Higher Order Thinking Questions

+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 1Is 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) (AI-2008)

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,CO2 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 ,CO2) 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 0 (ii) 2 Al + O2  Al2O3 ΔfG = -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 . At 2273 K:Mg(s) + ½ O2(g)  MgO(s) ; ΔfG (MgO(s)) = -341 kJ/mol at 2273K -----(3)

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 +heatNa2SO4 + 2HNO3 4HNO3+ heat4NO2+2H2O+O2 Cu+4HNO3Cu(NO3)2 +2NO2 +2H2O 2NO2[A] ═(cool)=> N2O4[B] ANS-3# P4(white Phosphorous)[A] +very dil KOHPH3[C] P4(white Phosphorous)[A] +heatRed- 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.H2SO411H2O +12C (black subs.) (b) Test tube(ii):2NaBr +3H2SO42NaHSO4 +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) +FeCl3Fe(OH)3 +3 NH4Cl Fe(OH)3 + 3HNO3  Fe(NO3)3( soluble) +3H2O ANS-6: 3Ca + N2Ca3N2 [B] [A] Ca3N2 + 6 H2O3 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 . 12# NF3 is more stable but NCl3 is less stable .

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. H2SO4 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 3px23py13pz1 (Ground state) ; S  3s2 3px13py13pz1 3d1 (excited state) Sp2 hybridisation in Sulphur . one half filled 3pz1 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 5# H2O < H2S < H2Se < H2Te ( Acid Strength and Reducing Character) --- BDE

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 28# Stability—ClO4ˉ > ClO3ˉ > ClO2 ˉ >ClOˉ

UNIT-8:d-Block Elements 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 t2g 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 ) 0 2+ Q.5 # The E (M /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 C2O4 2‾ . Ans- Since Clˉ of Hydrochloric acid is oxidized to chlorine by MnO4ˉ . so , actual amount of KMnO4 required to oxidize C2O4 2‾ couldnot be determined.

II# 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#EXPLAIN ? 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– 10 11# Silver atom has completely filled d orbitals (4d ) 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 pielectrons(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

region: [Ni(H2O)6]2+ [Ni(NH3)6]2+, [Ni(NO2)6]4–,

visible

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 sp3d2 forming outer orbital complex.

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]+ + [CoCl (en)2(NH3)]

UNIT 11:

ALCOHOLS PHENOLS AND ETHERS

(vii)

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 N2O4[B] ANS-3# P4(white Phosphorous)[A] +very dil KOHPH3[C] P4(white Phosphorous)[A] +heatRed- 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.H2SO411H2O +12C (black subs.) (b) Test tube(ii):2NaBr +3H2SO42NaHSO4 +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) +FeCl3Fe(OH)3 +3 NH4Cl Fe(OH)3 + 3HNO3  Fe(NO3)3( soluble) +3H2O ANS-6: 3Ca + N2Ca3N2 [B] [A] Ca3N2 + 6 H2O3 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 . 12# NF3 is more stable but NCl3 is less stable . 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. H2SO4 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 3px23py13pz1 (Ground state) ; S  3s2 3px13py13pz1 3d1 (excited state) Sp2 hybridisation in Sulphur . one half filled 3pz1 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 5# H2O < H2S < H2Se < H2Te ( Acid Strength and Reducing Character) --- BDE 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 28# Stability—ClO4ˉ > ClO3ˉ > ClO2 ˉ >ClOˉ

UNIT-8:d-Block Elements 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 t2g 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 ) 0 2+ Q.5 # The E (M /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 C2O4 2‾ .

Ans- Since Clˉ of Hydrochloric acid is oxidized to chlorine by MnO4ˉ . so , actual amount of KMnO4 required to oxidize C2O4 2‾ couldnot be determined.

II# 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#EXPLAIN ? 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– 10 11# Silver atom has completely filled d orbitals (4d ) 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 pielectrons(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 sp3d2 forming outer orbital complex. 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]+ [CoCl (en)2(NH3)]+

(vii)

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