Nitrogen & Phosphorus Compounds - Free MCQ Practice Test with solutions,

In solid state, N₂O₅ exists as nitronium nitrate 
In   nitrogen is sp and in  , nitrogen is sp².

Pure phoshine is not inflammable but due to contamination of P₂H₄ and P₄ vapours, it becomes inflammabl.

Acidic Character of P oxyacids is due to P—H bonds in the acid. Correct reducing property order is

Number of P – O bonds = 12 
Number of pair of electron = 16 

1. Ammonia (NH₃):

   • The lone pair of electrons on nitrogen is readily available for protonation.
   • The resulting ammonium ion (NH⁴⁺​) is highly stable, making NH₃ the most basic.

2. Hydrazine (N₂H₄):

   • Hydrazine has two nitrogen atoms with lone pairs, but protonation forms the hydrazinium ion (N₂H₅⁺​), where the positive charge is shared between two nitrogens.
   • This reduces stability, so N₂H₄ is less basic than ammonia.

3. Hydroxylamine (NH₂OH):

   • The -OH group exerts an electron-withdrawing inductive effect (-I), decreasing the electron density on nitrogen.
   • This makes the lone pair less available for protonation, so NH₂OH is the least basic.

Final Order:

NH₃ ​> N₂​H₄ ​> NH₂​OH

PBr does not split in the same fashion. The anion is not possible due to large size of Br-atoms. Hence, it splits differently.

H₃PO₂, H₃PO₃ and H₃PO₄ are mono, di and tribasic acids respectively.

P₄O₁₀,POCI₃ and cone. H₂SO₄ are act as dehydrating agent.

Sodium and potassium nitrate on heating gives nitrite and oxygen but not NO₂

Option (B) is correct: the Hoffmann bromamide reaction (Hofmann degradation) converts a primary amide into the corresponding primary amine with loss of the carbonyl carbon.

The reaction for acetamide is:

CH₃CONH₂ + Br₂ + 4 NaOH → CH₃NH₂ + Na₂CO₃ + 2 NaBr + 2 H₂O

In this transformation the amide nitrogen becomes the amine nitrogen (CH₃NH₂) while the carbonyl carbon is removed and eventually appears as carbonate (Na₂CO₃). The required reagents are Br_{2 and NaOH.}

Brief note on other options: the Carbylamine reaction converts a primary amine to an isocyanide and does not convert an amide to an amine; the Gabriel phthalimide synthesis is a method to obtain primary amines from alkyl halides; Stephen's reaction reduces nitriles to aldehydes. None of these give methylamine directly from acetamide.

Therefore, (B) Hoffmann bromamide reaction is the correct choice.

Phosphinic acid is a monoprotic acid as it contains one P−OH bond while phosphonic acid is a diprotic acid as it contains two P−OH bonds.

Due to the presence of H in A and B , they show reducing properties.

i. Haber's Process

• This process is used for the manufacture of ammonia.

• Reaction: Nitrogen reacts with hydrogen under high temperature and pressure in presence of iron catalyst to form ammonia.

• Match: p. Manufacture of ammonia

ii. Le-blanc Process

•  

  1. Sodium chloride is reacted with sulfuric acid to form sodium sulfate and hydrogen chloride gas.

  2. Sodium sulfate is then reduced using carbon to form sodium sulfide.

  3. Sodium sulfide is reacted with calcium carbonate to finally get sodium carbonate.

• Sulfuric acid is used in this process.

• Match: q. Sulphuric acid is used

iii. Birkeland-Eyde Process

• This process is used for the manufacture of nitric acid.

• It involves passing air through an electric arc to produce nitrogen oxides, which are further processed to get nitric acid.

• Match: r. Manufacture of nitric acid

iv. Solvay Process

• Another process used to manufacture sodium carbonate, but instead of sulfuric acid, it uses ammonia and carbon dioxide.

• It's a more modern and efficient process compared to the Le-blanc process.

• Match: s. Manufacture of sodium carbonate

N₂O₅ , N₂O and NH₄NO₃ have no bond between nitrogen atoms while others have.

Cr, Fe, Ni and Al are passive towards cone . HNO₃.

Equivalent weight, E₁ (oxidation) = M/4 and
Equivalent weight, E₂ (reduction) 
= M /4 
Equivalent wight of H₃PO₂ in the disproportionation reaction

Molecular nitrogen is less reactive than that of oxygen because nitrogen has high dissociation energy in comparison to oxygen so, its reactivity is less. Also bond length of nitrogen is shorter than oxygen because of the presence of triple bond between nitrogen atoms. So, both are true but not correct explanation.  

Due to e^- withdrawing capacity of fluorideion, it withdraws. from nitrogen in NF₃ make it weakerlig and while presence of e^- donating methyl group makes the nitrogen in N(CH₃)₃ a strong ligand. In aqueous medium, NF₃ furnishes fluorideion.