The balanced equation for the formation of HCI is,
H₂(g) + CI₂(g) \(\rightarrow\) 2 HCI (g)
As per the stoichiometric equation, under given conditions,
To produce 2 moles of HCI, 1 mole of chlorine gas is required.
To produce 44.8 litres of HCI, 22.4 litres of chlorine gas are required.
\(\therefore\) To produce 11.2 litres of HCI,

= 5.6 litres of chlorine are required.

The balanced chemical equation is,
\({ CaCO }_{ 3 }\left( s \right) \overset { \triangle }{ \longrightarrow } { CaO }_{ (s) }+{ CO }_{ 2 }(g)\)
As per the stoichiometric equation,
1 mole (100g) CaC0₃ on heating produces 1 mole CO₂

At STP, 1 mole of CO₂ occupies a volume of 22.7 liters
At STP, 50 g of CaCO₃ on heating produces,

= 11.35 liters of CO₂

The balanced stoichiometric equation for the formation of ammonia is
N₂(g) + 3H₂ (g) \(\rightarrow\) 2NH₃ (g)
As per the stoichiometric equation,
To produce 2 moles of ammonia, 3 moles of hydrogen are required.
\(\therefore\)  to produce 10 moles of ammonia,

= 15 moles of hydrogen are required.

CH_4(g) + 2O₂ \(\rightarrow\) CO₂ + 2H₂O
16g                          (2x18)g
As per stoichiometric equation,
16 g of methane produces 36 g of H₂O
\(\therefore\) 32 g of methane will produce = \(\frac { 36 }{ 16 } \times 32=72\) g of water.

The oxidation state of CI in ClO₂^- is +3. So, chlorine can get oxidised as well as reduced and can act as reductant and oxidant.
The disproportionation reaction of ClO₂^- is
\(\begin{matrix} +1 \\ 3Cl{ O }_{ 2 }^{ - } \end{matrix}\longrightarrow { Cl }^{ - }+\begin{matrix} +5 \\ Cl{ O }_{ 3 }^{ - } \end{matrix}\)
In CIO₄^- , CI is in its highest oxidation state, So it can only be an oxidant.

Electronic configuration of oxygen

ஃ 8^th electron present in 2p_x orbital and the quantum numbers are
n = 2,l = 1,m₁ = either + 1 or -1 and s = -1/2
Electronic configuration of chlorine 

15^th electron present in 3P_z orbital and the quantum numbers are n = 3, l = 1, m₁ = either +1 or -1 and ms = +1/2

\(\triangle x.\triangle p\ge \frac { h }{ 4\pi } \)
\(\triangle x.\triangle p\ge 5.28 \times{ 10 }^{ -35 }Kg{ m }^{ 2 }{ s }^{ -1 }\)
\(\triangle x.(m\triangle v)\ge 5.28 \times{ 10 }^{ -35 }Kg{ m }^{ 2 }{ s }^{ -1 }\)
Given \(\triangle\)v = 0.1%
v = 2.2 x 106 ms^-1
m = 9.1 x 10^-31Kg
\(\triangle\)v = \(\frac{0.1}{100}\times2.2\times{10}^{6}ms^{-1}\)
= \(2.2\times{10}^{6}ms^{-1}\)
\(\therefore \triangle x\ge \frac { { 5.28\times 10 }^{ -35 }{ Kgm }^{ 2 }{ s }^{ -1 } }{ 9.1\times { 10 }^{ -31 }Kg\times 2.2\times { 10 }^{ 3 }m{ s }^{ -1 } } \)
\(\\ \triangle x\ge 2.64\times { 10 }^{ -8 }m\)

Circumference of the orbit - 2\(\pi\)r         .... ...(1)
Circumference of the orbit of H - atom (n = 1) - n\(\lambda\),  .........(2)
(1) = (2) \(2 \pi r=\lambda \text { (or) } 2 \pi r=\frac{\mathrm{h}}{\mathrm{mv}}\)

In the periodic table vertical columns are called groups.
Elements in the same vertical column possess similar number of electrons in the outer orbitals.
∴ Elements a and c belongs to group 18
Elements b and d belongs to group 13

Z = 118; [₈₆Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶
In the periodic table the element with Z = 118 is located in p-block,
Period no = 7 (as n = 7 for valence shell)
Group no. = 18 (group no = 10+ ns electrons + np electrons) (n - outer most shell)

(i) Electronic configuration is the arrangement of electrons in an atom. The outermost electron shell is often referred to as the "valence shell"determines the chemical properties.
(ii) Ionization energy and electron affinity is the amount of energy released or required in pulling out or adding an electron to a neutral atom. So both depend on electronic configuration of the element.

Acidity : The number of hydroxyl ions present in one mole of a base is known as the acidity of the base.
Basicity : The number of replaceable hydrogen atoms present in a molecule of the acid is referred to as its basicity.

Cl(g) + e^- ⟶Cl^-(g)    AH = 348 kJ mol^-1
For one mole (35.5g) 348 kJ is released.
\(\therefore \text { For } 17.5 \mathrm{~g} \text { chlorine, } \frac{348 \mathrm{~kJ}}{35.5 \mathrm{~\not g}} \times 17.75 \mathrm{~\not g} \text { energy leased. }\)
∴ The amount of energy released =\(\frac { 348 }{ 2 } \) =174 kJ

Atomic number: 120
IUPAC temporary symbol: Unbinilium
IUPAC temporary symbol: Ubn
Possible electronic configuration :[Og] 8s²

\(\underset { 2\quad mol }{ { 2H }_{ 2 } } +\underset { 1\quad mol }{ { { O }_{ 2 } } } \rightarrow \underset { 2\quad mol }{ { 2H }_{ 2 }O } \)
Number of mole of H₂ taken = \(\frac{3}{2}\) = 1.5
Number of mole of O₂ taken = \(\frac{29}{32}\) = 0.9062
\(\therefore\) 2 mole of H₂ react with 1 mole of O₂
\(\therefore\) 1.5 mol of H₂ will react with = \(\frac{1}{2}\times \) 1.5 = 0.75 mole of O₂
Available mole of O₂ (0.9062) > required mol of O₂ (0.75)
Thus, O₂ is in excess and therefore, H₂ will be the limiting reagent.

Molar Mass of HCl = 1 + 35.5 = 36.5 
Basicity of HCl = 1
\(\therefore\) equivalent mass of hcl = \(\frac { 36.5 }{ 1 } =36.5g\) eq^-1

(i) \(\sigma \) = 1 x 0.35  + 18 x 0.85 + 10 x 1 = 25.65
(ii) \(\sigma \) = 9 x 0.35  +18 x 1.0 = 21.15

The mass of an atom (10^-27 kg) is too small to be measured directly. They are measured with reference to a standard atom, Hence they are called as relative atomic masses.

In a homogeneous mixture, the composition of the constituents are uniform throughout whereas in a heterogeneous mixture, the composition is not uniform throughout.

(iii) \(\overset { 0 }{ \underset { \underset { 2e^{ - } }{ \downarrow } }{ Cu } } { O }_{ 7 }+H\overset { +5 }{ \underset { { 1e }^{ - } }{ \underset { \uparrow }{ N } } } { O }_{ 3 }\longrightarrow \overset { +2 }{ Cu } \left( { No }_{ 3 } \right) _{ 2 }+\overset { +4 }{ N } { O }_{ 2 }+{ H }_{ 2 }O\)
Cu +2HNO₃ \(\longrightarrow \)  Cu(NO₃)₂ + NO₂ + H₂O
Cu + 2HNO₃ + 2HNO₃ \(\longrightarrow \) Cu(NO₃)₂ + 2NO₂ + 2H₂O
Cu + 4HNO₃ \(\longrightarrow \) Cu (NO₃)₂ + 2No₂ + 2H₂O

Mol.mass = 3(H) + 1(B) + 3(0)
= 3(1) + 1(11) + 3(16)
= 3 + 11 + 48 = 62

i) urea [CO(NH₂)₂]
Mol.mass = 1 (C) + 2(N) + 4(H) + 1(0)
= 1(12) + 2(14) + 4(1) + 1(16)
 = 12 + 28 + 4 + 16 = 60  
ii) Acetone [CH₃ COCH₃]
Mol.mass = 3(C) + 6(H) + 1(0)
= 3(12) + 6(1) + 1(16)
= 36 + 6 + 16 = 58
iii) Boric Acid [H₃ BO₃]
Mol.mass = 3(H) + 1(B) + 3(0)
= 3(1) + 1(11) + 3(16)
= 3 + 11 + 48 = 62
iv) Sulphuric  Acid [H₂ SO₄]
Mol.mass = 2(H) + 1(S) + 4(0)
= 2(1) + 1(32) + 4(16)
= 2 + 32 + 64 = 98

Step - 1 : To find atoms undergoing change in O.N.
K₂Cr₂O₇ + HI ⟶ KI + Crl₃ + H₂O + I₂
Step - 2 : To find the total increase and decrease in O.N.
K₂Cr₂O7 + CrI₃  (decrease of 3 unit / atom = Total decrease = 6 units / 2 atom)
HI ⟶ I₂ (increase of 1 unit / atom = total increase 1 x 6 = 6)
Step - 3 : To balance the total increase and decrease in O.N, multiply HI by 6.,
K₂Cr₂O₇ + 6 HI ⟶ KI + Crl₃ + H₂O + I₂
Step - 4 : To balance all atoms other than '0' and 'H'
K₂Cr₂O₇ + 6 HI ⟶ 2KI + 2Crl₃ + H₂O + 3I₂
This makes 14 iodine atoms on RHS. (These iodide ions do not undergo any change in O.N). Hence to balance the iodine atoms add 8HI to LHS.
i.e., K₂Cr₂O₇ + 14 HI ⟶ 2KI + 2CrI₃ + H₂O + 3I₂
These, the oxygen atoms are balanced by making 7H₂O as RHS.
K₂Cr₂O₇ + 14 HI ⟶ 2KI + 2CrI₃ + 7H₂O + 3I₂
The hydrogen atoms are balanced by themselves.
Hence the balanced equation is
K₂Cr₂O₇ + 14HI ⟶ 2KI + 2Crl₃ + 7H₂O + 3I₂

Step-1: To find atoms undergoing change in O.N.
\(\overset { +2-2 }{ CuO } +\overset { -3+1 }{ { NH }_{ 3 } } \rightarrow \overset { 0 }{ Cu } +\overset { 0 }{ { N }_{ 2 } } +\overset { +1-2 }{ { H }_{ 2 }O } \)
Step-2: To find total increase and decrease in O.N.
CuO ➝ Cu (decrease of 2 unit per atom)
NH3 ➝ N2 (increase of3 unit per atom)
Total decrease = 2 x 3 = 6
Total increase = 3 x 2 = 6
Step-3: To balance the total decrease and increase in O.N, multiply two by 3 and NH3 by 2.
3CuO + 2NH₃ ⟶ Cu + N₂ + H₂O
Step-4: To balance all atoms other than 'H' and 'O'
3CuO + 2NH₃ ⟶ 3Cu + N₂ + H₂O
Step-5: To balance 'O' atoms
3CuO + 2NH₃ ⟶ 3Cu + N₂ + 3H₂O
hydrogen atoms balance by themselves.
Hence the final equation is 
3CuO + 2NH₃ ⟶ 3Cu + N₂ + 3H₂O

\(\overset { o }{ Z } n\longrightarrow { Z }n^{ 2+ }\)
\(\overset { +5 }{ N } { O }_{ 3 }^{ - }\longrightarrow \overset { 2+ }{ NO } \)
(1) \(\Rightarrow \) Zn \(\rightarrow\) Zn²⁺ +2e^-                                  .....(3)
(2) \(\Rightarrow \) \({ NO }_{ 3 }^{ - }+{ 3e }^{ - }+{ 4H }^{ + }\longrightarrow NO+{ 2H }_{ 2 }O\)......(4)

\(\overset { +3 }{ C } _{ 2 }{ O }_{ 4 }^{ 2- }\longrightarrow \overset { +4 }{ C } { O }_{ 2 }\)
\(\overset { +6 }{ Cr } _{ 2 }{ O }_{ 7 }^{ 2- }\longrightarrow { Cr }^{ 3+ }\)
(1)  \(\Rightarrow \) \({ C }_{ 2 }{ O }_{ 4 }^{ 2- }\longrightarrow { 2CO }_{ 2 }+{ 2e }^{ - }\)
\({ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }\longrightarrow { 2Cr }^{ 3+ }+{ 7H }^{ 2 }O\)

Half reactions are:
\(\overset { +7 }{ M } { nO }_{ 4 }^{ - }\longrightarrow { Mn }^{ 2+ }\)
and \({ Sn }^{ 2+ }\longrightarrow { Sn }^{ 4+ }\)
(1) \(\Rightarrow \) \({ MnO }_{ 4 }^{ - }+{ 8H }^{ - }+5e^{ - }\longrightarrow { Mn }^{ 2+ }+{ 4H }_{ 2 }O\)
(2) \(\Rightarrow \) \({ Sn }^{ 2+ }\longrightarrow { Sn }^{ 4+ }+{ 2e }^{ - }\)

ii) 

iii) 

iv) 

Empirical formula = Na₂ SH₂₀ O₁₄ 
\(\left[ \begin{matrix} { Na }_{ 2 }{ SH }_{ 20 }{ O }_{ 14 } \\ =(2\times 23)+(1\times 32)+(20\times 1)+14(16) \\ =46+32+20+234 \\ =322 \end{matrix} \right] \)
n = \(\frac { molar\quad mass }{ caluclated\quad empirical\quad formula\quad mass } =\frac { 322 }{ 322 } =1\)
Molecular formula = Na₂ SH₂₀O₁₄
Since all the hydrogen in the compound are present as water
\(\therefore \) The molecular formula is Na₂ SO₄ 10H₂O.

Empirical Formula = C₆ H₆O
n = \(\frac { molar\ mass }{ calculated\ eprirical\ formula\ mass } \)
= \(\frac { 2\times \ vapour\ density }{ 94 } \frac { 2\times 47 }{ 94 } =1,\) 
Molecular formula (C₆H₆O)  x 1 = C₆H₆O.

Screening effect: The repulsive force between the inner shell electrons and the valence electrons leads to a decrease in the electrostatic attractive forces acting on the valence electrons by the nucleus. Thus, the inner shell electrons act as a shield between the nucleus and the valence electrons. This effect is called shielding effect.
Pauling's scale: Pauling, he assigned arbitrary value of electronegativities for hydrogen and fluorine as 2.2 and 4.0 respectively. Based on this the electronegativity values for other elements can be calculated using the following expression.
\(({ X }_{ A }-{ X }_{ B })=0.182\sqrt { E_{ AB } } -({ E }_{ AA }*{ E }_{ BB })^{ 1/2 }\)
Where E_AB' EAA and E_BB are the bond dissociation energies of AB, A₂ and B₂ molecules respectively. The electronegativity of any given element is not a constant and its value depends on the element to which it is covalently bound. The electronegativity values play an important role in predicting the nature of the bond.

The word Aufbau in German means 'building up'. In the ground state of the atoms, the orbitals are filled in the order of their increasing energies. That is the electrons first occupy the lowest energy orbital available to them.
Once the lower energy orbitals are completely filled, then the electrons enter the next higher energy orbitals. The order of filling of various orbitals as per the Aufbau principle which is in accordance with (n + l) rule.

(c) Assertion is true but reason are false