DAV CENTENARY PUBLIC SCL PASCIM ENCLAVE GRUP 17 ELEMENTS Q1. With what neutral molecule is - isoelectronic? Ans. F Q2. Give the formula of the noble gas species that is isostructrural with (a) I4 - (b) IBr2 - (c) Br3 - Ans. Try yourself. (ACCRDING T VSEPR TERY) Q3. Which among the following is the strongest oxidizing agent? 4 -, Br4 -, I4 - Ans. I4 - Q4. What do you mean by interhalogen compounds? Ans. The binary compounds formed by the two different halogens are known as interhalogen compounds. They may be represented by a general formula AXn (where n = 1, 3, 5, 7). Q5. Write the one chemical method for the preparation of fluorine. Ans. K2MnF6 + 2SbF5 2KSbF6 + MnF3 + ½ F2 Q6. Why SiF6 2- is known Si6 2- is not? Ans. Size of F is smaller than therefore, in SiF6 2- has lesser steric repulsions. Moreover, the interaction of lone pair electrons of F with Si is stronger than that of the lone pairs of chlorine. Q7. Electrolysis of KBr (aq) gives Br2 at anode but that of KF (aq) does not give F2. Give reason for disparity in behaviour. Ans. The stantdard electrode potential of fluorine is highest, hence F - ions (from KF) cannot be oxidized to fluorine by any oxiising agent. (Which always has lower E 0 value). Q8. Identify X in the following chemical reaction: 2 + 2X - 2 - + X2. Ans. X may be Br - or I - Q9. Give an example of the compound in which the oxidation state of chlorine atom is +7. Ans. Sodium perchlorate (Na4). Q10. Arrange the hydrides of halogens in the order of increasing acid strength. Ans. F < < Br < I Q11. What kind of bond is expected between oxygen and fluorine in oxygen fluoride? Ans. Covalent type of bond, because there is very small difference in the electronegativity values of F and. Q12. Arrange 4, I4 and Br4 in order of increasing oxidizing ability. Ans. 4 < I4 < Br4 (because the standard electrode potential, X4 - X3 - is highest in case of Br4) Q13. Arrange the halogens in order of bond dissociation energies. Ans. I2 < Br2 < F2 < 2 Q14. Name the anhydride of hypochlorous acid? Ans. 2. Q15. Name the anhydride of perchloro acid. Ans. 27. Q16. Which oxides of chlorine are paramagnetic? DAV CENTENARY PUBLIC SCL, PASCIM ENCLAVE
Ans. 2 and 3. Q17. Arrange the following oxyacids in increasing order of acid strength and decreasing order of oxidizing power. 4, 3, 2, Ans. < 2 < 3 < 4 (increasing order of strength) > 2 > 3 > 4 (decreasing order of oxidizing power) Q18. Draw the structures of chloric (I) acid, chloric (III) acid, chloric (V) acid, chloric (VII) acid. Ans. chloric (I) acid chloric (V) acid chloric (III) acid chloric (VII) acid Q19. Why F is a weaker acid than I. Ans. F (aq) has extensive hydrogen bonding hence dissociation is weak. Whereas I is weakly hydrogen bonded and is almost fully dissociated. Q20. Electron affinity of fluorine is less than that of chlorine. Why? Ans. Electron affinity of fluorine is less than that of chlorine. The low value of electron affinity of fluorine is probably due to the electron electron repulsion in relatively compact 2p orbitals of fluorine atom due to its small size. Q21. What are inter halogen compounds? Why most of the inter halogens are most reactive than molecular halogens? Ans. The halogens combine themselves to form a number of covalent compounds known as interhalogen compounds. This happens due to the difference in their electronegativities. Most of the inter halogen compounds are strong oxidizing agents. They are more reactive than halogens, because the polarity is developed due to the large difference in their electronegativities. Q22. Why do you add KF and exclude moisture in the electrolysis of F for the manufacture of fluorine? Give reasons. Ans. Addition of KF makes F more electroyte i.e., it makes F more conducting in molten state. Fluorine has high affinity for hydrogen. Because of this fluorine reacts with water to liberate oxygen. F2 + 2 2F - + 2 + + ½ 2 Q23. Arrange the following according to the property mentioned against each: (i) F,, Br, I in the order of increasing acid strength. (ii) 2S, 2, 2Te, 2Se in the order of decreasing thermal stability. (iii) 4, Br4, I4 in the order of increasing oxidizing ability Ans. (i) F < < Br < I (ii) 2S > 2 > 2Se > 2Te (iii) 4 < I4 < Br4 Q24. Arrange the following in the decreasing order of property indicated against each: (i) F2, 2, Br2, I2 (decreasing bond strength ) (ii) As23, 2, Ge2, Ga23 (increasing acidity) (iii) N3, P3, As3, Sb3 (decreasing base strength) (iv) 2, 2S, 2Se, 2Te (increasing acidic strength) (v) 2, 2S, 2Se, 2Te (decreasing boiling point) Ans. (i) 2 > F2 > Br2 > I2 (ii) Ga23 < As23 < Ge2 < 2 DAV CENTENARY PUBLIC SCL, PASCIM ENCLAVE
(iii) N3 > P3 > As3 > Sb3 (iv) 2 < 2S < 2Se < 2Te (v) 2 > 2Te > 2Se > 2S Q25. ydrogen fluoride has higher boliling point than hydrogen chloride. Why? Ans. ydrogen fluoride has higher boiling point than hydrogen chloride. This is due to the presence of strong intermolecular hydrogen bonding between F molecules, but there is no such strong hydrogen bonding in. Q26. SF6 is known but S6 is not known. Give reason. Ans. Due to small size of S, six large atoms cannot be accommodated around S atom. But small six F atoms can be easily accommodated around S atom to form SF6. Moreover, because of low electronegativity of, it cannot easily cause promotion of electrons in S to form S (VI). Q27. State the trends observed in case of the oxidizing property of the members of the halogen family. Ans. Since all the halogens have a strong tendency to accept electrons, they act as strong oxidizing agents. There oxidizing power decreases from fluorine to iodine. Since fluorine is the strongest oxidizing agent in the series, it will oxidize other halide ions to halogens in solutions or dry state. Q28. Why is it that anhydrous aluminium fluoride has a higher melting point than anhydrous alumimium chloride? Ans. Anhydrous aluminium fluoride is an ionic compound while anhydrous aluminium chloride is a covalent compound, therefore, the melting point of anhy. AlF3 is greater than anhy. Al3. Q29. Draw the structures of 2 and 27. Ans. Q30. F3 exists but F3 does not. Explain. Ans. Chlorine has vacant d orbitals, thus one p electron on promotion to d orbital gives 3s 2 3p 4 3d 1 (three unpaired electrons) which forms F3. Fluorine cannot form F3 due to the absence of d orbitals. Q31. Despite its lower electron affinity fluorine is a stronger oxidizing agent than chlorine. Explain. Ans. The reduction potential of fluorine is higher than the other halogens, therefore despite having lower value of electron affinity, fluorine is a stronger oxidizing agent. 1 2 0 F2 e aq F ( aq) E 2.87V Q32. P5 is known but PI5 is not known. Why? Ans. Due to small size of atom, five atoms can be accommodated around P atom. But I is of large size and therefore, five I atoms cannot be accommodated around P atom. As a result, P I bonds are weak and prefer to form PI3 rather than PI5. Q33. N3 gets readily hydrolysed while NF3 does not. Why? Ans. In N3, has vacant d orbitals to accept lone pair of electrons present on oxygen atom of water molecule. Q34. SF6 is known but S6 is not known. Explain. Ans. Due to small size of S, six large atoms cannot be accommodated around S atom. But small six F atoms can be easily accommodated around S atom to form SF6. Moreover, because of low electronegativity of, it cannot easily cause promotion of electrons in S to form S (IV). DAV CENTENARY PUBLIC SCL, PASCIM ENCLAVE
Q35. alogen have maximum negative electron gain enthalpy in the respective periods of the periodic table. Why? Ans. The halogens have the smallest size in their respective periods and therefore, high effective nuclear charge. Moreover, they have only one electron less than the stable noble gas configuration. Therefore, they have strong tendency to accept one electron to acquire noble gas electronic configurations and hence have maximum negative electron gain enthalpy in their respective periods. Q36. Fluorine exhibits only -1 oxidation state where as other halogens exhibit positive oxidation states also such as +1, +3, +5, +7. Why? Ans. Fluorine is most electronegative element and cannot exhibit any positive oxidation states. n the other hand, the other halogens are less electronegative and therefore, can exihibit positive oxidation states. They also have vacant d orbitals and hence can expand their octets and show +1, +3, +5, +7 oxidation states also. Q37. when reacts with finely divided iron forms ferrous chloride and not ferric chloride. Why? Ans. reacts with finely divided iron and produces 2 gas. Fe + 2 Fe2 + 2 Liberation of hydrogen prevents the formation of ferric chloride. Q38. Explain why inspite of nearly same electronegativity, oxygen forms hydrogen bonding while chlorine does not. Ans. xygen has smaller size than chlorine. The smaller size of oxygen favours hydrogen bonding. In other words, though electronegativity of is same as that of, it does not form hydrogen bonding because of its larger size. Q39. Explain why fluorine forms only one oxoacid F? Ans. Due to small size and high electronegativity, fluorine cannot act as central atom in higher oxoacids. Q40. Why are pentahalides more covalent than trihalides? Ans. In pentahalides, the oxidation state is more (+5) than in trihalides (+3). As a result of higher positive oxidation state of central atom, they have larger polarizing power and can polarize the halide ion to a greater extent than in the corresponding trihalide. Since larger the polarization, larger is the covalent character, therefore, pentahalides are more convalent than trihalides. Q41. With the increase in oxidation no. of a particular halogen atom, the acidic character of corresponding oxoacid Increases. Explain. ( < 2 < 3 < 4) Ans. ere oxygen is more electronegative than chlorine. The more the oxygen atom bonded to the chlorine atom more the electrons will be pulled away from the bond, and the more the bond will be weakened. Thus 4 requires the least energy to break the bond and form +. ence 4 is the strongest acid. The strength of acid is also explained on the basis of stability of ion formed. Thus the stability will be 4 - > 3 - > 2 - > ence the acidic strength will also decreases from 4 to. Q42. Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidizing power of F2 and 2. Ans. F2, is stronger oxidizing agent than 2. This can be explained on the basis of bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy. The process of oxidizing behaviour may be expressed as: ½ X2 (g) X (g) Δdiss. X (g) X - (g) Δeg X - (g) X - (aq) Δhyd The overall tendency for the change (i.e., oxidizing behaviour) depends upon the net effect of three steps. As energy is required to dissociate or convert molecular halogen into atomic halogen, the enthalpy change for this step is positive. n the other hand, energy is released in step (II) and (III), therefore, enthalpy change for these steps is negative. Now although fluorine has less negative electron gain enthalpy, yet ist is stronger oxidizing agent because of low enthalpy of dissociation and very high enthalpy of hydration. In other words, large amount of energy required in step (I) overweigh the smaller energy released in step (II) for fluorine. As a result, the Δ overall is more negative for F2 than for 2. ence, F2 is stronger oxidizing agent than 2. DAV CENTENARY PUBLIC SCL, PASCIM ENCLAVE
IMPRTANT REACTINS F GRUP 17 1. The relative oxidizing power of halogens can be illustrated by their reactions with water. Fluorine oxidizes water to oxygen whereas chlorine and bromine react with water to form corresponding hydrohalic and hypohalous acids. The reaction of iodine with water is non spontaneous. In fact, I - can be oxidized by oxygen in acidic medium; just the reverse of the reaction observed with fluorine. 2F2 + 22 4 + + 4F - + 2 X2 + 2 X + X (where X = or Br) 4I - + 4 + + 2 2I2 + 22 2. Laboratory preparation of 2 By heating manganese dioxide with conc. Mn2 + 4 Mn2 + 2 + 22 owever, a mixture of common salt and conc. 2S4 is used in place of. 4Na + Mn2 + 42S4 Mn2 + 4NaS4 + 22 + 2 By the action of on potassium permanganate. 2KMn4 +16 2K + 2Mn2 + 82 + 52 Manufacturing of 2 Deacon s process :- By the oxidation of hydrogen chloride gas by atmospheric oxygen in the presence of Cu2 (catalyst) at 723 K. Cu2 4 + 2 22 + 22 723K Electrolytic process :- Chlorine is obtained by the electrolysis of brine (conc. Na solution). Chlorine is liberated at anode. It is also obtained as a by product in many chemical industries. 3. With excess ammonia, chlorine gives nitrogen and ammonium chloride whereas with excess chlorine, nitrogen trichloride (explosive) is formed. 8N 3 6N N 3 2 4 2 ( excess ) N N 3 2 3 3 ( excess ) 4. With cold and dilute alkalies chlorine produces a mixture of chloride and hypochlorite but with hot and concentrated alkalies it gives chloride and chlorate. 2Na Na Na ( Cold & Dilute ) ( ot & Conc.) 2 2 6Na 3 5Na Na 3 This reaction is an example of disproportionation 2 3 2 reaction. 5. With dry slaked lime 2 gives bleaching powder 2Ca()2 + 22 Ca()2 + Ca2 + 22 6. The composition of bleaching powder is Ca()2. Ca2. Ca()2. 22 7. Chlorine water on standing loses its yellow colour due to the formation of and. ypochlorous acid () so formed, gives nascent oxygen which is responsible for oxidizing and bleaching properties of chlorine. 8. xidising action of chlorine 2FeS4 + 2S4 + 2 Fe2(S4)3 + 2 Na2S3 + 2 + 2 Na2S4 + 2 S2 + 22 + 2 2S4 + 2 I2 + 62 + 52 2I3 + 10 9. Laboratory preparation of 420K Na + 2S4 NaS4 + DAV CENTENARY PUBLIC SCL, PASCIM ENCLAVE
823K Na2S4 + NaS4 + Na gas can be dried by passing through concentrated sulphuric acid but I can t. 10. Interhalogen compounds are very useful fluorinating agents. F3 and BrF3 are used for the production of UF6 in the enrichment of U. 235 92 U(s) + 3F3 (l) UF6 (g) + 3F (g) DAV CENTENARY PUBLIC SCL, PASCIM ENCLAVE