test date: 28.11.2015 (saturday) test time: 12:30 pm to 02 ...€¦ · 02.11.2015 · a. i, ii,...

Dr. Sangeeta Khanna Ph.D 1 CHEMISTRY COACHING CIRCLE G:\+2 Grand test-6.doc

Test Date: 28.11.2015 (Saturday)

Test Time: 12:30 PM to 02:15 PM

Test Venue:

Lajpat Bhawan, Madhya Marg, Sector 15-B, Chandigarh

Dr. Sangeeta Khanna Ph.D


(+2) 28.11.2015

READ INSTRUCTIONS CAREFULLY

1. The test is of 1 hour 30 minutes duration.

2. The maximum marks are 271.

3. This test consists of 74 questions.

4. Keep Your mobiles switched off during Test in the Halls.

(Single Correct Choice Type) Negative Marking [-1]

This Section contains 42 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONLY ONE is correct. Marks: 42 × 4 = 168

[At. No. Sc = 21; Ti = 22; V = 23;Cr = 24; Mn = 25; Fe = 26; Co = 27; Ni = 28; Cu = 29; Zn = 30; Ag = 47; Pt = 78; Au = 79; Hg = 80; Pd = 46] 1. A complex whose IUPAC name is not correctly written is

Complex Name a. [Fe(C5H5)2] Bis (cyclopenta dienylo) iron (II) b. [Cr(C6H6)2] Bis (benzene) chromium (O) c. [CoCl2(H2O)4].2H2O Tetraaquadichloridocobalt(II)-2-water d. [Zn(NCS)4]

-2 Tetrathiocyanato-S-zincate (II) A,D

2. Which of the following statements is incorrect about Werner’s theory? a. Primary valency is the same thing as oxidation state b. Secondary valency is the same thing as co – ordination number c. Secondary valencies are satisfied only by negative ions d. Secondary valencies are directional whereas primary valencies are non – directional C 3. Which of the following ligand gives chelate complexes?

a. SCN– b. C2O2

4 c. pyridine d. NH2 - 3HN

B

Sol. C2O2

4 is a symmetrical bidentate ligand

4. One among the following complex ions will not show optical activity:

a. [Pt(Br) (Cl) (l) (NO2) (C5H5N) (NH3)]– b. cis-[Co(en)2Cl2]

+

c. [Co(en)(NH3)2Cl2]+ d. Trans-[Cr(NH3)4Cl2]

+

D

5. The IUPAC name for the compound K[SbCl5Ph] is:

a. potassium chlorophenyl antimonite (V) b. potassium pentachlorido phenyl antimonate (V)

c. potassium pentachloro benzylantimonate (V) d. None of these

B

6. [Ni(CN)4]2– and [NiCl4]

2– have similarity but not in:

a. Magnetic moment b. C.N. and O.N. c. Structure d. Both a and c

D

Sol. cdiamagneti planar, Square ; )CN(Ni 24

icparamagnet l,Tetrahedra ; NiCl 24

C.N. = 4 and O.N. = 2 in both the cases



7. In the separation of Cu2+ and Cd2+ in 2nd group of qualitative analysis of cations, tetrammine copper (II) sulphate and tetrammine cadmium (II) sulphate react with KCN to form the corresponding cyano complexes. Which one of the following pairs of the complexes and their relative stability enables the separation of Cu2+ and Cd2+?

a. K3[Cu(CN)4]; more stable and K2[Cd (CN)4] ; less stable

b. K2[Cu(CN)4] : less stable and k2 [Cd (CN)4] : more stable c. K2[Cu(CN)4] : more stable and k2 [Cd (CN)4] : less stable d. K3[Cu(CN)4] : less stable and K2[Cd(CN)4] : more stable A Sol. K3[Cu(CN)4] is more stable than K2[Cd(CN)4] and therefore, Cu2+ ions are not precipitated while Cd2+

ions are precipitated. 8. Co-ordination compounds have great importance in biological systems. In this context, which of the

following statements is incorrect?

a. Cyanocobalamin is Vitamin B12 and contains cobalt b. Haemoglobin is the red pigment of blood and contains iron c. Chlorophyll is the green pigment in plants and contains calcium d. Carboxypeptidase – A is an enzyme and contains zinc C Sol. Chlorophyll has Mg, not Ca. 9. The value of ‘spin only’ magnetic moment for one of the following configuration is 2.84 BM. The correct

one is:

a. d4 (in strong field) b. d4 (in weak ligand field) c. d3 ( in weak as well as in strong fields) d. d5 (in strong ligand field)

A Sol. 10. The spin magnetic moment of cobalt in the compound K2[Co(SCN)4] is

a. 3 bohr magneton b. 8 bohr magneton

c. 15 bohr magneton d. 24 bohr magneton

C Sol. Cobalt in [Co(SCN)4]

2– is present as Co2+. It involves the following hybridization scheme in the complex:

Three unpaired electrons produce magnetic moment of

BBB 15 3(5) )2n(n

11. Given

[Ag(NH3)2]+ Ag(NH3)

++ NH3; 401 104.1K

Ag(NH3)+ Ag++ NH3; 40

2 103.4K

The instability constant of the complex Ag(NH3)2 is equal to:

a. 7.14 × 103 b. 2.33 × 103 c. 6.02 × 10–8 d. 1.66 × 107 C Sol. The standard instability constant refers to the reaction [Ag(NH3)2]

+ Ag+ + 2NH3

2 unpaired electrons

3d 4s

4p

Co+2 =

sp3 hybridization



])NH(Ag[

]NH][Ag[

]NH(Ag[

]NH][)NH(Ag[

)NH(Ag[

]NH][Ag[K

3

3

23

33

23

230

ins

= )103.4)(104.1(KK 4402

01

= 6.02 × 10–8

12. Which of the following ligands is called -acceptors ? CO CN– NO+ (I) (II) (III)

a. I, II, III correct b. I, II only correct c. II, III only correct d. III only correct A

13. The number of isomers possible in the square planar complex [PtClBrlCN]– would be a. 2 b. 3 c. 4 d. 6 D

14. A complex MA3X3, where A and X are unidentate ligands may give

a. two geometrical isomers b. three geometrical isomers c. two geometrical isomers which can be resolved into a pair of enantiomers d. two geometrical isomers one of which can be resolved into a pair of enantiomers. A

Sol. MA3X3 can exist as two geometrical isomers 15. Which of the following facts about the complex [Cr(NH3)6]Cl3 is wrong?

a. The complex is paramagnetic b. The complex is an outer orbital complex c. The complex gives white precipitate with silver nitrate solution d. The complex involves d2sp3 hybridisation and is octahedral in shape. B

16. The magnetic moment (spin only) of [NiCl4]2– is:

a. 5.46 BM b. 2.82 BM c. 1.41 BM d. 1.82 BM B

17. Which can exist both as diastereoisomer and enantiomer?

a. [Pt(en)3]4+ b. [Pt(en)2ClBr]2+ c. [Ru(NH3)4Cl2]

0 d. [PtCl2Br2]0

B 18. Number of isomeric forms (constitutional and stereisomers) for [Rh(en)2(NO2)(SCN)]+ are:

a. Three b. Six c. Nine d. Twelve D

19. For transition metal octahedral complexes, the choice between high spin and low spin electronic configurations arises only for:

a. d1 to d3 complexes b. d4 to d7 complexes c. d7 to d9 complexes d. d1, d2 & d8 complexes B

20. The relative stability of the octahedral complexes of Fe(III) over Fe(II) with the bidentate ligands, (i) HO – CH2 – CH2 – OH, (ii) HO – CH2 – CH2 – NH2, (iii) H2N – CH2CH2–NH2, (iv) H2N–CH2 – CH2 – SH, follows the order

a. (i) > (ii) > (iii) > (iv) b. (ii) > (i) > (iv) > (iii) c. (iii) > (ii) > (iv) > (i) d. (iv) > (i) > (iii) > (ii) C



21. When a metal is in its low oxidation state, the metal-carbon bond in M-CO is stronger than the metal-

chloride bond in M – Cl, because,

a. chloride is a donor and carbon monoxide a acceptor

b. chloride is a acceptor and carbon monoxide a donor

c. chloride is a donor and the carbon monoxide is both a donor as well as a acceptor

d. chloride is both the donor as well as acceptor and the carbon monoxide is both and acceptor

C

22. Which of the following complex compounds will exhibit highest paramagnetic behaviour?

(At. No. Ti = 22, Cr = 24, Co = 27, Zn = 30)

a. [Zn(NH3)6]2+ b. [Ti(NH3)6]

3+ c. [Cr(NH3)6]3+ d. [Co(NH3)6]

3+

C

23. K3CoF6 is high spin complex. What is the hybrid state of Co atom in this complex?

a. sp3d b. sp3d2 c. d2sp3 d. dsp2

B

Sol. Hybrid state is sp3d2.

24. Match the List I and II and pick the correct matching from the codes given below

List I List II

(Complex) (Structure and magnetic moment)

(a) [Ag(CN)2]– 1. Square planar and 1.73 B.M.

(b) [Cu(CN)4]3– 2. Linear and zero

(c) [Fe(CN)6]3– 3. Octahedral and zero

(d) [Cu(NH3)4]2+ 4. tetrahedral and zero

(e) [Fe(CN)6]4– 5. octahedral and 1.73 B.M.

a. a – 2, b – 4, c – 5, d – 1, e – 3 b. a – 5, b – 4, c – 1, d – 3, e – 2

c. a – 1, b – 4, c – 4, d – 3 , e - 5 d. a – 1, b – 3, c – 4, d – 2 , e – 5

A

25. An organic compound ‘A’ with molecular formula C6H12O6 forms a yellow crystalline solid with

C6H5 – NH – NH2 and on reduction with sodium gives a mixture of sorbitol and mannitol. The

compound ‘A’ is

a. Glucose b. Fructose c. Mannose d. Sucrose

B

Sol. Since the compound ‘A’ forms osazone, so it may be glucose or fructose. Since the compound gives

sorbitol and mannitol on reduction, so it must be fructose not glucose.

26. Glucose is said to have –CHO group but it does not react with

a. Br2 water b. Fehling solution c. CH3OH d. NaHSO3

D

Sol. Glucose does not react with NaHSO3

27. In the given polypeptide:

Arg – Tyr – Met – Gly

N-terminus amino acid is

a. Tyrosine b. Methionine c. Arginine d. Glycine

C

Sol. By convention N-terminus is on the left side.



28. Match the Columns:

List – I (Complex ion) List – II (Magnetic moment , in B.M)

(I) [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br i. Coordination Isomerism

(II) [Co(NH3)5NO2]Cl2 and [Co(NH3)5ONO]Cl2 ii. Hydrate Isomerism

(III) [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl]Cl2.H2O iii. Ionisation Isomerism

(IV) [Co(NH3)6] [Cr(CN)6] and [Cr(NH3)6] [Co(CN)6] iv. Linkage Isomerism

I II III IV I II III IV a. (iii) (iv) (ii) (i) b. (iii) (iv) (i) (ii) c. (iii) (ii) (iv) (i) d. (iii) (i) (ii) (iv) A 29. In the extraction of silver, the reactions involved are

Roasted silver ore + CN– + H2O 2O X + OH–

X + Zn Y + Ag The species X and Y, respectively, are: a. [Ag(CN)2]

–, [Zn (CN)4]2– b. [Ag(CN)2]

– , [Zn(CN)6]2–

c. [Ag(CN)4]3–, [Zn(CN)4]

2– d. [Ag(CN)4]– , [Zn(CN)4]

2– A Sol. The reactions involved are

4Ag + 8CN– + 2H2O + O2 4 [Ag(CN)2]– + 4OH–

Ag2])CN(Zn[Zn])CN(Ag[2)Y(

242

30. The correct structure of ethylenediaminetetraacetic acid (EDTA) is

a. b. c. d.

C 31. Which of the following polymer has weakest intermolecular forces

a. Elastomer b. Fibre c. Thermoplastic d. Thermosetting A

32. HDP (High density polythene) is prepared by which of the following mechanism

a. Free radical polymerisation b. Cationic polymerisation c. Via Complex formation d. Condensation C

33. Which of the following polymer has PDI = 1

a. Nylon b. Terylene c. Teflon d. Rubber D

34. Which of the chemical Reagent is used to confirm straight chain of six carbon in mannose & fructose

a. NaBH4 b. Br2 water c. Red P + HI d. (CH3CO)2O C

HOOC – CH2

HOOC – CH2

N – CH = CH – N

CH2 – COOH

CH2 – COOH HOOC

HOOC

N – CH2 – CH2 – N

COOH

COOH

HOOC – CH2

HOOC – CH2

N – CH2 – CH2– N

CH2 – COOH

CH2 – COOH

HOOC – CH2

H

N – CH – CH – N

H

CH2 – COOH

CH2 COOH

CH2

HOOC



35. In an amino acid, the carboxyl group ionizes at pKa = 2.34 and ammonium ion at pKa = 9.60. The isoelectric point of the amino acid is at pH

a. 5.97 b. 2.34 c. 9.60 d. 6.97

A

Sol. Isoelectric point = 2

pkpk2a1a

= 5.97

36. An organic compound is found to contain C, H and O as elements. It is soluble in water, chars with

conc. sulphuric acid and also gives Blue colour with iodine. The organic compound should be

a. Glucose b. Fructose c. Starch d. Meso-tartaric acid

C

Sol. Charring with conc. H2SO4 and purple colour with Molish reagent shows it to be a carbohydrate.

As Fehling solution test is negative, it could not be glucose and fructose. Since it gives blue colour with

iodine, so it is starch which forms a blue starch –iodine complex.

37. Among the following complexes, which has a magnetic moment of 5.9 BM?

Ni(CO)4, [Fe(H2O)6]2+, [Co(NH3)6]

3+, [Mn(Br)4]2– [At. No. Mn = 25; Fe = 26, Co = 27; Ni = 28]

a. Ni(CO)4 b. [Fe(H2O)6]2+ c. [Co(NH3)6]

3+ d. [Mn(Br)4]2–

D

Sol. Mn2+ has d5 configuration and has 5 unpaired electrons so it has magnetic moment,

)2n(n

BM 5.92 M B )25(5

38. A 2-L solution contains 0.04 mol of each of [Co(NH3)5SO4]Br and [Co(NH3)5Br]SO4. To one litre of this

solution, excess of AgNO3 is added. To the remaining solution, excess of BaCl2 is added. The amounts

of precipitated salts respectively are:

a. 0.01 mol, 0.01 mol b. 0.01 mol, 0.02 mol c. 0.02 mol, 0.01 mol d. 0.02 mol, 0.02 mol

D

Sol. We will have

mol 02.0

34533mol 02.0

453 AgBrNO]SO)NH(Co[AgNOBr]SO)NH(Co[

mol 02.042532

mol 02.0453 BaSOCl]Br)NH(Co[BaClSO]Br)NH(Co[

39. Among the following metal carbonyls, the C – O bond order is lowest in:

a. [Mn(CO)6]+ b. [Fe(CO)5] c. [Cr(CO)6] d. [V(CO)6]

–

D

40. Given the following data about the absorption maxima of several complex ions, what is the order of 0 (splitting energy) for these ions:

Complex ion max.

(A) [CoCl6]3– 700

(B) [Co(CN)6]3- 400

(C) [Co(NH3)6]3+ 600

a. A > C > B b. B > A > C c. B > C > A d. A > B > C C



41. HIO4 splits glucose and fructose into formic acid and formaldehyde. Ratios of formic acid and formaldehyde from glucose and fructose are

a. 5/1 and 3/2 b. 5/1 and 4/2 c. 5/1 and 5/1 d. 4/2 and 4/2 A

Sol. Glucose 4HIO 55HCOOH + HCHO

Fructose 4HIO 54 HCOOH + 2 HCHO

1

5

glucose from deFormaldehy

glucose from acidFormic

2

3

fructose from deFormaldehy

fructose from acidFormic

42. Among I : Tollen’s reagent, II : Fehling solution, III : Br2 water, IV : Sanger’s reagent, which can be used to distinguish between an aldose and ketose?

a. I only b. III only c. II, III and IV d. I, II, III B

Sol. Since Br2 is a mild oxidizing agent, it oxidizes only aldehyde group (of aldose) and not ketone (of ketoses). Thus, when a small amount of Br2 water is added to a monosaccharide, the reddish brown colour of Br2 will disappear only if the monosaccharide is an aldose and there is no change if it is a ketose. Both aldoses and ketoses (glucose and fructose) will reduce Tollen’s reagent and Fehling solution.

Sanger’s reagent is used for N – terminal analysis of proteins.

SECTION – B (ASSERTION & REASON TYPE)

This Section contains 10 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONLY ONE is correct. (Negative Marking) 10 × 3 = 30 Marks

(a) If both A and R are correct and R is explanation of A (b) if both A and R are correct and R is not the explanation of A

(c) if A is correct R is wrong. (d) Both A & R is wrong

1. Assertion: A sulphate ion is a bidentate ligand and can also act as monodentate. Reason: Many a times mutidentate ligands do have flexidentate character

a. (a) b. (b) c. (c) d. (d) A

2. Assertion: Chelates are relatively more stable than non chelated complexes. Reason: Complexes containing ligands which can be easily replaced by other ligands are called labile complexes. a. (a) b. (b) c. (c) d. (d) B

3. Assertion: The number of unpaired electron in Ni(CO)4 is zero Reason: In this compound 4s-electrons of Ni atom enter the inner d-orbitals to facilitate the sp3-hybridization in Ni atom. a. (a) b. (b) c. (c) d. (d) A

4. Assertion: NF3 is weaker ligand than N(CH3)3 Reason: NF3 ionises to give F– ions in aqueous solution.

a. (a) b. (b) c. (c) d. (d) C



5. Assertion: The [Ni(en)3]Cl2(en = ethylenediamine) has lower stability than [Ni(NH3)6]Cl2 Reason: In [Ni(en)3]Cl2 the geometry of Ni is trigonal bipyramidal a. (a) b. (b) c. (c) d. (d) D

6. Assertion: Only cis [Pt(NH3)2Cl2] reacts with oxalic acid (H2C2O4) to form [PtCl2(ox)]2– not the trans isomer. Reason: The oxalate ion is bidentate ligand which occupies adjacent positions only. a. (a) b. (b) c. (c) d. (d) A

7. Assertion: Under the influence of a strong field ligand, d7-system will have only one unpaired electron either in coordination number six or four.

Reason: 1g

6g2et is the electronic configuration for both cases.

a. (a) b. (b) c. (c) d. (d) C

Sol. In strong field tetrahedral will not be possible d7 t2g6eg

1 ; t2g6eg

0 4p1

8. Assertion: (+) – Glucose reacts with Fehling’s solution and phenyl hydrazine, but not with NaHSO3. Reason: NaHSO3 cannot cleave the ring structure a. (a) b. (b) c. (c) d. (d) A

Sol. Both statement -1 and -2 are true and the statement -2 is true explanation for statement -1.

9. Assertion: Glucose reacts with CH3OH/HCl(g), to give - and -methyl glucosides Reason: C1 in ring structure of glucose is chiral. a. (a) b. (b) c. (c) d. (d) A

Sol. Both statement -1 and -2 are true and the statement-2 is true explanation for statement -1. 10. Assertion. Amino acids reacts with DNFB (Sanger’s Reagent) to give the product which is yellow in

colour. Reason. Generally aminoacids react with two molecules of nin-hydrin to give a pigment which is purple in colour.

a. (a) b. (b) c. (c) d. (d) B

SECTION – C (PASSAGE TYPE) Negative Marking

This Section contains 2 passage. Each of these questions has four choices A), B), C) and D) out of which ONLY ONE is correct. (Mark only one) (7 × 3 = 21 Marks)

Passage- 1 Ni+2 ions form the following complexes.

I ComplexNi KCN2

II ComplexNiexcess

KCl2

1. Which of the following is not true for complex formed by reaction of NiCl2 with DMG?

a. It is diamagnetic, square planar complex. b. It is a trans isomer c. Co. No. & oxidation number of Ni in this complex is 4 and +2. d. It has 4 hydrogen bonds & donor site of ligand is four nitrogen atoms D



2. Nickel ion involves

a. dsp2 hybridization in both complexes b. sp3 hybridization in both complexes c. dsp2 hybridization in cyano complex and sp3 hybridization in chloro complex d. sp3 hybridization in cyano complex and dsp2 hybridization in chloro complex C

3. Consider that coordination number of Ni2+ in 4.

NiCl2 + KCN(excess) A (cyano complex)

NiCl2 + conc. HCl (excess) B (chloro complex) The IUPAC name of A and B are:

a. potassium tetracyanonickelate (II), potassium tetrachloronickelate (II) b. tetracyanopotassiumnickelate (II), tetrachloro potassiumnickelate (II) c. tetracyanonickel (II), tetrachloronickel (II) d. potassiumtetracyanonickel (II) potassiumtetrachloronickel (II) A

Sol.

(II)] nickelatetetracyano potassium[A

422 KCl2])CN(Ni[KKCN4NiCl

In the second case, the complex formed is K2[NiCl4]. Its IUPAC name is potassium tetrachloronickelate (II).

4. Predict the magnetic nature of (cyanocomplex) (I) and (chlorocomplex) (II):

a. both are diamagnetic b. I is diamagnetic and II is paramagnetic with one unpaired electron

c. I is diamagnetic and II is paramagnetic with two unpaired electrons

d. both are paramagnetic

C

Passage -2

Disaccharide (A) H/O2H

(X) + Fructose

Disaccharide (B) H/O2H

2X

Disaccharide (C) Galactose)X(H

OH2

YX 3HNO

Fructose O2H/HgNa

Sorbitol + Mannitol

5. Milk sugar is

a. A b. B c. C d. X C

6. Y has

a. two –OH group b. three - OH groups c. two –COOH groups d. one-COOH group C

7. Sorbitol and mannitol are

a. Anomers b. C2 epimers c. C4-epimers d. Enantiomers B



Sol. 5 – 7

)Fructose(OHC)ecosGlu(OHCOHC 6126)Y(

6126H/O2H

(Sucrose) )A(112212

)ecosGlu(OHC2OHC)X(

6126H/O2H

)Maltose(B112212

)Galactose(

6126)ecosGlu(X

6126H/O2H

112212 OHCOHCOHC

.

(A) (Sucrose) is a non-reducing sugar due to the absence of a free –CHO group. X is glucose. C(Lactose) is known as milk sugar.

Glucose 3HNO

acid) Saccharic(Y4 COOH)CHOH(HOOC

Y has two –COOH groups. Sorbitol and Mannitol differ in configuration at C2 so they are C2 epimers

SECTION – D More than One Answer (No Negative Marking) This Section contains 6 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONE OR MORE may be correct. 6 × 4 = 24 Marks

1. Which of the following statement(s) is/are correct about stability of chelates?

a. as the number of rings in complex increases, stability of complex (chelate) also increases

b. a chelate having five membered ring is more stable if it contains double bonds

c. a chelate having six membered ring is more stable if it does not contain double bonds

d. Chelating ligands are atleast bidentate ligands

A, D

Sol. Only a and d are correct.

A chelate with five membered ring is more stable if they do not contain double bonds while the chelates

of six membered rings are more stable if they contain double bonds.

2. Which of the following statements is correct?

a. Cu2+ and Cd2+ ions can be separated by the formation of their cyano complexes b. For d8, d9 and d10 metal ions, there is only one way to fill the t2g and eg. levels so that there are no

low spin versus high spin octahedral complexes. c. The octahedral crystal field splitting is 4/9 of the tetrahedral splitting d. The complex [Pt(NH3) (Br) (NH2OH) (C5H5N)]+ has three geometrical isomers A, B, D 3. Which of the following statements is CORRECT?

a. The stability constant of [Co(NH3)6]3+ is greater than that of [Co(NH3)6]

2+ b. The cyano complexes are far more stable than those formed by halide ions c. The stability of halide complexes follows the order I < Br < Cl for 3d transition metals d. The stability constant of [Cu(NH3)4]

2+ is greater than that of [Cu(en)2]2+

A,B,C 4. The complex [Fe(H2O)5NO]2+ ion has a magnetic moment of 3.87 B.M. Which is true for this complex:

a. Fe in this complex exists in +1 oxidation state and nitrosyl as ON

(nitrosonium ion) b. The complex is octahedral in geometry as attained by sp3d2 hybridization c. There are three unpaired electrons in the central atom which is due to transfer of odd electron of

NO to Fe2+. d. The complex is an octahedral low spin complex with Fe in +2 oxidation state A,B,C



Sol. Magnetic moment = .electrons) (unpaired 3 n B.M 87.3)2n(n The complex may be regarded

as a high spin d7 complex of Fe+ and NO+.

5. Which of the following compounds is resolvable into d or forms?

a. [Zn(acac)2]

0 b. [Cu(gly)2] c. d. [Zn(gly)2] A,C,D or A,D

6. Which of the following is/are reducing sugar

a. galactose b. amylopectin c. Mannose d. Lactose A,C,D

SECTION – E (MATRIX MATCH TYPE) No Negative Marking

This Section contains 2 questions. Each question has four choices (A, B, C and D) given in Column I and five statements (p, q, r, and s) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. (2 × 8 = 16 Marks)

1. Match the complexes in Column I with their properties listed in Column II. Indicate your answer by

darkening the appropriate bubbles. [At. No. Pt = 78; Co = 27; Ni = 28] Column I Column II (A) [Co(NH3)4(H2O)2]Cl2 (P) geometrical isomers (B) [Pt(NH3)2Cl2] (Q) paramagnetic (C) [Co(H2O)5Cl]Cl (R) diamagnetic (D) [Ni(H2O)4Cl2]Cl2 (S) metal ion with +2 oxidation state A p, q, s; B p, s, r; C q, s; D q, p Sol. (a) [Co(NH3)4 (H2O)2] Cl2 can show geometrical isomers (choice p) It is also paramagnetic (choice q) because the metal ion is Co(II) with electronic configuration. In the complex, Co(II) contains one unpaired electron. (choice s). (b) [Pt(NH3)2Cl2] involves square planar geometry. It can exhibit geometrical isomers (choice p).

Pt in this complex has +2 oxidation state (choice s). Its electronic configuration in the complex is 5d8 with no unpaired electron and thus the complex is diamagnetic (choice r).

(c) [Co(H2O)5Cl] Cl has metal in +2 oxidation state (choice s) and thus has one unpaired electron which makes the complex paramagnetic (choice q)

(d) [Ni(H2O)4]Cl2]Cl2 has metal in +2 oxidation state (choice s). Its configuration in the complex is

Co

OH2

NH3

OH2

NH3 H3N

H3N

Co

OH2

NH3 H3N

H3N NH3

OH2

cis-isomer trans-isomer

3d 4s 4p 4d

From the ligands

Pt

NH3

Cl NH3

Cl

Pt

H3N

Cl NH3

Cl

cis-isomer trans-isomer

3d6 4s 4p 4d

From the ligands

Be O

O

C6H4

2 C

O



It has two unpaired electrons and is that paramagnetic (choice q). Hence, the correct-bubbled diagram is as follows. 2. Match the pair of complex compounds with the properties that are same in them. [At. No. Pt = 78]

Column – I Column – II (A) [Ti(H2O)6]

+3 & [Mn(CN)6]–4 (P) Hybridisation (In outer & inner complex it is taken same)

(B) [Ni(CN)4]–2 & [Ni(CO)4] (Q) Magnetic moment value

(C) [Cu(gly)2] & [Pt(gly)2] (R) Effective atomic number (D) K3[Fe(CN)6] & K4[Fe(CN)6] (S) Shape

Sol. A P,Q,S; B Q; C P, S; D P, S A: [Ti(H2O)6]

+3 ; d1; sp3d2; n = 1 EAN = 22 – 3 + 12 Octahedral [Mn(CN)6]

–4 ; d5; d2sp3; n = 1 EAN = 25 – 2 + 12 B: [Ni(CN)4]

–2 ; d8 ; dsp2 ; n = 0 ; EAN = 28 – 2 + 8 [Ni(CO)4] ; d

10 ; sp3 ; n = 0 ; EAN = 28 + 8 C: [Cu(gly)2] ; d

9 ; dsp2 ; n = 1; EAN = 29 – 2 + 8 [Pt(gly)2] ; d

8 ; dsp2; n = 0; EAN = 78 – 2 + 8

SECTION – VI (Integer Type) No negative Marking. This Section contains 6 questions. The answer to each question is a single digit integer ranging from 0

to 10. The correct digit below the question number in the ORS is to be bubbled. 7 × 3 = 21 Marks

1. The possible number of stereisomers for the formula [Ma3b2]

n± is _______, Sol. 6

2. The hapticity or (-nita) of the organic ligand in the following complex is . Sol. 6 3. The number of unpaired electrons in the t2g set of d orbitals in the complex [Co(H2O)3F3] is ______. Sol. 2; Co+3 = d6 ; t2g4eg2 ;

4. What is the number of H2O molecule present as a ligand in Blue vitriole. Sol. 4; [Cu(H2O)4]SO4.H2O 5. How many unpaired electrons will be present in [Ni(gly)2] complex. Sol. 2, Tetrahedral It is a tetrahedral complex with configuration d8. 6. How many of the following polymers are chain growth polymer

Teflon; PMA; Polystyrene; Neoprene; PVC; Buna-S; Nylon; Terylene, Bakelite; PHBV Sol. 6; [Condensation polymers are step growth & addition polymers are chain growth] 7. How many lone-pair are present in monomer Melamine of Melamine polymer. Sol. 6

Mo(CO)3

M

a

a

a

b

b

;

M

b

b

a

a

a

;

M

a

a

b

a

b

; M

b

b

a

a

a

M

a

b

a

a

b

;

; M

a

b

a

b

a

NH2

NH2 H2N N

N N

test date: 28.11.2015 (saturday) test time: 12:30 pm to 02 ...€¦ · 02.11.2015 · a. i, ii,...

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