Solved Board Paper For CBSE Class 12 Chemistry

When the new forces of interaction between the components are stronger than those in the pure components, then non-deal solutions show negative deviations.

Alloy like copper dissolved in gold.

1-Bromo-5,5-dichloropent-2-ene

= 180

Hence, the alkene is CH₃-CH₂-CH=CH-CH₃.

Rate = k [N₂O₅]

k = rate/[N₂O₅]

k = 2.4 x 10^-4/0.36

k = 6.67 x 10^-4 s^-1

(i) At the time of setting, the sun is at horizon. The light emitted by the Sun has to travel a relatively longer distance through the atmosphere. As a result, blue part of light is scattered away by the particulate in the atmosphere causing red part to be visible.

(ii) Cottrell's smoke precipitator, neutralises the charge on unburnt carbon particles, coming out of chimney and they get precipitated and settle down at the floor of the chamber.

(iii) Fe(OH)₃ sol is positively charged which is coagulated by negatively charged Cl^– present in sodium chloride solution.

(i) Since S_N2 reactivity decreases with increase in size of alkyl groups (-R) and C-Br bond is weaker than C-Cl bond, C-Br bond breaks easily. Therefore, the order of S_N2 reactivity in the given compound will be –
(CH₃)₂CHCl < CH₃CH₂Br < CH₃Cl < CH₃Br

(ii) The boiling point of haloalkanes decreases with increase in branching of carbon, while increases as the number of halogen atoms increases. According to all these points, the order of boiling points of the given compounds will be given as:

(iii) Carbon tetrachloride (CCl₄) has symmetrical structure, thus, it has zero dipole moments. In CHCl₃ (Chloroform), the resultant of two C-Cl dipoles is opposed by the result of C-Cl and C-H bond. Thus, it has 1.03D dipole moment. In case of dichloroethane (CH₂Cl₂), the resultant of dipole moments of two C-Cl bond is reinforced by the resultant of dipole moments of two C-H bond. Therefore, dichloroethane has the highest dipole moment.

(a) Chlorobenzene is slightly soluble in water. Although it is a polar molecule, but due the presence of large hydrocarbon part (benzene ring) hydrogen bonds formed between water and chlorobenzene are not as stronger as hydrogens bond present between water molecules. As a result, chlorobenzene is slightly soluble in water.

(b) (i) 2-Bromopropanoic acid is an optically active compound because it has one chiral centre and has non-superimposable mirror images.
(Where * indicate one chiral carbon)
2-Bromopropanoic acid

(ii) Differences between enantiomers and diastereomers:

hydrophobic hydrophilic

(b) CH₃(CH₂)₁₅ — N(CH₃)Br⁻

hydrophobic hydrophilic

(c) CH₃(CH₂)₆ — COO(CH₂CH₂O)_nCH₂CH₂OH

(i) Chemical name: Thiamine

Chemical formula: (C₁₂H₁₈N₄SOCl₂)

Main sources: Best source of thiamine is dried Brewer’s yeast. Also found in whole grain cereals and bread, nuts, pulses etc.

(ii) Beriberi (paralysis of legs) and loss of appetite are caused by deficiency of vitamin B1.

i)They are molecules of large size.

ii)They have lyophobic property.

Multimolecular colloids:-

i) They are formed by the aggregation of large number of atoms or molecules which have diameter less than 1nm.

ii) They have lyophilic property.

Associated colloids:-

i) They are formed by the aggregation of large number of ions in concentrated solution

ii) They contain both lyophilic and lyophobic groups

CaCO₃ CaO + CO₂

Cu²⁺(aq) + H₂(g) Cu(s) + 2H⁺(aq)

(a)

(b) In aniline, the lone pair pair of electrons on the N- atom is delocalised over the benzene ring. As a result, electron density on the nitrogen decreases. In contrast, in CH₃NH₂ , +I-effect of CH₃ increases the electron density on the N- atom. Therefore, aniline is a weaker base than methylamine and hence its pK_b value is higher than that of methylamine.

(i)

(ii) IUPAC name of CH₃–CH₂–NH–CH₃ is N-methylethanamine.

IUPAC name of CH₃–CH₂–N(CH₂CH₃)–CH₃ is N-ethyl-N-methylethanamine.

(i) Faraday' first law of electrolysis

Mass of any substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passed through the electrolyte

Faraday' second law of electrolysis

When same quantity of electricity is passed through solutions of different electrolytes connected in series, the weights of the substances produced at the electrodes are directly proportional to their equivalent weights.

(ii) conductance is reciprocal of resistance, so conductance of the solution = 1/20 = 0.05 ohm^-1.

(iii) For a weak electrolyte there is a very large increase in conductance with dilution near infinite dilution because as the concentration of the weak electrolyte is reduced more of it ionise.

(iv) Dissociation constant of a weak electrolyte can be calculated by the formula

K_c = c²/1 - where is degree of dissociation of the weak electrolyte.

(i) In an electrochemical cell, salt bridge is used to maintain the electrical neutrality of the solutions in two half cells. It completes the electrical circuit by allowing the ions to flow from one solution to other without mixing the two solutions.

(ii) The absolute value of the electrode potential of a single electrode cannot be measured because oxidation and reduction half cell cannot take place alone. To overcome this difficulty, we use a reference electrode whose electrode potential is arbitrarily assigned a value.

Example: Standard or Normal Hydrogen Electrode (SHE or NHE). Its electrode potential is taken as 0.00 V at zero Kelvin.

To determine the electrode potential of any electrode, a cell is set up with SHE as one of the electrodes. Since, the EMF of the SHE electrode is arbitrarily taken as 0.00 V, the EMF of the cell will directly give the electrode potential value of the other electrode.

(iii) The arrangement of various elements in order of their increasing values of electrode potentials is known as electrochemical series.

(a) Positive sign of electrode potential value represents the reduction potential. This indicates that greater the value of reduction potential, more easily the substance is reduced. It is said to be a stronger oxidising agent.

Thus, according to electrochemical series, F₂ has the highest reduction potential (strongest oxidising agent) and Li⁺ ion has the lowest reduction potential, thus it is the weakest oxidizing agent in the series.

(b) A metal with greater oxidation potential can displace metals with lower oxidation potential from their salt solution.

For example: Decreasing order of oxidation potentials of these metals are:

Mg > Zn > Fe > Cu > Ag

Hence, each metal can displace metals on its right from the salt solutions.

(i) Due to small size, the lone pair of electrons on oxygen atoms repels the bond pair of O-O bond to a greater extent than the lone pair of electrons on the sulphur atoms in S-S bond. As a result, S-S bond (bond energy = 213 kJ/mole) is much stronger than O-O (bond energy = 138 kJ/mole) bond and hence, sulphur has much stronger tendency for catenation than oxygen. Further, as the size of atom increases down the group from sulphur to polonium, the strength of element-element bond decreases. Therefore, tendency for catenation decreases accordingly.
Example: H-S-S-S-S-H (Polysulphides)
H-O-O-H (Polyoxides)

(ii) Nitrite ion shows two resonating structures. This explains that two bonds are equivalent and similar to that of double bond which has a shorter bond length.

In case of nitrate ion, there are three oxygen atoms sharing two single bonds and one double bond with nitrogen. Due to this, N-O bond length is similar to that of single bond which has high value of bond length. Hence, nitrate ion has larger N-O bond length than nitrite ion.

(iii) PCl₅ has trigonal bipyramidal structure in both liquid and gaseous state. In this structure, two axial P-Cl bonds (219 pm) are larger than three equatorial P-Cl bonds (204 pm) because greater bond pair-bond pair repulsion exists in axial bonds. Hence, all five P-Cl bonds are not equivalent.

(iv) (a) Structure of Pyrophosphoric acid

(b) Structure of boron trifluoride

b) 2NH₄Cl + Ca(OH)₂ = 2NH₃ + 2H₂O +CaCl₂

c) N in NH₃ has a lone pair of electron that it donates to the Cu²⁺ ions to form linkage with the metal and it leads to the formation of a deep blue colored complex which helps in the detection of Cu²⁺ ions.

Cu²⁺(aq) + 4NH₃(aq) = [Cu(NH3)₄]²⁺(aq)

(blue) (deep blue)