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HIGHER ORDER THINKING QUESTIONS (HOTS)
By A.k.panda,P.C.Panda,Rishi Verma,Umesh kumar, ,Satish Rajput UNIT-5:-SURFACE CHEMISTRY
ANSWER THE FOLLOWING :
1# “colloid is not a substance but a state of substance”. OR Same substance can act both as colloids and crystalloids
2# A sol is prepared by the addition of excess of KI to AgNO3 solution. Which one of the following will be most effective for its coagulation? [ Fe(CN)6]-4 , Al +3, PO4-3 , Mg +2 (2M)
3# Three gases A, B and C have critical temperatures 630 K, 33 K and 190 K respectively. Arrange them in increasing order of adsorption on activated charcoal.
4# What are the conditions when Tyndall effect can be observed? ( 2 M)
5# Why artificial rain can be cause by throwing common salt on the clouds?(1M) 6# What is CMC? What is its value for soap?(1M)
7 What is ZSM-5? Where it is used?(2 M)
8#100 mL of a colloidal solution is completely precipitated by the addition of 5 mL of 1 M NaCl solution. Calculate its coagulation value.( 2M )
9# Action of soap is due to emulsification and micelle formation. Comment.(2M) 10# Why is it necessary to remove CO when ammonia is obtained by Habber’s Process 11 # Why is the ester hydrolysis slow in the beginning and becomes faster after some time? 12# Why is it essential to wash the precipitate with water before estimating it quantitatively 13# What charge is likely to develop on the colloidal particles in the following cases ?
(a) A sol is prepared by addition of excess of AgNO3 solution in KI solution (b) Ferric chloride is added to NaOH
(c) If FeCl3 is added to excess of hot water (d) when KI solution is added to AgNO3 solution
(e) when AgNO3 solution is added to KI solution
14 # In which of the following does adsorption occurs and why ? (a) silica gel placed in atmosphere get saturated with water
(b) Anhydrous CaCl2 placed in the atmosphere gets saturated with water
ANSWER-1# The given statement is true . This is because the same substance may exist as colloid under certain conditions and as crystalloid under certain other conditions .
For example , NaCl in water behaves as a crystalloid while in benzene , it behaves as a colloid .
Similarly , dilute soap solution behaves like a crystalloid while concentrated soap solution behaves as a colloid ( associated colloid) .
It is size of the particle which matters , i.e. , the state in which the substance exists . If the size of the particle lies in the range of 1nm to 1000 nm , it is in the colloidal state
2# the addition of excess of KI to AgNO3 solution will result into negative sol AgI/I¯ , so Al +3 will be effective for coagulation
3# B < C < A . higher the critical temperature ,higher will be the Adsorption as van der waals forces are stronger near the critical temperature .
4# Tyndall effect is observed only when the following two conditions are satisfied.
(i) The diameter of the dispersed particles is not much smaller than the wavelength of the light used; and
(ii) The refractive indices of the dispersed phase and the dispersion medium differ greatly in magnitude.
5# Clouds are aerosols having small droplets of water suspended in air. Sometimes, the rainfall occurs when two oppositely charged clouds meet. It is possible to cause artificial rain by throwing electrified
sand or spraying a sol carrying charge opposite to the one on clouds from an aeroplane.
6#The formation of micelles takes place only above a particular concentration called critical micelle concentration (CMC).
For soaps, the CMC is 10–4 to 10–3 mol L–1. These colloids have both lyophobic and lyophilic parts. Micelles may
contain as many as 100 molecules or more.
7# Zeolite sieve of microporosity-5 . It converts alcohols directly into gasoline(petrol) by dehydrating them to give a mixture of hydrocarbons. Molecular formula – Hx[(AlO2)x(SiO2)96-x].16H2O
8# 50 .
9# A micelle consists of a hydrophobic hydrocarbon – like central core. The cleansing action of soap is due to the fact that soap molecules form micelle around the oil droplet in such a way that hydrophobic part of the stearate ions is in the oil droplet and hydrophilic part projects out of the grease droplet like the bristles. Since the polar groups can interact with water, the oil droplet surrounded by stearate ions is now pulled in water and removed from the dirty surface. Thus soap helps in emulsification and washing away of oils and fats. The negatively charged sheath around the globules prevents them from coming together and forming aggregates.
Figure:-(a) Grease on cloth
(b) Stearate ions arranging around the grease droplet and (c) Grease droplet surrounded by stearate ions (micelle formed)
10# CO acts as a poison for the catalyst ( finely divided Fe) in Haber’s Process . Acitivity increases if it is used in finely divided state . 11# H+ which is generated from one of the product ( from acetic acid ) acts as a catalyst for the reaction [ a case of autocatalysis] .
Hence the reaction becomes faster after sometime as production of H+ increases as time passes .
12# Some amount of the electrolyte mixed to form precipitate remains adsorbed on the surface of the particles of the precipitate . Hence , it is essential to wash the precipitate with water to remove the sticking electrolytes( or any other impurities) before estimating it quantitatively .
13# (a) positive sol i.e. , [AgI] Ag+ (b) Negative sol i.e. , [Fe(OH)
3] OHˉ (c) positive sol (d) positive sol AgI/Ag+ (e) negative sol AgI/I¯
14#(a) adsorption (b) Absorption
Unit-6: GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS ANSWER THE FOLLOWING :
Q.1# (a) Metal A is extracted by pyrometallurgy process, metal B by electrolytic refining and metal C by simply heating the ore. Arrange these metals in the increasing order of reactivity.(1M)
(b) What does a steep increase in the slope of a line on Ellingham diagram indicate?(1M) ( c) Why is Chalcopyrite roasted and not calcined during recovery of copper? ( 1M) (d) Why copper matte is put in silica lined converter?
(e) Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction? (f) Out of C and CO, which is a better reducing agent at 673 K ?
(g) Copper can be extracted by hydrometallurgy but not zinc. Explain. (h) Out of C and CO, which is a better reducing agent for ZnO ?
(i) The value of ΔfG0 for formation of Cr2 O3 is – 540 kJmolˉ1and that of Al2 O3 is – 827 kJmolˉ 1Is the reduction
of Cr2O3 possible with Al ?
(j) Why is the reduction of a metal oxide easier if the metal formed is in liquid state at the temperature of
reduction?
(k) Although thermodynamically feasible, in practice, magnesium metal is not used for the reduction of alumina in the metallurgy of aluminium. Why ?
(l) At a site, low grade copper ores are available and zinc and iron scraps are also available. Which of the two scraps would be more suitable for reducing the leached copper ore and why?
(m) The reaction, Cr2 O3 + 2 Al →Al2 O3 + 2 Cr (ΔG0 = – 421 kJ)
is thermodynamically feasible as is apparent from the Gibbs energy value. Why does it not take place at room temperature?
(n) Copper and silver lie below in the electrochemical series, yet they are found in combined state in the form of sulphide ores. Why? (2M)
2# Free energy of formation (ΔfG) of MgO(s) and CO(g) at 1273K and 2273K are given below (Sample
paper-2)
ΔfG (MgO(s)) = -941 kJ/mol at 1273K ΔfG (MgO(s)) = -341 kJ/mol at 2273K
ΔfG (CO(g)) = - 439 kJ/mol at 1273K ΔfG (CO(g)) = - 628 kJ/mol at 2273K
On the basis of above data ,predict the temperature at which carbon can be used as a reducing agent for MgO(s) . Q.3 # The choice of reducing agent in a particular case depends on thermodynamic factor . How far do you agree with this statement ? Support your opinion with two examples
Q.4# What chemical principle is involved in choosing a reducing agent for getting the metal from its oxide ore ? Consider the metal oxides , Al2O3 and Fe2O3 and justify the choice of reducing agent in each case . (3M)
Q.5 # The standard Gibbs energy of reactions at 1773K are given below: ( 3M) (AI-2005)
C + O2 CO2 : ΔrG0 = -380 kJ/mol ; 2C +O2 2CO : ΔrG0 = -500 kJ/mol
Discuss the possibility of reducing Al2O3 and PbO with carbon at this temperature.
4Al + 3O2 2Al2O3 : ΔrG0 = -22500 kJ/mol ; 2Pb + O2 2PbO : ΔrG0 = -120 kJ/mol
6# How is the concept of coupling reactions useful in explaining the occurrence of a non-spontaneous reaction ? (2M)(AI-2005&06)
ANSWER-UNIT-6 Answer--1:-- (a) C< A < B
(b) The steep increase in thje slope of a line due to change in phase, i.e. solid to liquid . The temperature at which such change occurs is indicated by increase in slope on positive side .For example, the abrupt increase in Zn to ZnO plot shows the melting of Zn metal .
(c) chalcopyrite is a sulphide ore , so it is roasted in presence of oxygen so as to convert into oxide ore for easy reduction . Carbonate ores are calcined .
(d) Copper matte contains Cu2S and FeS .In the converter FeS gets converted into FeO when hot air blast is blown
into the converter.Silica helps in removal of FeO impurity as slag. 2FeS + 3 O2 2FeO + 2 SO2
FeO (impurity) + SiO2 (silica) FeSiO3 (slag)
(e) The reducing agent forms its oxide when the metal oxide is reduced. The role of reducing agent is to provide
ΔG0 negative and large enough to make the sum of ΔG0 of the two reactions (oxidation of the reducing agent and
reduction of the metal oxide) negative.
Carbon is the poor reducing agent for sulphide ores wheras it is good reducing agent for oxide ores .
( f) At 673K , the ΔG0 vs T line for CO,CO
2 is lower than that of C, CO line . Carbon monoxide is the better
reducing agent at 673 K
(g) Metals (such as Cu) occupying lower positions in the electrochemical series can be extracted by
hydrometallurgy because the metal ions(Mn+) of such metals can be easily reduced by treatment with some more
electropositive metal.
Metals occupying higher positions in the electrochemical series cannot be extracted by hydrometallurgy because the metal ions(Mn+) of such metals are difficult to reduce. Zn has large negative reduction potential.
(h) The line for ΔG0 ( CO ,CO
2) lies above the line for ΔG0 ( Zn ,ZnO) at the temperature employed for reduction
of Zinc oxide.Hence reduction of ZnO with CO is not possible as the ΔrG0 for this reaction would be positive
ZnO + CO +1673K Zn + CO2 ΔrG0 = + Ve ( not feasible)
On the other hand , the line for ΔG0 ( C ,CO) Vs T lies below the line for ΔG0 ( Zn ,ZnO) .Hence ,coke can
reduce Zinc oxide to zinc as the ΔrG0 for this reaction would be negative and hence the reaction is feasible .
ZnO + C +1673K Zn + CO ΔrG0 = - Ve (feasible)
Answer (i) # (i) 2 Cr + 3/2 O2 Cr2 O3 ΔfG0 = -540 kj/mol
(ii) 2 Al + O2 Al2O3 ΔfG0 = -827 kj/mol
Equation (ii) – (i) , we get 2Al + Cr2O3 Al2O3 + 2Cr ΔrG0 =[ -827 –(-540)] kJ = - 287 kJ
Since the ΔrG0 for the reduction of Cr2 O3 with Al is negative therefore, the reaction is possible (j) The entropy is higher if the metal is in liquid state than when it is in solid state. The value of entropy change (ΔS) of the reduction process is more on +ve side when the metal formed is in liquid state and the metal oxide
being reduced is in solid state. Thus the value of ΔG0 becomes more on negative side and the reduction becomes
easier.
(k) Temperatures above the point of intersection of Al2O3 and MgO curves, magnesium can reduce alumina. But
the temperature required would be so high that the process will be uneconomic and technologically difficult.
(l) Zinc being above iron in the electrochemical series (more reactive metal is zinc), the reduction will be faster in case zinc scraps are used. But zinc is costlier metal than iron so using iron scraps will be advisable and
advantageous
(m) Certain amount of activation energy is essential even for such reactions which are thermodynamically feasible, therefore heating is required.
Answer-2:- - The desired equation is MgO(s) + C(s) Mg (s) + CO(g) ,. ΔrG = ?
At 1273 K :- Mg(s) + ½ O2(g) MgO(s) ; ΔfG (MgO(s)) = -941 kJ/mol at 1273K ---(1)
C(s) + ½ O2 (g) CO(g) ; ΔfG (CO(g)) = - 439 kJ/mol at 1273K ---(2)
Equation (2) – (1) ΔrG = ΔfG (CO(g)) - ΔfG (MgO(s)) = - 439 kJ/mol -(-941 kJ/mol) = + 502 kJ/mol
Since ΔrG is +ve this reduction is NOT FEASIBLE at this tempetrature .
C(s) + ½ O2 (g) CO(g) ; ΔfG (CO(g)) = - 628 kJ/mol at 2273K ---(4)
Equation (4) – (3) ΔrG = ΔfG (CO(g)) - ΔfG (MgO(s)) = - 628 kJ/mol -(-341 kJ/mol) = - 287 kJ/mol
Since ΔrG is -ve (NEGATIVE) this reduction is FEASIBLE at this tempetrature . So Carbon can acts as reducing agent at this temperature – i.e at 2273K
Answer-3—The thermodynamic factor helps us in choosing a suitable reducing agent for the reduction of particular metal oxide to metal . The feasibility of thermal reduction can be predicted on the basis of ΔG0 Vs T
plots for the formation of oxides, known as Ellingham diagram . from the diagram it can be predicted that metals for which the ΔfG0 ( of their oxides) is more negative can reduce those metal oxides for which standard free
energy of formation of their respective oxides is less negative .
In other words , a metal will reduce the oxides of the other metals which lie above it in the Ellingham Diagram because the ΔfG0 of the combined redox reaction will be negative by an amount equal to the difference in ΔfG0 of
the two metal oxides .
For example, both Al and Zn can reduce FeO to Fe but Fe cannot reduce ZnO to Zn and Al2O3 to Al .
Similarly , Carbon can reduce ZnO to Zn but not CO . thus the choice of reducing agent in a depends on
thermodynamic factor .
UNIT-7: p-Block Elements
I# IDENTIFY THE FOLLOWING:
1 # An element ‘A’ exists as a yellow solid in standard state. It forms a volatile hydride ‘B’ which is a foul
smelling gas and is extensively used in qualitative analysis of salts. When treated with oxygen, ‘B’ forms an oxide ‘C’ which is colourless, pungent smelling gas. This gas when passed through acidified KMnO4 solution,
decolourizes it. ‘C’ gets oxidized to another oxide ‘D’ in the presence of a Heterogeneous catalyst.
Identify A, B, C, D and also give the chemical equation of reaction of ‘C’ with acidified KMnO4 solution and for
conversion of ‘C’ to ‘D’.
2# When conc. sulphuric acid was added to an unknown salt present in a test tube, a brown gas (A) was evolved. This gas intensified when copper turnings were also added into this tube. On cooling, the gas ‘A’ changed into a colourless gas ‘B’.
(a) Identify the gases A and B. (b) Write the equations for the reactions involved.
3# A translucent white waxy solid ‘A’ on heating in an inert atmosphere is converted in to its allotropic form (B). Allotrope ‘A’ on reaction with very dilute aqueous KOH librates a highly poisonous gas ‘C’ having rotten fish smell. With excess of chlorine ‘A’ forms ‘D’ which hydrolysis to compound ‘E’. Identify compounds ‘A’ to ‘E’ . 4# Concentrated sulphuric acid is added followed by heating to each of the following test tubes labelled (i) to (v)
Identify in which of the above test tube the following change will be observed. Support your answer with the help of a chemical equation.
(a) Formation of black substance (b) Evolution of brown gas (c) Evolution of colourless gas
(d) Formation of brown substance which on dilution becomes blue
5 # A gas “X” is soluble in water . Its aq. Solution turns red litmus blue with excess of aq. CuSO4 solution it
give s deep blue colour with FeCl3 solution a brownish ppt. soluble in HNO3 is obtained. Identify gas”X” and
write reactions for changes observed .
6# Element “A” burns in N2 to give a compound “B” which is ionic in nature .” B” on reaction with water gives
‘C” and “D”.The solution of “C” becomes milky on passing CO2 through it. Identify the compound A ,B ,C ,D .
7# An orange solid “A” on heating gives a green residue”B”, water vapours and a colourless gas “C” . The gas “C” in dry condition is passed over heated Mg to give solid “D” which further reacts with water to form “E” which gives dense white fumes with HCl . Identify the compound A ,B ,C ,D and E and write the reactions involved .
8# A colourless inorganic salt “A” decomposes at 2500C to form “B” and “C” leaving no residue . The oxide “C”
is liquid at room temperature and is neutral towards litmus while “B” is neutral oxide and is used as anaesthetic in minor dental surgery . When Phosphorous burns in excess of –B produces compound which is a strong
dehydrating agent which when treated with HNO3 produces acid “D” and a gas “E” Identify A” to “E” and write
the reactions.
I# ANSWERS OF IDENTIFY THE FOLLOWING:
ANS-1#
S8[A] +8H2 + heat 8H2S[B]8H2S + 12O2 8SO2 [C] ;
5SO2 + 2KMnO4(pink)/H+ +2H2O K2SO4 +2MnSO4(Colourless solution)
2SO2 +O2 + Pt/heat 2SO3 (D)
ANS-2# 2NaNO3 +H2SO4 +heatNa2SO4 + 2HNO3
4HNO3+ heat4NO2+2H2O+O2
Cu+4HNO3Cu(NO3)2 +2NO2 +2H2O
2NO2[A] ═(cool)=> N2O4[B]
ANS-3# P4(white Phosphorous)[A] +very dil KOHPH3[C]
P4(white Phosphorous)[A] +heatRed- P(red Phosphorous)[B]
P4(white Phosphorous)[A] +Cl2(excess)PCl5[D]
PCl5[D] +H2O(excess)H3PO4[E]
ANS-4#(a) Test tube(i): C12H22O11+Conc.H2SO411H2O +12C (black subs.)
(b) Test tube(ii):2NaBr +3H2SO42NaHSO4 +SO2+Br2(brown gas)
(c) Test tube(v): KCl +H2SO4 +heat KHSO4+HCl(colourless gas)
(d) Test tube(iii): Cu +2 H2SO4 CUO(brown subs)
On dilution: CuO(brown subs) +2 dil.H2SO4 CuSO4(blue substance)
(e) Test Tube(iv) : S(yellow powder) + conc.H2SO4 ----SO2 +H2O
ANS-5: NH3(g) +H2O NH4OH(aq) [basic and turn red litmus blue]
3 NH4OH(aq) + CuSO4 [Cu(NH3)4]SO4 +4H2O
(deep blue)
3 NH4OH(aq) +FeCl3Fe(OH)3 +3 NH4Cl
Fe(OH)3 + 3HNO3 Fe(NO3)3( soluble) +3H2O
ANS-6: 3Ca + N2Ca3N2 [B]
[A]
Ca3N2 + 6 H2O3 Ca(OH)2 [C] + 2NH3 [D]
Ca(OH)2 +CO2 CaCO3 + H2O
ANS-7: (NH4)2Cr2O7 +heat N2 + Cr2O3 +
(Orange solid-A) [C] [green solid B ] N2 +3Mg +heat Mg3N2 [D]
Mg3N2 [D] +6H2O 3 Mg(OH)2 +2NH3 [E]
NH3 [E] +HCl NH4Cl (dense white fumes)
ANS-8:NH4NO3[A]+heat at 2500C N2O(g) [B] + 2 H2O [C]
10 N2O +P4 10 N2 + P4O10
10 P4O10 is a dehydrating agent
P4O10 + 4 HNO3 4 HPO3 [D] + 2 N2O5 (g)
II#
Explain Why ? / Give reason ? / Account for the following ?
1# HBr and HI can’t be prepared by treating metal bromides or iodides with conc. H2SO4 .
2# Why NO2 readily forms a dimmer, whereas ClO2 does not though both are odd electron molecule.
3# Write the balanced chemical equation for the reaction of Cl2 with hot and concentrated NaOH. Is this reaction a
disproportionation reaction? Justify.
4# SCl6 is not known but SF6 is known .
5# In trimethylamine, the nitrogen has a pyramidal geometry whereas in trisilylamine,it has a planar 6# SF6 is resistant to hydrolysis whereas SF4 is readily hydrolysed.
7# NCl3 gets hydrolysed to form NH3 and HOCl while PCl3 gets hydrolysed to give H3PO3 and HCl . Explain .
8# In the solid state , PCl5 is in ionic nature . Explain.
9# I2 form I3− but F2 does not formF3− .
10# NO is paramagnetic in gaseous state but diamagnetic in the solid and liquid state. 11# Yellow ppt. of Sulphur disappear when boiled with sodium sulphite .
13# SH6 and PH5 are not formed but SF6 and PF5 are formed .
14# KHF2 is known but compound of formula KHCl2 or KHBr2 are not known . Why ?
15# Iodine is more soluble in KI solution than in water . 16# HF is stored in wax -coated bottle .
17# HCl is not used to make the medium acidic in titrations involving KMnO4 .
18 # Addition of Cl2 to KI solution gives it a brown colour but excess of Cl2 turns it colourless.
19# CN− ion is known but CP− is not ?
20# Comment on the nature of two S—O bonds formed in SO2 molecule. Are the two S—O bonds in this
molecule.
21# Why anhydrous CaCl2 is not used to dry up Ammonia ?
22# What happens when Zinc is treated with Conc.H2SO4 .
23# Why conc.H2SO4 is not used for drying of H2S gas .
24# When HCl reacts with finely powdered iron , it forms ferrous Chloride and not ferric chloride . 25# Conc.HNO3 turns yellow on exposure to sunlight .Why ?
ANSWER-II# Explain Why ? / Give reason ? / Account for the following ?
ANS-1# Because Br- and I - will get oxidized by conc. H
2SO4 to respective Br2 and I2 .
ANS-2# odd electron molecule often get dimerise in order to pair the electrons but ClO2 does not .Thisis thought to be because the odd electron is delocalized.
ANS-3# 6 NaOH ( hot and conc.)+ 3Cl2 5 NaCl + NaClO3 + H2O . yes it a disproportionation reaction because
the OS of Cl2 is Zero and is oxidized to +5 (in NaClO3) and reduced to -1 (in NaCl) . Due to simultaneous
oxidation and reduction occurs of the same species i.e Cl2 .
ANS-4# SF6 is known because of small size and more electronegativity of Fluorine which can form strong bond
and hence the energy released can compensate to promote electrons in vacant d-orbitals of S which is not possible in case of Chlorine . Due to large size of Cl which cannot be accommodated around Sulphur and stability will also be less due to inter electronic repulsion between lone pairs of Chlorine ( whose size is large as compared to F) ANS-5# Due to presence of Vacant d-orbitals in Si , it can form Pл-dл bond with nitrogen . so reduced to SP2
hybridisation and becomes plannar with three sigma bonds . Which is not possible in case of trimethylamine as Carbon does have vacant d-orbital.
ANS-6# SF6 is octahedral ( which is symmetrical) where six F-atoms surrounding”S” protect it from attack of H2O molecules.
So ,SF6 is sterically protected then SF4 towards the even thermodynamically favorable hydrolysis reaction.
Also F-atoms do not have Vacant d-orbitals to accommodate electrons from water molecules . This makes SF6
chemically inert. Due to this it suppresses internal charges hence used in high voltage generators(gaseous insulators, and switch gears.
ANS-7# The Cl of NCl3 has vacant d-orbital which can accommodate electrons from water molecules and get
hydrolysed to produce HOCl and NH3 . But in case of PCl3 , it is the vacant d- orbital of P can accommodate
electrons from water molecules and get hydrolysed to give H3PO3 and HCl . [P—O bond strength > Cl—O] ANS-8# Due to unequal bond length of equatorial and axial bond that is why it is reactive . So, in solid state (in close range) , it can be stabilized by forming [PCl4]+ and [PCl6]- which are symmetrically stable tetrahedral and
octahedral str. respectively .
ANS-9# Due to presence of vacant d-orbital in Iodine it can accept the electrons of iodide ion .
ANS-10# NO is an odd electron molecule (paramagnetic) . In solid and liquid state , dimer is formed and there is no unpaired electrons and act as diamagnetic .
ANS-11# Na2SO3 + S ( yellow solid ) Na2 S2O3 ( sodium thiosulphate)
ANS-12# Because of small size and more electronegativity of Fluorine which can form strong bond with Nitrogen and becomes stable as compared to large size Chlorine .
ANS-13# It is because of two reason why hydrogen cannot form SH6 and PH5 though SF6 and PF5 are formed
Reason-1: Hydrogen is less electronegative element as compared to fluorine . S—H bond is weaker than S—H bond .
Reason-2: The enthalpy of atomization of H—H is very high as compared to F—F . High enthalpy of dissociation can not be compensated by energy released during bond formation .
ANS-14# KHF2 contains HF2− ion which is stable due to formation os H-bonding between H—F and F− . This is
possible because of high electronegativity of F ( F—H …..H) − on the other hand , Cl and Br cannot form
hydrogen bonds due to low electronegativity .
ANS-15# Due to formation of polyiodide ion .KI +I2 K+ [I3]− . Ion –dipole interaction between polyiodide
ion and water responsible for solubility .
ANS-16# Due to formation of complex called flurosilicic acid , H2[SiF6]
ANS-17# Chloride gets oxidized to chlorine by conc. Sulphuric acid ( an oxidizing agent) ANS-18# Cl2 + KI ( solution) KCl + I2 ( brown ) . Iodide reduces chlorine to chloride ion.
The I2 formed will react with excess Cl2 in presence of water to form HIO3 which makes it colourless .
ANS-19# Nitrogen being small size can form stable and effective Pл-Pл bond ( in case of cyanide ion) but which is not possible in C≡P− . P is larger size which cannot form stable and effective Pл-Pл bond .
ANS-20# S 3s2 3p
x23py13pz1 (Ground state) ; S 3s2 3px13py13pz1 3d1 (excited state)
Sp2 hybridisation in Sulphur . one half filled 3p
z1 can form Pл-Pл bond with half-filled 2p orbital of one
Oxygen and one half filled 3d1 of Sulphur with half-filled 2p orbital of another Oxygen can form Pл-dл bond .
And due to resonance the bond lengths are equal.
ANS-21# Because it t forms CaCl2.8NH3 , addition compound .
ANS-22# Zn reduces Sulphate to SO2 and ultimately colloidal sulphur .As Conc. H2SO4 is an oxidizing agent . ANS-23# H2S + Conc.H2SO4 ( an oxidizing agent ) Sulphur
ANS-24# Its reaction with iron produces H2
Fe + 2 HCl FeCl2 + Cl2 . Liberation of hydrogen creates a reducing environment and prevents the formation
of ferric chloride .
ANS-25# In presence of light nitric acid decomposes to form NO2 (brown) , O2 and water .
II # Arrange the Following in increasing order against the properties mentioned
:-1# Bond Dissociation Enthalpy:- (a) Br—Br , I—I , Cl—Cl , F—F (b) H—I , H—F, H—Br,H—Cl (c) O—H, H—Te, H—Se, H—S. (d) N—N, P—P, As—As
2# Acid strength:- (a) H—I , H—F , H—Br , H—Cl (b) HF, CH4 , H2O , NH3
(c) H2O, H2Te , H2Se , H2S
3# Base Strength:- BiH3 , NH3 , AsH3 , SbH3 , PH3
4# Thermal Stability:- (a) H2O , H2Te , H2Se , H2S (b) PH3 , BiH3 , AsH3 , SbH3 , NH3
5# Oxidizing Ability:- (a) Cl2 , F2 , I2 , Br2 (b) HCl , HNO3 ,H2SO4 ,H3PO4 (c) HClO3 ,HClO , HClO2 ,HClO4 (d) F2 ,Cl2 ,O2 , O3
6# Catenation property:- (a) As , N, P , Sb (b) Se ,S , Te ,O (c) Si , Sn , C , Ge 7# Bond Angle:- (a) H2Se , H2O, H2S ,H2Te (b) PH3 , BiH3 , AsH3 , SbH3 , NH3
8# Boiling Point :- (a) H2S , H2O , H2Te , H2Se (b) PH3 , BiH3 , AsH3 , SbH3 , NH3
9# Stability:- (a) Pb++ , C++ , Sn++ , Si++ (b) As3+ , Bi3+ , Sb3+ (c) Ga+ , Tl+ ,Al+, In+
10 # Covalent Character :- (a) M—Cl , M—Br , M—I , M—F (b) Cr2O3 , CrO, CrO3
(c) P2O5,Sb2O5, As2O5 (d) BeCl2, MgCl2 ,CaCl2, BaCl2
11# Acid Strength:- (a) HOClO2 , HOClO , HOCl ,HOClO3 (b) HOCl , HOI ,HOBr
12# Acidic Character -- (a) N2O, N2O5, N2O3 ,NO , N2O4 (b) ClO2 , Cl2O7 ,Cl2O , Cl2O6 (c) HNO2 & HNO3 (d) H2SO3 &H2SO4
(e)GeO2 ,ClO2 ,As2O3 ,Ga2O3 (f) P2O5 ,SO3 , N2O5 , CO2 , SiO2
(g) Al2O3 ,CaO, Cl2O7 ,SO3 (h) BF3 ,BBr3 , BCl3
13# Volatility:- H2O , H2Te , H2Se, H2S
14# Electron Gain Enthalpy :- (a) I , Br , Cl , F (b) N , O, P ,S (c) F, Cl , O , S 15 # Ionisation Enthalpy:- (a) O , N , F , C (b) Ar , Ne , He , Xe , Kr
16# Electronegativity:- (a) Cl ,F, Br, I (b) O , N , F , C 17# Stability:- Fˉ , Iˉ , Clˉ , Brˉ
18# Reducing properties: (a) H2O, H2Te , H2Se , H2S (b) H3PO4 , H3PO2 , H3PO3
CHECK-UP LIST:- ARRANGEMENT (in terms of properties mentioned.)
1# I—I < F—F < Br—Br < Cl—Cl (Bond Dissociation Enthalpy)—Inter-electronic repulsion 2# H—F < H—Cl < H—Br < H—I (Acid strength)—Lower BDE of HI,large size of I 3# M—I < M—Br < M—Cl < M—F ( Ionic Character)—Fajan’s Rule—Lager polarizability of Iˉ 4# BiH3 < SbH3 < AsH3 < PH3 < NH3 ( Base Strength) – small size of N – High electron density in
Ammonia
6# H2Te < H2Se < H2S < H2O ( Thermal Stability) ---BDE
7# H2O < H2Te < H2Se < H2S (Volatility)--- H-Bond and Vander waal’s force
8# H2S < H2Se < H2Te < H2O ( Boiling Point) -- H-Bond and Vander waal’s force
9# PH3 < AsH3 < NH3 < SbH3 < BiH3 (Boiling Point) -- H-Bond and Vander waal’s force
10# I2 < Br2 < Cl2 < F2 (Oxidizing Ability)
Electron gain enthalpy, hydration enthalpy, dissociation enthalpy . 11# H2Te < H2Se < H2S < H2O
( Bond Angle )--- Size of central atom , electronegativity, repulsion of bond pairs. 12# HClO4 < HClO3 < HClO2 < HClO ( Oxidizing Power)
13# HOCl < HOClO < HOClO2 < HOClO3
( Acid Strength) – Stability of its conjugate base , charge dispersal , Oxidation states. 14 # HOI < HOBr < HOCl ( Acid Strength)
---Stronger the O—X bond – Weaker the O—H bond – More the acidic character.
15# Cl2O < ClO2 < Cl2O6 < Cl2O7 ( Acid Strength)-- Higher oxidation states, covalent character
16# ClO4ˉ < BrO4ˉ > IO4ˉ or BrO4ˉ > IO4ˉ > ClO4ˉ (Oxidizing Power)
17# Ga2O3 < GeO2 < As2O3 < ClO2 (Acidic Character)
18# BF3 < BCl3 < BBr3 (Acidic Character)—Effective 2p—2p overlap in BF3 reduces the electron deficiency
of B , make it less acidic.
19# Pb < Sn = Ge < Si < < C ( Catenation Property) – Sigma bond strength 20# C++ < Si++ < Sn++ < Pb++ ( Stability) ---- Inert pair effect
21# N2O < NO < N2O3 < N2O4 or NO2 < N2O5 (Acidic Character)
– Higher oxidation states, covalent character 23# HCl < H2SO4 < HNO3 (Oxidizing Acid)
24# (i) N < P > As > Sb (ii) O < S > Se > Te ( catenation property) --- ( P , N ) ; ( O ; S ) 25# (a) HNO2 < HNO3 (b) H2SO3 < H2SO4 (Acidic character)
26# Iˉ < Brˉ < Clˉ < Fˉ (Stability)
27# Reactivity---ClF3 > BrF5 > IF7 > ClF7 > BrF3 >IF5 > BrF > IF3 >IF
UNIT-8:d-
B
lock
E
lements
Q.1 # Why is Cr2+ strongly reducing and Mn3+ strongly oxidising when both have d4 configuration. Ans:- Cr2+ is reducing as its configuration changes from d4 to d3, the latter having a half-filled t
2g level .On the
other hand, the change from Mn2+ to Mn3+ results in the half-filled (d5) configuration which has extra stability.
Q.2 # Explain why Cu+ ion is not stable in aqueous solutions?
Ans--Cu+ in aqueous solution underoes disproportionation, i.e., 2Cu+(aq) → Cu2+(aq) + Cu(s)
The E0 value for this is favourable. Cu++ has higher hydration enthalpy than Cu+
Q.3# Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised. Ans— Co(III) has greater tendency to form coordination complexes than Co(II) . Due to higher charge by radius ratio and also due to greater CFSE( Crystal Field Stabilisation Energy) . Co (III) Has d6 configuration .
In case of of strong ligand , It is t2g 6 eg0 . So , Co(II) changes to Co(III) i.e. easily oxidized .
Q.4 # Mn+2 compounds are more stable than Fe+2 toward oxidation to their + 3 state .(AI-06)
OR ,
Why E 0 value for the Mn 3+ /Mn 2+ is much more positive than Fe 3+ /Fe 2+ ? Ans-- Mn 3+ + e-
——
> Mn 2+(d4 system , less stable ) (d5 system , more stable ) – extra stable due to half-filled confN
Fe3+ + e-
——
> Fe 2+(d5system , more stable ) (d6 system , less stable )
Q.5 # The E0(M2+/M) value for copper is positive (+0.34V). What is possibly the reason for this? (Hint: consider its high ΔaH0 and low ΔhydH0)
Ans— The reduction potential value depends on relative value of three enthalpies viz ΔaH0 , ΔiH0 and - ΔhydH0 .
The copper has high ΔaH0 and ΔiH0 . Therefore high enthalpy required to transform Cu(s) to Cu2+(aq) is not
balanced by its low ΔhydH0 .
Q.6 # Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
Ans-- Because of small size and high electronegativity oxygen or fluorine can oxidise the metal to its highest oxidation state. Oxygen has the ability to form multiple bond with transition metals .
Q.7 # Which is a stronger reducing agent Cr2+ or Fe2+ and why ?
Ans-- Cr2+ is stronger reducing agent than Fe2+ Reason: d4 → d3 occurs in case of Cr2+ to Cr3+But d6 → d5 occurs in
case of Fe2+ to Fe3+ . In a medium (like water) d3 is more stable as compared to d5 (see CFSE)
Q.8 # Why is the E0 value for the Mn3+/Mn2+ couple much more positive than that for Cr3+/Cr2+ or Fe3+/Fe2+? Explain.
Ans--Much larger third ionisation energy of Mn (where the required change is d5 to d4) is mainly responsible for this. This also explains why the +3 state of Mn is of little importance.
Q.9 # Give reason for it?. Ni+2 compounds are thermodynamically more stable than Pt+2 compounds whilst Pt(IV) are relatively more stable than Ni(IV) compounds . OR, K2PtCl6 is well known cpds. but
corresponding Ni-compound is not known .
Ans--The oxidation state of Pt in K2 Pt Cl6 is +4 .The sum of the first four ionization energies (IE1+IE2 +IE 3+IE 4)
of pt is less than those of Ni.
Q.10# E0 for Mn+3 / Mn+2 couple is more positive than for Fe+3 / Fe+2 or Cr+3 / Cr+2 .
Ans-- Much larger third Ionisation enthalpy of Mn ( where the required change is d5 to d4) is mainly responsible for this . this also explains why the +3 state of Mn is of little importance .
Q.11 #.Why is HCl not used to acidify a permanganate solutions in volumetric estimation of Fe+2 or C 2O42‾ .
Ans- Since Clˉ of Hydrochloric acid is oxidized to chlorine by MnO4ˉ . so , actual amount of KMnO4 required to
oxidize C2O42‾ couldnot be determined
.
I I# IDENTIFY
THE
FOLLOWING:
1# A mixed oxide of iron and chromium FeO.Cr2O3 is fused with Sodium Carbonate in presence of air to form a
yellow coloured compound (A). On acidification the compound (A) forms an orange coloured compound (B) which is a strong oxidizing agent.
(i) Identify the compounds (A) and (B)
(ii) Write balanced chemical equations for each step.
2 # A green chromium compound (A) on fusion with alkali gives a yellow compound (B) which on acidification gives an orange coloured compound (C) . “C” on treatment with NH4Cl gives an orange coloured product (D) ,
which on heating decomposes to give back (A) . Identify A,B,C,D .Write equations for reactions .
3# (a) A blackish brown coloured solid (A) when fused with alkali metal hydroxides in presence of air produces a dark green compound (B), which on electrolytic oxidation in alkaline medium gives a dark purple compound (C). Identify (A), (B) and (C) and write balanced chemical equations for the reactions involved.
(b) What happens when an acidic solution of the green coloured compound (B) is allowed to stand for some time? Give the equation of the reaction involved. What is this type of reaction called? (Hint: MnO42- changes to
MnO4-)
4# (A) reacts with H2SO4 to form purple coloured solution (B) which reacts with KI to form colourless compound
(C). The colour of (B) disappears with acidic solution of FeSO4. With concentrated H2SO4 (B) forms (D) which
can decompose to give a black compound (E) and
O2. Identify (A) to (E) and write equations for the reactions involved.
5# When an orange coloured crystalline compound ‘A’ was heated with common salt and concentrated H2SO4, an
orange red coloured gas ‘B’ was evolved. The gas ‘B’ on passing through NaOH solution gave an yellow solution C (i) Identify A,B and C. (ii) Write balanced chemical equation involved in the reactions.
6# Two gases (A) and (B) turns acidified K2Cr2O7 solution green. Gas(A) turns lead acetate paper black and when
passed through gas(B) in aq. Solution , yellowish turbidity appears . Identify A and B and explain.
7# Explain why a green solution of K2MnO4 turns purple and a brown solid is precipitated when CO2 gas is
bubbled into the solution .
UNIT-9: CO-ORDINATION COMPOUNDS I # E X P L A I N ?
1# Why only Transition metals are known to form Pi-complexes
2#A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2– is colourless. Explain. 3#Though CO is a weak lewis base yet it forms a number of stable metal carbonyls .Explain
4# What will be the correct order for the wavelengths of absorption in the visible region for the following: and why [Ni(NO2)6]4–, [Ni(NH3)6]2+, [Ni(H2O)6]2+ ?
5# Explain why [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer orbital complex.
6# The spin only magnetic moment of [MnBr4]2– is 5.9 BM. Predict the geometry of the complex ion ?
7# A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2– is colourless. Explain.
8# [Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2– is diamagnetic. Explain why?
9# [Fe(CN)6]4– and [Fe(H2O)6]2+ are of different colours in dilute solutions. Why?
10# Amongst the following, which is the most stable complex and why ?
(i) [Fe(H2O)6]3+ (ii) [Fe(NH3)6]3+ (iii) [Fe(C2O4)3]3– (iv) [FeCl6]3–
11# Silver atom has completely filled d orbitals (4d10) in its ground state. How can you say
that it is a transition element?
12# The ability of oxygen to stabilize the metal in higher oxidation states exceeds that of fluorine.
13# Why is that while making an aq. solution of FeCl3 lab. Reagent some HCl solution is also added besides FeCl3
ANSWERS TO I # E X P L A I N ?
ANS-1: The transition metals have vacant d-orbitals which can accept electron pairs of ligand containing pi-electrons(H2C=CH2 ,C5H5− , C6H6 )
ANS-3# CO is a pi acceptor ligand which forms stable synergic bond due to back bonding
ANS-4# NO2− is a strong field ligand acc. to spectrochemical series , so it energy of absorption is in higher energy range .
H2O < NH3 < NO2−( spectrochemical series) The correct order for the wavelengths of absorption in the visible
region: [Ni(H2O)6]2+ [Ni(NH3)6]2+, [Ni(NO2)6]4–,
Ans-5:- In the presence of NH3, the 3d electrons pair up leaving two d orbitals empty to be involved in d2sp3 hybridisation forming inner
orbital complex in case of [Co(NH3)6]3+. In [Ni(NH3)6 ]2+, Ni is in +2 oxidation state and has d8 configuration, the hybridisation involved is
Ans-6:- Since the coordination number of Mn2+ ion in the complex ion is 4, it will be either tetrahedral (sp3 hybridisation) or square
planar (dsp2 hybridisation). But the fact that the magnetic moment of the complex ion is 5.9 BM, it should be tetrahedral in shape rather
than square planar because of the presence of five unpaired electrons in the d orbitals.
ANS:-7# Cyanide being the strong ligand , the orbital splitting energy is high in case of square plannar splitting in [Ni(CN)4]2–
The electrons are paired up during the electron fill-up in case of square plannar splitting , the transition of electrons not occurs in visible spectrum so it is colorless. In case of [Ni(H2O)6]2+ it is green , as water is a weak ligand in Octahedral
splitting , The unpaired electron can take transition in visible spectrum and the complementary colour is green.
ANS-8# [Ni(CN)4]2– forms square plannar (dsp2) complex with no unpaired electron so ,it is diamagnetic .
[Cr(NH3)6]3+ forms octahedral (d2sp3) complex with unpaired electron so ,it is paramagnetic .
ANS-9# Due the change in the nature of ligand , strong field and weak field , the orbital splitting energy changes . So ,the energy of absorption changes and hence , the complementary colour changes.
ANS-10# (iii) [Fe(C2O4)3]3– , due to presence of bidentate ligand . II# Explain these experimental results .
1 . Aqueous copper sulphate solution (blue in colour)gives: (a) a green precipitate with aqueous potassium fluoride , and
(b) a bright green solution with aq. potassium chloride . Explain these experimental results .
2.What is the coordination entity formed when excess of aq. KCN is added to an aq.solution of CuSO4 ?Why is it
that no precipitate of copper sulphide is obtained when H2S(g) is passed through the solution ?
3.A coordination compound has the formula CoCl3.4NH3 .It does not liberate NH3 but forms a precipitate of AgCl
on treatement with AgNO3 solution .Write the formula and IUPAC name of the complex
4# 0.1 M solution of a complex obtained by mixing CoCl3 and NH3 in the molar ratio 1:4 froze at -0.3720C . What
is the formula of the complex . (Kf = 1.86 0C C/m)
ANSWERS TO II : Explain these experimental results
Ans-3 :- As it does not liberate any ammonia ,it means all ammonia are acting as ligand . To form octahedral stable complex two more Cl- will also act as ligand . So only one Cl- will act as counter ion and link through primary linkage . Becxause it forms AgCl precipitate when the complex treated with AgNO3
The Complex is – [Co(NH3)4Cl2]Cl -- tetraamminedicholridocobalt(III)chloride
Ans-4 :- ∆Tf (theoretical) = Kf . m = 1.86 ×0.1 = 0.1860
∆Tf (obs) = 0.3720
i = ∆Tf (obs) /∆Tf (theoretical) = 0.3720 /0.1860 =2
The complex dissociate to form two ions [Co(NH3)Cl2]+ and Cl−
III# Write the IUPAC Name and draw the structures of all the isomers with this formula of complex. (optical & geometrical)
(i) [Co(en)2Cl2]Cl (ii) [PtCl(NH3)5]Cl3 (iv) [Co(OX)3]3‾ (v) [CoCl2(en)2]+ (vi) [CoCl2(en)(NH3)2]+ (vii)
[CoCl (en)2(NH3)]+
UNIT 11: ALCOHOLS PHENOLS AND ETHERS
1. Unlike phenols, alcohols are easily protonated. Explain. (1M)2. Why absolute alcohol can not be obtained from rectified spirit by simple distillation? (1M) 3. Explain why is o- nitro phenol more acidic than o-methoxy phenol? (1M)
4. Out of two isomeric aromatic compounds X,Y with formula C7H7OH only X gives purple
colour with neutral ferric chloride.Identify X and Y. (1M)
1. ‘A’ gives positive lucas test in 5 minutes. Its molecular mass is 60u. It reacts with PBr3 to give B which when treated with benzene in presence of AlCl3 gives compound C.
Write structural formula of A,B and C. (2M) 2. An organic compound A having formula C6H6O gives a characteristic colour with
Aq. FeCl3 solution.When A is treated with CO2 and NaOH at 400K under pressure B is obtained.
The compound B on acidification gives C which reacts with acetyl chloride to form D which is a Popular pain killer. Deduce strutures of A,B,C and D. (3M)
ANSWERS
1. In alcohols the lone pair of electrons on oxygen are localised and hence are easily available
For protonation.But in phenols the lone pair of electrons on oxygen are delocalised over the benzene ring due to resonance and hence are not easily available for protonation.
2. Rectified spirit contains 95% ethyl alcohol and 5% water and forms an azeotropic mixture Which distils at a constant temperature of 351.13K.
3. –NO2 group is an electron withdrawing group.So nitro phenol is more acidic than phenol
-OCH3 group is an electron releasing group so it is less acidic than phenol. Hence o-nitro
Phenol is more acidic than ortho methoxy phenol. 4. X is Cresol (o,mor p)
Y is Benzyl Alcohol
5. A is Secondary alcohol wiyh molar mass 60g/mol Therefore A is CH3-CH-CH3
│ OH
UNIT 12: ALDEHYDES, KETONES AND CARBOXYLIC ACIDS
1. Compound A (C4H8Cl2) is hydrolysed and then dehydrated to form a compound B C4H8O
which forms an oxime with NH2OH and gives negative tollen’s test. Write the structure
of A and B.
2. Arrange the following
i) in increasing order of acid strength
CH2ClCOOH, CHCl2COOH, CH3COOH, CCl3COOH
ii) in increasing order of reactivity towards HCN CH3CHO, CH3COCH3, HCHO, C2H5COCH3
3. Identify the compounds A B and C in the following reactions
4. Complete the following
CH3COOH Cl2/P ………….. NaCN ……….. H2O/H+ ……… Heat …..
5. Explain Why
i) Di-tert butyl ketone does not give ppt with NaHSO3 where as acetone does. ii) Benzoic acid does not give Friedel Craft’s reaction
UNIT 13 (Amines)
1. Give reason
a) p-methoxy aniline is a stronger base than aniline while p-nitroaniline is a weaker base than aniline. (1M)
b) Aromatic primary amines can’t be prepared by Gabriel Phthalamide synthesis.(1M) c) CH3CONH2 is a weaker base than CH3CH2NH2.(1M)
d) Carbon-nitrogen bond length is aromatic amines is shorter than in aliphatic amines. e) Aniline does not undergo Friedal-Craft reaction.(1M)
2. . i) Arrange in decreasing order of pkb values
C2H5NH2, C6H5NHCH3, (C2H5)2 NH, C6H5NH2 (1M)
ii) Arrange in increasing order of basic character Aniline, p-nitro aniline, p-toulidine
3. The compound C5H13N is optically active and reacts with HNO2 to give
C5H11OH and liberates N2 gas. Identify the compound and write the
reactions involved. (2M) 4. Name A,B,C in the following equation
a) CH3CH2COOH → A + Heat → B in presence of Br2/KOH → C
b) A Br2/KOH B HNO2 C Red P/I2 CH3I
ANSWERS
1. a) Methoxy group is electron releasing group whereas Nitro group is electron withdrawing group therefore in p-methoxy aniline the electron density on N- atom of –NH2 group increases making it more basic but in p-nitro
aniline the electron density on N- atom of –NH2 group is decreased making
it less basic.
b) Aromatic primary amines can not be prepared by Gabriel Phthalimide Synthesis because aryl halides do not under go nucleophillic substitution with the anion formed by phthalimide on reaction with KOH.
c) Due to resonance in CH3CONH2 , the lone pair of electrons in CH3CONH2 is delocalised .
Due to this electron density on N-atom in acetamide decreses. In CH3CH2NH2 due to +I effect of
CH3CH2 group electron density on N- atom increses . As a result CH3CONH2 is a weaker base than
CH3CH2NH2
d) It is because in aromatic amines C-N bond acquires some double bond character and thus becomes shorter than in aliphatic amines.
e) AlCl3 used as catalyst in Friedelcrafts reaction forms a salt with aniline
.Due to this nitrogen aquires positive charghe . Thus it acts as strong deactivating group for electrofilic reaction. Hence anilie does not give Foriedelcraft reaction.
2. i. C6H5NH2 > C6H5NHCH3> C2H5NH2>(C2H5)2NH
ii. p-nitroamiline<amiline<p-tolidine 5) i. A=CH3CH2COONH4
B= CH3CH2CONH2
C= CH3CH2NH2
ii. A=CH3CONH2
B=CH3NH2
C=CH3OH
ORGANIC REASONING QUESTIONS ( for practice)
Q.1 # Amino group is o,p-directing for aromatic electrophilic substitution . Why does aniline on nitration gives a substantial amount of meta-nitroaniline ?
Q.2# Aniline does not undergo Friedel-Crafts reaction (alkylation and acetylation) (PN-395) Q.3# Diazonium salts of aromatic amines are more stable than those of aliphatic amines. Q.4# Gabriel phthalimide synthesis is preferred for synthesising primary amines.
Q. 5 # Why cannot aromatic primary ammines be prepared by Gabriel phthalimide synthesis . Q.6# Why sec- & tert- alkyl halides are not used during Williamson’s Synthesis .
Q.7 # Although haloalkanes are usually prepared from alcohols , haloarenes can’t be prepared from phenols.
Q.8 # Give reason for the higher boiling point of ethanol in comparison to methoxy methane ? Q.9# Preparation of ethers by acid dehydration of secondary and tertiary alcohols is not a suitable method . Give reason . Q.10# Why is sulphuric acid not used during the reaction of alcohols with KI?
ANSWERS
ANS-1-- in the strongly acidic medium, aniline is protonated to form the anilinium ion which is meta directing. That is why besides the ortho and para derivatives, significant amount of meta derivative is also formed.
ANS-2-Aniline does not undergo Friedel-Crafts reaction (alkylation and acetylation) due to salt formation with aluminium chloride, the Lewis acid, which is used as a catalyst. Due to this, nitrogen of aniline acquires positive charge and hence acts as a strong deactivating group for further reaction. ANS-3- Primary aliphatic amines form highly unstable alkyldiazonium salts . Primary aromatic amines form arenediazonium salts which are stable for a short time in solution at low temperatures (273-278 K). The stability of arenediazonium ion is explained on the basis of resonance.
ANS-4- Because we obtain pure primary amines in Gabriel phthalimide synthesis which is not possible in ammonolysis .
ANS-5- Aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide. Due to resonance it developes partial double bond character between aromatic carbon and nitrogen.
Ans-6:- In case of secondary and tertiary alkyl halides, elimination competes over substitution. If a tertiary alkyl halide is used, an alkene is the only reaction product and no ether is formed. For example, the reaction of CH3ONa with (CH3)3C–Br gives exclusively 2-methylpropene. It is because
alkoxides are not only nucleophiles but strong bases as well. They react with alkyl halides leading to elimination reactions
Ans-7:- Because the carbon of phenyl group is sp2 hybridised and there is a partial double bond character due to resonance
Ans-8 :- Ethanol get associated with H-bonding but methoxy methane does not .
ANS-9--The method is suitable for the preparation of ethers having primary alkyl groups only. The alkyl group should be unhindered and the temperature be kept low. Otherwise the reaction favours the formation of alkene. The reaction follows SN1 pathway when the alcohol is secondary or tertiary. However, the
dehydration of secondary and tertiary alcohols to give corresponding ethers is unsuccessful as elimination competes over substitution and as a consequence, alkenes are easily formed.
Answer-10-H2SO4 cannot be used along with KI in the conversion of an alcohol to an alkyl iodide as it
converts KI to corresponding HI and then oxidizes it to I2.
Organic CONVERSION With Word Problems (5 Marks) FOR PRACTICE IDENTIFY THE COMPOUNDS A,B,C etc. by akp
Q.1# An organic compound (A) with molecular formula C8H8O forms an orange-red precipitate(B) with 2,4-DNP reagent and gives
yellow precipitate( C ) on heating with iodine in the presence of sodium hydroxide along with a colourless compound( D ) . ( A ) neither reduces Tollens’ or Fehlings’ reagent, nor does it decolourise bromine water or Baeyer’s reagent. On drastic oxidation of ( A ) with chromic acid, it gives a carboxylic acid (E) having molecular formula C7H6O2. Deduce the structures of organic compounds (A) to (E) and
explain the reactions involved. (TBQ and Sample Paper-1)
Q.2# An organic compound with the molecular formula C9H10O forms 2,4-DNP derivative, reduces Tollens’ reagent and undergoes
Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound. (TBQ)
Q.3# An organic compound (A) (molecular formula C8H16O2) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and
an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene. Write equations for the reactions involved. (TBQ)
Q.4# An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It
does not reduce Tollens’ reagent but forms an addition compound with sodium hydrogensulphite and give positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound. (TBQ)
Q.5# Primary alkyl halide C4H9Br (a) reacted with alcoholic KOH to give compound (b). Compound (b) is reacted with HBr to give (c)
which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d), C8H18 which is different from the compound
formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions. (TBQ)
Q.6# An organic compound (A) with molecular formula C5H8O2 is reduced to n-pentane on treatment with Zn-Hg/ HCl . ‘A’ forms a
dioxime with hydroxylamine and gives a positive iodoform test and Tollen’s test.Identify the compound ‘A’ and deduce its structure.(
Q.7# An organic compound (A) with molecular formula C8H8O gives a positive DNP and iodoform test . It does not reduce Tollens’ or
Fehlings’ reagent and does not decolourise bromine water also. On oxidation with Chromic acid(H2CrO4) , It gives carboxylic acid (B)
with a molecular formula C7H6O2 . Deduce the structures of organic compounds (A) and (B). (SP-2)(3 M)
Q.8 # An unknown aldehyde ‘A’ on reacting with alkali gives a β-hydroxy aldehyde , which losses water to form an unsaturated aldehyde
, 2-butenal . Another aldehyde ‘B’ undergoes disproportionation reaction in the presencence of conc.alkali to form products ‘C’ and ‘D’ . ‘C’ is an aryl alocohol with a formula C7H8O.
(i) Identify A and B (ii) Write the sequence of reactions involved (iii) Name the product , when B reacts with Zinc amalgam and hydrochloric acid . (SP-3) ( 5 Marks)
Q.9# A compound ‘X’ (C2H4O ) on oxidation gives ‘Y’ (C2H4O2) . ‘X’ undergoes haloform reaction.On treatment with HCN ‘X’ forms a
product ‘Z’ which on hydrolysis gives 2-hydroxy propanoic acid .
(i) Write down structures of ‘X’ and ‘Y’ (ii) Name the product when ‘X’ reacts with dil NaOH (iii) Write down the equations for the reactions involved. (SP-3) ( 5 Marks)
Q.10 # An optically active compound having molecular formula C7H15Br reacts with aqueous KOH to give a racemic mixture of products
.Write the structures of the compounds .
Additional Questions for practice …………
Q.11# A compound (A) with molecular formula C5H12O , on oxidation forms compound (B) with molecular formula C5H10O .The
compound (B) gives iodoform test ,but does not reduce ammoniacal silver nitrate solution.(B) on reduction with amalgamated zinc and conc.HCl gives compound (C)with molecular formula C5H12. Identify (A),(B),(C), write down the chemical reaction involved . (3) Q.12# An organic compound (A) with a molecular formula C3H6O2 when treated wit h excess of NH3 and then heated formed B
(C3H7ON). B on treatment with P2O5 formed C with molecular formula C3H5N. C on complete acid hydrolysis formed A . C on reduction
with complex metal hydride formed a basic nitrogeneous compound D with molecular formula C3H9N . D on treatment with HNO2
formed an alcohol ‘E’ (C3H8O).E on oxidation gave A . Give structures of A to E and write reactions . ( 5 Marks)
Q.13# An organic compound “A” has a molecular formula C3H7O2N on reaction with Fe and Conc.HCl give a compound “B” of
molecular formula C3H9N . Compound “B” on treatment with NaNO2 and HCl give another compound “C” of molecular formula
C3H8O.The compound “C” gives effervescence with Na.On oxidation with CrO3, the compound “C” gives a saturated aldehyde
containing three carbon atoms .Deduce the structures of “A”, “B” and “C” and write the chemical equations for the reactions involved. Q.14# A compound ‘A’ of molecular formula C7H7NO. On treatment with Br2 and NaOH it gives B .B gives a foul smelling substance C
on treatment with CHCl3 and KOH . B on treatment with NaNO2 and HCl at 00C and mixing with alkaline phenol give a red dye
D.Identify A to D
Q.15 # An organic compound “A” which has characteristic odour, on treatment with NaOH forms two compounds “B” and “C” .Compound “B” has a molecular formula C7H8O which on oxidation gives back compound “A’.Compound “C” is the sodium salt of an
acid .”C” when heated with sodalime yields an aromatic hydrocarbon”D”.Deduce the structure A,B,C,D .
Q.16# An organic compound “A” has a molecular formula C5H10O.It does not reduce Fehling’s Solution but forms a bisulphite
compound .It also positive iodoform test.What is the possible structure of “A”? Explain your reasoning which helped to arrive at the structure.
Q.17 # A compound “A” has a molecular formula C5H10O gave a positive 2,4-DNP test but a negative Tollen’s test. It was oxidized to
carboxylic acid “B” with molecular formula C3H6O2 when treated with alkaline KMnO4 under vigorous condition .Sodium salt of ‘B”
gave a hydrocarbon “C” on Kolbe’s electrolytic reduction Identify A,B and C and write chemical equations for the reaction. Q.18 # An organic compound ‘A’ having molecular formula ,C6H6O gives a characteristic colour with aqueous FeCl3 solution.’A’on
treatment with CO2 and NaOH at 400K under pressure gives ‘B’ which on acidification gives a compound ‘C’ . ‘C’ reacts with acetyl
chloride to give ‘D’ which is a popular pain killer .Deduce the structures of A,B,C and D . .
ORGANIC DISTINGUISH BETWEEN PAIRS OF COMPOUNDS FOR CBSE
(2 MARKS)
GIVE SIMPLE CHEMICAL TEST TO DISTINGUISH BETWEEN FOLLOWING PAIRS OF COMPOUNDS
PRIORITY-I PRIORITY-II
1# Benzyl chloride and Chlorobenzene
1 # Vinyl Chloride and Allyl Chloride
2# Methanol and Ethanol 2# Formic acid and Acetic acid .
3# Phenol and Benzoic acid 3# Phenol and Ethanol
4# Ethanol and Ethoxyethane 4# Phenol and cyclohexanol
5# Propanal and Propanone 5# Propanoyl chloride and Propanoic acid.
6# Ethylamine and Aniline. 6# Benzamide and p-Aminobenzoic acid .
7# Ethylamine and N-ethylaniline.
7# Propanal and Diethyl ether.
8# t-butylalcohol and Butan-2-ol .
8# Acetaldehyde and Benzaldehyde.
Methylcarbylamine
10# Nitromethane and Methyl nitrite
10# Butanol and 2-methyl propan-2-ol 11# Benzoic acid and Ethyl benzoate . 11# Propanal and butanone
12# Acetophenone and Benzophenone. 12# Isopropyl alcohol and propanol 13# Benzalehyde and Acetophenone.
14# Pentanone -2 and Pentanone-3. 15# Ethanal and Propanal
16# Methylamine and dimethylamine 17# Aniline and benzylamine
18# Aniline and N-methylaniline 19# Methyl acetate and Ethyl acetate. 20# Formaldehyde and Acetaldehyde
R E A G E N T S TO BE USED ***********
1# Iodoform test [POSITIVE TEST for ethanal, ethanol, --2—one , -2-ol , terminal CH3CO group, acetophenone ] ( NaOH + I2 )
OR (Na 2CO3 + I2 ) OR NaOI (Iodoform tests )--- yellow ppt. of iodoform forms.
2# Test for phenol
(a)Neutral Ferric Chloride solution Phenol gives violet colour complex with neutral FeCl3.
(b) Bromine water test:Phenol and Aniline readily decolourises bromine water giving a white ppt. of 2,4,6-tribromophenol /2,4,6-tribromoaniline
3# Test for Aldehydes and Formic acid
(a)Tollen’s Reagent (ammoniacal AgNO 3 solution)--- silver mirror appears.
(b) Fehling’s solution(alkaline sol. Of CuSO4 + sod.pot.tartarate)---- redish brown ppt. of cuprous oxide forms.
4# TEST FOR ALCOHOLS[ Na metal test] -- hydrogen gas evolved (effervescence) [ether does not responds this test] 5# TEST FOR CARBOXYLIC ACIDS [NaHCO3 solution ]----carbon dioxide evolves with acids. (effervescence)
6# ISOCYANIDE TEST OR CARBYLAMINE TEST ( test for primary amine)
CHCl3 + alcoholic KOH solution----an offensive smell of isocyanide results when treated with primary amine .
7# AZODYE TEST [Test for primary aromatic amines]
( NaNO2 +HCl) and beta-napthol (diazo test)----an orange or red dye will appear with primary aromatic amine
8# Hinsberg’s reagent ( benzene sulphonyl chloride and KOH ) ----To distinguish p-,s-,t- amines
Both 10 and 20 Amine reacts with benzene sulphonyl chloride ,the reaction product of 10 amine soluble in KOH due to presence of
acidic hydrogen but 20 Amine does not. 30 Amine does not react at all with benzene sulphonyl chloride
09# TEST FOR CHLORIDE ION [aq.NaOH and AgNO3 test ] -- white ppt. of AgCl forms due to free chloride ion
[ In a compound when Cl attached with SP3 carbon responds this test—easy for SN reaction]
10# Lucas Test ( anh. ZnCl2 + conc.HCl )---( turbidity appears immediately for tert-alcohol, after 5 minutes for sec –alcohol , after
heating with primary alcohol )
11# Acidic hydrolysis of ester + Iodoform test. ( to distinguish methyl acetate and ethyl acetate ) 12 # Reduction ( LiAlH4 ) and carbylamine test . (for nitrile and isonitrile)
JUST FOR PRACTICE ---- ORGNIC CHEMISTRY ( EXAM SPECIAL) by A.K.Panda
1 # Explain why ? Give reason
(i) Treatment of C6H5CHO and HCN gives a mixture of two isomers which cannot be separated even by
very careful fractional distillation .Why ? (1)
(ii) Why is vinyl chloride less reactive than ethyl chloride towards nucleophile ? (1) (iii)Why formaldehyde does not respond Aldol condensation reaction ? (1)
(iv) Why CF3C+HCH3 is much less stable than isopropyl cation? (1)
(v) Explain why this takes place with anti-Markownikoff’s regiochemistry (1) CF3CH=CH2 + HCl CF3CH2CH2Cl
(vi) Why does the reactivity of NH2 get reduced in acetanilide ? (1)
(vii) Amino group is o,p-directing for aromatic electrophilic substitution . Why does aniline on nitration gives a substantial amount of meta-nitroaniline ? (1)
(viii) There are two –NH2 groups in semicarbazide. However, only one is involved in the formation of
semicarbazones. (1)
(ix) Haloalkanes react with KCN to form alkyl cyanides as main product while AgCN forms isocyanides as the chief product. Explain. (1)
(x) Why sec- & tert- alkyl halides are not used during Williamson’s Synthesis . (1) (xi) Why N,N-Diethylbenzenesulphonamide is not soluble in alkali . (1)
(xii) Which compound will react faster in SN2 reaction with OHˉ and why ? (CH2=CHBr and CH2=CH—CH2Br )
(xiii) Although phenoxide ion has more number of resonating structures than carboxylate ion , carboxylic acid is a
stronger acid than phenol . Why? (1)2#(A) Give the structures of the major product(s) which are obtained when
(i) 2-methyl-2-methoxypropane is treated with hydroiodic acid .
(ii) Methyl magnesium bromide is treated with CO2 and subsequently hydrolysed .
(iii) Salicylic acid is acetylated
(iv) butan-1-ol is treated with Jones reagent (v) Hex-4-ene-nitrile is treated with DIBAL-H
(vi) ethanoyl chloride is treated with diethyl cadmium .
(vii) benzene diazonium chloride treated with DMA in presence of a base . (viii) Methanamine treated with benzoyl chloride .
(ix) Propanoic acid is treated with NABH4 .
(x) 1-bromo-1-methylcyclohexane is treated with alcoholic KOH (xi) Cyclohexene is treated with hot acidified KMnO4 .
(xii) Cyclohehanone is treated with semicarbazide .
(xiii) Benzaldehyde is treated with Acetophenone in presence of dilute NaOH . (B) How do products differ when ethyl bromide reacts separately with (i) aq . KOH and alc. KOH (ii) KNO2 and AgNO2 (iii) KCN and AgCN
3# Write chemical test to distinguish between
(i) Formic acid and acetic acid (ii) C6H5CH2NH2 and C6H5NH2
( iii) Propanoyl chloride and Propanoic acid. (iv) Vinyl Chloride and Allyl Chloride 4# Write the mechanism of
(i) Hydration of propene to yield propanol . (ii) Ethanamine is converted into N-Ethylethanamide . (iii) 1-propoxypropane from propan-1-ol (iv) Reaction of HI with methoxy ethane
5 # Give the IUPAC Name of
(i) Methyl vinyl ketone . (ii) Benzophenone (iii) Sulphanilic caid (iv) Aspirin (v) Terepthalic acid (vi) Adipic acid (vii) Phenetole (viii) Propiophenone (ix) DDT (x) Freon
UNIT-14# BIOMOLECULES PART-I : Answer the following :
1# An optically active compound having molecular formula C6H12O6 is found in two isomeric forms(A) and (B) .When (A) and (B) are dissolved in water they show the following equilibrium: (A)--- equilibrium mixture ---(B)
[α] = 1110 52.20 19.20
(i) What are such isomers called ? (ii) Can they be called enantiomers ? Justify your answer . (iii) Draw the cyclic structure of isomer (A) .
2# An optically active amino acid (A) can exist in three forms depending on the PH of the medium.If the molecular formula of (A) is C3H7NO2 then write (i) Structure of compound (A) in aqueous medium. What are such ions called ?
(ii) In which medium will the cationic form of compound (A) exist ?
(iii) In alkaline medium, towards which electrode will the compound (A) migrate in electric field ?
3 # When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained unlike DNA ? What does this fact suggest about the structure of RNA?
4# What are reducing and non-reducing sugars ? Give two examples. .What is the structural feature characterizing reducing sugars ?
5# Write down the structures and names of the products obtained when D-Glucose is treated with (a) HI (b) Br2 Water (c) HNO3 (d) NH2OH (e) Acetic Anhydride (f) HCN (g) ammoniacal AgNO3
6# What is the significance of D and (+) in α-D(+)-glucopyranose
7 # What are the expected products hydrolysis of (i) Lactose (ii) sucrose (iii) Maltose 8# What type of bonding helps in stabilising the α-helix structure of proteins?
