(i) \(2 \mathrm{MnO}_{4}^{-}+5 \mathrm{NO}_{2}^{-}+6 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Mn}^{2+}+5 \mathrm{NO}_{3}^{-}+3 \mathrm{H}_{2} \mathrm{O}\) 
(ii) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 e^{-} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\) 

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.

Out of the two, (a) cis - [CrCl₂(ox)²]^3- is chiral (optically active).

Ce³⁺ has electronic configuration [Xe] 4f¹.
Spin only formula:
Magnetic moment
\(\sqrt { 4S\left( S+1 \right) } -\sqrt { 4\times \frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } +1 \right) } \)
\(\sqrt { 2\times \frac { 3 }{ 2 } } =\sqrt { 3 } \)
1.732 B.M

Ti³⁺, V³⁺, Mn²⁺, Fe³⁺ and Co²⁺ are coloured due to presence of unpaired electrons, they can undergo d-d transitions others  Cu⁺, Sc³⁺ are colourless due to absence of unpaired electrons.

Triamminedichloridonitrosyl platinum (IV) bromide.

(i) Electronic configuration of \(\mathrm{Ti}:[\mathrm{Ar}] 3 d^2 4 s^2\)
Electronic configuration of \(\mathrm{Ti}^{4+}\)is [Ar]. It is stable noble gas configuration.
Electronic configuration of \(\mathrm{Ti}^{2+}[\mathrm{Ar}] 3 d^2\)
Electronic configuration of \(\mathrm{Ti}^{3+}:[\mathrm{Ar}] 3 d^1\)
As \(\mathrm{Ti}^{4+}\) acquires nearest noble gas configuration on loss of 4 electrons, thus it is more stable than the \(\mathrm{Ti}^{3+}\) and \(\mathrm{Ti}^{2+}\),
(ii) In case of transition elements, ions of the same charge in a given series show progressive decrease in radius with increasing atomic number.
As with increase in atomic number, the new electron enters in d-orbitals and expected to increase in atomic size, but due to poor shielding effect of d-orbitals, the electrostatic attraction between nucleus and outermost orbital increases and hence, the ionic radii decreases.
(iii) Zinc is a soft metal as compared to iron and chromium because it has low enthalpy of atomisation (due to the absence of unpaired electron), whereas, iron and chromium have high enthalpy of atomisation (due to the presence of unpaired electrons), which account for their hardness.

(a) Mn ²⁺ compounds are more stable due to half-filled d-orbitals. Fe²⁺compounds are comparatively less stable as they have six electrons in their orbitals. So, they tend to lose one electron from Fe²⁺and get stable 3d⁵ -configuration in Fe³⁺. Therefore, comparatively high positive value of E ° for Mn ³⁺/ Mn ²⁺indicates the stability of Mn ²⁺(d⁵) whereas comparatively low value for Fe³⁺/Fe²⁺ indicates the extra stability of Fe³⁺(d⁵ ).
(b) Energy required to convert metallic crystal into individual atom is enthalpy of atomisation. In transition row elements it first increases and reaches to maximum upto middle element and then decreases. This is because of strong inter atomic interaction due to unpaired electron. Greater the number of unpaired electron, stronger will be bonding and thus enthalpy of atomisation will also be more. Since iron has more unpaired electron than copper therefore its enthalpy of atomisation is more.
(c) The metal ions with partially or incomplete filled d-orbitals will be coloured in aqueous solution. While the metal ions having either empty or completely filled d-orbitals are colourless. The colour will be due to d-d transition of electrons. Thus, the outer electronic configuration of metal ions are Sc³⁺: 3d^o, Ti³⁺: 3d¹ Hence, among the given ions, Ti³⁺ will exhibit colour in aqueous solution while Sc³⁺ will be colourless.

Double salts dissociate completely into their constituent simple ions when dissolve in water, i.e they lose their identity in aqueous solution
e.g KCl.MgCl₂.6H₂O ⟶ K⁺+Ma²⁺+3Cl^-+6H₂O
While coordination compounds do not dissociate into simple ions, i.e. they retain their identity both in solid state and in aqueous solutions.
(ii) The IUPAC name of the following given complexes are:
(a) K₃[Fe(C₂O₄)₃]
Potassiumtrioxalatoferrate (III)
(b) [Pt(NH₃)₆]Cl₄: Hexaammineplatinum (IV) chloride
(iii) The structure of cis-isomer of [Co(NH3))4Cl₂]⁺ ion is.

(i) Tetrachloridonickelate(II) ion
(ii) sp³ hybridisation
(iii) tetrahedral
The different between the two sets of energy levels when the five degenerate d-orbitals split in the presence of an electrical field of ligands is called crystal field splitting energy.In the octahedral field, it is represented as \({ \Delta }_{ 0 }\) .
Electronic configuration of d⁴
(i) When  \({ \Delta }_{ 0 }>P;{ { t }_{ 2g } }^{ 4 }\)           
(ii) \({ \Delta }_{ 0 }>P;{ { t }_{ 2g } }^{ 3 }{ { e }_{ g } }^{ 1 }\)

(i) x + (0) + (-1) + (0) = + 2 or x = + 3
(ii) x - 4 = -2 or  x = + 2
(iii) x + 0 + 3(-1) = 0 or x = + 3
(iv) x + 2(-1) + 0 = + 1 or x = + 3
(v) [Fe(CN)₆]^3-, x + 6(-1) = - 3 or x = + 3

(a) E values for Cr³⁺/Cr²⁺is negative (- 0.41 V) and for Mn³⁺/Mn²⁺is positive (+1.57V).Thus, Cr²⁺ can undergo oxidation and, therefore, is reducing agent. On the other hand, Mn(III) can undergo reduction, and therefore, acts as an oxidizing agent.
(b) In the presence of complexing agents, cobalt gets oxidized from,+2 to +3 state because Co (III) is more stable than Co (II).
(c) After the loss of ns electrons, a d1 electron can easily be lost to give a stable configuration. Therefore, the elements having d¹ configuration are either reducing or undergo disproportionation.

(a) 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.
(b) The complete ionic equation will be
\(3 \mathrm{MnO}_4^{2-}+4 \mathrm{H}^{+} \rightarrow \mathrm{MnO}_2+2 \mathrm{MnO}_4^{-}+2 \mathrm{H}_2 \mathrm{O}\)
\(\mathrm{MnO}_4^{2-}\) undergoes disproportionation reaction to give MnO₂ and \(\mathrm{MnO}_4^{-}\)
(c) As a result, the elements of the chromium group have extremely high melting points. Manganese has a half-filled d-orbital configuration, which means that all of its electrons spin in the same direction. Mn has a lower melting point because it has fewer interatomic forces of attraction, which are easier to break.

(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) Mn shows the highest number of oxidation states because its valence shell electronic configuration is 3d⁵4s². As 3d and 4s are closer in energy, it has maximum number of electrons to lose or share. Hence, it shows all the oxidation states from +2 to + 7.
(ii) Chromium has highest melting point among all the given elements.
(iii) Scandium shows only +3 oxidation state
(iv) In the + 3 oxidation state, Mn is a strong oxidising agent because in Mn³⁺ ion, Mn exists in 3d⁴ configuration which is less stable and it can reduce to Mn²⁺ giving a more stable 3d⁵ configuration.Hence, it acts as a strong oxidising agent.

Preparation of KMnO₄ It is prepared by the fusion of pyrolusite ore (MnO₂) with an alkali metal hydroxide and an oxidising agent like KNO₃. Dark green K₂MnO₄ is obtained which on disproportionation in neutral or acidic solution gives potassium permanganate.
\(\underset{Pyrolusite}{2MnO_4}+4KOH+O_2⟶\underset{Potassium\ manahanate\\ ( Green\ mass)}{2KMnO_4}+2H_2O\) 
The green mass can be extracted with water and then oxidized either electrolytically or by passing chlorine/ozone into the solution.
Electrolytic oxidation
\(K_{2} M n O_{4} \leftrightarrow 2 K^{+}+M n O_{4}^{2-}\) 
\(H_{2} O \leftrightarrow H^{+}+O H^{-}\) 
At anode, manganate ions are oxidized to permanganate ions.
\(M n O_{4}^{2-} \leftrightarrow M n O_{4}^{-}+e^{-}\) 
Green       Purple
Oxidation by chlorine
\(2 \mathrm{~K}_{2} \mathrm{MnO}_{4}+\mathrm{Cl}_{2} \rightarrow 2 \mathrm{KMnO}_{4}+2 \mathrm{KCl}\) 
\(2 \mathrm{MnO}_{4}^{2-}+\mathrm{Cl}_{2} \rightarrow 2 \mathrm{MnO}_{4}^{-}+2 \mathrm{Cl}^{-}\) 
Oxidation by ozone
\(2 \mathrm{~K}_{2} \mathrm{MnO}_{4}+\mathrm{O}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{KMnO}_{4}+2 \mathrm{KOH}+\mathrm{O}_{2}\) 
\(2 \mathrm{MnO}_{4}^{2-}+\mathrm{O}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{MnO}_{4}^{2-}+2 \mathrm{OH}^{-}+\mathrm{O}_{2}\) 
(i) Acidified KMnO₄ solution oxidizes Fe (II) ions to Fe (III) ions i.e., ferrous ions to ferric ions.
\(\mathrm{MnO}_{4}^{-}+8 H^{+}+5 e^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}\) 
\(\left.F e^{2+} \rightarrow F e^{3+}+e^{-}\right] \times 5\) 
------------------------------------------------------------------------------------
\(\mathrm{MnO}_{4}^{-}+5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+4 \mathrm{H}_{2} \mathrm{O}\) 
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(ii) Acidified potassium permanganate oxidizes SO₂ to sulphuric acid.
\(\left.\mathrm{MnO}_{4}^{-}+6 \mathrm{H}^{+}+5 e^{-} \rightarrow \mathrm{Mn}^{2+}+3 \mathrm{H}_{2} \mathrm{O}\right] \times 2\) 
\(\left.2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{SO}_{2}+\mathrm{O}_{2} \rightarrow 4 H^{+}+2 S O_{4}^{2-}+2 e^{-}\right] \times 5\) 
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\(2 \mathrm{MnO}_{4}^{-}+10 \mathrm{SO}_{2}+5 \mathrm{O}_{2}+4 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{SO}_{4}^{2-}+8 \mathrm{H}^{+}\) 
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(iii) Acidified potassium permanganate oxidizes oxalic acid to carbon dioxide.
\(\left.\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+} 5 e^{-} \rightarrow \mathrm{Mn}^{2-}+4 \mathrm{H}_{2} \mathrm{O}\right] \times 2\) 
\(\left.\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow 2 \mathrm{CO}_{2}+2 e^{-}\right] \times 5\) 
------------------------------------------------------------------------------------------------
\(2 \mathrm{MnO}_{4}^{-}+5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_{2}+8 \mathrm{H}_{2} \mathrm{O}\) 
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(i) (c): Spectrochemical series:
\((\mathrm{I}^{-}<\mathrm{Br}^{-}<\mathrm{SCN}^{-}<\mathrm{Cl}^{-}<\mathrm{S}^{2-}<\mathrm{F}^{-}<\mathrm{OH}^{-}<\mathrm{C}_{2} \mathrm{O}_{4}^{2-}<\mathrm{O}^{2-}< \mathrm{H}_{2} \mathrm{O}<\mathrm{NCS}^{-}<\mathrm{EDTA}^{4-}<\mathrm{NH}_{3}\)(ii) (b)
(iii) (a) : When Δ_o > P, the electrons paired up in the t_2g level rather than going to the e_g level, so
when Δ_o> P : \(t_{2 g}^{4} e_{g}^{0}\)
and Δ_o < P : \(t_{2 g}^{3} e_{g}^{1}\)
(iv) (d): Fe²⁺: 3d⁶ ⇒\(t_{2 g}^{4} e_{g}^{2}\)
(Since, F^- is a weak field ligand)
Hence four unpaired electrons are present.
Magnetic moment (μ)\(=\sqrt{n(n+2)}=\sqrt{4(4+2)}=4.9 \mathrm{~B} \cdot \mathrm{M}\)