned with the identification of a substance, the elucidation of its structure and quantitative analysis of its composition.

· It is an interdisciplinary branch of science which deals with various disciplines of chemistry such as inorganic, organic, physical, industrial and biochemistry

INSTRUMENTAL METHODS

The methods dependent upon measurement of an electrical property and those based upon determination of the extent to which radiation is absorbed or upon assessment of the intensity of emitted radiation, all require the use of a suitable instrument, e.g. Polarograph, spectrophotometer etc., and in consequence such methods are referred to as instrumental methods.

The growth of instrumental methods (or) analysis is related to the developments in the field of electronics, so instrumental methods of chemical analysis have become the backbone of experimental chemistry.

ANALYTICAL TECHNIQUES –SPECTROSCOPY

Spectroscopy is the branch of science dealing with the study of interaction of electromagnetic radiation with

matter.

It is a powerful tool available for the study of atomic and molecular structure and it is also used in the

analysis of most of the samples.

Types of Spectroscopy:-

The study of spectroscopy is divided into two types. They are,

1. Atomic spectroscopy

2. Molecular spectroscopy.

Atomic Spectroscopy

It deals with the interactions of electromagnetic radiation with atoms.

Molecular Spectroscopy

It deals with the interaction of electromagnetic radiation with molecules.

Differences between atomic spectra and molecular spectra

Atomic Spectra Molecular Spectra

1 It occurs from the interaction of atoms and electromagnetic radiation.

It occurs from the interaction of molecules and electromagnetic radiation

2. It is a line spectra It is a complicated spectra.

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3.It is a due to electronic transition in an elementIt is due to vibration, rotational and electronic transition in a molecule

Electromagnetic radiation (or) Electromagnetic energy (or) Radiant energy

Electromagnetic radiation is produced due to the interaction between the electric field and the magnetic field.

Electromagnetic radiation is divided into number of regions according to their wave length.

It can be represented as, λ (lambda)

Wave length (λ)

It is the distance between two successive crests of a wave.

It is denoted by ‘λ’. Wave length is expressed in meters and to express very short wave

lengths nanometre (nm,10-9m), picometer (pm,10-12m) or ‘non SI’ Units Angstrom (Ao,10-10m)

λ

1 second

Frequency (ע)

It is the number of waves crossing a point in unit time. It is denoted by, ע

Frequency is expressed in s-1(or) Hertz, (Hz).

Velocity of light ( c )

The product of the wavelength and the frequency is a constant called the velocity of light

(or) speed of light. That is,

c = λ

Where,

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c - speed of light;

;frequency - ע

λ - wave length

c = 2.998 X 108 ms-1

c = 3 X 108 ms-1

Wave number

It is the reciprocal of the wave length. It is denoted by ע

Wave number is expressed in units of per centimeter,

Electromagnetic spectrum (EMS)

The arrangement of all types of electromagnetic radiations in the increasing order of their wavelength or decreasing order of their frequency is known as ‘Electromagnetic Spectrum’.

S. No. Type of radiation

Wave

Length Frequency

1 Gamma rays 10-11 1019

2 X ray 10-9 10-17

3 UV 10-7 1015

4 Visible 10-6 1014

5 IR 10-5 1013

6 Micro waves 10-3 1011

7Radio waves (low energy) 102 106

PRINCIPLE OF ELECTRONIC SPECTROSCOPY

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When a beam of polychromatic light is passed through a prism or grating it splits up into seven different colour. The set of colours thus obtained is called a spectrum.

The complete spectrum may extend from gamma rays of wave length 10-13 m to radiowaves of wavelength 105m.

Classification of spectra

There are two main classes of spectra namely,

1. Absorption Spectrum.

2. Emission Spectrum.

Absorption spectrum

When white light (ie.having all the wave length) is passed through an absorbing substance and then observed through a spectroscope, it is found that certain colours (or) wave lengths are missing and dark lines appear at their places. The spectrum so obtained is called absorption spectrum.

Emission spectrum

When the light emitted by a substance is passed through a prism and examined directly with a spectroscope, the spectra obtained is referred to as emission spectrum.

Absorption Laws

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These are,

There are two laws which govern the absorption of light by the molecules.

1. Lambert’s law

2. Beer’s law.

Lambert’s Law

When a beam of monochromatic light is passed through a solution of anabsorbing substance, the rate of decrease of intensity of radiation (‘dI’) with thickness of the absorbing solution (‘dx’) is directly proportional to the intensity of incident radiation (I). Mathematically, the law is expressed as,

Where,

k = absorption co-efficient.

On integrating the above expression between the limits,

I = Io at x = 0 and I = I at x = l

We get

-ln I/ Io = Kl

ln Io/ I = Kl

Kl = ln Io/ I

By taking natural logarithm,

Kl = 2.303 log Io / I

Beer’s Law

When a beam of monochromatic light is passed through a solution of an absorbing substance, the rate of decrease of intensity of radiation (‘dI’ ) with thickness of absorbing solution (‘dx’) is directly proportional to the intensity of incident radiation (I), as well as concentration of the solution( c).

Mathematically, it is expressed as

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Where,

K = Absorption co-efficient.

On integrating the above expression between the limits,

I = Io at x = 0

I = I at x = l

We get,

-ln I/ Io = KCl

ln Io/ I = KCl

By taking natural logarithms,

2.303 log Io/ I = KCl

log Io/ I = K / 2.303 . cl

where,

log Io/ I = εcl

ε = K / 2.303, is called the molar absorption coefficient and

log Io/ I = A, is called the absorbance

A = εcl

This equation is known as Beer – Lambert’s Law

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Limitations of Beer – Lambert’s Law

This law is not obeyed if the radiation used in polychromatic.

1. It is applicable only for dilute solutions.

2. It is not applied to suspensions.

3. Deviation may occur, if the solution contains impurities.

UV – Visible Spectroscopy:

Absorption radiation in the UV (wave length range = 200 – 400 nm) and Visible (wave length range = 400 – 750 nm) regions of the electromagnetic spectrum results transitions between electronic levels. This is due to the larger energy change corresponds to 100 – 100,000 kJ/mol which cause simultaneous change in vibrational and rotation energies.

Generally energy change due to electronic transition is greater than that of vibrational and rotational transitions. Hence the UV and Visible spectra of simple molecules exhibit narrow absorption peaks.

Permitted energy levels in UV and Visible regions:

The vibrational and rotational fine structure lines are not observed during the spectrum is run in solution. This is because of the physical interactions between sample organic molecule and the solvent molecule, cause collisional broadening of the lines.

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Electronic absorption spectrum of a solution of benzene in hexane:

The above spectra of benzene shows larger peaks at 250 nm due to the presence of П electrons and smaller peaks & troughs indicate vibrations of molecules.

Types of electrons in organic molecules involving in transitions:

1. σ electrons.

2. П electrons.

3. Non – bonding electrons.

According to Molecular orbital theory, the interaction of atomic orbitals leads to the formation of “bonding” and “anti – bonding” molecular orbitals. The relative energies of bonding, anti – bonding and non – bonding molecular orbitals are given in the following diagram.

Bonding and antibonding orbitals.

Molecules absorb radiation from UV – Visible region and undergo various transitions. During this transition, an electron from one of the filled σ, П or non – bonding orbitals get excited to vacant σ* or П* orbitals. Corresponding to possible excitations, there are various transitions are possible as follows.

These transitions are classified into two types as follows:

1. Allowed transition:

2. Forbidden transition:

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The order of transition is

Out of the above possible transitions, the last three ones account for the absorption in 200

– 800 nm region of electromagnetic radiation and first three transitions requires much higher energy, thus the molecules with n or П electrons give rise to characteristic spectra in the region 200 – 800 nm of electromagnetic region.

Chromophores:

The structural units of the compound having n or П electrons, absorbs selective wavelength of UV – Visible radiation are called chromophores.

Example: - N=N- , C=C, C=O, etc.

Auxochromes:

The polar groups with lone pair of electrons support the intensity of chromophores are called auxochromes.

Example: .. ..

-O – H , - O – R , etc...

Bathochromic shift:

Absorption and intensity shifts:

It is also called red shift. The substitution of a selective group in a molecule makes the absorption to longer wavelength is called bathochromic shift.

Example: Alkyl substitution on olefins.

Hypsochromic shift:

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It is also called blue shift. The substitution of a selective group in a molecule make the absorption to shorter wavelength is called hypsochromic shift. 

Example: Chlorine substitution on olefins.

Hyperchromic effect:

The substitution of a selective group in a molecule causes increase in the intensity of absorption maximum of the molecule, and then the effect is called hyperchromic effect. 

Example: Methyl substitution on benzene.

Hypochromic effect:

The substitution of a selective group in a molecule causes decrease in the intensity of absorption maximum of the molecule, and then the effect is called hypochromic effect. Example: Chlorine substitution on benzene.

UV – Visible Spectrophotometer: Instrumentation:

Components:

Block diagram for a UV – Visible spectrophotometer.

Radiation source:

Hydrogen or deuterium lamps are used. It provides stable, continuous and sufficient intensity.

Filter:

It is also called monochromator. It permits the radiation of required wavelength only.

Cell:

It is a transparent and uniformly constructed container which contains either sample solution or reference

solvent.

Detectors:

It converts the absorbed radiation into current. There are three types of detectors, viz., Barrier layer cell,

photo multiplier tube and photo cell.

Recorder: 

It converts the signal reaches to itself into spectrum of a molecule.

Working:

The radiation from the source is passed through the monochromator where it splitted into two equal beams,

one half is passed into the sample cell and another half is passed into the reference cell containing solvent.

The detector will measure the comparison of intensities of beam of light. If the sample absorbs light then the

intensity of sample beam is less than the intensity of reference beam. It will be recorded as a signal in

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recorder. The instruments gives output graph (absorption spectrum) of a plot of the wavelength verses

absorbance (A) of

the light at each wavelength, where A = log (I0/I).

  UV – Visible spectra : (a) benzene in ethanol              (b) naphthalene in methanol.

Applications of UV spectroscopy:

1. It is used to determine the structure of vitamins, detecting steric hindrance, study rates of reactions and determine the dissociation constants of acids and bases from the change of absorption spectra with pH.

2. It is used to determine the dissociation energy of a molecule accurately from the wavelength.

3. It provides the information regarding moment of inertia, vibrational frequency and interatomic distances of diatomic molecules.

4. It is used to identify the cis and trans isomers of a compound from absorption spectra.

5. It is used to know the purity of a compound.

6. It is used to determine the structure of organic compounds. For example,

isatin can be assigned following two possible structures:

However, the corresponding structures of two possible methyl ethers are known to us.

On comparison of the spectra of isatin with that of two methyl isomers (III) and (IV), the spectrum of isatin is similar to that of N-methyl ether (III). Hence isatin is assigned the structure (I).

7. It is used in quantitative analysis to determine the concentration of unknown sample by using Beer – Lambert’s equation, A = E c l.

IR SPECTROSCOPY:

The spectra of a molecule arised in IR region (12500 cm–1– cm–1 )

due to the absorption of energy and transition occurs between different vibrational levels. Hence it is called vibration spectroscopy.

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All types of molecules cannot interact with IR radiation. Only those molecules which exhibit change in dipole moment during a vibration can exhibit IR spectra.

Evidently, the homo-nuclear diatomic molecules like H2, O2, N2, Cl2, etc do not show change in dipole moment during vibration. Consequently, these do not exhibit IR spectra. The hetero- nuclear diatomic / polyatomic molecules like HCl, BeCl2, NH3, CH4, CO2, C6H6, etc shows change in dipole moment and thus they exhibit IR spectra.

The IR spectral region at 1400 cm–1 – 700 cm–1 gives rich, intense and clear absorption bands for all functional groups in the organic compounds. This region is called finger – print region. It is used to identify the functional group present in the organic compound, Identify the molecule and find out the characteristics of the molecule.

The IR spectral region at 4000 cm–1 – 600 cm–1 gives intense absorption bands associated with bending and stretching vibrations of particular functional group in organic compounds. This region is called group frequency region. It is used to identify the types of functional groups present in organic molecules.

The molecules have certain number of vibrational modes. It can be calculated using the following formulae.

(a)For a linear molecule, No of fundamental vibrational mode = 3n – 5

(b)For a non – linear molecule, No of fundamental vibrational mode = 3n – 6

where n = number of atoms in a molecule.

Molecule HCl BeCl2 NH3 CH4 CO2 C6H6

FVM 1 4 6 9 4 30

Stretching and bending vibrations in water molecule:

Water molecule has non – linear structure. It has three fundamental vibrational modes which are corresponding to the frequencies 3652 cm–1 (Symmetric stretching vibration), 3756 cm–1 (Asymmetric stretching vibration) and 1596 cm–1 (Bending vibration) respectively.

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Generally stretching frequency is greater than bending frequency, because more energy is required to stretch the bond than to bend. All the above three vibrations are IR active and giving IR spectra at various frequencies. Hence IR active molecule undergoes change in dipole moments.

Stretching and bending vibrations in carbon dioxide molecule:

Carbon dioxide molecule has linear structure. It has four fundamental vibrational modes which are corresponding to the frequencies 1340 cm–1 (Symmetric stretching vibration), 2350 cm–1 (Asymmetric stretching vibration) and twice 666 cm–1 (In plane bending vibration and Out of plane bending vibration) respectively.

In symmetrical stretching, both bonds are shortened or elongated to the same extent. Hence there is

no change in dipole moment. So it is IR inactive.

In asymmetrical stretching, one of the bonds is shortened and the other is elongated. Hence there is

change in bond length and dipole moment. So it is IR active.

In bending, both in-plane and out of plane bending involves variation of bond angle.

Hence there is change in bond angle and dipole moment. So it is IR active.

Even we have three active vibrations at 2350 cm–1,666 cm–1 and 666 cm–1 respectively, we get two absorption band only, One at 2350 cm–1 and another one at 666 cm–1

Components:

Radiation source:

The Nernst glower (Oxides of Zr, Y and Er) is heated to 1500 radiation, which is used as radiation source.

C to give IR

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Optical prism:

 It is also called mirror, which is used to reflect the radiation on filter.

Filter:

It is also called monochromator, which sent the individual frequencies to the detector.

Amplifier:

 It amplifies the current received from the detector.

Motor:

 It drives the wedge.

Recorder:

 It draws the IR spectrum, based on the movement of wedge.

Working:

The IR radiation from radiation source is splitted into two equal half beams; one half is passed into the sample cell and another half is passed into the reference cell containing solvent respectively. Then these beams are fall on the mirror and reflected to the monochromator, where the selective radiation is sent to the detector. The radiation received by the detector is converted into current. It is amplified and coupled to the motor which drives a wedge. Based on the movement of the wedge the recorder draws the absorption bands on the chart. Finally, we get a spectrum as a graph of Transmittance verses wave number from the IR spectrophotometer.

Applications of IR :

1. It is used to identify the presence of functional groups in organic compounds.

For example, IR spectra of both benzaldehyde and acetophenone shows absorption peak at 1700 cm–1

This indicates that the presence of keto group

(C=O) in both the compounds.

2. It is used to detect the presence of impurities in organic compounds, by comparing the IR spectra of the pure (shows actual absorption bands) and impure (shows extra absorption bands) compounds.

3. It is used to distinguish inter and intra molecular hydrogen bonding in organic compounds.

4. It is used to study the molecular symmetry, dipole moment, structure, bond angle and bond length, etc. of various organic and inorganic compounds.

5. It is used to distinguish positional isomers of organic compounds.

6. It is used in rapid quantitative analysis of mixture of compounds.

7. It is used to study the kinetics of a reaction.

Infrared absorption frequency for different groups:

C – H Stretching Alkynes 3300 cm-1

Aromatics 3020 – 3150 cm-1

Alkenes 3010 – 3100 cm-1

Alkanes 2850 – 2970 cm-1

C-C

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C C Alkynes 2100 – 2260 cm-1

C = C alkenes 1610 – 1680 cm-1

C = C aromatic 1500 – 1600 cm-1

C = C aromatics 1450 – 1600 cm-1

C = O Stretching: Carbonyl compounds, Ester, carboxylic acids 1690 – 1760 cm-1

C-O StretchingAlcohols, ethers, carboxylic aids esters 1050 -1300

O – H Stretching : Alcohols, phenols 3590 – 3650 com-1

Alcohols, phenols (H2 – bonding ) 3200 – 3550 cm-1

Carboxylic acid 3500 – 3650 cm-1

Carboxylic acid (H2 – bonding ) 2500 – 2700 cm-1

N – H Stretching: Amines and amides 3300 – 3500 cm-1

C – N stretching Amines and amides 1180 – 1360 cm-1

Selection rules:

The most fundamental requirement for IR activity is that a vibration must cause change in the dipole moment of the molecule for example

I R active:dipolemoment ≠ O

Ex: CO2, HCl, CH3Cl , C6 H6

I R inactive:dipolemoment = O

Ex: H2 , O2 , N2 (homo nuclear species)

If a molecule has a centre of symmetry, then the vibrational are centrosymmetric & are inactive in IR and active in Raman. The vibrations which are not cento symmetric are active in IR &inactive in raman

Number of fundamental vibrations:

Nuclear magnetic Resonance spectroscopy (NMR)

An NMR spectrum is the interaction of oscillating magnetic field of electromagnetic radiation and the magnetic field of the hydrogen nucleus or other nuclei when placed in external magnetic field. The source of energy in NMR is radio wave from 2.35 – 18.6 tesla this frequency in the 100-800MHz range. The radio wave has longer wave length, hence its frequency is low energy is also low. When these low energy radio waves interact with a molecule, the spins of the nuclei are changed

Principle:

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This technique is associated with nuclear spin. Since a proton is spinning, it generates circulate electric current, which in turn produces a magnetic field, so a spinning proton behaves like a tiny magnet.

The nuclear of a Hydrogen atom (Proton) behaves as a spinning bar magnet because it possesses both electric & magnetic spin.

When a proton is placed in external magnetic field (H0), it can align itself in one of two possible orientations with respect to the applied field.

i) Low energy alignment of the nucleus in which the magnetic field of the nucleus is in the same direction as the applied magnetic field.

ii) High energy orientation in which the two magnetic fields oppose each other.

The energy difference between two energy states is very low. A transition from low energy state to high energy state can be obtained by providing energy equal to the difference in energy (ΔE) between the two states.

The proton absorb energy and moves from one energy state to the other and this transition is called flipping of proton. When the quantum of energy (hv) of electromagnetic radiation matches with difference of energy between the two energy states at field strength, nucleus and radio frequency are in resonance. Absorption takes place and a signal is observed. All the protons don not absorb at the same frequency. It depends on applied magnetic field strength and protons electric field.

Nuclei with odd mass number of proton or odd number of neutrons or both give NMR spectra. Ex: 1H,13C,31P ,19F etc., because they have asymmetrical charge distribution.Spin quantum number of the nuclei will be ½, 3/2 , 5/2 etc.,

16O ,12C , 32S , 14N , 2H does not give NMR spectra because of symmetric charge distribution & their spin quantum number is a integral value.

Interpretation of NMR spectra:

A set of protons with identical electronic environment gives NMR signal at same position. These protons are called equivalent protons. Similarly, protons with different electronic environment with gives NMR signal at different positions. The protons are called non equivalent protons.

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Ex CH4 one signal

CH3-CO-CH3 one signal

CH3-CH2-CH3 two signals

CH3-CH2-CH2-OH four signals

Number of signals:Tells about how many types of protons are present in a molecule (equivalent & non equivalent)

1) Chemical shift (position of signals):

When a molecule is placed in a magnetic field, its elections are caused to circulate, and they produce induced / secondary magnetic field. Rotation of electrons about the protons generates a field in such a way that at the proton it opposes the applied field. Therefore, the proton is said to be shielded.

Rotation of electrons about the nearby nuclei generates a field. Which can either oppose or reinforce the induced field opposes at the proton. If the induced field opposes the applied field, the strength of proton is increased, and proton is said to be deshielded.

Shielding shifts the absorption up field and deshielding shifts the absorption down field. Such shifts in the position of NMR absorptions which arise due to the shielding and deshielding of proton by the electrons are called chemical shifts.

To calculate chemical shifts for various protons, signal of TMS ( tetra methyl silane) is taken as reference.

In this four methyl groups are equivalent. So, we get single sharp peak. TMS is also inert to chemicals. Highly volatile easily removed from the system A little amount of TMS is mix with the compound and then we take the NMR spectra. & TMS is

taken as the zero point.

Chemical shift is the difference in the absorption position of a sample proton & absorption position of a reference compound. It is represented by δ. For TMS δ = o

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chemical shift=V sample−¿V TMS

operating frequency∈megha cycle¿

T = 10-δ

Applications:

Chemical shift gives information about

The structure of the molecule Different functional groups. Position of different protons Tells about electronic environment of different types of protons.

Chemical shift for various types of protons

Intensities and multiplicity of signals

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The multiplicity of signal is depends on number of neighbouring protons. In order to find the multiplicity of signal, count the number of neighbouring protons (n) and add one to it (n+1)

Neighbouring protons multiplicity (n+1) Relative Intensity name0 1 1 singlet1 2 1:1 Doublet2 3 1:2:1 Triplet3 4 1:3:3:1 Quartet4 5 1:4:6:4:1 Pentet5 6 1:5:10:10:5:1 Sextet

Ex : CH3 – CH2 - Cl

a b a ---- 2 +1 = 3 triplet intensity

b ---- 3 +1 = 4 quartet

CH3 – CH2 – CH2 – Br

a b c

a 2+1 = 3 triplet

b 5+1 =6 sextet

c 2+1 = 3 triplet

Magnetic Resonance Imaging (MRI)

MRI is one of the most noticeable applications of NMR. Resonance of protons to produce proton density maps (or images) of the human body. MRI can be utilized to discriminate between healthy & diseased tissues of a body. It is based on the fact that the protons present with in water, lipids, fats etc., resonates at a given

frequency. Human body contains 75% of water, & each H2O molecule has two hydrogen nuclei, so images of

the different parts of the body can easily be taken. In a diseased condition of body part, the distribution of water, lipids, fats etc., hence by using MRI

one can detect the diseased part of the body.

Procedure:

For getting image in MRI, a varying magnetic field is applied across the body part under consideration. The protons in various regions of body come to resonance at different radio frequency & the intensity of signal is proportional to the number of protons at that magnetic field.

The body part is then rotated in to a different orientation & another projection is made.

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Finally, the data obtained from different projections is combined by computer to get three dimensional image of the body part.

MRI requires very small time for scanning a body part.

Applications:

For investigating the functioning of myocardium, heart etc., To identify the regions of excessive fat deposition in different body organs, blood vessels etc., For the analysis of blood. For estimation of fluorine concentration in different body parts.

TUTORIAL QUESTIONS:

Q1.Differentiate between Emission Spectra and Absorption Spectra

Q2. Elucidate the principle involved in IR spectra.

Q3. Discuss the applications of UV Spectra

Q4. Identify the following spectra and names the compound.

Q5. State the characterisation of UV Spectra.

Q6.Give the selection rules for IR spectra

Q7. Discuss the significance of Chromophores.

Q8 . Write the principle involved in UV spectra.

Q9. Write a brief note on the application of NMR in Magnetic Resonance Imaging.

Q10. Enlist the applications of IR spectra.

ASSIGNMENT QUESTION:

1) Define the term Molecular Spectroscopy.2) Write the expression in which energy changes in a molecule are specified?3) What are the different types of energies present in a molecucle?

a) Define the terms i) Frequency ii) Wavelength iii) Wavenumber of a molecule.b) Write a short note on Emission Spectra.c) Explain the representation of absorption spectra.

4) Differentiate between emission spectra and absorption spectra.5) Explain the principle involved in IR spectra.6) How are vibrational spectra of a diatomic molecucle derived?7) Explain the selection rules for IR Spectra8) Write a short note on applications of IR spectra.9) Write a short note on Sheilding and Desheilding of protons in HNMR spectra.10) Explain the principle involved in UV spectra.11) Explain the charecteristics of UV visible spectra.12) What are Chromophores? What is their significance?13) Mention the applications of UV Spectra.

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14) Explain the principle involved in NMR.15) Write a short note on application of NMR in MRI.

Code No : 1800BS08

MALLA REDDY ENGINEERING COLLEGE FOR WOMEN (Autonomous Institution – UGC Govt. of India) Permanently Affiliated to JNTUH, Approved by AICTE Accredited by NBA and NAAC with A grade– ISO 9001-2015Certified B.TECH I YEAR II Semester Regular Examinations, April 2019 ENGINEERING CHEMISTRY Time: 3 hours (ECE) Max Marks: 70

Note: This question paper Consists of 5 Sections. Answer FIVE Questions, Choosing ONE Question from each SECTION and each Question carries 14 marks.

SECTION-I

1. (a) Write down the molecular orbital energy level diagram of N2 (7M)(b) Write the crystal field splitting of [Ni(CO4)] complex. (7M) OR

2. (a) Explain crystal field splitting of d-orbitals in tetrahedral crystal fields. (7M)(b) What are Salient features of Crystal field theory? (7M)

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SECTION-II3. (a) Explain the estimation of hardness by complexometric method. (7M)

(b) Describe the formation of sludge and scale and how to remove them. (7M)

(OR)

4. (a) Describe desalination of brackish water by reverse osmosis process. (7M)(b) Explain softening of hard water by Ion exchange method. (7M)

SECTION-III5. (a) What are secondary batteries? Explain working of Lead-acid storage battery. (7M)

(b) Discuss electrochemical theory of corrosion. (7M)

(OR)

6. (a) Describe the construction of calomel electrode. (5M)(b) Explain water line corrosion. (4M)

(c) Write briefly about electroless plating. (5M)

SECTION-IV7. (a) Describe the classification of isomers. (7M)

(b) Explain the mechanism of SN2 reaction with suitable example. (7M)

(OR)

8. (a) Explain determination of optical activity by centre of symmetry. (7M)(b) Describe the structure, synthesis and pharmaceutical applications of Aspirin. (7M)

SECTION-V

9. (a) Give the principle of UV spectroscopy. Explain different energy transitions. (7M)(b) Discuss the principle and types of molecular vibrations. (7M)

(OR)

10. (a) State and derive Beer-Lambert’s law giving its applications. (7M)(b) What do you meant by chemical shift in NMR spectroscopy? (7M)

Code No : 1800BS08

MALLA REDDY ENGINEERING COLLEGE FOR WOMEN(Autonomous Institution – UGC Govt. of India)

Permanently Affiliated to JNTUH, Approved by AICTEAccredited by NBA and NAAC with A grade– ISO 9001-2015Certified

B.TECH I YEAR II Semester Advanced Supplementary Examinations, August 2019 ENGINEERING CHEMISTRY

Time: 3 hours (ECE) Max Marks: 70Note: This question paper Consists of 5 Sections. Answer FIVE Questions, Choosing ONE Question from each

SECTION and each Question carries 14 marks.

SECTION-I1. (a) Explain the molecular energy level diagram of O2 and F2 molecules (10M)

(b) Discuss the salient features of Crystal field theory (4M)OR

2. Discuss the crystal field splitting of transition metal ion d-orbitals in Tetrahedral and octahedral complexes. (14M)

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SECTION-II3. Differentiate between sludge and scale. Discuss their formation in boilers, ill-effects and methods to prevent

their formation. (14M)OR

4. (a) Describe in detail the softening of the hard water by Ion-exchange method (10M)(b) Write a short note on caustic embrittlement (4M)

SECTION-III5. (a) Derive Nernst’s equation for single electrode potential and explain the terms involved in it. Write its

applications. (10M)(b) What is an ion-selective electrode? Explain its working. (4M)

OR

6. (a) Explain the theories of electrochemical corrosion of iron with mechanisms (10M)(b) Discuss any four factors affecting corrosion with suitable examples (4M)

SECTION-IV7. (a) Write a note on structural and stereoisomers with suitable examples (7M)

(b) Explain the mechanism of SN2 reaction (7M)

OR8. (a) Explain with mechanism of “Markownikoff and anti Markownikoff’s rule (10M)

(b) Explain a method of synthesis of Paracetamol and its uses (4M)

SECTION-V9. (a) Write the wave number regions of the following functional groups in IR spectroscopy. i. -NH2, ii. -

COOH, iii. C=O, iv. C N, v. C C (10M)(b) Explain about Beer-Lamberts law (4M)

OR10. (a) Explain the types of electronic excitations and applications in UV-Visible spectroscopy

(10M)(b) Discuss the principle of NMR spectroscopy (4M)

UNIT WISE OBJECTIVE QUESTIONS

UNIT-1

FILL IN THE BLANKS

1. The probability of finding a particle in a given energy level is related to ________.2. The bonding characteristic of bonding molecular orbital σ is given by________.3. If all the electrons in a molecule are paired, the molecule becomes________.4. The stability of benzene ring is due to ________ of electrons.5. ________ is an example of elemental semi conduction.6. The crystal field stabilization energy (CFSE) is denoted by________.7. The anti bonding molecular orbitals of fluorine remain as ________ in HF molecule.8. The wave functions are always finite , single valued and ________.9. An electron can jump from one energy level to another level by ________.10. The cyclic molecules in which delocalization actually leads to stabilization are just ________.

MULTIPLE CHOICE QUESTIONS:

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1) An atomic orbitals containsa) Pairs of electronsb) single electronc) three paired electrond) two pairs of electron

2) One of the following is a lowest energy bonging molecular orbital a) σ b) π* c) π d) σ*

3) The bond order 3 indicates the presence of a) double bond in the molecule b) single bond in the molecule c) triple bond in the molecule d) conjugation of double and single bonds in the molecule

4) By introducing a trivalent impurity into silicon, one of the following type is produced a) positive holeb) dopingc) insulatord) conductor

5) If the Fermi energy gap is more the material becomes a) semiconductor b) doping c) insulator d) either conductor or semi conductor

6) During the formation of an octahedral complex , the energy of t2g set is decreased by a) 0.6Eo b) 1.0 Eo c) 0.4Eo d) 0.656Eo

7) The bond order for oxygen molecule is a) 3 b) 2 c) Zero d) 1

8) The orbital under the influence of magnetic field is said to be five fold degenerate is a) s b) p c) d d) f

9) One of the following molecule contain (4n+2)π electrons a) cyclopentadiene b) cylopentadienyl free radical c) cyclopentadienyl anion d) cyclopentadienyl cation

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10) The square planar complexes are a) high spin complexes b) medium spin complexes c) low spin complexes d) no spin complexes

UNIT-2

MULTIPLE CHOICE QUESTIONS:

1) Permanent hardness of water cannot be removed bya) Treatment with lime sodab) By permutit processc) Boiling d) Ion exchange process

2) Blow down operation causes the removal of a) scales b) sludgesc) acidityd) sodium chloride

3) Calgon is trade name given to a) Sodium silicate b) Sodium hexa meta phosphatec) Sodium meta phosphate d) Calcium phosphate

4) Caustic embrittlement can be avoided by using

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a) Sodium phosphate b) Hydrogenc) Ammonium hydroxide d) Sodium sulphate

5) Which is not the unit of hardness of watera) ppmb) epmc) degree Clarke d) none of these

6) Brakish water can be purified by using a) Lime soda processb) Permutit processc) Filtration d) Reverse osmosis method

7) The full name of EDTA a) Ethyl diamine tetraacetic acidb) Ethylene diamine tetraacetic acidc) Ethylene amine tetra aceticacid

8) Solubility of calcium sulphate in watera) Increases with increase in temperatureb) Decrease with increase in temperature c) Remains unaltered with increase in temperatured) Does not follow any regular trend with increase of temperature

9) Hardness of water is expressed in terms of equivalents of a) MgCO3

b) CaCO3

c) Na2CO3

d) K2CO3

10) The relation between mg/L and ppm isa) 1mg/L= 1ppmb) 10 mg/L= 1ppmc) 1mg/L= 10ppmd) 1mg/L= 106ppm UNIT-3MULTIPLE CHOICE QUESTIONS:

1. HCl is called an electrolyte because

a. Its molecules are made of electrically charged particlesb. It breaks up into ions when current is passed through itc. It ionizes when electric current is passed through itd. It ionizes when dissolved in a proper solvent.

2. Calomel electrode is constructed using a solution ofa. Saturated Kclb. Saturated CaCl2 c. Saturated MgCl2

d. Saturated NaCl 3. The standard oxidation potential of quinhydrone electrode at 250c is

a. 0.6994Vb. 0.8994Vc. -0.6994Vd. -0.8994V

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4. Volatile oxidation corrosion product of a metal is,a. Fe2O3

b. MoO3

c. Fe3O4

d. FeO5. Lower is PH, corrosion is,

a. Greaterb. Lowerc. Constantd. None of above

6. Electrochemical corrosion takes place on,a. Anodic area.b. Cathodic areac. Near cathoded. Near anode

7. Chemical formula of Rust is,a. Fe2O3

b. FeOc. Fe3O4

d. Fe2O3.XH2O8. Which of following metals could provide cathodic protection to Fe?

a. Al & Cub. Al & Znc. Zn & Cud. Al & Ni

9. Smaller the grain size, corrosion is,a. Greaterb. Lowerc. Constantd. Doesn’t affected

10. Process of corrosion enhanced by,a. AIR & Moistureb. Electrolytes in waterc. Metallic impuritiesd. Gases like CO2 & SO2

FILL IN THE BLANKS

11. The resistance of a metallic conductor _______ as the temperature is increased.12. Graphite is a ________ conductor.13. A device which converts electrical energy into chemical energy is ________.14. Any metal above hydrogen in electrochemical series can be easily oxidized. Hence they undergo

________ easily.15. Nernst equation for electrode reaction is ________.16. Pitting corrosion is an________ accelerated attack.17. In galvanic series, a metal high in series is more________.18. Hydrogen overvoltage is ________.19. Coating of Zn, Al and Cd on steel are ______ because their electrode potentials are lower.20. Green film of basic carbonate on surface of Cu contains CuCO3 and __ .

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UNIT 4FILL IN THE BLANKS:

1. A carbon atom having four different groups is known as ................

2. Only the ..............molecules show the phenomenon of ...........isomerism

3. d- and l-isomers of a compound are mirror images of each other and thus known as.................

4. The phenomenon of separating dl mixture of a compound into d- and l- isomers is known as...................

5. The R and S system of nomenclature is considered to be ..........

6. According to sequence rule –NH2 group has ............. priority over –COOH group

7................ conformation of n-butane is the least stable

8. Number of dibromo derivatives possible for propane are............

9. Restricted rotation is possible in....................... compounds

10. Optical isomers which are non-super imposable and do not has object mirror relationship are

called...................

MULTIPLE CHOICE QUESTIONS:1. Optical activity of a molecule is due to

a. lack of plane of symmetry

b. Presence of asymmetric carbon

c. Molecular asymmetry

2. Optical isomers have different

a. configurations

b. conformations

c. structure

3. Chiral molecules are those which are

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a. not super imposable on their mirror image.

b. are super imposable on their mirror image.

c. show geometrical isomerism.

d. are unstable molecules.

4. d- and l- tartaric acids are examples of

a. enantiomers

b. diastereomers

c. tautomers

5. the terms D and L indicate

a. dextro and leavo rotator

b. relative configuration

c. absolute configuration

6. the functional isomers of ethane nitrile is

a. ethyl carbylamines

b. methane nitrile

c. methyl carbylamines

d. propane nitrile

7. Functional isomers of ethers are

a. ketones

b. aldehydes

c. alcohols

d. esters

8. Which of the following compounds can exhibit geometrical isomerism?

a. 1-hexane

b. 2-methyl-2-pentene

c. 3-methyl-1-pentene

d. 2-hexene

9. Which of the following can exist as diastereomers

a. lactic acid

b. 1-butene

c. 2-butene

d. ethane

10. A racemic mixture contains equal ........... of d- and l- isomers

a. no of molecules

b. masses

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c. no of moles

d. all the above are correct

UNIT 5

MULTIPLE CHOICE QUESTIONS:1. When vibration motion is accompanied by a change in dipole moment of the molecules, one of the following spectra is shown.a)NMR b) Raman spectra c) microwave spectra d) IR2. The spectra arises due to the transitions induced between the nuclear spin energy levels of a molecule and applied is shown .a)IR b) U.V c) NMR d) NPF3. The number of signals given by the NMR spectra of CH3CHBr.CH2.CH3

a) 3 b) 1 c) 4 d)54.The reference used in NMR spectroscopy is a) tetra methyl silicon b) trimethyl silicon c) trimethyl silloxane d) tetra methyl siloxane5.The medical imaging technique used in radiology to produce pictures of the anatomy of human body is calleda) X-ray photoelectron microscope b) X-ray diffraction c) Low energy electron diffraction d) Raman spectrometer

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6.The NMR splitting of 1,1 –dibromo ethane (H3C-CHBr2 ) is as followa) doublet, quartet b) doublet , triplet c) triplet ,quartet d) singlet ,doublet 7.The ratio of the initial intensity of light and final intensity of light after passing through a sample is measured as a) transmission b) absorbance c) intensity d) Wave length 8.The chemical shifts of alcohols is shown at δ by a) 10-δ b δ−10 c) 10-τ d) T- τ9.The planks constant is given by the value a) 1.68×10-28 erg/sec b) 1.67×10-27 erg.secc) 1.67 ×10-28 erg/sec d) 1.67×10-27 erg.sec10.One of the following is a surface characterization technique a) X-ray photoelectron microscope b) X-ray diffraction c)MRI d) NMR

FILL IN THE BLANKS :1.The energy is emitted or absorbed by the matter in discrete amounts called ………..2.The lowest energy state of electrons is called ……………….3.The entire range over which electromagnetic radiation exists is called ……………4.A shielded proton requires higher ………………5.The number of signals given by CH3-CH2-CH2Cl is……………….6.The position of the signals tells us the ……………of each kind of protons .7.The number of signals tell us how many different kinds of ………………. Are there in a molecule .8.In MRI scanning the excited hydrogen atoms emit a …………..signal.9.The splitting of peak of –CH in CH3-CH Cl Br is ……...10.The environment of the absorbing protons is reflected b y…………….

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