SECTION- A
1. Arrange the following in the increasing order of their property indicated (Any 2)a. Phenol , benzene , benzaldehyde and benzoic acid (Boiling Point)
b. Methanol , butanol , octanol (Solubility In Water)
c. CH3CH2OH , CH3COOH , C6H5CH2COOH , NO2CH2COOH (Acidic Strength)
Ans. (a) Benzene < benzaldehyde < phenol < benzoic acid.
(b) octanol < butanol < methanol.
(c) CH3CH2OH < CH3COOH < C6H5CH2COOH < NO2CH2COOH
2. Define molar conductivity for solution of an electrolyte.Discuss the variaton of molar conductivity with change in by giving oncentration a graphical representation.
Ans. Molar conductivity of a solution at a dilution V is the conductance of all the ions produced from 1 mole of the electrolyte dissolved in V cm3 of the solution when the electrodes are one cm apart and the area of the electrodes is so large that the whole of the solution is contained between them. It is represented by ∆m. Molar conductivity increases with decrease in concentration. When concentration approaches zero, the molar conductivity is known as limiting molar conductivity and is represented by the symbol Ë°m. For strong electrolytes, Λ increases slowly with dilution and can be represented by the equation: Λm = E°m° – A c ½
3. Give reason to support the answer:
a. Formaldehyde does not take part in aldol condensation .
b. Aldehydes and ketones have lower boiling points than corresponding alcohols.
Ans.(a) Formaldehyde does not contain a-hydrogen atom. Therefore it does not take part in aldol condensation.
(b) It is due to weak molecular association in aldehydes and ketones arising out of the dipole- dipole interactions.
SECTION- B
4. Account for the following:a. Alkylamines are more basic than ammonia.
b. Methylamine in aq. medium gives reddish brown ppt. with FeCl3.
c. Electrophilic substitution in case of aromatic amines takes place more readily than benzene.
Ans.(a) Due to electron releasing inductive effect (+I) of alkyl group, the electron density on the nitrogen atom increases and thus, it can donate the lone pair of electrons more easily than ammonia.
(b) Methylamine being more basic than H2O, it accepts a proton from water liberating OH– ions.
(c) Aniline exists as a resonance hybrid of the following five structures :
OR
4. Convert the following :
a. Ethanamine to methanamine
b. Benzoic acid to aniline
c. Methylbromide to methylisocynide
Ans.
5. Answer the following questions:
a. Why a solution of [Ni(H2O)6]2+ is green while a solution of [Ni(CN)4]2- is colourless?
b. Write the formula and hybridisation of the compound: Potassium tetrachloridonickelate.
Ans.(a) [Ni(H2O)6]2+ is an outer orbital complex due to weak field ligand H2O and the presence of unpaired electrons undergoes d—d transition by absorbing red light and shows green colour while [Ni(CN)4]2- is an inner orbital complex and has no unpaired electrons hence colourless.
(b) Formula of the complex potassium tetrachloridonickelate (II) K2[NiCl4] and hybidisation is sp3
5. A coordination compound has electronic configuration of the central metal ion is t2g3eg2
a. Is the coordination compound a high spin or low spin complex?
b. Draw the crystal field splitting diagram for the above complex.
Ans.(a) This compound is high spin complex becouse electrons are going to higher energy level eg means pairing energy is more.
(b)
6. Accont for the following:
a. Zn , Cd , Hg are considered as d block elements but not as transition elements.
b. Generally there is an increase in density of elements from Titanium (z = 22) to Copper (z= 29) in the first series of transition elements.
c. Cu(I) ion is not known in aqous solution.
Ans(a)Elements like Zn, CD and Hg of the 12 column of the d block have completely filled d-orbital and hence are not considered as transition elements. Transition Elements are so named, indicating their positioning and transition of properties between, s and p block elements.
(b) (i) The density of elements from titanium to copper increase in the first series of transition elements. This is due to decrease in metallic radius coupled with increase in atomic mass results in a general increase in the density. (ii) Many transition metals and their components show catalytic properties.
(c) Cu2+(aq) is much more stable than Cu+(aq). This is because although second ionization enthalpy of copper is large but Δhyd (hydration enthalpy) for Cu2+(aq) is much more negative than that for Cu+(aq) and hence it more than compensates for the second ionization enthalpy of copper. Therefore, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation as follows : 2Cu+ → Cu2+ + Cu.
7. An organic compound (A) with molecular formula C8H1602 was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid also produced (B) . on dehydration (C) gives but-1-ene. Write the equations for the reactions involved.
Ans. Since the organic compound (A) with molecular formula (M.F.) C8H1602 upon hydrolysis with dil. H2S04 gives carboxylic acid (B) and the alcohol (C) therefore it must be an ester. Further since oxidation of (C) with chromic acid produces the acid (B), therefore both the carboxylic acid (B) and the alcohol (C) must’.contain the same number of carbon atoms.
8. a. What happens when a freshly precipitated Fe(OH)3 is shaken with little amount of dilute solution of FeCl3?
b. What are peptizing agents?
c. Differentiate between peptization and coagulation.
Ans:
When freshly precipitated Fe(OH)3 is shaken with the little amount of dilute solution of FeCl3 (electrolyte). Peptization process (ions of the electrolyte are adsorbed by the precipitate particles) takes place by converting the Fe(OH)3 precipitate into a colloidal solution positively charged solution of Fe(OH)3.
b. The process of converting a precipitate into colloidal sol by shaking it with dispersion medium in the presence of small amount of electrolyte. The electrolyte used in this process is called peptizing agent.
c.
9. What happens when:
a. Benzene diazonium chloride reacts with H3PO2
B. Aniline with Bromine
c. N-Methylaniline with mixture of chloroform and aq. KOH
9.a. Write the IUPAC name for the following organic compound:
C6H5NH-CH3
b. Complete the following:
i. C6H5N2Cl + C2H5OH → A
ii. C6H5NH2 + conc. H2SO4 → B
Ans.
10. Represent the cell in which the following reaction takes place. The value of Eo cell is 3.17 V .What is the value of Ecell?
Mg (s) + 2Ag+(0.0001M) ---------> Mg2+(0.130M) + 2Ag(s)
11.a. Manganese exhibits the highest oxidation state of +7 among the 3d series of transition elemements.Why?
b. Cr2+ is reducing in nature while with the same d-orbital configuration d4 Mn3+ is an oxidising agent. Give reason .
c. Unlike Cr3+ , Mn2+ , Fe3+ and the subsequent other M2+ ions of the 3d series of elements , the 4d and the 5d series metals generally do not form stable cationic species.
Ans. Manganese is the 3d series transition element that shows the highest oxidation state. The electronic configuration of Manganese is
Mn(25) = [Ar} 3d5 4s2
Manganese has the highest oxidation state because the number of unpaired electrons in the outermost shell is more, i.e. 3d5 4s2. If we consider all the transition metals the highest oxidation state is eight and the element which shows +8 oxidation state are Ruthenium (Ru) and Os(Osmium). (b) Cr2+ is reducing agent as its configuration changes from d4 to d3, when it is oxidized to Cr3+. The d3 configuration have a half-filled t2g level which is very stable. c. Because high enthalpies of atomisation of 4d and 5d series and high ionization enthalpies, the M.P. and B.P. of heavier transition elements are greater than those of first transition series which is due to stronger intermetallic bonding. Hence 4d and 5d series metals generally do not form stable cationic species.
OR
11. a. What is lanthanoid contraction ?
b. Zr (z=40) and Hf (z=72) have almost identical radii?
c. From element to element actinoid contraction is greater than the lanthanoid contraction. Give reason.
Ans.(a) The steady decrease in the size of lanthanide ions with the increase in atomic number is called lanthanide contraction.
(b) Zr and Hf have almost identical atomic radii due to lanthanoid contraction. Hf is post lanthanoid element. As a result of lanthanoid contraction, the atomic size of Hf is similar to Zr (element of previous period).
(c) Actinoid contraction is greater from element to element than lanthanoid contraction due to poor shielding of 5f electrons (in actinoids) as compared to shielding of 4f electrons (in lanthanoids). Due to this,the valence electrons of actinoids experience greater effective nuclear charge than that in lanthanoids.
SECTION- C
12. Read the passage given below and answer the question the follow.The term order of a reaction is the sum of exports of concentration terms appearing in the rate equation. It is an experimentally determinable quantity. It may be whole number, fraction, zero on even negative.Knowledge of order does not require knowledge of mechanism. Order of a ‘reaction’ may change with change in experimental condition namely pressure, temperature etc.
A 1st order reaction is a reaction whose rate determining step (r.d.s) involves only one molecule. Thus
the step is
A → Product.
A 2nd order reaction may be of two types:
2 A→ Product
or A + B → Product.
(a) Give the unit of rate constant of a reaction.
(b) In what condition order and molecularity of a reaction becomes equal.
(c) In which type of reaction order and molecularity are different.
(d) Write expression for rate constant of first order reaction. Mention unit of rate constant of a firstorder reaction.
OR
(d) Show that half life period of a first order reaction is independent of initial concentration of the reactant.
Ans. (a) Unit of rate constant = (Concentration)1–n Time–1
When n = order of reaction
(b) In elementary reaction, order and molecularity are same.
(c) In complex multistep reaction, order and molecularity are different.
(d) Rate = K [R]1, unit of k in first order reaction rate ,i.e. s-1.
OR
(d) The time during which initial concentration of reactant is reduced to half is called half life period. It is denoted as t1/2
We know that, for first order reaction, the rate constant (k) is-
k = (2.303/t) log(a/a-x) -------1
where a is initial concentration of reactant and (a-x) is concentration after time t.
From equation 1-
t = (2.303/k) log(a/a-x) -------2
Now, when, t = t1/2 then, x = a/2
Now putting the value of t and x in equation 2 we get-
t1/2 = (2.303/k) log2
or, t1/2 = 0.693/k --------3
From the equation 3 we see that half life period is inversely proportional to k and independent of initial concentration of the reactant.