In these complexes, Co (III) has six d-electrons. The first complex ion is diamagnetic and the second has paramagnetic character due to four unpaired electrons. In the [Co(NH₃)₆]³⁺complex, the two 3d-electrons get paired up with the other two leaving two vacant orbitals and these vacant orbitals get Sp³d² hybridized. In the second [CoF]^3- complex, the 3d electrons are not disturbed and the outer 4d-orbitals are used for hybridization.
  

\({ C }_{ 6 }{ H }_{ 5 }-\overset { \underset { | }{ Cl } }{ CH } -{ C }_{ 6 }{ H }_{ 5 }\) will get hydrolysed easily because carbocation formed will be stabilized by resonance effect of two phenyl groups, whereas in \({ C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }Cl\), he carbocation is stabilized by resonance effect of one phenyl group.

(i) In l-bromo-l-methylcyclohexane, the \(\beta \)-hydrogens on either side of the Br atom are equivalent, therefore, only 1- alkene is formed.

ii) 2-Boro-2-methylbutane has two different sets of equivalent \(\beta \)-hydrogens and hence, in principle, can give two alkenes (I and II). But according to SaytzetT rule, more highly substituted alkene (II), being mor~ stable, is the major product.

(iii) 3-Bromo-2, 2, 3-trimethylpentane has two different sets of \(\beta \)-hydrogens and hence, in principle, can give two alkenes (I and II). But according to SaytzetT rule, more highly substituted alkene (II), being more stable, is the major product.

(i)

(ii)
 
(iii)

(i)

(ii)
 
(iii)
 
(iv)

This is nucleophilic substitution reaction. KCN is an ionic compound, it breaks into K⁺ and CN^- in aqueous solution. It attacks 'C' of alkyl halide through carbon since C-C bond is more stable than C-N bond due to lack of polarity. However, AgCN is covalent compound, therefore, bond can form only throu9'h 'N' which has lone pair to donate for forming coordinate bond in isocyanide.

Due to stronger -I-effect of F than CI, CHF₃ should be more acidic than CHCI₃. But actually reverse is true. This is due to the reason that: CCI₃^- left after the removal of a proton from CHCI₃ is stabilized by resonance due to the presence of d-orbitals on CI but: CF₃^- left after the removal of a proton from CHF₃ is not stabilized by resonance due to the absence of d-orbitals on F.

A 50 : 50 mixture of two enantiomers of any optically active compound is called a racemic mixture. It is always optically inactive since rotation caused by the molecules of one enantiomer. For example, an equimolar mixture of (+)-2 bromobutane and (-) -2- bromobutane is called a racemic mixture.

A beam of light which was vibrations in only one plane is called plane polarised light.

Reactivity increases as the carbon-halogen bond dissociation energy decreases, i.e., CH₃F < CH₃CI < CH₃Br < CH₃I.

Because of bigger size and lower electronegativity, iodide ion can donate a pair of electrons more easily than bromide ion and hence iodide ion is a better nucleophile than bromide ion.

(i) Chirality: It is a geometrical property of a rigid object ( or molecules ) by which it gets such spatial arrangement of points or atoms that, the molecule becomes non-super imposable of its mirror image.
The chiral molecule of the object does not have any element of symmetry like, a mirror image, centre of inversion (i) etc.
3-bromopent-1-ene is represented as
(ii) Due to +I effect of alkyl groups the 2° carbonium ion CH₃—CH⁺ —CH₂—CH₃ derived from sec. butyl chloride is more stable than the 1° carbonium ion CH₃—CH₂—C H₂⁺ derived from n-propyl chloride. Therefore sec. butyl chloride gets hydrolyzed more easily than n-propyl chloride under SN1 conditions.
This is because; iodine is a better leaving group due to its large size. So, it will be released at a faster rate in the presence of an incoming nucleophile

p-Dichlorobenzene is symmetrical, therefore, first into crystal lattice more readily and has higher melting point as compared to o- and m- isomers.  Its solubility is higher because it is symmetrical and has more force of attraction with the solvent.

(4-tert-butyl-3-idoheptane)

CH₃−CH(Cl)CH₂CH₃ will be more easily hydrolysed because it will form secondary carbocation which is more stable than primary carbocation.

The coordination number of Fe is 6.

(i) Mg
(ii) Fe
(iii) Co
(iv) Pt

Hexaamminechromium (III) hexacyanocobaltate (III).

Diamminechlorido methanamine platinum(II) chloride.

Tetraamminedicyanocobalt (III) chloride.

(i) (a) Haemoglobin
(b) Chlorophyll
(ii) Applications of complex compounds.
(a) They are used in photography, Le. AgBr forms soluble complex with sodium thiosulphate rn photography.
(b) K[Ag(CN)₂] is used for electroplating of silver, K[Au(CN)₂] is used for gold plating.
(c) Some of ligands oxidise Co²⁺ to Co³⁺ ion.
(d) EDTA is used for estimation of Cla²⁺ and Mg²⁺ in hard water.
(e) Silver and gold are extracted by treating zinc with their cyanide complexes.
(f) Ni²⁺ is tested and estimated by DMG (dimethylglyoxime).

Chlorine withdraws electrons through inductive effect and releases electrons through resonance. Through inductive effect, chlorine destabilises the intermediate carbocation formed during the electrophilic substitution.
 
Through resonance, halogen tends to stabilise the carbocation and the effect is more pronounced at ortho- and para- positions. The inductive effect is stronger than resonance and causes net electron withdrawal and thus causes net deactivation. The resonance effect tends to oppose the inductive effect for the attack at ortho- and parapositions and hence makes the deactivation less for ortho- and paraattack. Reactivity is thus controlled by the stronger inductive effect and orientation is controlled by resonance effect.

When white light falls on the complex, some part of it is absorbed.  Greater the CFSE, greater is the energy absorbed or shorter is the wavelength absorbed \((E={hc\over\lambda })\).  The observed colour is the complementary colour of the colour absorbed.

(i) As isomer A reacts with AgNO₃ to give a white precipitate, CI must be present in the ionization sphere. As it does not react with BaCI₂, SO₄^2- must be present in coordination sphere.
Formula of A = [Co(NH₃)₅SO₄] CI (coordination no. of Co = 6) As reactions are reverse for isomer B, formula of B = [Co(NH₃)₅CI]S0₄
(ii) Ionization isomerism.
(iii) A = Pentaarnminesulphatocobalt (III) chloride ; B = Pentaamminechloridocobalt (ill) sulphate

\((i)\quad { CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ \underset { \overset { | }{ { CH }_{ 3 } } }{ C } } -{ CH }_{ 3 }\) 
 All the H atoms are equivalent.
\((ii)\quad \overset { a }{ { CH }_{ 3 } } -\overset { b }{ { CH }_{ 2 } } -\overset { c }{ \underset { n-Pentane }{ { CH }_{ 2 } } } -\overset { b }{ { CH }_{ 2 } } -\overset { a }{ { CH }_{ 3 } } \)   
Replacement of a, b atom gives different products.
\((iii)\quad \overset { a }{ { CH }_{ 3 } } -\underset { \overset { | }{ { CH }_{ 3 } } }{ \overset { b }{ { CH }_{ 2 } } } -\overset { c }{ { CH }_{ 2 } } -\overset { d }{ { CH }_{ 3 } } \) 
Replacement of a, b, c and d H atoms give different products.

We know that S_N2 reactions proceed with inversion of configuration at the stereocentre. If, however, two S_N2 reactions are carried out at the same stereocentre of a compound, retention of configuration will occur. Thus, (R)-2-octanethiol can be obtained from (R)-2-bromooctane by first reacting it with Nal in acetone and then with NaSH in ethanol

(i) HCN (PK_a 10) and NH3 (PK_a 11·25) are very weak: acids. Therefore, their conjugate bases, i.e., CN^- ion and NH₂ ^- are very strong bases. Since strong bases are bad leaving groups, their displacement in nucleophilic substitution reactions is never observed,
(ii) Iodide ion is a powerful nucleophile and hence reacts rapidly with RCI to form RI,
Further because 1- ion is a better leaving group than Cl- ion, therefore, RI is more rapidly hydrolysed than RCI to form ROH,
The I^- ion thus regenerated recycles in the above reaction thereby explaining its catalytic effect.

tert-Alkyl halides prefer to undergo dehydrohalogenation in presence of a strong base such as Na metal instead of undergoing Wurtz reaction as explained below:
\(\underset{tert-Butyl\ bromide}{(CH_3)_3C-Br}+2Na\longrightarrow\underset{tert-Butylsodium}{(CH_3)_3Na^+}+Na^+Br^-\)
Thus, only rand 2° alkyl halides undergo Wurtz reaction while 3° alkyl halides prefer to undergo dehydrohalogenation to form alkenes.