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Contents

• Group 15 elements
• Dinitrogen
• Ammonia
• Oxoacids of hydrogen – Nitric Acid
• Phosphorous
• Phosphine
• Phosphorous Halides
• Group 16 elements
• Dioxygen
• Oxides
• Ozone
• Sulphur
• Sulphur Dioxide
• Sulphuric Acid

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• Group 17 elements
• Hydrogen Chloride
• Oxoacids of Halogens
• Inter Halogen Compounds
• Group 18
• Xenon discussed in detail

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P –BLOCK�Group - 15

>The Group 15 elements are:

(Elements with atomic number)

Nitrogen (N) 🡪7

Phosphorus (P) 🡪15

Arsenic (As) 🡪33

Antimony (Sb) 🡪51

Bismuth (Bi) 🡪83

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>Configuration:

N 🡪1s² 2s² 2p³

(ns² np³)

The s-orbital in these elements is completely filled and p-orbital is half filled making their electronic configuration extra stable.

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• Size:

As we move from top to bottom, there’s an increase in the size because of the increase in the number of shells. From N to P, there’s an increase in the size but from As to Bi, only a small increase in the size because of the presence of completely filled d- and/or f-orbital in higher members.

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• Ionization Enthalpy:

It is defined as the amount of the energy required to remove the outermost electron. Since the size increases down the group, therefore ionization enthalpy decreases down the group.

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• Electronegativity:

It is the property to attract the shared pair of electron towards itself. Down the group, since the size increases, tendency to attract the shared pair of electron will decrease and hence electronegativity will decrease.

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Physical Properties:

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• All the elements of this group are polyatomic.
• Dinitrogen (N₂) is a diatomic gas while all others are solid.
• Nitrogen and Phosphorus are non-metals, Arsenic and Antimony are metalloids and Bismuth is a metal.
• Except Nitrogen, all other elements show allotropic nature.

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Chemical properties:

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Oxidation states:

• They show oxidation state of -3, +3 and +5.
• Tendency to exist -3 oxidation state decreases down the group due to increase in size and metallic character.
• Bismuth hardly shows -3 oxidation state.
• The stability of +5 oxidation state decreases down the group and that of +3 oxidation state increases. This is because of inert pair effect.
• The only well established +5 oxidation state in Bismuth is in BiF₅.

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Q. What do you understand by inert pair effect? What happens due to it?

Ans. The non-participation of inner line two s- electrons in the bond formation is known as inert pair effect. The inner lying s- electron pair does not participate in bonding and acts as

an inert pair. Because of this effect, tendency to lower oxidation state increases down the group.

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Q. Nitrogen is restricted to a maximum covalency of 4, why?

Ans. Because only four 2s and 3p orbitals are available for bonding.

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Q. Why the covalency of other elements of group 15 can be increased beyond 4?

Ans. Because they have vacant d-orbitals in their outermost shell.

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Q. Why and how does Nitrogen show anomalous behavior?

Ans. This is because of its:

• small size
• high electronegativity
• high ionization enthalpy
• non-availability of d-orbitals

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This is shown as follows:

It has a unique ability to form p∏-p∏ multiple bonds with itself and with other elements having small size and high electronegativity.

Catenation tendency is weaker in Nitrogen as compared to other elements of this group.

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Q. Why the heavier elements of group 15 do not form p∏-p∏ bonds?

Ans. Because their atomic orbitals are so large and diffuse that they cannot have effective overlapping.

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Q. Why is the single bond in N-N much weaker than single bond in P-P?

Ans. Because of high inter-electronic repulsion of the non-bonding electrons owing to the small bond lengths.

Because of the above reason, Catenation tendency in Nitrogen is much weaker as compared to other elements of this group.

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Reactivity towards Hydrogen

• Group 15 elements form hydrides of the formula EH₃ where E is the element of group 15.
• Hydrides show regular gradation in their properties i.e. stability decreases from NH₃ to BiH₃ because of decrease in bond dissociation enthalpy.
• Reducing character of hydrides increases i.e. NH₃ (ammonia) is the strongest reducing agent while BiH₃ is the weakest.
• PH₃ has lower boiling point than NH₃. PH₃ molecules are not associated through hydrogen bonding in liquid state.

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Reactivity towards Oxygen

• They form two types of oxides E₂O₃ and E₂O₅.
• The oxide in higher oxidation state (E₂O₅) is more acidic.
• Acidic character decreases down the group.
• Oxides of the type E₂O₃ of Nitrogen and Phosphorus are purely acidic, of Arsenic and Antimony are amphoteric and of Bismuth are Basic.

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Reactivity towards Halogen

• They form halides of the formula EX₃ and EX₅.
• Nitrogen does not form penta halide due to non-availability of the d-orbitals in the valence shell.
• Penta halides are more covalent than tri halides.
• All the tri halides except Nitrogen are stable. NF₃ is the most stable tri halide of Nitrogen.
• Tri halides except BiF₃ are covalent in nature.

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Reactivity towards Metals

• These elements react with metals to form their binary compounds exhibiting –3 oxidation state.

Examples:

Ca₃N₂ (Calcium Nitride)

Ca₃P₂ (Calcium Phosphide)

Na₃As₂ (Sodium Arsenide)

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DINITROGEN

Preparation:

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a) Lab preparation

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NH₄Cl(aq) + NaNO₂(aq) → N₂(g) + 2H₂O(l) + NaCl (aq)

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b) General

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(NH₄)₂Cr₂O₇ → N₂ + 4H₂O + Cr₂O₃

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c) For obtaining by Thermal Decomposition of Na or Ba azide

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Ba(N₃)₂ → Ba + 3N₂

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Properties

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• It has two stable isotopes: ¹⁴N and ¹⁵N.
• It has a very low solubility in water.
• It is quite inert at room temperature because of high bond enthalpy of N≡N bond.
• As temperature increases, reactivity increases
• It shows the following reactions:

6Li + N₂→ 2Li₃N

3Mg + N₂→ Mg₃N₂

N₂ (g) + 3H₂ (g) → 2NH₃ (g) [Haber’s process]

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AMMONIA

Preparation

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• Urea

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NH₂CONH₂ + 2H₂O → (NH₄)₂CO₃ → 2NH₃ + H₂O + CO₂

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• On small scale

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2NH₄Cl + Ca(OH)₂ →2NH₃ + 2H₂O + CaCl₂

(NH₄)₂SO₄ + 2NaOH → 2NH₃ + 2H₂O + Na₂SO₄

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• On large scale

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N₂(g) + 3H₂(g) → 2NH₃(g)

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Q. What are the conditions in accordance to Le Chatelier’s principle for the formation of Ammonia ?

Ans. ~Pressure of about 200 atm.

~Temperature of about 700 K.

~Catalyst- Iron oxide with small amounts of K₂O

and Al₂O₃

Q. Draw a flow chart for the production of NH₃ by Haber’s process.

Ans.

Properties

1. Structure of Ammonia:

It has pyramidal structure with one lone pair of electrons.

Hybridization present is sp³ but the tetrahedral angle gets reduced to 107.8° because of repulsion of lone pair of electrons.

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2. Ammonia gas is highly soluble in water and its aqueous solution is weakly basic due to the formation of OH– ions.

NH₃ (g) + H₂O (l) → NH₄⁺ (aq) + OH^– (aq)

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3. In water, Ammonia forms hydroxide which is knows as Ammonium Hydroxide.

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Q. What happens when Ammonia Hydroxide is made to react with Zinc Sulphate and Ferric Chloride respectively?

Ans. ZnSO₄ (aq)+ 2NH₄OH (aq) → Zn(OH)₂ (s) + (NH₄)₂SO₄ (aq)

(White ppt)

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2FeCl₃ (aq) + NH₄OH (aq) → Fe₂O₃.xH₂O + NH₄Cl (aq)

(Brown ppt)

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Q. How are the metal ions like Cu²⁺ and Ag⁺ detected using Ammonia?

Ans. a) Cu²⁺ (aq) + 4NH₃ (aq) → [Cu (NH₃)₄]²⁺(aq)

(Blue) (Deep blue)

b) Ag⁺ (aq) + Cl^- (aq) → AgCl (s)

(Colourless) (White ppt)

AgCl (s) + 2NH₃ → [Ag (NH₃)₂] Cl

(White ppt) (Colourless)

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Q. From the reaction, what typical characteristic of NH₃ can we gather?

Ans. The presence of lone pair of electrons on the Nitrogen atom of the Ammonium molecule makes it a Lewis base. It donates the electron pair and forms linkage with metal ions and this finds an application in detection of metal ions.

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Q. Draw a table of the various oxides of Nitrogen.

Ans.

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Q. Draw a table which gives the structure of oxides of Nitrogen.

Ans.

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Q. Why does NO₂ dimerise?

Ans. The structure of NO₂ is

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The structure of NO₂ as depicted above contains odd number of valence electrons. It behaves as a typical odd molecule. On dimerisation, it is converted to stable N₂O₄ molecule having even number of molecules.

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OXOACIDS of Nitrogen

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NITRIC ACID

Preparation of Nitric Acid

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1. In the laboratory, nitric acid is prepared by heating KNO₃ or NaNO₃ and concentrated H₂SO₄ in a glass retort.

NaNO₃ + H₂SO₄ → NaHSO₄ + HNO₃

2. On a large scale it is prepared mainly by Ostwald’s process.

4NH₃ (g) + 5O₂ (g) → 4NO (g) + 6H₂O (g)

3NO₂ (g) + H₂O (l) → 2HNO₃ (aq) + NO (g)

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Structure of Nitric acid:

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Properties

1. In the gaseous state, it exists as a planar molecule.

2. It behaves as a strong acid giving hydronium and nitrate iona.

HNO₃(aq) + H₂O(l) →H₃O⁺(aq) + NO₃^–(aq)

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3. Concentrated nitric acid is a strong oxidising agent and attacks most metals except noble metals such as gold and platinum.

3Cu + 8 HNO₃(dilute) → 3Cu(NO₃)₂ + 2NO + 4H₂O

Cu + 4HNO₃(conc.) → Cu(NO₃)₂ + 2NO₂ + 2H₂O

4Zn + 10HNO₃(dilute) → 4 Zn (NO₃)₂ + 5H₂O + N₂O

Zn + 4HNO₃(conc.) →Zn (NO₃)₂ + 2H₂O + 2NO₂

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4. Some metals (e.g., Cr, Al) do not dissolve in concentrated nitric acid because of the formation of a film of oxide on the surface.

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5. Concentrated nitric acid also oxidizes non-metals

I₂ + 10HNO₃ →2HIO₃ + 10 NO₂ + 4H₂O

C + 4HNO₃ →CO₂ + 2H₂O + 4NO₂

S₈ + 48HNO₃(conc.) →8H₂SO₄ + 48NO₂ + 16H₂O

P₄ + 20HNO₃(conc.) →4H₃PO₄ + 20 NO₂ + 4H₂O

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Brown ring test of Nitrates�

When to a Nitrate solution, freshly prepared Ferrous Sulphate solution is added and then conc. H₂SO₄ acid along the side of the test tube is added, then a brown ring at the interphase is formed.

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NO₃ + 3Fe²⁺ + 4H⁺ →NO + 3Fe³⁺ + 2H₂O

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[Fe (H₂O)₆ ]²⁺ + NO → [Fe (H₂O)₅ (NO)]²⁺ + H₂O

(Brown)

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PHOSPHOROUS

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White Phosphorus

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• It is a white waxy solid.
• It is poisonous.
• Insoluble in water but soluble in Carbon disulphide (CS₂).
• It glows in dark (chemiluminescence).

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• It dissolves in boiling NaOH solution in an inert atmosphere giving Phosphine and Sodium hypophosphite.

P+ 3NaOH + 3H₂O →PH₃ + 3NaH₂PO₂

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• It is less stable and therefore, more reactive than the other solid phases because of angular strain in the P₄ molecule.
• It readily catches fire in air to give dense white fumes of P₄O₁₀.
• It exists as discrete tetrahedral P₄ molecule.

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Red Phosphorus

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• It is obtained by heating white phosphorus at 573K in an inert atmosphere for several days.
• It has iron grey lustre.
• It is odourless, nonpoisonous.
• Insoluble in water as well as in carbon disulphide.
• It does not glow in the dark.
• Chemically, red phosphorus is much less reactive than white phosphorus because it is a polymeric in nature consisting of a number of P₄ tetrahedral units joined together.

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Black phosphorus:

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• It has two forms α-black phosphorus and β-black phosphorus.

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α-black phosphorus

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~α is formed when Red phosphorus is heated in a sealed tube at 803K.

~It can be sublimed in air.

~It has opaque, monoclinic or rhombohedral crystals.

~It does not oxidize in air.

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β-black phosphorus

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~β is prepared by heating white phosphorus at 473 K under high pressure.

~It does not burn in air upto 673K.

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PHOSPHINE ( PH₃ )

Preparation:

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Ca₃P₂ + 6H₂O →3Ca(OH)₂ + 2PH₃

Ca₃P₂ + 6HCl →3CaCl₂ + 2PH₃

Laboratory preparation:

P₄ + 3NaOH (conc.) + 3H₂O → PH₃ + 3Na H₂PO₂

(Sodium hypophosphite)

When pure, it is non inflammable but becomes inflammable

owing to the presence of P₂H₄ or P₄ vapours. To purify it from the impurities,

it is absorbed in HI to form phosphonium iodide which on treating with KOH

gives off phosphine.

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PH₄I + KOH → KI + H₂O + PH₃

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Properties

• In the pure state, it is non-inflammable but becomes flammable owing to the presence of P₂H₄ or P₄ vapour.

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• It is a colourless gas with rotten fish smell

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• Highly poisonous.

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• It explodes in contact with traces of oxidizing agents like HNO₃, Cl₂ and Br₂ vapours.

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• It is slightly soluble in water.

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• It decomposes in water in presence of light giving red phosphorus and H₂.

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• It is weakly basic like Ammonia and gives Phosphonium compounds like acids.

PH₃ + HBr → PH₄Br

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• It interacts with Phosphonic acid and mercuric chloride

3CuSO₄ +2PH₃ →Cu₃P₂ + 3H₂SO₄

3HgCl₂ + 2PH₃ → Hg₃P₂ + 6HCl

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PHOSPHOROUS HALIDES

It forms two types of halides, PX₃ and PX₅.

In PX₃ (X = F, Cl, Br, I)

PX₅ (X = F, Cl, Br)

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It is obtained by reacting white phosphorus with thionyl chloride.

P₄ + 8SOCl₂ → 4PCl₃ + 4SO₂ + 2S₂Cl₂

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It is obtained by dry chlorine with white phosphorous on heating.

P₄ + 6Cl₂ → 4PCl₃

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Structure:

Phosphorus trichloride�

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(shape – pyramidal)�

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Properties

• It is a colourless oily liquid.
• Hydrolyses in the presence of moisture.

PCl₃ + 3H₂O → H₃PO₂ + 3HCl

• It reacts with organic compounds containing –OH group such as CH₃COOH, C₂H₅OH.

3CH₃COOH + PCl₃ → 3CH₃COCl + H₃PO₃

3C₂H₅OH + PCl₃ → 3C₂H₅Cl + H₃PO₃

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PHOSPHOROUS PENTACHLORIDE

Preparation

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• Phosphorus pentachloride is prepared by the reaction of white phosphorus with excess of dry chlorine.

P₄ + 10Cl₂ → 4PCl₅

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• It can also be prepared by the action of SO₂Cl₂ on phosphorus.

P₄ + 10SO₂Cl₂ → 4PCl₅ + 10SO₂

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Properties

• It is a yellowish white powder
• It hydrolysis in moist air to POCl₃ and finally gets converted to phosphoric acid.

PCl₅ + H₂O → POCl₃ + 2HCl

POCl₃ + 3H₂O → H₃PO₄ + 3HCl

• When heated, it sublimes but decomposes on stronger heating.

PCl₅ → PCl₃ + Cl₂

• It reacts with organic compounds containing –OH group converting them to chloro derivatives.

C₂H₅OH + PCl₅ → C₂H₅Cl + POCl₃ + HCl

CH₃COOH + PCl₅ → CH₃COCl + POCl₃ + HCl

• Finely divided metals on heating with PCl₅ give corresponding chlorides.

2Ag + PCl₅ → 2AgCl + PCl₃

Sn + 2PCl₅ → SnCl₂ + 2PCl₃

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Structure of PCl₅ with bond formation (in gaseous and liquid state)�

• There are two types of bonds: 2 are axial and other three are equatorial.
• → Axial PCl is longer than equatorial PCl bond length because of the repulsion.
• → In the solid state, it exists as ionic solid PCl₄, PCl₆ with cationic part as tetrahedral and anionic part is octahedral.

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• Q. What indicates the basicity of the oxoacids of Phosphorus?
• Ans. The presence of Hydrogen atoms which are directly attached to PO indicate the basicity.

Oxoacids of phosphorus

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Structures of Oxoacids

The acids which contain P-H bond have strong reducing properties

Hypophosphorous acid is a good reducing agent as it contains two P-H bonds and reduces silver nitrate to silver

The P-H bonds are not ionisable to give H⁺ and do not play any role in basicity

Only those hydrogen atoms which are attached with oxygen in P-OH form are ionisable and cause basicity

H₃PO₃ and H₃ PO₄ are dibasic and tribasic respectively as the structure have two P-OH bonds and three P-OH bonds respectively

GROUP 16

includes:

1. Oxygen (O)
2. Sulphur (S)
3. Selenium (Se)
4. Tellurium (Te)
5. Polonium (Po)

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Anomalous behavior of Oxygen

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IONIZATION ENTHALPY

Ionisation enthalpy decreases down the group. It is due to increase in size. However, the elements of this group have lower ionisation enthalpy values compared to those of Group15 in the corresponding periods.

This is due to the fact that Group 15 elements have extra stable half-filled p orbitals electronic configurations.

Q. Why water is a liquid and H2S is a gas though oxygen� and sulphur belong to the same group? �

Ans. Oxygen atom has a small size and high electronegativity, therefore, it is capable of forming hydrogen bonds which is not possible in H₂S. Due to the presence of hydrogen bonding it has a compact structure and is present in liquid state.

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Q. Why does the stability of +4 oxidation state increase

down the group while that of +6 decrease ?

Ans. The general outer electronic configuration of 16 group elements is ns²np⁴ . Therefore, the common oxidation state would be +4 and +6 . Stability of +4 will increase due to the inert pair effect.

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Q. Why does oxygen limit its covalency to 4 ? OR Why

cannot oxygen expand its covalency beyond 4 while

the other elements can do so?

Ans. This is due to the absence of the d-orbitals in the oxygen

atom.

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Reactivity towards Hydrogen

Reactivity towards Hydrogen

• All the elements combine with hydrogen to form hydrides of the type H₂E
• Acidic character increases from top to bottom because of decrease in bond dissociation enthalpy down the group
• Thermal stability decreases down the group because of the increase in the bond length
• Except water all hydrides have reducing property and this increases down the group.

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Reasoning Questions

Q: H₂S is less acidic than H₂Te. Why?

A: Due to the decrease in bond (E–H) dissociation enthalpy down the group, acidic character increases.

Q: H₂Te is the strongest reducing agent amongst the group 16 hydrides. Why?

A: Down the group bond enthalpy decreases making the loss of H easier, hence reducing property increases. Te is the last member, hence the strongest reducing agent.

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Physical properties of hydrides

Reactivity towards Oxygen

• All the elements form oxides of the formula EO₂ and EO₃
• Reducing properties of dioxides decreases down the group
• All oxides are acidic in nature .

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Reasoning Questions

Q: Reducing property of dioxide decreases from SO₂ to TeO₂ . Why?

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Q: SO₂ is reducing while TeO₂ is an oxidising agent. Why?

Reactivity towards Halogens

• They form a large number of halides of the formula EX₆, EX₄, EX₂
• The stability of the halides decreases down the group, in the order F > Cl > Br > I ( Stability decreasing )
• Among Hexahalides, Hexa fluorides are the only stable halides. They are gaseous in nature and have octahedral structure.
• Sulphur hexafluoride, SF₆ is exceptionally stable for steric reasons.

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DIOXYGEN ( o₂ )

Preparation of Dioxygen

1. From KClO₃

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2. From Oxides (Thermal decomposition of oxides of metals)

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3. From decomposition of hydrogen peroxide

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4. On a large scale it is prepared by electrolysis of water

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Properties of O₂

• Dioxygen is odourless and colourless
• It is paramagnetic inspite of having ‘even’ number of electrons
• It reacts with nearly all metals and

non-metals except gold, platinum and some noble gases.

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Some of the common reactions of Dioxygen are

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OXIDES

• Oxides are a binary compound of oxygen with another element is called oxide
• Oxides can be simple (MgO, Al₂O₃) or mixed (Pb₃O₄, Fe₃O₄)
• Another way of classification of simple oxides is-

a) Acidic

b) Basic

c) Amphoteric and

d) Neutral

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ACIDIC OXIDE

~An oxide which combines with water to give an acid is called an Acidic Oxide

E.g.. : SO₂, CO₂, N₂O₅

~Generally non-metallic oxides are acidic

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BASIC OXIDE

~The oxides which give a base on reaction with water are called Basic Oxides

E.g.. : Na₂O, CaO, BaO

~Generally, metallic oxides are basic in nature

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AMPHOTERIC OXIDE

~These are oxides which exhibit a dual behavior i.e. they show properties of both acidic and basic oxides

E.g.. : Al₂O₃

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NEUTRAL OXIDE

~Oxides which are neither acidic nor basic

E.g.. : CO, NO, N₂O

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OZONE

• Ozone is an allotropic forms of oxygen
• It is formed from atmospheric oxygen in the presence of sunlight
• Ozone protects the earth’s surface from excessive concentration of ultraviolet (UV) radiations
• It is a pale blue gas, dark blue liquid or violet-black solid
• Ozone is thermodynamically unstable and gives dioxygen and nascent oxygen on decomposition and because of this nascent oxygen, it is a powerful oxidising agent
• Two chemical reactions to show the oxidising nature of ozone are

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Reasoning Question

Q: Ozone is thermodynamically unstable with respect to oxygen . Why?

A: Decomposition of ozone into oxygen results in :

• the liberation of heat (ΔH is negative) and
• an increase in entropy (ΔS is positive).

This results in large negative Gibbs energy change

Q: Why does O3 act as a powerful oxidising agent?

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Q: How is O₃ estimated quantitatively?

A: When ozone reacts with an excess of potassium iodide solution , iodine is liberated which can be titrated against a standard solution of sodium thiosulphate. This is a quantitative method for estimating O3 gas.

• The structure of ozone is as follows

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• Q: All the O-O bonds in ozone are equivalent. Explain why?

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• Ozone saves us from the UV radiations but there is a depletion in the ozone concentration in the upper layer of the atmosphere due to supersonic jet airoplanes and use of freons which are used in the aerosol sprays and as refrigerants .

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SULPHUR

• Sulphur exists in numerous allotropes of which two are the most important which are – yellow rhombic (∞) and monoclinic (ꞵ)

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• At room temperature, Rhombic is stabler and it transforms to monoclinic sulphur when heated above 369 K
• Rhombic Sulphur is

~yellow in colour

~melting point - 385.8 K

~specific gravity – 2.06

~formed on evaporating the solution of Roll Sulphur in

carbon disulphide

~insoluble in water and readily soluble in carbon disulphide

• Monoclinic sulphur

~melting point – 393 K

~specific gravity – 1.98

~soluble in carbon disulphide

~prepared by melting rhombic sulphur in a dish and cooling

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• α sulphur is stable below 369 K and transforms into β sulphur above this temperature. At 369 K both the forms are stable, hence this temperature is called –

Transition Temperature

• Both Rhombic and Monoclinic Sulphur have S₈ molecules. They both have puckered or typical forms. Several other modifications of sulphur containing 6 – 20 atoms per ring have been prepared.

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QUESTION

• Which form of sulphur shows paramagnetic behaviour ?
• In vapour state sulphur partly exists as S2 molecule which has two unpaired electrons in the antibonding π * orbitals like O2 and, hence, exhibits paramagnetism.

SULPHUR DIOXIDE

Methods of Preparation:

• General Method

S (s) + O₂ 🡪 SO₂ (g)

• Lab Method

SO₃^2- (aq) + 2H⁺ 🡪 H₂0 (l) + SO₂ (g)

• Industrial Method

4FeS₂ (g) + 11O₂ (g) 🡪 2Fe₂O₃ (s) + 8SO₂ (g)

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Properties:

• Colourless gas
• Pungent smell
• Highly soluble in water
• When passed through water forms sulphurous acid (H₂SO₃)
• When passed through NaOH

2NaOH + SO₂ 🡪 Na₂SO₃ + H₂O

Na₂SO₃ + H₂O + SO₂ 🡪 2NaHSO₃

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Properties continued….

• When made to react with chlorine it gives sulphuryl chloride

SO₂ (g) + Cl₂ (g) 🡪 SO₂Cl₂ (l)

• When reacted with oxygen in presence of vanadium (V) pentoxide forms sulphur dioxide

2SO₂ (g) + O₂ (g) ----🡪 2SO₃ (g)

• It behaves as Reducing agents
• Converts Fe³⁺ 🡪 Fe²⁺

2Fe³⁺ + SO₂ + 2H₂O 🡪 2Fe²⁺ + SO₄^2- + 4H⁺

• Decolourises acidified KMnO4 solution

5SO₂ + 2MnO₄^- + 2H2O 🡪 5SO₄^2- + 4H⁺ + 2Mn²⁺

• It is a detection of SO₂
• Its angular in shape

Resonating structures:

V2O5

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QUESTIONS

• Show with the help of reactions, that the behaviour of sulphur dioxide is very similar to that of carbon dioxide.
• What happens when sulphur dioxide is passed through an aqueous solution of Fe(III) salt?
• How is the presence of SO2 detected ?
• Comment on the nature of two S–O bonds formed in SO2 molecule. Are the two S–O bonds in this molecule equal ?

OXOACIDS OF SULPHUR

SULPHURIC ACID

Preparation of Sulphuric acid –CONTACT PROCESS

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1. Burning of sulphur or sulphide ore in air to generate SO₂.

S + O2 -----🡪 SO2

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2. Conversion of SO₂ to SO₃ by the reaction with oxygen in the presence of a catalyst V₂O₅.

V₂O₅

2SO₂ (g) + O₂ (g) ----🡪 2SO₃ (g) + H

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3. Absorption of SO₃ in H₂SO₄ to give Oleum (H₂S₂O₇)

SO₃ + H₂SO₄ 🡪 H₂S₂O₇ (oleum)

H₂S₂O₇ + H₂O 🡪 H₂SO₄

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Q. Give a flow chart for the formation/manufacture of� H₂SO₄.

Ans. The flow diagram is as follows

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Properties �

a) Colourless, dense, oily liquid

b) Dissolves in water with large quantity of heat evolved. This is the reason why concentrated acid must be added slowly into water with constant stirring.

c) Chemical activities/chemical reaction/chemical properties of sulphuric acid are a result of following characteristics

• low volatility
• strong acidic character
• strong affinity for water
• ability to act as an oxidizing agent

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Q. How do you prove that it is a dibasic acid ?

Ans. This acid forms two series of salts: Normal sulphates and acid

sulphates (Bisulphates).

It can also be proved by following two reactions :

H₂SO₄ (aq) + H₂O (l) 🡪 H₃O⁺ (aq ) + HSO₄^- (aq)

HSO₄^- (aq) + H₂O (l) 🡪 H₃O⁺ (aq) + SO₄^2- (aq)

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d) Because of its low volatility it is used in manufacture of more volatile acids from their volatile salts

2MX + H₂SO₄ 🡪 2HX + M₂SO₄ (X= F, Cl, NO₃)

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e) Concentrated H₂SO₄ is a strong dehydrating agent

🡪It is used to dry wet gases provided; they do not react with it.

🡪It is used to remove water from organic compound like charring of carbohydrates.

C₁₂H₂₂O₁₁ ------🡪 12C + 11H₂O

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f) Strong oxidizing agent property of hot concentrated H₂SO₄ is intermediate between phosphoric and nitric acid.

Cu + 2H₂SO₄ (conc.) 🡪 CuSO₄ + SO₂ + 2H₂O

C + 2H₂SO₄ (conc. ) 🡪 CO₂ + 2SO₂ + 2H₂0

3S + 2H₂SO₄ (conc. ) 🡪 3SO₂ + 2H₂O

H₂SO₄

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QUESTIONS

• Why is Ka₂<<Ka₁ for H2SO4 in water?

USES

Sulphuric acid is a very important industrial chemical. A nation’s industrial strength can be judged by the quantity of sulphuric acid it produces and consumes. It is needed for the manufacture of hundreds of other compounds and also in many industrial processes. The bulk of sulphuric acid produced is used in the manufacture of fertilisers (e.g., ammonium sulphate, superphosphate). Other uses are in:

(a) petroleum refining (b) manufacture of pigments, paints and dyestuff intermediates (c) detergent industry (d) metallurgical applications (e.g., cleansing metals before enameling, electroplating and galvanising (e) storage batteries (f) in the manufacture of nitrocellulose products and (g) as a laboratory reagent.

Group 17

includes:

1. Fluorine ( F )
2. Chlorine ( Cl )
3. Bromine ( Br )
4. Iodine ( I )
5. Astatine ( At ) ( Radioactive )

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PROPERTIES

• Halogens are coloured - F2, has yellow, Cl2 , greenish yellow, Br2, red and I2, violet colour.
• One curious anomaly is the smaller enthalpy of dissociation of F2 compared to that of Cl2

QUESTIONS

Q: Why does F2 have a low F-F bond dissociation enthalpy?

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Q: Why are halogens coloured?

This is due to absorption of radiations in visible region which results in the excitation of outer electrons to higher energy level. By absorbing different quanta of radiation, they display different colours.

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PHYSICAL PROPERTIES

Q. Why do halogens have maximum negative electron gain enthalpy in their respective periods ?

Ans. Because of the small size and high effective nuclear charge

They readily accept one electron to change their configuration from ns²np⁵ to ns²np⁶.

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Q. Although electron gain enthalpy of fluorine is less negative as compared to chlorine but F₂ is stronger oxidizing agent than chlorine (Cl₂)

Ans. Due to two reasons

🡪 Due to low enthalpy of dissociation of F-F bond.

🡪 High hydration enthalpy of F-

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Q. Why does Fluorine exhibit only one oxidation state of -1 while others of the family exhibit besides -1,+1,+3,+5 and +7 oxidation states ?

Ans . Fluorine atom has no d-orbitals in its valence shell and therefore, cannot expand its octet. Being the most electronegative it exhibits only -1 oxidation state. While the others do not face this problem.

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GROUP 1 [Aashi Sharma, Abhishree, Akshita, Antara, Dafiya]

Q. What are the causes of anomalous behaviour of fluorine ?

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GROUP 2 [Deeksha, Diya , Fiza, Hiba , Keiara ]

Q. Why are halogens coloured?

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GROUP 3 [Krithika, Nikitha, ritu , Rushda , Sai Bhavika]

Q. Why does fluorine have low bond dissociation enthalpy?

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GROUP 4 [Sanjana, Sanjanaa Sriram, sarah, Saumya, Savannah]

Q. Why is electron gain enthalpy of chlorine more negative than that of fluorine?

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GROUP 5 [Shamita, Shehzeen, Shreya Shree, Syeda Zainab, Tanisha]

Q. Explain why fluorine is the strongest oxidizing agent amongst group 17.

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Reactivity of Hydrogen towards Halogens.

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1. They form Hydrogen Halides. H₂+X₂ 🡪 2HX
2. Stability of Halides decreases down the group. HF is most stable.
3. Acid strength increases down the group. HF is most reactive.
4. Acid strength 🡪 HF < HCl < HBr < HI

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Q. Hydrogen Fluoride is liquid while all other hydrogen halide are gases.

Ans.

Fluorine being the most electronegative element is capable of forming hydrogen bonds and hence it is in a more compact state and i.e. liquid.

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PHYSICAL PROPERTIES OF HX

Reactivity of oxygen towards halogens:

1. When oxygen combines with halogens, it forms oxides except when oxygen and Fluorine combine together the compound formed called oxygen fluoride..
2. Fluorine and oxygen forms two compounds OF₂ and O₂F₂. OF₂ is thermally stable. 🡪 Both of them are good fluorinating agents
3. Chlorine forms oxide with the formula Cl₂O, ClO₂, Cl₂O₆, Cl₂O₇. ClO₂ is a very good bleaching agent for paper pulp and textile and in water treatment.
4. Bromine oxides are the least stable halogen oxides which exist only at low temp. They are powerful oxidizing agents. Common formulae for Bromine oxides are Br₂O, BrO₂, BrO₃.
5. Iodine oxides are insoluble solids and decompose on heating. Their oxides have the formula I₂O₄, I₂O₃ and I₂O₇. I₂O₅ is a strong oxidising agent and is used in the estimation of carbon monoxide.

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Reactivity of Halogens towards metals.

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• Halogen and metals form Metal Halides. For example, Magnesium with Bromide form Magnesium Bromide.

Mg + Br₂ 🡪 MgBr₂

• Ionic character decreases down the group. That means MF is more ionic.

MF> MCl > MBr >MI

• If a metal exhibits more than one oxidation state than the halide in higher oxidation state is more covalent i.e. SnCl₄, PbCl₄, SbCl₄ and HF₆ are more covalent than SnCl₂, PbCl₂, SbCl and LiF₄.

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CHLORINE

Methods for the preparation of Chlorine.

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• MnO₂ + 4HCl 🡪 MnCl₂ + Cl₂ + 2H₂0
• 4NaCl + MnO₂ + 4H₂SO₄ 🡪 MnCl₂ + 4NaHSO₄ + 2H₂O + Cl₂
• 2KMnO₄ + 16 HCl 🡪 2KCl + 2MnCl₂ + 8H₂O + 5Cl₂

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Q. Give the methods of manufacture of chlorine.

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Deacon’s process

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• When HCl gas is treated with atmospheric oxygen in presence of CuCl₂ as catalyst at 723K we get chlorine gas.
• 4HCl + O₂ 🡪 2Cl₂ + 2H₂O

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Electrolytic Process

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• It is prepared by electrolysis of Brine i.e. concentrated NaCl solution. Chlorine is obtained at anode.

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Properties of chlorine

• Greenish yellow gas with pungent and suffocating smell
• Soluble in water
• Heavier in air
• It reacts with metals and non-metals to form chlorides

2Al + 3Cl₂ 🡪 2AlCl₃

2Na + Cl₂ 🡪 2NaCl

2Fe + 3Cl₂ 🡪 2FeCl₂

P₄ + 6Cl₂ 🡪 4PCl₃

S₈ + 4Cl₂ 🡪 4S₂Cl₂

• It has high affinity for hydrogen. Therefore, it combines with compounds containing hydrogen and hydrogen itself to give HCl.

H₂ + Cl₂ 🡪 2HCl

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Q. Give two or three reactions to show that chlorine is very fond of hydrogen.

Ans. H₂ + Cl₂ 🡪 2HCl

H₂S + Cl₂ 🡪 2HCl + S

C₁₀H₁₆ + 8Cl₂ 🡪 16HCl + 10C

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• Reaction of ammonia with chloride

8NH₃ + 3Cl₂ 🡪 6NH₄Cl + N₂

(excess)

NH₃ + 3Cl₂ 🡪 NCl₃ + 3HCl

(excess)

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• Reaction of ammonia alkalis

2NaOH (cold , dilute) + Cl₂ 🡪 NaCl + NaOCl + H₂O

6NaOH (hot, conc) + 3Cl₂ 🡪 5NaCl + NaClO₃ + 3H₂O

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• Reaction with dry slaked lime

2Ca(OH) ₂ + 2Cl₂ 🡪 Ca(OCl) ₂ + CaCl₂ + 2H₂O

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Q. Give the formula of Bleaching powder and equation for its equation . Ca(OCl) ₂ . CaCl₂. Ca(OH)₂. 2H₂0

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Q. What happens to chlorine water on long standing ?

Ans.

• It looses its yellow colour due to the formation of HCl and HOCl. This HOCl gives nascent oxygen which is responsible for oxidizing and bleaching properties of chlorine.

Cl2 + H2O ----🡪 HCl + HOCl

HOCl -----🡪 HCl + (O)

Q. Give equation showing the oxidation properties of chlorine.

Ans. 2FeSO₄ + H₂SO₄ + Cl₂ 🡪 Fe₂ (SO₄) ₃ + 2HCl

Na₂SO₃ + Cl₂ + H₂O 🡪 Na₂SO₄ + 2HCl

SO₂ + 2H₂O + Cl₂ 🡪 H₂SO₄ + 2HCl

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Q. Give the equation to show the bleaching property of chlorine water.

Ans. Cl₂ + H₂O 🡪 2HCl + O

Coloured substance + O 🡪 colourless substance

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Q. Where is chlorine water used for bleaching purposes ? and what type of Bleaching action it has ?

Ans. It bleaches vegetable or organic matter in presence of moisture. It has permanent bleaching effect.

HYDROGEN CHLORIDE

Preparation in Lab :

• NaCl + H₂SO₄ 🡪 NaHSO₄ + HCl
• NaHSO₄ + NaCl 🡪 Na₂SO₄ + HCl

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Properties :

1. Colourless gas with pungent smell
2. Extremely soluble in water HCl (g) + H₂O (l) 🡪 H₃O⁺ (aq) + Cl^- (aq)
3. In aqueous solution it is called Hydrochloric acid.
4. It reacts with ammonia NH₃ + HCl 🡪 NH₄Cl
5. It reacts with ammonia to give ammonium chloride (dense white fumes )
6. When three parts of concentrated HCl are mixed with one part of concentrated HNO₃ we get aqua regia. This is used in dissolving noble metals. Au + 4H⁺ + NO^3- + 4Cl^- 🡪 AuCl^4- + NO + 2H₂O 3Pt + 16 H⁺ + 4NO^3- + 18Cl^- 🡪 3PtCl₆^2- + 4NO + 8H₂O
7. Hydrochloric acid decomposes salts of weaker acids

Na₂CO₃ + 2HCl 🡪 2NaCl + H₂O + CO₂

NaHCO₃ + HCl 🡪 NaCl + H₂O + CO₂

Na₂SO₃ + 2HCl 🡪 2NaCl + H₂O + SO₂

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Oxoacids of Halogens�

Due to high electronegativity and small size, chlorine forms only one oxoacid HOF.�

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STRUCTURE OF OXOACIDS

Interhalogen compounds

When one Halogen combines with other Halogen atom the compound formed is known as Interhalogen compounds can have the following formula

XX’, XX’₃, XX’₅ and XX’₇ where X is the larger halogen than X

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Interhalogen compounds

• Most of the portion already covered.
• Interhalogen compounds are prepared by direct combination of the two halogens. The product formed depends upon the conditions given:

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• As the ratio between radii of X and X′ increases, the number of atoms per molecule also increases.
• Their physical properties are intermediate which means that the of constituent’s halogens except their melting point and boiling point are little higher than expected.
• Interhalogen compounds are more reactive than halogen except Fluorine because X-X’ bond in interhalogens is weaker than X-X bond in halogens except F-F’ bond.
• All of them undergo hydrolysis
• XX’ + H₂O → HX’ + HOX
• Their structure can be explained VSEPR Theory.
• XX₃ has bent ‘T’ shape.
• XX₅ → square pyramidal
• IX₇ → pentagonal bipyramidal

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Deduce the molecular shape of BrF3 on the basis of VSEPR theory.

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The central atom Br has seven electrons in the valence shell. Three of these will form electron-pair bonds with three fluorine atoms leaving behind four electrons. Thus, there are three bond pairs and two lone pairs. According to VSEPR theory, these will occupy the corners of a trigonal bipyramid. The two lone pairs will occupy the equatorial positions to minimise lone pair-lone pair and the bond pairlone pair repulsions which are greater than the bond pair-bond pair repulsions. In addition, the axial fluorine atoms will be bent towards the equatorial fluorine in order to minimise the lone-pair-lone pair repulsions. The shape would be that of a slightly bent ‘T’.

REASONING QUESTIONS

• Why is ICl more reactive than I₂ BUT F₂ is more reactive than ClF ?

GROUP 18

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• Helium (He)
• Neon (Ne)
• Argon (Ar)
• Krypton (Kr)
• Xenon (Xe)
• Radon (Rn)

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1. All noble gases except radon occur in nature.

2. Radon is obtained as a decay product of Radium

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3. The general configuration is ns²np⁶ and are noble gases because they generally do not interact with most of the elements.

4. Due to stable configuration, they have very high ionization enthalpy.

5. They are monotonic, colorless and sparingly soluble in water.

6. They have low melting and boiling points because of only type of inter-atomic interaction in these elements is weak dispersion forces.

7. Helium has lowest boiling point of 4.2K and has an unusual property of diffusing through most commonly used gas materials like rubber, glass or plastics.

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10. Neil Bartlett, prepared a red compound which is formulated as O₂⁺PtF₆^–. He, then realised that the first ionisation enthalpy of molecular oxygen (1175 kJmol–1) was almost identical with that of xenon (1170 kJ mol–1). He made efforts to prepare same type of compound with Xe and was successful in preparing another red colour compound Xe⁺PtF₆^– by mixing PtF6 and xenon.

11. Xenon forms three fluorides under three specific conditions

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12. Another method of preparation of XeF₆ is

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14. They are powerful fluorinating agents and are readily hydrolised by water.

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There structures are deduced from VSEPR theory

XeF₂ - linear

XeF₄ – square planar

XeF₆ – distorted octahedral

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15. Xenon fluorides react with fluorides to form cationic species and fluoro anions.

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Hydrolysis of XeF₄ and XeF₆

1. The hydrolysis is shown as follows

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2. Partial hydrolysis of XeF₆ gives the following

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3. XeO₃ is a colorless explosive solid with pyramidal structure

4. XeF₄ is a colorless volatile liquid with square pyramidal structure

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• Uses :

Helium is a non-inflammable and light gas. Hence, it is used in filling balloons for meteorological observations. It is also used in gas-cooled nuclear reactors. Liquid helium (b.p. 4.2 K) finds use as cryogenic agent for carrying out various experiments at low temperatures. It is used to produce and sustain powerful superconducting magnets which form an essential part of modern NMR spectrometers and Magnetic Resonance Imaging (MRI) systems for clinical diagnosis. It is used as a diluent for oxygen in modern diving apparatus because of its very low solubility in blood.

Neon is used in discharge tubes and fluorescent bulbs for advertisement display purposes. Neon bulbs are used in botanical gardens and in green houses.

Argon is used mainly to provide an inert atmosphere in high temperature metallurgical processes (arc welding of metals or alloys) and for filling electric bulbs. It is also used in the laboratory for handling substances that are air-sensitive.

There are no significant uses of Xenon and Krypton. They are used in light bulbs designed for special purposes.

Q. Why are noble gases least reactive?

OR

Why are they considered inert?

Ans. 1. Because of their noble gas configuration i.e. ns²np⁶

2. High ionization enthalpy and more positive electron gain enthalpy.

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Q. What made us believe that inert or noble gases can also interact?

Ans. 1. Neil Bartlett observed the reaction of a noble gas.

2. He prepared a red compound with the formula O₂⁺PtF₆^-

3. This compound made him realize that the first ionization enthalpy of molecular oxygen (1175 KJ/mol) was almost identical with that of Xenon (1170 KJ/mol)

4. Therefore, he made compound with Xenon and was successful in making another red coloured compound.

Xe⁺PtF₆^- by mixing PtF₆ and Xenon

5. This is how many compounds related to Xenon have been synthesized.

Q: What inspired N. Bartlett for carrying out reaction between Xe and PtF6?

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Q: Does the hydrolysis of XeF6 lead to a redox reaction?

A: No, the products of hydrolysis are XeOF4 and XeO2F2 where the oxidation states of all the elements remain the same as it was in the reacting state.

Q: Why is helium used in diving apparatus?

Q: How are XeO3 and XeOF4 prepared?

Q: Give the formula and describe the structure of a noble gas species which is isostructural with:

(i) ICl₄^– (ii) IBr₂^– (iii) BrO₃^–

Q: Why do noble gases have comparatively large atomic sizes?

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Q: List the uses of neon and argon gases