Inorganic chemistry

C H A P T E R

I Group 15 Elements : The Nitrogen Family

1. In group 15 of the Periodic Table, the elements, nitrogen (₇N), phosphorus (₁₅P), arsenic (₃₃As), antimony (₅₁Sb) and bismuth (₈₃Bi) are present. 2. (a) The elements of this group can exhibit various oxidation states ranging between –3 to + 5.

(b) Maximum covalency of Nitrogen is four because it does not have d- orbitals to expand its covalency.

3. The atomic (covalent) and ionic radii (in a particular oxidation state) of the elements of nitrogen family (group 15) are smaller than the corresponding elements of carbon family (group 14).

There is a considerable increase in covalent radius from N to P. However, from As to Bi, only a small increase is observed.

4. Nitrogen displays a great tendency to form

pp – pp multiple bonds with itself as well as with

carbon and oxygen. The tendency to exhibit

pp – pp multiple bonding decreases as we move

down the group.

5. Group 15 elements are more electronegative than group 14 elements. Electronegativity decreases on moving down the group from N to Bi. 6. All elements of group 15 form gaseous hydrides

of the type MH₃.

(a) Th e basic str ength of th e hydr ides decreases as we down the group. Thus, NH₃ is the strongest base.

NH₃ > PH₃ > AsH₃ > SbH₃

(b) The thermal stability of the hydrides decreases as the atomic size increases,. 7. N cannot form NX₅ because of non-availability

of d-orbitals. Bi cannot form a BiX₅ because of

reluctance of 6s electrons of Bi to participate in bond formation.

8. Nitrogen forms a number of oxides. The rest of the members (P, As, Sb and Bi) of the group form two types of oxides: E₂O₃ and E₂O₅.

II Group 16 Elements

9. In group 16 of the Periodic Table, elements, oxygen (₈O), sulphur (₁₆S), selenium (₃₄Se), tellurium (₅₂Te) and polonium (₈₄Po) are present. 10. The elements have the electronic configuration

ns2_np4 _{for their valence shells. The first element} of the group 16 differs in its chemical behaviour from that of other members of the group due to its small size and high electronegativity. 11. The metallic character increases with increase in

atomic number. The first four elements are non-metallic in character. Non-non-metallic character is strongest in O and S, weaker in Se and Te while Po is metallic.

12. Atomic and ionic radii increases from top to bottom, due to increase in the number of shells. 13. Ionisation enthalpy decreases down the group, due to increase in size. Elements of group 16 have lower ionisation enthalpy values as compared to group 15 in the corresponding periods. This is due to the fact that group 15 elements have extra stable h alf-filled p-orbital electr on ic configurations.

14. Oxygen has less negative electron gain enthalpy than sulphur due to compact nature of oxygen atom.

15. Next to F, O has highest electronegativity value amon g th e elemen ts. Within the group, electronegativity decreases with increase in atomic number.

16. The tendency for catenation decreases as we go down the group.

17. All the elements of the group form volatile hydrides.

(a) The volatility increases from water to hydrogen sulphide and then declines. This is evident in their boiling point. Increasing order of boiling points of hydrides is H₂S < H₂Se < H₂Te < H₂O. Down the group boiling point increases because of increase in molecular weight which increases the van der Waal’s forces of interaction. H₂O has abnormally high b.p. due to hydrogen bonding.

(b) The thermal stability of the hydrides decreases in the order:

H₂O > H₂S > H₂Se > H₂Te > H₂Po.

(c) The strength of the hydrides as acids increases in the order:

H₂O < H₂S < H₂Se < H₂Te.

18. All the elements of group 16 form binary halides. 19. (a) S, Se and Te form a number of oxo-acids. Among the oxo-acids of S, sulphuric acid is most important.

(b) Sulphurous acid (H₂SO₃) and thiosulfuric acid (H₂S₂O₃) are unstable and cannot be isolated. They exist only in aqueous solutions or in the form of their salts. III Group 17 Elements

20. The group 17 of the Periodic Table contain fluorine (₉F), chlorine (₁₇Cl), bromine (₃₅Br), iodine (₅₃I) and astatine (₅₅At).

21. Electronic configuration is ns2 _np5 _{for valence} shells.

22. Halogens have the smallest atomic radii in their respective periods due to maximum effective nuclear charge.

23. The first ionisation energies are relatively high but decreases down the group. Iodine can lose an electron and form I+ _ion.

24. Electron affinity varies as Cl > F > Br > I 25. F is the most electronegative element known.

Electronegativity decreases down the group. 26. Halogens are good oxidising agents. The

oxidising power decreases down the group. 27. (a) Reactivity varies as F₂ > Cl₂ > Br₂ > I₂

(b) Order of ionic character in M – X bond is M – F > M – Cl > M – Br > M – I

28. Strength of hydrohalic acids varies as: HF < HCl < HBr < HI

The order of B.P is HCl < HBr < HI < HF 29. (a) Hypohalous acids are all weak acids and

exist in solution only. Acid strength decreases down the group.

HOCl > HOBr > HOl

(b) Acid strength increases as the number of O-atoms increases for a given halogen atom. HOCl < HClO₂ < HClO₃ < HClO₄ IV Group 18 Elements : The Noble Gases

30. In group 18 of the Periodic table, elements helium (₂He), neon (₁₀Ne), argon (₁₈Ar), krypton (₃₆Kr), xenon (₅₄Xe) and radon (₈₆Rn) are present. They are collectively called as noble gases.

31. Noble gases are located at the end of each period. Their valence shell orbitals are fully occupied. 32. They are monoatomic and are sparingly soluble

in water.

33. Xe forms fluorides XeF₂, XeF₄ and XeF₆. 34. XeO₃ is trigonal pyramidal in shape whereas

XeOF₄ is square pyramidal. 35. Uses:

(a) He is a non-inflammable gas, lighter than air, therefore, used in filling balloons for meteorological observations.

(b) Ar is used to provide an inert atmosphere.

7.1. Why are pentahalides more covalent than trihalides ?

Ans. The group 15 elements have 5e–1_{s in their valence}

shell. It is difficult to lose 3e–1_{s to form E}3+ _and

even more difficult to lose 5e–1_{s to form E}5+_.

Thus, they have very little tendency to form ionic compounds. Further, since the elements in +5 state have less tendency to lose e–1_{s than in the}

+3 state, elements in +5 state have more tendency to share e–1_{s and hence pentahalides are more}

covalent than trihalides.

7.2. Why is BiH₃ the strongest reducing agent amongst all the hydrides of Group 15 elements?

Ans. This is because as we move down the group, the

size increases, as a result, length of E–H bond increases and its strength decreases, so that the

bond can be broken easily to release H₂ gas. Hence, BiH₃is the strongest reducing agent. 7.3. Why is N₂ less reactive at room temperature?

Ans. Due to presence of triple bond between two

N-atoms (N º N), the bond dissociation energy of N₂ is very high. As a result, N₂ becomes less reactive at room temperature.

7.4. Mention the conditions required to maximise the yield of ammonia.

Ans. Ammonia is prepared by Haber's process as given

below : N₂(g) + 3H₂(g) 700 K, 200 × 10 Pa5 Fe O + K O + Al O2 3 2 2 3 Mo (promoter) 2NH₃(g) D_fH° = – 92.4 kJ mol–1

According to Le Chatelier's principle, to maximise the yield of ammonia, high P an d T ~ 700 K should be used. The catalyst increases the rate of reaction and Mo promoter increases the efficiency of Fe catalyst.

7.5. How does ammonia react with a solution of Cu2+_?

Ans. Cu2+(aq) + 4NH₄OH(aq) ® 3 4 2+ tetrammine copper (II) ion (deep blue)

[Cu(NH ) ]

+ 4H₂O 7.6. What is the covalence of nitrogen in N₂O₅ ?

Ans. In N₂O₅, each N-atom has four shared pairs of e–1s as shown: O N O O N O O O N O O N O O

Thus, the covalency of N is 4.

7.7. Bond angle in PH₄+ is higher than that in PH₃ . Why?

Ans. P in PH₃ is sp3–hybridized with 3 bond pairs and one lone pair around P. Due to stronger lp–bp repulsions than bp–bp repulsions, tetrahedral angle decreases from 109°28' to 93.6°. As a result, PH₃ is pyramidal.

In PH₄+, there are 4 bp's and no lone pair. As a result, there are only identical bp–bp repulsions so that PH₄+ assumes tetrahedral geometry and the bond angle is 109°28'.

Hence, bond angle of PH₄+ > bond angle of PH₃ 7.8. What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO₂ ? Ans. P₄ + 3NaOH + 3H₂O 2 CO atm D ¾¾® ₃ Phosphine PH + 2 sod. hypo-phosphite 3NaHPO 7.9. What happens when PCl₅ is heated?

Ans. PCl₅ ¾¾®D PCl₃ + Cl₂

On heating, the less stable axial bonds break to form PCl₃.

7.10. Write a balanced equation for the hydrolytic reaction of PCl₅ in heavy water

Ans. PCl₅+D O₂ ¾¾®POCl₃+2DCl

3 2 3 4

POCl +3D O¾¾®D PO +3DCl

5 2 3 4

PCl +4D O¾¾®D PO +5DCl 7.11. What is the basicity of H₃PO₄?

Ans. H₃PO₄ is tribasic as shown below :

O P HO

OH OH

Due to three ionizable P–OH

bonds, H₃PO₄ is tribasic. 3 4 2 3 2 4 H PO +H OƒH O++H PO -2 2 4 2 3 4 H PO-+H OƒH O++HPO -2 3 4 2 3 4 HPO -+H OƒH O++PO -7.12. What happens when H₃PO₃ is heated?

Ans. On heating, H₃PO₃ disproportionates to form PH₃ and H₃PO₄ with O.S. of – 3 and + 5.

3 –3 5

3 3 3 3 4

Orthophosphorous Phosphine Orthophosphoric

acid acid

H PO PH H PO

+ D +

¾® +

7.13. List the important sources of sulphur.

Ans. Sulphur mainly occurs in the combined states in

earth's crust in the form of sulphates and sulphides.

Sulphates : gypsum (CaSO₄.2H₂O); epsom (MgSO₄.7H₂O); baryte (BaSO₄), etc.

Sulphides : Galena (PbS); zinc blende (ZnS); copper pyrites (CuFeS₂); iron pyrites (FeS₂), etc. Traces of sulphur occur as H₂S and in organic materials such as eggs, proteins, garlic, onion, mustard, hair and wool.

7.14. Write the order of thermal stability of the hydrides of Group 16 elements.

Ans. The thermal stability of hydrides of group 16

elements decreases down the group. This is because down the group, size of the element (M) increases, M-H bond length increases and thus, stability of M-H bond decreases so that it can be broken down easily. Hence, we have order of thermal stability as

H₂O > H₂S > H₂Se > H₂Te > H₂Po 7.15. Why is H₂O a liquid and H₂S a gas ?

Ans. Due to high electronegativity of O than S, H₂O undergoes extensive intermolecular H-bonding. As a result, H₂O exists as an associated molecule in which each O is tetrahedrally surrounded by four H₂O molecules. Therefore, H₂O is a liquid at room temperature.

On the other hand, H₂S does not undergo H-bonding. It exists as discrete molecules which are held together by weak van der waals forces of attraction. A small amount of energy is required to break these forces of attraction. Therefore, H₂S is a gas at room temperature. 7.16. Which of the following does not react with

oxygen directly? Zn, Ti, Pt, Fe

Ans. Pt being a noble metal does not react with oxygen

directly. In contrast, Zn, Ti and Fe are active metals and hence they react with oxygen directly to form their oxides.

7.17. Complete the following reactions:

(i) C₂H₄ + O₂ ® (ii) 4Al + 3 O₂®

Ans. (i) C H_{2 4}+3O₂¾¾¾®Heat 2CO₂+2H O₂ (ii) 4Al 3O+ ₂¾¾¾®Heat 2Al O_{2 3}

7.18. Why does O₃ act as a powerful oxidising agent?

Ans. On heating, O₃ readily decomposes to give O₂ and nascent oxygen.

Heat

3 2

O ¾¾¾®O +O(nascent oxygen) Since nascent oxygen is very reactive, therefore, O₃ acts as a powerful oxidising agent.

7.19. How is O₃ estimated quantitatively?

Ans. When O₃ is treated with excess of KI solution buffered with borate buffer (pH = 9.2), I₂ is liberated quantitatively.

2I–(aq) + H₂O(l) + O₃(g) ¾® 2OH–(aq) + I₂(s) + O₂(g) The I₂ thus liberated is titrated against a standard solution of sodium thiosulphate using starch as an indicator.

2 2 3 2 2 4 6

2Na S O + ¾¾I ®Na S O +2NaI

7.20. What happens when sulphur dioxide is passed through an aqueous solution of Fe(III) salt?

Ans. SO₂ acts as a reducing agent and reduces aqueous solution of Fe (III)salt to Fe (II) salt.

2 2 2 4 SO +2H O¾¾®SO -+4H++2e -3 2 2Fe++2e-¾¾®2Fe + 3 2 2 2 2 4 2Fe++SO +2H O¾¾®2Fe++SO -+4H+ 7.21. Comment on the nature of two S–O bonds formed

in SO₂ molecule. Are the two S–O bonds in this molecule equal ?

Ans. SO₂ exists as an angular molecule with OSO bond angle of 119.5°. It a resonance hybrid of two canonical forms :

S

O O

S

O O

Due to resonance, the two p-bonds are equal. 7.22. How is the presence of SO₂ detected ?

Ans. SO₂ is a pungent smelling gas. It can be detected by two test :

(i) SO₂ turns pink colour of KMnO₄ to colourless due to reduction of MnO₄– _{to Mn}2+

– 4 (Pink) 2MnO + 5SO₂ + 2H₂O ¾® 2+ (colourless)2Mn + 5SO₄2– + 4H+ (ii) It turns orange colour of acidified K₂Cr₂O₇ to green due to reduction of Cr₂O₇2– to Cr3+

2 2 7 (orange) Cr O - +3SO +2H₂ + ¾® 3+ (green) 2Cr + 3SO₄2– + H₂O 7.23. Mention three areas in which H₂SO₄ plays an

important role.

Ans. (i) It is used in the manufacture of fertilizers such as (NH₄)₂SO₄, calcium superphosphate. (ii) It is used as an electrolyte in storage

batteries.

(iii) It is used in petroleum refining, detergent industry and in the manufacture of paints, pigments and dyes.

7.24. Write the conditions to maximise the yield of H₂SO₄ by Contact process.

Ans. The main step in the production of H₂SO₄ is :

2 2 3 SO ( )g +O ( )g ƒ2SO ( );g 1 fH 196.6kJmol -D ° =

-The reaction is exothermic, reversible and forward reaction proceeds with decrease in

volume. According to Le-Chateliers' principle, to maximise the yield of SO₃ and hence H₂SO₄, a low T, a high pressure and V₂O₅ is used as a catalyst.

7.25. Why is K

a2 << Ka1 for H2SO4 in water ?

Ans. Being a dibasic acid, H₂SO₄ ionized in two stages as follows: 2 4 2 3 4 H SO (aq)+H O( )l ®H O (+ aq) HSO (+ - aq) K_a1 > 10 2 4 2 3 4 HSO (- aq)+H O( )l ¾¾®H O (+ aq) SO+ -(aq) K_a2 = 1.2 × 10–2 K

a2 << Ka1 because negatively charged HSO4–

has much less tendency to donate a proton to H₂O as compared to neutral H₂SO₄.

7.26. Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F₂ and Cl₂.

Ans. F₂ is a much stronger oxidising agent than Cl₂. The oxidising power of a species depends upon the electrode potential. Higher is the electrode potential value, greater is the oxidising power. Electrode potential, in turn, depends upon 3 factors :

(i) bond dissociation energy (BDE) (ii) electron gain enthalpy (EGE) (iii) hydration energy (HE)

Although the EGE of fluorine is less negative than that of chlorine, the BDE of F₂ is much lower than that of Cl₂ but HE of F– _{ion is much higher}

than that of Cl– _{ion. It compensates the effect of}

other two (BDE and EGE). As a result, the electrode potential of F₂ is higher than that of Cl₂ and hence F₂ is a much stronger oxidising agent than Cl₂.

7.27. Give two examples to show the anomalous behaviour of fluorine.

Ans. Two examples to show the anomalous behaviour

of fluorine are :

(i) Due to non-availability of d-orbitals in its valence shell, fluorine cannot expand its octet and thus it shows only –1 O.S. On the other hand, other halogens due to presence of d-orbitals show positive O.S. of +1, +3, +5 and +7 besides O.S. of – 1.

(ii) Due to its small size, the three lone pair of e–1s on each F-atom in F–F molecule, repel the bond pair. As a result, F–F BDE is lower than that of Cl–Cl bond.

7.28. Sea is the greatest source of some halogens. Comment.

Ans. Sea water contains Cl–, Br– and I– of Na, K, Mg and Ca, but mainly NaCl. Dried up sea beds contain NaCl and carnallite, KCl. MgCl₂.6H₂O. Certain seaweeds contain upto 0.5% of iodine as NaI and chile saltpetre (NaNO₃) contains upto 0.2% of NaIO₃. Thus, sea is a great source of halogens.

7.29. Give the reason for bleaching action of Cl₂.

Ans. In presence of moisture or in aqueous solution,

Cl₂ liberates nascent oxygen.

2 2

Cl +H O¾¾®2HCl [O]+

nascent oxygen

This nascent oxygen brings about the oxidation of coloured substance to colourless substances.

Coloured Substances + O ® Colourless

substance

Thus, the bleaching action of Cl₂ is due to oxidation.

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

Ans. Phosgene and mustard gas are two poisonous

gases which can be prepared from Cl₂.

(i) 2 _charcoalhv 2 Phosgene CO Cl+ ¾¾¾¾®COCl (ii) 8 2 2 2 sulphur monochloride (boiling) S +4Cl ¾¾® 4S Cl CH₂ CH2 + S₂Cl₂ CH₂Cl CH2 — S — CH2 + S CH₂Cl Mustard gas 7.31. Why is ICl more reactive than I₂?

Ans. ICl is more reactive than I₂ because I–Cl bond is weaker than I–I bond. Consequently, ICl breaks easily to form halogen atoms which readily bring about the reactions.

7.32. Why is helium used in diving apparatus?

Ans. Because of its low solubility (as compared to

N₂) in blood, a mixture of oxygen and helium is used in diving apparatus used by deep sea divers. 7.33. Balance the following equation:

XeF₆ + H₂O ® XeO₂F₂ + HF.

Ans. XeF₆+2H O₂ ¾¾®XeO F_{2 2}+4HF

7.34. Why has it been difficult to study the chemistry of radon?

Ans. Because radon is a radioactive element with a

short half life of 3.82 days. This makes the study of chemistry of radon difficult.

7.1 Discuss the general characteristics of Group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity. Sol. In group 15 of the Periodic Table, the elements,

nitrogen (₇N), phosphorus (₁₅P), arsenic (₃₃As), antimony (₅₁Sb) and bismuth (₈₃Bi) are present. The elements of this group can exhibit various oxidation states ranging between –3 to + 5. Negative oxidation state will be exhibited when they combine with less electronegative element and positive oxidation state will be exhibited with more electronegative element. Positive oxidation state becomes more favourable as we more down the group due to increasing metallic character & electropositivity. Although due to inert pair effect the stability of +5 state will also decrease. The only stable compound of Bi (V) is BiF₅. The atomic (covalent) and ionic radii (in a particular oxidation state) of the elements of nitrogen family (group 15) are smaller than the corresponding elements of carbon family (group 14). On moving down the group, the covalent and ionic radii (in a particular oxidation state) increase with increase in atomic number. There is a considerable increase in covalent radius from N to P. However, from As to Bi, only a small increase is observed.

As the size increases on moving down the group, the ionisation enthalpy increases. The ionisation enthalpy of nitrogen group elements is more than the corresponding elements of oxygen group. This is because of more stable half filled outermost p- subsh ell of n itr ogen group elements. Electronegativity decreases down the group with increase in atomic size.

7.2 Why does the reactivity of nitrogen differ from phosphorus?

Sol. N₂ exist as a diatomic molecule containing triple bond between two N-atoms. Due to the presence of triple bond between the two N-atoms, the bond dissociation energy is large (941×4 kJ mol–1_{). As} a result of this, N₂ is inert and unreactive whereas, phosphorus exists as a tetratomic molecule, containg P – P single bond. Due to the presence

of single bond, the bond dissociation energy is weaker (213 kJmol–1) than N º N triple bond (941× 4 kJ mol–1_{) and moreover due to presence} of angular strain in P₄ tetrahedra. As a result of this, phosphorus is much more reactive than nitrogen.

7.3 Discuss the trends in chemical reactivity of group 15 elements.

Sol. Hydrides : All elements of group 15 form gaseous hydrides of the type MH₃.

In all the hydrides the central atom is sp3 hybridized and their shape is pyramidal due to presence of lone pair of electrons.

(a) Th e basic str ength of th e hydr ides decreases as we move down the group. Thus, NH₃ is the strongest base.

NH₃ > PH₃ > AsH₃ > SbH₃

(b) The thermal stability of the hydrides decreases as the atomic size increases, i.e., the M – H bond strength decreases which means reducing character increases. (c) In the liquid state, the molecules of NH₃ are

associated due to hydrogen bonding. The molecules of oth er hydrides ar e not associated.

(d) NH₃ is soluble in water whereas other hydrides are insoluble.

(e) All the hydrides, except NH₃, are strong reducing agents and react with metal ions (Ag+_{, Cu}2+_{, etc.) to form phosphides,} arsenides or antimonides.

Halides : The elements of group 15 form two series of halides MX₃ and MX₅.

(a) All the elements of the group form trihalides. The ionic character of trihalides increases as we move down the group. Except NCl₃ all the trihalides are hydrolysed by water. This is due to the absence of d-orbitals in nitrogen.

(b) PF₃ is not hydrolysed because fluorine being more electronegative than oxygen forms more stable bonds with phosphorus than P – O bonds.

(c) N cannot form NX₅ because of non-availability of d-orbitals. Bi cannot form BiX₅ because of reluctance of 6s electrons of Bi to participate in bond formation.

(d) The hybridisation of M in MX₃ is sp3 _and shape is pyramidal. M in MX₅ is sp3_{d as} hybridised and shape is trigonal pyramidal. The axial bonds in MX₅ are weaker and lon ger, So MX₅ are less stable an d decompose on heating eg:

5 3 2

PCl ¾¾®D PCl +Cl Oxides :

(a) Nitrogen forms a number of oxides. The rest

of the members (P, As, Sb and Bi) of the group form two types of oxides: E₂O₃ and E₂O₅.

(b) The reluctance of P, As, Sb and Bi to enter into pp – pp multiple bonding leads to cage structures of their oxides and they exist as dimers, E₄O₆ and E₅O₁₀.

(c) The basic nature of the oxides increases with increase in atomic number of the element. Thus, the oxides of nitrogen (except N₂O and NO), P (III) and As (III) are acidic, Sb (III) oxide is amphoteric and Bi (III) oxide is basic.

7.4 Why does NH₃ form hydrogen bond but PH₃ does not?

Sol. Nitrogen has an electronegativity value 3×0, which is much higher than that of H (2×1). As a result, N – H bond is quite polar and hence NH₃ undergoes intermolecular H – bonding.

- - - H - - - H — - - - H — H — -- -- -- H N N N H H H H d

–

–

–

Phosphorus have an electronegativity value 2×1. Thus, P – H bond is not polar and hence PH₃ does not undergo H – bonding.

7.5 How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions involved.

Sol. In laboratory, nitrogen is prepared by heating an equimolar aqueous solution of ammonium chloride and sodium nitrite. As a result of double decomposition reaction, ammonium nitrite is formed. Ammonium nitrite is unstable and decompose to form nitrogen gas.

NH₄Cl (aq) + NaNO₂ (aq)¾¾®

NH₄NO₂ (aq) + NaCl (aq) NH₄NO₂ (aq)_¾¾¾Heat_®N

2 (g) + 2H2O (l)

7.6 How is ammonia manufactured industrially? Sol. Commercially, by Haber’s process.

N₂ (g) + 3H₂ (g) * D ˆˆˆ† ‡ˆˆˆ2NH₃ (g) 1 fH 46.1kJ mol -D = -* iron oxide, K₂O, Al₂O₃

The optimum conditions for the production of NH₃ are pressure of 200 atm and temperature of 100 K.

7.7 Illustrate how copper metal can give different products on reaction with HNO₃.

Sol. On heating with dil HNO₃, copper gives copper nitrate and nitric oxide.

3Cu + 8HNO₃ (dil)_¾¾¾Heat_®

3Cu (NO₃)₂ + 4H₂O + 2NO With concentrated HNO₃, copper gives NO₂ instead of NO.

Cu + 4HNO₃(conc.)_¾¾¾Heat_®

Cu(NO₃)₂ + 2H₂O + 2NO₂ 7.8 Give the resonating structures of NO₂ and N₂O₅. Sol. Resonating structures of NO₂ are:

.. .. .. .._{O O}.. — — _— N _N + . _. . . .. .. _{..O O..} — — —

Resonating structures of N₂O₅ are:

N N N N .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. O O O O O O O O — — —— — — —— — — — — _— — — _— .. .. .. .. O O – ₊ + – – – + + N N .. .. .. .. .. .. .. .. .. O O O O —— —— — — — — .. .... O – – + +

7.9 The HNH angle value is higher than HPH, HAsH and HSbH angles. Why?

[Hint : Can be explained on the basis of sp3 hybridisation in NH₃ and only s-p bonding between hydrogen and other elements of the group].

Sol. In all these cases, the central atom is sp3 hybridized. Three of the four sp3 _{orbitals form} three s-bonds, while the fourth contains the lone pair of electrons. On moving down from N to Sb, the electronegativity of the central atom goes on decreasing. As a result of this, bond pairs of electrons lie away and away from the central atom. This is because of the force of repulsion between the adjacent bond pairs goes

on decreasing and the bond angles keep on decreasing from NH₃ to SbH₃. Thus, bond angles are in the order:

HNH > HPH > HAsH > HSbH (107 8°)× (93 6 )× ° (91 8 )× ° (91 3 )× ° 7.10 Why does R₃P = O exist but R₃N = O does not

(R = alkyl group)?

Sol. Nitrogen does not contains d-orbitals. As a result, it cannot expand its covalency beyond four and cann ot for m pp – dp multiple bonds. In constrast, P contains the d-orbitals, and can expand its covalency beyond 4 and can form

pp – dp multiple bonds.

Hence R₃P = O exist but R₃N = O does not. 7.11 Explain why NH₃ is basic while BiH₃ is only

feebly basic.

Sol. In both NH₃ and BiH₃, N and Bi have a lone pair of electrons on the central atom and hence

should behave as Lewis bases. But NH₃ is much more basic than BiH₃. Since the atomic size of N is much smaller than that of Bi, therefore, electron density on N-atom is much higher than that on Bi-atom. Thus, the tendency of N in NH₃ to donate its lone pair of electrons is much more in comparison to tendency of Bi in BiH₃. Hence, NH₃ is more basic than BiH₃.

7.12 Nitrogen exists as diatomic molecule and phosphorus as P₄. Why?

Sol. Nitrogen exists as a diatomic molecule having a triple bond between the two N-atoms. This is due its small size that it forms pp – pp multiple bonds with itself and with carbon /oxygen as well. On the other hand, phosphorus due to its larger size does not form multiple pp – pp bonds with itself. It prefers to form P – P single bonds and hence it exists as tetrahedral P₄ molecule. 7.13 Write main differences between the properties of white phosphorus and red phosphorus.

Sol.

Property White Phosphorus Red Phosphorus

(i) State Translucent Brittle, substance

(ii) Colour White gets yellowish on

exposure to light Red

(iii) Odour Garlic like odour Odourless

(iv) Hardness Soft like wax and can be

cut by knife Hard

(v) Poisonous nature Poisonous Non- poisonous

(vi) Solubility _{Soluble in CS2 Insoluble in CS2}

(vii) Chemiluminescence Glows in dark Dose not glow in dark.

(viii) Density 1.8 2.1

(ix) Reactivity Very reactive Less reactive

(x) Action of oxygen Burns with greenish glow to form P4O10

Combines with O2 only on heating to form

Structure of white and red phosphorus are given below:

P P

P

(i) Structure of white phosphorus 60

P

P

P P

(ii) Structure of red phosphorus

P P P P P P P P P

7.14 Why does nitrogen show catenation properties less than phosphorus?

Sol. The extent of catenation depends upon the strength of the element – element bond. The N – N bond strength (159 kJ mol–1_{) is weaker} than P – P bond strength (213 kJ mol–1_{). Thus,} nitrogen shows less catenation properties than phosphorus.

7.15 Give the disproportionation reaction of H₃PO₃. Sol. On heating, H₃PO₄ undergoes self - oxidation reduction, i.e., disproportionation to form PH₃.

+3 –3 +5

3 3 3 3 4

Phosphorus acid Phosphine Orthophosphoric acid

4H PO ¾®D PH + 3H PO

7.16 Can PCl₅ act as an oxidising as well as a reducing agent. Justify.

Sol. The oxidation state of P in PCl₅ is + 5. Since P has five electrons in its valence shell, therefore, it cannot donate electron and cannot increase its oxidation state beyond + 5, Thus, PCl₅ cannot act as a reducing agent. It can act as oxidizing agent by itself undergoing reduction.

+5 0 +3 +1 2 5 3 P Cl + H ¾¾®PCl + 2HCl 0 +5 +1 +3 5 3 2Ag + PCl ¾¾®2 AgCl + PCl

7.17 Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation state and hydride formation.

Sol. (1) Electronic configuration: O (At. no. = 8) = [He] 2s2 _2p4 S (At. no. = 16) = [Ne] 3s2 _3p4 Se (At. no. = 34) = [Ar] 3d10 _4s2 _4p4 Te (At. no. = 52) = [Kr] 4d10 _5s2 _5p4 Po (At. no. = 84) = [Xe] 4f14 _5d10 _6s2 _6p4 Thus, all these elements have the same ns2

np4 _{(n = 2 to 6) valence shell electronic} configuration, hence are justified to be placed in group 16 of the Periodic Table. (2) Oxidation state : Two more electrons are

needed to acquire the nearest noble gas configuration. Thus, the minimum oxidation state of these elements should be – 2. O and to some extent S show – 2 oxidation state. Ot h er elemen t being more electropositive than O and S, do not show negative oxidation state. As these contain six electrons, thus, maximum oxidation state shown by them is + 6. Other oxidation state shown by them are + 2 and + 4. O do not show + 4 and + 6 oxidation state, due to the absence of d-orbitals.

Thus, on the basis of maximum and minimum oxidation states, these elements are justified to be placed in the same group 16 of the periodic table.

(3) Hydride formation: All these elements share two of their valence electrons with 1s– orbital of hydrogen to form hydrides of the general formula EH₂, i.e., H₂O, H₂S, H₂Se, H₂Te and H₂Po. Thus, on the basis of hydride formation, these elements are justified to be placed in the same group 16 of the Periodic Table.

7.18 Why is dioxygen a gas but sulphur a solid? Sol. Due to the small size and high electronegativity,

oxygen forms pp – pp multiple bonds. As a result, oxygen exists as diatomic (O₂) molecules. These molecules are held together by weak van der Waal’s forces of attraction which can be overcome by collisions of the molecules at room temperature. Therefore, O₂ is a gas at room temperature. Due to its bigger size and lower electronegativity, sulphur does n ot form

pp – pp multiple bonds. It prefers to form S – S

single bonds. S – S single bond is stronger then O – O single bond. Thus, sulphur has higher tendency for catenation than oxygen. Due to higher tendency for catenation and lower

tendency for pp – pp multiple bonds sulphur exits as octa-atomic (S₈) molecule. Due to bigger size, the force of attraction holding the S₈ molecules together are much stronger which cannot be overcome by collisions of molecules at room temperature. Therefore, sulphur is solid at room temperature.

7.19 Knowing the electron gain enthalpy values of O®O– and O®O2– _{as – 141 and 702 kJ mol}–1 respectively, how can you account for the formation of a large number of oxides having O2– _{species and not O}–_?

Sol. Let us consider the reaction of oxygen with monopositve metal, we can have two compounds MO(O in -1 state) and M₂O (O in -2 state). The energy r equir ed for for mation of O–2 _is compensated by increased coulombic attraction between M+ _{and O}–2_{. Coulombic force of} attraction, F_A is proportional to product of charges on ions i.e.

1 2 2 µ A q q F

r where q1 and q2 are charges on ions

and r is distance between ions. Same logic can be applied if metal is dispositive.

7.20 Which aerosols deplete ozone?

Sol. Aerosols like chlorofluorocarbons (CFC’s), i.e., freon (CCl₂F₂), depletes the ozone layer by supplying Cl• _{free radicals which convert O}

3 to O₂. h 2 2 2 Freon CCl F ( )g ¾¾n®gCl( )g +gCClF ( )g •Cl (g) + O₃ (g)¾¾® ClO•(g) + O₂ (g) ClO•(g) + •O (g)¾¾®•Cl (g) + O₂ (g)

7.21 Describe the manufacture of H₂SO₄ by contact process?

Sol. Preparation of sulphuric acid:

By Contact Process: Burning of sulphur or sulphide ores in presence of oxygen to produce SO₂. Catalytic oxidation of SO₂ with O₂ to give SO₃ in the presence of V₂O₅.

2SO₂ (g) + O₂ (g)_{¾¾¾® 2SO}V O_{2 5} 3 (g)

Then SO₃ made to react with sulphuric acid of suitable normality to obtain a thick oily liquid called oleum.

SO₃(g) + H₂SO₄(_{l )}¾¾® H₂S₂O₇(_{l )}

Then oleum is diluted to obtain sulphuric acid of desired concentration.

H₂S₂O₇(_{l ) + H2}O(_{l )}¾¾® 2H₂SO₄(_{l )} The sulphuric acid obtained by contact process is 96-98 % pure.

7.22 How is SO₂ an air pollutant?

Sol. (1) SO₂ dissolves in moisture present in air to form H₂SO₄ which damages building materials especially marble (acid – rain). CaCO₃ + H₂SO₃¾¾®CaSO₃ + H₂O + CO₂ (2) It corrodes metals like Fe and steel. It also brings about fading and deterioration of fabrics, leather, paper, etc., and affecting the colour of paints.

(3) Even in low concentration ( 0 03 ppm); × , it has damaging effect on the plants. If exposed for a long time, i.e., a few days or weeks, it slows down the formation of chlorophyll i.e., loss of green colour. This is called chlorosis. (4) It is strongly irritating to the respiratory track. It cause throat and eye irritation, resulting into cough, tears and redness in eyes. It also cause breathlessness and effects larynx i.e., voice box.

7.23 Why are halogens strong oxidising agents? Sol. The halogens are strong oxidising agents due to

low bon d dissociation en thalpy, h igh electronegativity and large negative electron gain enthalpy.

7.24 Explain why fluorine forms only one oxoacid, HOF.

Sol. Cl, Br and I form four series of oxo acids of general formula HOX, HOXO, HOXO₂ and HOXO₃. In these oxo-acids, the oxidation states of halogens are + 1, + 3, + 5, and + 7 respectively. However, due to high electronegativity, small size and absence of d-orbitals, F does not form oxo-acids with + 3, + 5 and + 7, oxidation states. It just forms one oxo-acid (HOF).

7.25 Explain why inspite of nearly the same electronegativity, nitrogen forms hydrogen bonding while chlorine does not.

Sol. Both nitrogen (N) and chlorine (Cl) have electronegativity of 3.0. However, only nitrogen is involved in the hydrogen bonds (e.g., NH₃) and not chlorine. This is due to smaller atomic size of nitrogen (atomic radius =70 pm) as compared to chlorine (atomic radius = 99) pm). therefore, N can cause greater polarisation of N–H bond than Cl in case of Cl—H bond. Consequently, N atom is involved in hydrogen bonding and not chlorine.

7.26 Write two uses of ClO₂.

Sol. (1) ClO₂ is an excellent bleaching agent. It is 30 times stronger bleaching agent then the Cl₂. It is used as a bleaching agent for paper pulp in paper industry and in textile industry. (2) ClO₂ is also a powerful oxidising agent and chlorinating agent. It acts as a germicide for disinfecting water. It is used for purifying drinking water.

7.27 Why are halogens coloured?

Sol. The h alogens ar e coloured because their molecules absorb light in the visible region. As a result of which their electrons get excited to higher energy levels while the remaining light is transmitted. The color of halogens is the color of this transmitted light.

7.28 Write the reactions of F₂ and Cl₂ with water. Sol. 2F₂ (g) + 2H₂O (l)¾¾®4H+ _{(aq) + 4F}– _{(aq) + O}

2 (g) 3F₂ (g) + 3H₂O (l)¾¾®6H+ _{(aq) + 6F}– _{(aq) + O}

3 (g) Cl₂ (g) + H₂O (l)¾¾®HCl (aq) + HOCl (aq) F₂ oxidises water, whereas Cl₂ undergoes disproportion in water.

7.29 How can you prepare Cl₂ from HCl and HCl from Cl₂? Write reactions only.

Sol. ₂

Oxidising agent

MnO + 4HCl¾¾® MnCl₂ + Cl₂ + 2H₂O

We can also used KMnO₄, K₂Cr₂O₇, etc., in place of MnO₂.

H₂ + Cl₂¾¾¾¾¾¾®Diffused sunlight 2HCl

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

Sol. N. Bartlett observed that PtF₆ reacts with O₂ to give an compound O₂+ _[PtF 6] –_. PtF₆ (g) + O₂ (g)¾¾®O₂+ _[PtF 6] –

Since the first ionization enthalpy of Xe (1170 kJ mol–1_{) is fairly close to that of O}

2 molecule (1175 kJ mol–1_{), he thought that PtF}

6 should also oxidise Xe to Xe+_{. This inspired Bartlett to} carryout the reaction between Xe and PtF₆. When PtF₆ and Xe were made to react, a rapid reaction took place and a red solid, Xe+_[PtF

6] – was obtained. Xe + PtF₆¾¾¾®278 K Xe+ _[PtF 6] –

7.31 What are the oxidation states of phosphorus in the following: (i) H₃PO₃ (ii) PCl₃ (iii) Ca₃P₂ (iv)Na₃PO₄ (v) POF₃ Sol. (i) H₃PO₃ 3 (+1) + x + 3 (– 2) = 0 \ x = + 3 (ii) PCl₃ x + 3 (– 1) = 0 x = + 3 (iii) Ca₃P₂ 3 (+ 2) + 2x = 0 x = – 3 (iv) Na₃PO₄ 3 (+ 1) + x + 4 (– 2) = 0 x = + 5 (v) POF₃ x + 1 (– 2) + 3 (– 1) = 0 x = + 5.

7.32 Write balanced equations for the following: (i) NaCl is heated with sulphuric acid in the

presence of MnO₂.

(ii) Chlorine gas is passed into a solution of NaI in water. Sol. (i) 2 4 4 2 2 2 2 2 2 4 2 4 2 2 NaCl H SO NaHSO HCl] 4 4HCl MnO MnCl Cl 2H O

4NaCl MnO 4H SO MnCl 4NaHSO Cl 2H O

+ ¾¾® + ´

+ ¾¾® + +

+ + ¾¾® + + +

i.e Cl– _{is oxidized by MnO} 2

(ii) Cl₂ (g) + 2NaI (aq)¾¾®2NaCl (aq) + I₂ (s) i.e. I– _{is oxidized by Cl}

2.

7.33 How are xenon fluorides XeF₂, XeF₄ and XeF₆ obtained?

Sol. XeF₂ , XeF₄ and XeF₆ are obtained by direct reaction between Xe and F₂ as follows:

673 K, 1bar 2 Ni tube 2 excess Xe( ) + F ( )g g ¾¾¾¾®XeF ( )s 873 K, 7 bar 2 4 (1n 1:5 ratio) Xe ( )g +2 F ( )g ¾¾¾¾®XeF ( )s 573 K, 60 – 70bar 2 6 (1n 1:20 ratio) Xe ( )g +3F ( )g ¾¾¾¾¾¾®XeF ( )s 7.34 With what neutral molecule is ClO–

isoelectronic? Is that molecule a Lewis base? Sol. ClO– _{has 17 + 8 + 1 = 26 electrons.}

Also, OF₂ has (8 + 2 × 9) = 26 electrons. and ClF has (17 + 9) = 26 electrons.

Out of these, ClF can act as Lewis base. The chlorine atom has three lone pair of electrons which it donates to form compounds like ClF₃, ClF₅ and ClF₇

7.35 How are XeO₃ and XeOF₄ prepared? Sol. (i) 6XeF₄ + 12H₂O¾¾¾¾¾Hydrolysis®

4Xe + 2XeO₃ + 24HF + 3O₂ XeF₆ + 3H₂O¾¾¾¾¾Hydrolysis®XeO₃ + 6HF (ii) XeF₆ + H₂O ¾¾¾¾¾Hydrolysis®XeOF₄ + 2HF 7.36 Arrange the following in the order of property

indicated for each set:

(i) F₂ , Cl₂, B r₂, I₂ – increasing bond dissociation enthalpy.

(ii) HF, HCl, HBr, HI – increasing acid strength.

(iii) NH₃ , PH₃, AsH₃, SbH₃, BiH₃ – increasing base strength.

Sol. (i) Bond dissociation enthalpy decreases as the bond distance increases from F₂ to I₂ due to increase in the size of the atom, on moving from F to I.

F – F bond dissociation enthalpy is smaller then the Cl – Cl and even smaller than Br – Br. This is because F atom is very small and have large electron-electron repulsion among the lone pairs of electrons in F₂ molecule where they are much closer to each other than in case of Cl₂. The increasing order of bond dissociation enthalphy is

I₂ < F₂ < Br₂ < Cl₂

(ii) Acid strength of HF, HCl, HBr and HI depends upon their bond dissociation enthalpies. Since the bond dissociation enthalpy of H – X bond decreases from H – F to H – I as the size of atom increases from F to I.

Thus, the acid strength order is HF < HCl < HBr < HI

The weak acidic strength of HF is also due to H-bonding due to which release of H+ becomes difficult.

(iii) NH₃, PH₃, AsH₃, SbH₃ and BiH₃ behaves as Lewis bases due to the presence of lone pair of electrons on the central atom. As we move from N to Bi, size of atom increases. Electron density on central atom decreases and hence the basic strength decreases from NH₃ to BiH₃. Thus basic strength order is

BiH₃ < SbH₃ < AsH₃ < PH₃ < NH₃

7.37 Which one of the following does not exist ? (i) XeOF₄ (ii) NeF₂

(iii) XeF₄ (iv) XeF₆

Sol. NeF₂ does not exist. This is because the sum of first and second ionization enthalpies of Ne are much higher than those of Xe. Consequently, F₂ can oxidise Xe to Xe2+ _{but cannot oxidise Ne to} Ne2+_.

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

(i) ICl₄– (ii) IBr₂– (iii) BrO₃– Sol. (i) ICl₄–_{: In ICl}

4

–_{, central atom I has seven} valence electrons and one due to negative charge. Four out of these 8 electrons are utilized in forming four single bonds with four Cl atoms. Four remaining electrons constitutes the two lone pairs. It is arranged in square planar structure. ICl₄– _{has 36} valence electrons. A noblegas species having 36 valence electrons is XeF₄ (8 + 4 × 7 = 36). XeF₄ is also square planar.

Cl Cl Cl Cl .. .. I (ii) IBr₂–_{: In IBr}

2

–_{, central atom I has eight} electrons. Two of these are utilized in forming two single bonds with two Br atom. Six remaining electrons constitutes three lone pairs. It is arranged in linear structure.

Br Br .. .. .. I

IBr₂– _{has 22 valence electrons. A noble gas} species having 22 valence electrons is XeF₂ (8 + 2 × 7 = 22).

XeF₂ is also linear.

(iii) In BrO₃– _{ion the central Br atom has 8 valence} electrons (7 +1). Out of these, it shares 4 with two atoms of O forming Br = O bonds. Out of the remaining four electrons, 2 are donated to the third O atom which accounts for its negative charge. The remaining 2 electrons constitute one lone pair. In order

to minimise the force of repulsion, the structure of BrO₃– _{ion must be pyramidal.} BrO₃– _{ion has (7 + 3 × 6 + 1) = 26 valence} electrons and is isoelectronic as well as iso-structural with noble gas species XeO₃ which has also 26(8 + 3 × 6) electrons.

Br O O O – • • Xe O O • • O

7.39 Why do noble gases have comparatively large atomic sizes?

Sol. This is because noble gases have only van der Waal’s radii while others have covalent radii. van der Waal’s radii are larger than covalent radii.

7.40 List the uses of neon and argon gases. Sol. Uses of Neon

Neon is used in discharge tubes and fluorescent bulbs for advertisement display purposes. Glow of different colours ‘neon signs’ can be produced by mixing neon with other gases. Neon bulbs and used in botanical gardens and in green houses.

Uses of Argon

Argon is used mainly to provide an inert atmosphere in high temperature metallurgical processes such as arc welding of metals and alloys. In the laboratory, it is used for handling substance which are air sensitive.

It is used in filling incandescent and fluorescent lamps where its presence retards the sublimation of the filament and thus increases the life of the lamp.

It is also used in “neon signs” for obtaining lights of different colours.