Test : Chemistry Class 12 Mini Mock - 5 NEET MCQs & solutions - Free

a) Any aldehyde gives secondary alcohol on reduction

• Aldehydes (RCHO) are reduced to primary alcohols (RCH₂OH) using reducing agents like NaBH₄ or LiAlH₄. Secondary alcohols are obtained from ketones, not aldehydes.
• Incorrect.

b) Reaction of vegetable oil with H₂SO₄ gives glycerine

• Vegetable oils are triglycerides. Hydrolysis (saponification) with a base (like NaOH) or enzymatic processes yields glycerine (glycerol) and fatty acids. H₂SO₄ is not typically used for this purpose; it may cause other reactions like sulfonation.
• Incorrect.

c) C₂H₅OH, iodine with NaOH gives iodoform

• The iodoform test requires a compound with a methyl ketone group (RCOCH₃) or a compound that can be oxidized to one, like ethanol (C₂H₅OH). Ethanol reacts with I₂ and NaOH to form iodoform (CHI₃) via oxidation to acetaldehyde and further reaction.
• Correct.

d) Sucrose on reaction with NaCl gives invert sugar

• Sucrose hydrolyzes to invert sugar (a mixture of glucose and fructose) under acidic conditions (e.g., with HCl) or enzymes (invertase). NaCl, a neutral salt, does not catalyze this reaction.
• Incorrect.

Correct answer: c) C₂H₅OH, iodine with NaOH gives iodoform.

Percentage volume of oxygen = 21%. (Given)
∵ 100 ml of air contains = 21ml of O₂
∴ Volume of oxygen in one litre of air

Therefore no. of moles 

(∵ volume of 1 litre of gas at S.T.P. is 22400 ml)

Aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis as it  is used for the preparation of aliphatic primary amines and not aromatic primary amines.

When phthalimide is treated with aqueous or ethanolic potassium hydroxide, it forms its potassium salt. This salt, when heated with an alkyl halide and then subjected to alkaline hydrolysis, yields the corresponding primary amine.

Aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.


So, the assertion is wrong.
Reason: 
Aryl halides do not undergo nucleophilic substitution with anion formed by phthalimide.

So, the reason is wrong.

Hence, both assertion and reason are wrong.

The correct answer is option B.

The I.U.P.A.C. name for Fe(C₅H₅)₂ is bis(η⁵ -cyclopentadienyl) iron(II) 
The oxidation state of iron is +2 and is written in parenthesis in roman numerals. Two cyclopentadienyl ligands are coordinated to Fe. The prefix bis indicates 2. η⁵ indicates that the cyclopentadienyl ligands are penta coordinates.

 Solid iodine is a molecular solid.

The correct answer is option A

Kohlrausch's law states that the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the conductances of the anions and cations. If a salt is dissolved in water, the conductivity of the solution is the sum of the conductances of the anions and cations.
Hence, at infinite dilution the ionic conductivity of ions is additive.

Reduction potential means to accept electrons to reduce oneself.
 A + e^- → A^- ∆E_reduction = +ve value
Since, the reduction potential is negative, it means that the reaction will reverse to make ∆E value +ve. So the reaction becomes,
A → A⁺ + e^- 
This becomes oxidation of A. So oxidation of A will be easy. 

Catalyst: Substances which alter  the rate of a chemical reaction and themselves remain chemically and quantitatively unchanged  after the reaction are known as catalysts and the phenomenon is known as catalysis.

La (OH)₃ is more basic than Li (OH)₃. In lanthanides the basic character of hydroxides decreases as the ionic radius decreases.

Photography is an example of photochemistry used in our daily life.

In DNA, four bases have been found. They are adenine (A), guanine (G), cytosine (C) and thymine (T). The first three of these bases are found in RNA also but the fourth is uracil (U). RNA contains cytosine and uracil as pyrimidine bases while DNA has cytosine and thymine. Both have the same purine bases i.e. guanine and adenine.

Based on electro chemical series, oxidising power of
F₂ > Cl₂ > Br₂ > I₂

On passing Cl₂ in to a solution containing KF, Kl and KBr,
2KBr + Cl₂ → 2KCI + Br₂ (orange)
2KI + Cl₂ → 2KCI + l₂ (violet)
2KI + Br₂ → 2KBr + l₂ (violet)
Br₂ formed also oxidises Kl to l₂ (violet)

The carbon with which the Br is bonded with another 3 carbon atoms. So haloalkane is Tertiary Haloalkane.

E_a  >  ΔH

 (IV) is most reactive as it is benzylic as well as electron withdrawing effect of — NO₂ further increases the reactivity. (Ill) is least reactive due to resonance effect resulting in partial double bond character between carbon and chlorine. (I) is less reactive in S_N2 reaction than II, due to greater steric hindrance in (I).

CFSE for [CoCI₄]^2- will be 
i.e.  = 8000cm^-1

Alcohols have higher boiling points than aldehydes.

Se is non-conductor in dark but acts as conductor when exposed to light.

The best method for the separation of naphthalene and benzoic acid from their mixture is sublimation because it is applicable for those organic compounds which pass directly from solid to vapour state on heating and vice versa on cooling.
In these compounds naphthalene is volatile and benzoic acid is non-volatile due to the formation of dimer via hydrogen bonding (intermolecular).

• Ketones are organic compounds featuring a carbonyl group (C=O) bonded to two carbon atoms.
• During catalytic hydrogenation, the carbonyl group is reduced, adding hydrogen atoms.
• This process converts the ketone to an alcohol.
• The carbonyl carbon in a ketone is attached to two carbon groups, resulting in a secondary alcohol upon reduction.
• Therefore, the reduction of ketones yields secondary alcohols.
• Option A: secondary alcohols is the correct answer.

Urea (46.6%N). % of N in other compound are : ( NH₄)₂ SO₄ = 21.2%; CaCN₂ = 35.0% and NH₄ NO₃ = 35.0%

The Lanthanides are transition metals from Atomic numbers 58 (Ce) to 71(Lu).
Hence the electron configulation becomes : (n –2) f ^{1– 14} (n – 1) s²p⁶ d^{0 – 1} ns².

According to Raoult's law, the relative lowering in vapour pressure of a dilute solution is equal to the mole fraction of the solute present in the solution.

 mole fraction of solute

This reaction is not proceed in forward direction.

By the following equilibrium

Cyanide ion is transferred to organic phase otherwise KCN is insoluble in organic solvents.

From the given options we find option (a) is correct. The oxidising power of halogens follow the order F₂ > Cl₂ > Br₂ > I₂ . Option (b) is incorrect because it in not the correct order of electron gain enthalpy of halogens. The correct order is Cl₂ > F₂ > Br₂ > I₂ . The low value of F2 than Cl2 is due to its small size. Option (c) is incorrect. The correct order of bond dissociation energies of halogens is Cl₂ > Br₂ > F₂ > I₂ . Option (d) is correct. It is the correct order of electronegativity values of halogens. Thus option (b) and (c) are incorrect.

Stratosphere layer strongly absorb harmful UV-radiations (λ. 255 nm).

or  K_c = Antilog 15.57 = 3.7 × 10¹⁵ ≈ 4 × 10¹⁵

The cyclopentadienyl ligand, C₅H₅^-, is a negative ligand commonly used in organometallic chemistry. It is notably found in the compound ferrocene.

This makes option B the correct answer.