Limitations of Bohr's model of an atom:
(i) It could not explain spectrum of multi-electron atoms.
(ii) It could not explain Zeeman and Stark effects.
(iii) It could not explain shape of molecules.
(iv) It was not in accordance with Heisenberg's uncertainty principle.
Quantum Mechanical Model: It was developed on the basis of Heisenberg's uncertainty principle and dual behaviour of matter. Main features of this model are given below :
(i) The energy of electrons in an atom is quantized i.e. can only have certain values.
(ii) The existence of quantized electronic energy levels is a direct result of the wave-like properties of electrons.
(iii) Both, the exact position and velocity of an electron in an atom cannot be determined simultaneously.
(iv) The orbitals are filled in increasing order of energy. All the information about the electron in an atom is stored in orbital wave function.\(\Psi \)
(v) From the value of\(\Psi \)²  at different points within atom, it is possible to predict the region around the nucleus where electron most probably will be found.

Estimation of halogens: It involves oxidising the organic substance with fuming nitric acid in the presence of silver nitrate. The halogen of the substance is thus converted to silver halide which is separated and weighed:
Weight of organic compound = W gm
weight of silver halide = x g.
% of halogen = \(\frac{\text{At.wt.of halogen}\times100x}{\text{Mol.wt of silver halide}\times w}\)
Estimation of sulphur: The organic substance is heated with fuming nitric acid but no silver nitrate is added. The sulphur of the substance is oxidised to sulphuric acid which is then precipitated as barium sulphate by adding excess of barium chloride solution. From the weight of BaS04 so obtained the percentage of sulphur can be calculated.
% of sulphur = \(\frac{32(At.weight\ of\ S)}{233(mol\ weight\ of\ BaSO_4)}\times\frac{weight\ of\times100\ BaSO_4}{weight\ of\ organic\ compound}\)
Estimation of phosphorous: The organic substance is heated with fuming nitric acid whereupon phosphorous is oxidised to phosphoric acid. The phoshoric acid is
precipitated as ammonium phosphomolybdate, (NH₄)₃ PO₄·12MoO₃, by the addition of ammonia and ammonium molybdate solution which is then separated, dried and weighed.
% of P = \(\frac{31\times w_1\times 100}{1877\times w}\)
where Molar mass of (NH₄)₃ PO₄.12MoO₃ = 1877 g
If phosphorous is estimated as Mg₂P₂O₇
% of P = \(\frac{62\times w_1\times100}{222\times w}\)

Sodium bicarbonate (NaHCO₃)  on hetaing composes to produces CO₂ gas which whwn pases thrugh lime water, turns milky.
2NaHCO₃ \(\xrightarrow[]{\Delta}\)Na₂CO₃ + CO₂ + H₂O
Ca(OH)₂(aq) + CO₂(s)\(\rightarrow \)CaCO₃(s) + H₂O(l)
But Na₂CO₃ (Sodium carbonate is stable to heat does not decompose to heat).

Water free from ca²⁺ and Mg²⁺ ions is called soft water.

\({ H }_{ 2 }O_{ 2 }\)can act as an oxidising as well as a reducing agent both in acidic and basic medium
(i) Oxidising agent in acidic medium
\(2{ Fe }^{ 2+ }(aq)+2{ H }^{ + }(aq)+{ H }_{ 2 }{ O }_{ 2 }(aq)\rightarrow 2{ Fe }^{ 3+ }(aq)+2{ H }_{ 2 }O(l)\)
Here, \({ Fe }^{ 2+ }\) get oxidised to \({ Fe }^{ 3+ }\)
(ii) Oxidising agent in acidic medium
\(Mn^{ 2+ }(aq)+6H^{ + }(aq)+5{ H }_{ 2 }O(aq)\rightarrow 2Mn^{ 2+ }(aq)+8{ H }_{ 2 }O(l)+5O_{ 2 }(g)\)
Here, \(Mn^{ 2+ }\)is oxidised to \(Mn^{ 4+ }\)
(iii) Reducing agent in acidic medium
\(2MnO_{ 4 }^{ - }(aq)+6H^{ + }(aq)+{ 5H }_{ 2 }O_{ 2 }(aq)\rightarrow 2Mn^{ 2+ }(aq)+8{ H }_{ 2 }O(l)+5O_{ 2 }(g)\)
Here, oxidation state of Mn is reduced from +7 to +2
(iv)Reducing agent in basic medium
\({ I }_{ 2 }(s)+{ H }_{ 2 }O_{ 2 }(aq)+2OH^{ - }(aq)\rightarrow 2I^{ - }(aq)+2{ H }_{ 2 }O(l)+O_{ 2 }(g)\)
her, oxidation state of I is reduced from zero to -1.

Balance increase and decrease in oxidation number
\({ I }_{ 2 }+{ 10NO }_{ 3 }^{ - }\longrightarrow { 10NO }_{ 2 }+{ 2IO }_{ 3 }^{ - }\)
Balance charge by writing 8H⁺ in LHS of the equation.
\({ I }_{ 2 }+{ 10NO }_{ 3 }^{ - }+{ 8H }^{ + }\longrightarrow { 10NO }_{ 2 }+{ 2IO }_{ 3 }\)
Balance H-atoms by writing 4H₂O in RHS of the equation.
\({ I }_{ 2 }+{ 10NO }_{ 3 }^{ - }+{ 8H }^{ + }\longrightarrow { 10NO }_{ 2 }+{ 2IO }_{ 3 }^{ - }+{ 4H }_{ 2 }O\)
Oxygen atoms are automatically balanced.
This representation a balanced redox reaction.

A heat capacity of water = heat capacity of calorimeter, the heat gained by water = heat lost by calorimeter
\(=125\times (296.5-286.4)\times 4.184 \ J=5282J=5.282kJ\)

P_t = 25.21 atm

Cl-atom has empty d-orbitals and it acquires excited state at the time of bonding when electrons from 3p-orbitals are promoted to 3d-orbitals.

In excited state Cl-atom can exhibit a covalency of three. Hence, ClF₃ is possible. F-atom cannot expand its octet due to absence of empty d-orbitals in 2nd energy shell.
Hence, it cannot exhibit covalency more than 1. Therefore, FCl₃ is not possible.

(i)  Ni = + 2, Pt = + 4
(II) Platinum forms more stable complexes in +4 state due to its higher stability than +2 state.

Molarity \(=\frac { moles \ of \ AlCl_{ 3 } }{ volume \ of \ solution } \times 1000\)
Moles of AlCl₃ \(=\frac { molarity \ \times \ volume \ of \ solution }{ 1000 } \)
Now, 1 molecule of AlCl₃ contains 3 Cl^- ions.
1 mol of AlCl₃ contains 3 mol of Cl^- ions.
0.0025 mol of AlCl₃ contains Cl^- ions \(=0.025\times 3\)
\(=0.075\ mol\)
1 mol of Cl^- ions \(=6.022\times { 10 }^{ 23 }\) Cl^- ions.
0.075 mol of Cl^- ions \(=6.022\times { 10 }^{ 23 }\times 0.075\)
\(=4.5\times 10^{ 22 }\)

The electron pairs involved in the bond formation are known as bond pairs or shared pairs.

In NaBH₄, H is present as hydride ion. Therefore, its oxidation number is -1. Thus,
\(\overset { +1 }{ Na } \overset { x }{ B } \overset { -1 }{ { H }_{ 4 } } \) 
\(\therefore\) 1 (+1) + x + 4 (-1) = 0 or x = +3
Thus, the oxidation number of B in NaBH₄ = + 3.

Alkenes are rich source of loosely held \(\pi \) electrons, due to which they show electrophilic addition reaction. Electrophilic addition reactions of alkenes are accompanied by large energy changes so these are energetically favourable than that of Electrophilic substitution reactions.
In special conditions alkenes also undergo free radical substitution reactions.
In arenes during electrophilic addition reactions aromatic character of benzene ring is destroyed while during electrophilic substitution reactions of areans are enegetically more favourable than that of electrophilic addition reaction.
That's why alkenes prefer to undergo electrophilic addition reaction while arenes prefer electrophilic substitution reactions.

For exhibiting cis-trans (or geometrical isomerism, a molecule must fulfil the following condition.
It must have atlest one double bond.

(a) C–Br, since Br is more electronegative than H, (b) C–O, (c) C–O

\(HF\rightleftharpoons { H }^{ + }+{ F }^{ - }\)
K_a (acid) x K^b (conjugate base) = 1.47 x 10 ^-11
\({ K }_{ b }{ (F }^{ - })=\frac { 1.0\times { 10 }^{ -14 } }{ 6.8\times { 10 }^{ -14 } } =1.47\times { 10 }^{ 11 }\)

For the combination of N₂ and O₂ to form NO, the standard Gibbs energy of formation, is +ve
1/2N₂(g)+1/2O₂(g)→NO(g)(Δ_fG^∘NO = +86.7 kJmol⁻¹)
Therefore, this reaction is non-spontaneous under the standard conditions and hence N₂ and O₂ do not combine.

0.022

2M

(i) Both react with nitrogen to form nitrides.
(ii) Both react with O₂ to form monoxides
(iii) Both the elements have the tendency to form covalent compounds.
(iv) Both can form complex compounds.

Volume of acid taken = 50 mL of 0.5 M H₂SO₄
= 25 mL of 1.0 M H₂SO₄
Volume of alkali used for neutralisation of excess acid 
= 60 mL of 0.5 M NaOH
= 30 mL of 1.0 M NaOH
H₂SO₄ + 2NaOH\(\longrightarrow \)Na₂SO₄ + 2H₂O
1 mole of H₂SO₄ = 2 moles of NaOH
Hence 30 mL of 1.0 M NaOH
= 15 mL of 1.0 M H₂SO₄
\(\therefore \) Volume of acid used by ammonia = 25 - 15 = 10 mL
% of nitrogen = 1.4 x N₁ x Vol. of acid used/w
(where N₁ = Normality of acid and w = mass of the organic compound taken)
% of nitrogen = (1.4 x 2 x 10/0.5) = 56.0.

First member of each group of representative elements ( i.e. s and p-block elements ) shows anomalous behaviour due to
(i) small size
(ii) high ionisation enthalpy
(iii) high electronegativity and
(iv) absence of d- orbitals.
For example in s-block elements, lithium shows anoimalous behaviour from rest of the alkali metals.
(i) Compounds of lithium have significant covalent character. While compounds of other alkali metals are predominantly ionic.
(ii) Lithium reacts with nitrogen to form lithium nitride while other alkali metals do not form nitrides.
In p-block elements, first member of each group has four orbitals, one 2s- and three 2p-orbitals in their valence shell. So, these elements show a maximum covalency of four ehile other members of the same group or different group show a maximum covalency beyond four due to availability of vacant d-orbitals.