(a) Calculation for molar of the gas
10.0 L of the given gas at STP weigh = 11.6 g
\(\therefore \) 22.4 L of the given gas at STP will weigh
\(\frac { 11.6\times 22.4 }{ 10 } =25.984\quad g\)
Molar mass = 25.984 = 26 \({ mol }^{ -1 }\)
(b) Empirical formula mass (CH) = 12 + 1 = 13
\(\therefore \) \(n=\frac { molecular \ mass }{ empirical \ formula \ mass } =\frac { 26 }{ 13 } =2\)
Hence, molecular formula 
\(=n\times CH=2\times CH={ C }_{ 2 }{ H }_{ 2 }\)

(a) All the given species have a same number of electrons  \(({ 10e }^{ 1 })\)Therefore, all are isoelectronic species.
(b) The ionic radii of isoelectronic species decreases with increase in atomic number (as magnitude of the nuclear charge increase with increase in atomic number) 
Therefore, their ionic radii increase in the order.\(\underset { z=13 }{ { Al }^{ 3+ } } <\underset { 12 }{ { Mg }^{ 2+ } } <\underset { 11 }{ { Na }^{ + } } <\underset { 9 }{ { F }^{ - } } <\underset { 8 }{ { O }^{ 2 } } <\underset { 7 }{ { N }^{ 3- } } \)

\(\mathrm{NH}_4 \mathrm{NO}_3(s) \longrightarrow \mathrm{N}_2 \mathrm{O}(g)+2 \mathrm{H}_2 \mathrm{O}(l) \)
\( \Delta H_{\text {Reaction }}=\Delta H_{\text {Products }}-\Delta H_{\text {Reactants }} \)
\( =\Delta H_{\mathrm{N}_2 \mathrm{O}}+2 \times \Delta H_{\mathrm{H}_2 \mathrm{O}}-\Delta \mathrm{H}_{\mathrm{NH}_4 \mathrm{NO}_3} \)
\( =81.46+2(-285.8)-(-367.57) \)
\( \Delta H=-122.560 \mathrm{~kJ} \)
\( \Delta H=\Delta E+\Delta n R T \)
\( -122560=\Delta E+1 \times 8.314 \times 298 \)
\( \Delta E=-125037 \mathrm{~J} \)
\(=-125.037 \mathrm{~kJ}\)

Let the total mass of the gaseous mixture = 100 g.
Mass of CO = 90.55 g
And, mass of CO₂= (100 – 90.55) = 9.45 g
Now, the number of moles of CO, nCO\(=\frac{90.55}{28}\)=3.234mol
Number of moles of CO­₂,nCO₂\(=\frac{9.45}{44}\)=0.215mol
The partial pressure of CO,
pCO=nCOnCO+nCO2×ptotal
\(= \frac{3.234}{3.234+0.215}×1 \)= 0.938 atm
Partial pressure of CO₂,
pCO₂ = nCO₂nCO+nCO₂×ptotal
\(=\frac{ 0.215}{3.234+0.215}×1\)=0.062atm
Kp=P2COPCO2 = (0.938)20.062=14.19
For the given reaction,
Δng= 2 –1 = 1
We know that,
KP=KC(RT)Δn
14.19= KC(0.082×1127)1
\(KC=\frac{14.19}{0.082×1127}\)
= 0.153

(a) 2, 2, 4-trimethylpentane 
Condensed formula: (CH₃)₂CHCH₂C (CH₃)₃
Bond line formula:

(b) 2-hydroxy-1, 2, 3-propanetricarboxylic acid
Condensed Formula:  (COOH)CH₂C(OH) (COOH)CH₂(COOH)
Bond line formula:

The functional groups present in the given compound are carboxylic acid (-COOH) and alcoholic (-OH) groups.
(c) Hexanedial Condensed Formula: (CHO) (CH₂)₄ (CHO)
Bond line Formula:

The functional group present in the given compound is aldehyde (-CHO).

According to de Broglie equation,\(\lambda =\frac { h }{ mv } \)
Mass of electron = 9.1 x 10^-31 kg; Planck's constant = 6.626 x 10^-34 kgm²s^-1
Velocity of electron = 1% of speed of light = 3.0 x 10⁸ x 0.01 = 3 x 10⁶ me-1
Wavelength of electron (\(\lambda \)) =\(\frac { h }{ mv } =\frac { (6.626x10^{ -34 }kgm^{ 2 }s^{ -1 }) }{ (9.1x10^{ -31 }kg)x(3x10^{ 6 }ms^{ -1 }) } \)
= 2.43 x 10^-10 m.

The increasing order is
HI < I₂ < ICI < HIO₂ < KIO₃

In case of hydrogen atom, there is no neutron. It consists of only one proton

This is because of intermixing of 2s and 2p_z orbitals because of their close proximity. Due to intermixing, \({ \sigma 2p }_{ z }\) molecular orbital becomes higher in energy than  \({ \pi 2p }_{ x }\) and  \({ { \pi 2p } }_{ y }\)  molecular orbitals.

Covalent bond is formed by mutual sharing of electrons. The shared pair of electrons present between the bonded atoms are called bond pairs of electrons and unshared pair (non-bonding) electrons are called the lone pairs of electrons. e.g., ammonia, NH₃ contains 3 bond pairs and 1 lone pair of electrons, whereas CH₄ contains only 4 bond pairs.

No, the oxidation number of an element in any species is an apparent charge on the atom which it appears to have acquired when all other atoms in the species are removed as ions.

Orbitals having same energy belonging to the same subshell.

This is the simplest form of chromatography. Here a strip of paper acts as an adsorbent. It is based on the principle which is partly adsorption. The paper is made of cellulose fibres with molecules of water adsorbed on them. This acts as stationary phase. The mobile phase is the mixture of the components to be identified prepared in a suitable solvent.

It is peroxyacetyl nitrate.

Catenation: The property to form chains or rings not only with single bonds but also with multiple bonds with itself is called catenation.
For example, carbon forms chains with (C - C) single bonds and also with multiple bonds (C = C or C \(\equiv \) C).

At high pressure, the real gas shows large deviations from ideal behaviour as the curves are far apart.