Solution

Set of d-orbitals which is used by central metal during formation of  $$MnO_4^-$$ is $$d_{xy}, d_{yz}, d_{xz}$$.
In spherical field of four oxide ligands, the five d orbitals are degenerate.
They split into $$t_{2g}  and  e_g$$ levels in tetrahedral field.
The three $$t_{2g}$$ orbitals mix with 4s orbital to form $$d^3s$$ hybrid orbitals.

SOLUTION

For electrolysis of Al₂O₃
(2/3)Al₂O₃ → (4/3)Al + O₂
Oxidation state of O₂ in Al₂O₃ is -2 and reduce to 0 In left hand side of above equation number of 'O' are 6
Thus (2/3)× 6 ×(-2)=4 electrons are accepted by Oxygen or n=4
Now ΔG=-nFE
-827,000=-4 × 96500 × V
V=2.14 V

When methane gas is treated with chlorine in the presence by sunlight, one hydrogen of methane replaced by the chlorine atom and forms methyl chloride. The mechanism involved in this reaction is free radical mechanism. So it is an example of free radical substitution reaction.

Solution

Acetophenone being ketone does not react with Tollen's reagent to give silver mirror

Solution

Phenol reacts with bromine in carbon disulphide at low temperature to give. In nonpolar solvents or solvents of low polarity phenol does not ionize to yield phenoxide ion. Therefore, electophile Br+ has no option but to attack the less active benzene ring of phenol. 

Reaction can be limited to monohalogenation if halogenation is carried out in non polar solvant like CS2, CCl4 or in a solvent of low polarity such as CHCl3 at a low temperature 

In nonpolar solvents or solvents of low polarity phenol does not ionize to yield phenoxide ion. Therefore, electophile Br+Br+ has no option but to attack the less active benzene ring of phenol.

SOLUTION

The most stable free radical is C₆H₅CHCH₃ because of resonance.

SOLUTION

In bcc
distance closest approach = a(√3/2)
173 = a(√3/2) a ≈200 pm