The crystal lattices of transition metals have vacant interstitial sites into which small sized nonmetal atoms (H, C, N, B) can easily fit, resulting in the formation of compounds known as interstitial compounds. These are usually non-stoichiometric and are neither typically ionic nor covalent. For example, TiC, Mn₄H, Fe₃H, TiH1.7"etc. They have chemical properties similar to that of the parent metal but different physical properties such as hardness, electrical conductivity, etc.

It is due to resonance as shown below the partial positive charge on carbonyl carbon atom is reduced therefore, these cannot undergo nulephilic addition reactions.

HCOOH < CICH₂COOH < CCl₃COOH < CF₃COOH.

Add CHCI₃ and alc. KOH,C₆H₅ - NH₂ gives foul smell of isocyanide whereas C₆H₅ - NH - CH₃ does not.  

(i) \(\underset { Aniline }{ C_{ 6 }H_{ 5 }NH_{ 2 } } +CHCl_{ 3 }+\underset { (alc) }{ 3KOH } \quad { \longrightarrow }\underset { Phenyl\quad carbylamine }{ C_{ 6 }H_{ { 5 } }N_{ \rightarrow }^{ = }C } +3KCl+3H_{ 2 }O\)
(ii) \(\underset { Benzene\\ diazonium\\ Chloride }{ C_{ 6 }H_{ 5 }N_{ 2 }Cl } +\underset { Hypo\\ phosphorus\quad \\ acid }{ H_{ 3 }PO_{ 2 } } +H_{ 2 }O\quad { \longrightarrow }\underset { Benzene }{ C_{ 6 }H_{ 6 } } +\underset { Phosphoric\\ acid }{ H_{ 3 }PO_{ 3 } } +N_{ 2 }+HCl\)

It is because hydration energy and lattice energy of Cu²⁺ is more than that of Cu⁺.

(i) Ligands play an important role in exhibiting the colour to the complex. \(\mathrm{H}_2 \mathrm{O}\) is a weak field ligand, which causes small d-d splitting and correspondingly complex give green colour, whereas, ethane-1, 2diamine (en) is a strong field ligand due to which d-d splitting is increased and correspondingly complex give blue colour.
(ii) Formula : \(\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{NCH}_2 \mathrm{CH}_2 \mathrm{NH}_2\right)_3\right]_2\left(\mathrm{SO}_4\right)_3\) Hybridisation : \(d^2 s p^3\)

(i) Mn³⁺ /Mn²⁺ has large positive E^o value. Hence, Mn³⁺ can be easily reduced to Mn²⁺ because Mn²⁺ has half-filled electronic configuration, so it is stable and Mn³⁺ is least stable. Therefore, it is a good oxidising agent.
(ii) There is decreasing negative electrode potentials of Mn²⁺/ M in the first transition series due to increasein the sum of IE₁and IE_2. It shows that in general, the stability of +2 oxidation state decreases from left to right. Exceptions are Mn and Zn in which the greater stability of +2 state for Mn is due to half-filled d-subshell(d⁵) in Mn2+ and that of Zn is due to completely filled d-subshell (d¹⁰) in Zn ²⁺. 
(iii) It is because oxygen can form multiple bonds, whereas fluorine can only form single bonds with metals.

 (i) Conversion of Propanone to Propane:

 (ii) Conversion of Benzoyl chloride to benzaldehyde:

(iii) On treatment with dilute alkali, ethanol produces 3-hydroxybutanal gives But-2-enal on hheating.

(i) Hexaamminecobalt (Ill) chloride.
(ii) Tetrachloridro nickelate (II)ion.
(ill) Potassium hexacyanoferrate (Ill)

The aliphatic carboxylic acids are stronger acids than phenols. The difference in the relative acidic strengths can be understood if we compare the resonance hybrids of carboxylate ion and phenoxide ion.
(i) The electron charge in the carboxylate ion is more dispersed in comparison to the phenoxide ion since 'there are two electronegative oxygen atoms in carboxylate ion as compared to only one oxygen atom in phenoxide ion.
(ii) Carboxylate ion is stabilised by two equivalent resonance structures in which negative charge is on more electronegative O atom. But phenoxide ion has non-equivalent resonance structures in which the negative charge is also on less electronegative carbon atom. 
Therefore, the carboxylate ion is relatively more stable as compared to phenoxide ion. Thus, the release of H⁺ ion from carboxylic acid is comparatively easier or it behaves as a stronger acid than phenol.

Gabriel phthalimide synthesis: It is the process in which pure aliphatic primary amines are prepared by reaction of potassium phthalimide  with alkyl halide, followed by hydrolysis.

(a)\(4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \longrightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2\)
\(2Na_2CrO_4 + 2H+ \longrightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O\)
\(Na_2Cr_2O_7 + 2KCI \longrightarrow K_2Cr_2O_7 + 2NaCI\)
(b)\(2MnO_2 + 4KOH + O_2 \xrightarrow{\Delta} 2K_2MnO_4 + 2H_2\)
\(\underset{Magnate\ ion}{MnO_4^{2-}}\xrightarrow{electrolysis}\underset{Permagnate\ ion}{MnO_4^-}+e^-\)

(a) Ce(III) ion have the outer electronic configuration 4 f¹5d⁰6s⁰ which can easily lose the electron from 4f orbital to acquire the stable configuration, i.e. 4f⁰5d⁰6s⁰ and form Ce(IV) ion.
(b) The E° (Mn²⁺/Mn) value is -1.18 V, which is highly negative as compared to other neighbouring d-block elements due to extra stability of half-filled orbitals of Mn²⁺ (i.e. 3d⁵).
(c) Greater is the number of unpaired electrons stronger is the resultant bonding hence greater is the enthalpy of atomisation.
In 3d series elements, Zn has the lowest enthalpy of atomisation, it is due to the absent of unpaired electrons in ns and (n - 1) d shells.
(d) When sodium chromate is acidified than orange colour solution of sodium dichromate is formed.
\(2 \mathrm{Na}_2 \mathrm{CrO}_4+2 \mathrm{H}^{+} \longrightarrow \underset{\substack{\text { Sodium dichromate } \\ \text { (Orange) }}}{\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7}+2 \mathrm{Na}^{+}+\mathrm{H}_2 \mathrm{O}\)
(e) Zn, Hg and Cd have completely filled d-orbitals in the ground state as well as in their common oxidation states. Thus, formation of strong metallic bond is not possible.
(f) Permanganate titration is not carried out in the presence of HCI because if HCl is used, the oxygen produced from the reaction of KMnO₄ and HCl is partly utilised in oxidising HCl to Cl₂, which itself acts as an oxidising agent and partially oxidises the reducing agent.
(g) In lower oxidation states, transition metals behave like metals and metal oxides are basic in nature. Thus, in lower oxidation states, transition metal oxides are basic.
As the oxidation state increases, its metallic character decreases due to decrease in size, thus, ít becomes less metallic or more non-metallic. Oxides of a non-metal may be acidic or neutral.
Thus, in higher oxidation states, transition metal oxids are amphoteric or acidic.

(b) (CH₃)₂NH > CH₃NH2 > (CH₃)₂N
(c) Add NaNO₂ and conc. HCI. Cool it to 0-5^o C. Add alkaline solution of phenol. C₆H₅NH₂ gives orange azo dye, whereas CH₃NH₂ does not.

(i) (a) Metal carbonyls: Metal carbonyls are the compounds in which carbon monoxide (CO) acts as the ligand. The metal-carbon bond in metal carbonyls possesses both \(\sigma\) and \(\pi\) characters, e.g. N(CO)₄[nickeltetracarbonyl]. Its IUPAC name is tetra carbonyl nickel(0).
(b) Chelate: An inorganic metal complex in which there is a close ring of atoms caused by attachment of a ligand to a metal atom at two points. An example is the complex ion formed between ethylene diamine and cupric ion,[Cu(NH₂CH₂CH₂NH₂)₂]²⁺.

(ii) (a) Add BaCl₂ solution. [Co(NH₃)]₅Cl]SO₄ will give white ppt. of BaSO₄.  
 \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{SO}_{4}+\mathrm{BaCl}_{2}(a q) \)\( \longrightarrow\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}+\mathrm{BaSO}_{4} \downarrow \)
                                                                    \(\text { (White ppt.) }\)
Add AgNO₃(aq) solution [Co(NH₃)₅SO₄] Cl will give white ppt. of AgCl.
\( {\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{Cl}+\mathrm{AgNO}_{3}(a q)} \) \(\longrightarrow \mathrm{AgCl}+\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{NO}_{3} \)
                                                                                          \(\text { (White ppt) }\)
(c) Fe³⁺ (4s⁰3d⁵) In strong field \( \Delta_{0}>P t_{2 g}^{5} e g^{0} \)
In weak field \(\Delta_{0}

(i) Acetylation      
Acetyl group is introduced in an organic compound.  Reagent is acetyl chloride or acetic anhydride.
Bases such as pyridine and dimethylaniline are used to neutralize acid (HCl or acetic acid).
(ii) Cannizzaro reaction
Aldehydes (not having alpha hydrogen atoms) undergo self oxidation-reduction (disproportionation) reaction. One molecule is oxidized to carboxylic acid and other molecule is reduced to alcohol. Concentrated alkali is the reagent.
(iii) Cross aldol condensation  
Aldol condensation between different carbonyl compounds (aldehdyes and ketones). If both reactants contain alpha hydrogen atom, four different products can be obtained.
(iv) Decarboxylation
Loss of CO₂​ from carboxylic acids to form hydrocarbons. For this, sodium salts of carboxylic acids are heated with soda lime.

(i) (a) Due to the comparatively smaller size of the metal ions, their high ionic charges and the availability of vacant d-orbitals for bond formation, transition metals form a  large number of complex compounds.
(b) As oxidation number (or oxidation state) of an element increases ionic character decreases. In general, the oxides in lower oxidation states of metals are basic and in their higher oxidation state, the oxides are amphoteric.
In lower oxidation state of the metal, some of the valence electrons of the metal atom are not involved in bonding. Hence, it can donate electrons and behave as a base. In higher oxidation state, valence electrons are involved in bonding and hence, electrons are not available for donation. Instead, their effective nuclear charge is high and hence they behave as acids.
(c) Mn³⁺(3d⁴) is less stable than Mn²⁺(3d⁵) because Mn²⁺ has stable half-filled configuration. Cr³⁺ has stable 3d³(t³_2g) configuration, therefore, Cr³⁺ cannot be reduced to Cr²⁺. That's why, EO value for the Mn³⁺ / Mn²⁺ couple is much more positive than Cr³⁺ /Cr²⁺. In other words, Mn³⁺ is a strong oxidising agent.
(ii) Similarity Both lanthanoids and actinoids exhibit +3 oxidation state predominantly.Difference Lanthanoids have less tendency towards complex formation while actinoids have greater tendency towards complex formation.

\([Mn(CN)_6]^{3-}\)
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.

(a): Under acidic condition, aniline gets protonated to anilinium ion (-NH₃⁺ group). This is deactivating and m-directing group. Thus, controlled nitration of aniline mainly gives m-nitroaniline.

(a): A half-filled or fully-filled orbital is more stable than incompletely filled orbital.

(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) (CH₃)₃N, N, N-dimethyl methanamine.
(b) 
 a foul smelling compound.
(c) 

(d) Add NaNO₂ and cone. HCI. Cool it to 0 to 5° C. Then add alkaline solution of phenol. Aniline gives orange dye where as ethyl amino does not.
(e) \(\begin{array}{ll} \mathrm{A}^{\prime} \text { is } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} & { }^{\prime} \mathrm{B}^{\prime} \text { is } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}_{2}^{+} \mathrm{Cl}^{-} \end{array}\)
(f) 'A' is C₆H₅CN, 'B' is C₆H₅COOH
(g) It is used as cationic detergents used in hair conditioners and shampoo.