Jp Xii Physical&inorganic Chemistry (10) - Prev Chaps + Co-ordination Compound.pdf

PHY./INO. CHEMISTRY

TARGET : JEE (Main + Advanced) 2016 Course : VIJETA (JP)

NO. 15 & 16

Date : 01-06-2015 TEST IN FORMATI ON

DATE : 07.06.2015

CUMULATIVE TEST (CT) - 1 (ADVANCED)

Syllabus : Solution & Colligative Properties, Solid State, Coordination compound (up to VBT), Thermodynamics, Ionic equilibrium and s-Block Element. (All ChemInfos and Handouts till date) This DPP is to be discussed in the week (01-06-2015 to 06-05-2015)

DPP No. # 15 (JEE-MAIN) Total Marks : 61

Max. Time : 38 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.15 ChemINFO : 4 Questions ('–1' negative marking) Q.17 to Q.20

1.

(3 marks, 2 min.) (4 marks, 2 min.)

[45, 30] [16, 08]

Enthalpy of solution of CsBr(s) is 10 kJ.mol1. If the enthalpies of hydration of Cs+(g) and Br(g) are 475 and 655 kJ.mol, what should be the lattice energy of CsBr(s) in kJ.mol1 ?

foy;u dh ,UFkSYih 10 kJ.mol1 ;fn Cs+(g) rFkk Br(g) dh ty;kstu ÅtkZ Øe'k% 475 rFkk 655 kJ.mol gS rc CsBr(s) dh tkyd ÅtkZ kJ.mol1 esa D;k gksxh \ CsBr(s) ds

(A*) 1120

2.

(B) 1130

(C) 1140

(D) 1150

In the reaction, xVO + yFe2O3  FeO + V2O5. What is the value of x and y respectively ? (A) 1, 1 (B*) 2, 3 (C) 3, 2 (D) None of these

vfHkfØ;k xVO + yFe2O3  FeO + V2O5 esa x o y ds eku Øe'k% fuEu gSA Sol.

3.

(A) 1, 1 (B*) 2, 3 xVO + yFe2O3  FeO + V2O5 3H2O + 2VO  V2O5 + 6H+ + 6e– v.f.=3 – 2e + 2H+ + Fe2O3  2FeO + H2O v.f.=3 So, x = 2 and y = 3.

(D) buesa

ls dksbZ ugh

How many grams of NaCl must be dissolved in 225 g of water to yield a solution having the same boiling point as that containing 20 g glucose (M – 180) and 30 g sucrose (M – 342) in 225 g of water? (Na – 23, Cl – 35.5) 225 g ty esa NaCl dh fdruh ek=kk foys; dh tkuh pkfg, fd izkIr foy;u dk DoFkukad] 225 g ty esa 20 g Xywdkst (M – 180) rFkk 30 g lwØkst (M – 342) j[kus (A*) 5.8 (C) 12.3

4.

(C) 3, 2

For an equilibrium reaction A(g) + B (g) C (g), H =+ ve an increase in temperature would cause (A*) an increase in the value of Keq (C) no change in the value of Keq.

okys foy;u ds leku gksrk gS (Na – 23, Cl – 35.5) (B) 11.6 (D) None of these buesa

ls dksbZ ugha

(B) a decrease in the value of Keq (D) a change in Keq which cannot be qualitatively predicted.

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PAGE NO.- 1

,d lkE; vfHkfØ;k A(g) + B (g) C (g), H =+ ve ds fy,] rkieku esa o`f) fuEu esa ls fdldk dkj.k gksxkA (A*) Keq ds eku esa o`f) dkA (B) Keq ds eku esa deh dkA (C) Keq ds eku esa dksbZ ifjorZu ugha gksxkA (D) Keq esa ifjorZu dk tks fd xq.kkRed :i ls ifjdfyr ugha fd;k tk ldrk gSaA Sol.

log

If H = + ve K2 > K1, 5.

1 1  –   T1 T2 

k2 ΔH = k1 2.303R

(T2 > T1) Hence (A) is correct.

If the shortest wavelength of H atom in Lyman series is x. Then longest wavelength in Balmer series of He+ is :

gkbMªkstu ijek.kq dh ykbeu Js.kh dh lcls NksVh rjaxnS/;Z 'x' gS] rc He+ ds ckej Js.kh dh lcls yEch rjaxnS/;Z gSA (A*) Sol.

9x 5

(B)

36 x 5

(C)

x 4

(D)

5x 9

Shortest wevelength of H atom in Lyman series correspond to transition n =  to n = 1. 

1 = (1)2 × R x

  1 – 1   ( )2   

1 = R. ........(1) x Longest wavelength of He+ ion in Blamer series correspond to transition n = 3 to n = 2. 

1 2 y = (2) × R ×

 1 1   –  ( 2 )2 (3 ) 2   

1 5 y = 4 × R × 36

........(2)

Dividing (1) by (2), we get y= 6.

Sol.

9x Ans. 5

The density of water at 4°C is 1.0 × 103 kg m–3. The volume occupied by one molecule of water is approximately 4°C ij ikuh dk ?kuRo 1.0 × 103 kg m–3 gSA ikuh ds ,d v.kq }kjk ?ksjk x;k yxHkx vk;ru gS& (A*) 3.0 × 10–23 mL (B) 6.0 × 10–22 mL (C) 3.0 × 10–21 mL (D) 9.0 × 10–23 mL 3 3 d = 1.0 × 10 kg/m = 1 g/mL 18

mass of one water molecule = N g (NA = Avogadro number) A 

d=

= 7.

m V

 18

6  10 23

V=

18 m = N  1 mL d A

mL = 3.0 × 10–23 mL

The compressibility factor of He is He dk lEihM~;rk xq.kkad gS (A) Linearly increasing with volume (C) Equal to 1

(B*) Linearly increasing with pressure (D) Linearly decreasing with pressure

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PAGE NO.- 2

(A) vk;ru

ds lkFk js[kh; :i ls c
For He  P (Vm–b) = RT or

PVm Pb =1+ =Z RT RT

 Z is linearly increasing with pressure 8.



Z,

nkc ds lkFk js[kh; :i ls c
Given the following reaction at equilibrium N2(g) + 3H2 (g) 2NH3 (g) Some inert gas is added at constant volume. Predict which of the following facts will be affected ? (A) More of NH3(g) is produced. (B) Less of NH3(g) is produced (C*) No affect on the degree of advancement of the reaction at equilibrium. (D) Kp of the reaction is increased.

fuEu vfHkfØ;k lkE; ij nh x;h gS % N2(g) + 3H2 (g)

2NH3 (g)

fu;r vk;ru ij blesa dqN vfØ; xSl feyk;h x;hA fuEu esa ls] mu rF;ksa dks crkb;s] tks dh çHkkfor gksax s % (A) NH3(g) vf/kd mRiUu gksrh gSA (B) NH3(g) de mRiUu gksrh gSA (C*) lkE; ij vfHkfØ;k o`f) dh ek=kk (degree of advancement of the reaction) çHkkfor ugha gksrh gSA (D) vfHkfØ;k ds Kp esa o`f) gksrh gSA Sol.

At constant volume addition of inert gas do not change concentration of any of the species involved in reaction So, equilibrium is unaffected.

9.

It is known that atoms contain protons, neutrons and electrons. If the mass of neutron is assumed tobe half of its original value whereas that of electron is assumed to be twice of this original value. The atomic mass of C612 will be : (A) Twice (B) 75% less (C*) 25% less (D) one-half of its

ijek.kq esa izksVkWu] U;wVªkWu o bysDVªkWu mifLFkr gksrs gS ;fn U;wVªkWu dk nzO;eku mlds okLrfod eku dk vk/kk rFkk bysDVªkWu dk nzO;eku mlds okLrfod eku dk nqxuk eku fy;k tk;s rks C612 dk ijek.kq Hkkj gksxkA (A) nqxuk (B) 75% de (C*) 25% de (D) mlds okLrfod eku dk vk/kk Sol.

1 u. 2 mass of electron is negligible. So doubling it will not affect the atomic mass. mass of neutron =

 New atomic mass = 6 × 1u + 6 ×

 % Decrease = 10.

3u × 100 = 25% 12u

1 u = 9u 2

 Decrease in atomic mass = 12u – 9u = 3u.

Ans.

To a solution of 20 ml of 0.1 M acetic acid, a solution of 0.1M NaOH is added from burette. If ‘r’ is the ratio of

[salt] , at what rate is the pH changing with respect to r when 5 ml of the alkali have been added? [acid] [

20 ml, 0.1 M ,flfVd

vEy ds foy;u esa] C;wjsV ls 0.1M NaOH dk foy;u feyk;k tkrk gSA ;fn ‘r’ [ gS] rks r ds lkis{k pH fdl nj ls ifjofrZr gksrh gS] tc 5 ml {kkj feyk;k tkrk gSA (A) 3

(B)

2.303 3

(C*)

3 2.303

] , dk vuqikr ]

1 (D) 3 x2.303

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PAGE NO.- 3

Sol.

When 5 ml of alkali has been added then

 salt  5 1  =   acid  15 3

Now pH = pKa + log r pH = pKa + 11.

n r 2.303

So

d (pH) 1 3 1 =0+ × = . dr 2.303 2.303 r

All the following complex ions are found to be paramagnetic : P : [FeF6]3– ; Q : [CoF6]3– ; R : [V(H2O)6]3+ The correct order of their paramagnetic moment (spin only) is :

;

S : [Ti(H2O)6]3+

;

S : [Ti(H2O)6]3+

fuEu lHkh ladqy vk;u vuqpqacdh; ik, x, gSa & P : [FeF6]3–

Q : [CoF6]3–

;

;

R : [V(H2O)6]3+

buds vuqpqacdh; vk?kw.kksZ ¼dsoy pØ.k½ dk lgh Øe gS & Sol.

(A*) P > Q > R > S (B) P < Q < R < S (C) P = Q = R = S (D) P > R > Q > S (A) On the basis of number of electrons the correct order is P > Q > R > S. bysDVªksuksa dh la[;k ds vk/kkj ij lgh Øe P > Q > R > S gSA Complex No. of unpaired electrons.

ladqy (P) (Q) (R) (S) 12.

v;qfXer bysDVªksuksa dh la[;k

[FeF6]3– [CoF6]3– [V(H2O)6]3+ [Ti(H2O)6]3+

5 4 2 1.

The crystal field-splitting for Cr3+ ion in octahedral field increases for ligands I–, H2O, NH3 , CN– and the order is : v"VQydh; {ks=k esa Cr3+ ds fy;s fØLVy&{ks=k foikVu] I–, H2O, NH3 , CN– fyx.Mksa ds fy;s c
Sol.

13.

(B) CN– < I– < H2O < NH3

(C) CN– < NH3 < H2O < I– (D) NH3 < H2O < I– < CN– Increase Order of ligands Strength I– < H2O < NH3 < CN– fyxs.M lkeF;Z dk c<+rk gqvk Øe I– < H2O < NH3 < CN– gSA In which of the following configurations will there be the possibility of both para and diamagnestism, depending on the nature of the ligands?

fuEu esa ls fdl foU;kl esa] fyxs.M dh izd`fÙk ij fuHkZj djrs gq,] vuqpqEcdRo o izfrpqEcdRo nksuksa dh lEHkkouk gksxh& Sol.

(A) d7 SFL WFL

14.

6

2, 2, 2 2g 2, 1, 1 2g

d =t 6

d =t

eg

0, 0

eg

1, 1

(B) d3 – diamagnetic

(C*) d6

(D) d5

¼izfrpqEcdh;½ – Paramagnetic.¼vuqpqEcdh;½

The complex for which the calculation of crystal field splitting can be most easily done, by knowing its absorption spectrum, will be

og ladqy] ftlds fy, vo'kks"k.k LisDVªe dks tkudj cgqr vklkuh ls fØLVy {ks=k foikVu dk ifjdyu fd;k tk ldrk gSa] gksxk& Sol.

(A) [TiCl6]2– (B) [Fe(H2O)6]2+ Since this is a d1 system.

(C*) [Ti(CN)6]3–

(D)[CoF6]3–

pw¡fd ;g d1 ra=k gSA

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PAGE NO.- 4

15.

The correct order for the CFSE (numerical value) for the following complexes is Complex P Q R S 3– 3– 3+ Formula [CoF6] [Co(CN)6] [Co(NH3)6] [Co(H2O)6]3+ fuEu ladqyksa ds fy, CFSE (vkafdd eku) dk lgh Øe gS &

ladqy lw=k Sol.

16.

P

Q 3–

R 3–

S 3+

[CoF6] [Co(CN)6] [Co(NH3)6] [Co(H2O)6]3+ (A) P > Q > R > S (B*) Q > R > S > P (C) S > R > P > Q (D) R > Q > P > S CFSE depends on the strength of ligands which follows order CFSE ds vuqlkj fyxs.M lkeF;Z dk Øe fuEUk izdkj ls fuHkZj djrk gSA CN– > NH3 > H2O > F– . On the basis of nature of ligands the correct order is Q > R > S > P. fyxs.M dh izd`fr ds vk/kkj ij lgh Øe Q > R > S > P gSA STATEMENT -1 : For cyclic process U = 0 STATEMENT -2 : Work & heat are significant only when the system undergoes a change. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B*) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True dFku -1 : pfØ; izØe (cyclic process) ds fy, U = 0

dFku -2 : tc ra=k esa ifjorZu (change of state) gksrk gS rc gh dk;Z rFkk Å"ek lkFkZdrk j[krs gSaA (A) dFku&1 lR; gS] dFku&2 lR; gS ; dFku&2, dFku&1 dk lgh Li"Vhdj.k gSA (B*) dFku&1 lR; gS] dFku&2 lR; gS ; dFku&2, dFku&1 dk lgh Li"Vhdj.k ugha gSA (C) dFku&1 lR; gS] dFku&2 vlR; gSA (D) dFku&1 vlR; gS] dFku&2 lR; gSA

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PAGE NO.- 5

ChemINFO-1.11

COORDINATION COMPOUNDS

Daily Self-Study Dosage for mastering Chemistry

Crystal Field Splitting Diagrams of d-orbitals for various Coordination Geometries

In absence of ligand all 5 d-orbitals of the central metal ion are degenerate i.e. they have the same energy. The electrons in the d-orbitals and those in the ligand repel each other due to repulsion between like charges. Thus, the electorns in d-orbitals which are closer to the ligands will have a higher energy than those further away which results in the d-orbitals splitting in energy. The magnitude of splitting (D) is affected by the following factors: 1. the nature of the metal ion. 2. the metal’s oxidation state. A higher oxidation state leads to a larger splitting. 3. the arrangement of the ligands around the metal ion. 4. the nature of the ligands surrounding the metal ion. Stronger ligands cause greater splitting in energy levels of d-orbitals. The following diagram shows the crystal field splitting of d-orbitals in tetrahedral, octahedral, tetragonal and square planar geometries for the same metal-ligand combination.

Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 17.

Which of the following relations is not correct? (A*) o < t

(B) o  (9/4)t

(C) sp > t

18.

Select the correct statement? (A) In tetrahedral field z2 lies higher in energy than xy (B) Splitting in square planar field is less than that in tetrahedral field (C) In square planar field five d-orbitals are split into 2 different enrgy levels. (D*) In octahedral field t2g orbital lie lower in energy than eg orbitals

19.

Which d-orbital has the highest energy in square planar field? (A*) x2 – y2 (B) z2 (C) xy

20*.

Magnitude of  depends on (A*) the metal’s oxidation state (C*) number of ligands surrounding the metal

(D) sp = 1.30

(D) yz

(B*) nature of the metal (D*) strength of ligand

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PAGE NO.- 6

ChemINFO-1.11

COORDINATION COMPOUNDS

Daily Self-Study Dosage for mastering Chemistry

fofHkUu milgla;kstu T;kferh;ksa ds fy, d-d{kdksa dk fØLVy {ks=k foikVu fp=k

fyxs.M dh vuqifLFkfr esa dsUnzh; /kkrq vk;u ds 5 d-d{kd leHkza'k gksrs gSa vFkkZr~ ;s leku ÅtkZ j[krs gSaA d-d{kdksa esa bysDVªkWu rFkk leku vkos'kksa ds e/; izfrd"kZ.k ds dkj.k os bysDVªkWu tks fyxs.M esa gSa] ,d nwljs dks izfrdf"kZr djrs gSaA vr% d-d{kdksa esa bysDVªkWu tks fyxs.Mksa ds lehi gSa] fyxs.Mksa ls nwj okys bysDVªku dh vis{kk mPp ÅtkZ j[ksaxsA ftldk ifj.kke d- d{kdksa esa ÅtkZ dk foikVu gksrk gSA foikVu () dk ifjek.k fuEu dkjdksa ls izHkkfor gksrk gS : 1. /kkrq vk;u dh izd`frA 2. /kkrqvksa dh vkWDlhdj.k voLFkk] mPp vkWDlhdj.k voLFkk eas vf/kd foikVu gksrk gSA 3. /kkrq vk;u ds pkjksa vksj fyxs.Mksa dh O;oLFkk 4. /kkrq vk;u dks ?ksjs gq, fyxs.Mksa dh izd`frA izcy fyxs.Mksa ds dkj.k d-d{kdksa ds ÅtkZ ry esa vf/kd foikVu gksrk gSA ;g fp=k leku /kkrq fyxs.M la;kstu ds fy, prq"Qydh;] v"VQydh;] prq"ddks.kh; o oxZ leryh; T;kferh;ksa esa d-d{kdksa ds fØLVy {ks=k foikVu dks n'kkZrk gSA

Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 17.

fuEu esa ls dkSulk lEcU/k lgh ugha gS? (A*) o < t

(B) o  (9/4)t

(C) sp > t

(D) sp = 1.30

18.

lgh dFku dk p;u dhft,A (A) prq"Qydh; {ks=k esa z2 ,xy dh vis{kk mPp ÅtkZ j[krs gSaA (B) oxZlery {ks=k esa foikVu] prq"Qydh; {ks=k esa foikVu ls de gksrk gSA (C) oxZ lery {ks=k esa ikap d-d{kdksa dk] nks fHkUu ÅtkZ ryksa esa foikVu gksrk gSA . (D*) v"V Qydh; {ks=k esa t2g d{kd] eg d{kdksa dh vis{kk fuEu ÅtkZ j[krs gSaA

19.

oxZlery {ks=k esa dkSuls d-d{kd vf/kdre ÅtkZ j[krs gS? (A*) x2 – y2

20*.

(B) z2

 dk ifjek.k fuHkZj djrk gSA (A*) /kkrqvksa dh vkWDlhdj.k voLFkk ij (C*) /kkrq dks ?ksjs gq, fyxs.Mksa dh la[;k

(C) xy

ij

(B*) /kkrq dh izd`fr ij (D*) fyxs.M ds lkeF;Z

(D) yz

ij

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PAGE NO.- 7

DPP No. # 16 (JEE-ADVANCED) Total Marks : 65

Max. Time : 39 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.11 Multiple choice objective ('–1' negative marking) Q.12 to Q.14 Integer type Questions ('–1' negative marking) Q.15 ChemINFO : 4 Questions ('–1' negative marking) Q.16 to Q.19

1.

Sol.

(3 (4 (4 (4

marks, 2 min.) marks, 2 min.) marks 3 min.) marks, 2 min.)

[33, [12, [04, [16,

22] 06] 03] 08]

A gaseous mixture contains CO2(g) and H2(g) in a 2.2 : 1 ratio by mass. The ratio of the number of molecules of CO2(g) and H2(g) is ,d CO2(g) vkSj H2(g) ds feJ.k ds nzO;eku dk vuqikr 2.2 : 1 gSA CO2(g) vkSj H2(g) ds v.kqvks dh la[;k dk vuqikr gSA (A) 5 : 2 (B*) 1 : 10 (C) 10 : 1 (D) 1 : 5 Let mass of CO2 = 2.2 x g and mass of H2 = x g 

number of moles of CO 2 = number of moles of H2 =



2.2 x 44

x 2

Ratio of number of molecules of CO 2 to number of molecules of H2 =

2.2 x x 2. 2 x 2 1 ÷ = × = 44 2 44 x 10

= 1 : 10 2.

Select correct statement(s):

lgh dFku pqfu;s& (A) we can condense vapour simply by applying pressure (B) to liquefy a gas one must lower the temperature below T C and also apply pressure (C) at Tc, there is no distinction between liquid and vapour state, hence density of the liquid is nearly equal to density of the vapour (D*) all the statements are correct statement (A) ge nkc dks iz;qDr dj ok"i dks la?kfur dj ldrs gSA (B)

xSl dks nzohd`r djus ds fy, rkieku dks TC ls de djuk pkfg, rFkk lkFk gh nkc Hkh iz;qDr djuk pkfg,A (C) Tc ij nzo rFkk ok"i voLFkk esa dksbZ foHksn ugha gksrk gS vr% nzo dk ?kuRo] ok"i ds ?kuRo ds yxHkx cjkcj jgrk gSA (D*) lHkh dFku lgh dFku gSA 3.

. Van der Waals equation is not applicable below critical temperature because it doesn’t accurately give the details of all the values of volume at all pressure. . Above critical temperature, all the three roots of volume can be real for Van der Waals equation. . At critical point slope of the isotherm is maximum. Which of the above statement (s) is true : (A)  (B*)  &  (C) , &  (D)  and  . Økafrd

rki ds uhps okUMj okWYl lehdj.k ykxw ugha gksrh D;ksafd ;g lHkh vk;ru ds ekuksa ds ckjs esa lHkh nkc ij lgh ugha crkrhA . Økafrd rki ds Åij vk;ru ds lHkh rhuks oxZ okUMj okWYl lehdj.k ds fy;s okLrfod gSA . Økafrd fcUnq ij lerkih; oØ dk
Below critical temp. gas exist as liquid form so, Vander waals Eq. is not applicable At critical point slope of iso therm is max. (zero) and below this critical point slope is negative.

4.

For the reaction N2O4 (g) equilibrium constant Kp is :

Kp / p (A)  = 4  K / p p

2NO2 (g), the relation connecting the degree of dissociation () of N2O4(g) with

Kp (B)  = 4  K p

 Kp / p   (C*)  =    4  Kp /P 

1/ 2

 Kp /p (D) =   4  Kp

   

1/ 2

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PAGE NO.- 8

vfHkfØ;k N2O4 (g) 2NO2 (g) ds fy,] N2O4(g) ds fo;kstu fd ek=kk () dks lkE; fu;rkad Kp ds lkFk lEcfU/kr djus okyk lEcU/k (relation) gS % Kp / p (A)  = 4  K / p p

4pα2 1 – α2

Kp (B)  = 4  K p

 a2 (k p + 4p) = k p

 Kp / p   (C*)  =    4  Kp /P 

1/ 2

 Kp /p (D) =   4  Kp

   

1/ 2

1/ 2  kp     p     =  kp    4  4   

Sol.

kp =

5.

Equimolar solution of AOH and BOH are prepared. The I.P. of A and B are 5.1 and 13.0eV respectively. The E.N. of A and B are 0.9 and 3.2 respectively. Reviewing the data pick out the wrong statement. (A) Solution of BOH will give effervescence with sodium carbonate. (B) Phenolphthalein will give pink colour with AOH solution. (C*) BOH solution will show a PH > 7. (D) Treatment of NH4Cl with AOH will lead to liberation of ammonia. AOH rFkk BOH dk leeksyj foy;u rS;kj fd;k x;kA A rFkk B dk vk;uu foHko Øe'k% 5.1 rFkk 13.0eV gSA A rFkk B

dh fo|qr _.kkRedrk 0.9 rFkk 3.2 gSA bu vadks ds vk/kkj ij vlR; dFku dk p;u djsaA (A) BOH dk ?kksy lksfM;e dkcksZusV ds lkFk cqncqnkgV iSnk djrk gSaA (B) QsukW¶FkSyhu AOH ds ?kksy ds lkFk xqykch jax çnku djrk gSaA (C*) BOH dk ?kksy PH > 7 nsrk gSA (D) NH4Cl dk AOH ds lkFk veksfu;k nsrk gSaA 6.(a)

Spin only magnetic moment of a complex having CFSE = – 0.6 0 and surrounded by weak field ligands can be: (A) 1.73 BM (B) 4.9 BM (C*) both (A) & (B) (D) None of these ,d ladqy] ftldk CFSE = –0.6 0 gS rFkk tks nqcZy {ks=k fyxsaM ls f?kjk gS] dk dsoy pØ.k pqEcdh; vk?kw.kZ gks ldrk

gSA Sol. (b)

Sol.

(A) 1.73 BM (B) 4.9 BM (C*) (A) rFkk (B) nksuks (D) buesa The options can give CFSE = – 0.6 0 with weak field ligands  d4 and d9. nqcZy {ks=k fyxs.M ds lkFk CFSE = – 0.6 0 d4 rFkk d9 esa nh tk ldrh gSA

ls dksbZ ugha

Which of the following statements is not correct? (a) [Ni(H2O)6]2+ and [Ni(NH3)6]2+ have same value of CFSE (b) [Ni(H2O)6]2+ and [Ni(NH3)6]2+ have same value of magnetic moment (A*) Only a (B) Only b (C) Both a and b

(D) None of these

fuEu esa ls dkSulk dFku lgh ugha gSA (a) [Ni(H2O)6]2+ rFkk [Ni(NH3)6]2+ ds CFSE ds eku leku gSA (b) [Ni(H2O)6]2+ rFkk [Ni(NH3)6]2+ ds pqEcdh; vk?kw.kZ ds eku leku gSA (A*) dsoy a (B) dsoy b (C) a o b nksukas

(D) buesa

ls dksbZ ugh

Ammonia is a stronger field ligand than water. So, CFSE of [Ni(NH3)6]2+ is greater than [Ni(H2O)6]2+

veksfu;k ty dh vis{kk izcy {ks=k fyxs.M gS blfy, [Ni(NH3)6]2+ dh CFSE] [Ni(H2O)6]2+ ls vf/kd gksrh gSA 7.

Nitroprusside ion is : A : [FeII(CN)5NO+]2– and not (A) estimating the concentration of iron (C*) measuring the magnetic moment

B : [FeII(CO)5NO]2+. A and B can be differentiated by. (B) measuring the concentration of CN – (D) thermally decomposing the compound.

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PAGE NO.- 9

ukbVªksizqlkbM vk;u A = [FeII(CN)5NO+]2– rFkk u fd B = [FeII(CO)5NO]2+ A rFkk B dks fuEu ds }kjk foHksfnr fd;k tk ldrk gS (A) vk;ju dh lkanzrk ds ifjdyu }kjk (B) CN– dh lkanzrk ds ekiu }kjk (C*) pqEcdh; vk?kw.kZ ds ekiu }kjk (D) ;kSfxd ds Å"eh; fo;kstu }kjk Sol.

In A, NO+ has no unpaired e–s. So, complex is diamagnetic but in B, NO is odd e– molecule, having one odd e–. So, B is paramagnetic. A, NO+ dksbZ v;qfXer bysDVªksu ugha j[krk gS] blfy, ladqy izfrpqEcdh; gSa] ijUrq B esa NO fo"ke bysDVªksu v.kq gSa] ,d

fo"ke bysDVªksu j[krk gSa] blfy, B vuqpqEcdh; gSA 8.

All the following complexes shows a decrease in their weights when placed in a magnetic balance. Then, which of the these has square planar geometry :

fuEufyf[kr lHkh ladqy] pqEcdh; rqyk esa j[kus ij Hkkj esa deh n'kkZrs gSA rc buesa ls dkSulk oxkZdkj leryh; T;kferh j[krk gSA (A) Ni(CO)4 (B*) K[AgF4] (C) Na2[Zn(CN)4] (D) None of these buesa ls dksbZ ugha Sol.

Diamagnetic complexes shows decrease in weight when placed in magnetic balance. Ni(CO)4  Tetrahedral & diamagnetic K [AgF4]  Square planar & diamagnetic Na2[Zn(CN)4]  Tetrahedral & diamagnetic

gy %

izfrpqEcdh; ladqy pqEcdh; rqyk esa j[kus ij vius Hkkj esa dgh n'kkZrs gSA Ni(CO)4  prq"Qydh; rFkk izfrpqEcdh; K [AgF4]  oxkZdkj lery rFkk izfrpqEcdh; Na2[Zn(CN)4]  prq"Qydh; rFkk izfrpqEcdh;

9.

It is given that a complex formed by one Ni2+ ion and some Cl– ions and some PPh3 molecules does not show geometrical isomerism and its solution does not show electrical conductance. Then, which is correct about the complex : (A) It is square planar (B*) It is tetrahedral (C) It is diamagnetic (D) none of the above is correct

fn;k x;k gS fd ,d Ni2+ vk;u rFkk dqN Cl– vk;u rFkk dqN PPh3 v.kq ls cuk ,d ladqy T;kferh; leko;ork iznf'kZr ugha djrk gS rFkk bldk foy;u fo|qr pkydrk ugha n'kkZrk gSA rc ladqy ds ckjs esa dkSulk dFku lgh gSA (A) ;g lery oxkZdkj gSA (B*) ;g prq"Qydh; gSA (C) ;g izfrpqEcdh; gSA (D) mijksDr esa ls dksbZ lgh ugha gSA Sol.

It is not showing geometrical isomerism

;g T;kfefr leko;ork iznf'kZr ugha djrk gSA 10.

 Tetrahedral & paramagnetic. blfy, prq"Qydh; rFkk vuqpqEcdh;

gSA

The green coloured complex K2[Cr(CN)4(NH3) (NO)] is paramagnetic and its paramagnetic moment (spin only) is 1.73 B.M. Which of the following is correct about it : (i) Its IUPAC name is Potassium amminetetracyanonitrosylchromate () (ii) Its IUPAC name is Potassium amminetetracyanonitrosoniumchromate () (iii) Hybridisation state of chromium is sp3d2 (iv) It cannot show geometrical isomerism (v) Hybridisation state of chromium is d2sp3 (vi) It can show linkage isomerism gjs jax dk ladqy K2[Cr(CN)4(NH3) (NO)] vuqpqacdh; gS rFkk bldk vuqpqacdh; vk?kw.kZ ¼dsoy pØ.k½ 1.73 B.M. gSA fuEu

esa ls blds lanHkZ esa D;k lgh gSA (i) bldk IUPAC uke iksVSf'k;e ,EehuVsVªklk;uksukbVªksflyØksesV () gSA (ii) bldk IUPAC uke iksVSf'k;e ,EehuVsVªklk;uksukbVªkslksfu;eØksesV () gSA (iii) Øksfe;e dh ladj.k voLFkk sp3d2 gSA (iv) ;g T;kferh; leko;ork iznf'kZr ugha dj ldrk gSA (v) Øksfe;e dh ladj.k voLFkk d2sp3 gSA (vi) ;g ca/kuh leko;ork iznf'kZr dj ldrk gSA (A) (ii), (iii),(iv)

(B) (i),(iii), (vi)

(C) (i), (v)

(D*) (ii), (v), (vi)

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PAGE NO.- 10

Sol.

The complex is actually

okLrfod ladqy gSA    ( ) k 2 Cr (CN) 4 (NH3 ) (N O ) in which Cr() is d2sp3 hybridised with one unpaired electron.      () k 2 Cr (CN)4 (NH3 ) (N O)  

11.

esa Cr() ,d v;qfXer bysDVªksu ds lkFk d2sp3 ladfjr gSaA

When 200g of a 20% solution were cooled part of the solute precipitated, the concentration of solution become 12%. The mass of precipitaed substance is

tc 20% foy;u ds 200g foy;u dks B.Mk fd;k x;k] rc foys; dk dqN Hkkx vo{ksfir gks tkrk gS] rFkk foy;u dh lkUnzrk 12% gks tkrh gSA vo{ksfir inkFkZ dk Hkkj fuEu gS& Sol.

(A) 16.16 (B) 17.17 (C*) 18.18 20% solution means, 100 g of solution contains 20 of solute  Amount of solute present in 200 g of solution = 20 × 2 = 40 g Mass of solvent = 200 – 40 = 160 g Solution produced = 12% i.e. 12 g of solute in 88 g of the solvent Mass of solute after cooling =

(D) 19.19

12 × 100 = 21.82 g 88

Mass of solute ppt = 40 – 21.82 = 18.18 g Sol.

20% foy;u

dk vFkZ gS] 100 g foy;u] 20 foys; dks j[krk gSA  200 g foy;u esa mifLFkr foys; dh ek=kk = 20 × 2 = 40 g foyk;d dk nzO;eku = 200 – 40 = 160 g iz'ukuqlkj cuk;k x;k foy;u = 12% vFkkZr~ 12 g foys;] 88 g foyk;d esa mifLFkr gSA 12

B.Mk djus ds i'pkr~ foys; dk nzO;eku = 88 × 100 = 21.82 g vo{ksfir foys; dk nzO;eku = 40 – 21.82 = 18.18 g 12.*

Which are correct statements ? (A*) [Ag(NH3)2]+ is linear with sp hybridised Ag+ ion. (B*) NiCl42– , VO43– and MnO4– have tetrahedral geometry (C*) [Cu(NH3)4]2+ , [Pt(NH3)4]2+ and [Ni(CN)4]2– have dsp2 hybridisation of the metal ion (D*) Fe(CO)5 have bipyramidal structure with dsp3 hybridised iron. dkSuls lgh dFku gS ? (A*) [Ag(NH3)2]+ , sp ladfjr Ag+ vk;u

js[kh; vkd`fr dk gS (B*) NiCl , VO4 rFkk MnO leprq"Qydh; T;kfefr j[krs gS (C*) [Cu(NH3)4]2+ , [Pt(NH3)4] rFkk [Ni(CN)4]2– , /kkrq vk;u dk ladj.k dsp2 j[krs gS (D*) Fe(CO)5 , dsp3 ladfjr vk;ju ds lkFk f}fijkfeMy lajpuk j[krk gS (A) H3N  Ag+ NH3 (linear & sp hybridisation) ¼jsf[k; rFkk sp ladj.k½ 2– 4

Sol.

3–

– 4 2+

(B)

prq"Qydh; (C) [Cu(NH3)4]2+ 

prq"Qydh;

prq"Qydh;

Cu2+  3d9 , 4s0

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PAGE NO.- 11

[Ni(CN)4]2–

Ni2+  3d8 , 4s0

[Pt(NH3)4]2+ 4d & 5d metals always forms square planar with 4 ligands. 4d rFkk 5d /kkrq esa 4 fyxs.M ds lkFk oxkZdkj lery lajpuk cukrh (D) Fe(CO)5 ; Fe  3d6, 4s2

gSaA

Pairing

  by SFL

13._*

Which of the following statements is/are correct : (A*) Among NaF & NaBF4, NaF is relatively less soluble in water. (B*) Larger negative value of lattice enthalpy is a favourable condition for formation of ionic bond. (C*) In general, ionic compounds do not conduct electricity in solid state. (D) The water solubility of alkalline earth metal carbonates increases on moving down the group. fuEu esa ls dkSulk@dkSuls dFku lgh gS : (A*) NaF ,oa NaBF4 esa

ls] NaF ty esa vkisf{kd :i ls de ?kqyu'khy gksrk gSA ,UFkSYih dk cM+k _.kkRed eku vk;fud ca/k fuekZ.k ds fy, vuqdwy ifjfLFkfr gSA (C*) lkekU;r;k] vk;fud ;kSfxd Bksl voLFkk esa oS|qr /kkjk dk pkyu ugha djrs gSaA (D) {kkjh; e`nk /kkrq ds dkcksZusVksa dh tyh; foys;rk] oxZ esa uhps dh vksj tkus ij c<+rh gSA (B*) tkyd

Sol.

g y-

(A) Larger the size difference between ions, greater will be the solubility. (B) Larger negative value of lattice enthalpy is a favourable condition for formation of ionic bond as larger release of energy causes greater stability of lattice. (C) In general, ionic compounds conduct electricity in molten state and aqueous medium. (A) vk;uksa ds e/; vkdkj esa ftruk T;knk vUrj gksxk] foys;rk mruh gh T;knk gksxhA (B) tkyd

,UFkSYih dk cM+k _.kkRed eku vk;fud ca/k fuekZ.k ds fy, vuqdqy ifjfLFkfr gksrh gS] D;ksafd cgqr vf/kd ÅtkZ eqDr gksus ds dkj.k tkyd (lattice) vis{kkd`r vf/kd LFkk;h gksrh gSA (C) lkekU;r;k] vk;fud ;kSfxd xfyr voLFkk ,oa tyh; ek/;e esa oS|qr /kkjk dk pkyu djrs gSaA

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PAGE NO.- 12

14.*

Which of the following do not exist ?

fuEu esa ls fdldk vfLrRo ugha gS \ Sol.

(A*) SH6 (B*) HFO4 (C*) FeI3 (D) HClO3 (A) With hydrogen sulphur does not undergo sp3d2 hybridisation because of larger difference in energies between s, p and d-orbitals. Sulphur show +6 oxidation state with highly electronegative elements like O and F. (B) As fluorine is smaller and more electronegative than oxygen. (C) I– being stronger reducing agent reduces Fe3+ to Fe2+.

Sol.

lYQj] gkbMªkstu ds lkFk sp3d2 ladj.k ugha cukrk D;ksa fd buds e/; s, p rFkk d-d{kdksa esa ÅtkZ dk vUrj vf/kd gksrk gSA (A) lYQj vf/kd fo|qr _.kkRed rRo tSls O ,oa F ds lkFk +6 vkWDlhdj.k voLFkk fn[kkrk gSA (B) ¶yksjhu] vkWDlhtu dh rqyuk esa vf/kd fo|qr_.kkRed gksrk gS rFkk ¶yksjhu dk vkdkj NksVk gksrk gSA (C) I– izcy vipk;d gksrk gS tks fd Fe3+ dk vip;u Fe2+ esa dj nsrk gSA

15.

What is the degree of dissociation of trichloroacetic acid, if its 1.0 m solution has a freezing point of K – 2.53 °C? Kf = 1.86 molal VªkbDyksjks ,flVhd ,flM dh fo;kstu dh ek=kk D;k gS] ;fn bldk 1.0 m foy;u fgekad fcUnq – 2.53 °C j[krk gS Kf = K 1.86 . molal

Ans.

= 0.36

Sol.

Tf = i Kf m 2.53 = i × 1.86 × 1 i = 1.36 = 1 +  

= 0.36.

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PAGE NO.- 13

ChemINFO-1.12 Daily Self-Study Dosage for mastering Chemistry

COORDINATION COMPOUNDS Coordination Compounds in Extraction of Gold and Silver

Cyanide process or Mac-Arthur-Forrest process: Some important extraction processes of metals, like those of silver and gold, make use of complex formation. Au and Ag combine with cyanide in the presence of oxygen and water to form a water soluble complex ion [Au(CN)2]– and [Ag(CN)2]–, respectively. Au and Ag can be separated in pure metallic form from this solution on reduction with zinc. Step 1: The impure ore is crushed into fine pieces. Step 2: The finely grind ore is mixed with a dilute solution of NaCN or KCN in the presence of air (for O 2). Gold or Silver particles from the ore are oxidised (Au  Au+ or Ag  Ag+) into water soluble complex. 4 M(s) + 8 CN–(aq)+ 2 H2O(aq) + O2(g)  4 [M(CN)2]–

(aq)

+ 4 OH–(aq)

(M= Ag or Au)

Step 3: The remaining solids in the ore are then separated from the solution by filtration. Step 4: Zinc dust is then added to the filtred cyanide solution. The zinc displaces (reduces) the metal ion from the complex (Au+  Au or Ag+  Ag), and gold or silver metal is precipitated out of the solution. 2 [M(CN)2]–(aq) + Zn(s)  2 M(s) + [Zn(CN)4]2–(aq)

(M= Ag or Au)

Reactions for Gold: 4 Au(s) + 8 NaCN + O2 + 2 H2O  4 Na[Au(CN)2](aq) + 4 NaOH 2 Na[Au(CN)2](aq) + Zn(s)  2 Au(s) + Na2[Zn(CN)4](aq) Reactions for Silver: 4 Ag(s) + 8 NaCN + O2 + 2 H2O  4 Na[Ag(CN)2](aq) + 4 NaOH 2 Na[Ag(CN)2](aq) + Zn(s)  2 Ag(s) + Na2[Zn(CN)4](aq) Note: Cyanide process is an economical method for extracting gold from low-grade ore. But, due to the highly poisonous nature of cyanide, which causes environmental hazard and pollution, this process is banned in a number of countries. Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 16.

What is the IUPAC name of the complex formed in the cyanidation of gold? (A*) Sodium dicyanidoaurate(I) (B) Sodium dicyanidoaurate(III) (C) Sodium dicyanidogold(I) (D) Sodium dicyanidogold(III)

17.

Which of the following is true for the complex of silver formed in its extraction using cyanide process? (A) Effective Atomic Number of Ag is 86 (B*) The geometry of the complex is linear (C) Coordination number of Ag is 4 (D) It is an organomettalic compound

18*.

Select incorrect options for cyanide process. (A) It is also called as Mac-Arthur-Forrest process (B) It is used for extraction of silver and gold (C) It is banned in some countries (D*) It is an expensive method for extraction of Ag and Au from thier low-grade ores

19.

Select correct options for cyanide process (A) it involves formation of water-insoluble complex of Ag and Au (B*) air is used to oxidise the metal in NaCN solution (C) Zn is used in one of the steps as oxidising agent (D) Ag and Au are oxidised to thier +3 oxidation state

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ChemINFO-1.12 Daily Self-Study Dosage for mastering Chemistry

COORDINATION COMPOUNDS xksYM rFkk flYoj ds fu"d"kZ.k esa milgla;kstu ;kSfxd

lk;ukbM izØe ;k ekd&vksZFkj&QkjsLV izØe : /kkrqvksa ds dqN izeq[k fu"d"kZ.k çdeksa esa ladqy fuekZ.k dk mi;ksx gksrk gS] tSls xksYM rFkk flYoj ds fu"d"kZ.k esa A vkWDlhtu rFkk ty dh mifLFkfr eas Au rFkk Ag lk;ukbM vk;u ls la;ksftr gksdj ty foys; ladqy vk;u Øe'k% [Au(CN)2]– rFkk [Ag(CN)2], cukrs gSaA bl foy;u esa Au rFkk Ag dks 'kq) /kkfRod :i eas i`Fkd djus ds fy, ftad ds lkFk vipkf;r djrs gSaA in 1: v'kq) v;Ld dks lw{e VqdM+ksa esa rksM ysrs gSaA in 2: vUr eas lw{e vk;Ld dks ok;q dh mifLFkfr eas NaCN ;k KCN ds ruq foy;u esa fefJr djrs gSaA v;Ld ls xksYM rFkk flYoj d.k ty foys; ladqy eas vkWDlhd`r (Au  Au+ or Ag  Ag+) gks tkrs gSaA 4 M(s) + 8 CN–(aq)+ 2 H2O(aq) + O2(g)  4 [M(CN)2]–

(aq)

+ 4 OH–(aq)

(M= Ag or Au)

in 3: v;Ld esa cps Bksl dks fuL;anu }kjk foy;u ls i`Fkd dj fy;k tkrk gSA in 4: fQj fuL;afnr lk;ukbM foy;u esa ftad [email protected] dks feykrs gSa ftad] ladqy ls /kkrq vk;u dks foLFkkfir (Au+  Au or Ag+  Ag), dj

nsrk gS rFkk xksYM ;k flYoj /kkrq] foy;u ls vo{ksfir gks tkrh gSA

2 [M(CN)2]–(aq) + Zn(s)  2 M(s) + [Zn(CN)4]2–(aq)

(M= Ag or Au)

xksYM ds fy;s vfHkfØ;k: 4 Au(s) + 8 NaCN + O2 + 2 H2O  4 Na[Au(CN)2](aq) + 4 NaOH 2 Na[Au(CN)2](aq) + Zn(s)  2 Au(s) + Na2[Zn(CN)4](aq)

flYoj ds fy;s vfHkfØ;k: 4 Ag(s) + 8 NaCN + O2 + 2 H2O  4 Na[Ag(CN)2](aq) + 4 NaOH 2 Na[Ag(CN)2](aq) + Zn(s)  2 Ag(s) + Na2[Zn(CN)4](aq)

uksV : lk;ukbM izØe] fuEu dksfV v;Ld ls xksYM fu"d"kZ.k ds fy, de [kphZyk izØe gS] ijUrq lk;ukbM dh vR;Ur tgjhyh izd`fr ftlls okrkoj.k nwf"kr rFkk [krjs esa gks ldrk gS] ds dkj.k bl izØe dks dbZ ns'kksa esa izfrcfU/kr dj fn;k x;k gSA Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. 16.

xksYM ds lk;ukbMhdj.k ls cuus okys ladqy dk IUPAC uke D;k gS? (A*) lksfM;e MkbZlk;ukbMksvkWjsV (I) (B) lksfM;e MkbZlk;ukbMksvkWjsV(III) (C) lksfM;e MkbZlk;ukbMksxksYM(I) (D) lksfM;e MkbZlk;ukbMksxksYM(III)

17.

tc lk;ukbM izØe }kjk flYoj dk fu"d"kZ.k gksrk gS rc blls cus ladqy ds fy, fuEu esa ls dkSulk dFku lgh gS \ (A) Ag dk izHkkoh ijek.kq Øekad 86 gSA (B*) ladqy dh vkd`fr js[kh; gSA (C) Ag dh leUo; la[;k 4 gSA (D) ;g ,d dkcZ/kkfRod ;kSfxd gSA

18*.

lk;ukbM izØe ds fy, xyr dFku dk p;u dhft,A (A) ;g ekd&vksZFkj&QkjsLV izØe Hkh dgykrk gSA (B) ;g xksYM rFkk flYoj ds fu"d"kZ.k ds fy, iz;qDr gksrk gSA (C) dqN ns'kksa esa ;g izØe izfrcaf/kr gSA (D*) Au rFkk Ag dks muds fuEu dksfV v;Ldksa ls fu"df"kZr djus ds fy, ;g egaxk izØe gSA

19.

lk;ukbM izØe ds fy, lgh fodYi dk pquko dhft,A (A) ;g Ag rFkk Au ds ty&vfoys; ladqy ds fuekZ.k esa Hkkxhnkjh j[krk gSA (B*) NaCN foy;u esa /kkrq dks vkWDlhd`r djus ds fy, ok;q dk mi;ksx gksrk gSA (C) inksa esa ls ,d in esa Zn dk mi;ksx vkWDlhdj.k vfHkdeZd ds :i esa fd;k tkrk gSA (D) Ag rFkk Au viuh +3 vkWDlhdj.k voLFkk esa vkWDlhd`r gks tkrs gSaA Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PTC024029

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