(i)From the electronic configuration of the 3d transition series, we can see that, as we go from left to right, the number of unpaired electrons increases from Sc to Cr and decreases thereafter. So the paramagnetic nature of the 3d transition series increaes till Cr and then decreases regularly.
(ii)Transition metals have vacant d-orbitals and because of this their valency is variable so they can be used as oxidising agent as well as reducing agent. also they have large surface area so they can act as a good catalyst.
(iii)The transition metals form interstitial compounds because there are vacant spaces in the lattice of transition metals which can be filled by small atoms like H,C,N etc.

Positive deviation from Raoult's law occurs when the total vapour pressure of the solution is more than corresponding vapour pressure in case of ideal solution. 
\(P={P}_{{A}}+{P}_{{B}}>{P}_{{A}}^{\circ} {X}_{{A}}+{P}_{{B}}^{\circ} {X}_{{B}}\)
Negative deviation from Raoult's law occurs when the total vapour pressure of the solution is less than corresponding vapour pressure in case of the ideal solution.
\({P}={P}_{{A}}+{P}_{{B}}<{P}_{{A}}^{\circ} {X}_{{A}}+{P}_{{B}}^{\circ} {X}_{{B}}\)
For positive deviation from Raoult's law, Δ _{mix ,}H has a positive sign.
For negative deviation from Raoult's law, Δ _{mix .}H has a negative sign.

\(2NH_x\rightarrow N_2+3H_2\)
\(Rate=-\frac{s[NH_3]}{dt}=\frac{d[N_2]}{dt}=\frac{1}{3}\frac{d[H_2]}{dt}\)
For zero order reaction, rate=k
\(\therefore\ -\frac{1}{2}\frac{s[NH_3]}{dt}=\frac{d[N_2]}{dt}=\frac{1}{3}\frac{d[H_2]}{dt}\)
\(2.5 \times 10^{-4}\ mol L^{-1}\ s^{-1}\)
Rate of production of N₂ =\(\frac{d[N_2]}{dt}\)  
\(2.5 \times 10^{-4}\ mol L^{-1}\ s^{-1}\)
Rate of production of N₂ =\(\frac{d[N_2]}{dt}=3\times2.5\times10^{-4}\)
\(=7.5 \times 10^{-4}\ mol\ L^{-1}\ s^{-1}\)  

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

(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) \({ \Delta }_{ 0 }>P;{ { t }_{ 2g } }^{ 3 }{ { e }_{ g } }^{ 1 }\)

(a) Lanthanoid contraction. In the lanthanoids, there is regular decrease in the size of atoms and ions with increase in atomic number. This decrease of ionic and atomic radii in the lanthanoid elements is called lanthanoid contraction. For example, the ionic radii decrease from Ce³⁺(111pm) to Lu³⁺ (93 pm).
\(Ti\rightarrow Zr\rightarrow Hf\) 
\(V\rightarrow Nb\rightarrow Ta\) 

869.6 s.

166.16 min.

0.46 volt

32.76 S cm² mol^-1, 1.85 \(\times\) 10^-5

0.25 Molal aqueous solution to urea means that
moles of urea = 0.25 mole
mass of solvent (NH₂CONH₂) = 60 g mol^-1
\(\therefore\) 0.25 mole of urea = 0.25 x 60 = 15g
Mass of solution = 1000+15 = 1015g = 1.015 kg
1.015 kg of urea solution contains 15g of urea
\(\therefore\)2.5 kg of solution contains urea = 15/1.015 x 2.5 = 37 g

Unidentate Ligands: Those ligands which can form only one coordinate bond are called unidentate ligands, e.g. NH₃ , CN^- Didentate Ligands: Those ligands which can form two coordinate bonds are called bidentate ligands,\(\overset { { COO }^{ - } }{ \underset { { COO }^{ - } }{ | } } \)  (ox), H₂NCH₂CH₂NH₂(en). Ambidentate Ligands: Those ligands which can form bond through either of the two donor atoms are called ambidentate ligands, e.g. CN^- , NO₂^-  etc.

(i) Ambidentate ligand: Ligands which can ligate (link) through two different atoms present in it are called ambidentate ligands, e.g.\({ { NO }_{ 2 } }^{ - },{ SCN }^{ - },{ CNO }^{ - },{ CN }^{ - },{ { NO }_{ 2 } }^{ - }\) can link through 'N' as well as oxygen while SCN^- can link through'S' as well as 'N' atom.
(ii) Denticity: The number of ligating (linking) atoms present in ligand is called denticity of a ligand, e.g.
CN^- has 1,\(\overset { { COO }^{ - } }{ \underset { { COO }^{ - } }{ | } } \) has 2,
H₂N-CH₂-CH₂-NH-CH₂-CH₂-NH₂ has 3.
H₂NCH₂CH₂NHCH₂CH₂NHCH₂CH₂NH₂  has 4.
[EDTA]^3-  has 5, [EDTA]^4- has 6 denticity.

(i) +3 (III)  
(ii) dsp, octahedral
(iii) paramagnetic
(iv) Two geometrical isomers
(v) Yes, there may be optical isomer also due to presence of polydentate ligand.
(vi) Dichlorido bis-(ethane 1, 2-diamine) Iron (III)

\({ { \wedge }^{ ° } }_{ m }\left( { CaCl }_{ 2 } \right) ={ \lambda }^{ ° }\left( { Ca }^{ 2+ } \right) +2{ \lambda }^{ ° }\left( { Cl }^{ - } \right) \)
\(=119.0S{ cm }^{ 2 }{ mol }^{ -1 }+2\times 76.3S{ cm }^{ 2 }{ mol }^{ -1 }=271.6S{ cm }^{ 2 }{ mol }^{ -1 }\)
\({ { \wedge }^{ ° } }_{ m }\left( { MgSO }_{ 4 } \right) ={ \lambda }^{ ° }\left( { Mg }^{ 2+ } \right) +2{ \lambda }^{ ° }\left( { { SO }_{ 4 } }^{ 2- } \right) \)
\(=106 \ S{ \ cm }^{ 2 }{ \ mol }^{ -1 }+160 \ S{ \ cm }^{ 2 }{ \ mol }^{ -1 }=266 \ S{ \ cm }^{ 2 }{ \ mol }^{ -1 }\)

(a) Cr₂O₇^2-+14H⁺+6e^-\(\longrightarrow \)2Cr³⁺ +7H₂O
(b) CrO₄^2-+2H⁺\(\rightleftharpoons \)2HCrO₄^-\(\rightleftharpoons \)Cr₂O₇^2-+H₂O
(c) MnO₄^-+2H₂O+3e^- \(\overrightarrow { medium } \)MnO₂ +4OH^-

(a)\((i) Cr_2O_7^{2-} + 14 H^+ + 6Fe^{2+}\longrightarrow\ 2Cr^{3+} + 6Fe^{3+} + 7H_2O\)
\((ii) 2 MnO_4^- + 10 I^- + 16 H^+\longrightarrow 2Mn^{2+} + 8H_2O + 5I_2\)
(b) Mn²⁺: 3d⁵4s⁰ has S unpaired electrons.It is highly paramagnetic and it is attracted by magnet.

500 ml of 0.15 M CuSO₄ solution contains \(\frac { 500\times 0.15 }{ 1000 } \)
Mass of Cu = 0.075 x 63.5 = 4.7625 g; Eq. wt. of Cu2⁺ =\(\frac{63.5}{2}\)=31.75
m = Z \(\times\) l \(\times\) t
4.7625 = \(\frac{31.75}{96500}\)\(\times\)0.15 \(\times\) t
t= \(\frac { 4.7625\times 96500 }{ 31.75\times 0.15 } \) = 96500 sec = \(\frac { 96500 }{ 60\times 60 } \) = 26.80 hours.

\(\frac { dx }{ dt } =k{ \left[ A \right] }^{ 1 }{ \left[ B \right] }^{ 0 }\)
\(\\ From \ expt. \ I\)
\(2.0\times { 10 }^{ -2 }=k\left[ 0.1 \right] ,\)
\(k=2.0\times { 10 }^{ -1 }{ min }^{ -1 }\)
\(=0.2{ \ min }^{ -1 }=2\times { 10 }^{ -1 }{ min }^{ -1 }\)
\(In \ expt. \ II\)
\(\frac { dx }{ dt } =k{ \left[ A \right] }^{ 1 }\)
\(4\times { 10 }^{ -2 }=2\times { 10 }^{ -1 }\times \left[ A \right]\)
\(\Rightarrow \ \left[ A \right] =0.2 \ mol \ { L }^{ -1 }\)
\(In \ expt. \ III\)
\(Rate=\frac { dx }{ dt } =k{ \left[ A \right] }\)
\( =1.2\times 0.4\)
\(=8\times { 10 }^{ -2 }mol \ { L }^{ -1 }{ min }^{ -1 }\)
\(In \ expt.IV\)
\(2.0\times { 10 }^{ -2 }=0.2\times \left[ A \right]\)
\(\left[ A \right] =0.1mol \ { L }^{ -1 }\)

\(Molarity \ (M)=\frac{\text { Mass of solute } / \text { molar mass }}{\text { Volume of solution in litres }} \\\)
M = 0.15 M = 0.15 mol L^-1 ;
Molar mass of solute = 7 x 12 + 6 x 1 x 2 x 16 = 122 g mol^-1;
Volume of solution = 250 mL = 0.25 L.
\(\left(0 \cdot 15 \mathrm{~mol} \mathrm{~L}^{-1}\right)=\frac{\text { Mass of solute }}{\left(122 \mathrm{~g} \mathrm{~mol}^{-1}\right) \times(0 \cdot 25 \mathrm{~L})}\)
Mass of solute = (0.15 mol L^-1) x (122 g mol^-1) x (0.25 L) = 4.575 g