The acidity of a compound depends on the stability of the conjugate base. An acetic acid conjugate base is resonance stabilized hence it is the most acidic. 

The phenol is more stable than alcohol and water because the phenol conjugate base is resonance stabilized. 

The structure is as follows:
 

The conjugate base of water is more stable than ethanol because CH₃CH₂ group destabilized the negative charge on the oxygen atom. 

The stability of the conjugate base is as follows: 

The acidity order is as follows:
e 
 

In the Lucas test, primary, secondary, and tertiary alcohols are distinguished on the basis of their reactivity with conc. $HCl + ZnCl_2$ , known as Lucas Reagent. 

Tertiary alcohol gives immediate white turbidity with Lucas reagent, secondary alcohol gives white turbidity in 5 minutes, and primary alcohol does not react with Lucas reagent at room temperature. 

The general reaction is as follows:

Hence,  Statement I is correct but statement II is incorrect. 

NaBH₄ is a mild reducing agent and reduces. Aldehyde into the primary alcohol and ketone into the secondary alcohol. In the given compound ketone group is present, thus, NaBH₄ reduces the ketone group into an alcohol group. 

In the second step, the molecule undergoes a dehydration reaction, and an alkene is obtained as a product. 

The reactions are as follows:

Thus, option first is the correct answer. 
 

Explanation:
The first reaction is a reduction reaction. When acid reacts with a reducing agent then alcohol is obtained as a product. 
Here, B₂H₆ and LiAlH₄ can be used as reducing agents. 

The second reaction is also known as the hydroboration-oxidation reaction. In this reaction, alkene reacts with diborane gives trialkyl boranes as addition product. This is oxidized to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide. 

Thus, in both reactions X is B₂H₆. 

Thus, an option first is the correct answer. 
 

When primary alcohol reacts with a strong oxidizing agent then an acid is formed as a product. When primary alcohol reacts with the mild oxidizing agent then aldehyde is formed as a product. 

Here, CrO₃ – H₂SO₄, K₂Cr₂O₇ + H₂SO₄, and KMnO₄ + KOH/H₃O⁺ are strong oxidizing agents thus, converts alcohol into acid. 

But Cu, 573 K, and CrO₃₊ (CH₃CO)₂O are the weak oxidizing agent and reduce primary alcohol to aldehyde. 

Thus, option third is the correct answer.