Explore topic-wise InterviewSolutions in Current Affairs.

(i) Charge is quantised that is charge Present on any object will always be the integral multiple of charge present on each electron.

Q = ne

(ii) Charge is conserved

According to Krichhoff's junction rule Total incoming current = Total outgoing current 2A + 7A + 5A = 3A + i

14A = 3A + i

i = 14A - 3A = 11A.

Correct answer is (a,c,d)

Correct answer is (a,b)

Correct answer is (a,b,d)

(i) (c)

(ii) (b)

(iii) (b)

(iv) (a)

Correct answer is (B) 0.1 Molal

Correct answer is (A) Mol/litre

(b) It forms a ppt which is soluble with alkali

It forms a ppt which is soluble with alkali only in case of primary amines.

The compounds will act as strongest reducing agent are given below

(d) SbH₃

(b) negative

Aldhyde compound vapour is passed through Cu-tube at 300°C produces.

(b) exothermic compound 

Sodium is heated with ammonia gas, the produced compound is NaNH₂

Answer is (d) Clemensen's Reaction

Tollen’s reagent is used for detecting Aldehyde.

Dettol consists of   Chlroxylenol + terpeneol

Lanthanoids 

1. Differentiating electron enters in 4f orbital 

2. Binding energy of 4f orbitals are higher 

3. They show less tendency to form complexes 

4. Most of the lanthanoids are colourless 

5. They do not form oxo cations

6. Besides +3 oxidation states lanthanoids show +2 and +4 oxidation states in few cases.

Actinoids 

1. Differentiating electron enters in 5f orbital 

2. Binding energy of 5f orbitals are lower 

3. They show greater tendency to form complexes

 4. Most of the actinoids are coloured. 

5. E.g : U³⁺ (red), U⁴⁺ (green), UO₂²⁺ (yellow) 

6. They do form oxo cations such as UO₂²⁺ NpO₂²⁺ etc. 

7. Besides +3 oxidation states actinoids show higher oxidation states such as +4, +5, +6 and +7

The compound [X] is  R – NC

In the Ellingham diagram, the formation of carbon monoxide is a straight line with negative slope. In this case AS is positive as 2 moles of CO gas is formed by the consumption of one mole of oxygen. Hence, CO is more stable at higher temperature.

Metallic oxides such as Cr₂O₃ can be reduced by an aluminothermic process. In this process, the metal oxide is mixed with aluminium powder and placed in a fire clay crucible. To initiate the reduction process, an ignition mixture (usually magnesium and barium peroxide) is used.

BaO₂ + Mg ⟶ BaO + MgO

During the above reaction a large amount of heat is evolved (temperature upto 2400°C, is generated and the reaction enthalpy is - 852 kJ mol) which facilitates the reduction of Cr₂O₃ by aluminium power.

Cr₂O₃ + 2Al + Δ ⟶ 2Cr+ Al₂O₃ 

Answer is (d) Schiff's base

Nitrogen has a triple bond between its two atoms because of its small size and high electronegativity. Phosphorus P₄ has single bond, that is why it is tetraatomic.

Inert gases have very high ionisation enthalpy because they have stable electronic configuration. Therefore they have no tendency to loss electrons. Thus the valency of inert gases are zero.

Standard electrode potential: The potential of an electrode. When the concentration of all the species involved in a half cell is unity is known as standard electrode potential. It is denoted by E°.

E = E° when [inⁿ⁺] = 1 mol litre^-1

• The nature of adsorbate can influence the adsorption. Gases like NH₃, CO₂ are easily liquefiable as have greater Van der Waals forces of attraction and hence readily adsorbed due to high critical temperature. 
• But permanent gases like H₂ . N₂ can not be easily liquefied and having low critical temperature and adsorbed slowly.

I. Lewis acids: 

(i) Lewis acid is a species that accepts an electron pair. 

(ii) Lewis acid is a positive ion (or) an electron deficient molecule. 

(iii) Example, Fe²⁺ , CO₂ , BF₃ , SiF₄ etc… 

II. Lewis bases: 

(i) Lewis base is a species that donates an electron pair. 

(ii) Lewis base is an anion (or) neutral molecule with atleast one lone pair of electrons. 

(iii) Example, NH₃ , F^- , CH₂ = CH₂ CaO etc….

In the molten state, ionic solids dissociate to give free ions and hence can conduct electricity. However in the solid state, as the ions are not free to move but remain held together by strong electrostatic forces of attraction, so they cannot conduct electricity in the solid state.

Acid rain:

The moisture or rain water containing harmful acids falls and collects itself in tank, ponds, lakes etc. This is called acid rain. These pollutants due SO₂, NO₂ etc. dissolves in water.

SO₂ + H₂O → H₂SO₄

SO₃ + H₂O → H₂SO₄

3NO₂ + H₂O → 2HNO₃ + NO↑

Two important uses of formalin.

(i) It is used to preserve biological specimens.

(ii) It is used for the synthesis of bakelite.

3d and 4d transition series contains only ten elements because d orbits can contain maximum ten electrons.

1. Diborane can be conveniently prepared in the laboratory by the oxidation of sodium borohydride with iodine. 

2NaBH₄ + I₂ → B₂H₆ + 2NaI + H₂ 

2. In BCl₃ , the central boron atom contains only six electrons. Hence, it has the tendency to accept electrons and acts as a Lewis acid.

HCl is stronger acid because it has big size and smaller electronegativity than HF.

1. S-H bond energy is less than that of O-H bond energy. So H⁺ can be easily generated from H₂S.

2. When HCl is treated with H₂O it undergoes hydrolysis as per the following reaction and dissolves. 

HCl + H₂O → H₃O + Cl^-

Fe is a d-orbital element and show properties of transition metals like variable valency, coloured ion, complex formation etc.

But sodium is s-block element. It does not show the properties of transition metal.

K₃ [Fe(CN)₆] → Potassium Hexa-cyanoferrate (III)