Role of Chemistry in Environmental Protection and Economic Development

Guess Paper – 2009 Class – X Subject – Chemistry (P- Block Elements) Kalra sir 09460472649 [email protected] com Group 15 elements: |Symbol |Atomic |Electronic configuration | | |number | | | |7 |[He]2s22p3 | |N | | | |P |15 |[Ne] 3s23p3 | |As |33 |[Ar]3d104s24p3 | |Sb |51 |[Kr]4d105s25p3 | |Bi |83 |[Xe]4f 145d106s26p3 |

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General introduction, electronic configuration, occurrence, oxidation states, trends in physical and chemical properties; nitrogen – preparation, properties and uses; compounds of nitrogen: preparation and properties of ammonia and nitric acid, oxides of nitrogen (structure only); Phosphorous-allotropic forms; compounds . of phosphorous: preparation and properties of Phosphine, halides (PCl3, PCl5) and oxoacids (elementary idea only) Group 16 elements: Symbol |Atomic |Electronic configuration | | |number | | | |8 |[He]2s22p4 | |O | | | |S |16 |[Ne] 3s23p4 | |Se |34 |[Ar]3d104s24p4 | |Te |52 |[Kr]4d105s25p4 | |Po |84 |[Xe]4f 145d106s26p4 |

General introduction, electronic configuration, oxidation states, occurrence, trends in physical and chemical properties; Dioxygen: preparation, properties and uses; simple oxides; Ozone. Sulphur – allotropic forms; compounds of sulphur: preparation, properties and uses of Sulphurdioxide; sulphuric acid: industrial process of manufacture, properties and uses, oxoacids of Sulphur (structures only).

Group 17 elements: |Symbol |Atomic number |Electronic configuration | |F |9 |[He]2s22p5 | |Cl |17 |[Ne] 3s23p5 | |Br |35 |[Ar]3d104s24p5 | |I |53 |[Kr]4d105s25p5 | |At |85 |[Xe]4f 145d106s26p5 |

General introduction, electronic configuration, oxidation states, occurrence, trends in physical and chemical properties; compounds of halogens: preparation, properties and uses of chlorine and hydrochloric acid, Interhalogen compounds, oxoacids of halogens (structures only). Group 18 elements: Symbol |Atomic number |Electronic configuration | |He |2 |1s2 | |Ne |10 |[He]2s22p6 | |Ar |18 |[Ne] 3s23p6 | |Kr |36 |[Ar]3d104s24p6 | |Xe |54 |[Kr]4d105s25p6 | |Rn |86 |[Xe]4f 145d106s26p6 | General introduction, electronic configuration. Occurrence, trends in physical and chemical properties, uses. 1.

Account for the following: (Group 15 elements) (i) There is a considerable increase in covalent radius from N to P. However, from As to Bi only small increase in covalent radius is observed. Ans: This is due to the presence of completely filled d and/or f orbital in heavier members. (ii) Ionization enthalpy decreases down the group 15. Ans: Due to gradual increase in atomic size. (iii) The ionization enthalpy of the group 15 elements is much greater than that of group 14 elements in the corresponding periods. Ans: Because of the extra stable half-filled p orbital electronic configuration and smaller size. iv) Nitrogen exists as diatomic molecule and phosphorus as P4. Ans: Because N2 is Diatomic molecules, hence weak Vander Waal’s force of attraction thus is a gas whereas P4 is tetraatomic hence Stronger Vander Waal’s force of attraction thus it is solid. (v) NH3 is basic while BiH3 is only feebly basic. Ans: NH3 is basic due to smaller size & high electro negativity of Nitrogen. (vi) The HNH angle value is higher than HPH, HAsH, and HSbH angles. Ans: (vii) R3P=O exist but R3N=O does not. Ans: Due to the absence of d orbitals in valence shell of nitrogen.

Because of inability of Nitrogen to expand its covalency beyond four, nitrogen cannot form d ? –p ? bond (viii) Nitrogen shows catenation properties less than phosphorus. Ans:Because N-N bond is weaker than the single P-P bond. & strong p? –p? overlap in Nitrogen (ix) Ammonia has higher boiling point than Phosphine. Ans: Ammonia (NH3) form hydrogen bond but Phosphine (PH3) does not (x) H3PO3 is diprotic acid. Ans:Due to presence of two ionisable –OH group. (Draw the structure of H3PO3) (xi) Oxides of nitrogen have open chain structures while those of phosphorus have closed chain or cage structures.

Ans: Nitrogen has ability to form multiple bonds involving p(-p( overlap. (xii) A nitrogen atom has five valence electrons but it does not form the compound NCl5. Ans: Because of absence of d-orbitals it can’t expand its covalency from 3 to 5. (xiii) Nitrogen is fairly inert gas. Ans: Nitrogen exists as triply bonded diatomic non polar molecule. Due to short internuclear distance between two nitrogen atoms the bond strength is very high. It is, therefore, very difficult to break the bond. (xiv) All the bonds in the molecules of PCl5 are not equal.

Ans: PCl5 has a trigonal bipyramidal shape in the gas space. A trigonal bipyramidal is an irregular structure in which some bond angles are 90 degree and others of 120 degree resulting in unequal P-Cl bond lengths PCl5 + H2O( POCl3 + 2HCl & PCl5 (PCl3 + Cl2 (xv) In solid state PCl5 exists as Ionic compound. Ans: Since solid phosphorous pentachloride exists as [PCl4]+ [PCl6]- and hence exhibit some ionic character. [PCl4]+ is tetrahedral and the anion, [PCl6]–octahedral. (xvi) Nitrogen does not form pentahalides.

Ans:Nitrogen with n = 2, has s and p orbitals only. It does not have d orbitals to expand its covalence beyond four. That is why it does not form pentahalide. (xvii) NH3 is a good complexing agent. Ans: Because nitrogen has lone pair of electrons which it can donate to form co-ordinate bond (xviii) Nitrogen shows anomalous behaviour. Ans: Nitrogen differs from the rest of the members of this group due to its smaller size, high electronegativity, high ionisation enthalpy and non-availability of d orbitals (xix) Pentavalent Bismuth is a strong oxidizing agent.

Ans:Bi3+ is more stable than Bi5+ due to inert pair effect (xxi) The first ionization energy of nitrogen is greater than oxygen. Ans: Because of the extra stable half-filled p orbitals electronic configuration of nitrogen. (xxii) NCl3 gets hydrolysed easily while NF3 does not. Ans: In NCl3, Cl has vacant d-orbital but in NF3, F does not have to accept lone pair of electrons donated by O2 atoms of H2O (xxiii)PH3 has lower boiling point than NH3. Ans: PH3 molecules are not associated through hydrogen bonding in liquid state. That is why the boiling point of PH3 is lower than NH3. xxiv)Pentahalides of group 15 are more Covalent than trihalides Ans:Higher the positive oxidation state of central atom, more will be its polarizing power which, in turn, increases the covalent character of bond formed between the central atom and the other atom. (xxv)N2 is less reactive at room temperature. Ans: Because of strong p? –p? overlap Nitrogen has triple bond between two nitrogen atoms N? N which has high bond dissociation enegy. So it is less reactive (xxvi)NH3 act as Lewis base. Ans: Nitrogen atom in NH3 has one lone pair of electrons which is available for donation.

Therefore, it acts as a Lewis base (xxvii)Bond angle in PH4+ is higher than that in PH3. Ans: Both are SP3 hybridised. In PH4+ all the four orbitals are bonded whereas in PH3 there is a lone pair of electrons on P, which is responsible for lone pair-bond pair repulsion in PH3 reducing the bond angle to less than 109° 28?. (xxviii) NO2 dimerises to form N2O4 .Ans:NO2 contains odd number of valence electrons. It behaves as a typical odd molecule. On dimerisation, it is converted to stable N2O4 molecule with even number of electrons. (xxix) NO2 is coloured but N2O4 is colourless. Ans.

NO2 has unpaired electrons therefore it absorbs light from visible and radiate brown colour whereas N2O4 doesnot have unpaired electrons so it does not absorb light from visible region. (xxx) H3PO2 and H3PO3 act as as good reducing agents while H3PO4 does not. Ans: In H3PO2, two H atoms are bonded directly to P atom & in H3PO3 one H atom is bonded directly to P atom which imparts reducing character to the acid, whereas in H3PO4 there is no H atom bonded directly to P atom (xxxi) NH3 form hydrogen bond but PH3 does not. Ans: Due to smaller size & high electronegativity of Nitrogen. xxxii) CN? ion is known but CP? ion is not known. Ans: Account for the following: (Group 16 elements) (i)Elements of group 16 generally show lower value of first ionization enthalpy compared to the corresponding periods of group 15. Ans: Due to extra stable half-filled p orbitals electronic configurations of Group 15 elements, larger amount of energy is required to remove electrons compared to Group 16 elements. (ii) Tendency to show –2 oxidation state diminishes from Sulphur to polonium in group 16. Ans:The outer electronic configuration of group 16 elements is ns2 np4.

These elements therefore have the tendency to gain two electrons to complete octet. Since elctronegativity and I. E. decrease on going down the group, tendency to show –2 oxidation state diminishes. (iii) Oxygen generally exhibit oxidation state of –2 only whereas other members of the family exhibit +2, +4, +6 oxidation states also. Ans:Oxygen is a electronegative element thus exhibit oxidation state of –2. .Other members of the family have d orbitals and therefore, can expand their octets and show + 2, + 4, + 6 oxidation states also. (iv)There is large difference between the melting point of Oxygen & Sulphur.

Ans:The large difference between the melting and boiling points of oxygen and sulphur may be explained on the basis of their atomicity; oxygen exists as diatomic molecule (O2) whereas sulphur exists as polyatomic molecule (S8). (v) SF6 is known but SH6 is not known. Ans: Because of very high electronegativity if fluorine , sulphur exhibits its maximum oxidation stateof +6 in SF6. SH6 is not formed because hydrogen is feebly electronegative. (vi) Oxygen is anomalous in many properties. Ans:The anomalous behaviour of oxygen,is due to its small size and high electronegativity (vii) SF6 is not easily hydrolysed.

Ans: Since sulphur is hexa-covalent it cannot expand its covalency. . Sulphur hexafluoride, SF6 is exceptionally stable for steric reasons. (viii) SCl6 is not known but SF6 is known. Ans: Because of high electronegativity of fluorine sulphur exhibits its maximum oxidation state (+6) in SF6 (ix) The acidic strength of hydrides of group 16 varies in the order H2O< H2S < H2Se < H2Te. Ans:The increase in acidic character can be explained in terms of decrease in bond (H–E) dissociation enthalpy down the group. (x) H2S is less acidic than H2Te. Ans:Due to the decrease in bond (E–H) dissociation enthalpy down the group, cidic character increases. (xi) H2O is a liquid and H2S is a gas. Ans: Because of small size and high electronegativity of oxygen, molecules of water are highly associated through hydrogen bonding resulting in its liquid state. (xii) Ozone (O3) act as a powerful oxidising agent. Ans: Due to the ease with which it liberates atoms of nascent oxygen (O3 > O2 + O), it acts as a powerful oxidising agent. (xiii) Sulphur vapours exhibits paramagnetism Ans :In vapour state sulphur partly exists as S2 molecule which has two unpaired electrons in the antibonding ? orbitals like O2 and, hence, exhibits paramagnetism. (xiv)Dioxygen is a gas but Sulphur is a solid. Ans: Because O2 is Diatomic molecules, hence weak Vander Waal’s force of attraction thus is a gas whereas S8 is octaatomic hence Stronger Vander Waal’s force of attraction thus it is solid. (xv) H2O has higher boiling point than H2S. Ans:Due to presence of Intermolecular hydrogen bonding in H2O. (xvi) Inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not. Ans:Oxygen has smaller size than chlorine. Smaller size favours hydrogen bonding

Account for the following: (Group 17 elements) (i)Halogens have the smallest atomic radii in their respective periods. Ans:The halogens have the smallest atomic radii in their respective periods due to maximum effective nuclear charge. (ii)Halogens have maximum negative electron gain enthalpy in the respective periods of the periodic table. Ans:Halogens have the smallest size in their respective periods and therefore high effective nuclear charge. As a consequence, they readily accept one electron to acquire noble gas electronic configuration. (iii) Halogens are coloured. Ans:Halogens are 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. For example, F2, has yellow, Cl2 , greenish yellow, Br2, red and I2, violet colour. (iv) Bond dissociation energy of F2 is less than Cl2. Ans:Due tolarge electron-electron repulsion among the lone pairs in F2 molecule where they are much closer to each other than in case of Cl2. (v)Although electron gain enthalpy of Fluorine is less negative as compared to Chlorine, Fluorine is stronger oxidizing agent than chlorine.

Ans:It is due to (i) low enthalpy of dissociation of F-F bond (ii) high hydration enthalpy of F-. (vi) Halogens are strong oxidizing agent. Ans:Due to strong tendency to accept electrons (vii) Fluorine exhibit oxidation state of –1 only whereas other halogens exhibit +1, +3, +5, +7 oxidation states also. Ans:Fluorine is the most electronegative element and cannot exhibit any positive oxidation state. Other halogens have d orbitals and therefore, can expand their octets and show + 1, + 3, + 5 and + 7 oxidation states also. (viii) Fluorine is anomalous in many properties. (I.

E, Bond dissociation, etc) Ans:The anomalous behaviour of fluorine is due to its small size, highest electronegativity, low F-F bond dissociation enthalpy, and non availability of d orbitals in valence shell. (ix)Most of the reactions of fluorine are exothermic. Ans:Due to the small and strong bond formed by it with other elements) (x)Fluorine forms only one oxoacid, HOF. Ans:Due to high electronegativity and small size F does not form HOFO,HOFO2&HOFO3 in which oxdn. no. will be +3,+5 &+7 (xi)The acidic strength of hydrogen halides varies in the order HF H–I. (xii) HI in aqueous solution is strong acid than HF.

Ans:Due to low bond (H–X) dissociation enthalpy in HI (xiii) HF is the weakest acid among hydrohalo-acids inspite of the fact that fluorine is most electronegative. Ans:Due to highest bond (H–X) dissociation enthalpy in HF (xiv) Metal fluorides are more ionic than its chlorides. Ans:Greater the difference in electronegativity, more will be its ionic character, fluorine being more electronegative than chlorine (xv) HF has higher boiling point than HCl. Ans:Due to presence of intermolecular hydrogen bonding in HF (xvi) HF is least volatile whereas HCl is most volatile among all halogens.

Ans:HF is associated with intermolecular hydrogen bonding, so it is least volatile whereas HCl is gas and has least surface area, therefore minimum Vander waals forces of attraction and minimum boiling point (xviii)The boiling point of hydrogen halides varies in the order: HF > HI>HBr>HCl. Ans:The boiling point of HF is highest due to intermolecular hydrogen bonding then HI>HBr>HCl. ( DecreasingVanderwaals forces) (xix) Interhalogen compounds are more reactive than related elemental halogens. Ans:In general, interhalogen compounds are more reactive than halogens due to weaker X–X1 bonding than X–X bond.

Thus, ICl is more reactive than I2. (xx). ICl is more reactive than Cl2. Ans:Due to weaker I-Cl bonding than Cl-Cl bond. Thus, ICl is more reactive than I2. (xxi) The acidic strength of oxoacids of halogens varies in the order HOF>HOCl>HOBr>HOI. Ans:Greater the electronegativity of halogen in oxoacids greater is acidic strength of oxoacids. (F>Cl>Br>I) (xxii) The acidic strength of oxoacids of halogens varies in the order: HClO4> HClO3> HClO2> HClO Ans:Greater the oxidation state of halogen in oxoacids greater is acidic strength of oxoacids. +7>+5>+3>+1) (xxiii) OF2 should be called oxygen fluoride and not fluorine oxide. Ans:OF2 should be called oxygen fluoride and not fluorine oxide. Since oxygen is less electronegative than fluorine, OF2 should be called oxygen diflouride. (xxiv) ClF3 exists but FCl3 does not. Ans:Due to unavailability of d-orbitals in fluorine atom it cannot expand its valence shell. Therefore it is unable to form FCl3 whereas Chlorine has vacant d-orbitals. Hence it can promote one of the 3 p electrons to the 3-d sub shell and shows +3 oxidation state and forms ClF3 (xxv)Cl2 bleaches a substance permanently but SO2 does it temporarily.

Ans:Cl2 bleaches a substance permanently because it is due to Oxidation but SO2 does it temporarily because it is due to reduction. (xxvi)When HCl reacts with finely powdered iron, it forms ferrous chloride and not ferric chloride. Ans:Its reaction with iron produces H2. Fe + 2HCl (FeCl2 +H2 . Liberation of hydrogen prevents the formation of ferric chloride. (xxvii) Chlorine water has both oxidizing and bleaching properties. Account for the following: (Group 18 elements) (i)The elements of group 18 are known as noble gases.

The elements present in Group 18 have their valence shell orbitals completely filled and, therefore, react with a few elements only under certain conditions. Therefore, they are now known as noble gases. (ii)Noble gases are mostly chemically inert. Their inertness to chemical reactivity is attributed to the following reasons: (i) The noble gases except helium (1s2) have completely filled ns2p66 electronic configuration in their valence shell. (ii) They have high ionisation enthalpy and more positive electron gain enthalpy. (iii)Noble gases have comparatively largest atomic sizes.

In noble gases we can measure only Van der Waals radii which are larger than covalent radii. (iv)Noble gases exist as monatomic. Noble gases have stable electronic configuration, therefore, they cannot form covalent bonds. So they exist as monoatomic (v)Noble gases exhibit very high ionization enthalpy. Due to stable electronic configuration these gases exhibit very high ionisation enthalpy. (vi)Noble gases exhibit large positive values of electron gain enthalpy. Since noble gases have stable electronic configurations, they have no tendency to accept the electron and therefore, have large positive values of electron gain enthalpy. vii)Noble gases have very low boiling points Noble gases being monoatomic have no interatomic forces except weak dispersion forces and therefore, they are liquefied at very low temperatures. Hence, they have low boiling points. .(viii) Of the noble gases only Xenon is known to form real chemical compounds. The ionization energy of xenon is relatively low and therefore, it is possible to excite the paired electrons from np orbitals to nd subshell (ix) Noble gases form compounds with fluorine and oxygen only.

Because fluorine and oxygen are strong oxidizing agents(most electronegative elements) (x) Helium is used for inflating aeroplane tyres & filling balloons for meteorogical observations. Helium is a non-inflammable and light gas (xi)Helium is used in diving apparatus. Because of its very low solubility in blood. (xii)It has been difficult to study the chemistry of radon. Radon is radioactive with very short half-life which makes the study of chemistry of radon difficult. (xiii) Xenon readily forms compounds but Krypton does not form compounds easily.

Xenon has lower ionization energy than Krypton, therefore, Xe forms compounds (xiv) He and Ne does not form compounds with fluorine. Due to non-availability of vacant d-orbital (xv) Xe does not forms compounds such as Xe F3 and XeF5. By the promotion of one, two or three electrons from filled p-orbital to the vacant d-orbital in the valence shell, 2,4 or 6 half filled orbitals are formed. Thus Xe can combine only with even number of fluorine and not odd. 2. Answer the following: (Group 15 elements) (i) A translucent white waxy solid (A) on heating in an inert atmosphere is converted to its allotropic form (B).

Allotrope (A) on reaction with very dilute aqueous KOH liberates a highly poisonous gas (C) having rotten fish smell. With excess of chlorine forms (D) which hydrolyses to compound (E). Identify compounds (A) to (E). Ans: (ii) Discuss the general characteristics of group 15 elements with reference to their electronic configuration, oxidation state, atomic size, and ionization enthalpy and electro negativity. Ans: Refer to page Number 167 of NCERT textbook. (iii)Nitrogen is anomalous in many properties. Why? Give two examples to show that Nitrogen is anomalous in many properties.

Ans: Nitrogen differs from the rest of the members of this group due to its smaller size, high electronegativity, high ionisation enthalpy and non-availability of d orbitals. Nitrogen has unique ability to form p ? -p ? multiple bonds with itself and with other elements having small size and high electronegativity. Nitrogen exists as a diatomic molecule with a triple bond (one s and two p) between the two atoms. Consequently, its bond enthalpy is very high. (iv)Arrange the following in increasing order of the property indicated: a) NH3, PH3, AsH3, SbH3, BiH3 (Thermal stability) ) NH3, PH3, AsH3, SbH3, BiH3 (Bond dissociation enthalpy) c) NH3, PH3, AsH3, SbH3, BiH3 (Reducing character) d) NH3, PH3, AsH3, SbH3, BiH3 (Basic character) Ans: (a)NH3> PH3 > AsH3 >SbH3 >BiH3 (b) NH3< PH3 < AsH3 BiH3 (v) Write the balanced equations involved in the laboratory preparation of Dinitrogen. Ans (a) In the laboratory, dinitrogen is prepared by treating an aqueous solution of ammonium chloride with sodium nitrite. NH4Cl (aq) + NaNO2 (aq) > N2 (g) + 2H2O (l) + NaCl (aq) (b) It can also be obtained by the thermal decomposition of ammonium dichromate. NH4)2Cr2O7 > N2 + 4H2O + Cr2O3 (c) By the thermal decomposition of sodium or barium azide. Ba (N3)2 > Ba + 3N2 (vi) How is Dinitrogen commercially produced? Ans: Refer to page Number 168 of NCERT textbook (vii)Write the reaction of: Thermal decomposition of sodium azide. Ans: 2NaN3> 2Na +3N2 (viii) How is Ammonia manufactured industrially? Ans: Refer to page Number 170 of NCERT textbook (ix)Mention the conditions required to maximize the yield of ammonia? Ans: Refer to page Number 170 of NCERT textbook (x) How is Ammonia obtained on small scale? Ans: Refer to page Number 170 of NCERT textbook xi)Draw the structure of Ammonia? Ans: Refer to page Number 171 of NCERT textbook (xii)Write down the names, reaction involved in preparation, physical appearance and chemical nature of N2O, NO, N2O3, NO2, N2O4, & N2O5. Ans: Refer to page Number 172 of NCERT textbook (xiii) Draw the structure of N2O, NO, N2O3, NO2 , N2O4, & N2O5. Ans: Refer to page Number 171 of NCERT textbook (xiv)Why does NO2 dimerise. Ans: NO2 contains odd number of valence electrons. It behaves as a typical odd molecule. On dimerisation, it is converted to stable N2O4 molecule with even number of electrons. xv)What is the covalence of nitrogen in N2O5? Ans: From the structure of N2O5 it is evident that covalence of nitrogen is four. (xvi) How is nitric acid obtained on large scale by Ostwald process? Ans: Refer to page Number 174 of NCERT textbook. (xvii) Write the balanced equations involved in the laboratory preparation of nitric acid. Illustrate how copper metal can give different products on reaction with HNO3. Ans: Refer to page Number 174 of NCERT textbook. (xviii) Draw the structure of nitric acid in the gaseous state. Ans: Refer to page Number 174 of NCERT textbook. xix)Metals like Cr, Al do not dissolve in nitric acid. Why Ans: Metals (e. g. , Cr, Al) do not dissolve in concentrated nitric acid because of the formation of a passive film of oxide on the surface. (xx)Describe Brown ring Test. Ans: The brown ring test for nitrates depends on the ability of Fe2+ to reduce nitrates to nitric oxide, which reacts with Fe2+ to form a brown complex. The test is usually carried out by adding dilute ferrous sulphate solution to an aqueous solution containing nitrate ion, and then carefully adding concentrated sulphuric acid along the sides of the test tube.

A brown ring at the interface between the solution and sulphuric acid layers indicate the presence of nitrate ion in solution. NO3 – + 3Fe2+ + 4H+ > NO + 3Fe3+ + 2H2O [Fe(H2O)5]2+ + NO > [Fe (H2O)5 (NO)]2+(brown) + H2O (xxi) Give some uses of nitric acid? Ans: Refer to page Number 175 of NCERT textbook (xxii)Write main differences between the properties of white phosphorus and red phosphorus. Ans: Refer to page Number 175 of NCERT textbook (xxiii) Write the balanced equations involved in the preparation of PH3 (Phosphine). Ans: P4 +3NaOH+ 3H2O( PH3 +NaH2PO4 (xxiv)In what way can it be proved that PH3 is basic in nature?

Ans: PH3 reacts with acids like HI to form PH4I which shows that it is basic in nature. PH3 + HI ( PH4I . Due to lone pair on phosphorus atom, PH3 is acting as a Lewis base in the above reaction. (xxv) What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO2? Ans: Same as (xxiii) (xxvi) Write the balanced equations involved in the preparation of PCl3 & PCl5. Ans: Refer to page Number 177 of NCERT textbook (xxvii) Draw the structure of PCl5 in the gaseous state and in the solid state Ans: Refer to page Number 177 of NCERT textbook xxviii)Why does PCl3 fume in moisture? Ans: PCl3 hydrolyses in the presence of moisture giving fumes of HCl. PCl3 + 3H2O (H3PO3 +3HCl (xxix)Are all the five bonds in PCl5 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. (xxx) Write the name and draw the shape of following of following H3PO4, H3PO3, H3PO2, H3P2O7 & (HPO3)3. Ans: Refer to page Number 179 of NCERT textbook (xxxi)What is the basicity of H3PO4, H3PO3, H3PO2?

Ans: Basicity of H3PO4=3, H3PO3=2, H3PO2=1 Only those H atoms which are attached with oxygen in P–OH form are ionisable and cause the basicity. Thus, H3PO3 and H3PO4 are dibasic and tribasic, respectively as the structure of H3PO3 has two P–OH bonds and H3PO4 three. (xxxii)What happens when H3PO3 is heated? Ans: Refer to page Number 180 of NCERT textbook (xxxiii)Why does the reactivity of nitrogen differ from phosphorus? Ans: N2 is less reactive than phosphorus Because of strong p? –p? overlap Nitrogen has triple bond between two nitrogen atoms N? N which has high bond dissociation enegy.

So it is less reactive (xxxiv) Give the disproportionation reaction of H3PO3. Ans: 4H3PO4 > 3H3PO4 + PH3 (xxxv) Can PCl5 act as an oxidising as well as reducing agent? Justify Ans: (xxxvi) How do you account for the reducing behaviour of H3PO2 on the basis of its structure? Ans: In H3PO2, two H atoms are bonded directly to P atom which imparts reducing character to the acid. Answer the following: (Group 16 elements) 1. Write the name and draw the shape of following of following: H2SO4, H2SO3,H2S2O8 and H2S2O7. Ans: Refer to page Number 189 of NCERT textbook. 2. Draw the structure of Sulphur in S8 (rhombic form) and S6 form.

Ans: S8 ring in both the forms is puckered and has a crown shape and in cyclo-S6, the ring adopts the chair form. Refer to page Number 187 of NCERT textbook 3. Give two examples to show that Oxygen is anomalous in many properties. Ans: The anomalous behaviour of oxygen is due to its small size and high electronegativity one typical example of effects of small size and high electronegativity is the presence of strong hydrogen bonding in H2O which is not found in H2S. 4. List the important sources of Sulphur. Ans:Combined sulphur exists primarily as sulphates such as gypsum CaSO4. H2O, epsom salt MgSO4. 7H2O, baryte BaSO4 and sulphides such as galena PbS, zinc blende ZnS, copper pyrites CuFeS2. Traces of sulphur occur as hydrogen sulphide in volcanoes. Organic materials such as eggs, proteins, garlic, onion, mustard, hair and wool contain sulphur. 5. Write the order of thermal stability of the hydrides of group 16 elements. Ans: H2O>H2S > H2Se > H2Te> H2Po 6. Write the balanced equations involved in the laboratory preparation of Dioxygen. (Hint: FromKClO3, Ag2O, HgO, Pb3O4, PbO2, H2O2) Ans: Refer to page Number 184 of NCERT textbook. 7. How is Dioxygen manufactured?

Ans: On large scale it can be prepared from water or air. Electrolysis of water leads to the release of hydrogen at the cathode and oxygen at the anode. Industrially, Dioxygen is obtained from air by first removing carbon dioxide and water vapour and then, the remaining gases are liquefied and fractionally distilled to give dinitrogen and dioxygen. 8. Give some uses of Dioxygen? Ans: (i) In normal respiration and combustion processes (ii) oxygen is used in oxyacetylene welding, in the manufacture of many metals, particularly steel. (iii)Oxygen cylinders are widely used in hospitals, high altitude flying and in mountaineering. iv)The combustion of fuels, e. g. , hydrazines in liquid oxygen, provides tremendous thrust in rockets. 9. How are simple oxides classified, explain with examples? Ans: Simple oxides can be classified on the basis of their acidic, basic or amphoteric character. An oxide that combines with water to give an acid is termed acidic oxide (e. g. , SO2, Cl2O7, CO2,) The oxides which give a base with water are known as basic oxides (e. g. , Na2O, CaO, BaO) Some metallic oxides exhibit a dual behaviour. They show characteristics of both acidic as well as basic oxides. Such oxides are known as amphoteric oxides.

They react with acids as well as alkalies. There are some oxides which are neither acidic nor basic. Such oxides are known as neutral oxides. Examples of neutral oxides are CO, NO andN2O. 10. Aluminium oxide (Al2O3) shows amphoteric character . Justify with chemical reactions. Ans: Al2O3 + 6HCl + 9H2O( 2[Al(H2O)6]3+ + 6Cl- Al2O3+ 6NaOH + 3H2O( 2Na3[Al(OH)6] 11. How is Ozone prepared? Ans: When a slow dry stream of oxygen is passed through a silent electrical discharge, conversion of oxygen to ozone (10%) occurs. The product is known as ozonised oxygen. 3O2 > 2O3 ? H0 (298 K) = +142 kJ mol–1 12.

Draw the structure Ozone (O3). Ans: Refer to page Number 186 of NCERT textbook 13. (a)What are the threats to Ozone layer? Ans: Experiments have shown that nitrogen oxides (particularly nitric oxide) combine very rapidly with ozone and there is, thus, the possibility that nitrogen oxides emitted from the exhaust systems of supersonic jet aeroplanes might be slowly depleting the concentration of the ozone layer in the upper atmosphere. Another threat to this ozone layer is probably posed by the use of freons which are used in aerosol sprays and as refrigerants. 13(b)How is Ozone (O3) estimated quantitatively?

Or On addition of ozone gas to KI solution, violet vapours are obtained Ans: When ozone reacts with an excess of potassium iodide solution buffered with a borate buffer (pH 9. 2), iodine is liberated which can be titrated against a standard solution of sodium thiosulphate. This is a quantitative method for estimating O3 gas. 14. Which form of Sulphur shows paramagnetic behaviour? Ans: S2 is the dominant species and is paramagnetic like O2. 15. Write differences between monoclinic Sulphur (? ) and rhombic Sulphur (? ). Ans: Refer to page Number 187 of NCERT textbook 16.

Write the balanced equation involved in the laboratory preparation and industrial preparation of Sulphur Dioxide. Ans: Refer to page Number 188 of NCERT textbook 17. Draw the structure of Sulphur Dioxide. Comment on the nature of two S-O bonds formed in SO2 molecule. Are the two S-O bonds in this molecule equal? Ans: Refer to page Number 188 of NCERT textbook 18. What happens when:(a)Sulphur Dioxide is passed through an aqueous solution of Fe (III) salt? (b) Sulphur Dioxide reacts with acidified potassium permanganate (c) Sulphur Trioxide is passed through water. (d)Concentrated H2SO4 is added to calcium fluoride.

Ans:(a)2Fe3+ + SO2 +2H2O (2Fe2+ + SO42- + 4H+ (c)CaF2 + H2SO4( CaSO4 + 2HF (b)5SO2 + 2MnO4- +2H2O(5SO42- +4H+ +2Mn2+ (d)SO3 + H2O ( H2SO4 19. How is the presence of Sulphur Dioxide (SO2) detected? Ans: When moist, sulphur dioxide behaves as a reducing it decolourises acidified potassium permanganate(VII) solution. 5SO2 + 2MnO4- +2H2O(5SO42- +4H+ +2Mn2+ 20. Explain how sulphuric acid is manufactured by the contact process. Give reactions also. Ans: Sulphuric acid is manufactured by the contact process which involves three steps: (i)Burning of sulphur or sulphide ores in air to generate SO2.

S8 +8O2 [pic] 8SO2 (ii) Conversion of SO2 to SO3 by the reaction with oxygen in the presence of a catalyst (V2O5), 2SO2 + O2 [pic] 2SO3 (iii)Absorption of SO3 in H2SO4 to give Oleum (H2S2O7) SO3 + H2SO4[pic] H2S2O7 (iv)Dilution of oleum with water gives H2SO4 of the desired concentration. H2S2O7 + H2O [pic] 2H2SO4 21. Write the conditions to maximize the yield of H2SO4 by contact process. Ans: Low temperature and high pressure are the favourable conditions for maximum yield. But the temperature should not be very low otherwise rate of reaction will become slow. 22.

Mention the areas in which Sulphur Dioxide (SO2) and sulphuric acid (H2SO4) plays an important role. Ans: Refer to page Number 188 & 191 of NCERT textbook 23. How is Sulphur Dioxide (SO2) an air pollutant? Ans: It dissolves in rain water and produces acid rain. 24. Which aerosols deplete ozone? Ans:Freons 25. Draw the structure of SF6 & SF4. Ans:Try Yourself 26. Give chemical reaction in support of each of the following; a) Sulphuric acid (H2SO4) is a strong dehydrating agent b) Sulphuric acid (H2SO4) is a strong oxidising agent. c) Sulphur Dioxide is reducing agent.

Ans: (a) C12H22O11 [pic]12C + 11H2O (b) Cu +2H2SO4 (Conc. ) [pic] CuSO4 + SO2 + 2H2O d) 5SO2 + 2MnO4- +2H2O(5SO42- +4H+ +2Mn2+ 27. Concentrated sulphuric acid is added followed by heating to each of (i) to (v) (i)Cane sugar (ii) Sodium Bromide (iii) Copper turnings (iv) Sulphur powder (v) Potassium Chloride . Identify in which of the above the following change will be observed. Support your answer with the help of a chemical equation. (a) Formation of black substance(b) Evolution of brown gas (c) Evolution of colour less gas (d) Formation of brown substance which on dilution becomes blue. e) Disappearance of yellow powder along with evolution of colourless gas. 28. When conc. H2SO4 was added into an unknown salt present in a test tube, a brown gas (A) was evolved. This gas intensified when copper turnings were also added into this test-tube. On cooling, the gas (A) changed into a colourless gas (B). (a) Identify the gases A and B. (b) Write the equations for the reactions involved. Answer the following: (Group 17 elements) (i)Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F2 and Cl2.

Ans:F2 has more oxidising power than Cl2. Because standard electrode potential value of F2 is higher than Cl2 . Standard electrode potential value are dependent on the parameters bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy. (ii)Give two examples to show that Fluorine is anomalous in many properties. Ans: Ionisation enthalpy, electronegativity, enthalpy of bond dissociation and electrode potentials are all higher for fluorine than expected from the trends set by other halogens. Also, ionic and covalent radii, m. p. and b. p. and electron gain enthalpy are quite lower than expected.

The anomalous behaviour of fluorine is due to its small size, highest electronegativity, low F-F bond dissociation enthalpy, and non availability of d orbitals in valence shell. (iii)Sea is the greatest source of some halogens. Comment Ans:Sea water contains chlorides, bromides and iodides of sodium, potassium, magnesium and calcium. (iv)Give the reason for bleaching action of Cl2. Ans:It is a powerful bleaching agent; bleaching action is due to oxidation. Cl2+ H2O ( 2HCl + O Coloured substance + O > Colourless substance,Bleaching effect of chlorine is permanent. v)Name two poisonous gases which can be prepared from chlorine gas. Ans: Poisonous gases such as phosgene (COCl2), tear gas (CCl3NO2), mustard gas (vi) Write the balanced equation involved in the laboratory preparation of chlorine gas. (Hint: from MnO2 and KMnO4) Ans:By heating manganese dioxide with concentrated hydrochloric acid. MnO2 + 4HCl > MnCl2 + Cl2 + 2H2O However, a mixture of common salt and concentrated H2SO4 is used in place of HCl. 4NaCl + MnO2 + 4H2SO4 > MnCl2 + 4NaHSO4 + 2H2O + Cl2 (ii) By the action of HCl on potassium permanganate. KMnO4 + 16HCl > 2KCl + 2MnCl2 + 8H2O + 5Cl2 (vii) How is chlorine gas manufactured by Deacons process and Electrolytic process? Ans: (i) Deacon’s process: By oxidation of hydrogen chloride gas by atmospheric oxygen in the presence of CuCl2 (catalyst) at 723 K. 4HCl +O2 > 2Cl2 + 2H2O (Kalra sir 09460472649 [email protected] com) (ii) Electrolytic process: Chlorine is obtained by the electrolysis of brine (concentrated NaCl solution). Chlorine is liberated at anode. It is also obtained as a by–product in many chemical industries. viii) Give five uses of chlorine gas. Ans:It is used (i) for bleaching wood pulp (required for the manufacture of paper and rayon), bleaching cotton and textiles, (ii) in the extraction of gold and platinum (iii) in the manufacture of dyes, drugs and organic compounds such as CCl4, CHCl3, DDT, refrigerants, etc. (iv) in sterilizing drinking water and (v) preparation of poisonous gases such as phosgene (COCl2), tear gas (CCl3NO2), mustard gas (ClCH2CH2SCH2CH2Cl). (ix) Write the balanced equation involved in preparation of Hydrogen Chloride.

Ans: In laboratory, it is prepared by heating sodium chloride with concentrated sulphuric acid. NaCl + H2SO4 > NaHSO4 + HCl at (420K) NaHSO4 + NaCl) > Na2SO4 + HCl at ( 823K) (x) Give uses for each of the following ;(i) ClO2 (ii)I2O5 (iii)HCl (iv)Interhalogen compounds. Ans: Try Yourself (xi)Deduce the molecular shapes of following on the basis of VSEPR theory. (a)BrF5 (b) ClF5 (c) IF7 (d) BrF3 (e) ClF3 (f) I3- Ans: Try Yourself (xii) Write the name and draw the shape of following of following: HClO4, ClO4-HClO3, HClO2, and HClO. Ans: Refer to page number: 200 of NCERT Textbook xiv)Write the reactions of F2 and Cl2 with water. Ans: Refer to page number: 195 of NCERT Textbook (xv) Give chemical reaction in support of each of the following; i) Chlorine gas is Oxidising agent. Ans: SO2 + 2H2O + Cl2 > H2SO4 + 2HCl ii) Fluorine is a stronger oxidizing agent than chlorine. Ans: F2 + 2X– > 2F– + X2 (X = Cl, Br or I) (xvi)How can you prepare Cl2 from HCl and HCl from Cl2. Write reactions only? Ans: Try Yourself (xvii)Write the balanced chemical equation for the reaction of Cl2 with hot and concentrated NaOH. Is this reaction a disproportionation reaction?

Justify. Ans:3Cl2 + 6NaOH > 5NaCl + NaClO3 + 3H2O, Yes, chlorine from zero oxidation state is changed to –1 and +5 oxidation states. Answer the following: (Group 18 elements) (i)How are XeF2 XeF4, XeF6, XeO3, XeOF4 prepared? Ans:(i)Xe (g) + F2 (g) (XeF2(s) Temp= 673K, Pressure=1bar (xenon in excess) (ii)Xe (g) + 2F2 (g) (XeF4(s) Temp= 873 K, Pressure=7 bar (1:5 ratio) (iii)Xe (g) + 3F2 (g) ( XeF6(s) Temp= 573 K, Pressure=60 -70bar (1:20 ratio) (iv)XeF6 +3H2O ( XeO3 + 6HF (v)XeF6 +H2O ( XeOF4 + 2HF (ii)Draw the shape of following of following: XeF2 XeF4, XeF6, XeO3, and XeOF4.

Ans: Refer to page Number NCERT Book: 205 (iii)Give the formula and describe the structure of a noble gases species which is isostructural with (i) BrO3- (ii) ICl4-(iii) IBr2- Ans: (i) XeO3 (ii) XeF4 (iii) XeF2 (iv)Give two uses for each of the following: (i) Helium (ii) Neon (iii) Argon Ans:Helium is used in filling balloons for meteorological observations. 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 (v) Which compound led to discovery of compounds of noble gases? Ans: O2+PtF6- (vi)Does the hydrolysis of XeF6 lead to a redox reaction? Ans: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. (vii) What inspired N. Bartlett for carrying out reaction between Xe and PtF6?

Ans In March 1962, Neil Bartlett, then at the University of British Columbia, observed the reaction of a noble gas. First, he prepared a red compound which is formulated as O2+PtF6-. He, then realised that the first ionization 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. After this discovery, a number of xenon compounds mainly with most electronegative elements like fluorine and oxygen, have been synthesized. . Complete the following equations: (Group 16 elements) 1. Ca + O2 [pic] 2. Al+ O2 [pic] 3. P4+ O2 [pic] 4. ZnS + O2[pic] 5. CH4+ O2 [pic] 6. SO2+ O2[pic] 7. HCl+ O2 [pic] 8. CaO+ H2O[pic] 9. PbS+ O3[pic] 10. Cu + H2SO4 (Conc. ) ( 11. C + H2SO4 (Conc. )( 12. S + H2SO4 (Conc. )( 13. MX + H2SO4 (Conc. )( Complete the following equations: (Group 17 elements) (i) NaCl + MnO2 + H2SO4 ( (ii)MnO2 + HCl ( (iii)KMnO4 + HCl ( (iv) NaCl + H2SO4 ( (v)Cl2 + NaOH (cold & dilute) ( (vi)Cl2 + NaOH (Hot & Concentrated) ( (vii)Ca (OH)2 + Cl2 ( (viii)NH3 (excess) + Cl2( (ix) NH3 + Cl2 (excess)( (x)Al + Cl2( xi) Na + Cl2( (xii) Fe+ Cl2( (xiv) P4+ Cl2( (xv) S8+ Cl2( (xvi) H2+ Cl2( (xvii) H2S+ Cl2( (xviii) C10H16+ Cl2( (xix) H2O+ Cl2( (xx)F2 + X-( (xxi) Cl2 + X-( (xxii) Br2 + X-( (xxii)NaI + Cl2( Complete the following equations: (Group 18 elements) Kalra sir 09460472649 [email protected] com (a)6XeF4 + 12H2O(4Xe + 2XeO3 + 24 HF + 3 O2 (b) XeF6 +3H2O ( XeO3 + 6HF (c) 2XeF2 + 2H2O (2Xe (g) + 4HF + O2(g) (d) XeF4+O2F2( XeF6 + O2 (e) XeF2+ PF5([XeF]+ [PF6]– (f) XeF4 + SbF5([XeF3]+ [SbF6]– (g)XeF6 +MF( M+[XeF7]– (M = Na, K, Rb or Cs) Kalra sir 09460472649 [email protected] com

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