1. PERIODIC TABLE 1. Sodium Bismuthet is a strong oxidising agent.
® Sodium Bismuthet ® NaBiO3
\ Bismuth is in +5 oxidising state
Valance shell configuration of ® Bio 6s2 6p3
2. How many d-electrons are present in Titanic Sulphate ?
® Titanic Sulphate ® Ti2(SO4)3
\ Ti is in +3 oxidising state
Tio ® 3d2 4s2
Ti3+ ® 3d1
Number of d-electron = 1
3. How many d-electrons are present in Potassium Permanganate ?
® Potassium Permanganate ® KMnO4
Mn is in +7 oxidising state.
\ Mno ® 3d5 4s2
\ Mn+7 ® 3d o 4so
\ Number of d-electron = 0
4. What is the valance shell configuration of Mercurous ion (Hg+) ?
® Hgo ® 5d10 6s2
Hg+ ® 5d10 6s1

3.  s-block elements :-
group - 1 ® alkali metals
group - 2 ® alkaline earth metals
 d-block elements :-
These are called Transition Elements, because it transits metal to non-metal character.
 p-block elements :-
These are called Representative Elements. Major inorganic chemistry is done by p-block.
 O-group :-
These are called Bridging Elements, because it bridges non-metal to metal.
TRANSITION ELEMENTS
 Definition :-
In ground state or in stable oxidation state, if penultimate (before ultimate / n-1) d-orbital is incompletely filled, these elements are called Transition elements.

4.  Example :-
Feo ® 3d6 4s2
Fe+2 ® 3d6 (Penultimate)
Fe+3 ® 3d5
2) Feo ® 3d10 4s1
Cu+2 ® 3d9 (stable, Penultimate)
Cu+ ® 3d10
3) Zno ® 3d10 4s2
Zn+2 ® 3d10
[ * Zn IS NOT A TRUE TRANSITION ELEMENT--- why ?
® Zn violates the defination of transition elements. That’s why, in true sense it is not a transition element. But it exhibits COMPLEX FORMATION REACTION which is one of the most important properties of so-called transition elements. That’s why Zn is considered as transition element.]

5.  Special properties of Transition Elements :-
 Complex Formation Reaction
 Magnetism
 Colour
1) Complex Formation Reaction :-
a) Ligand :
A species having lone pair / negative charge / p-cloud which are able to co-ordinate with central metal ion is called ligand.
example ---

6. b) Co-ordination Number / CN :
The number of ligands per metal ion to for a stable geometry of any complex is called co-ordination number / C.N.
example ---
c) Complex compound :
The species where the identity of parent ion is lost, is called perfect complex compound.

7. example ---
INF: Cd (CN)2-4 can’t obey the definition of transition element so it is imperfect complex but Cu(CN) 2-4 is a perfect complex

8.  Criteria of transition elements to exhibit complex formation reaction :-
Presence of lower energetic (n-1) partially filled d-orbitals is the cause to accommodate ligands, forming complex.
example ---
Fe(CN) [strong ligand]
2) For full-filled (n -1) shells or for weak field ligands, Zeff. of nucleus sufficiently increases which helps to accommodate ligands in outer shells.
example -
[ Fe(H2O)6] [weak ligand]

9. 3) Transition elements are not so reactive, so their complexes are easily isolated.
[ (a) If inner d-orbital is utilised in bonding, the complex is called inner sphere / inner orbital complex / low spin complex.
(b) If outer d-orbital is used in bonding, the complex is called outer sphere / outer orbital complex / high spin complex ]
2) Magnetism :-
Electrons have two types of motions --- (i) orbital motion and (ii) spinning motion, i.e. orbital magnetic moment (L) and spinning magnetic moment (S) are combined to produce net magnetic moment (T).
 T = L + S

10. It can be proved that,
After the formation of complex, orbital motion of electron is quenched / reduced. So, we may neglect orbital magnetic moment (L). Now, observed magnetic moment may be considered to arise only from unpaired spins. The spin-only magnetic moment s may be written.

11. example ---
1. Say 10 grams of potassium ferrocyanide (K4[Fe(CN)6]) and potassium ferricyanide (K3[Fe(CN)6]) are supplied to you in separate test tube. How will you identify which is what without any loss of mass ?

12. 2. Calculate magnetic moment of Ti2(SO4)3.
* Orbital Contribution in Magnetic Moment :-
In certain cases, mobs is slightly greater than mcal . It can be explained by considering orbital contribution.
Those electronic configuration where three planes (xy, yz, xz) are involved, i.e. electronic configuration is written in three ways, leads to same amount of orbital contribution. But it has no quantitive calculation or co-relation. That’s why, d1, d2, d5 low spin etc. system exhibit slightly higher magnetic moment but d3, d6 low spin exhibit normal mS (spin-only magnetic moment).

13. So, overall it is diamagnetic compound. [NO+ ® cataionic ligand ]
Exception :-
Na2 [ FeII(CN)5 (NO)+]
Normally Nitric Oxide (NO) is a neutral oxide having 11th odd electron in its MOS. In case of 3d metal ion like Fe3+, odd electron of NO is fully transferred to 3d orbital of Fe3+ (Ferric ion) which is energitically allowed. That’s why in Na2 [Fe(CN)5NO] (Sodium Nitropruside), Fe is in +2 oxidising state, and NO is in +1 oxidising state.
Feo ® 3d6 4s2
Fe2+ ® 3d6
So, overall it is diamagnetic compound. [NO+ ® cataionic ligand ]

14. 2) Na2 [ RuIII(CN)5 (NO) ]
Ruthenium belonging to 4d series, 11th electron of NO is not transferred to 4d due to energy consideration level. That’s why, in the anologous Ruthenium complex, Ru bears +3 oxidation state and NO is neutral. Hence it is paramagnetic.
3) Fe2O3 mobs is subnormal (is slightly lower) than the normal value --- why ?
In Fe2O3 cluster, there is anti ferro-magnetic coupling or super exchange interraction through Fe O Fe3+. Thus paramagnetic value of over all compound reduces.

15. Para, Ferro, Ferri, Anti-ferro Magnetism Molecular Magnetic Theory
¹ Diamagnetic substance. If m = O for individual obs molecule, it is called diamagnetic substance
m » O

16. Consider different orientations of molecular magnets, which is controlled by temperature.
1) At a particular temperatue, if all molecular mnagnets are randomly distributed having a finite magnetic moment, that substance is called paramagnetic substance.
2) At a particular temperature, if all molecular magnets are alligned along the same direction resulting highest magnetic moment, then the substance is called strong paramagnetic substance or ferro magnetic substance.
3) At a particular temperature, if major molecular magnets are alligned in same direction and minor molecular magnets are aligned in opposite direction, i.e. overall magnetic moment reduces, then this sort of substance is called weak paramagnetic substance or ferri magnets substance.
4) At a particular temperature, if 50% molecular magnets are in same direction and rest 50% are in opposite direction, overall magnetic moment is exactly zero or almost zero. This sort of substance is called anti-ferro magnetic substance.

17. Example ----
[N.B. :- If all electrons are paired up within the atom or within the molecule resulting m = 0, this called diamagnetic substance. But if same or different molecules are separately para-magnetic, but after combination where a new compound is not generated, magnetic moment lends to zero ( m ® 0). This sort of substance is called antiferro magnetic substance.
Example ----
For Tetra Thyazil (S4N4), ( m ® 0) due to S ... S anti-ferro magnetic coupling or super exchange process. ]

18. 3) Colour :-
We know that after the formation of complex, 5-d-orbitals are splited in two catagories - lower 3 and upper 2. If any one or more electrons from lower level is excited to upper d-level, it absorbs some amount of energy, the un-absorbed light causes colour.
Intensity of colour ®
(i) Spin Selection Rule ®
Electron transition takes place without change in spin.
D s = sfinal (sf) ---- sinitial (sL) = 0
[ * Maganus sulphate MnSO4) is faint pink or pale flesh coloured -- why ?
Mno ® 3d5 4s2
Mn2+ ® 3d5

19. In MnSO4, Mn is in +2 oxidation state.
In MnSO4, Mn2+ belongs to d5 system and it is high spin complex because SO4 = is weak field legand. Here, each of the d-orbitals is singly occupied. So, no electron is shifted / excited to upper d-level from lower d-level, because spin selection rule is violated. That’s why it cause faint pink colour.
Many Mn2+ compounds are off white or pale / flesh coloured.
Ex --- MnSO4, [Mn(H2O)6]2+]
* Boltzmann distribution :-
According to Boltzmann distribution, all molecules are not excited at a time, that’s why colour is continued.
In case of strong ligand like CN-, spin selection rule is followed.

20. Þ(ii). Laporte ‘orbital’ Selection Rule ----
Þ(ii) Laporte ‘orbital’ Selection Rule Electron transition takes place within different l-valued orbitals. D l ¹ 0 D l = ± 1 So, all d-d transitions are Laporte - forbidden (L-forbidden), since the change in l = 0
Relaxation -I :
For non-centro-symmetric complex, i.e. mixed ligated complex (mixture of ligands), Laporte-allowed transition takes place.

21. Relaxation II :
If some s or p character are involved in bonding d-d transitions are considered as impure d-d transitions and it is Laporte allowed i.e. intensity increases.
Mixing of this kind occurs in complexes which do not possess a centre of symmetry, for example tetrahedral complex (Td) or asymmetrically substituted octahedral complexes. Thus [MnBr4]2- which is tetrahedral and [Co(NH3)5Cl]2+ which is octahedral but non-centro-symmetric are both coloured.

22. For tetrahedral complexes, no d-orbitals are directly involved to bind the ligands. But dxy, dyz, dxz orbitals are more closer to ligands resulting reverse spliting of d-orbitals i.e. upper 3, lower 2.
S and three p orbitals i.e. total four orbitals of central metal ion are responsible for the binding of four ligands, resulting tetrehedral geometry, i.e. s and p are involved in bonding, but transition of electron takes place in between d-orbitals. Laporte concluded that these transition

23. This conversation is used for the detection of water ; as well as it is used in Silica jel to identify the nature of atomosphere --- is it moisty or not.]
1) LMCT / Ligand ® Metal chagre transfer ---
(a) n+ ® high charge (low size)
(b) L ® polarisable / deformable / larger size

24. If metal ion is of high charge and legand electrons are strongly attracted towards metal ions, at the cost of some amount of energy. The unabsorbed light causes colour.
Example ----
MnO4--- (permanganate) violet
Cr2O7--- (dichromate) orange
CrO72--- (chromate) yellow
HgS black / red (vermillion)
HgO black
CoS, NiS black
MnS pink / black
CdS yellow
As2S yellow
Sb2S orange

25. 2) MLCT/Metal ® Ligand charge transfer ---
The example of MLCT is very few, because ligand is not generally able to accept electron from metal. There are certain ligands which exhibit both donation as well as acceptance of electron. There are called ∏-acidic ligand.
3) NMCT / Metal-metal charge transfer ---

26. Chemical Periodicity
1) Periodic Properties :
Those properties which are systematically varied along period or group are called periodic properties.
Example --- atomic radius, electronegativity, ionisation potential etc.
2) Non-periodic Properties :
Those properties which are not systematically varied along group or period are called nonperiodic properties.
Example --- radioactivity magnetism

27. a) Primary Periodicity :-
(i) Atomic size ---
From left to right in a period, electrons are added in the same quantum level, again protons are increased within nucleus. So, contraction takes place, i.e. size decreases.
Along group, electrons are added in new quantum level, that’s why gradually size increases.
(ii) Electronegativity (E.N.)
(iii) Electron Affinity (E.A.)

29. (iv) Oxidising and Reducing properties -----
Mention the relative electronegativity, electron affinity and oxidising power of F, Cl, Br, I in aquas medium and mention the reasons.
Ans. 1) E.N F > Cl > Br > I
In a group electronegativity decreases on moving down the group. This is due to the effect of increased atomic radius.
2) E.A Cl > F > Br > I
Valence shell electronic configuration of halogen is ns2np5. For F, n = 2, for I n = 5, i.e. electron density in the valence shell of F is maximum, i.e. surface electron density is maximum. As electron affinity is a measure of attraction of electron, size-factor is pre- dominating, but as electron is totally added on the valence shell, electron-electron repulsion is to be considered (In case of E.N. as it partial attraction, no repulsion point of view is considered).
Cl being slightly larger than F, repulsion factor is vanished and considering both these effects, the relative order of E.A. is ---
Cl > F > Br > I

30. 3) O.P. --- F > Cl > Br > I
Electronic concept of oxidising agent is the species should accept electron.
E.A. of Cl is greater than that of F. It implies that Cl should be stronger oxidising agent than F. But really in aquas medium, F2 is stronger oxidising agent than Cl2. This observation reveals the fact that oxidising power is not dependent on E.A. only, but there are some another factor also. This is best explained by Born-Haber Cycle.
This cycle tells us oxidising power of X2 (EOX) depends on three energy factors --- Bond energy (B.E.), Electron affinity (E.A.), Hydration energy (D Hhydration). Neglecting B.E. factor, DHhydration of F(g)---- is so high (negative) that overall, EOX of F2 maximises.

31. [N.B.:1) Hydration Energy (D Hhydration) :-
The amount of energy released when 1 mole gaseous species is hydrated, is called hydration energy.
Li+, Na+, K+, Rb+, Cs+
Li+ is the smallest cation. So its charge density maximises. So huge water molecules surround Li+ in multimolecular layer, i.e. relative order of DHhydration is ®
Li+ > Na+ > K+ > Rb+ > Cs+
2) In perfectly anhydrous gaseous state where DHhydration factor is ignored, only E.A. plays the important role, i,e, gas-phase oxidising power is ®
Cl2 > F2 > Br2 > I2
3) Born-Haber Cycle is valid when initial states and final states are state functions, i.e. they depend on state parameters (temperature, pressure, catalyst, composition), but not path-function.
Redox potential is a path function. That’s why its additivity rule is not applied. ]

32. Ionisation Potential / Ionisation Enthalpy :-
The minimum amount of energy required to release the losely bounded valance shell electron from an isolated gaseous atom is called ionisation potential or ionisation energy of the atom.
Units :- electron-volts per atom (eV/atom) kilo Joules per mole of atoms (kJ mol---1 )
1 eV per atom = kJ mol---1

33. * Factor on which I.P. depends ---
1) Charge of cation is directly proportional to I.P. with increasing charge, nuclear attraction towards valance shell increases, that’s why I.P. increases.
2) Atomic size is inversely proportional to I.P. with increasing size, nuclear attraction towards valance shell reduces. That’s why lower amount of energy is required to ionise it.
3) Penetrating power or elipticity of the valance shell orbital is directly proportional to I.P., i.e. the relative order of ionisation potential and penetrating power is ® s > p > d > f

34. 4) Half-filled or full-filled stabilisation in valance shell configuration is directly proportional to I.P.
Example ----
2s22p1 2s22p2 2s22p3 2s22p4
B C N O
Size Order : B > C > N > O
I.P. : B < C < O < N
Size of oxygen is smaller than Nitrogen. So, I.P. of oxygen should be greater than Nitrogen. But valance shell configuration of Nitrogen is 2s22p3, i.e. 2p3 is half-filled stabilized. So, to ionise it, some extra amount of energy is required. That’s why I.P.I of Nitrogen is greater than that of Oxygen.
5) If screening effect of inner electron decreases, nuclear attraction towards valance shell electron increases, i.e. I.P. increases.

35. Screening Effect and Slater’s Rules :-
For multi-electronic system, there are three types of forces :-
1) Nucleus - valance shell electron attraction
2) Nucleus - inner electrons attraction
3) Inner electron - valance shell electron repulsion.
So, we see inner electrons (es) look like a negative screen so that nuclear attraction towards valance shell reduces. This phenomenon of inner es is called Screening / Shiedling Effect.

36. Slater’s Rules for the calculation of screening constant (s) :-
1) Electrons are grouped as below :-
(1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p)
2) Electrons in the right hand side of nth group contribute nothing towards total screening.
3) Except the last electron in nth group, the rest electrons contribute unit each, empirically by Slater (Exception for He, it is 0.33) s = 0.33
4) For (n-1) group, each electron contributes 0.85 unit.
5) For (n-2) group or lower groups, each electron contribute totally, i.e. 1 unit each.
6) For nd or nf shell, it follows 0.35 rule, but rest electrons contribute totally.

37. Example ----
(1) (1s)2 (2s, 2p)8 (3s, 3p)8
s = 7 x x x 1.0 = 11.25
z* = zeff. = z ---- s = = 6.75
(2) (1s)2 (2s, 2p)8 (3s, 3p)8 (3d)10
s = 9 x x 1.0 = 21.15
z* = z ---- s = = 6.85

41. What is the historically first innert gas compound and how is it related to I.P. ?
Ans. The first ionisation energy (I.P.I) of molecular oxygen is almost identical to that of Xenon (Xe). [I.P.I for O2 O2+ is 1165 kJ mol---1, I.P.I for Xe Xe+ is 1170 kJ mol---1]
Again the compound of molecular Oxygeniol cation O2+ Pt F6--- (Di oxygenial Hexafluroplatinate) is well knwon. From this analogy, Bartlett isolated the first innert gas compound i.e. Xenial Hexafluoroplatinate [Xe+PtF6--- ]
b) Secondary Periodicity :-
Inert Pair Effect / Inert s-pair Effect / “G-2” Rule :-
On going from top to bottom, in a group, 2 unit valancy decreases from the respective group valancy and it is mostly prominent in last periodic elements. This phenomenon is called Inert Pair Effect / Inert s-pair Effect / “G-2” Rule (‘Group Valancy - 2’ rule)

42. Group - II
* Theory of Inert Pair Effect :-
(1) In terms of screening effect
n = 3 3s p | 3do
n = 4 Zeff (3d10)Ü (4s p1---6) | 4do
n = 5 (3d10 4d10) (5s p1---6) | 5do
n = 6 Zeff (3d10 4d10 5d10 4f14) Ü (6s2 6p1---6)

43. In 3rd period, 3d orbit starts as outer vacant shell
In 3rd period, 3d orbit starts as outer vacant shell. In 4th period, there is inner 3d10 configuration having weak screening effect, i.e. Zeff of nucleus increases which strongly pulls the valance (4s, 4p) electrons but this attraction is not isotropic (equal). It is rather anisotropic. 4s2 valance shell electrons are strongly pulled compared to 4p because of most penetrating behaviour of 4s. So, chemical innertness of 4s2 valance pair increases, i.e. highest oxidaiton state of 4th periodic elements is reluctantly (unwillingly) stabilised.
In 5th period, there are 3d10, 4d10 inner electrons, which exhibit weaker screening i.e. Zeff increases, but increasing size effect balances the strong pulling character of nucleus. That’s why normal observation is noted.
In 6th period, there are huge d and f inner electrons which exhibit weakest screening. So Zeff sufficiently increases and it plays important role. Zeff becomes so high that 6s2 valance pair is almost chemically innert, i.e. not involved in chemical bonding. So 2 unit valancy reduces from the respective group valancy. This innertness is not an intrinsic (basically innert) innertness, it is rather kinetic one.

44. So, we see, in 4th period innert pair effect starts, in 5th period it is vanished, but in 6th period it is mostly prominent.
Mercurous ion written as Hg22+, why not Hg+ ?
Hgo ® 5d10 6s2 Hg+ ® 5d10 6s1
Valance shell configuration of mercurous ion is 6s1 which is strongly pulled by nucleus. That’s why its energy reduces, so mercurous-mercurous (Hg+ --- Hg+) or 6s s1 covalent bonding is favoured. Hence it is occupied as dimer.
(2) Relativistic Effect ®
We know, velocity if electron (v) is directly proportional to z.
or, V µ Z
or, V = Z au (atomic unit)
For 6s2 Mercury, V = 80 au
Again velocity of light c is around 154 au

45. We know, the relativistic mass eqyation is
putting all these data we have meff » m0 for mercury, i.e. we see for heavier elements having atomic number 80 or onwards, effective mass sufficiently increases from its rest mass. Again we know, radius is inversely proportional to mass.
So, with increasing mass, effective radius decreases. So these electrons are not easily involved in bonding, i.e. chemical innertness is observed.

46. Application of Inert Pair Effect
1) SO2, SeO2, TeO2 --- Mention and explain the relative oxidising power.
Ans. From top to bottom, i.e. from s to Tes, bond energy decreases, so the relative oxidising power should be TeO2 > SesO2 > SO2. But Se belonging to 4th period, itss +4 oxidations state is inherently unstable due to the spresence of inner 3d10 e5. That’s why SeO2 always exhibits a strong afinity to have lower oxidation state, i.e. SeO2 is the strongest oxidising agent. Hence the relative oxidising power is ® SeO2 > TeO2 > SO2.
2) ClO4- (perchlorate), BrO4- , IO Mention and explain the relative oxidising power.
Ans. From top to bottom, i.e. from Cl to I, bond energy decreases, so the relative oxidising power should be ClO4- < BrO4- < IO4- . But Br belonging to 4th period, its +7 oxidation state is inherently unstable due to the presence of inner 3d10 e5. That’s why BrO4- always exhibits a strong affinity to have lower oxidation state, i.e. BrO4- is the strongest oxidising agent. Hence the relative oxidising power is ClO4- < IO4- < BrO4-.

47. 3) Al, Ga, In ------ Mention and explain the relative electron affinity.
Ans. From top to bottom, i.e. from Al to In, surface electron density reduces, so the relative E.A. should be Al > Ga > In.
4) PCl5, AsCl5, SbCl Mention and explain the relative stability.
Ans. From top to bottom, i.e. from P to Sb bond energy decreases. So the relative oxidising power should be PCl5 < AsCl5 < SbCl5. But As belonging to 4th period, +5 oxidation state of As is inherently unstable due to the presence of inner 3d10 es. So the relative oxidizing power is ® AsCl5> SbCl5 >PCl5. So the relative stability is ® PCl5 >SbCl5 > AsCl5.
5) CCl4 exists, but not PbCl why ?
Ans. Pb belonging to 6th period, +4 oxidation state of Pb is inherently unstable due to the presence of inner d and f electrons i.e. the innert pair effect. So PbCl4 does not exist. But C belonging to 2nd period, inert pair effect is not observed. So, CCl4 exists.

48. 6) CO2 is moderate oxidising agent, but PbO2 is a strong oxidising agent ---- why ?
Ans. Pb belonging to 6th period, +4 oxidation state of Pb is inherently unstable due to the presence of inner d and f electrons i.e. the inert pair effect. So it always exhibit strong affinity to have lower oxidation state. So PbO2 is a strong oxidising agent. But as C belonging to 2nd period, no innert pair effect is observed. So, CO2 is moderate oxidising agent.
7) NaNO3 is a moderate oxidising agent, but NaBiO3 is a strong oxidising agent ---- why ?
Ans. Bi, belonging to 6th period, +5 oxidation state of Bi is inherently unstable due to the innert pair effect. So it always exhibit a strong affinity to have lower oxidation state. So, NaBiO3 is a strong oxidising agent. But as N belonging to 2nd period, no innert pair effect is observed. So, NaNO3 is moderate oxidising agent.

49. 9) BiCl4 does not exist ---- why ?
Ans. Bi belonging to 6th period, +5 oxidation is inherently unstable due to the inert pair effect. So Bi always exhibits strong affinity to have lower oxidation state. So it does not exist.

50. 10) Mercury is inert ---- why ?
Ans. Valence shell configuration of Mercury is 6s1. Hg belonging to 6th period, 6s1 electron is strongly pulled by nucleus due to inert pair effect. So, Mercury is inert.
11) Au may exist as Au why ?
Ans. Au belongs to 6th period. So 6s2 electrons are strongly pulled by the nucleus, due to inert pair effect. So, Au exhibits strong affinity to have lower oxidation state. So when it reacts more electro- positive metal like Cs, it may exist as Au- .

51. 1. Oxygen is gas, diatomic, but sulphur is polyatomic solid ---- why ?
O --- O sigma bond distance is shorter than S --- S sigma bond distance. That’s why 2pp --- 2pp overlap is best possible compared to 3pp --- 3pp of sulphur, i.e. oxygen is actate stabilised by the formation of double bond. As there is no chance to double bond in case of sulphur, to have actate stability, it is polymerised. Thus sulphur is solid, polyatomic and oxygen is diatomic gas.

52. 2. Nitrogen is diatomic gas, buts phosphorus is polytomic solid ---- why ?
N --- N sigma bond distance is shorter than P --- P sigma bond distance. That’s why 2pp --- 2pp overlap is best possible compared to 3pp --- 3pp of phosphorus, i.e. Nitrogen is octate stabilised by the formation of triple bond. As there is no chance to triple bond in case of phosphorus, to have actate stability, it is polymerised. Thus phosphorus is polyatomic but nitrogen is diatomic gas.

53. 3. Trimethylamine [(CH3)3N] and Trisilymanine [(SiH3)3N] --- compare relative Lewis basicity.
Si contains vacant 3d, so Nitrogen lone pair is involved in dp --- pp bonding with this vacant 3d orbital. So, availablity of Nitrogen lone pair decreases. That’s why Lewis basicity decreases.
Methyl groups exhibit strongs +I effect (Induction effect), so Nitrogen lone pair density increases. So it easily donates to Lewis acid.
So, Lewis basicity order is ® (CH3)3N > (SiH3)3N [ Lone pair acceptor is called Lewis acid. Lone pair doner is called Lewis base.

54. 4. Methyl isothiocyanate (CH3 --- NCs) and Silyl isothiocyanate (SiH3 --- NCs) --- compare the sketal structure.
Repulsive force between the lone pair of nitrogen and bond pairs makes the angular structure of CH3 --- NCS.
Si contains vacant 3d, so Nitrogen lone pair is inovlved in dp --- pp bonding with this vacant 3d orbital. So, SiH3 --- NCS has linear structure.

55. [ N.B. ---
CN cyanide ion
OCN cyanate ion
SCN thiocyanate ion
SeCN selenocyanate ion
NCN cyanamide ion
N ozide ion
ONC Fulminate ion
NC --- isocyanide
NCO --- isocyanate
NCS isothiocyanate ion ]

56. 5. CH4, SiH4 --- Compare boiling point.
Ans. Normal molecular weight is directly proportional to boiling point. That’s why boiling point of Silane (SiH4) is greater than that of Methane (CH4).
6. CCl4, SiCl4 --- Compare boiling point.
Ans. Molecular weight of SiCl4 is greater than that of CCl4. Still boiling point reverses.
In SiCl4, Si contains vacant 3d orbital. Cl contains filled p orbital. So there is a good chance of dp --- pp bonding. So bond length reduces and surface area decreases. So, Van Der Waal’s Force decreases. So, boiling point decreases.

57. [ N.B. --- Van Der Waal’s Force / London Dispersive Force :-
For non-polar molecules, overall electron cloud is considered as a symmetrical distribution. But within very short interval of time, say sec, if we are able to calculate the polarity of a molecule, it is found that all non-polar molecules are polar. But if we allow huge time, overall polarity vasnishes. This time-dependent polarity is called Fluctuating / Instanteneous Polarity. This causes opposite polarity of the neighbouring molecule and a very weak intermolecular force is created, which is called Van Der Waal’s Force / London Dispersion Force - ]

58. Hydrolysis :-
1) “D-pathway” / Desociative / Ionic Pathway :-
2) “A – pathway” / Associative pathway :-

59. 3. CCl4 + H2O 
SiCl4 + H2O 
Write the reactio
SiCl4 + 4H2O = H4SiO4 + 4HCl
(silicic acid) / Si(OH)4
SiCl4 contains vacant 3d - orbital in Si and Cl. But Si is more electropositive than Cl. So, water-attack takes place through Si, via associative pathway Ultimately four Cl groups are substituted by four OH groups resulting Silicic Acid (H4SiO4) and 4HCl.
C contains no vacant d, so there is no chance of water-attack. Cl contains vacant d, but it has negative polarity. So CCl4 is resistant to hydrolysi

60. 4. NCl3 + H2O 
PCl3 + H2O 
Write the reactio
PCl3 + 3H2O = H3 PO3 + 3HCl
In PCl3 , both central p and Cl contain vacant d. But p is more electropositive than Cl. That’s why water-attack favours through phosphorus, resulting phosphorus acid and HCl.
NCl3 + 3H2O ==== NH3 + 3HOCl
In NCl3, N contains no vacant space, but Cl contains vacant space. Again N-Cl being almost non-polar (E.N. same ® 3.0), water- attack through Cl takes place. That’s why, NH3 and HOCl are resulted.

61. [ N.B. ---
H3PO3 ® 2H+ + HPO3--
Na2HPO3 ® normal salt ]
5. CCl4 + H2O 
NCl3 + H2O
Write the reactio
In CCl4, central C contains no vacant space, but through Cl due to negative polarity of Cl. It is contrary to NCl3 case.

63. To dichloro derivative of carbon, if we add water, we are ended with Ketone. But in case of Si-derivative expected Silicon compound analogous to Ketone is not synthesised, because Si is larger than C. So OH groups are four apart from each other. So chance of water elimination is very very low. This intermidiate is polymerised resulting Silicones / Polyalkylsiloxane.
[ N.B. -- Silicones --
Silicones have a wide variety of commercial uses as fluids, oils, elastomers (rubbers) and resins.]
Diagonal Relationship
There are certain diagonally placed elements which exhibit some identical chemical properties. This similarity is weaker than group- similarities and it is called Diagonal Relationship.

64. In 2nd and 3rd period for left-handed three lighter diagonal pair, this relatioship is mostly obeyted. It can be explained by there relative ionic potential (f) values.
For another diagonal pair i.e. for heavier diagonal pair, due to different screening factor. f values significantly deviates. That’s why diagonal relationship is not observed.

65. Al M---  Al(OH)3 ¯
(white gel - like precipitable / p.p.t.)
Be M---  Be(OH)2 ¯
Al + NaOH  NaAlO2 + H2
Be + NaOH  Na2BeO2 + H2
(Sodium berilate)
Different scales of Electronegativity
1) Pauling’s Scale :-
Consider A---B bond where, A, B two dissimilar atoms of different electronegativity. Let bond energies of A --- A, B --- B and A --- B be represented as EA --- A, EB---B, EA---B respectively.
It may be seen that the energy of A - B bond is almost always greater than the geometric mean of the energies of A --- A and B --- B bond.

66. Bond energies are expressed in eV
Bond energies are expressed in eV. Extra bond energy can be expressed in keal mol---1 (C.G.S) or in kJ mol---1 (S.I.)
E.N. of Fluorine is 4.0 and E.N. of Hydrogen is 2.1. These are used as standard E.N. to determine E.N. of other elements.

67. 2) Mulliken’s Scale :-
In this scale, E.N. of any atom is considered as the difference of ionisation potential and electron affinity of that atom.
I.P. and E.A. values are expressed in eV. If these are expressed in Kilo - callories, then it will be
Pauling and Mulliken values of electronegativity are related as

68. 3) Allred and Rochow Scale :-
Here electronegativity of an element is considered as the electrostatic attraction of nucleus towards the electron at a distance equal to the covalent radius (rcov.)
[N.B. --- Here Zeff is calculated by applying Slater’s rules. But the point to be noted that last electron is to be counted in calculation.]
d-contraction and f-contraction
In d-block and f-block, if we run from left to right in a period, electrons are entered into the d-orbitals or f-orbitals which are low screening. So, obviously Zeff increases, and size decreases in appreciable rate. This phenomenon is known as d-contraction / Lanthanide contraction and f-contraction / Actinide contraction.