(CH₃)₂C = C (CH₃)₂

CH₃ CH (OH) CH₂CH₃
2R-OH+ 2Na ⟶ 2RONa + H₂ ↑
2 moles of alcohol gives 1 mole of H₂ which occupies 22.4 L at 273 K and 1 atm number of moles of alcohol
= 2 moles of R -OH/22.4L of H₂ x 560 ml
= 0.05 moles
No of moles = mass/molar mass = m/M
= 3.7/0.05 = 74 g mol^-1
General formula for (R-OH) Cn H_{2n + 1}OH
n(12) + (2n + 1) (1) + 16 + 1 = 74
14n = 74 - 18
14n = 56
n = 56/14 = 4
The  "2" alcohol which contains 4 carbon is CH₃ - CH (OH) CH₂ - CH₃

a) Zn²⁺ (d¹⁰ ⇒  diamagnetic)
b) Co³⁺ (d⁶ Low spain ⇒  t_2g⁶ e⁰_g ; diamagnetic)
c) Ni²⁺  (d⁸ Low spain ⇒  t_2g⁶ e²_g ; paramagnetic)
d) [Ni(CN)₄]²⁺ (dsp² ; square planar, diamagnetic)

Three isomers. If we consider any one of the ligands as reference (say Py), the arrangement of other three ligands (NH₃, Br^- and Cl^-) with respect to (Py) gives three geometrical isomers.

Ti⁴⁺ (d⁰ ⇒  0BM)
Co²⁺ (d⁷ spain free ⇒  t_2g⁵, e²_g; n = 3; μ = 3.9BM)
Cu²⁺  (d⁹ Low spain ⇒  t_2g⁶, e³_g; n = 1; μ = 1.732BM)
Ni²⁺  (d⁸ Low spain ⇒  t_2g⁶, e²_g; n = 2; μ = 2.44 BM)

Molecular formula: MSO₄Cl.6H₂O
Formation of white precipitate with Barium chloride indicates that SO^2-₄ ions are outside the coordination sphere, and no precipitate with AgNO₃ solution indicates that the Cl^- ions are inside the coordination sphere. Since the coordination number of M is 6.Cl^- and 5 H₂O are ligands, remaining 1 H₂O molecular and SO^2-₄ are in the outer coordination sphere.

MnO^-₄ + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O

Mn²⁺ ⇒ 3d⁵ contains 5 unpaired electrons
n = 5,
\( \sqrt{n(n+ 2)} \) BM
\(= \sqrt{5(5+ 2)} = \sqrt{35} = 5.92 BM\)

1. Fe²⁺ the electronic configuration is 3d⁶.
2. Fe³⁺ the electronic configuration is 3d⁵.
3. So it has exactly half filled stable electronic configuration. Hence Fe³⁺ is more stable than Fe²⁺.
4. For manganese Mn²⁺. the electronic configuration is 3d⁵ and that of Mn³⁺ is 3d⁴. Here Mn²⁺ has exactly half-filled stable electronic configuration.

(i) Transition metals have number of unpaired electron. They are involved in metallic bonding. Hence they show high melting point.
(ii) As we move from left to right along the transition metal series melting point first increases reach a maximum value and then decreases as the d-electrons pair up and become less available for bonding.

Electronic configuration of Ti = 3d²4s²
Electronic configuration of Ti³⁺ =3d¹
Hence number of unpaired electron = 1
Spin only magnetic moment \((\mu)=\sqrt{\mathrm{n}(\mathrm{n}+2)}\)
= \(\sqrt{1(1+2)} \)
= \(\sqrt{3}\) 
=1.732 BM
Electronic configuration of \(\mathrm{Mn}=3 \mathrm{~d}^{5} 4 \mathrm{~s}^{2}\)
Electronic configuration of \(\mathrm{Mn}^{2+}=3 \mathrm{~d}^{5}\)
Hence number of unpaired electrons = 5
Spin only magnetic moment 
\((\mu) =\sqrt{5(5+2)}\)
= 5.92 BM

a. 3MnO₄^2- + 4H⁺ ⟶  2MnO₄^- + MnO₂ + 2H₂O
(Manganate ion) (Permanganate ion)  Manganese dioxide
b. C₆H₅CH₃ \(\overset { acidified }{ \underset { KMnO_{ 4 } }{ \longrightarrow } } \) C₆H₅COOH
 Toluene                         Benzoic Acid
c. 2MnO₄^- + 10Fe²⁺+16H⁺ \(\underrightarrow { { 8H }^{ + } } \)  2Mn²⁺+ 10Fe³⁺  + 8H₂O
d. 2KMnO₄ \(\overset { \triangle }{ \underset { Red\ hot }{ \longrightarrow } } \) K₂MnO₄ + MnO₂ + O₂
(Potassium Permanganate)    (Potassium Manganate)
e. Cr₂O₇^2- + 6I^- + 14H⁺ \(\underrightarrow { { (O) }}\) 2Cr³⁺ + 3I₂ + 7H₂O
                 (Iodide ion)                        Iodine
f. Na₂Cr₂O₇ + 2KCl ⟶ K₂Cr₂O₇ + 2NaCl
(Sodium dichromate)    (Potassium dichromate)

Lanthanoid contraction:
As we move across 4f series, the atomic and ionic radii of lanthanoids show gradual decrease with increase in atomic number. This decrease in ionic size is called lanthanoid contraction.
Effects (consequence) of lanthanoid contraction:
1. Basicity difference:
As we move from Ce³⁺ to Lu³⁺, the basic character of Ln³⁺ ions decrease. Due to the decrease in the size of Ln³⁺ ions, the ionic character of Ln -OH bond decreases (covalent character increases) which results in the decrease in the basicity.
2. Similarities among lanthanoid:
In the complete f - series only 10 pm decrease in atomic radii and 20 pm decrease in ionic radii is observed because of this very small change in radii of lanthanoids, their chemical properties are quite similar.
3. The elements of the second and third transition series resemble each other more closely than the elements of the first and second transition series.

(i) When excess of liquid ammonia is added to an aqueous solution of copper sulphate gives tetra ammine copper (II) sulphate is formed.
CUso₄ + 4NH₃ \(\rightarrow\) [Cu(NH₃)₄]SO₄
(ii) The co-ordination entity is [Cu(NH₃)₄]²⁺

(i) This is also called as salt isomerism.
(ii) This type of isomers arises when an ambidentate ligand is bonded to the central metal atom/ion through either of its two different donor atoms. In the below mentioned examples, the nitrite ion is bound to the central metal ion Co³⁺ through a nitrogen atom in one complex and through oxygen atom in other complex.
\(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{NO}_{2}\right)\right]^{2+}\)

[Mn(CN)₆]^3-
Mn - 3d³ 4s², Mn³⁺ - 3d⁴
Electronic configuration in isotropic field.

No of paired electrons (n'p) = 0
Therefore E_Iso = 0
High spin complex:
Ligand field: Electronic configuration: \(t_{2g}^{3}\) \(e_{g}^{1}\)

CFSE = {[3(-0.4) + 1(0.6)] \(\Delta\)_o + 0 \(\times\) P} - 0
=(-1.2 + 0.6) \(\Delta\)_o = -0.6 \(\times\) 38500
CFSE = -23,100 cm^-1.
Low spin complex:
Ligand field: Electronic configuration: \(t_{2g}^{4}\) \(e_{g}^{0}\)

CFSE = {[4(-0.4) + 0(0.6)] \(\Delta\)₀ + 1 \(\times\) P} - (0)
= (-1.6)  \(\Delta\)₀ + P
= - 0.6 \(\times\) 38,500 + 28,000
= -61,600 + 28,800
= -32,800 cm^-1
High negative CFSE value indicates that low spin complex is favoured.

(v) CH₂ = CH – CH(Cl)–O–CH₃ : 3 - methoxy,3- choloro propene (mixed ether)
(vi) dibenzyl ether CH₆CH₅ -O-CH₂-C₆H₅ : benzoxy toluene + Simple ether
(vii) vinyl allyl ether : ethoxy  2 - propene
CH₂ = CH –O– CH₂–CH–CH₂(mixed ether)

Mechanism:
This mechanism undergoes Saytzeff rule.
During intramolecular dehydration, if there is a possibility to form a carbon - carbon double bond at different locations, the preferred location is the one that gives the more (highly) substituted alkene i.e., the stable alkene.

[Ni (H₂O)₆]²⁺

It has two unpaired electrons. So there is d-d transition. Hence it is green coloured.
[Ni(CN)₄]^2-

There is no unpaired electrons. So it is colourless, as there is no d-d transition.

Copper exhibits +1 oxidation state most frequently Cu (29) - electronic configuration 3d¹⁰4s¹ copper ready to lose outer most one electron to attain the stable full filled electronic configuration. Hence it exhibits +1 oxidisation state.

As we move from left to right in a transition metal series, the ionization enthalpy increases as expected. This is due to increase in the nuclear charge corresponding to the filling of d electrons. The increase in first ionisation enthalpy with increase in atomic number along a particular series is not regular. The added electron enters (n-1) d orbital and the inner electrons act as a shield and decrease the effect of nuclear charge on valence ns electrons. Therefore, it leads to variation in the ionization energy value.

(i) The actual position of Lanthanides in the periodic table is at group number 3 and period number 6. However, in the sixth period after lanthanum, the electrons are preferentially filled in inner 4f sub shell and these fourteen elements following lanthanum show similar chemical properties.
(ii) Similarly the fourteen elements following actinium resemble in their physical and chemical properties. Hence they are placed separately bottom of the modern periodic table.

IUPAC defines transition metal as an element whose atom has an incomplete d-sub shell or which can give rise to cations with an incomplete d-sub shell. They occupy the central position of the periodic table, between s and p-block elements.
Examples: Fe, Cu, Ag, Au