Van der Waals forces are weak intermolecular forces that are dependent on the distance between atoms or molecules. These forces arise from the interactions between uncharged atoms/molecules.

For example, Van der Waals forces can arise from the fluctuation in the polarizations of two particles that are close to each other

Electron gain enthalpy of an element may be defined as the energy released when a neutral isolated gaseous atom accepts an extra electron to form the gaseous negative Ion i.e. anion.It is denoted by

 Δ eg H.

Greater the amount of energy released in the above process, higher is the electron gain enthalpy of the element. The electron gain enthalpy of a element is a measure of the firmness or strength with which an extra electron is bound to it.

Electron gain enthalpy is measured in electron volts per atom or kJ per mole.

The process of adding an electron to the atom can be either exothermic or endothermic.

Energy is released when an electron is added to the atom. Therefore, the electron gain enthalpy is negative.

The electron gain enthalpy for halogens is highly negative because they can acquire the nearest stable noble gas configuration by accepting an extra electron.

Noble gases have large positive electron gain enthalpy because the extra electron has to be placed in the next higher principal quantum energy levels there by producing highly unstable electronic configuration.

After the addition of 1 electron, the atoms becomes negatively charged and the second electron is to be added to a negatively charged Ion. But the addition of second electron is opposed by the electrostatic repulsion and hence the energy has to be supplied for the addition of second electron. The second electron gain enthalpy of an element is positive.

When an electron is added to oxygen atom to form  O¯ ion ,energy is released.But when another electron is added to O¯ ion to form O2- ion, the energy is absorbed to overcome the strong electrostatic repulsion between the negatively charged  O¯ ion and second electron being added.

Phosphorus is a 15th group element and it forms the most stable pentavalent compound like P2​O5​. because available of d-electrons. Also, on moving down the group, the stability of the +5 oxidation state decreases. 

Thus, phosphorus form the most stable pentavalent compounds.

Oxidation States

• The number allotted to an element in a compound representing the number of electrons lost or gained by an atom of the element of the compound is called oxidation state.

For example, the electron configuration of copper is [Ar] 3d¹⁰ 4s¹. It attains noble gas configuration by losing one electron. The energy required to lose one more electron is very less and hence copper loses 2 electrons and forms Cu²⁺ ion. Therefore copper exhibits +1 and +2 oxidation state. But +2 oxidation states are more common.

It forms compounds like CuCl₂ and also with oxygen like CuO. In both the cases the oxidation state of Cu is +2.

• Transition elements exhibit varying oxidation states due to the minor energy difference between ns and (n -1) d orbitals.
• Along with ns electrons, (n -1) d electrons takes part in bonding. But due to the availability of few electrons for bonding Scandium does not show variable oxidation states.
• Due to presence of more d electrons, zinc has less orbital available for bonding and hence does not exhibit varying oxidation state.
• Among d-block elements the elements belonging to 8^th group exhibit maximum oxidation state.
• Among the elements of 3d –series Manganese belonging to 7^th group exhibits maximum oxidation state.
• Among the elements of 4d-Series Ruthenium belonging to 8th group exhibits maximum oxidation state.
• Among the elements of 5d-Series Osmium belonging to 8th group exhibits maximum oxidation state.
• The oxidation number of a free element is always 0.
• Oxidation number of (group I) elements like Li, Na, K, Rb, Cs is +1.
• Oxidation number of (group II) elements like Be, Mg, Ca, Sr, Ba is +2.
• Oxidation number of oxygen is -2.
• For example, oxidation state of Phosphorous in the compound HPO₃^2- can be calculated by the following method:

Oxidation state of H = +1

Oxidation state of O = -2

Oxidation state of O₃ = 3(-2)  [Since it has 3 atoms of oxygen.]

Overall oxidation state of the compound = -2

Let P represent the oxidation state of Phosphorous.

Therefore,

HPO₃^2- = +1+P+3(-2) = -2

• P = +3

Given that,

Time = 25 min

Percentage = 25 %

A first order reaction takes 25 minutes for 25% decomposition

We need to calculate the value of k

Using formula of k

Put the value into the formula

When 75% of the reaction will be completed.

We need to calculate the value of 

Using formula for 

Put the value into the formula

Hence, The time is 120.5 min.

According to the valence bond theory, Electrons in a molecule occupy atomic orbitals rather than molecular orbitals. The atomic orbitals overlap on the bond formation and the larger the overlap the stronger the bond.

Applications of Valence Bond Theory

• The maximum overlap condition which is described by the valence bond theory can explain the formation of covalent bonds in several molecules.
• This is one of its most important applications. For example, the difference in the length and strength of the chemical bonds in H₂ and F₂ molecules can be explained by the difference in the overlapping orbitals in these molecules.
• The covalent bond in an HF molecule is formed from the overlap of the 1s orbital of the hydrogen atom and a 2p orbital belonging to the fluorine atom, which is explained by the valence bond theory.

Limitations of Valence Bond Theory

The shortcomings of the valence bond theory include

• Failure to explain the tetravalency exhibited by carbon
• No insight offered on the energies of the electrons.
• The theory assumes that electrons are localized in specific areas.
• It does not give a quantitative interpretation of the thermodynamic or kinetic stabilities of coordination compounds.
• No distinction between weak and strong ligands.
• No explanation for the colour exhibited by coordination compounds.

There are four main factors that can affect the reaction rate of a chemical reaction:

1. Reactant concentration. Increasing the concentration of one or more reactants will often increase the rate of reaction. This occurs because a higher concentration of a reactant will lead to more collisions of that reactant in a specific time period.

2. Physical state of the reactants and surface area. If reactant molecules exist in different phases, as in a heterogeneous mixture, the rate of reaction will be limited by the surface area of the phases that are in contact. For example, if a solid metal reactant and gas reactant are mixed, only the molecules present on the surface of the metal are able to collide with the gas molecules. Therefore, increasing the surface area of the metal by pounding it flat or cutting it into many pieces will increase its reaction rate.

3. Temperature. An increase in temperature typically increases the rate of reaction. An increase in temperature will raise the average kinetic energy of the reactant molecules. Therefore, a greater proportion of molecules will have the minimum energy necessary for an effective collision .

t's called Wurtz response on the off chance that you wanna gaze upward or something

Methyl bromide experiences Wurtz response to give ethane,

(C₂H₆). 2CH₃Br + 2Na = C₂H₆ + 2NaBr.

This is the Wurtz response, it frames ethane.

Dynamic metals and natural halides are a hazardous blend.

At the point when methyl bromide is treated with sodium within the sight of dry ether, ethane is framed.

This response is known as the Wurtz response.

Colloidal solutions of gold prepared by different methods are of different colours because of different diameters of colloidal gold particles. The colour of colloidal solutions depends upon the size of the colloidal particles.

Laboratory Preparation of Nitric Acid

In the laboratory, nitric acid can be prepared by heating sodium or potassium nitrate with concentrated sulphuric acid to about 423-475 K.

NaNO₃ + H₂SO₄ ——> NaHSO₄ + HNO₃

Anhydrous nitric acid can be obtained by distillation of concentrated aqueous nitric acid with P₄0₁₀.

Manufacture of Nitric Acid
 

Nitric acid is commonly manufactured by Ostwald process in which it is prepared by the catalytic oxidation of ammonia by atmospheric oxygen. The reaction is carried out at about 500 K and 9 x 10⁵ Pa (9 bar) pressure in the presence of Pt or Rh gauge as catalyst.

4NH₃(g) + 50₂(g)——> 4NO(g) + 6H₂0(g)    ΔH =- 90.2 kJ

Pt/Rh gauge,  500K, 9 bar

Nitric oxide thus formed combines with oxygen to form nitrogen dioxide.

2NO(g) +  O₂ (g) ——> 2 NO₂ (g) 

Nitrogen dioxide so formed, dissolves in water to give nitric acid.

3NO₂ (g) + H₂O(l) —–> 2HNO₃(aq) + NO(g)

Dilute nitric acid is further concentrated by dehydration with concentrated sulphuric acid to get about 98% acid.

Properties of Nitric Acid

Physical Properties

1) Pure nitric acid is a colourless liquid.

2) It has boiling point 355.6 K and freezing point 231.4 K.

3) laboratory grade nitric acid contains about 68% of HNO₃ by mass and has a specific gravity of 1.504.

4) The impure acid is generally yellow due to the presence of nitrogen dioxide as impurity. Nitric acid containing dissolved nitrogen dioxide is known as fuming nitric acid.

5) It has a corrosive action on skin and produces painful blisters.

Chemical Properties

(1) Acidic character: It is one of the strongest acids because it is highly ionised in aqueous solution giving hydronium and nitrate ions.

2HNO₃(aq) +H₂O (l) ——> H₃O⁺ + NO₃¯(aq)

It turns blue litmus red. It forms salts with alkalies, carbonates and bicarbonates.

NaOH + HNO₃ —-> NaNO₃ + H₂O

Na₂CO₃ +  HNO₃ —-> 2NaNO₃ + H₂O + CO₂

NaHCO₃ + HNO₃ —-> NaNO₃ + H₂O + CO₂

(2) Action on metals: With the exception of gold and platinum, nitric acid attacks all metals forming a variety of products. The product depends upon the nature of metal, the concentration of acid and temperature.

(A) Metals that are more electropositive than hydrogen (Mg, Al, Mn, Zn, Fe, Pb, etc.). In this case nascent hydrogen is liberated which further reduces nitric acid.

M + 2HNO₃ ——> M(NO₃)₂ + 2H

HNO₃ + H —-> Reduction product + H₂O

The principal product is NO₂, with conc. HNO₃, N₂O with dil. HNO₃, and ammonium nitrate with very dil. HNO₃.

For example: Zn reacts as:

Laboratory Preparation of Nitric Acid

In the laboratory, nitric acid can be prepared by heating sodium or potassium nitrate with concentrated sulphuric acid to about 423-475 K.

NaNO₃ + H₂SO₄ ——> NaHSO₄ + HNO₃

Anhydrous nitric acid can be obtained by distillation of concentrated aqueous nitric acid with P₄0₁₀.

Manufacture of Nitric Acid
 

Nitric acid is commonly manufactured by Ostwald process in which it is prepared by the catalytic oxidation of ammonia by atmospheric oxygen. The reaction is carried out at about 500 K and 9 x 10⁵ Pa (9 bar) pressure in the presence of Pt or Rh gauge as catalyst.

4NH₃(g) + 50₂(g)——> 4NO(g) + 6H₂0(g)    ΔH =- 90.2 kJ

Pt/Rh gauge,  500K, 9 bar

Nitric oxide thus formed combines with oxygen to form nitrogen dioxide.

2NO(g) +  O₂ (g) ——> 2 NO₂ (g) 

Nitrogen dioxide so formed, dissolves in water to give nitric acid.

3NO₂ (g) + H₂O(l) —–> 2HNO₃(aq) + NO(g)

Dilute nitric acid is further concentrated by dehydration with concentrated sulphuric acid to get about 98% acid.

Properties of Nitric Acid

Physical Properties

1) Pure nitric acid is a colourless liquid.

2) It has boiling point 355.6 K and freezing point 231.4 K.

3) laboratory grade nitric acid contains about 68% of HNO₃ by mass and has a specific gravity of 1.504.

4) The impure acid is generally yellow due to the presence of nitrogen dioxide as impurity. Nitric acid containing dissolved nitrogen dioxide is known as fuming nitric acid.

5) It has a corrosive action on skin and produces painful blisters.

Chemical Properties

(1) Acidic character: It is one of the strongest acids because it is highly ionised in aqueous solution giving hydronium and nitrate ions.

2HNO₃(aq) +H₂O (l) ——> H₃O⁺ + NO₃¯(aq)

It turns blue litmus red. It forms salts with alkalies, carbonates and bicarbonates.

NaOH + HNO₃ —-> NaNO₃ + H₂O

Na₂CO₃ +  HNO₃ —-> 2NaNO₃ + H₂O + CO₂

NaHCO₃ + HNO₃ —-> NaNO₃ + H₂O + CO₂

(2) Action on metals: With the exception of gold and platinum, nitric acid attacks all metals forming a variety of products. The product depends upon the nature of metal, the concentration of acid and temperature.

(A) Metals that are more electropositive than hydrogen (Mg, Al, Mn, Zn, Fe, Pb, etc.). In this case nascent hydrogen is liberated which further reduces nitric acid.

M + 2HNO₃ ——> M(NO₃)₂ + 2H

HNO₃ + H —-> Reduction product + H₂O

The principal product is NO₂, with conc. HNO₃, N₂O with dil. HNO₃, and ammonium nitrate with very dil. HNO₃.

Uses of Nitric Acid

(i) It is used in the manufacture of ammonium nitrate for fertilizers.

(ii) It is used in the manufacture of sulphuric acid by lead chamber process.

(iii) It is used in the manufacture of explosives such as trinitro toluene (TNT), nitroglycerine, picric acid, etc.

(iv) It is used in the manufacture of dyes, perfumes and silk.

(v) It is used for the manufacture of nitrates for use in explosive and pyrotechnics.

(vi) It is used in picking of stainless steel and etching of metals.

(vii) It is also used as an oxidiser in rocket fuels.

(viii) It is used in the purification of gold and silver as aqua regia.

This is a simulation of the Brownian motion of a big particle (dust particle) that collides with a large set of smaller particles (molecules of a gas) which move with different velocities in different random directions.

the acid is usually manufactured using a method known as the contact process. Earlier, in this process, a chemical element known as platinum was first used as a catalyst. Later, an inorganic compound known as vanadium oxide replaced platinum due to cost constraints.

In any case, today we will be looking at the contact process in detail.

Contact process for manufacturing of sulphuric acid:

Steps involved in the manufacturing of sulphuric acid are as stated below:

1. Preparation of sulphur dioxide.
2. Conversion of sulphur dioxide into sulphur trioxide.
3. Conversion of sulphur trioxide formed into concentrated H₂SO₄.

a chemical reaction also occurs when ammonia dissolves in water. In aqueous solution, ammonia acts as a base, acquiring hydrogen ions from H₂O to yield ammonium and hydroxide ions. In contrast, the ammonium ion acts as a weak acid in aqueous solution because it dissociates to form hydrogen ion and ammonia.