Class 12th

Initially, after protonation followed by loss of water, secondary carbocation is formed, further Hydride shift leads to 3° carbocation.

Homoleptic complexes have all ligands identical. Potassium trioxalatoaluminate (III) is K? [Al (C? O? )? ] which has only oxalate ion as ligand. All others have more than one type of ligands.

The greater stability of Cu²? (aq) rather than Cu? (aq) is due to the much more negative ΔhydH of Cu²? (aq) than Cu? , which more than compensates for the second ionisation enthalpy of Cu.

BF? behaves as Lewis acid due to incomplete octet in valence shell of Boron. Hence it can accept a lone pair of electrons.

When the reactants and the catalyst are in different phases, then the catalysis is known as heterogenous catalysis.

1. N? (g) + 3H? (g) - (Fe (s)-> 2NH? (g) (Heterogenous catalysis)

2. 2SO? (g) + O? (g) - (NO (g)-> 2SO? (g) (Homogenous catalysis)

3. C? H? O? (aq) + H? O (l) → C? H? O? (aq) + C? H? O? (aq) (Homogenous catalysis)

4. NO (g) + O? (g) → NO? (g) + O? (g)

A reaction can have zero activation energy, for example, •CH? + •CH? → CH? − CH? has zero activation energy.

Let the new rate be, R'.

A' = [3 A], B' = [B]

R' = K [A']² [ B']

= K [3 A]² [ B]

= 9 K [ A]² [ B]

= 9 * R

Hence, rate would become nine times.

Free energy is extensive property, and E°cell is an intensive property. ΔrG is depends on the ' n ' which is number of electron transferred in the reaction ΔrG = −nFE°

Conductivity = Conductance * Cell constant

Conductivity = (1/ resistance )

Cell constant = Conductivity * Resistance

Cell constant = 0.0210 * 60 = 1.26Cm? ¹

Since, atom B forms CCP structure. Therefore, there will be 4-B atoms.

Now, atom A occupies 1/3 of tetrahedral voids.

Hence, number of A atoms = 1/3 * 8 = 8/3

The correct formula of the compound = A? /? B?

= A? B?

x + y = 2 + 3 = 5