iit-...when we move down the group in periodic table, size of element increases, electronegativity...

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2. Section-A 30ONLY ONE Q.1 to Q.10 Q.11 to Q.3

Section-B 10ONE or MORE than ONE

2 marks.

-C 20 Q.41 to Q.50 1 MarkQ.51 to Q.60 2 Marks

1 mark 1/3 marks2 marks 2/3 marks

contains Multiple Choice Questions (MCQ). Each question has 4 choices (a), (b), (c) and (d), for its answer, out of which is correct. From carries 1 Marks and 0 carries 2 Marks each.

3. contains Multiple Select Questions(MSQ). Each question has 4 choices (a), (b), (c) and (d) for its answer, out of which is/are correct. For each correct answer you will be awarded

4. Section contains Numerical Answer Type (NAT) questions. From carries each and carries each. For each NAT type question, the value of answer in between 0 to 9.

5. In all sections, questions not attempted will result in zero mark. In Section–A (MCQ), wrong answer will result in negative marks. For all questions, will be deducted for each wrong answer. For all questions, will be deducted for each wrong answer. In Section–B (MSQ),there is no negative and no partial marking provision. There is no negative marking in Section –C (NAT) as well.

Regn. No.

E: [email protected], W : www . careerendeavour.inJIA SARAI



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SECTION-A : MULTIPLE CHOICE QUESTIONS (MCQ’s)

Q.1 to Q.10 : Carry 1 Mark each.

1. In photosynthesis, the predominant metal present in the reaction centre of photosystem-II is(a) Zn (b) Cu (c) Mn (d) Fe

Soln. Mn is the predominant metal present in the reaction centre of photosystem-II in photosynthesis.Correct option is (c)

2. The order of reactivity of following transition metals is(a) Ir3+ > Rh3+ > Co3+ (b) Rh3+ > Ir3+ > Co3+

(c) Co3+ > Rh3+ > Ir3+ (d) Rh3+ > Co3+ > Ir3+

Soln. When we move down the group in periodic table, size of element increases, electronegativity decreases asa result of which reactivity of transition metal decreasesReactivity order is Co3+ > Rh3+ > Ir3+

Correct option is (c)3. Which of the following have more bond angle

(a) CH4 (b) SnCl2 (c) NH3 (d) H2OSoln. Order of bond angle is sp > sp2 > sp3

CH4 sp3, SnCl2 sp

2, NH3 sp3, H2O sp

3

So, highest bond angle obtained in SnCl2Correct option is (b)

4. What is the bond order of the N2+

(a) 122

(b) 2 1 (c)1 22 (d) 1 2

Soln. When we draw MOELD of N2+ -bonding molecular orbital contain 4 electrons which make 2-bond

-bonding molecular orbital contain one electron which make 12 bond, total bond order = 12

2

Correct option is (a)5. Experimentally which of the ion have high ionic conductance

(a) K+ (b) Rb+ (c) Cs+ (d) Na+

Soln. Experimentally, down the group ionic conductance of ions increases so Cs+ has high ionic conductance.Correct option is (c)

6. Number of B–O bond in sodiumperoxoborate ion [Na2B2O4(OH)4].6H2O is(a) 2 (b) 4 (c) 8 (d) 6

Soln. When we draw the structure of sodiumperoxoborate the number of B–O bond comes out to be 8.

O O

B

OO

B

OH

OHHO

HO

2Na+ 6H2O

2–1

2

3

4

5

6 7

8

Correct option is (c)



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7. In the given reaction the product A is

B3N3H6 + 9H2O AA(a) H3BO4 (b) H3BO2 (c) H3BO5 (d) H3BO3

Soln. In the given chemical reaction

B3N3H6 + 9H2O 3H3BO3 + 3NH3 + 3H2Correct option is (d)

8. What is the magnetic moment of Gd3+ ion(a) 6.93 BM (b) 5.91 BM (c) 7.93 BM (d) 6.63 BM

Soln. Gd = [Xe] 4f7, 5d1 , 6s2

Gd3+ = 4f7

Number of unpaired electron (n) = 7

( 2) 7(7 2)n n

63 7.937 BM Correct option is (c)

9. What is the type of spinels of given compounds NiFe2O4 and NiCr2O4(a) Normal, inverse (b) Normal, normal (c) Inverse, inverse (d) Inverse, normal

Soln. Calculate CFSE of spinelsNormal spinels: The divalent ions occupy the tetrahedral voids, whereas trivalent ions occupy the octahedralvoids. e.g. NiCr2O4(Ni

2+, Cr3+)Inverse spinels: The divalent ions occupy the octahedral voids, whereas half of trivalent ions occupy thetetrahedral voids. e.g. NiFe2O4(Ni

2+, Fe3+)Correct option is (d)

10. The structure of [CB9H10AsCo(n5 – Cp)] compound is

(a) nido (b) arachno (c) closo (d) hyperclosoSoln. On solving [CB9H10AsCo(

5 – Cp)], it give structure of B12H14 which is closoC = BH,As = BH2, [Co(

5 – Cp)], 9 + 5 = 14 electrons related to four electrons carbon of 4 electrons BH unitB9H10 + BH + BH2 + BH B12H14 closoCorrect option is (c)

Q.11 to Q.30 : Carry 2 Marks each.

11. Zinc metal in carbonic anhydrase is soften by(a) Coenzyme (b) Apoenzyme(c) Cofactor (d) Stella cyanins

Soln. Apoenzyme soften the zinc metal in carbonic anhydraseCorrect option is (b)

12. Lithium has the most negative standard electrode potential of any element in the periodic table because of(a) high ionization energy (b) high lattice energy(c) low lattice energy (d) high hydration energy

Soln. Because of high hydration energy of lithium, it has most negative standard electrode potential.Correct option is (d)



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13. Correct combination of metal, number of CO ligands and charge for M-CO complex (complex follow 18e–Rule) is [M(CO)x(NMe2)(AsPh2)(SiMe3)2(Br)](a) M = Cr, x = 4, z = 0 (b) M = Rh, x = 3, z = –1(c) M = Co, x = 2, z = 0 (d) M = Fe, x = 2, z = +1

Soln. When M = Co, x = 2, z = 0 combination complex, follow 18 e– Rule.Correct option is (c)

14. Number of M–M bond in given complex, which obey the 18e– Rule[(CO)2Rh( -Cl)2Rh(CO)2](a) 1 (b) 3 (c) 2 (d) 4

Soln. Total number of M – M bond = 18 18 2 32 2

2 2n VSE

n = Number of metalVSE = Valence Shell ElectronCorrect option is (c)

15. Arrange the following in decreasing order of M–C bond length(A) Ir(CO)(Cl)[PPh3]2 (B) Ir(CO)(Cl)[PEt3]2 (C) Ir(CO)(Cl)[(PC6F5)3]2(a) C > A > B (b) B > C > A (c) B > A > C (d) A > B > C

Soln. Electron withdrawing order of ligand isP(C6F5)3 > PPh3 > PEt3More electron withdrawing nature causes less synergic bonding which leads to the more M–C bond length.Correct option is (a)

16. [Ir(Cl)(CO)(NO)(PPh3)2]BF4NO ligand in the given complex is(a) linear, one electron donor (b) bent, one electron donor(c) bent, three electrons donor (d) linear, three electrons donor

Soln. [Ir(Cl)(CO)(NO)(PPh3)2]+

Let number of electron donated by number of ligand is x.9 + 1 + 2 + x + 4 – 1 = 18x + 15 = 18

3x So, number of ligand donate three electron forms linear geometryCorrect option is (d)

17. The hybridization of complex (i) and (ii) are(i) [XeF6] (ii) [TeCl6]

2–

(a) sp3d3 and p3d3 (b) sp3d2 and p3d3 (c) p3d3 and sp3d3 (d) p3d3 and sp3d2

Soln. Lone pair present in [XeF6] involve in hybridisation to form sp3d3, but lone pair present in [TeCl6]

2– notinvolve in hybridisation hence involve p3d3 hybridization.Correct option is (a)

18. The ratio of and -bond in a given compound S4N4F4 is(a) 4 : 1 (b) 3 : 1 (c) 2 : 1 (d) 3 : 2

Soln. When we draw the structure of S4N4F4



5

S N S

N

SNS

N

F

FF

F 4-bond12-bond

Ratio = 12 3:14

Correct option is (b)19. The increasing order of rate of ligand exchange is:-

(a) [PF6]–, SF6, [SiF6]

2–, [AlF6]3– (b) [AlF6]

3–, [SiF6]2–, [PF6]

–, SF6(c) SF6, [PF6]

–, [SiF6]2–, [AlF6]

3– (d) [AlF6]3–, SF6, [PF6]

–, [SiF6]2–

Soln. Higher the oxidation state of central atom, slower the rate of ligand exchange.Hence correct answer is (c)

20. In the given ionic compound, what is the correct order of colour intensity(i) MnO4

– (ii) ReO4– (iii) TcO4

–

(a) (i) > (ii) > (iii) (b) (ii) > (iii) > (i) (c) (iii) > (i) > (ii) (d) (i) > (iii) > (ii)Soln. Down the group colour intensity decreases

MnO4– > TcO4

– > ReO4–

Correct option is (d)21. Fe4[Fe(CN)6]3 show deep colour, because of

M = Metal, L = Ligand, CT = Charge Transfer, I = Intervalence(a) M–LCT (b) L–LCT (c) M–MCT (d) I–CT

Soln. The complex in which the metal are present in two different oxidation state show intervalence charge transferwhich involve a transfer of a electron between two metal ions having different oxidation state.Correct option is (d)

22. In the structure of CuSO4 .5H2O (blue withdrawal) number of H2O in hydrated form is(a) Five (b) Four (c) One (d) Three

Soln. Only one water molecule are in hydrated form.Correct option is (c)

23. Which form best adduct with BF3 in aqueous phase(a) (CH3)3N (b) (CH3)2NH (c) (SiH3)3N (d) (SiH3)2O

Soln. In aqueous phase basicity order(CH3)2NH > (CH3)3NH(SiH3)3N and (SiH3)2O species involve back bonding. Hence, lone pair on nitrogen and oxygen is not easilyavailable. So (CH3)2NH form best adduct with BF3[(CH3)2NH BF3] adductCorrect option is (b)



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24. In the hypervalent structure of ICl3, total number of non-bonding electron pairs are(a) 15 (b) 20 (c) 25 (d) 30

Soln. Hypervalent structure of ICl3 is

ICl

I

Cl

Cl

Cl

Cl

Cl

Total number of non-bonding electron pair = Total non-bonding electrons of Cl + Total non-bonding of I= 16 + 4 = 20Correct option is (b)

25. A ten membered cyclic silicates is formed containing four sodium ions and some Zn ion this cyclic silicatescontains hydrated H2O molecules which are present in 1 : 3 in mole ratio. What is number of Zn metal ionin complete formula.(a) 2 (b) 4 (c) 3 (d) 5

Soln. The formula will formed Na4Zn3[Si5O15].9H2O three zinc (Zn2+) required to complete the formula.

Correct option is (c)26. Which one of the following hybridization does not involve during hydrolysis of PCl5

(a) sp3 (b) sp3d2 (c) sp2 (d) sp3dSoln. Hydrolysis of PCl5 gives H3PO4, POCl3 which are not sp

2 hybridised.Correct option is (c)

27. The Following process gives a polymer 2H OCu polymerisation3 300ºCCH Cl Si A B C The (A) can also be formed by CH3MgCl and SiCl4 in 1 : 1 ratio. Then the type of polyer (C) is(a) Cross linked polymer (b) Linear chain polymer(c) Dimer (d) Cyclic polymer

Soln. polymerization Cross-Linked polymer

Correct option is (a)28. Arrange the following compound in decreasing order of their bond angle

BeCl2, BF3, (C2H5)2O, SnCl2, PH3, H2O(a) BeCl2 > BF3 > SnCl2 > (C2H5)2O > H2O > PH3(b) BeCl2 > SnCl2 > BF3 > (C2H5)2O > PH3 > H2O(c) BF3 > BeCl2 > (C2H5)2O > SnCl2 > PH3 > H2O(d) BF3 > BeCl2 > SnCl2 > (C2H5)2O > H2O > PH3

Soln. Bond angle defined by (i) using VSEPR theory(ii) Order of Bond angle sp > sp2 > sp3

(a) BeCl2 spBF3 sp

2

(C2H5)2O sp3

SnCl2 sp2

PH3 No hybridization least bond angleH2O sp

3

BeCl2 > BF3 > SnCl2 > (C2H5)2O > H2O > PH3Correct option is (a)



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29. In the given compound, which one of the following compound have N–O–N bond.N2O3, N2O4, N2O5, N2O2(a) N2O4 (b) N2O5 (c) N2O2 (d) N2O3

Soln. N2O5, maximum oxidation state = Number of valence electronOxidation state of N2O5 = maximum oxidation state of N2O52 10 0x 5x N2O5 has calculated oxidation state equal to maximum oxidation state so it containN–O–N bond.Correct option is (b)

30. An aqueous solution of a sodium salt on reaction with sulfanilic acid and -naphthylamine followed by thereaction with Zn, a red colour solution is obtained. The sodium salt is(a) NaCl (b) NaNO2 (c) Na2HPO4 (d) NaNO3

Soln. 2Zn Zn 2e

3 2 2NO 2H e NO H O

NH2

SO3H

HNO2–2H2O

•CH3COOH N

SO3H

N • OOCCH3

NH2

NN NH2HO3S

Diazodye (red)

Correct option is (d)

SECTION-B : MULTIPLE SELECT QUESTIONS (MSQ’s)

Q.31 to Q.40 : Carry 2 Marks each.

31. In the given compound, the number of isolobal compounds is/are(a) [Mn(CO)5], CH3 (b) Fe(CO)4, O(c) Co(CO)3, R2Si (d) [Mn(CO)5], R–S

Soln. Isolobal compound have,

(a) [Mn(CO)5] CH3

7 + 10 = 17 7

(b) [Fe(CO)4] O

8 + 8 = 16 6

(d) [Mn(CO)5] R–S

7 + 10 = 17 7

Correct options are (a), (b) and (d)32. For the [Ni(CN)4]

2– complex, the correct statements is/are(a) Thermodynamically unstable (b) Thermodynamically more stable(c) Kinetically labile (d) Kinetically inert

Soln. 24Ni CN

thermodynamically more stable because of more CFSE and kinetically more labile because

more available space to interact with ligand (less steric hinderance).Correct options are (b) and (c)



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33. The reaction of [Cd(H2O)6]2+ with Br– to give [Cd(Br)4]

2–, the reaction occur in number of step, theequilibrium constant of each steps are K1, K2, K3, K4(a) K2 > K3 (b) K3 > K4(c) K4 > K3 (d) K1 > K3

Soln. Generally the first dissociation constant higher than second but in case of 2

2 6Cd H O

K4 > K3

becauase of number of molecule in product more than reactant as a result entropy of reaction increasesG H T S .

1 2 3K K K but 4 3K K

4 2K2 3 4 23Cd H O Br Br CdBr 3H O

34. The correct statement about hemocyanin is/are(a) Cu(I) present in deoxyhemocyanin(b) Cu(II) present in oxyhemocyanin(c) It bind O2 reversibly(d) Inactive site for hemocyanin is Cu (II) blue 3-Imidazole Ring

Soln. Correct options are (a),(b),(c)35. Which of the following statement correct for the following coordination compound [Co4(CO)12]

(a) Nine terminal ligand (b) Eight terminal ligand(c) Three bridging ligand (d) Eight terminal and three bridging ligand

Soln. 4 12Co CO

Co

C

OC

OC Co

Co

C

Co

OC COCO

CO

CO

OO

OC CO

C

O

Three bridge and 9 terminal.36. The three complex Cobaltocene, Nickelocene and Ferrocene are given the correct statement of

stability is/are(i) [Fe( 5 – C5H5)2] (ii) [Co(

5 – C5H5)2](iii) [Ni( 5 – C5H5)2](a) (i) > (iii) (b) (iii) > (i) (c) (ii) > (i) (d) (i) > (ii)

Soln. Only ferrocene follow 18 electron, rule.Correct options are (a) and (d).



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37. In the hydrolysis of XeF6, which of the following product will form(a) XeOF4 (b) XeO2F2 (c) XeO3 (d) XeF4

Soln. In the hydrolysis of XeF6, oxidation state of centeral atom remains same.Correct options are (a), (b) and (c)

38. For the given complex(i) [Ni(H2O)6]

2+ (ii) [Mn(H2O)6]2+ (iii) [Cr(H2O)6]

3+

(iv) [Ti(H2O)6]3+, the ideal octahedral geometry will observed in

(a) (i) (b) (ii) (c) (iii) (d) (iv)Soln. Octahedral complex does not undergoes any type of geometric distortion contain ideal octahedral geometry.

Correct options are (a), (b) and (c)

39. The oxides of carbon, phosphorus, sulphur, chloride are given CO2, P2O3, SO2, ClO2, the correct orderof acidic nature is/are(a) CO2 > P2O3 > SO2 (b) ClO2 > SO2 > P2O3(c) SO2 > P2O3 > CO2 (d) SO2 > ClO2 > P2O3

Soln. On moving left to right in periodic table acidity of oxide increases the correct order isClO2 > SO2 > P2O3 > CO2

40. In case of octahedral dissociative mechanism, the rate of overall substitution reaction:(a) Depends on concentration of original complex(b) Independent on concentration of original complex(c) Independent on concentration of incoming ligands(d) Depends on concentration of incoming ligand

Soln. Dissociative mechanism of octahdral complex is SN1 reaction

6 5ML Y ML Y L Rate = K[ML6]

1

First order.Hence, correct options are (a) and (c).

SECTION-C : NUMERICAL ANSWER TYPE (NAT’s)

Q.41 to Q.50 : Carry 1 Mark each.

41. DMG combine with Ni2+ ion to give Ni-complex having number of type of ring is ...........................Soln. Two type of rings are formed

C NH3C

CH3C N

O

O H O

N C CH3

H O

N C CH3Ni2+



10

42. The bond order of O22+ molecule is ........................................

Soln.

2pz

2pz*

2px 2py

2px 2py* *

2p 2p

Bond order = 3

In 22O molecular orbital diagram of 2p molecular orbital contain six bonding electron and zero anti-bonding

electron. The bond order comes out to be 3.

43. Number of O–Ni–O bond in the given EDTA complex [Ni(EDTA)]2– is ...................................

Soln. Structure of 2Ni EDTA complex contain six O–Ni–O bond

Ni

O

OC

O C CH2CH2

O

C CH2

N

CCH2

N

H2C

CH2

O

O

O

O

44. The given complex [( 3 – C5H5)Cr(CO)n(CH3)] obey 18e– Rule, the value of “n” is .........................

Soln. 3 5 5 3nC H Cr CO CH 3 + 6 + 2×n + 1 = 18Therefore, n = 4

45. The reaction of H2 with wilkinson catalyst result in cis addition, the metal oxidation state change by ............Soln. Catalytic cycle of cis addition of H2 with wilkinson catalyst, metal oxidation state change by +2.46. The number of P–O–P bond in H12P10O31 is ...............................Soln. H12P10O31 is a acyclic compound.

Number of P–O–P bond = 10–1= 9n = number of posphorous atom in acyclic compound.



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47. Total number of species having peroxy linkage among perbenzoic acid, Persulphuric acid, Perchloric acid,Perboric acid is/are ...................................

Soln. Only perchloric acid does not contain any peroxylinkage, the structure of perchloric acid is

O Cl

O

O

OH

Hence, correct answer is (3)

48. Some species given below:AlCl3, GaBr3, BCl3, BH3Number of species does not exist in dimeric form ...........................

Soln. The species having back bonding does not undergo dimeric form

Cl B

Cl

ClHence, correct answer is (3)

49. During hydrolysis of P4O10. How many water molecules required ...........................Soln. Structure of P4O10

P

O

OO P

P

O

O

O

P

O

O

O O

6H2O molecule required for hydrolysis of P4O10.Hence, correct answer is (6)

50. The number of unpaired electron (s) present in the species [Fe(H2O)5(NO)]2+ which is formed during brown

ring test is ..............................

Soln. 22 5

7

Fe H O NO

Fe 3d

3d 4s

Three unpaired electronHence, correct answer is (3)



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Q.51 to Q.60 : Carry 2 Marks each.51. Number of isostructural species is/are ..................................

I3+, BF3, O3, H2O, SnCl2, CH4, XeO2F2, IF2

+, IF4+

Soln. 3 3 2 2 2I , O , H O, SnCl , IF are isostructural.

Correct answer is (5)52. The overall charge present on the cyclic silicate anion [Si3O9]

n– is ..........................

Soln.

O

O

Si

O

Si

Si

O

O

OO

OO

six negative charge present in [Si3O9]n–

53. 100 ml unknown solution of K2Cr2O7 is titrated with excess KI solution the liberated iodine required 0.1M, 100ml standard hypo solution for complete titration the concentration of K2Cr2O7 is .......................

Soln. Equivalent of K2Cr2O7 = Equivalent of hypo

6100 100MX 0.1 11000 1000

2 2 70.1M K Cr O 0.0166

54. Radioactive decay of 13153 I to give

z54 Xe , z is ...................................

Soln. Radioactive decay of 13153 I gives 13154 Xe

131 13183 54I Xe

55. In the given reaction form product P

O OM

Ph

(Grubb's secondgeneration catalyst)

P

the number of sp2 hybridised carbon present in the product P is .......................

Soln.O O

MPh

(Grubb's secondgeneration catalyst)

O6

7

89

4

51

23

ProductNine carbon are sp2-hybridized



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56. During hydroformylation reaction, the precursor molecule converted to true catalyst, the number of COligand present in true catalyst is ...............................

Soln. 2H ligand dissociation

2 oxidative cleavage8 4 3Pr ecatalystPrecursor True catalyst

Co CO HCo CO HCo CO

Number of CO ligand in true catalyst is three.57. An old piece of wood has 25.6% is much C14 as ordinary wood today has, the age of wood (in years)

is ........... (t1/2 of C14 = 5760 years).

Soln. Suppose the amount of C14 present in the wood originally (same which the wood today has) = No

The amount of C14 present now in the old wood (Nt) = 25.6 a 0.256a100

The time t in which C14 changed from a to 0.256a will then be given by

2.303 Not logk 0.256 No

4 1

½

0.693 0.693k 1.203 10 yeart 5760

4

2.303 1t log 11329 years1.203 10 0.256

58. Number of species undergoes back bonding are ....................................CO2, NH3, NF3, NCl3, AlCl3, CCl4, PF3, PCl5, BCl3, CH3NCS, CF3

–, BBr3.Soln. Only NCl3, PF3, BCl3, undergoes back bonding.59. In the following species

BCl3, AlCl3, BF3, SiF4, SiCl4. The number of species undergoes partial hydrolysis ....................Soln. Partial hydrolysis takes place only in those compound like BF3, SiF4 , which on hydrolysis give stable

product 4BF and 26SiF

respectively..

60. Total number of coordination isomer possible for [Co(NH3)6] [FeCl6] is .............................Soln. Coordination isomer are six

3 66Co NH FeCl

3 5 35Co NH Cl FeCl NH

3 2 4 34 2Co NH Cl FeCl NH

3 4 2 32 4Co NH Cl FeCl NH

3 5 3 5Co NH Cl FeCl NH

6 3 6CoCl Fe NH

END



14

IIT-JAM CHEMISTRY(CY) 15-01-2017

TEST SERIES - 3

INORGANIC CHEMISTRY

ANSWER KEY

SECTION-A

1. (c) 2. (c) 3. (b) 4. (a) 5. (c)6. (c) 7. (d) 8. (c) 9. (d) 10. (c)11. (b) 12. (d) 13. (c) 14. (c) 15. (a)16. (d) 17. (a) 18. (b) 19. (c) 20. (d)21. (d) 22. (c) 23. (b) 24. (b) 25. (c)26. (c) 27. (a) 28. (a) 29. (b) 30. (d)

SECTION-B

31. (a),(b),(d) 32. (b),(c) 33. (a),(c),(d) 34. (a),(b),(c)35. (a),(c) 36. (a),(d) 37. (a),(b),(c) 38. (a),(b),(c)39. (b),(c) 40. (a),(c)

SECTION-C

41. (2) 42. (3) 43. (6) 44. (4) 45. (2)46. (9) 47. (3) 48. (1) 49. (6) 50. (3)51. (5) 52. (6) 53. (0.015 to 0.017) 54. (131) 55. (9)56. (3) 57. (11329) 58. (3) 59. (2) 60. (6)

iit-...when we move down the group in periodic table, size of element increases, electronegativity...

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