Lanthanoids primarily show three oxidation states (+2, +3, +4). Among these oxidation states, +3 state is the most common. Lanthanoids display a limited number of oxidation states because the energy difference between 4f, 5d, and 6s orbitals is quite large. On the other hand, the energy difference between 5f, 6d, and 7s orbitals is very less. Hence, actinoids display a large number of oxidation states. For example, uranium and plutonium display +3, +4, +5, and +6 oxidation states while neptunium displays +3, +4, +5, and +7. The most common oxidation state in case of actinoids is also +3.

Benzene, on reaction with benzoyl chloride gives benzophenone through an intermediate benzoylinium cation.

This is an example of Friedel-Crafts acylation reaction.

No (E_cell can be zero but not \({ E }_{ cell }^{ 0 }\) ). Further, \({ \Delta G }^{ 0 }=-nF{ E }_{ cell }^{ 0 }\) . As \({ E }_{ cell }^{ 0 }\neq 0\) ,therefore, \({ \Delta }_{ r }{ G }^{ 0 }\) also cannot be zero.

As the size of the element increases down the group, the E-H bond dissociation energy decreases and hence E-H bond breaks more easily. Thus, the thermal stability. of the hydrides of group 16 elements decreases down the group, i.e., \({ H }_{ 2 }O>{ H }_{ 2 }S>{ H }_{ 2 }Se>{ H }_{ 2 }Te>{ H }_{ 2 }Po.\)

Benzoyl peroxide
    

Define Raoult's law for binary solutions may be defined as: at a given temperature, for a solution of two volatile liquids the partial vapour pressure of each component of the solution is equal to the product of the vapour pressure of the pure component and its mole fraction in solution. Thus,
\(P_A=P_A \quad{ }^{\circ} \times x_a \text { and } P_B=P_B \quad{ }^{\circ} \times x_B\)

The reaction is of second order.

The excess energy (over and above the average energy of the' reactants) which must be supplied to the reactants to undergo chemical reaction is called activation energy. It is the difference between threshold energy and average energy of reactants.
Activation energy = Threshold energy - Average energy of reactants.
(i) Activation energy decreases by the use of a catalyst.
(ii) Activation energy decreases by rise in temperature.

Transition metals form a large variety of alloys. This is due to the fact that transition metals have almost similar sizes and therefore, the atoms of one metal can easily substitute the atoms of other metals in their lattices. When we cool the mixture solution of two or more transition metals, solid alloys are generally formed. For example, Mn dissolves in molten iron to form manganese steel. The alloys formed are hard, having high melting points. These are also resistant to corrosion in comparison to the parent metals. Tungsten is also used for preparing stainless steel.

(i) Since size of S is more than that of 0, the distance between central atom and hydrogen atoms is more in \({ H }_{ 2 }S\)  than in \({ H }_{ 2 }O\). Therefore, bond dissociation enthalpy of \({ H }_{ 2 }S\) is less and bond cleavage is more easy than in \({ H }_{ 2 }O\) . Therefore, \({ H }_{ 2 }S\)  is more acidic than \({ H }_{ 2 }O\).
(ii) In the resonance structures of \({ NO }_{ 2 }^{ - }\), two bonds are sharing a double bond, while in \({ NO }_{ 3 }^{ - }\), three bonds4 are sharing a double bond.
As a result, the bond in \({ NO }_{ 2 }^{ - }\)will be shorter than in \({ NO }_{ 3 }^{ - }\) and therefore, N-O bond length in \({ NO }_{ 2 }^{ - }\)  is shorter than N-O in \({ NO }_{ 3 }^{ - }\).
(iii)The larger tendency of oxygen to stabilise the higher oxidation state is because of the tendency of oxygen to form multiple bonds with metal.

(i) [Co(NH₃)₄(H₂O)Cl]Cl₂
(ii) K₂[Zn(OH)₄]
(iii) K₃[Al(C₂O₄)₃]
(iv) [CoCl₂(en)₂]⁺
(v) [Ni(CO)₄]

Due to strong electron-donating effect of the -NH₂  group, the electron density increases at the 0-, positions. Further, when aniline is treated with Br₂ the Br⁺ attacks the benzene ring at o- and p-positions to form carbocation intermediates which are stabilized not only by the usual resonance of the Henzene ring. but also by the -NH₂ group as shown below: In case of o-bromination, the carbocation intermediate is stabilized not only by usual resonating structure (I, III and IV) but is also stabilized by resonance structure (II) in which the lone pair of electrons on the N. atom interacts with the positively charged carbon of the ring. Similarly, in case of p-bromination, the carbocation is stabilized not only by the usual resonating structures (V, VI, and VIII) but is also stabilized by the resonance structure (VII) in which the lone pair of electrons on the N-atom interacts with the positively charged carbon of the ring. These additional resonating structures (II and VII) increase the stability of the carbocation to such an extent that bromination occurs instantaneously at the p- and two. o- positions giving 2, 4, 6-tribromoaniline.

Elevation in boiling point may be calculated from the relation,

\(\triangle { T }_{ b }=\frac { { K }_{ b }\times { \omega }_{ B }\times 1000 }{ M_{ B }\times \omega _{ A } } \)
\(\\ { \omega }_{ B }=0.052g,{ \omega }_{ A }=80.2g,{ K }_{ b }=5.2K{ m }^{ -1 },{ M }_{ B }=180\)
\(\therefore \triangle { T }_{ b }=\frac { 5.2\times 0.052\times 1000 }{ 180\times 80.2 } =0.0187\)
Boiling point of water = 373 K
Boiling point of solution = 373 + 0.0187 = 373.0187 = 373.02 K
Now, depression in freezing point,
\(\triangle { T }_{ f }=\frac { { K }_{ b }\times { \omega }_{ B }\times 1000 }{ M_{ B }\times \omega _{ A } } \)
\(\\ { \omega }_{ B }=0.052g,{ \omega }_{ A }=80.2g,{ K }_{ f }=1.86K{ m }^{ -1 },{ M }_{ B }=180\)
\(\therefore \triangle { T }_{ f }=\frac { 1.86\times 0.052\times 1000 }{ 180\times 80.2 } =0.067 \)
Freezing point of water = 273 K
Freezing point of solution = 273 - 0.067 = 272.933K.

Graphite is a good conductor of electricity due to the presence of free electrons within its layers. But in diamond, no free electrons are present and therefore, it is a bad conductor of electricity. Therefore, diamond is not used as an anode.

Depression in freezing point is related to the molality, therefore, the molality of the solution with respect to ethylene glycol \(=\frac{\text { moles of ethylene glycol }}{\text { mass of water in kilogram }}\)
\(\text {Moles of ethylene glycol }=\frac{45 \mathrm{~g}}{62 \mathrm{~g} \mathrm{~mol}^{-1}}=0.73 \mathrm{~mol}\)
\(\text {Mass of water in } \mathrm{kg}=\frac{600 \mathrm{~g}}{1000 \mathrm{~g} \mathrm{~kg}^{-1}}=0.6 \mathrm{~kg}\)
\(\text {Hence molality of ethylene glycol }=\frac{0.73 \mathrm{~mol}}{0.60 \mathrm{~kg}}=1.2 \mathrm{~mol} \mathrm{~kg}^{-1}\)
Therefore freezing point depression,
ÄT_f = 1.86 K kg mol^–1 x 1.2 mol kg^–1 = 2.2 K
Freezing point of the aqueous solution = 273.15 K – 2.2 K = 270.95 K

Two types of emulsions are as follows:

Application of Emulsions:
(i) In medicine: The various pharmaceuticals and cosmetics are available in liquid form such as codliver oil etc., and some creams and ointment are emulsions of water in oil type.
(ii) Cleaning action of soap: This action is based on the formation of oil in water type emulsions

Step I
Calculation of the quantity of charge passed:
We know that,
Watt = Ampere x Volt
Ampere =\(\frac { Watt }{ Volt } =\frac { 100 }{ 110 } \)
Now charge = Current x Time
=\(\left( \frac { 100 }{ 110 } amp \right) \times 4\times 60\times 60\)
= 13091 C
Step II
Calculation of mass of Cadmium deposited:
The cathodic reaction is
Cd²⁺_(aq)+2e^-\(\rightarrow\)Cd_(s)
112.2g 2 x 96500C
2 x 96500 C of charge deposited Cd = 112.2 g
13091 C charge will deposite Cd
\(\frac { 112.2g }{ 2\times 96500C } \times 13091C\)
= 7.61g

Comparing oxidation potential of iron with those of metals A, B and C, iron has higher oxidation potential than C only. Hence, when coating of C is broken, rusting will become faster.

(i) HOCI in presence of an acid generates the reactive electrophile, chloronium ion «r, which attacks benzene to give chlorobenzene (A).
(ii) Chlorobenzene (A) on treatment with NaNH2/liq. NH₃ undergaes-dehydrohalogenation via benzyne to afford aniline (B).
(iii) Aniline (B) on treatment with HBF4 forms the corresponding salt anilinium tetrafluoroborate (C) which on heating with NaNO₂ undergoes Balz-Schiemann reaction through the intermedium formation of benzenediazonium tetrafluoroborate to afford fluorobenzene (D).

The main limitation of Williamson's synthesis is that for preparing unsymmetrical ethers, the halide used should preferably be primary because secondary and tertiary alkyl halides may form alkenes as major products due to elimination process. For 2° and 3° alkyl halide, elimination competes over substitution. If a 3° alkyl halide is used, an alkene is the only product and no ether is formed. For example, if we want to prepare ethyl tert-butyl ether, then we should take ethyl bromide and sodium tert-butoxide.

However, if we use tert-butyl bromide and sodium ethoxide as reactants, then the major products would be 2-methyl propene (iso-butylene) and ethanol.

Similarly, alkyl aryl ethers (phenolic ethers) can be easily prepared by using sodium phenoxide and alkyl halides. For example,

However, aryl halide and sodium alkoxide cannot be used for preparing phenolic ethers because aryl halides are less reactive towards nucleophilic substitution reactions than alkyl halides.

To begin with both ethanol and phenol react with PCI₅ to give the corresponding chlorophosphate esters.
Since in ethanol, C-O bond has only single bond character, therefore, nucleophilic attack by Cl^- ion occurs easily on chlorophosphate ester (I) to give chloroethane. However, in case of phenol, C-O bond has some double bond character due to resonance and hence difficult to break. Instead nucleophilic attack by C₆H₅OH occurs twice on the chlorophosphate ester (II) to form intermediate (III). Further, nucleophilic attack by C₆H₅OH on III does not occur due to steric hindrance. Instead (III) undergoes hydrolysis by water to form triphenylphosphate.

In \({ NCI }_{ 3 }\) CI has vacant d-orbitals to accept the lone pair of electrons donated by oxygen atom of \({ H }_{ 2 }O\)  molecules. But in \({ NF }_{ 3 }\), F does not have vacant d-orbitals. Therefore, \({ NCI }_{ 3 }\) undergoes hydrolysis but \({ NF }_{ 3 }\) does not.
\({ 2NCI }_{ 3 }+{ 3H }_{ 2 }\longrightarrow { NH }_{ 3 }+{ 3HOCI }\) 
\({ NF }_{ 3 }+{ H }_{ 2 }O\longrightarrow \) No reaction

\(\)

\(\Delta { T }_{ f }=\frac { { K }_{ f }{ W }_{ b }\times 1000 }{ { M }_{ b }\times { W }_{ a } } \)
\(0.383=\left( \frac { 3.83\times 2.56 }{ M\times 100 } \right) \times 1000\)
M = 256
S \(\times \) x = 256
32 \(\times \) x = 256
x = 8
(b) (i) If we place blood cells in solution containing 1.2% sodium chloride solution, they shrink.
(ii) It we place blood cells in solution containing 0.4% sodium chloride solution, they swell.

(i) Iron is present in + 3 oxidation state as Fe (III).
(ii) Iron hydroxide i.e., Fe(OH)₃ is insoluble in water and is therefore, not assimilated in the soil.
(iii) Fe(III)-EDTA complex is readily absorbed by the soil as it is water soluble.
(iv) The addition of this complex compound in proper amount makes plants healthy and the yield of fruits gets increased. But its deficiency also affect adversely so we will provide Fe-EDTA complex.