Falange en la nit del feixisme

1. DEL DESERT A L’OASI

1.1. Falange en la nit del feixisme

NCERT QUESTIONS

1. Explain the bonding in coordinatiExplain the bonding in coordination compounds in terms of Won compounds in terms of W ernererner’’

s postulates. s postulates.

2. FeSOFeSO₄₄ solution mixed with (NH solution mixed with (NH₄₄))₂₂SOSO₄₄ solution in 1 : 1 molar ratio gives the test of Fe solution in 1 : 1 molar ratio gives the test of Fe²⁺²⁺ ion but CuSO ion but CuSO₄₄ solution solution mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of Cu

mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of Cu²⁺²⁺ ion. Explain why? ion. Explain why?

3. Explain with two examples each Explain with two examples each of the of the following: coordination entityfollowing: coordination entity, ligand, coordination number, coordination, ligand, coordination number, coordination polyhed

polyhedron, homoleptic ron, homoleptic and heteroleptic.and heteroleptic.

4. What is What is meant by unidentate, didentate and ambidentate ligands? Give two examples for each.meant by unidentate, didentate and ambidentate ligands? Give two examples for each.

5. Specify the oxidatiSpecify the oxidation numbers of on numbers of the metals in the following coordination entities :the metals in the following coordination entities : (i) [Co(H

(i) [Co(H₂₂O)(CN)(en)O)(CN)(en)₂₂]]²⁺²⁺ (ii) [CoBr(ii) [CoBr₂₂(en)(en)₂₂]]⁺⁺ (iii) [PtCl(iii) [PtCl₄₄]]²²^–^– (iv) K

(iv) K₃₃[Fe(CN)[Fe(CN)₆₆]] (v) [Cr(NH(v) [Cr(NH₃₃))₃₃ClCl₃₃]]

6. Using IUPUsing IUPAC norms write the formAC norms write the formulas for the fulas for the following :ollowing :

((ii) ) TTeettrraahhyyddrrooxxoozziinnccaattee((IIII)) ((iiii) ) PPoottaassssiiuum tm teettrraacchhlloorriiddooppaallllaaddaattee((IIII)) ((iiiiii) ) DDiiaammmiminneeddiicchhlloorriiddooppllaattiinnuumm((IIII)) ((iivv) ) PPoottaassssiiuum m tteettrraaccyyaannoonniicckkeellaattee((IIII)) ((vv) ) PPeennttaaaammmiminneenniittrriittoo--OO--ccoobbaalltt((IIIIII)) ((vvii) ) HHeexxaaaammmiminneeccoobbaalltt((IIIIII) ) ssuullpphhaattee (v

(viiii) Po) Pottaasssisiuum trm trii((ooxxaallaattoo)c)chhrroomamattee((IIIIII)) (v(viiiiii) He) Hexxaaaammmmiinneeppllaattiinnuum(m(IIVV)) ((iixx) ) TTeettrraabbrroommiiddooccuupprraattee((IIII)) ((xx) ) PPeennttaaaammmmiinneenniittrriittoo--NN--ccoobbaalltt((IIIIII)) 7.

7. Using IUPUsing IUPAC norms write AC norms write the systematic names the systematic names of the fof the following:ollowing:

(i) [Co(NH

(i) [Co(NH₃₃))₆₆]Cl]Cl₃₃ (ii) [Pt(NH(ii) [Pt(NH₃₃))₂₂Cl(NHCl(NH₂₂CHCH₃₃)]Cl)]Cl (iii) [Ti(H

(iii) [Ti(H₂₂O)O)₆₆]]³⁺³⁺ (iv) [Co(NH(iv) [Co(NH₃₃))₄₄Cl(NOCl(NO₂₂)]Cl)]Cl (v) [Mn(H

(v) [Mn(H₂₂O)O)₆₆]]²⁺²⁺ (vi) [NiCl(vi) [NiCl₄₄]]²²^–^– (vii) [Ni(NH

(vii) [Ni(NH₃₃))₆₆]Cl]Cl₂₂ (viii) [Co(en)(viii) [Co(en)₃₃]]³⁺³⁺

(ix) [Ni(CO) (ix) [Ni(CO)₄₄]]

8. List various types of isomerism possible for coordination compounds, giving an example of each.List various types of isomerism possible for coordination compounds, giving an example of each.

9. How many geometrical isomers How many geometrical isomers are possible in the fare possible in the following coordination entities?ollowing coordination entities?

(i) [Cr(C

(i) [Cr(C₂₂OO₄₄))₃₃]]³³^–^– (ii) [Co(NH(ii) [Co(NH₃₃))₃₃ClCl₃₃]]

10.

10. Draw the structures of optical isomers of :Draw the structures of optical isomers of : (i) [Cr(C

(i) [Cr(C₂₂OO₄₄))₃₃]]³³^–^– (ii) [PtCl(ii) [PtCl₂₂(en)(en)₂₂]]²⁺²⁺ (iii) [Cr(NH(iii) [Cr(NH₃₃))₂₂ClCl₂₂(en)](en)]⁺⁺ 1

11.1. Draw all the isomers (geometrical and optical) of :Draw all the isomers (geometrical and optical) of : (i) [CoCl

(i) [CoCl₂₂(en)(en)₂₂]]⁺⁺ (ii) [Co(NH(ii) [Co(NH₃₃)Cl(en))Cl(en)₂₂]]²⁺²⁺ (iii) [Co(NH(iii) [Co(NH₃₃))₂₂ClCl₂₂(en)](en)]⁺⁺ 12.

12. Write all the geometrical isomers of [Pt(NHWrite all the geometrical isomers of [Pt(NH₃₃)(Br)(Cl)(py)] and how many of these )(Br)(Cl)(py)] and how many of these will exhibit optical isomers?will exhibit optical isomers?

13.

13. Aqueous copper sulphate solution (blue in colour) gives :Aqueous copper sulphate solution (blue in colour) gives : (i) a green precipitate with aqueous potassium fluoride and (i) a green precipitate with aqueous potassium fluoride and

(ii) a bright green solution with aqueous potassium chloride. Explain these experimental results.

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Page No. # 40 Page No. # 40 14.

14. WhaWhat is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of coppert is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H

sulphate? Why is it that no precipitate of copper sulphide is obtained when H₂₂S(g) is passed through thisS(g) is passed through this solution?

solution?

15.

15. Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory :Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory : (i) [Fe(CN)

(i) [Fe(CN)₆₆]]⁴⁴^–^– (ii) [FeF(ii) [FeF₆₆]]³³^–^– (iii) [Co(C(iii) [Co(C₂₂OO₄₄))₃₃]]³³^–^– (iv) [CoF(iv) [CoF₆₆]]³³^–^– 16.

16. Draw figure to show the splitting of d orbitals in an octahedral crystal field.Draw figure to show the splitting of d orbitals in an octahedral crystal field.

17.

17. What is What is spectrochemical series? Explain the difference between a weak spectrochemical series? Explain the difference between a weak field ligand and a strong ffield ligand and a strong field ligand.ield ligand.

18.

18. What is What is crystal field splittincrystal field splitting energy? How does the magnitude ofg energy? How does the magnitude of

" "

₀₀ decide the actual configuration of d orbitals decide the actual configuration of d orbitals in a coordination entity?

in a coordination entity?

19.

19. [Cr(NH[Cr(NH₃₃))₆₆]]³⁺³⁺ is paramagnetic is paramagnetic while [Ni(CN)while [Ni(CN)₄₄]]²²^–^– is diamagnetic. Explain why? is diamagnetic. Explain why?

20.

20. A solution of [Ni(HA solution of [Ni(H₂₂O)O)₆₆]]²⁺²⁺ is green but a is green but a solution of [Ni(CN)solution of [Ni(CN)₄₄]]²²^–^– is colourless. Explain. is colourless. Explain.

21.

21. [Fe(CN)[Fe(CN)₆₆]]⁴⁴^–^–and [Fe(Hand [Fe(H₂₂O)O)₆₆]]²⁺²⁺ are of different colours in dilute solutions. Why? are of different colours in dilute solutions. Why?

22.

22. Discuss the nature of bonding in metal carbonyls.Discuss the nature of bonding in metal carbonyls.

23.

23. Give the oxidation state, d orbital occupation and coordination numGive the oxidation state, d orbital occupation and coordination number of the central mber of the central m etal ion in the followingetal ion in the following complexes:

complexes:

(i) K

(i) K₃₃[Co(C[Co(C₂₂OO₄₄))₃₃]] ((iiiiii) ) ((NNHH₄₄))₂₂[CoF[CoF₄₄]] ((iiii) ) cciiss--[[CCrr((eenn))₂₂ClCl₂₂]]CCll ((iivv) ) [[MMnn((HH₂₂O)O)₆₆]SO]SO₄₄ 24.

24. Write down the Write down the IUPIUPAC name for AC name for each of the following complexes each of the following complexes and indicate the oxidation state, electronicand indicate the oxidation state, electronic configuration and coordination number.

configuration and coordination number. Also give stereochemistry and magnetic moment of Also give stereochemistry and magnetic moment of the complex :the complex : (i) K[Cr(H

(i) K[Cr(H₂₂O)O)₂₂(C(C₂₂OO₄₄))₂₂].3H].3H₂₂OO ((iiii))[[CCoo((NNHH₃₃)5Cl-]Cl)5Cl-]Cl₂₂ (iii) CrCl

(iii) CrCl₃₃(py)(py)₃₃ (iv) Cs[FeCl(iv) Cs[FeCl₄₄]] ((vv))KK₄₄[Mn(CN)[Mn(CN)₆₆]]

25.

25. What is What is meant by stability of a coordination compound in solution? State the factors which govern stability ofmeant by stability of a coordination compound in solution? State the factors which govern stability of complexes.

complexes.

26.

26. What is meant by the chelate effect? Give an example.What is meant by the chelate effect? Give an example.

27.

27. Discuss briefly giving an example in each case the role of Discuss briefly giving an example in each case the role of coordination compounds in :coordination compounds in : ((ii) b) biioollooggiiccaal sl syysstteemmss ((iiii) ) mmeeddiicciinnaal l cchheemmiissttrry y aanndd

(iiiii) a) ananallyytiticacal cl chehemimiststryry (i(ivv) ) exexttraractctiionon/m/meetatallllururggy y of of memetatalsls..

28.

28. How many ions are produced from the complex Co(NHHow many ions are produced from the complex Co(NH₃₃))₆₆ClCl₂₂ in solution? in solution?

((ii) ) 66 ((iiii) ) 44 ((iiiiii) ) 33 ((iivv) ) 22 29.

29. Amongst the following ions which one has the highest magnetic momAmongst the following ions which one has the highest magnetic mom ent value?ent value?

(i) [Cr(H

(i) [Cr(H₂₂O)O)₆₆]]³⁺³⁺ (ii) [Fe(H(ii) [Fe(H₂₂O)O)₆₆]]²⁺²⁺ (iii) [Zn(H(iii) [Zn(H₂₂O)O)₆₆]]²⁺²⁺

30.

30. The oxidation number of cobalt in K[Co(CO)The oxidation number of cobalt in K[Co(CO)₄₄] is :] is :

((ii) ) + + 11 ((iiii) ) + + 33 ((iiiiii)) – –

1 1

(iv)(iv) – –

3 3

31.

31. Amongst the following, the most stable complex is :Amongst the following, the most stable complex is : (i) [Fe(H

(i) [Fe(H₂₂O)O)₆₆]]³⁺³⁺ (ii) [Fe(NH(ii) [Fe(NH₃₃))₆₆]]³⁺³⁺ (iii) [Fe(C(iii) [Fe(C₂₂OO₄₄))₃₃]]³³^–^– (iv) [FeCl(iv) [FeCl₆₆]]³³^–^– 32.

32. What will be the correct order What will be the correct order for the wavelengths of absorption in the visible region for the following :for the wavelengths of absorption in the visible region for the following : [Ni(NO

[Ni(NO₂₂))₆₆]]⁴⁴^–^–,, [[NNii((NNHH₃₃))₆₆]]²⁺²⁺,, [[NNii((HH₂₂O)O)₆₆]]²⁺²⁺??

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Page No. # 41 Page No. # 41

EXERCISE # 1 EXERCISE # 1

PART # I PART # I A-1.

A-1. (A)(A) A-2.A-2. (B)(B) A-3.A-3. (D(D)) A-4.A-4. (D(D)) A-5.A-5. (A)(A) A-6.A-6. (D(D)) A-7.A-7. (B)(B) A-8.

A-8. (A)(A) A-9.A-9. (C(C)) B-1.B-1. (C(C)) B-2.B-2. (C(C)) B-3.B-3. (A)(A) B-4.B-4. (B)(B) B-5.B-5. (D(D)) B-6.

B-6. (C(C)) B-7.B-7. (B)(B) B-8.B-8. (B)(B) B-9.B-9. (C(C)) B-10.B-10. (B)(B) B-11.B-11. (A)(A) B-12.B-12. (C(C)) B-13.

B-13. (D(D)) B-14.B-14. (C(C)) C-1.C-1. (A)(A) C-2.C-2. (C(C)) C-3.C-3. (A)(A) C-4.C-4. (A)(A) C-5.C-5. (C(C)) C-6.

C-6. (B)(B) C-7.C-7. (C(C)) C-8.C-8. (A)(A) C-9.C-9. (B)(B) C-10.C-10. (B)(B) C-11.C-11. (B)(B) C-12.C-12. (C(C)) C-13.

C-13. (D(D)) C-14.C-14. (C(C)) D-1.D-1. (D(D)) D-2.D-2. (A)(A) D-3.D-3. (A)(A) E-1.E-1. (D(D)) E-2.E-2. (C(C)) E-3.

E-3. (A)(A) E-4.E-4. (A)(A) E-5.E-5. (A)(A) E-6.E-6. (C(C)) E-7.E-7. (D(D)) E-8.E-8. (C(C)) E-9.E-9. (C(C)) E-10.

E-10. (C(C)) E-11.E-11. (C(C)) F-1.F-1. (C(C)) F-2.F-2. (B)(B) F-3.F-3. (D(D)) G-1.G-1. (A)(A) G-2.G-2. (C(C)) G-3.

G-3. (A)(A) G-4.G-4. (C(C)) G-5.G-5. (A)(A) G-6.G-6. (C(C)) G-7.G-7. (D(D)) G-8.G-8. (C(C)) G-9.G-9. (B)(B) G-10.

G-10. (D(D)) G-11.G-11. (D(D)) G-12.G-12. (C(C)) G-13.G-13. (C(C)) G-14.G-14. (C(C)) H-1.H-1. (C(C)) H-2.H-2. (B)(B) H-3.

H-3. (B)(B) H-4.H-4. (C(C)) H-5.H-5. (C(C)) I-1.I-1. (D(D)) I-2.I-2. (D(D)) I-3.I-3. (C(C)) I-4.I-4. (A)(A) I-5.

I-5. (D(D)) I-6.I-6. (C(C)) I-7.I-7. (D(D))

PART # II PART # II 1.

1. (B)(B) 2.2. (D(D)) 3.3. (A)(A) 4.4. (B)(B) 5.5. (D(D)) 6.6. (C(C)) 7.7. (B)(B) 8.

8. (D(D)) 9.9. (A(A – –

p, q, r) ; (B p, q, r) ; (B

– –

p, p, s); s); (C (C

– –

q, r) ; (D q, r) ; (D

– –

q, r) q, r)

10.

10. (A(A – –

p, r, s, t) ; (B p, r, s, t) ; (B

– –

q) ; (C q) ; (C

– –

q) ; (D q) ; (D

– –

p, r, s) p, r, s)

11.11. (A(A – – p,s) ; (B p,s) ; (B – –p, s, t) ; (Cp, s, t) ; (C – – s, t) ; (D s, t) ; (D – – p, s) p, s) 12.

12. (A)(A) 13.13. (C(C)) 14.14. (A)(A) 15.15. (B)(B) 16.16. (A)(A) 17.17. (B)(B) 18.18. (C(C)) 19.

19. (A)(A) 20.20. (D(D)) 21.21. (B)(B) 22.22. (C(C)) 23.23. FalseFalse 24.24. TrueTrue 25.25. FalseFalse 26.

26. FalseFalse 27.27. TrueTrue 28.28. FalseFalse 29.29. TrueTrue 30.30. TrueTrue 31.31. TrueTrue 32.32. FalseFalse

EXERCISE # 2 EXERCISE # 2

PART PART

# #

I I 1.

1. (A)(A) 2.2. (A)(A) 3.3. (A)(A) 4.4. (C(C)) 5.5. (A)(A) 6.6. (D(D)) 7.7. (B)(B) 8.

8. (C(C)) 9.9. (B)(B) 10.10. (D(D)) 111.1. (B)(B) 12.12. (D(D)) 13.13. (B)(B) 14.14. (A)(A) 15.

15. (D(D)) 16.16. (A)(A) 17.17. (A)(A) 18.18. (C(C)) 19.19. (C(C)) 20.20. (A)(A) 21.21. (C(C)) 22.

22. (D(D)) 23.23. (C(C)) 24.24. (C(C)) 25.25. (B)(B) 26.26. (C(C)) 27.27. (D(D)) 28.28. (D(D)) 29.

29. (C(C)) 30.30. (ACD)(ACD) 31.31. (BCD)(BCD) 32.32. (BD)(BD) 33.33. (ABD)(ABD) 34.34. (BD)(BD) 35.35. (CD)(CD) 26.

26. (ABC)(ABC) 37.37. (AD)(AD) 38.38. (BCD)(BCD) 39.39. (ABCD)(ABCD) PART PART

# #

II II 1.

1. (c)(c) [Fe(CO)[Fe(CO)₅₅]],, PPeennttaaccaarrbboonnyylliirroonn((00)) (d

(d)) [Fe(C[Fe(C₂₂OO₄₄))₃₃]]³³^–^–,, TTrriiooxxaallaattooffeerrrraattee((IIIIII)) OORR TTrriiss((ooxxaallaattoo))ffeerrrraattee((IIIIII)) (e

(e)) [Cu(NH[Cu(NH₃₃))₄₄]SO]SO₄₄,, TTeettrraaaammmmiinneeccooppppeerr((IIII) s) suullpphhaattee (f

(f)) Na[Cr(OH)Na[Cr(OH)₄₄]],, SSooddiiuum m tteettrraahhyyddrrooxxiiddoocchhrroommaattee((IIIIII)) (g

(g)) Co(gly)Co(gly)₃₃,, TTrriiggllyycciinnaattooccoobbaalltt((IIIIII)) OORR TTrriiss((ggllyycciinnaattoo))ccoobbaalltt((IIIIII)) (h

(h)) [Fe(H[Fe(H₂₂O)O)₅₅(SCN)](SCN)]²⁺²⁺,, PPeennttaaaaqquuaatthhiiooccyyaannaattoo – –

S S

– –

iron(III) iron(III)

(i)

(i) KK₂₂[HgI[HgI₄₄]],, PPoottaassssiiuum m tteettrraaiiooddiiddoommeerrccuurraattee((IIII)) (j)

(j) Co[Hg(SCN)Co[Hg(SCN)₄₄]],, CCoobbaalltt((IIII) t) teettrraatthhiiooccyyaannaattoo – –

S S

– –

mercurate(II) mercurate(II)

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Page No. # 42 Page No. # 42 (k)

(k) FeFe₄₄[Fe(CN)[Fe(CN)₆₆]]₃₃,, IIrroonn((IIIIII) h) heexxaaccyyaanniiddooffeerrrraattee((IIII)) (l)

(l) KK₃₃[Co(NO[Co(NO₂₂))₆₆]],, PPoottaassssiiuum m hheexxaanniittrriittoo – –

N N

– –

cobaltate(III) cobaltate(III)

(m)

(m) [Ni(dmg)[Ni(dmg)₂₂]],, BBiiss((ddiimmeetthhyyllggllyyooxxiimmaattoo))nniicckkeell((IIII)) (n

(n)) KK₂₂[PtCl[PtCl₆₆]],, PPoottaassssiiuum m hheexxaacchhlloorriiddooppllaattiinnaattee((IIVV)) (o

(o)) NaNa₂₂[Fe(CN)[Fe(CN)₅₅NONO⁺⁺]],, SSooddiiuum m ppeennttaaccyyaanniiddoonniittrorossoonniiuumfmfeerrrraattee((IIII)) (p

(p)) [Fe(H[Fe(H₂₂O)O)₅₅(NO(NO⁺⁺)]SO)]SO₄₄,, PePentntaaaaququananititrorososoniniumumirironon(I(I) ) susulplphahatete (q

(q)) [Cu(CN)[Cu(CN)₄₄]]³³^–^–,, TTeettrraaccyyaanniiddooccuuppeerraattee((II)) 2.

2. ((aa)) DDiiaammmmiinneettrriiaaqquuaahhyyddrrooxxiiddoocchhrroommiiuumm((IIIIII) n) niittrraattee [[CCrr((NNHH₃₃))₂₂(H(H₂₂O)O)₃₃(OH)](NO(OH)](NO₃₃))₂₂ ((bb)) TTeettrraakkiiss((ppyyrriiddiinnee))ppllaattiinnuumm((IIII) ) tteettrraapphheennyyllbboorraattee((IIIIII)) [[PPtt((PPyy))₄₄][B(ph)][B(ph)₄₄]]₂₂

((cc)) DDiibbrroommiiddootteettrraaccaarrbboonnyylliirroonn((IIII)) [[FFee((BBrr))₂₂(CO)(CO)₄₄]]

(d)) TTetretraammiaamminecnecobaobalt(lt(IIIIII)-)-

33

-amido--amido-

33

-hydroxidobis(ethylenediamine or ethane-1, -hydroxidobis(ethylenediamine or ethane-1, 2-diamine)cobalt(III) chloride2-diamine)cobalt(III) chloride

((ee)) AAmmmmoonniiuum m ddiiaammmmiinneetteettrraakkiiss((iissootthhiiooccyyaannaattoo))cchhrroommaattee((IIIIII)).. ((NNHH₄₄)[Cr(NH)[Cr(NH₃₃))₂₂(NCS)(NCS)₄₄]]

((ff)) PPeennttaaaammmmiinneeddiinniittrrooggeennrruutthheenniiuumm((IIII) ) cchhlloorriiddee [[RRuu((NNHH₃₃))₅₅NN₂₂]Cl]Cl₂₂ ((gg)) BBiiss((ccyyccllooppeennttaaddiieennyyll))iirroonn((IIII)) [[FFee((

1 1

⁵⁵_–_–

_C _C

5 5HH₅₅))₂₂]]

((hh)) BBaarriiuum m ddiihhyyddrrooxxiiddooddiinniittrriittoo--OO--ooxxaallaattoozziirrccoonnaattee((IIVV)) BBaa[[ZZrr((OOHH))₂₂(ONO)(ONO)₂₂(ox)](ox)]

((ii)) TTeettrraappyyrriiddiinneeppllaattiinnuumm((IIII) ) tteettrraacchhlloorriiddooppllaattiinnaattee((IIII)) [[PPtt((ppyy))₄₄][PtCl][PtCl₄₄]]

(j)) TTetretraammaammineineaquaquacoacobalbalt(It(III)-II)-

33

-cyanidotetraamminebromidocobalt(III) -cyanidotetraamminebromidocobalt(III) [(NH[(NH₃₃))₄₄(H(H₂₂O)CoO)Co – –

CN CN

– –

Co(NH Co(NH

₃₃))₄₄Br]Br]⁴⁺⁴⁺

3.. ((aa)) ii < i < iv < iii.ii < i < iv < iii. (b(b)) ((ii) ) 66 ((iiii) ) 22 ((iiiiii) ) 11 4.

4. CCoommpplleexx GGeeoommeettrryy HHyybbrriiddiissaattiioonn NNuummbbeer r oof f uunnppaaiirreed d eelleeccttrroonnss((nn) ) MMaagg. . mmoommeenntt CN =2

CN =2

((aa)) [[AAgg((NNHH₃₃))₂₂]]⁺⁺ LLiinneeaarr sspp 00 00 ((bb)) [[CCuu((CCNN))₂₂]]^–^– LLiinneeaarr sspp 00 00 ((cc)) [[AAuuCCll₂₂]]^–^– LLiinneeaarr sspp 00 00

CN = 4 CN = 4

((dd)) [[PPttCCll₂₂(NH(NH₃₃))₂₂]] SSqquuaarre e PPllaannaarr ddsspp²² 00 00 ((ee)) [[ZnZn(C(CN)N)₄₄]]²²^–^– TTeettrraahheeddrraall sspp³³ 00 00 ((ff)) [[CCuu((CCNN))₄₄]]³³^–^– TTeettrraahheeddrraall sspp³³ 00 00

((gg)) [[MMnnBBrr₄₄]]²²^–^– TTeettrraahheeddrraall sspp³³ 55 55..9922BBMM ((hh)) [[CCuu(N(NHH₃₃))₄₄]]²⁺²⁺ SSqquuaarre e PPllaannaarr ddsspp²² 11 11..7733BBMM ((ii)) [[CCooII₄₄]]²²^–^– TTeettrraahheeddrraall sspp³³ 33 33..8877BBMM

CN = 6 CN = 6

((jj)) [[MMnn((CCNN))₆₆]]³³^–^– OOccttaahheeddrraall dd²²spsp³³ 22 22..8833BBMM ((kk)) [[CCrr(N(NHH₃₃))₆₆]]³⁺³⁺ OOccttaahheeddrraall dd²²spsp³³ 33 33..8877BBMM ((ll)) [[FFee((CCNN))₆₆]]³³^–^– OOccttaahheeddrraall dd²²spsp³³ 11 11..7733BBMM (m

(m)) [I[Ir(r(NHNH₃₃))₆₆]]³⁺³⁺ OOccttaahheeddrraall dd²²spsp³³ 00 00

((nn)) [[VV((CCOO))₆₆]] OOccttaahheeddrraall dd²²spsp³³ 11 11..7733BBMM ((oo)) [[FFee((HH₂₂O)O)₆₆]]²⁺²⁺ OOccttaahheeddrraall sspp³³dd²² 44 44..9900BBMM ((pp)) [[MMnnCCll₆₆]]³³^–^– OOccttaahheeddrraall sspp³³dd²² 44 44..9900BBMM

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Page No. # 43 Page No. # 43 5.

5. (i) H(i) H₁₂₁₂OO₆₆ClCl₃₃CrCr A should be [Cr(H

A should be [Cr(H₂₂O)O)₆₆]Cl]Cl₃₃ because it is not reacting because it is not reacting with Hwith H₂₂SOSO₄₄ if there would have some moles of water outer the if there would have some moles of water outer the coordination sphere then it will be

coordination sphere then it will be reacting with Hreacting with H₂₂SOSO₄₄ (B) weight of H

It means one mole of H₂₂O in B complex outer the coordination sphereO in B complex outer the coordination sphere B = [Cr[H

(ii) In both complexes chromium is in +3 oxidation state. Chromium with 3d

(ii) In both complexes chromium is in +3 oxidation state. Chromium with 3d³³ configuration has 3 unpaired configuration has 3 unpaired electrons with weak field as well as strong field ligand. So, the hybridisation scheme is as follow :

electrons with weak field as well as strong field ligand. So, the hybridisation scheme is as follow :

((iiiiii))

33

= = nn((nn7722)) = = 1515 ((iivv)) EEAAN N = = 2244 – –

3 + 12 = 33 3 + 12 = 33

(v) (v) YYes, both have two ions es, both have two ions per formula unit.per formula unit.

[Cr(edta)] ta)] ^–^–enantiomenantiomersers

–

no higher S_n_n axis); [Pt (dien)Cl] axis); [Pt (dien)Cl]⁺⁺ has a plane of symmetry and hence is achiral. has a plane of symmetry and hence is achiral.

(b)

(b) CarbonylhyCarbonylhydridobis(trimethylphosphine)irridium(I).dridobis(trimethylphosphine)irridium(I).

!!

r is in +1 r is in +1 oxidation state; 5doxidation state; 5d⁸⁸ configuration has higher CFSE and configuration has higher CFSE and thus the complex is thus the complex is square planar.square planar.

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Page No. # 44 Page No. # 44 [[

!!

rH(CO)(PMerH(CO)(PMe₃₃))₂₂]]

5dd 66ss 66pp

dsp

dsp²²hybhybridridisaisatiotionn Geometry = Square planar

Geometry = Square planar

Magnetic moment = O (all electrons are

Magnetic moment = O (all electrons are paired).paired).

8. cis, trans and optical isomers are possible.cis, trans and optical isomers are possible.

9. (a(a)) There aThere are threre three conse constitutitutiontional isal isomersomers (i) [Ru(NH

(i) [Ru(NH₃₃))₅₅(NO(NO₂₂))]]CCll ((iiii) ) [[RRuu((NNHH₃₃))₅₅Cl](NOCl](NO₂₂) or [Ru(NH) or [Ru(NH₃₃))₅₅Cl]ONO (iii)Cl]ONO (iii) [Ru(NH[Ru(NH₃₃))₅₅ ONO]Cl ONO]Cl (i) & (ii) are ionisation isomers

(i) & (ii) are ionisation isomers (i) & (iii) are linkage isomers (i) & (iii) are linkage isomers

((bb)) ((ii)) ((iiii)) ((iiiiii))

((iivv)) ((vv)) ((vvii))

10.

10. Diastereisomers are stereisomers which are not enatiomers.Diastereisomers are stereisomers which are not enatiomers.

(a (a))

Both cis and trans isomers do not s

Both cis and trans isomers do not show optical activhow optical activity because of the presence of plane and centre of ity because of the presence of plane and centre of symmetries.symmetries.

(b) It will not exhibit geometrical isom

(b) It will not exhibit geometrical isomerism as it erism as it exists only in one form as exists only in one form as given belowgiven below..

(c) In tetrahedral geometry all positions are adjacent to each other so it will not exhibit geometrical isomerism.so it will not exhibit geometrical isomerism.

(d) In square planar geometr

(d) In square planar geometry there is plane of symmetry. So it does not show optical isomerism.y there is plane of symmetry. So it does not show optical isomerism.

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Page No. # 45 Page No. # 45 (e) It will not exhibit geometrical isom

(e) It will not exhibit geometrical isomerism as it erism as it exists only in one form as exists only in one form as given belowgiven below..

Co Co³⁺³⁺

SCN SCN

en en

SCN SCN

SCN SCN SCN

SCN

(f) Cr(NH

(f) Cr(NH₃₃))₂₂(H(H₂₂O)O)₂₂ClCl₂₂]]⁺⁺ is of Mais of Ma₂₂bb₂₂cc₂₂type which has following isomeric forms.type which has following isomeric forms.

(aa)(bb)(cc) (aa)(bb)(cc) (aa)(bc)(bc) (aa)(bc)(bc) (bb)(ac)(ac) (bb)(ac)(ac) (cc)(ab)(ab) (cc)(ab)(ab) (ab)(ac)(bc) (ab)(ac)(bc)

(g (g))

11.1. (a) No ;(a) No ; (b) Y(b) Yes ; (c) Yes ; (c) Yes ; (d) Yes ; (e) Yes ; (d) Yes ; (e) Yes ; (f) es ; (f) No.No.

12.

8 8

Electronic distribution in a tetrahedral dElectronic distribution in a tetrahedral d⁶⁶ ion ion t

t_2g_2g

eg eg

"_t_t Number of unpaired electrons = 4Number of unpaired electrons = 4

8

Electronic distribution in square planar dElectronic distribution in square planar d⁷⁷ ion ion

dz dz²² dyz dyz dzxdzx

Number of unpaired electron = 1 Number of unpaired electron = 1 d

d_x_x22_–_– y y22

dxy dxy

13.

13. 2121 14.14. 55 15.15. 22 16.16. 44 17.17. 8484 18.18. 1212 19.19. 00 20.

20. 55 21.21. 88 22.22. 33 23.23. 00 24.24. 55 25.25. 22 26.26. 00 27.

27. 55 28.28. 55

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Page No. # 46 Page No. # 46

EXERCISE # 3 EXERCISE # 3

PA PART # RT # II 1.

1. In [Co(NHIn [Co(NH₃₃))₆₆]]³⁺³⁺, the oxidation state of cobalt is +3 and coordination number , the oxidation state of cobalt is +3 and coordination number is 6. Sois 6. So Co

Co³⁺³⁺ i ionon NH

NH₃₃ is stronger field ligand. So it compels for the pairing of electrons. T is stronger field ligand. So it compels for the pairing of electrons. Thenhen [Co(NH

[Co(NH₃₃))₆₆]]³⁺³⁺

d²²spsp³³ - hybridisation - hybridisation Thus with 6 coordination number, the complex [Co(NH

Thus with 6 coordination number, the complex [Co(NH₃₃))₆₆] is octahedral as giv] is octahedral as given below.en below.

In [Ni(CN)

In [Ni(CN)₄₄]]²²^–^–, the oxidation state of nickel is +2 and , the oxidation state of nickel is +2 and coordination number is 4. Socoordination number is 4. So Ni

Ni²⁺²⁺

CN^–^– is stronger field ligand. So it compels for the pairing of electrons. T is stronger field ligand. So it compels for the pairing of electrons. Thenhen [Ni(CN)

[Ni(CN)₄₄]]²²^–^–

dsp

dsp²² - hybridisation - hybridisation Thus with 4 coordination number four, the complex is

Thus with 4 coordination number four, the complex is square planar as given belowsquare planar as given below

In [Ni(CO)

In [Ni(CO)₄₄], the oxidation state of ], the oxidation state of nickel is zero and the coornickel is zero and the coordination number is four. Sodination number is four. So Ni (O)

Ni (O)

CO is strong field ligand. So it compels for the pairing of electrons; so nickel under goes rearrangem

CO is strong field ligand. So it compels for the pairing of electrons; so nickel under goes rearrangem ent. Thenent. Then [Ni(CO)

[Ni(CO)₄₄]]

sp³³ - hybridisation - hybridisation Thus with 4 coordination number, the complex is tetrahedral

Thus with 4 coordination number, the complex is tetrahedral as given belowas given below..

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Page No. # 47 Page No. # 47 2.

2. (A)(A) 3.

3. As form(A) gives white precipitate with AgNOAs form(A) gives white precipitate with AgNO₃₃ solution and precipitate is readily soluble in dilute aqueous solution and precipitate is readily soluble in dilute aqueous ammonia, the com

ammonia, the complex must be having the Clplex must be having the Cl^–^– ion in the ionisation sphere. Hence(A) must be having the formula ion in the ionisation sphere. Hence(A) must be having the formula [Cr(NH

[Cr(NH₃₃))₄₄ClBr]ClBr]⁺⁺ClCl^–^–..

Ag⁺⁺ + Cl + Cl^–^–

$ $

AgCl AgCl

6 6

(white). (white).

AgCl + 2NH

AgCl + 2NH₃₃

$ $

[Ag(NH [Ag(NH₃₃))₂₂]]⁺⁺ClCl^–^– (soluble complex). (soluble complex).

Similarly form(B) gives

Similarly form(B) gives pale yellow precipitate with Apale yellow precipitate with AgNOgNO₃₃ and precipitate is soluble in concentrated ammonia, and precipitate is soluble in concentrated ammonia, the complex must be having the Br

the complex must be having the Br^–^– in the ionisation sphere. Hence(B) must be having the formula [Cr(NH in the ionisation sphere. Hence(B) must be having the formula [Cr(NH₃₃))₄₄ClCl₂₂]]⁺⁺BrBr^–^–..

Ag⁺⁺ + Br + Br^–^–

$ $

AgBr AgBr

6 6

(pale yellow). (pale yellow).

AgBr + 2NH

AgBr + 2NH₃₃

$ $

[Ag(NH [Ag(NH₃₃))₂₂]]⁺⁺ Br (soluble complex). Br (soluble complex).

In both complexes, the chromium is the central metal ion with +3 oxidation state. In both, the ammonia is a In both complexes, the chromium is the central metal ion with +3 oxidation state. In both, the ammonia is a strong field ligand so it compels for pairing of electrons. T

strong field ligand so it compels for pairing of electrons. Thus,hus,

[Cr(NH

[Cr(NH₃₃))₄₄ClBr]ClBr]⁺⁺ o orr [[CCrr((NNHH₃₃))₄₄ClCl₂₂]]⁺⁺

As i

As it cont contaitains thns three uree unpanpaireired eld electectronrons,s, So,So,

33

= = 33((337722)) = 3.872 B.M. = 3.872 B.M.

4. In[NiClIn[NiCl₄₄]]²²^–^– nickel is in+2 oxidation state and Cl nickel is in+2 oxidation state and Cl^–^– is weak field ligand. So, is weak field ligand. So,

[Ni(Cl) [Ni(Cl)₄₄]]²²^–^–

33

= = nn((nn7722)) = = 22((227722)) = = 22..882 2 BB..MM.. ;; n n = = NNoo. . oof f uunnppaaiirreed d eelleeccttrroonnss..

Hence with coordination number four, the structure is Hence with coordination number four, the structure is

Tetrahedral Tetrahedral

In[Ni(CN)

In[Ni(CN)₄₄]]²²^–^– nickel is in+2 oxidation state and nickel is in+2 oxidation state and CNCN^–^– is strong field ligand, So it compels for pairing of electrons. is strong field ligand, So it compels for pairing of electrons.

Then, Then,

[Ni(CN) [Ni(CN)₄₄]]²²^–^–

As all electrons are paired so diamagnetic.

Hence with coordination number four, the structure is Hence with coordination number four, the structure is

Square planar Square planar

5. Let n is the number of unpaired electron in the chromium ion.Let n is the number of unpaired electron in the chromium ion.

Since

33

= = nn((nn7722)) oorr 11..773 3 == nn((nn7722)) B B..MM.. oorr 11..7733

× 1.73 = n × 1.73 = n

²² + 2n. + 2n.

Heennccee n n = 1= 1..

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Page No. # 48 Page No. # 48 As the CN

As the CN^–^– and NH and NH₃₃ are strong fields ligands, they compel are strong fields ligands, they compel for pairing of electrons. So,for pairing of electrons. So,

[Cr(NO)(CN)

[Cr(NO)(CN)₄₄(NH(NH₃₃)])]²²^–^–==

Hence, the oxidation state of chro

Hence, the oxidation state of chromium is +1 (having 3dmium is +1 (having 3d⁵⁵ configuration). So accord configuration). So according to charge on the complexing to charge on the complex NO should be NO

NO should be NO⁺⁺ and the structure of this complex is octahedral with d and the structure of this complex is octahedral with d²²spsp³³ hybridisation as given below hybridisation as given below

According to IUPAC nomenclature its name is According to IUPAC nomenclature its name is ::

Potassium amminetatracyanidonitrosoniumchromate(I)

Potassium amminetatracyanidonitrosoniumchromate(I) OROR Potassium amminetatracyanidonitrocyliumchromate(I).

Potassium amminetatracyanidonitrocyliumchromate(I).

6. (D(D)) 7.7. (C(C)) 8.

8. NiNi²⁺²⁺ + 2dmg + 2dmg # # # # ^NH^NH # # ⁴⁴^OH^OH # # $$ [Ni(dmg) [Ni(dmg)₂₂]]

6 6

(bright red). (bright red).

It acquires stability through chelation and intra molecular H-bonding.

In [Ni(dmg)

In [Ni(dmg)₂₂] the nickel is in +2 oxidaiton state and to have square planar geom] the nickel is in +2 oxidaiton state and to have square planar geometry because of chelation theetry because of chelation the pairing of electrons takes place. So

pairing of electrons takes place. So

[Ni(DMG) [Ni(DMG)₂₂]]

As all electrons are paired, so complex is diamagnetic. Nickel with coordination number four will have the As all electrons are paired, so complex is diamagnetic. Nickel with coordination number four will have the structur

structure as given belowe as given below..

CH33 – – ^{C = N}^{C = N} H

H33CC – – ^{C = N}^{C = N}

N = C

N = C – – ^CH^CH33

N = C

N = C – – ^CH^CH33

Ni Ni O --- H O --- H – – ^O^O

O – – H --- O H --- O rosy red ppt rosy red ppt

+2 +2

9. (A)(A) 1

100.. ((aa)) Fe Fe³⁺³⁺

)) Excess Excess ((

SCN SCN

# # # # # # $ $ #

#

⁰⁰ blood red[Fe((H blood red[Fe((H₂₂O)O)₅₅(SCN)](SCN)]²⁺²⁺ (A (A))

# # # #

^F^F⁰⁰

# #

⁽⁽^excess^excess

# # # #

⁾⁾

$ $

colourless(B) [Fe(F colourless(B) [Fe(F₆₆)])]³³^–^– + SCN + SCN^–^– + + 5H5H₂₂O.O.

(A)

(A) Pentaaquathiocyanato-S-iroPentaaquathiocyanato-S-iron(III) n(III) ; ; (B) (B) HexafluoridofeHexafluoridoferrate(III)rrate(III) (b)

(b) FeFe³⁺³⁺CFSE electron configuration, tCFSE electron configuration, t_2g_2g^{1, 1, 1}^{1, 1, 1}ee_g_g^1,1^1,1; as F; as F^–^– being weak field ligand does not compel for pairing of being weak field ligand does not compel for pairing of

In document LA FALANGE INQUIETA (página 36-42)

1. DEL DESERT A L’OASI

1.1. Falange en la nit del feixisme

NCERT QUESTIONS

s postulates. s postulates.

" "

1 1

3 3

EXERCISE # 1 EXERCISE # 1

p, q, r) ; (B p, q, r) ; (B

p, p, s); s); (C (C

q, r) ; (D q, r) ; (D

q, r) q, r)

p, r, s, t) ; (B p, r, s, t) ; (B

q) ; (C q) ; (C

q) ; (D q) ; (D

p, r, s) p, r, s)

EXERCISE # 2 EXERCISE # 2

# #

# #

S S

iron(III) iron(III)

S S

mercurate(II) mercurate(II)

N N

cobaltate(III) cobaltate(III)

33

33

1 1

C C

33

CN CN

Co(NH Co(NH

33

3 + 12 = 33 3 + 12 = 33

!!

!!

8 8

8

8

EXERCISE # 3 EXERCISE # 3

$ $

6 6

$ $

$ $

6 6

$ $

33

33

33

× 1.73 = n × 1.73 = n

6 6

# # # # # # $ $ #

#

# # # #

# #

# # # #

$ $

_C _C