132.Equivalance point of potentiometric titration
is noted by plot of
1.EMF Vs Volume of titrant added
2.∆E Vs Volume of titrant added
3. ∆2E /∆V2 Vs Volume of titrant added
4.All of these
Ans:-3
133.Equivalance point EMF of titration of
Fe2+ Vs Ce4+ during potentiometric
titration is
1.Zero volts
2.1.1 volts
3. 0.69 volts
4.1.45 volts
Ans:-2
134.Equivalance point of a Redox titration
by plotting ∆E/ ∆V vs. titrant volume, is
noted with
1.Peak
2.Sudden change in EMF
3. Parabolic curve
4.Constant EME portion
Ans:-1
135.When mixture of HCL and Phosphoric acid
is titrated with standard alkali, the first
equivalence point is obtained between pH___
1.
2.
3.
4.
1-2
4-6
7-8
8-10
Ans:-2
136.When mixture of HCl and Acetic acid
is titrated with standard alkali, the first
equivalence point corresponds to
1.Compete neutralization of HCl and half
neutralization of acetic acid
2. Compete neutralization of HCl
3. Compete neutralization of Acetic acid
4. half neutralization of acetic acid
Ans:-2
137.When mixture of HCl and CH3COOH
is titrated with standard alkali, the second
equivalence point corresponds to
1.Compete neutralization of HCl and half
neutralization of acetic acid
2. Compete neutralization of HCl and CH3COOH
3. Compete neutralization of CH3COOH
4. half neutralization of CH3COOH
Ans:-2
138.Which region lies between visible and
microwave region in the electromagnetic
spectrum
1.X-ray
2. Radio waves
3. Infra red
4. Ultra-violet
Ans:-3
139.Useful region of UV used in spectral
analysis comprises of radiation having
wavelength____
1.10-200 nm
2. 200-400 nm
3. 400-800 nm
4. 100-200 nm
Ans:-2
140.Absorption of energy in the wavelengths of
UV visible region by a compound containing
chromophore, is utilized for,
1.Vibration of bonds
2. Rotation of molecules
3. Breaking of multiple bonds
4. Excitation of electron from bonding to anti
bonding orbitals
Ans:-4
141.False statement for electromagnetic
spectrum___
1.Velocity of electromagnetic rays is same in
vacuum
2. Velocity of electromagnetic rays decreases after
passing through a matter
3. Frequency of all electromagnetic rays is same
4. An electromagnetic rays has both electrical and
magnetic components.
Ans:-3
142.Relation between wavelength,
frequency and velocity of electromagnetic
rays is
1.V=λ/υ
2. V=λ x υ
3. V=υ/λ
4. V= 1/λ
Ans:-2
143.The correct statement from the following is
1. λmax Chromophore is increases when attached to
auxochrome
2. Absorption spectrum of a compound is the graph of
absorbance Vs. Wavelength
3. Absorption spectrum of a compound is the graph of
absorbance Vs. Concentration
3. In the beers law, concentration is assumed to be
constant
4. Velocity of all electromagnetic radiation is same in
vacuum.
a)1,2,4
b) 2,3,5
c) 1,2,5
d)1,2,3
Ans:-3
144.Energy in the photons of ___ wave is
lowest among the following
1.X-ray
2. Infra red
3. Ultra violet
4. Visible
Ans:-2
145.The wavelength region used to identify
a functional group in organic compound is
1. Infra red
2. Ultra violet
3. Visible
4. Microwave
Ans:-1
146.The energy required for following
function in molecule, is in the order,
1. Electronic excitation > Vibrational excitation>
Rotational excitation
2. Electronic excitation >Rotational excitation
>Vibrational excitation
3. Vibrational excitation>Rotational excitation
>Electronic excitation
4. Vibrational excitation >Electronic excitation >
Rotational excitation
Ans:-1
147.In potentiometry sharp end point of
titration is obtained by plotting differential
plot of
1. pH Vs. Volume of titrant added
2. ∆E/ ∆V Vs. Volume of titrant added
3. ∆ pH Vs. ∆V
4. EMF Vs. Volume of titrant added
Ans:-2
148.In potentiometric titration end point from the
graph of EMF of cell Vs. Volume of titrant added
is determined by considering
1. The maxima
2. The minima
3. The point of inflection
4. The point of intersection with Y-axes
Ans:-3
149.In potentiometric redox titration
between Fe2+ and Ce4+ the overall reaction
is
1. Fe2+ + Ce4+
2. Fe3+ + Ce4+
3. Fe2+ + Ce3+
4. Fe3+ + Ce3+
Ans:-1
Fe3+ + Ce3+
Fe2+ + Ce3+
Fe3+ + Ce4+
Fe2+ + Ce4+
150.The first differential derivative plot of
a potentiometric titration gives a sharp end
point from the graph as
1. Point of intersection on Y- axis
2. The minima of curve
3. The point of inflection
4. The maxima of curve
Ans:-1
151. In Potentiometric redox titration
between Fe2+ and Ce4+ at equivalence
point
1. Fe2+ and Fe3+ ions are present
2. Ce3+ and Fe3+ ions are present
3. Ce3+ and Ce4+ ions are present
4. Fe3+ and Ce4+ ions are present
Ans:-2
152. In which method a cell is constructed using
at least one electrode reversible with respect to
one of the ion taking part of the reaction?
1. pH metry
2. Conductometry
3. Potentiometry
4. Colorimetry.
Ans:-3
153. Potential of a cell during Redox titration of Fe2+
and Ce4+can be calculated using the formula
1. E1= 0.75 +0.0591log[Ce4+]/[Ce3+] and
E2= 1.45 +0.0591log[Fe3+]/[Fe2+]
2. E1= 1.45 +0.0591log[Fe3+]/[Fe2+] and
E2= 0.75 V+0.0591log[Ce4+]/[Ce3+]
3. E1= 0.75 +0.0591log[Fe3+]/[Fe2+] and
E2= 1.45 +0.0591log[Ce4+]/[Ce3+]
4. E1= 0.75 +0.0591log[Ce3+]/[Ce4+] and
E2= 1.45 +0.0591log[Fe2+]/[Fe3+]
Ans:-3
154. In Potentiometric titration, when in 100 ml 0.1N
Fe2+ 50 ml Ce4+ added calculate EMF of cell if
E0Fe=0.75 V
1.
2.
3.
4.
0.65 V
0.69 V
3. 0.75 V
1.1 V
Ans:-3
155. In Potentiometric titration, when in 100 ml 0.1N
Fe2+ 90 ml Ce4+ added calculate EMF of cell if
E0Fe=0.75 V
1.
2.
3.
4.
0.806 V
0.75 V
0.867 V
0.927 V
Ans:-1
156. In Potentiometric titration, when in 100 ml 0.1N
Fe2+ 101 ml Ce4+ added calculate EMF of cell if
E0Ce=1.45 V
1.
2.
3.
4.
1.1 V
1.332 V
1.391 V
0.9 V
Ans:-2
157. Select the correct equation which will be used
for calculating cell EMF before equivalence point,
when Fe2+ ions are titrated with Ce4+
potentiometrically
1. E= Ecell+0.0591log[Fe3+]/[Fe2+]
2. E= Ecell+0.0591log [Fe2+] /[Fe3+]
3. E= Ecell+0.0591log[Ce3+]/[Ce42+]
4. E= Ecell+0.0591log[Ce4+]/[Ce3+]
Ans:-1
158. Match the following
1. UV spectroscopy
A. Wheatstone bridge
2.Conductometry
B. No visual indicators
3.pH metry
C. Colored solution
4.Potentiometer
D. Electronic transition
1) 1-D
2) 1-C
3) 1-A
4) 1-C
Ans:-1
2-A
2-B
2-B
2-A
3-B
3-A
3-C
3-D
4-C
4-D
4-D
4-B
159. Match the following
k. Lamberts law
1. A  Concentration
l. Beers law
2. A  thickness x Concentration
m.Kohlrausch law
3. 0 = + + n. Lamberts-Beers law 4. A  thickness
1) k-4
2) k-4
3) k-4
4) k-2
Ans:-2
l-2
l-1
l-1
l-1
m-3
m-3
m-2
m-3
n-1
n-2
n-3
n-4
160. Choose correct relation from the following the
following
1.
2.
3.
4.
E vibration > Erotational > Eelectronic
Erotational > Eelectronic > E vibration
Eelectronic > E vibration > Erotational
Eelectronic > Erotational> E vibration
Ans:-3
161. Rotational energy level transition occurs
1)In infrared region of electromagnetic spectrum
2)In Microwave region of electromagnetic spectrum
3)In UV region of electromagnetic spectrum
4)In visible region of electromagnetic spectrum
Ans:-2
162. Vibrational energy level transition occurs
1)In infrared region of electromagnetic spectrum
2)In Microwave region of electromagnetic spectrum
3)In UV region of electromagnetic spectrum
4)In visible region of electromagnetic spectrum
Ans:-1
163. Electronic energy level transition occurs
1)In infrared region of electromagnetic spectrum
2)In Microwave region of electromagnetic spectrum
3)In UV region of electromagnetic spectrum
4)In visible region of electromagnetic spectrum
Ans:-3
164. Lamberts law is……..
1)ln I0/It  Path length
2)ln I0/It  Concentration
3)Absorbance  Concentration x path length
4)Transmittance  Concentration
Ans:-2
165. Spectrophotometer is based on…….. Absorption law
1)Kohlrausch law
2)Beer-Lamberts law
3)Ohms law
4)Faradays law
Ans:-2
166. The combined lamberts- beers law is
1)ln I0/It  Path length
2)ln I0/It  Concentration
3)Absorbance  Concentration x path length
4)Transmittance  Concentration
Ans:-3
167. Spectroscopic method of analysis are based on
measurement of electromagnetic radiation…… by the
sample
1) Emitted
2) Absorbed
3) Both 1 &2
4) None of the above
Ans:-3
168. Electronic transition occur by……. Radiation
1) Infra red
2) Microwave
3) Visible
4) UV
Ans:-4
169. Material of sample holder should….
1) Partially absorbed the radiation
2) Not allow the radiation to pass through it
3) React with either solvent or sample
4) Be transparent to sample
Ans:-4
170. Absorbance is……
1) A= ln (1/T)
2) A = Log10(1/T)
3) A = Log10(T)
4) A=Loge(1/T)
Ans:-2
171. Law that govern colorimetry is
1) Lamberts law
2) Beers law
3) Both 1 & 2
4) None of these
Ans:-3
172. Mathematical form of Lamberts Beers
law is
1) A=bC
2) A  C
3) A  b
4) A  
Ans:-1
173. According to Beers law, absorbance of
solution is
1) A=bC
2) A  C
3) A  b
4) A  
Ans:-2
174. Transmittance is defined by the
equation
1) T=I/I0
2) -A =-log T
3) T= A-10
4) T= 10-bC
Ans:-1
175. Unit of molar extinction coefficient is
1) Lit mole1 cm-1
2) Lit mole1 cm1
3) Lit mole-1 cm-1
4) Lit mole-1 cm1
Ans:-3
176. Accordingly to the Beers-Lamberts law,
intensity of beam of monochromatic
radiation decreases…..
1) Linearly
2) Cyclically
3) Exponentially
4) In the multiple of wavelength
Ans:-3
177. The correct equation for Transmittance
is …..
1) A=bC
2) A = Log10(I0/It )
3) T= 10-bC
4) T= A-10
Ans:-3
178. In the Equation A= bC,  is called
1) Absorption coefficient
2) Extinction coefficient
3) Desorption coefficient
4) Excitation coefficient
Ans:-2
179. According to Beers law as
concentration of sample increases
arithmetically, amount of transmitted light
1) Decreases arithmetically
2) increases arithmetically
3) Decreases exponentially
4) increases exponentially
Ans:-3
180. According to Beers law, absorbance of
solution is proportional to….. ,Provided
that…… of medium of remains constant
1) Concentration, thickness
2) Thickness, Concentration
3) Concentration , Temperature
4) Length, Breadth
Ans:-1
181.The light source used in UV
spectroscopy is
1) Halogen lamp
2) Helium lamp
3) Hydrogen lamp
4) Na-vapour lamp
Ans:-1
182.Monochromator used in UV
spectroscopy is
1) Grating
2) Prism
3) Laser
4) Glass plate
Ans:-1
183.Quantitative analysis of compound by
UV- Visible spectroscopy involves plot of
1) Absorbance Vs. Concentration
2) Absorbance Vs. Wavelength 
3) Wavelength  Vs. Concentration
4) Absorbance Vs. Path length
Ans:-1
184. An UV- spectrum of compound is
plot of
1) Absorbance Vs. Wavelength 
2) Absorbance Vs. Concentration
3) Transmittance Vs. Concentration
4) Energy Vs. Concentration
Ans:-1
185. The detector used in UV- Visible
spectrophotometer is
1) Phototube
2) Photovoltaic cell
3) Photomultiplier tube
4) All of these
Ans:-4
186. The detector used in UV- Visible
spectrophotometer is
1) Phototube
2) Photovoltaic cell
3) Photomultiplier tube
4) All of these
Ans:-4
187. Chromophore is a functional group
containing multiple bond capable of
absorbing UV radiation…..
1) Below 200 nm
2) At 200 nm
3) Above 200 nm
4) Above 800 nm
Ans:-3
188. Chromophore is a functional group
containing multiple bond capable of
absorbing UV radiation…..
1) Below 200 nm
2) At 200 nm
3) Above 200 nm
4) Above 800 nm
Ans:-3
189. When absorption maxima of a
compound shifts to longer wavelength is
called as
1) Hypsochromic shift
2) Bathochromic shift
3) Hypo chromic shift
4) Hyper chromic shift
Ans:-2
190. When absorption intensity of a
compound is increases the effect is known as
1) Red
2) Hyper chromic effect
3) Hypo chromic effect
4) Hypsochromic effect
Ans:-2
191. When absorption intensity of a
compound is decreases the effect is known
as
1) Hypsochromic effect
2) Bathochromic effect
3) Hypo chromic effect
4) Hyper chromic effect
Ans:-3
192. When wavelength of absorption of a
compound is shift to longer wavelength, it
1) Contains conjugation of C=C
2) Contains heteroatom
3) Contains cyclic structure
4) Has no conjugation
Ans:-1
193. When wavelength of absorption of a
compound is shift to shorter wavelength, the
shift is due to……
1) Auxochrome
2) Chromophore
3) No interaction with solvent
4) Conjugation in structure
Ans:-1
194. Which of the following is not a
Chromophore
1) C=C
2) C=O
3) C-N
4) N=O
Ans:-3
195. Which of the following is a
auxochrome
1) C=O
2) -OH
3) C=N
4) C=C
Ans:-2
196. Which of the following is not a
Auxochrome
1) C=O
2) -NH2
3) -OH
4) -Cl
Ans:-2
197. In Hypsochromic effect…..
1)max decreases
2)max increases
3)Absorption intensity decreases
4) Absorption intensity increases
Ans:-1
198. In Hypo chromic effect…..
1)max decreases
2)max increases
3) Absorption intensity decreases
4)Absorption intensity increases
Ans:-3
199. In Bathochromic effect…..
1)max decreases
2)max increases
3)Molar absorption coefficient decreases
4)Molar absorption coefficient decreases
Ans:-2
200. In Hyper chromic effect…..
1) max decreases
2) max increases
3) Absorption intensity decreases
4) Absorption intensity increases
Ans:-4
201. Instead of taking aniline in ethanol, if
aniline is dissolved in dil. HCl , the fact about
UV absorption is……….
1) max decreases
2) max increases
3) Absorption intensity decreases
4) Absorption intensity increases
Ans:-1
202. Increase in length of conjugation in
compound causes….. In max and …. In molar
absorption coefficient.
1) increases , decreases
2) decreases, decreases
3) decreases, increases
4) increases, increases
Ans:-4
203.Attaching an auxochrome causes…..
max and …. In molar absorption coefficient.
1) increases , decreases
2) decreases, decreases
3) decreases, increases
4) increases, increases
Ans:-4
204.Excitation of electron in molecule is
caused by the absorption of …….. radiation
1) increases , decreases
2) decreases, decreases
3) decreases, increases
4) increases, increases
Ans:-4
205…… and ….. Are the forbidden electronic
transition
1. π – π* , σ - σ*
2. n – π* , π – π*
3. σ– π* , π– σ*
4. n– σ* , n – π*
Ans:- 3
206.Energy of electron is lowest in its ……
molecular orbital
1. σ
2. π
3. π*
4. σ*
Ans:- 1
207.π – π* electronic transition will not possible in
1. Benzene
2. Ethylene
3. Formaldehyde
4. Ethane
Ans:- 4
208.an isolated functional group capable of absorbing
radiations above 200 nm is called
1. Chromophore
2. Auxochrome
3. Hypochrome
4. Hypsochrome
Ans:- 1
209.which of the following is a Chromophore ?
1. C-OH
2. C-Br
3. C=O
4. -NH2
Ans:- 3
210.which of the following is not a Chromophore ?
1. C=C
2. C=O
3. –N=N4. -C-OH
Ans:- 4
211.Compound containing Chromophore is called as
1. Chromophene
2. Auxigen
3. Auxochrome
4. Chromogen
Ans:- 4
212. Which of the following statement is true in case
of auxochrome
1. Auxochrome absorb radiation in UV region
2. Auxochrome absorb radiation and bring about π –
π* transition
3. Auxochrome does not absorb radiation in UV
region
4. Auxochrome absorb radiation and bring about n–
π* transition
Ans:- 4
213. which of the following is not a Auxochrome ?
1. –CH3
2. –CH=CH3. -NHCH3
4. -OH
Ans:- 2
214. Change in absorption wavelength and
absorption intensity of a chromophore is caused
due to
1. Non-conjugate structure
2. Solvent
3. Auxochrome
4. High temperature
Ans:- 3
215. Absorption of UV radiation by a compound
causes
1. Rotation of bonds
2. Electronic excitation
3. Vibration of bonds
4. Breaking of bonds
Ans:- 2
216. Match the following
1.Bathochromic shift
2.Hyperchromic shift
3.Hypsochromic shift
4. Hypochromic shift
1) 1-E
2. 1-H
3. 1-F
4. 1-G
Ans:- 3
2-F
2-E
2-G
2-E
E.Absorption intensity decreases
F. max shift to longer wavelength
G. Absorption intensity increases
H. max shift to lower wavelength
3-H
3-G
3-H
3-H
4-G
4-F
4-E
4-F
217. Match the following
1.Bathochromic shift
2.Hyperchromic shift
3.Hypsochromic shift
4. Hypochromic shift
1) 1-F
2. 1-F
3. 1-H
4. 1-G
Ans:- 2
2-H
2-G
2-G
2-E
E. Introduction of group that distort geometry
F. Change of solvent
G. Presence of auxochrome
H. removal of conjugation
3-E
3-H
3-F
3-F
4-G
4-E
4-E
4-H
218. Aniline in acidic medium shows…… shift
due…..
1) Blue shift, Due to loss of conjugation
2. Red shift, Due to loss of conjugation
3. Red shift, Due to presence of conjugation
4. Hyperchromic shift, Due to presence of
conjugation
Ans:- 1
219.Minimum energy required for the transition
1. σ - σ*
2. π – π*
3. n – π*
4. n– σ*
Ans:- 3
220.Which of the following transition is not
possible
1. σ - σ*
2. n– σ*
3. n – π*
4. σ - n*
Ans:- 4
220.Which of the following transition is not
possible
1. σ - σ*
2. n– σ*
3. n – π*
4. σ - n*
Ans:- 4