Hydrogen peroxide is used to restore the white colour which was lost due to the reaction of hydrogen sulphide in air with the white pigment Pb₃(OH)₂(CO₃)₂ to form black colored lead sulphide (PbS).
Hydrogen peroxide oxidises black coloured lead sulphide to white coloured lead sulphate, there by restoring the colour.
PbS + 4H₂O₂ \(\rightarrow\)  PbSO₄ + 4H₂O

(i) Water reacts with SiCl₄ to give silica.
SiCl₄ + 4H₂O ⟶ Si(OH)₄ + 4HCI
Si(OH)₄ \(\overset { \Delta }{ \longrightarrow } \) SiO₂ + 2H₂O
Silica
(ii) Water reacts with P4010 to give orthophosphoric acid.
P₄O₁₀ + 6H₂O ⟶ 4H₃PO₄
Orthophosphoric acid

(a) The energy required to remove the most loosely held electron from an isolated gaseous atom is called as ionization energy.
(b) (i) Variation in a period:
On moving across a period from left to right, the ionization enthalpy value increases. This is due to the following reasons:
1. Increase of nuclear charge in a period
2. Decrease of atomic size in a period
Because of these reasons, the valence electrons are held more tightly by the nucleus, thus ionization enthalpy increases.
Hence, ionization energy is a periodic property.
(ii) Variation in a group:
As we move from top to bottom along a group, the ionization enthalpy decreases. This is due to the following reasons:
1. A gradual increase in atomic size.
2. Increase of screening effect on the outermost electrons due to the increase of number of inner electrons.
Hence, ionization enthalpy is a periodic property.

Anomalies of Mendeleev's periodic table
(i) Some elements with similar properties were placed in different groups whereas some elements having dissimilar properties were placed in same group, but iodine (127) was placed in VII group.
Example: Tellurium (127.6) was placed in VI group.
(ii) Some elements with higher atomic weights were placed before lower atomic masses in order to maintain the similar chemical nature of elements. This concept was called inverted pair of elements concept.
Example: ⁵⁹₂₇Co and ^58.7 ₂₈Ni
(iii) Isotopes did not find any place in Mendeleev's periodic table.
(iv) Position of hydrogen could not be made clear.
(v) He did not leave any space for lanthanides and actinides which were discovered later on.
(vi) Elements with different nature were placed in one group, Example: Alkali metals and coinage metals were placed together:
(vii) Diagonal and horizontal relationships were not explained.

Electronic Configuration of Aluminium
\(\underbrace { { Al }^{ 13 }{ 1s }^{ 2 } }_{ (n-2) } \underbrace { 2s^{ 2 }{ 2p }^{ 2 } }_{ (n-1) } \underbrace { { 3s }^{ 2 }{ 3p }^{ 1 } }_{ n } \)

∴ Effective nuclear charge = Z - S = 13 - 9.5
(Z_eff)_Al =3.5
Electronic Configuration of chlorine
\(\underbrace { { 1s }^{ 2 } }_{ (n-2) } \underbrace { 2s^{ 2 }{ 2p }^{ 2 } }_{ (n-1) } \underbrace { { 3s }^{ 2 }{ 3p }^{ 5 } }_{ n } \)

∴ Effective nuclear charge = Z - S = 17 - 10.9
(Z_eff)_cl = 6.1
(Z_eff)_cl > (Z_eff)_Al and hence r_clAl

2H₂ + O₂ ➝ 2H₂O
The isotope of hydrogen which has one proton and one neutron in its nucleus is Deuterium.
2D₂ + O₂ ➝ 2D₂O
The product is heavy water (Deuterium oxide). H₂O and D₂O have same chemical properties but the
reaction velocity of D2O is slightly less due to the difference in the mass number of the isotopes known as isotopic effect. Deuterium is heavier than protium so reacts slowly.

(i) The element (A) is hydrogen reacts with carbon monoxide to form methanol (B).
CO + 2H₂ \(\overset{Cu}{\rightarrow}\) CH₃OH
         (A)           (B)
(ii) Hydrogen reacts with nitrogen to form ammonia (C). 
\({ N }_{ 2 }+3{ H }_{ 2 }\quad \overset { 380-{ 450 }^{ 0 }C }{ \underset { 200/atm\quad 1le }{ \rightleftharpoons } } 2\underset { (C) }{ { NH }_{ 2 } } \)

(i) A important peroxide is hydrogen peroxide (A).
(ii) H₂O₂ disproportionates to form oxygen and water.
\(\underset { (A) }{ { H }_{ 2 }{ O }_{ 2 } } \quad \longrightarrow { H }_{ 2 }O+\frac { 1 }{ 2 } { O }_{ 2 }\)
(iii) H₂O₂ reacts with FeSO₄ in acidic condition to form ferric sulphate (B).
2FeSO₄ + H₂SO₄ + H₂O₂ ➝ Fe₂(SO₄)₃ + 2H₂O
                                (A)           (B)
(iv) H₂O₂ reacts with KMnO₄ in basic condition and form Manganese dioxide (C) and potassium hydroxide (D).
2KMnO₄ + 3H₂O₂  ➝ 2MnO₂ + 2KOH + 2H₂O + 3O₂
                   (A)            (C)          (D)

(i) The universal solvent is water (A).
(ii) Water reacts with chlorine gas to form Hydrochloric acid (B) and Hypochlorous acid (C).
Cl₂ + H₂O ➝ HCl + HOCl
          (A)       (B)     (C)
(iii) Water reacts with silicon tetrachloride (SiCI₄) to give silicon dioxide (D).
SiCl₄ + 2H₂O  ➝ SiO₂ + 4HCl
               (A)          (D)     (B)

(i) Hydrogen bond :
When a hydrogen atom (H) is covalently bonded to a highly electronegative atom (F or °or N), the bond is polarized in such a way that the hydrogen atom is able to form a weak bond (electrostatic attraction) between the hydrogen atom of a molecule and the electronegative atom of a second molecule. The bond thus formed is called a hydrogen bond.
(ii) Intermolecular Hydrogen:
Intermolecular hydrogen bonds occur between two separate molecules.
They can occur between any numbers of like or unlike molecules as long as hydrogen donors and acceptors are present and in positions in which they can interact. Eg: Water, HF, etc, 
(iii) Intramolecular Hydrogen: 
This type of bond is formed between hydrogen atom and N, O or F atom of the same molecule.
This type of hydrogen bonding is commonly called chelation and is more frequently found in organic compounds. Eg: o-nitro phenol, salicylic acid, etc.

Z = 17
Z* = Z -S
= 17- [(0.35 \(\times\) No. of other electrons in nth shell) + (0.85\(\times\) No. of electrons in (n -1)^th shell) + (1.00 \(\times\) total number of electrons in the inner shells)]
= 17 - [(0.35 \(\times\) 6) + (0.85 \(\times\) 8) + (1 \(\times\) 2)]
= 17 - 10.9 = 6.1
Z* =  6.1

The electronic configuration of K atom is
K₁₉ = (1s²)(2s²p⁶)(3s²3p⁶)4s¹
Effective nuclear charge (Z*) = Z - S
Z* = 19 - [(0.85 \(\times\) No. of electrons in (n -1)t^h shell) + (1.00 total number of electrons in the inner shells)]
= 19-[0.85 \(\times\) (8) + (1.00 \(\times\) 10)]
Z* =  2.20

(i) Al₄C₃ + 12D₂O ➝ 4Al(OD)₃ + 3CD₄
(ii) CaC₂ + 2D₂O ➝ Ca(OD)₂ + C₂H₂
(iii)Mg₃N₂ + 6D₂O ➝ 3Mg(OD)₂ + 2ND₃
(iv) Ca₃P₂ + 6D₂O ➝ 3Ca(OD)₂ + 2PD₃

(i) Hydrogen liberates heat when it comes in contact with a metal. This property of hydrogen is used in atomic hydrogen welding and cutting torch.
(ii) Hydrogen reacts with unsaturated oil (Vegetable oil) in the presence ofNi catalyst. This property is used in the production of vanaspathi ghee.
(iii) Hydrogen combines with nitrogen in the ratio 3:1 in presence of Fe as catalyst and:
Mo as promoter. This property is used in the manufacture of ammonia . (Haber's process).
N₂ + 3H₂ \(\overset { 673k,200atm }{ \underset { Fe,Mo }{ \rightleftharpoons } } \) 2NH₃

(i) Electron configuration of nitrogen
(Z = 7) 1s² 2s² 2p³.
Electron configuration of oxygen
(2= 8) 1s² 2s² 2p⁴.
Nitrogen has a half filled electronic configuration which is much more stable than an incomplete p-orbital of oxygen which would need to give up one of it's electrons to attain the stability of nitrogen. Hence nitrogen would require more ionization energy to remove an electron from it's outer shell than oxygen.
(ii) Electron configuration of carbon
(Z = 6) 1s²2s²2p².
Electron configuration of Boron
(Z = 5) Is²2s²2p¹
The size of a carbon atom is smaller than boron So the valence electron of carbon has greater nuclear charge than that of boron. Hence the first I.E of carbon is greater than that of boron. However, the second ionization enthalpy of boron is higher than that of carbon. This is because after losing electron, Boron has a fully filled orbital (2s²) than carbon (2p¹). Fully filled orbitals have more stability than partially filled orbitals so greater amount of energy will be needed to remove an electron from boron. So in this case, the second I.E of boron is higher than that of carbon.
(iii) The electron affinities of Be, Mg and noble gases are almost zero because both Be (Z = 4; 1s²2s²) and Mg (Z = 12; Is²2s²2p⁶3s²) are having s orbital fully filled in their valence shell. Fully filled orbitals are most stable due to symmetry. Therefore, these elements would be having least tendency to accept electron. Hence, Be and Mg would be having zero electron affinity. Whereas N (Z = 7; 1s²2s²2px¹2py¹2pz¹ and P (Z = 15) Is² 2s² 2p⁶ 3s² 3p³ is having half filled 2p-subshell. Half filled sub shells are most stable due to symmetry (Hund's rule). Thus, nitrogen and phosphorous are having least tendency to accept electron. Hence, have low electron affinity.
(iv) Fluorine is highly electro negative in nature therefore as it gains the electron its octet become stable and releases the energy so exothermic. while in oxygen the addition of first electron is exothermic in nature but addition of second electron experiences high repulsive force. So needs extra external energy to enter outer shell, hence endothermic in nature.

r(K⁺)+r(Cl^-) = d(K^+-Cl-) = 3.14 Å.
The effective nuclear charge for K+ and CI- can be calculated as follows.
K+ = (1s²)          (2s²2p⁶)        (3s²3p⁶)
        inner shell    (n-1)th shell   n^th shell
Z*(K-) = Z-S
= 19 - [(0.35 x 7) + (0.85 x 8) + (1 x 2)]
= 19 - 11.25 = 7.75
Z*(Cl-) = 17- [(0.35 x 7) + (0.85 x 8) + (1 x 2)]
= 17-11.25 = 5.75
∴ \(\frac { r({ K }^{ + }) }{ r(Cl^{ - }) } =\frac { Z*(Cl^{ - }) }{ Z*(K^{ + }) } =\frac { 5.75 }{ 7.75 } \)=0.74
∴ r(K⁺) = 0.74 r(Cl^-)
Substitute (2) in (1)
0.74 r(Cl^-) + r(Cl^-) = 3.14 Å.
1.74 r(Cl^-) = 3.14 Å
r(Cl^-) = \(\frac { 3.14\overset { 0 }{ A } }{ 1.74 } \)=1.81.Å.

The element which occupies group number (16) and period number (2) is oxygen. (B) is D₂O which is used as a moderator in nuclear reactions.
So (A) must be deuterium, which is an isotope of hydrogen
    \(2\underset { (A) }{ { D }_{ 2 } } +{ O }_{ 2 }\rightarrow 2\underset { (B) }{ { D }_{ 2 }O } \)
So (B) is D₂O
(A) adds to (C) as follows :
\(3 \mathrm{D}_{2}+\mathrm{C}_{3} \mathrm{H}_{6} \rightarrow \mathrm{CH}_{3}-\mathrm{CH}-\mathrm{CH}_{2}\)
So (D) is 1,2 - dideutero propane. 

On moving diagonally across the periodic table, the second and third period elements show certain similarities. It is quite pronounced in the following pair of elements.

The similarity in properties existing between the diagonally placed elements is called diagonal relationship.

Variation along a period: Ionisation energy usually increases along a period. This is due to increase of nuclear charge and decrease in size as we move from left to right in a period.
Periodic variation in group: Ionisation energy decreases down a group. As we move down a group, the valence electron occupies new shells, the distance between the nucleus and the valence electron increases. So, the nuclear forces of attraction on valence electron decreases and hence ionisation energy also decreases down a group.

(i) Hydrogen has the electronic configuration of 1s1 which resembles with ns¹ general valence shell configuration of alkali metals and shows similarity with them as follows:
1. It forms unipositive ion (H⁺) like alkali metals (Na⁺,K⁺,Cs⁺)
2. It forms halides (HX), oxides (H₂O), peroxides (H₂O₂) and sulphides (H₂S) like alkali metals (NaX, Na₂O, Na₂O₂, Na₂S)
3. It also acts as a reducing agent.
4. lt is an electro positive element
However, unlike alkali metals which have ionization energy ranging from 377 to 520 kJ mol^-1, the hydrogen has 1.314 kJ mol^-1 which is much higher than alkali metals.
Like the formation of halides (X ^-) from halogens, hydrogen also has a tendency to gain one electron to form hydride ion(H⁺) whose electronic configuration is similar to the noble gas, helium. However, the electron affinity of hydrogen is much less than that of halogen atoms. Hence, the tendency of hydrogen to form hydride ion is low compared to that of halogens to form the halide ions as evident from the following reactions:
\( 1 / 2 \mathrm{H}_{2}+\mathrm{e}^{-} \rightarrow \mathrm{H}^{-} \Delta \mathrm{H}=+36 \mathrm{kcal} \mathrm{mol}^{-1} \)
\(1 / 2 \mathrm{Br}_{2}+\mathrm{e}^{-} \rightarrow \mathrm{Br}^{-} \Delta \mathrm{H}=-55 \mathrm{kcal} \mathrm{mol}^{-1}\)
Since, hydrogen has similarities with alkali metals as well as the halogens; it is difficult to find the right position in the periodic table. However, in most of its compounds hydrogen exists in +1 oxidation state. Therefore, it is reasonable to place the hydrogen in group 1 along with alkali metals as shown in the latest periodic table published by IUPAC.