Question 1

Among Fe^ 3+ , Pb^ 2+ , Cu^ 2+ and Mn^ 2+ , identify the one that gets precipitated out while passing H_2S in presence of NH_4OH as group reagent. The highest possible oxidation state of the correspon

Accepted Answer

The group reagent H_2S in the presence of NH_4OH is used for the precipitation of Group IV cations in qualitative analysis. Among the given ions, Mn^ 2+ belongs to Group IV and is precipitated as MnS. Pb^ 2+ and Cu^ 2+ belong to Group II and are precipitated by H_2S in an acidic medium. Fe^ 3+ belongs to Group III and is precipitated as Fe(OH)_3 by NH_4OH in the presence of NH_4Cl. The metal corresponding to the precipitated ion is Manganese (Mn). The electronic configuration of Mn is [Ar] 3d^5

Question 2

Among Fe^ 2+ , Fe^ 3+ , Cr^ 2+ and Zn^ 2+ , the ion that shows positive borax bead test and with highest ionisation enthalpy is:

Accepted Answer

The borax bead test is given by transition metal ions that have unpaired electrons and form coloured compounds. Zn^ 2+ has a completely filled 3d^ 10 configuration, making it colourless, so it does not give a positive borax bead test. The remaining ions and their electronic configurations are: Fe^ 2+ : [Ar] 3d^ 6 Cr^ 2+ : [Ar] 3d^ 4 Fe^ 3+ : [Ar] 3d^ 5 The ionisation enthalpy of an ion refers to the energy required to remove an electron from it. Fe^ 3+ has a highly stable half-filled d-subshell

Question 3

A salt with few drops of conc. HCl gives apple green colour in flame test. The group precipitate of the salt is dissolved in acetic acid and treated with K_2CrO_4 to give yellow precipitate. When the

Accepted Answer

An apple green colour in the flame test indicates the presence of Ba^ 2+ ion. The group precipitate of Ba^ 2+ is BaCO_3, which dissolves in acetic acid to form barium acetate. Upon treatment with K_2CrO_4, it gives a yellow precipitate of barium chromate (BaCrO_4), confirming the presence of Ba^ 2+ . When the sodium carbonate extract is heated with conc. HNO_3 and ammonium molybdate, the formation of a canary yellow precipitate indicates the presence of phosphate (PO_4^ 3- ) ion. The canary yell

Question 4

A student has been given 0.314 g of an organic compound and asked to estimate Sulphur. During the experiment, the student has obtained 0.4813 g of barium sulphate. The percentage of sulphur present in

Accepted Answer

Given: Mass of organic compound = 0.314 g, Mass of BaSO _4 obtained = 0.4813 g. \%S = 32 233 0.4813 0.314 100 = 32 0.4813 100 233 0.314 = 1540.16 73.162 = 21.05\%

Question 5

In the Group analysis of cations, Ba ^ 2+ ~\&~ Ca ^ 2+ are precipitated respectively as

Accepted Answer

In Group V of cation analysis, both Ba²⁺ and Ca²⁺ are precipitated by adding Na₂CO₃ or (NH₄)₂CO₃ solution. Ba²⁺ precipitates as BaCO₃ (white precipitate) and Ca²⁺ precipitates as CaCO₃ (white precipitate). Both form insoluble carbonates under these conditions, making the answer carbonate & carbonate.

Question 6

Consider three metal chlorides x , y and z, where x is water soluble at room temperature, y is sparingly soluble in water at room temperature and z is soluble in hot water. x , y and z are respectivel

Accepted Answer

Metal chloride solubility characteristics: x must be water soluble at room temperature, y must be sparingly soluble at room temperature, and z must be soluble in hot water but not in cold water. CuCl₂ is highly soluble in water (x). AgCl is notoriously insoluble in water and sparingly soluble at any temperature, classifying it as sparingly soluble (y). PbCl₂ exhibits unusual solubility behavior—it is sparingly soluble in cold water but its solubility increases with temperature, making it soluble

Question 7

By usual analysis, 1.00 g of compound ( X ) gave 1.79 g of magnesium pyrophosphate. The percentage of phosphorus in compound (X) is : (nearest integer) (Given, molar mass in g mol ^ -1 : O =16, Mg =24

Accepted Answer

Given: 1.00 g of compound (X) produces 1.79 g of Mg₂P₂O₇. Molar mass of Mg₂P₂O₇ = 2(24) + 2(31) + 7(16) = 222 g/mol. Moles of Mg₂P₂O₇ = 1.79 / 222 = 0.00806 mol. Since each mole of Mg₂P₂O₇ contains 2 moles of P, moles of P = 2 × 0.00806 = 0.01612 mol. Mass of P = 0.01612 × 31 = 0.4997 g ≈ 0.50 g. Percentage of phosphorus = (0.50 / 1.00) × 100 = 50%.

Practical Chemistry — Previous Year Questions & Solutions