In ground state, Ti has 23 electrons with electronic configuration 3d³4s².
The oxidation state of Ti in the given compound is +3.
Hence it will now have the configuration 3d². Since it has 2 unpaired electrons and has the ability to
undergo d-d transition, the given complex gives violet colour.

This reaction involves the preparation of ether by Williamson's synthesis. It takes place via S_N2 attack of an alkoxide ion on primary alkyl halide because here the alkyl halide used is primary. Moreover in method (i) the C-Br bond has some double bond character.

(i)

(ii)
 
(iii)

(i) Add Tollen's reagent. Propanal will give silver mirror whereas propanone will not react.
(ii) Add NaHCO₃ solution. Benzaldehyde will not react whereas benzoic acid will give brisk effervescence due to CO₂.

We know that, m = K_H \(\times\) P
\(\therefore\) 6.56 \(\times\) 10^-2g = K_H \(\times\) 1 bar ...(i)
\(\therefore\) 0 5.00 \(\times\) 10^-2 g = K_H \(\times\) P.........(ii)
K_H = 6.56 \(\times\) 10^-2/1 bar (from i)
K_H = 5.00 \(\times\) 10^-2//p bar (from ii)
\(\therefore \quad \frac{6.56 \times 10^{-2}}{1}=\frac{5.00 \times 10^{-2}}{p}\)
\(\therefore \quad P=\frac{5 \cdot 00}{6 \cdot 56}=0.762 \text { bar. }\)

(i) Haloalkanes react with Mg in dry ether to form Grignard reagents (RMgX), which react with water to form hydrocarbons.


(iii) Haloalkanes react with Na in dry ether to give hydrocarbons containing double number of C-atoms.

(i) Since compound (E) with molecular formula, C₃H₆0 does not reduce Tollens' reagent and Fehling's solution but forms 2, 4-dinitrophenylhydrazone, it must be a ketone. But the only possible ketone having the molecular formula, C₃H₆O is acetone or propanone. Thus, compound (E) is acetone or (propanone) CH₃COCH3·
(ii) Since acetone (E) is obtained by heating compound (8) with Ca(OH)₂ therefore, (B) must be acetic acid (ethanoic acid), CH₃COOH.
(iii) Since (D) on oxidation with KMn04 gives acetic acid (8), therefore, (D) must be ethyl alcohol (ethanol), CH₃CH₂OH.
(iv) Since acetic acid (8) and ethyl alcohol (D) are obtained by hydrolysis of (C) under acidic conditions, therefore, (C) must be ethyl acetate (ethyl ethanoate), CH₃COOC₂H₅
(v) Since ethyl acetate (C) and acetic acid (8) are obtained by treatment of compound (A) with ethyl alcohol, therefore, compound (A) must be acetic anhydride (ethanoic anhydride), (CH₃COO)₂O.
(vi) All the reactions involved in this problem can now be explained as follows

\([Mn(CN)_6]^{3-}\)
d²sp³, (ii) inner orbital complex, (iii) paramagnetic, (iv) 2.87 B.M.
(i) d²sp³ (ii) Inner orbital complex, (iii) diamagnetic, (iv) \(\mu=0\)
(i) d²sp³, (ii) Inner orbital complex, (iii) paramagnetic, (iv) 3.87 B.M.
(i) sp³d², (ii) outer orbital complex (iii) paramagnetic (iv) 4.9 B.M.

(i) Molar mass of CH₂Cl₂ = 12 × 1 + 1 × 2 + 35.5 × 2 = 85 g mol^–1
Molar mass of CHCl₃ = 12 × 1 + 1 × 1 + 35.5 × 3 = 119.5 g mol^–1
\(\text {Moles of } \mathrm{CH}_{2} \mathrm{Cl}_{2} \quad=\frac{40 \mathrm{~g}}{85 \mathrm{~g} \mathrm{~mol}^{-1}}=0.47 \mathrm{~mol}\)
\(\text {Moles of } \mathrm{CHCl}_{3} \quad=\frac{25.5 \mathrm{~g}}{119.5 \mathrm{~g} \mathrm{~mol}^{-1}}=0.213 \mathrm{~mol}\)
\(x_{\mathrm{CH}_{2} \mathrm{Cl}_{2}}=\frac{0.47 \mathrm{~mol}}{0.683 \mathrm{~mol}}=0.688\)
x_CHCl3 = 1.00 – 0.688 = 0.312
\(p_{\text {total }}=p_{1}^{0}+\left(p_{2}^{0}-p_{1}^{0}\right) x_{2}=200+(415-200) \times 0.688\)
= 200 + 147.9 = 347.9 mm Hg
(ii) y₁ = p₁/p_total, we can calculate the mole fraction of the components in gas phase (y₁).
\(p_{\mathrm{CH}_{2} \mathrm{Cl}_{2}}=0.688 \times 415 \mathrm{~mm} \mathrm{Hg}=285.5 \mathrm{~mm} \mathrm{Hg}\)
\(p_{\mathrm{CHCl}_{3}}=0.312 \times 200 \mathrm{~mm} \mathrm{Hg}=62.4 \mathrm{~mm} \mathrm{Hg}\)
\(y_{\mathrm{CH}_{2} \mathrm{Cl}_{2}}=285.5 \mathrm{~mm} \mathrm{Hg} / 347.9 \mathrm{~mm} \mathrm{Hg}=0.82\)
\(y_{\mathrm{CHCl}_{3}}=62.4 \mathrm{~mm} \mathrm{Hg} / 347.9 \mathrm{~mm} \mathrm{Hg}=0.18\)

(i) CH₂FCO₂H is a stronger acid.
(ii) CH₂FCO₂H is a stronger acid.
(iii) CH₃CHFCH₂CO₂H is a stronger acid.

A molecule or an ion which has only one donor atom to form one coordinate bond with the central metal atom is called unidentate ligand, e.g.Cl^- and NH₃.
A molecule or an ion which contains two donor atoms and hence forms two coordinate bonds with the central metal atom is called a didentate ligand, e.g.,
 \(\overset { { CH }_{ 2 }{ NH }_{ 2 } }{ \underset { { CH }_{ 2 }{ NH }_{ 2 } }{ | } } and\quad \overset { { COO }^{ - } }{ \underset { { COO }^{ - } }{ | } } \)
a molecule or an ion which contains two donor atoms but only one of them forms a coordinate bond at a time with the central metal atom is called ambidentate ligand, e.g. CN^- or NC^-and: NO^-₂ or ONO^-

Given, W₂ = 3.9 s. WI = 49 g; \(\Delta T_f\)= 1.62 K
M₂(benzoic acid) = 122 g mol^-1
K_f(benzene) = 4.9 K kg  mol^-1
\(\Delta T_f=\frac{i\times K_f \times W_2 \times 1000}{M_2 \times W_1}\)
\(i=\frac{\Delta T_f\times M_2 \times W_1}{K_f \times W_2 \times 1000}\)
\(i=\frac{1.62 K \times 122 g mol^{-1}\times 49 g}{4.9 K kg mol^{-1}\times 3.9 g \times 1000}\)
i = 0.50
Since, i < I, therefore solute (benzoic acid) undergoes association.

(i) Acidified K₂Cr₂O₇ (followed by hydrolysis with dil, H₂SO₄)
(ii) Pyridinium chlorochromate (PCC)
(iii) Aqueous bromine
(iv) Acidified or alkaline KMnO₄ (followed by hydrolysis with dil. H₂SO₄)
(v) 85% H₃PO₄ at 440 K
(vi) Ni/H₂ or NaBH₄ or LiAlH₄

A hydrocarbon with the molecular formula, C₅H₁₀ belongs to the group with a general molecular formula CnH2n. Therefore, it may either be an alkene or a cycloalkane. Since hydrocarbon does not react with chlorine in the dark, it cannot be an alkene. Thus, it should be a cycloalkane. Further, the hydrocarbon gives a single monochloro compound, C₅H₉Cl by reacting with chlorine in bright sunlight. Since a single monochloro compound is formed, the hydrocarbon must contain H−atoms that are all equivalent. Also, as all H−atoms of a cycloalkane are equivalent, the hydrocarbon must be a cycloalkane. 

(b) If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
Explanation:
Carbon tetrachloride rises to atmosphere and deplete the ozone layer. This depletion of ozone layer increases exposure of UV rays to human being which lead to increase of skin cancer, eye diseases and disorder with disruption of the immune system.

A) Assertion and reason both are correct statements and reason is correct explanation for assertion.
Explanation:
If a pressure larger than the osmotic pressure is applied to the solution side, the pure solvent flows out of the solution to the solvent through semi-permeable membrane and this phenomenon is called reverse osmosis.

(d): Alcohols undergo intermolecular H-bonding and hence, their boiling points are higher than those of hydrocarbons.
Among isomeric alcohols, boiling point decreases in the order: 1° > 2° > 3°.

(c) Carboxylate ion is more stable than phenoxide ion due to two equivalent resonating structures in which negative charge is at the O-atom. In phenoxide ion, negative charge is at the less electronegative carbon atom. Therefore, corresponding acid of the carboxylate ion, i.e. carboxylic acid is more acidic than phenol. Hence, (A) is correct but (R) is incorrect.

[Ni(CN)₄]^2- is square planar and dimagnetic.
\(\begin{equation} \begin{array}{l} \mathrm{Ni} \rightarrow 3 d^{8} 4 s^{2} \\ \mathrm{Ni}^{2+} \rightarrow 3 d^{8} 4 s^{0} \end{array} \end{equation}\)

\(\therefore\)This complex has no unpaired electron.
Both (A) and (R) are correct and (R) is the correct explanation of (A).

(a) Henry's law.
(b) P_gas = KH X _gas
Where P_gas = Partial pressure of gas
X_gas = Mole fraction of gas
K_H = Henry's law constant.
(c) Slope = K_H, Henry's law constant.
(d) Solubility of gas in liquid increase with increase in pressure. Solubility of gas in liquid increase with decrease in temperature.
(e) (i) Nature of gas
(ii) Temperature

(i) (a): It is an example of cross Cannizzaro reaction where aromatic aldehyde gets reduced to alcohol and aliphatic aldehyde gets oxidised to its sodium salt (both aldehydes must not contain any \(\alpha\)-hydrogen).

(ii) (c)
(iii) (a): The Cannizzaro product of given reaction yields 2, 2, 2-trichloroethanol.

(iv) (a): C-C bond is not formed in Cannizzaro reaction while other reactions result in the formation of C-C bond.