isotropic

When coordination number is eight, r⁺/r^- ratio lies between 0.225 to 0.414

MX₂

6.02\(\times\)10¹⁷mol^-1

Quartz glass (SiO₂)

6,6

Frenkel defect

The edge length of the crystal A⁺B⁺ is equal to the distance between A⁺ and B^- ions

8 tetrahedral voids within the unit cells.

Diamond

OF₂ is an oxide of fluorine

NaHSO₄

IF₃> BrF₃>ClF₃

F₂>Cl₂>Br₂>I₂ : Bond dissociation energy

Ar

+8

3

All the ions are coloured

Any one of these

\(Molar \ conductivity,{ \wedge }_{ m }=\frac { K\times 1000 }{ { c }_{ m }(Molarity) } ; \ Equivalent \ conductivity,{ \wedge }_{ eq }=\frac { K\times 1000 }{ { c }_{ m }(Normality) } \)
\(Eq.wt. \ of \ { Al }_{ 2 }{ \left( { SO }_{ 4 } \right) }_{ 3 }=\frac { Mol.wt }{ 6 } (Total+ve \ valency \ of \ Al \ in \ { Al }_{ 2 }{ \left( { SO }_{ 4 } \right) }_{ 3 }=6)\)
\(\therefore Normality=6\times Molarity\)
\(Hence, \ \frac { { \wedge }_{ m } }{ { \wedge }_{ eq } } =\frac { Normality }{ Molarity } =6or{ \wedge }_{ m }=6{ \wedge }_{ eq }\)
\(Alternatively, \ { \wedge }_{ eq }=\frac { { \wedge }_{ m } }{ { v }_{ + }{ z }_{ + } } =\frac { { \wedge }_{ m } }{ 2\times 3 } =\frac { { \wedge }_{ m } }{ 6 } \)

The metal ions (Zn ²⁺) formed by the loss of electrons will accumulate in one electrode and the negative ions \({ { (SO }_{ 4 } }^{ 2- })\) will accumulate in the other. Thus, the solutions will develop charges and the current stops flowing. Moreover, inner circuit is not completed.

Higher the standard reduction potential of a species, more easily it is reduced at the cathode. As Ag⁺ (aq) has greater standard reduction potential, therefore, the reaction that will occur at the cathode is
\({ Ag }^{ + }(aq)+{ e }^{ - }\longrightarrow Ag(s)\)

Differences between electrochemical cell and electrolytic cell.

\({ \wedge }^{ ° }\left[ { Al }_{ 2 }{ \left( { SO }_{ 4 } \right) }_{ 3 } \right] =2{ \lambda }^{ ° }\left( { Al }^{ 3+ } \right) +3{ \lambda }^{ ° }\left( { { SO }_{ 4 } }^{ 2- } \right)\)
\({ \wedge }^{ ° }\left[ { Al }_{ 2 }{ \left( { SO }_{ 4 } \right) }_{ 3 } \right] =858S{ cm }^{ 2 }{ mol }^{ -1 },\)
\( { \lambda }^{ ° }\left( { { SO }_{ 4 } }^{ 2- } \right) =106S{ cm }^{ 2 }{ mol }^{ -1 }\)
\(\therefore \ 858=2{ \lambda }^{ ° }\left( { Al }^{ 3+ } \right) +3\times 160\)
\(\\ or \ 2{ \lambda }^{ ° }\left( { Al }^{ 3+ } \right) =858-480=378\)
\(\therefore \ { \lambda }^{ ° }\left( { Al }^{ 3+ } \right) =\frac { 378 }{ 2 } =189S{ cm }^{ 2 }{ mol }^{ -1 }\)

(i) Mercury cell
(ii) Fuel cell
(iii) Lead storage cell
(iv) Dry cell

The particles of colloidal size formed due to aggregation of several ions or molecules with lyophobic as well as lyophilic particles are called micelles. The common micelles system is soap. The micelles behave as normal electrolytes at low concentrations but as colloids at higher concentrations. The micelles differ from normal colloidal solutions in the sense that in the normal solutions the particles are already in the colloidal range and no such aggregation or association takes place. On the other hand, the micelles become of colloidal size on aggregation.

The fraction of molecules having energy equal to or greater than activation energy,
Fraction of molecules \(={ e }^{ { -{ E }_{ a } }/{ RT } }\)
\(={ e }^{ \frac { -209.5\times 1000 }{ 8.314\times 581 } }\)
\(\\ ={ e }^{ -43.37 }=1.461\times { 10 }^{ -19 }\).

Here, \(k=2.418\times { 10 }^{ -5 }{ s }^{ -1 },{ E }_{ a }=179.9\quad kJ\quad { mol }^{ -1 },\quad T=546K.\)
\(or \ logA=logk+\frac { { E }_{ a } }{ 2.303RT } =log(2.418\times { 10 }^{ -5 }{ s }^{ -1 })+\frac { 179.9 \ kJ \ { mol }^{ -1 } \ }{ 2.303\times 8.314\times { 10 }^{ -3 }kJ \ { K }^{ -1 } \ { mol }^{ -1 }\times 546K } \)
\(\\=(-5+0.3834){ s }^{ -1 }+17.2081=12.5924{ s }^{ -1 }\)
\(A=Antilog(12.5924){ s }^{ -1 }=3.912\times { 10 }^{ 12 }{ s }^{ -1 }\)
According to Arrhenius equation,
\(k=A{ e }^{ { -{ E }_{ a } }/{ RT } }or \ ln \ k=lnA-\frac { { E }_{ a } }{ RT } or \ logk=logA-\frac { { E }_{ a } }{ 2.303RT }\)

\(\triangle\)_rG° = - nFE^o_{ceIl  ,} n = 6
= - 6 x 96500 C/mol  x  0.34 V
=  - 196860J /mol or - 196.860 kJ / mol
\(\therefore\) E^o_ceIl  = 0.059V / n  x  log K_c
log K_c  = \(\frac{0.34V \times 6}{0.059V}\) = 34.5762

\({p H}=3 \Rightarrow\left[{H}^{+}\right]={1 0}^{-3} {M}\)
now,
\({E}_{\text {cell }} ={E}^{\circ}-\frac{{0 . 0 5 9}}{4} \log \frac{\left[{F} {e}^{2+}\right]^{2}}{[{H}+]^{4} {P}_{\mathrm{O}_{2}}} \)
\(=1.67-\frac{0.059}{4} \log \frac{\left(10^{-3}\right)^{2}}{\left(10^{-3}\right)^{4} \times 0.1}=1.67-\frac{0.059}{4} \log 10^{2} \)
\(=1.67-\frac{0.059}{4} \times 7=1.67-0.105=1.565=1.57 \mathrm{~V} \)

\(k=1.15\times { 10 }^{ -3 }{ s }^{ -1 }\)
\({ \left[ R \right] }_{ 0 }=5g\quad \left[ R \right] =3g\)
\(k=\frac { 2.303 }{ t } \log { \frac { { \left[ R \right] }_{ 0 } }{ \left[ R \right] } } \)
\(t=\frac { 2.303 }{ 1.15\times { 10 }^{ -3 } } \log { \frac { 5 }{ 3 } }\)
\(=4.438\times { 10 }^{ 2 }s\)

Log K = log A - Ea/2.303RT
Log K’ = log A - Ea/2.303RT
Log K = log A - Ea/2.303RT
Log K’/K = Ea-Ea/2.303 RT = 9.8037
Or K’/K = 6.36 x 109
UR catalysted /UR uncatalysed = K’/K = 6.4 x 10 9

\(k=\frac{2303}{t} \log \frac{a}{a-x}\)
\( k=\frac{2.303}{600 \mathrm{~s}} \log \frac{0.04}{0.03} \Rightarrow k=4.8 \times 10^{-4}\)
\( \\ t_{1 / 2}=\frac{0.693}{k}=\frac{0.693}{4.8 \times 10^{-4}}=1.44 \times 10^{3} \mathrm{~s}\)

(i) It shows positive deviation.
It is due to weaker interaction between acetone and ethanol than ethanol-ethanol interactions.
(ii) Given : W_B  = 10g, W_s = 100 g, W_A = 90 g, M_B = 180 g.mol and d = 1.2 g/mL
\(M=\frac { Wt%\times density\times 10 }{ Mol.wt } \)
\(M=\frac { 10\times 1.2\times 10 }{ 180 } \)
= 0.66 M or 0.66 mol/L
\(m=\frac { { W }_{ B }\times 1000 }{ { M }_{ B }\times { W }_{ A }(in\quad g) } \)
\(m=\frac { 10\times 1000 }{ 180\times 90 } \)
= 0.61 m or 0.61 mol/kg
(or any other suitable method)

\(\Delta { T }_{ f }=\frac { { K }_{ f }{ W }_{ b }\times 1000 }{ { M }_{ b }\times { W }_{ a } } \)
\(0.383=\left( \frac { 3.83\times 2.56 }{ M\times 100 } \right) \times 1000\)
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.

\(k={0.693\over t_{1/2}}={0.693\over 2.1hr}=0.33hr^{-1}\)
x = 99% of a = 0·99 a
\(t={2.303\over k}log{a\over a-x}={2.303\over 0.33hr^{-1}}log{a\over a-0.99a}={2.303\over 0.33}log 10^2=13.69hours\)
(ii) Amount decomposed = 99% of 6.2 g =\({99\over 100}\times 6.2g=6.138g\)
1 mol NH₂NO₂ (63g) produce N₂O at STP = 22·4 L
6.138 g will produce N₂O at STP =\({22.4\over 63}\times6.138\ L=2.2176L\)

As.ΔH = - 33 kJ mol^-1, the reaction is exothermic. The activation energy diagram will be as shown in fig.
ΔH = E_f - E_b= - 33 kJ
k_f = Ae^-E_f/RT
kb = Ae^-E_b/RT
\(K_c={k_f\over k_b}=e^{(E_b-E_f)/RT}\)
In \(K_c={E_b-E_f\over RT}\ or\ log\ K_c={E_b-E_f\over 2.303RT}={30000\ J\ mol^{-1}\over 2.303(8.314JK^{-1}mol^{-1})300K}=5.2227\)
K_c = Antilog 5·2227 = 1·67 x 10⁵
Substituting \(E_b={31\over 20}E_f,\)We get
\(E_f-{31\over 20}E_f=-33\ or\ {-{11\over 20}}E_f=-33\ or\ E_f={33\times20\over 11}=60kJ\ mol^{-1}\)
E_b = E_f + 33 = 60 + 33 = 93 kJ mol^-1

Surface area per gram of gel = 561.8 m².

(i) \({ NH }_{ 4 }{ NO }_{ 2 }(s)\underrightarrow { heat } { N }_{ 2 }+{ 2H }_{ 2 }O\) 
                'A'                 'B'
(ii) \({ N }_{ 2 }+{ 3H }_{ 2 }\longrightarrow 2{ NH }_{ 3 }(g)\) 
                              'C' basic
(iii) \({ 4NH }_{ 3 }+{ 5O }_{ 2 }\longrightarrow 4NO+{ 6H }_{ 2 }O\) 
\(\\ 2NO+{ O }_{ 2 }\longrightarrow { 2NO }_{ 2 }\) 
                  'D' (part of acid rain)
\({ 3NO }_{ 2 }+{ H }_{ 2 }O\longrightarrow 2{ HNO }_{ 3 }+NO\)      

a) Silica gel is a good adsorbent and hence water vapour are adsorbed on its surface. Anhydrous CaCl₂ undergoes absorption because it combines with water molecules to form hydrated calcium chloride, CaCI₂. 2H₂O.
b) BF₃ acts as a strong Lewis acid and hence used as catalyst in industrial processes.

Soap stabilizes the emulsion between dirt, grease, and water. In soap, molecular ions get aggregated to form micelles which help in removing dirt and grease.

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.

(i) When acetone & chloroform are mixed together a hydrogen bond is formed between them which increases the intermolecular attraction between them & hence decreases the vapour pressure.
(ii) When a non volatile solute is added to a solvent, the vapour pressure of the resulting solution is lower than the vapour pressure of pure solvent.This is because, the surface area is partly occupied by non volatile solute molecules and partly by solvent molecules. Due to this, the rate of evaporation is lowered.
(iii) Molality of a solution is defined as the number of moles of solute dissolved per kg of solvent.
(C₆H₁₂O₆) = 10% by weight.
Thus in 100 g solution mass of glucose = 10 g and of water = 90 g
Number of moles of glucose = \(\frac{ molar \ mass}{mass}\)
n = \(\frac{10}{180}\) = \(\frac{1}{18}\)
Mass of solvent = 90; g = 0.09 kg
Molality, m = \(\frac{1}{18}\)/ 0.09
m = \(\frac{100}{18 \times9}\) = 0.617 molal ≈ 0.6 molal
The molality of a solution of glucose (C₆H₁₂O₆) in water is labelled as 10% by weight is  0.617 molal.
= 0.617 m

(ii) Calculation of elevantion in b.p. temperature (ΔT_b) NaCI dissociates in aqueous solution as NaCl (s)→Na⁺ (aq)+Cl⁻(aq)
i = 2, W_B = 15.0g, M_B = 58.44 g mol⁻¹,W_A = 250g = 0.25kg
K_b= 0.152 K kg mol^{−1
\(\Delta T_{b}=i \times K_{b} \times m=\frac{i \times K_{b} \times W_{B}}{M_{B} \times W_{A}}\)
\(\Delta T_{b}=\frac{2 \times\left(0.512 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right) \times(15.0 g)}{\left(58.44 \mathrm{~g} \mathrm{~mol}^{-1}\right) \times(0.25 K g)}=1.05 K\)}
Step II. Calculation of boilling point of solution (Tb)
^{\(T_{b}=T_{b}^{\circ}+\Delta T_{b}=(373+1.05) K=374.05 K \)}

(i) When concentration approaches zero, the molar conductivity is known as limiting molar conductivity.
The change in Λ_m with dilution is due to the increase in the degree of dissociation and consequently the number of ions in the total volume of the solution that contains 1 mol of electrolyte, hence Λ_m increases steeply.
(ii) \(E_{\text {cell }} =E_{\text {cell }}^{\circ}-\frac{0.059}{n} \log \frac{\left[\mathrm{Mg}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]} \)
\(=2.71 \mathrm{~V}-\frac{0.059}{2} \log \frac{0.1}{0.001} \)
\(=2.71 \mathrm{~V}-\frac{0.059}{2} \log 10^{2}=2.651 \mathrm{~V} \)

According to the question, the given values are 1 S crn^-1 = 100 S m^-1 = 1 S cm^-1 /100 Sm^-1 = 1 (unit conversion factor)                         

So, the graph between Am and C^1/2 is shown below

Λ_m^o = Intercept on the Λ_m axis = 124.0 S cm² mol^-1 (the value of Λ_m on extrapolation to zero concentration).

At anode : 2I^c−→I^c−₃ + 2e⁻
At cathode : 2Fe³⁺+ 2e⁻→ 2Fe²⁺
\(\text { At equlibrium, } E_{c e l l}=0 \)
\(E^{c-} \cdot c e l l=\left(E^{c-} \cdot \text { red }\right)_{c}-\left(E^{c-} \cdot \text { red }\right)_{a} \)
\(=E^{c-} \cdot\left(F e^{3+} \mid F e^{2+}\right)-E^{c-} \cdot\left(I_{3}^{c-} \mid 3 I^{c-}\right) \)
\(=0.78-0.54=0.24 V\)
\( E_{\text {cell }}=E^{c-} \text { -cell }-\frac{0.06}{2} \log K_{e q}\left(n_{\text {cell }}=2\right) \)
\(E^{c-} \cdot \text { cell }=0.03 \log K_{\text {eq }} \)
\(\log K_{e q}=\frac{E^{c-} \cdot \text { cell }}{0.03}=\frac{0.24 V}{0.03 V}=8.0 \)
\(K_{\text {eq }}=10^{8}\)

(i) Cell reaction: Sn + 2 H⁺ ➝ Sn²⁺+ H² (n = 2)
nernst eqn : \(E_{cell} =E ^0_{cell}-{0.0591\over 2}log{[Sn^{2+}]\over[H^+]^2}=0-(-0.14)-{0.0591\over 2}log{0.05\over (0.02)^2}\)
\(=0.14-{0.0591\over 2}log125=0.14-{0.0591\over2}(2.0969)=0.078V\)
(ii) Cell reaction: 2 Br^- + 2 H⁺ ⇾ Br₂ + H₂ (Note carefully)
nernst eqn : \(E_{cell}=E^0_{cell}-{0.0591\over 2}log{1\over [Br^-]^2[H^+]^2}\)
\(E_{cell}={(0-1.08)}-{0.0591\over 2}log{1\over(0.01)^2(0.03)^2}\)
\(=-1.08-{0.0591\over 2}log(1.111\times10^7)=-1.08-{0.0591\over 2}(7.0457)=-1.08-0.208=-1.288V\)

Number of moles of \(^{238} U=\frac{1}{238}\)
Number of moles of  \({ }^{206} P b=\frac{0.1}{206} \)
Applying the  relationship, t  \(=\left(\frac{2.303}{\lambda} \log \left[1+\frac{.^{206} P\\ b}{.^{238} U}\right]\right) =\frac{2.303}{0.693} \times 4.5 \times 10^{9}\)
\(\log \left[1+\frac{\frac{0.1}{206}}{\frac{1}{238}}\right]=7.098 \times 10^{8} years\)

(i)Preparation of chlorine by electrolytic method : Chlorine is obtained by the electrolysis of brine solutions (Conc. NaCl). Cl₂ gas is liberated at anode.
2NaCl→2 Na⁺+2Cl⁻
2 H₂O+2e⁻→2OH⁻+H₂ (cathode)
2 Cl⁻→Cl₂+2e⁻(anode)
(iii)The oxoacids of chlorine are HClO,HClO ₂ ​ ,HClO ₃ ​ ,HClO ₄ ​ . The increasing order of their oxidizing power is :
HClO ₄ ​ <HClO _{3 ​} <HClO ₂ ​ <HClO This is because the oxidizing power is inversely proportional to the thermal stability of acids for oxacids of chlorine. Here, the thermal stability of oxyacids is HClO ₄ ​ >HClO ₃ ​ >HClO ₂ ​ >HClO.
(iv)Helium is used as a coolant in nuclear reactors.
(v)It is a linear compound and angle of XeF₂ is 180 degrees.

(a) The high enthalpies of atomization are due to a large number of unpaired electrons in their atoms. Therefore, they have stronger interatomic interactions and hence, stronger bonding between atoms. Thus, they have high enthalpies of atomization.
(b) Most of the compounds of transition elements contain unpaired electrons in their (n-1) d-subshells. Therefore, they are paramagnetic in nature and are attracted by the magnetic field. The magnetic character is expressed in terms of magnetic moment. The larger the number of unpaired electrons in a substance, the greater is the paramagnetic character and larger is the magnetic moment. The magnetic moment is expressed in Bohr magneton abbreviated as B.M. For example, Ti²⁺ has 2 unpaired electrons and has less magnetic moment than Mn²⁺which has 3 unpaired electrons. Mn²⁺has 5 unpaired electrons and has maximum magnetic moment among the divalent transition metal ions because d-subshell can have a maximum of 5 unpaired electrons.
(c) Most of the transition metal ions are colored both in the solid state and in aqueous solutions.The color of these ions is attributed to the presence of incomplete (n - 1) d subshell. The electrons in these metal ions can be easily promoted from one energy level to another in the same d-subshell. The amount of energy required to excite the electrons to higher energy states within the same d-subshell corresponds to the energy of certain colors of visible light. Therefore, when white light falls on a transition metal compound, some of its energy corresponding to a certain color, is absorbed causing promotion of d-electrons. This is known as d-d transitions. The remaining colors of white light are transmitted and the compound appears colored. For example, hydrated cupric compounds absorb radiations corresponding to red light and the transmitted color is greenish blue (which is complementary color to red color). Thus, cupric compounds have greenish-blue color
(d) Some transition metals and their compounds act too good catalysts for various reactions. This is due to their ability to show multiple oxidation states. The common examples are Fe, Co, Ni, V, Cr, Mn, Pt, etc.The transition metals form reaction intermediates with the substrate by using empty d-orbitals. These intermediates give reaction paths of lower activation energy and therefore, increase the rate of reaction. 

(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.

(i) We should avoid making atom bombs because they lead to the lot of destruction and lot of national money is spent on it.
(ii) Nuclear radiation may cause mutation in genes which leads to genetic disorders.
(iii) U-235 is fissionable material.
(iv) Breeder reactors are used to produce nuclear fuelic fissionable material needed for nuclear reactions.

(i) (a) In p-block, elements show variable oxidation state. It increases as we move from left to right in the periodic table. The maximum oxidation state shown by p-block element is equal to the total number of valence electrons. Whereas in d-block, elements show different oxidation states because of incomplete d-subshell. The variable oxidation state is due to the participation of ns and (n -1) d electrons in bonding.
(b) In aqueous solution, Cu⁺ ion undergoes disproportionation reaction.
2Cu⁺ (aq) ⇾ Cu²⁺ + Cu(s)
The stability of Cu²⁺ ion in aqueopus solution is due to negative enthalpy of hydration. which more compensate for the IE₂ or Cu.
(c) When orange solution containing \(Cr_2O_7^{2-}\) ion is treated with an alkali, a yellow solution of  (Chromate ion) is obtained.
\(\underset{Dichromate\\(Orange)}{Cr_2O_7^{2-}}\xrightarrow{OH^-}\underset{chromate\ ion\\(Yellow)}{Cr_2O_4^{2-}}\)
(ii) Chemistry of actinoids is complicated as compared to the lanthanoids. The two reasons are:
(a) 5-orbital present in actinoids is more exposed to the outer environment while 4-f orbital present in lanthanoide are deeply buried.
(b) Lanthanoids show limited number of oxidation states as +2,+3 and + 4 (out of which + 3is most common). This is-due to the large energy gap between 4f and 5d subshells. The dominant oxidation state of actinoids is also + 3 but they show a number of other oxidation states also like uranium (Z = 92) and plutonium (Z = 94) show +3, + 4, + 5and +6, neptunium shows +3, + 4, + 5 and +7. This is due to small energy difference between 5f, 6d, and 7s subshells.

(i) Electronic configuration of Mn ²⁺= [Ar]¹⁸3d⁵
Electronic configuration of Fe²⁺ = [Ar]¹⁸3d⁶ 
Mn ²⁺ having half-filled d-orbitals will be more stable than Fe²⁺, as it has partially filled d-orbitals.
(ii) Zinc has completely filled d-orbitals, which limits its tendency to form metallic bonds. Thus, it requires least enthalpy to get atomised.
(iii) MO₃F is MnO₃F.
In MO₃F,
Let the oxidation state of M is x.
x+3 x (-2)+(-1) = 0
or, x = + 7, i.e. M is in oxidation state of +7. Hence, the given compound is MnO₃F. 
(iv) E^O Cr³⁺ ICr ²⁺ is negative (-0.4 V). It shows the stability of Cr³⁺ ions i.e. Cr³⁺ in solution cannot be reduced to Cr²⁺ ions. Mn³⁺ has high positive value of E^O. Thus, Cr³⁺ is the most stable, Mn³⁺ is least stable.
(v) The tendency to form complex compounds is due to:
i. Small size and high charge on metal ion 
ii. The availability of d orbitals for accommodating electrons donated by the ligand.