SECTION- A
1. Prove that for a first order reaction the time required for 99.9 % completion is about 10 times its half-life period.Ans.
Ans.
a. (CH3CH2)2NCH3
b. m−BrC6H4NH2
Ans. a. N-Ethyl-N-methylethanamine
b. 3-Bromobenzenamine or 3-bromoaniline
SECTION- B
4. . Following are the transition metal ions of 3d series:Ti3+, V2+, Mn3+, Cr3+(Atomic numbers : Ti = 22, V = 23, Mn = 25, Cr = 24)
Answer the following:
(a) Which ion is most stable in an aqueous solution and why?
(b) Which ion is a strong oxidising agent and why?
(c) Which ion is colourless and why?
Ans. Ti4+ = 1s22s22p63s23p6
V2+ = 1s22s22p63s23p63d3
Mn3+ = 1s22s22p63s23p63d4
Cr3+ = 1s22s22p63s23d3
(a) Ti4+ is most stable in an aqueous solution because of fully filled valence shell (3s23p6) configuration (noble gas configuration).
(b) Mn3+ is the strong oxidising agent as it oxidises other species it will reduce itself by taking an e– and will stabilise its configuration (3d5).
(c) Ti4+ is colourless due to absence of unpaired electrons (3s23p6).
OR
4. Complete the following equation :
(a) 2MnO4– + 16H+ + 5S2– →
(b) KMnO4 +heat ------->
(c) What is the general electronic configuration of transition elements?
Ans. (a) 2MnO4– + 16H+ + 5S2– —→ 2Mn2+ + 5S + 8H2O
(b) 2KMnO4 + heat ----> K2MnO4 + MnO2 + O2
(c) The general electronic configuration of transition elements is (n – 1)d1–10ns1-2
5. Answer the following:
(a) Write the formula of the following coordination compound - Iron (III) hexacyanoferrate (II)
(b) Predict the number of unpaired electron in [MnBr4]2–.
(c) Write the hybridisation and number of unpaired electrons in the complex [CoF6]3–. (Atomic No. of CO =27)
Ans. (a) Molecular formula of Iron(III) hexacyanoferrate(II) is Fe4[Fe(CN)6]3.
(b) [MnBr4]2–, Mn has (+ 2) oxidation state hence 4s03d5 configuration. Br– is a weak field ligand, five unpaired electrons. So, highly paramagnetic.
(c) Electronic configuration of Co3+ ion is: 3d6 Electronic configuration of sp3d2 hybridised (as F– is a weak field ligand) orbitals of Co3+, with six pairs of electrons from six F– ions:
6. Answer the following questions:
(a) What is zeta potential? Explain.
(b) Why colloidal solutions differ in colour?
(c) What is added to gasoline to decrease knocking?
Ans. (a) As colloidal particles adsorb a specific charge on their surface, to acquire either negative or positivecharge, they attract further oppositely charged ions from the dispersed medium which forms a second layer of mobile opposite charge which surrounds the first fix layer. Such a double layer of opposite charges is called Helmholtz-doublelayer.The mobile layer differences into the bulk of the liquid. The potential difference between the fixed layer and the mobile diffused layer is called electrokinetic potential or zeta potential.
(b) Colour of a colloidal solution depends upon the size and shape of dispersed phase and nature of dispersion medium. Hence, different colloidal solutions transmit different light of spectrum depending upon the variations and hence have different colours.
(c) Tetraethyl lead (TEL).
7. A first order reaction takes 20 minutes for 25% decomposition. Calculate the time when 75% of the reaction will be completed. (Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021)
Ans.
(a) Name the class of compounds being discussed. What is the hybridisation state of nitrogen in these class of compounds?
(b) What is ammonolysis?
(c) What are the compounds obtained on ammonolysis of ethyl chloride?
Ans. (a) The class of compounds here is ‘amines’. Hybridisation state of nitrogen in these class of compounds is sp3.
(b) The process of cleavage of C-X bond of an alkyl halide molecule by ammonia molecule is known as ammonolysis. The halogen atom of alkyl or benzyl halide gets replaced by an amino (–NH2) group. The primary amine thus obtained behaves as a nucleophile and in subsequent reactions, secondary amine, tertiary amine and quaternary ammonium salts are produced.
(c) After the formation of ethyl amine(primary amine),it goes further to form secondary amine,tertiary amine and quaternary ammonium salt respectively.
9. Account for the following:
(a) p-Nitrobenzoic acid has higher Ka than benzoic acid.
(b) Carboxylic acids have higher boiling points than alcohols.
(c) Acetone is soluble in water but benzophenone is not.
Ans. (a) Para nitro benzoic acid is more acidic than benzoic acid because of the NO2 functional group which is a electron withdrawing group. Due to which the electron density on the hydrogen atom of the carboxylic acid becomes low ,and it can be easily removed. Hence p-Nitrobenzoic acid is more acidic and have higher Ka value.
(b) Carboxylic acids have higher boiling point than alcohols due to more extensive association of carboxylic acid molecules through inter-molecular hydrogen bonding. The hydrogen bond formed by the carboxylic acids are stronger than those in alcohols because O−H bond in -COOH is more strongly polarised due to the presence of electron withdrawing carboxyl group in adjacent position than the O−H bonds of alcohols. Therefore, the boiling points of carboxylic acids particularly lower members, are higher than alcohol of comparable molecular masses.
(c) Acetone is highly soluble in water because it can undergo hydrogen bonding with water molecules because of the presence of polar carbonyl group. In case of benzophenone the carbonyl group is sterically hindered by two big phenyl group (C6H5COC6H5), hence, the carbonyl oxygen is masked and cannot participate in hydrogen bonding with water.
10. Answer the following questions:
(a) Name two oxo-metal onions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.
(b) What are the different oxidation states exhibited by the lanthanoids?
(c) Complexes of which transition metals are useful in the polymerisation of alkynes?
Ans.(a) MnO4− : Mn has +7 oxidation state and group number is 7.
CrO42− : Cr has + 6 oxidation state and group number is 6.
(b) In the lanthanide series, + 3 oxidation state is most common i.e., Ln(III) compounds are predominant.
However, + 2 and + 4 oxidation states can also be found in the solution or in solid compounds.
(c) Nickel (Ni) metal is useful in the polymerisation of alkynes.
11.(a) State the products of electrolysis obtained on the cathode and anode in the following:
(i) Dilute solution of H2SO4 with Pt electrodes.
(ii) Aqueous solution of AgNO3 with Ag electrodes.
(b) Write the cell formation and calculate the standard cell potential of the galvanic cell for the following reaction:
Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s) Also, calculate G° for the above reaction. Given: [E°Ag+/Ag = + 0.80 V, E°Fe3+/Fe2+ = 0.77 V] 1 F = 96500 C mol–1
Ans. (a) (i) Dilute solution of H2SO4 with Pt electrode have ions, H+, SO42– and OH–in it as Pt is an inert electrode.
At cathode (Reduction) : 2H+(aq) + 2e– → H2(g)
At anode (Oxidation) : 2OH-(aq) → O2(g) + 2H+(aq) + 4e–
SO42– will remain in solution as it has a higher discharge potential than OH– .So,H2 gas is evolved at cathode and O2 gas is evolved at anode. (ii) Aqueous solution of AgNO3 with Ag electrode.Ions present are: Ag+, H+, OH–, NO3– and Ag atom.
At cathode (Reduction): Ag+(aq) + e– → Ag(s)
At anode (Oxidation): Ag(s) → Ag+(aq) + e–
As Ag has highest discharge potential compared to all other ions hence it gets dissolved as Ag+ ion.Hence at cathode we get solid Ag deposit and at anode the silver electrode dissolves. (b) Pt, Fe2+ | Fe3+(aq) || Ag+(aq) | Ag(s)
E°cell = E°Ag+/Ag – E°Fe3+/Fe2+
= + 0.80 V – (+ 0.77 V) = 0.03 V
G° = – nFE°cell n = 1
G° = – 1 × 96500 C mol–1 × 0.03 V
= – 2895 J mol–1 = – 2.895 kJ/mol–1.
SECTION- C
12. Read the passage given below and answer the questions that follow: Both aldehydes and ketones are collectively termed as carbonyl compounds as they contain carbonyl group.Aliphatic aldehydes and ketones both are expressed by the general formula CnH2nO. For example both propanaldehyde (CH3CH2CHO) and action (CH3COCH3) can be represented by the molecular formula C3H6O.The simplest aromatic aldehyde is benzaldelyde.(a) What happens when acetaldehyde is reacted with a trace of H2SO4.
(b) Which type of aldehydes undergo aldol condensation?
(c) Write one chemical reaction to examplify cannizzaro reaction.
(d) Which aldehyde is a gas? Why it is soluble in water?
OR
(d) Write the mechanism of addition of HCN to >C = O group.
Ans. (a) Acetaldehyde on reaction with trace of H2SO4 gives a cyclic trimer compound known as paraldehyde.
(c)
OR
