I and III, I and IV, II and III and II and IV represent chain isomerism.
a) \({ CH }_{ 3 }CO{ CH }_{ 3 }\) 
b) \(H-CH={ CH }_{ 2 }\).

\({ H }_{ 2 }O_{ 2 }\) decomposes slowly on exposure to light and dust particles. In the presence of metal surfaces or traces of alkali present in glass containers, the traces of alkali present in glass containers, the decomposition of \({ H }_{ 2 }O_{ 2 }\) is catalysed. It is, therefore, stored in wax lined glass or plastic vessels in dark. Urea is added as a negative catalyst or stabiliser to check its decomposition.
\({ 2H }_{ 2 }O_{ 2 }(l)\overset { hv }{ \rightarrow } { 2H }_{ 2 }O(l)+{ O }_{ 2 }(g)\)

In H₂O, oxygen atom is sp³ hybridised with two lone pairs. The four sp³ hybridised orbitals acquire a tetrahedral geometry with two corners occupied by hydrogen atoms while other two by the lone pairs.
The bond angle is reduced to 104.5^o due to greater repulsive forces between lp - lp and the molecule thus acquires a V-shape or bent structure (angular structure).

In CO₂ molecule, carbon atom is sp-hybridised. The two sp hybrid are oriented in opposite direction forming an angle of 180^o \(O\overset { \pi }{ = } C\overset { \pi }{ = } O\\ \quad \quad \sigma \quad \quad \sigma \)
That's why H₂O molecule has bent structure whereas CO₂ molecule is linear.

ionization\(({ _{ 7 }N }={ 1s }^{ 2 },{ 2s }^{ 2 },{ 2p }^{ 3 }),\) enthalpy of nitrogen is greater than oxygen \(({ _{ 8 }O }={ 1s }^{ 2 },{ 2s }^{ 2 },{ 2p }^{ 4 }),\)due to extra stable half-filled 2p-orbitals. Similarly, ionization enthalpy of phosphorus is \(({ { _{ 15 }{ P } } }={ 1s }^{ 2 },{ 2s }^{ 2 },{ 2 }p^{ 6 },{ 3s }^{ 2 },{ 3p }^{ 3 })\) greater than sulfur on\(({ { _{ 16 }{ S } } }={ 1s }^{ 2 },{ 2s }^{ 2 },{ 2 }p^{ 6 },{ 3s }^{ 2 },{ 3p }^{ 4 })\) moving down the group, ionization enthalpy decreases with increasing atomic size. so, the increasing order of first ionization enthalpy is S.

We know that mass number of the element,
A = p + n = 81    .....(i)
Let the number of protons, p = x
Then, number of neutrons,
\(n=x+\frac { 31.7 }{ 100 } x=1.317x\)
(As number of neutrons are 31.7% more than the protons.)
Hence, from Eq.(i)
\(x+1.317x=81\)
or \(2.317x=81=\frac { 81 }{ 2.317 } =34.958\approx 35\)
Therefore, number of protons = 35 and the symbol is. \(_{ 35 }^{ 81 }{ Br }\) (Number of protons = atomic number)

In Bohr model, an electron is regarded as a charged particle moving in well defined circular orbits about the nucleus. An orbit can completely be defined only if both the position and the velocity of the electron are known exactly at the same time. This is not possible according to the Heisenberg uncertainty principle. Further more, the wave character of the electron is not considered in Bohr model.
Therefore, concept of movement of an electron in an orbit was replaced by the concept of probability of finding electron in an orbital due to de-Broglie concept of dual nature of electron and Heisenberg's uncertainty principle. the changed model is called quantum mechanical model of the atom.

Temperature corresponding to 60 mm \(\simeq \) 313 K.