P BLOCK CLASS XII STUDY MATERIAL

CHAPTER- 7

p-BLOCK ELEMENTS

POINTS TO REMEMBER

The general valence shell electronic configuration of p-block elements ns² np^1-6

GROUP 15 ELEMENTS

Group 15 elements ; N, P, As, Sb & Bi

General electronic configuration: ns²np³

PHYSICAL PROPERTIES

^Ø        Dinitrogen is a diatomic gas while all others are solids.

^Ø        N & P are non-metals. As & Sb metalloids & Bi is a metals. This is due to decrease in ionization enthalpy & increase in atomic size.

^Ø        Electronegativity decreases down the group.

CHEMICAL PROPERTIES

o Common oxidation states : -3, +3 & +5.

o        Due to inert pair effect, the stability of +5 state decreases down the group & stability of +3 state increases .

o        In the case of Nitrogen all Oxidation states from +1 to +4 tend to disproportionate in acid solution , e.g.:- 3HNO₂→HNO₃ + H₂O + 2NO

Anomalous behavior of Nitrogen:- due to its small size, high electronegativity, high ionization enthalpy and absence of d-orbital.

N₂ has unique ability to form pπ-pπ multiple bonds whereas the heavier element of this group do not form pπ–pπ bond because their atomic orbitals are so large & diffuse that they cannot have effective overlapping.

Nitrogen exists as diatomic molecule with triple bond between the two N atoms whereas other elements form single bonds in elemental state.

N cannot form dπ-pπ bond due to the non-availability of d-orbitals whereas other elements can.

TRENDS IN PROPERTIES

Stability - NH₃>PH₃>AsH₃>SbH₃>BiH₃

Bond Dissociation Enthalpy- NH₃>PH₃>AsH₃>SbH₃>BiH₃

Reducing character - NH₃<PH₃<AsH₃<SbH₃<BiH₃

Basic character- NH₃>PH₃>AsH₃>SbH₃>BiH₃

Acidic character- N₂O₃>P₂O₃>As₂O₃>Sb₂O₃>Bi₂O₃

DINITROGEN

PREPARATION

·      Commercial preparation –By the liquefaction & fractional distillation of air.

·      Laboratory preparation –By treating an aqueous solution NH₄Cl with sodium nitrate . NH₄Cl +NaNO₂→ N₂ + 2H₂O + NaCl

·      Thermal decomposition of ammonium dichromate also give N₂.   (NH₄)₂Cr₂O₇ → N₂ +4H₂O + Cr₂O₃

·      Thermal decomposition of Barium or Sodium azide gives very pure N₂.

PROPERTIES

At high temperature nitrogen combines with metals to form ionic nitride (Mg₃N₂) & with non-metals, covalent nitride.

AMMONIA- NH₃

PREPARATION

^Ø        In laboratory it is prepared by heating ammonium salt with NaOH or lime.   2NH₄Cl + Ca(OH)₂→2NH₃+2H₂O + CaCl₂

^ In large scale it is manufactured by Haber ’s process

                                                  N₂+3H_2→ 2NH₃

                             ∆H⁰= - 46.1kJ/mol

According to   Le chatelier’s  principle  the  favourable conditions are:-

Optimum temperature: 700 K  High pressure: 200 atm

                      Catalyst: Iron Oxides

                           Promoter: K₂O & Al₂O₃

PROPERTIES

Ammonia is a colorless gas with pungent odour.

Highly soluble in water.

In solids & liquid states it exists as an associated molecule due to hydrogen bonding which accounts for high melting & boiling points of NH₃

Trigonal Pyramidal shape of NH₃ molecule

Name		Formula			Oxidation state	Chemical nature
Nitrous oxide or		N2O			+1	Neutral
Laughing gas
Nitric oxide		NO			+2	Neutral
Dinitrogen trioxide		N2O3			+3	Acidic
Dinitrogen tetra oxide		N2O4or NO2			+4	Acidic
Dinitrogen pentaoxide		N2O5			+5	Acidic

Aqueous solution of ammonia is weakly basic due to the formation of OH^- ion .

ZnSO₄+ 2NH₄OH→Zn(OH)₂+ (NH₄)₂SO₄

Ammonia can form coordinate bonds by donating its lone on nitrogen, ammonia forms complexes.

CuSO₄+4NH₃     →   [Cu(NH₃)₄]₂SO₄

NITRIC ACID (HNO₃)

PREPARATION

Ostwald’s process it is based on catalytic oxidation of ammonia by atmospheric oxygen . The main steps are

1)       4NH3 + 5O2	Pt	4NO + 6H2O
	500k, 9 Bar

2)                2NO+O₂ → 2NO₂

3)      3NO₂ + H₂O  → 2HNO₃ +NO

PROPERTIES

(i)     Conc. HNO3 is a strong oxidizing agent & attacks most  of metals. 

(ii)   Cr & Al do not dissolve in HNO₃ because of the formation of a possive film of oxide on the surface.

(iii) It oxidizes non-metals like I₂ to HIO₃, C to CO₂ , S to H₂SO₄

(iv) Brown ring test is used to detect NO3^-.

PHOSPHOROUS

ALLOTROPIC FORMS:  White , red,  α-black & β-black .

White phosphorous is more reactive than red phosphorous because white P exists as discrete P₄ molecules, in red Phosphorus several P₄ molecules are linked to formed polymeric chain.

PHOSPHINE (PH₃)

Preparation: It is prepared in laboratory by heating white P with concentrated NaOH solution in an Inert atmosphere of CO₂.

P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂

PHOSPHOROUS HALIDES

Phosphorous forms two types of halides PX₃ & PX₅ (X=F,I,Br)

Trihalides have pyramidal shape and pentahalides have trigonal bipyramidal structure.

OXOACIDS OF PHOSPHOROUS

·      The acids in +3 oxidation state disproportionate to higher & lower oxidation. 4H₃PO₃→ 3H₃PO₄+PH₃

·      Acids which contains P-H bond have strong reducing properties.eg:-H₃PO₂, H₃PO₃ are reducing agents due to presence of P-H bonds.

·      Hydrogen atom which are attached with oxygen in P-OH form are ionisable and are responsible for the basicity of the acid .

GROUP-16 ELEMENTS (CHALCOGENS)

Group 16 Elements:- O,S,Se,Te,Po

General electronic configuration:ns²np⁴

Element           Occurrence

Oxygen                   Comprises 20.946% by volume of the atmosphere.

Sulphur                    As sulphates such as gypsum CaSO₄.2H₂O,Epsom salt MgSO₄.7H₂O and Sulphides

Such as galena PbS, Zinc Blende ZnS, Copper Pyrites CuFeS₂

ATOMIC & PHYSICAL PROPERTIES

·      Ionisation enthalpy decreases from oxygen to polonium.

·      Oxygen atom has less negative electron gain enthalpy than S because of the compact nature of the oxygen atom.However from the S onwards the value again becomes less negative upto polonium

·      Electronegativity gradually decreases from oxygen to polonium, metallic character increases from oxygen to polonium.

·      Oxygen & S are non-metals, selenium and telerium are metalloids. Po is a radioactive metal.

·      Oxygen is a diatomic gas while S, Se & Te are octa atomic S₈ ,Se₈ & Te₈ molecules which has puckered ’ ring’ structure.

CHEMICAL PROPERTIES

·      Common oxidation state:-  -2,+2,+4 &+6.

·      Due to inert effect,the stability of +6 decreases down the group and stability of +4 increases.

Oxygen exhibits +1 state in O₂F₂, +2 in OF₂.

Anamolous behavior of oxygen-due to its small size, high electronegativity and absence of d-orbitals.

TREND IN PROPERTIES

Acidic character-H₂O<H₂S<H₂Se<H₂Te

Thermal stability-H₂O>H₂S>H₂Se>H₂Te

Reducing character-H₂S<H₂Se<H₂Te

Boiling point-H₂S<H₂Se<H₂Te<H₂O

Reducing property of dioxides-SO₂>SeO₂>TeO₂

Stability of halides-F>Cl>Br>I

HALIDES

DI HALIDES: sp³ hybridisation but angular structure.

TETRA HALIDES: sp³d hybridisation-see-saw geometry SF₄

HEXA HALIDES: sp³d²,octahedral  SF₆

DIOXYGEN

Prepared by heating oxygen containing salts like chlorates, nitrates

2KClO₃       ^heat→ 2KCl+3O₂

OXIDES

A binary compound of oxygen with another element is called oxide. Oxides can be classified on the basis of nature.

·      Acidic Oxides :- Non-metallic oxides. Aqueous solutions are acids. Neutralize bases to form salts.Ex: SO₂, CO₂, N₂O₅ etc.

·      Basic Oxides: Metallic oxides. Aqueous solutions are alkalies. Neutralize acids to form salts.Ex:Na₂O, K₂O etc.

·      Amphoteric oxides:-some metallic oxides exhibit a dual behavior. Neutralize both

acids & bases to form salts.    Ex:-Al₂O₃, SbO₂, SnO, etc.

OZONE PREPARATION

Prepared  by subjecting cold, dry oxygen to silent electric discharge.

                   3O₂    →     2O₃

PROPERTIES

Due to the ease with which it liberates atoms of nascent oxygen, it acts as a powerful oxidizing agent. For eg:- it oxidises lead sulphide to lead sulphate and iodide ions to iodine.

PbS +4O₃→PbSO₄+4O₂

SULPHUR DIOXIDE

PROPERTIES

·    Highly soluble in water to form solution of sulphurous acid       

SO₂+H₂O→H₂SO₃

·  SO₂ reacts with Cl₂ to form sulphuryl chloride     SO₂+Cl₂→SO₂Cl₂

·      It reacts with oxygen to form SO₃ in presence of V₂O₅ catalyst    2SO₂+O₂→2SO₃

·      Moist SO₂ behaves as a reducing agent. It converts Fe(III) ions to Fe(II) ions & decolourises acidified potassium permanganate (VII) solution( It is the test for the SO₂ gas).

SULPHURIC ACID

PREPARATION

It is manufactured by contact process which involves 3 steps

1.      Burning of S or Sulphide ores in air to generate SO₂.

2.      Conversion of SO₂ to SO₃ in presence of V₂O₅ catalyst

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

PROPERTIES

1.  In aqueous solution it ionizes in 2 steps

H₂SO₄+H₂O→H₃O⁺+HSO₄^-

HSO₄^-+H₂O→H₃O⁺+SO₄^2-

2.  It is a strong dehydrating agent Eg:-charring action of sugar

C12H22O11  + H2SO4

12C +11H2O

3.  It is a moderately strong oxidizing agent.

Cu+2H₂ SO₄ (conc.) →CuSO₄+ SO₂ +2H₂O          

C+2H₂SO₄ (conc)  →   CO₂ + 2SO₂ + 2H₂O

GROUP 17 ELEMENTS(HALOGENS)

Group 17 elements: F,Cl,Br,I,At

General electronic configuration:ns²np⁵

Element	Occurrence
Fluorine	As insoluble fluorides(fluorspar CaF2,Cryolite and
	fluoroapattie)
Cl, Br,I	Sea water contains chlorides, bromides and iodides
	Of
	Sodium, potassium magnesium and calcium, but is
	mainly sodium chloride solution(2.5% by mass).
	Certain forms of marine life(various seaweeds)

ATOMIC & PHYSICAL PROPERTIES

i.            Atomic & ionic radii increase from fluorine to iodine.

ii.            Ionization enthalpy gradually decreases from fluorine to iodine due to increase in atomic size.

iii.            Electron gain enthalpy of fluorine is less than that of chlorine. It is due to small size of fluorine & repulsion between newly added electron &electrons already present in its small 2p orbital.

iv.            Electronegativity decreases from fluorine to iodine. Fluorine is the most electronegative element in the periodic table.

v.            The color of halogens is due to absorption of radiations in visible region which results in the excitation of outer electrons to higher energy level.

vi.            Bond dissociation enthalpy of fluorine is smaller than that of chlorine is due to electron-electron repulsion among the lone pair in fluorine molecules where they are much closer to each other than in case of chlorine. The trend: Cl-Cl>Br-Br>F-F>I-I.

CHEMICAL PROPERTIES

OXIDATION STATES:-

However, chlorine, bromine &iodine exhibit +1, +3, +5, +7 oxidation states also.

Fluorine forms two oxides OF₂ and O₂F₂. These are essentially oxygen fluorides because of the higher electronegativity of fluorine than oxygen.

Anomalous behavior of fluorine-

due to its small size, highest electronegativity, low F-F bond dissociation enthalpy and absence of d-orbitals.

TRENDS IN PROPERTIES

Oxidizing property –F₂>Cl₂>Br₂>I₂

Acidic strength- HF<HCl<HBr<HI

Stability & bond dissociation enthalpy- HF>HCl>HBr>HI

Stability of oxides of halogens- I>Cl>Br

Ionic character of halides –MF>MCl>MBr>MI

CHLORINE

PREPARATION

1.      MnO₂ +4HCl  →  MnCl₂+Cl₂+2H₂O

2.      4NaCl+MnO₂+4H₂SO₄  →^MnCl₂+4 NaHSO₄+2H₂O+Cl₂

3.      2KMnO₄+16HCl  → 2KCl+2MnCl₂+8H₂O+5Cl₂

4.      DEACON’S   PROCESS

4HCl+O₂—^CuCl_2→2Cl₂+2H₂O

5. By electrolysis of brine solution. Cl₂ is obtained at anode.

PROPERTIES

i.            With cold and dilute alkali Cl₂ produces a mixture of chloride and hypochlorite but with hot and concentrated alkalis it gives chloride and chlorate.

2NaOH+Cl_2→NaCl+NaOCl+H₂O 6NaOH+3Cl_{2 -------------------------→} 5NaCl+NaClO₃+3H₂O

ii.            With dry slaked lime it gives bleaching powder. 2Ca (OH) ₂+2Cl_{2     -----------------}→

Ca(OH)₂ +CaCl₂+2H₂O

iii.            It is a powerful bleaching agent; bleaching action is due to oxidation

Cl₂+H₂O→2HCl + [O]

Colored substance + (O) → colorless substance

iv.            Action of concentrated H₂SO₄ on NaCl give HCl gas. NaCl+H₂SO₄ ^420K  NaHSO₄+HCl

3:1 ratio of conc. HCl & HNO₃ is known as  aquaregia  & it is used for dissolving noble metals like Au and Pt.

OXOACIDS OF HALOGENS

Interhalogen compounds are prepared by direct combination of  halogens. Ex: ClF, ClF₃, BrF₅, IF₇

They are more reactive than halogens because X-X’ is weaker-X bondsin than X halogens (except F-F).

TYPE	STRUCTURE
XX’3	Bent T-shaped
XX’5	Square pyramidal
XX’7	Pentagonal bipyramidal

GROUP-18 ELEMENTS

GROUP-18 ELEMENTS: He, Ne, Ar, Kr, Xe & Rn General electronic configuration:ns²np⁶

Atomic radii- large as compared to other elements in the period since it corresponds to vander Waal radii.

Inert –due to complete octet of outermost shell, very high ionization enthalpy & electron gain enthalpies are almost zero.

The first noble compound prepared by Neil Bartlett was XePtF₆ & Xenon. O₂⁺PtF₆^-.led to the discovery of XePtF₆ since first ionization enthalpy of molecular oxygen (1175 kJmol^-1) was almost identical with that of xenon (1170kJmol^-1).

PROPERTIES

Xe+F₂      673K, 1bar   →  XeF₂

Xe (g) +2F₂(g) 873k, 7bar → XeF₄(s)

Xe (g) +3F₂(g)  573k, 60-70 bar  →  XeF₆(s)

XeF₆+MF →M⁺ [XeF₇]^-

XeF₂+PF5→[XeF]⁺[PF₆]^-

XeF₆+2H₂O→XeO₂F₂+4HF (partial hydrolysis)

SOLVED QUESTIONS

1 MARK QUESTIONS

1.      Ammonia has higher boiling point than phosphine. Why?

Ans-Ammonia forms intermolecular H-bond.

2.      Why does PCl₃ fume in moisture?

Ans- In the presence of (H₂O), PCl₃ undergoes hydrolysis giving fumes of HCl .               PCl₃ + 3H₂O→H₃PO₃ + 3HCl

3.      What Happens when H₃PO₃ is Heated ?

Ans- It disproportionate to give orthophosphoric acid and Phosphine.

4H₃PO_3→ 3H₃PO₄ + PH₃

4.Why H₂S is acidic and H₂O is neutral ?

Ans- The S –H bond is weaker than O –H bond because the size of S atom is bigger than that of O atom . Hence H₂S can dissociate to give H⁺ Ions in aqueous solution.

5.Name two poisonous gases which can be prepared from chlorine gas?

Ans-    Phosgene (COCl₂), tear gas (CCl₃NO₂)

6.Name the halogen which does not exhibit positive oxidation state .

Ans- Flourine being the most electronegative element does not show positive oxidation state .

7.Iodine forms I₃^-  but F₂ does not form F₃^-  ions .why?

Ans- Due to the presence of vacant d-orbitals , I₂ accepts electrons from I^- ions to form I₃^- ions , but due to absence of d-orbitals F₂ does not accept electrons from F-ions to form F₃^- ions.

8.Draw the structure of peroxosulphuric acid.

Ans- As per text book.

9.Phosphorous forms PCl₅ but nitrogen cannot form NCl₅. Why?

 Ans- Due to the availability of vacant d-orbital in P it is able to expand its octet.

2 MARK QUESTION (SHORT ANSWER TYPE QUESTION)

1. Why is HF acid stored in wax coated glass bottles?

Ans- This is because HF does not attack wax but reacts with glass. It dissolves SiO₂ present in glass forming hydrofluorosilicic acid.

SiO₂ +6HF→H₂SiF₆+2H₂O

2. What is laughing gas? Why is it so called? How is it prepared?

Ans- Nitrous oxide (N₂O) is called laughing gas, because when inhaled it produced hysterical laughter. It is prepared by gently heating ammonium nitrate.

NH₄NO₃→N₂O+2H₂O

3. Give reasons for the following:

(i)   Conc.HNO3 turns yellow on exposure to sunlight.

(ii)   PCl5 behaves as an ionic species in solid state.

Ans- 

(i)Conc HNO₃ decompose to NO₂ which is brown in colour & NO₂ dissolves in HNO₃ to it yellow.

(ii) It exists as [PCl₄]⁺ [PCl₆]^- in solid state.

4. What happens when white P is heated with conc. NaOH solution in an atmosphere of CO₂? Give equation.

Ans-Phosphine gas will be formed.

P₄+3NaOH+3H₂O→PH₃+3NaH₂PO₂

5. How is ozone estimated quantitatively?

Ans- When ozone reacts with an excess of potassium iodide solution

Buffered with a borate buffer (pH-9.2), Iodide is liberated which can be titrated against a standard solution of sodium thiosulphate . This is a quantitative method for estimating O₃ gas.

6. Are all the five bonds in PCl₅ molecule equivalent? Justify your answer.

Ans- PCl₅ has a trigonal bipyramidal structure and the three equatorial P-Cl bonds are equivalent, while the two axial bonds are different and longer than equatorial bonds.

7. NO₂ is coloured and readily dimerises. Why ?

Ans- NO₂ contains odd number of valence electrons.It behaves as a typical odd molecules .On dimerization; it is converted to stable N₂O₄ molecule with even number of electrons.

8. Write the balanced chemical equation for the reaction of Cl₂ with hot and concentrated NaOH .Is this reaction a dispropotionation reaction? Justify:

Ans- 3Cl₂+6NaOH→5NaCl+NaClO₃+3H₂O

Yes, chlorine from zero oxidation state is changed to -1 and +5 oxidation states.

 9. Account for the following.

(i)SF₆ is less reactive than SF₄.

(ii) Of the noble gases only xenon compounds are known.

Ans.

 (i)In SF₆ there is less repulsion between F atoms than In SF₄.

(ii)Xe has low ionisation enthalpy & high polarising power due to larger atomic size.

10. With what neutral molecule is ClO- Isoelectronic? Is that molecule a Lewis base?

Ans- ClF ,Yes, it is Lewis base due to presence of lone pair of electron.

 3 MARK QUESTIONS

 (i) why is He used in diving apparatus?

(ii)Noble gases have very low boiling points.Why?

 (iii)Why is ICl more reactive than I2?

Ans-

(i)It is not soluble in blood even under high pressure.

(ii)Being monoatomic they have weak dispersion forces.

(ii)I-Cl bond is weaker than I-I bond.

2.Complete the following equations.

 (i)XeF₄+H₂O →

(ii)Ca₃P₂+H₂O→

(iii)AgCl(s) +NH₃ (aq) →

Ans-  (i) 6XeF₄+12H₂O→4Xe+2XeO₃+24HF+3O₂

(ii)Ca₃P₂+6H₂O→3Ca(OH)₂+2PH₃

(iii)AgCl(s) +2NH₃ (aq) →[Ag(NH₃)₂]Cl(aq)

3. (i)How is XeOF₄ prepared ?Draw its structure.

(ii)When HCl reacts with finely powdered iron, it forms ferrous chloride and not ferric chloride .Why?

Ans-  (i)Partial hydrolysis of XeOF₄

  XeF₆ + H₂O→ XeOF₄ + 2HF Structure -square pyramidal.

(ii) Its reaction with iron produces H₂

Fe+2HCl→FeCl₂+H₂

Liberation of hydrogen prevents the formation of ferric chloride.

5 MARK QUESTION

1. Account for the following.

(i)Noble gas form compounds with F₂ &O₂ only.

 (ii)Sulphur shows paramagnetic behavior.

(iii)HF is much less volatile than HCl.

(iv) White phosphorous is kept under water.

(v)Ammonia is a stronger base than phosphine. 

Ans-

 (i) F₂&O₂ are best oxidizing agents.

(ii)In vapour state sulphur partly exists as S₂ molecule which has two unpaired electrons in the antibonding pi *orbitals like O₂ and, hence, exhibit paramagnetism.

(iii)HF is associated with intermolecular H bonding.

(iv)  Ignition temperature of white phosphorous is very low (303 K). Therefore on explosure to air, it spontaneously catches fire forming P₄O₁₀. Therefore to protect it from air, it is kept under water.

(v)Due to the smaller size of N, lone pair of electrons is readily available.

2. When Conc. H₂SO₄ was added to an unknown salt present in a test tube, a brown gas (A) was evolved. This gas intensified when copper turnings were added in to test tube. On cooling gas (A) changed in to a colourless gas (B).

(a) Identify  the  gases  ‘A’  and  ‘B’

(b)Write the equations for the reactions involved

Ans- The  gas whereas‘A’is NO₂.‘B’  is  N₂O₄

NaNO₃(Salt ) +  H₂SO₄ (conc.)  →NaHSO₄ + HNO₃

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

                                           Blue        Brown (A)

2NO₂ (On cooling) →           N₂O₄

                                        Colourless(B)

3. Arrange the following in the increasing order of the property mentioned.

(i)HClO, HClO₂, HClO₃, HClO₄ (Acidic strength)

(ii)As₂O₃, ClO₂, GeO₃, Ga₂O₃ (Acidity)

(iii)NH₃, PH₃, AsH₃, SbH₃ (H-E-H bond angle)

(iv)HF, HCl, HBr, HI (Acidic strength)

(v)MF, MCl, MBr, MI (ionic character)

Ans-(i)Acidic strength:HClO<HClO₂<HClO₃<HClO₄

(ii)Acidity: Ga₂O₃<GeO₂<AsO₃<CIO₂

(iii)Bond angle: SbH₃<AsH₃<PH₃<NH₃

(iv)Acidic strength: HF<HCl<HBr<HI

(v)Ionic character: MI<MBr<MCl<MF

ASSIGNMENTS

Very shot answer type questions:

1)   PH3 has lower boiling point than NH3. Explain.

2)   Why are halogens coloured.

3)   What are chalcogens?

4)   Which noble gas is Radioactive?

5)   Explain why fluorine always exhibit an oxidation state of - 1 only.

6)   Which compound led to the discovery of compounds of noble gas?

7)   Name the most electronegative element.

8)   Why is OF₆ compound not known?

9)   Why N₂ is not reactive?

10)    Ammonia acts as a ligand. Explain.

Short answer type questions:

1) White Phosphorous is more reactive than red phosphorous. Explain.

2)   Why do noble gases have comparatively large atomic sizes?

3)   Arrange in decreasing order of Ionic character

M –F, M –Cl, M –Br, M –I

4)   H₃PO₂ acid behaves as a mono basic acid .

5)   Arrange the following in the order of property indicated:

 a) AS₂O₃, ClO₂, GeO₂, Ga₂O₃__Increasing acidity

b) H₂O, H₂S, H₂Se, H₂Te__Increasing acid strength.

6)   Arrange in decreasing order of bond energy:

F₂, Cl₂, Br₂, I₂

7) Complete the following:

i)   HNO₃  + P₄O₁₀    →

ii)    IO₃ ^- + I ^-  +   H+   →

8)     Give the chemical reactions in support of following observations:

 a) The +5 oxidation state of Bi is less stable than +3 oxidation state.

 b) Sulphur exhibits greater tendency for catenation than selenium.

9)   How would you account for following?

i)   Enthalpy of dissociation of F₂ is much less than that of Cl₂.

ii)   Sulphur in vapour state exhibits paramagnetism.

10) Draw structures of following:

a)     H₂SO₅   

b)       XeF₄

LEVEL –III

1. Complete and balance:

i)   F₂ + H₂O (cold)→

ii)  BrO₃ ^- + F₂ + OH ^-    →

iii)   Li + N₂ (cold) →

iv)   NH₃ + NaOCl→

2)   Despite lower electron affinity of F₂, it is stronger oxidising agent than Cl₂. Explain.

3)Give reasons:

a)    Nitric oxide becomes brown when released in air.

b)    PCl₅ is ionic in nature and exist in the solid state.

4)   Which of the two is more covalent SbCl₃ or SbCl₅?

5)   Addition of Cl₂ to KI solution gives a brown colour but excess of it turns colourless. Explain.

VALUE BASED QUESTIONS

Q1 .You are staying near a fertilizer factory. In the middle of the night there is a leakage of ammonia which is detected by its smell. Within 10 minutes you find the smell is intolerable.

a. What would you do as first aid against this gas spill accident for self and neighbour? (2)

b. What value do you derive from this? (1)

Ans:

a. Ammonia is highly soluble in water. It is detected by its characteristic fishy

odour. Hence keep a wet kerchief on your nose to stop inhaling the gas.

Then help your neighbours with your suggestion.

b. Alertness to tackle disasters for society.

Q2.Manoj went to a paper industry. The manager, paper industry H₂O₂ insisted that H₂O₂   be used for bleaching instead of Chlorine in bleaching. Manoj had learnt that Cl₂ is also a bleaching agent

a. Then why is H₂O₂ used instead of Cl₂ (2)

b. What value do you derive from this? (1)

Ans:

a) H₂O₂ after bleaching, the product formed is water. While when using chlorine the byproduct is HCl.

b) Seeing the products of the reaction we should select the reagents, so that there is minimum pollution.

Q3.Nitrogen combines with hydrogen to form ammonia. N₂ + 3H₂ → 2NH₃.

Ammonia is the basic raw material for preparing fertilizers. Always associated with a refinery/ petrochemical industry we have a fertilizer industry.

a. Why? (2)

b. What is the value you derive from this. (1)

Ans:a.

 In the refinery / petrochemical industry hydrogen gas is evolved as a bi product.

Hence to recycle this hydrogen a fertilizer unit is established nearby.

b. Recycling of industrial waste keeps the environment clean.

Q4. At an exhibition a FORTUNE TELLER predicts your future. Ram and Shyam ran to get their fortune read. The fortune teller asked them to take a paper from the lot. He put the paper into a trough of water. Both the children read what was given in the paper.

a. Give a posssible reason for this. (2)

b. What value do you get from this? (1)

Ans:

a) The writing was done using solution of lead acetate. This had become

invisible after drying. The trough contained a solution of H₂S. Reaction of

H₂S with water gave a black precipitate of lead sulphide. Hence the

Writing becomes visible.

b) Knowledge is the antidote to fear and blind belief.

Q5. Ajay, a student of chemistry, was performing chemical reaction between sodium thiosulfate and HCI. He found that time required to appear turbidity increases when concentration of HCI or sodium thiosulfates or both decreases.

(a) Mention the reason for appearance of turbidity.

(b) Write the chemical reaction  involved.

(c) Mention the values associated with above experiment.

Ans :The precipitation of  S.

a)      Na₂S₂O₃ (aq) + 2HCl (aq) → 2NaCl(aq) + H₂O (l) + SO₂(g) + S(s)

b)      Knowledge of science.

_{FREQUENTLY ASKED QUESTIONS IN BOARD EXAMS}

_{(FOR SELF PRACTICE)}

Q.1Why is red phosphorus less reactive than white phosphorus?                                              1

Q.2 What is the basicity of H₃PO₂ acid and why?                                                                      1

Q.3Which is a stronger reducing agent, SbH₃ or BiH₃, and why?                                              1

Q.4Nitrogen is relatively inert as compared to phosphorus, why?                                             1

Q.5 Name two poisonous gases which can be prepared from chlorine gas.                               1                                                                                         

Q.6 XeF₂ is a linear molecule without a bent, why?                                                                   1   

Q.7 Complete the following chemical reaction equations-

(i)XeF₂ + H₂O   --------->      (ii)PH₃  + HgCl₂ --------->                                                             2

Q.8 Explain-  (i)Oxygen is a gas but sulphur a solid.  (ii)The halogens are coloured.                2                                                           

Q.9How are interhalogen compounds formed? What general compositions can be assigned to them?      2

Q.10 Explain-

(i)Electron gain enthalpy of fluorine with negative sign is less than that of chlorine.

(ii)The two oxygen oxygen bond lengths in ozone molecule are identical.                                 2

Q.11Explain-

(i)The acid strength order of compounds increases in the order-    PH₃ < H₂S <HCl

(ii )SF₆ is kinetically inert.                                                                                                           2

Q.12 Explain-          (i)O₃ acts as a powerful oxidising agent.

(ii)BiH₃ is the strongest reducing agent amongst all the hydrides of group 15 elements.           2                                                                              

Q.12 Explain-  (i)Boiling point of H₂O is higher than HF.  (ii)H₂S is more acidic than H₂O.  (iii)Fluorine does not exhibit any positive oxidation state.                                                         3

Q.13 How would you account for the following –   (i) H₂S is more acidic than H₂O.

(ii)The N-O bond in NO₂^- is shorter than the N-O bond in NO₃^- .

(iii) Both O₂ and F₂ stabilize high oxidation states but the ability of oxygen to stabilize the higher oxidation state exceeds that of fluorine.                                                                                      3                                                                                                  

Q.14 Explain following-  (i)NF₃ is an exothermic compound whereas NCl₃ is not.

(ii)All the bonds in SF₄ are not equivalent.  

(iii)Despite lower value of its electron gain enthalpy with negative sign, fluorine (F₂) is a stronger oxidising agent than chlorine (Cl₂).                                                                              3

Q.15 (a)Draw the structures of the following –  (i) H₂S₂O₈                       (ii) HClO₄

(b)How would you account for the following –  (i) NH₃ is a stronger base than PH₃.

(ii)In the structure of HNO₃, the N-O bond (121pm)is shorter than N-OH  bond(140 pm).

(iii) F₂ is stronger oxidising agent than Cl₂.                                                                                5

Q.16(a) Draw the structures of following-  (i) H₂S₂O₇                (ii)     HClO₃

(b)Explain the following observations-       (i) ICl is more reactive than I₂.

(ii) All the P-Cl bonds in PCl₅ are not equivalent.

(iii) Sulphur has a greater tendency for catenation than oxygen.                                                    5  

Q.17 (a) Draw structures-       (i) H₃PO₂    (ii) XeOF₄   

(b) Explain the following observations-

(i)The electron gain enthalpy of sulphur atom has a greater negative value than that of oxygen atom.     

 (ii)Nitrogen does not form pentahalides.

(iii)In aqueous solution HI is a stronger acid than HCl.                                                               5

Q.18 (a) Draw structures-         (i) BrF₃ /ClF₃                 (ii) N₂O₅/H₂SO₄

(b)Explain the following-   (i)No chemical compound of helium is known.

(ii)Bond dissociation energy of fluorine is less than that of chlorine.

(iii) Phosphorus has a greater tendency for catenation than nitrogen.                                          5                                                                                                   

Q.19 (a)Draw structures-                                                                                                               5

(i)(HPO₃)₃                                   (ii) H₄P₂O₅                 (iii) XeF₄

(b)Complete  chemical equations-

 (i) SO₃  + H₂SO₄ ------->              (ii) XeF₄ + H₂O --------->   (iii)  Cu + HNO₃ (dil) -------->          

 (iv)  XeF₄  + O₂F₂ -----> (v) NaOH (cold ,dilute) + Cl₂-----> (vi)  XeF₆+ H₂O ------------->                                                                                    

Q.20 (a) What happens when –

(i)Chlorine gas is passed through a hot concentrated solution of NaOH?

(ii)Sulphur dioxide gas is passed through an aqueous solution of a Fe(III) salt?

(b) Answer the following –

(i) What is the basicity of H₃PO₃ and why?

(ii) Why does fluorine not play the role of a central atom in inter halogen compounds?

(iii) Why do noble gases have very low boiling points?                                                                5

Q.21 Give reasons for the followings-

(a)(CH3)₂ P=O exists but (CH3)₂ N=O does not.

(b)Oxygen has less electron gain enthalpy with negative sign than sulphur.

(c)H₃PO₂ is a stronger reducing agent than H₃PO₃.  

(d) Why does NO₂ dimerise?  

(e)Why does NH₃  acts as a Lewis base?                                                                                         5

Q.22 Complete the following reactions-

(a) C + Conc.H₂SO₄   ----->

(b) XeF₄ + H₂O₂   --------->     

(c) P₄ + H₂O          ---------->    

(d)Ag + PCl₅                    ------>   

(e) CaF₂  +  H₂SO₄          ----->                                                                                                         5

Q23. Why is helium used in diving apparatus?                                                                            1

Q24.Draw the structure of (a) BrF₃ (b) XeF₄ (c) XeO₃                                                                3

Q25. Arrange the following in the order of the property indicated for each set                          3

a) F₂, Cl_2,Br_2, I₂  -increasing bond dissociation enthalpy

b ) HF, HCl, HBr, HI -increasing acid strength

c )NH_3,PH_3,AsH_3,SbH₃  -    decreasing base strength

Q25.  a) What are inter halogen compounds? Give examples also.

  b)How is the presence of SO₂ detected?

  c) Draw the structure  of noble gas species which is isoelectronic with BrO₃^-.                  3 

Q26. i)  Account for the following:                                                                                                      5

      a)Nitrogen shows catenation properties lesser than phosphorus.

      b) K_a2<<K_a1 for H₂SO₄ in water.  

      c) HI is the strongest reducing agent amongst the hydrogen halides.

ii)    Write the reactions for the following:

a)Sulphuric acid reacts with copper.

b) Ammonia reacts with  copper sulphate solution.

Q.27                                                                                                                                        5

i)What are the conditions, required to maximize the yield of sulphuric acid by Contact Process?

ii) Account for the following :

a)Enthalpy of dissociation for F₂ is smaller than that for Cl₂.     

b)Inter Halogen compounds more reactive than Halogens?        

iii)  Give equations that involve during the brown ring test.

 Q.28                                                                                                                                       2

           (I)         PH₃ has lower boiling point than NH₃, why?

(II)       Write balance equation: when ammonia is dissolved in water.

   Q.29  Draw the structure of                                                                                                             3

(1)   H₂ SO₃      (2) H₂SO₄        (3) H₂S₂O₇

Q.30 Explain each of the following.                                                                                       3

(1)   Nitrogen is much less reactive than phosphorus.

(2)   The solubility of + 5 oxidation state decreases drown in the group 15.

(3)   The bond angles (O – N – O) are not of the same value in NO₂^- and NO₂⁺

Q.31Give reasons for the following:

(i)               Aqueous solution of ammonia is slightly basic.

(ii)             The bond angle in PH₄⁺ higher than in PH_3.

(iii)           Axial bonds in PCl₅ longer than equatorial bonds.

Q.32Give reasons-

(i)     PF₅ is well known compound but NF₅ is not known.

(ii)   Fluorine shows abnormal behaviour.

(iii) The acid strength of acids increases in the order- HF < HCl <HBr <HI.

Q.33 Give reasons for the following-

(i)        Orthophosphorous acid is not tribasic acid.

(ii)      Valency of oxygen is generally two whereas sulphur shows valency of two, four and six.

(iii)    Phosphine has lower boiling point than ammonia.

SOME SOLVED BOARD EXAM PROBLEMS

1
Ans
2 Ans
3
Ans
4
Ans