The aromatic compounds apparently contain alternate double and single bonds in a cyclic structure, and resemble benzene in chemical behaviour. They undergo substitution reactions rather addition reactions. This characteristic behaviour is called Aromatic character or Aromaticity.
Conditions for Aromaticity:
(i) An aromatic compound is cyclic and planar.
(ii) Each atom in an aromatic ring has a p-orbital. These p-orbitals must be parallel so that a continuous overlap is possible around a ring.
(iii) The cyclic \(\pi\)-molecular orbital formed by the overlap of p-orbitals must contain (4n + 2) \(\pi\) electrons, where (n = 0, 1, 2, 3, 4 etc.)

The hydroxide formed in aqueous solution is Al(OH)₃

Sodium peroxide It is obtained by heating sodium metal (Which is obtained ny the electrolysis of molten NaCl), in excess of air. The initially formed sodium oxide reacts with more O2 to form Na₂O₂.   
4Na + O₂ \(\xrightarrow{\Delta}\)  2 Na₂O₂ 2Na₂O;
2Na₂O + O₂ + O₂\(\xrightarrow{\Delta}\)  2 Na₂O₂ ;

Kala chana , soya bean and sprouted lentils are the sources of iron potassium and proteins

Yes, because these are unsaturated fats and hence,good for health. They do not form cholesterol if taken in small quantity.

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

Use the following steps to solve out such problems.
Step I   Write the balanced equation
\({ NH }_{ 2 }CN(s)+\frac { 3 }{ 2 } { O }_{ 2 }(g)\rightarrow { N }_{ 2 }(g)+{ CO }_{ 2 }(g)+{ H }_{ 2 }O(l)\)
Step II  Calculate \({ \Delta n }_{ g }\)
Difference of moles of gaseous products and
reactants, \({ \Delta n }_{ g }={ n }_{ p }-{ n }_{ r }=2-\frac { 3 }{ 2 } =\frac { 1 }{ 2 } =0.5mol\)
Step III Calculate \(\Delta H\) by using the formula
\(\Delta H=\Delta U+{ \Delta n }_{ g }RT\)
\(\Delta H=-742.7 \ kj \ { mol }^{ -1 }\)
\( +(0.5mol\times 8.314\times { 10 }^{ -3 }kj \ { mol }^{ -1 }\times 298K)\)
During calculation always remember the units of different quantities, i.e. 
\(\Delta H,\Delta U\) and R must be the same.

According to Boyle's law,  \({ p }_{ 1 }{ V }_{ 1 }={ p }_{ 2 }{ V }_{ 2 }\)
\(1bar\times 2.27L=0.2bar\times { V }_{ 2 }\)
\(or \ { V }_{ 2 }=\frac { 1bar\times 2.27L }{ 0.2bar }\)
\( =11.35L\)
Since, balloon burst at 0.2 bar pressure, the volume of balloon should be less than 11.35 L.

(i) Group 1 elements are most reactive.
(ii) Group 18 elements are least reactive because they have stable electronic configuration.
(iii) It may cause high blood pressure and any other harm to the body. It disturbs the person physically and mentally also.
(iv) They are cool, calm, happy and help other people. They are more healthy than aggressive people.
(v) aggressive people pick up fight. They do not accept their fault. Sometimes, it leads to physical injury which put them in trouble. We should not involve in such act.

(i) Representation of electronic configuration is excluded by Hund's rule since electrons are not shown in separate orbitals of the same subshell with parallel spins.
(ii) Representation of electronic configuration is correct.
(iii) Representation of electronic configuration is excluded by the Aufbau principle  because the 2s subshell fills before the 2p.
(iv) Representation of electronic configuration is excluded by the Pauli exclusion principle since the two electrons in the 2s-orbital cannot have the same spin.

\(NaOH+{ HNO }_{ 3 }\longrightarrow { NaNO }_{ 3 }+{ H }_{ 2 }O\) 
Applying molarity equation
M₁V₁(acid) = M₂V₂ (base); \({ M }_{ 1 }\times 30=0.5\times 40\)
\({ M }_{ 1 }\times 30=20; \ { M }_{ 1 }=\frac { 20 }{ 30 } \) 

Let s be the solubility of Ag₂CrO_4.
 

Let be the solubility of AgBr.
 

Therefore, the ratio of the molarities of their saturated solution is 

Pass them through dilute cold KMnO₄ solution (purple) or Br₂ in CCl₄ solution (red). Propene will decolourise both the solutions but propane does not react.

(i) In solid form they are (anions & cations) closely packed, that is why they are not able to move.
(ii) In water they aquire mobility and can conduct electricity.
(iii) By this property these compounds can be applied in electroplating of all types of cells.

(a)

(b)

(c)

Protium - \(^1_1H\)
Deuterium - \(^2_1H\) or D
Tritium - \(^3_1H\) or T
Mass ratio of Protium: Deuterium: Tritium
= 1: 2: 3

By conventional method. CH₃COOH = 2x + 4 - 4 = 0 or x = 0
By chemical bonding method, C2 is attached to three H-atoms (less electronegative than carbon) and one-COOH group (more electronegative than carbon).
\(H\overset { 2 }{ - } \overset { \overset { H }{ I } }{ \underset { \overset { I }{ H } }{ C } } -\overset { \overset { 0 }{ II } }{ C } -OH\)
therefore, O.N. of C₂ = 3 (+1) + x + 1 (-1) = 0 or x = -2
C1 is, however, attached to one oxygen atom by a double bond, one .O.H (O.N. = -1) and one CH₃ (O.N. = +1) group, therefore, O.N. of C₁ = + 1 + x + 1
(-2) + 1 (-1) = 0 or x = +2

According to Dalton's atomic theory, an atom is the ultimate particle of matter which cannot be further divided.

For exhibiting cis-trans (or geometrical isomerism, a molecule must fulfil the following condition.
The groups attached to each double bonded carbon atom must be different.

The expression for the equilibrium constant, K_C for each of the following reactions are:
(a) \({ K }_{ c }=\frac { \left[ NO \right] ^{ 2 }\left[ { Cl }_{ 2 } \right] }{ \left[ NOCl \right] _{ 2 } } \)
(b) \({ K }_{ c }=\left[ { No }_{ 2 } \right] ^{ 4 }\left[ { O }_{ 2 } \right] \)
(because molar concentrations of pure solids are constant)
(c) \({ K }_{ c }=\frac { \left[ CH_{ 3 }COOH \right] \left[ { C }_{ 2 }{ H }_{ 2 }OH \right] }{ \left[ { CH }_{ 3 }COO{ C }_{ 2 }{ H }_{ 5 } \right] \left[ { H }_{ 2 }O(l) \right] } \)
(d) \({ K }_{ c }=\frac { 1 }{ \left[ { Fe }^{ 3+ } \right] .\left[ { OH }^{ 1 } \right] ^{ 3 } } \) because [Fe(OH)₃ (s) = 1]
(e) \({ K }_{ c }=\frac { \left[ { IF }_{ 5 } \right] ^{ 2 } }{ \left[ { F }_{ 2 } \right] ^{ 5 } } \)(because [I₂(s) = 1])

For one mole of an ideal gas, \({ C }_{ V }=\left( \frac { \triangle U }{ \triangle T } \right) _{ V }\)
\(or \ \triangle U={ C }_{ V }\triangle T\)
For an isothermal process, T is constant so that \(\triangle T=0\)
\(\therefore \)  \(\triangle U=0\)

(a) The electronic configuration of an element having 5 electrons in its outermost subshell should be ns²np⁵. This is the electronic configuration of the halogen group. Thus, the element can be F, Cl, Br, I, or At.
(b) An element having two valence electrons will lose two electrons easily to attain the stable noble gas configuration. The general electronic configuration of such an element will be ns². This is the electronic configuration of group 2 elements. The elements present in group 2 are Be, Mg, Ca, Sr, Ba.
(c) An element is likely to gain two electrons if it needs only two electrons to attain the stable noble gas configuration. Thus, the general electronic configuration of such an element should be ns²np⁴. This is the electronic configuration of the oxygen family.
(d) Group 17 has metalloid, non–metal, liquid as well as gas at room temperature. F and Cl are the gaseous states, Br is liquid state and I is solid state. Here, the F, Cl, Br and I are non-metals whereas, At is metalloid.

(a) ₁₅P = 1s², 2s², 2p⁶, 3s², 3p³. 3 unpaired electrons.
(b) ₁₄Si = 1s², 2s², 2p⁶, 3s², 3p². 2 unpaired electrons.
(c) ₁₄Cr = 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁵, 4s¹. 6 unpaired electrons.
(d) ₂₆Fe = 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁶, 4s². 4 unpaired electrons
(e) ₃₆Kr = 1s², 2s², 2p⁶, 3s², 3p⁶, 3d¹⁰, 4s², 4p⁶. No unpaired electrons.

Energy in second level, E₂ =-\(\frac { x }{ 2^{ 2 } } \)=-\(\frac { x }{ 4 } \)
Energy in third energy level, E₃= \(=-\frac { x }{ 3^{ 2 } } =-\frac { x }{ 9 } \)
Energy required for an electron to jump,
\(\left( E_{ 3 }-E_{ 2 } \right) =-\frac { x }{ 9 } +\frac { x }{ 4 } =\frac { 5x }{ 36 } \)

It is the phenomenon of an atom to form a strong covalent bond with the atoms of itself. Carbon shares the property of catenation to maximum extent because it is small in size and can form P\(\pi\)-P\(\pi\) multiple bonds to itself.

(a) Surface tension: It is defined as the force acting per unit length perpendicular to the line drawn on the surface. It's unit is Nm^-1.
(b) Surface tension of a liquid depends upon following factors.
(i) Temperature: Surface tension decreases with rise in temperature. As the temperature of the liquid increases, the average kinetic energy of the molecules increases. Thus, there is a decrease in intermolecular force of attraction which decreases the surface tension.
(ii) Nature of the liquid: Greater the magnitude of intermolecular forces of attraction in the liquid, greater will be the value of surface tension.

Due to small size, lithium does not form alums

(i) An aldehyde with molar mass of 44 u is ethanal, CH₃CH = O
(ii) Write two moles of ethanal side by side with their oxygen atoms pointing towards each other.
CH₃CH = O      O = CHCH₃
   Ethanal            Ethanal
(ii) Remove the oxygen atoms and join them by a double bond, the structure of alkene 'A' is

As required, but-2-ene has three C-C, eight C-H \(\sigma \)-bonds and one C-C \(\pi\)-bond.

(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).

392 g of sulphuric acid
Molar mass of H₂SO₄ = 2 x 1 + 32 + 4 x 16 = 98 g
98 g of sulphuric acid = 1 mol
392 g of sulphuric acid = 1 mol x\(\frac { 392 \ g }{ (98 \ g) } \) = 4 mol

q = 2.303 nRT log \(\frac { { V }_{ 2 } }{ V_{ 2 } } \) = 2.303\(\times \frac{10}{40}\times log\frac{5}{10}\)
=-103.635 cal
For isothermal expansion, \(\Delta\)U = 0  ...........(i)
W = \(\Delta\)U - q = 0 - (-103.635)
= +103.635 cal 
Also, when temperature is constant,  
p₁V₁ = p₂V₂ or pV = constant
\(\Delta\)H = \(\Delta\)E + \(\Delta\)(pV) = 0 + 0 = 0 ..........(ii)