inorganic solution for 3 year m.sc. entrance exam held on ... · m.sc. entrance test series 2017...

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M.Sc. Entrance Test Series 2017 - 2018

Inorganic Solution for 3rd Year M.Sc. Entrance

Exam held on 21st January, 2018

MULTIPLE TYPE QUESTION (MCQ) Q. 1 – Q.10 carry one mark each

Q1.

Ans: (b)

(2)

Ans: (c)

(3)

Ans: (d)

Q4.

Ans. (d) pH (HCOOH) = pH (CH3COOH)

[H3O+] = [H3O

+]2

[H3O+] = 2211 CKCK

K1C1 = K2C2

1

2

2

1

C

C

K

K

Given, 1

4

K

K

2

1

4

1

C

C

2

1

Q11.

Solution: (a) �̅�𝐻2+ =

1

𝜆𝐻2+ = R(

1

𝑛12 −

1

𝑛22) ........ (i)

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�̅�𝐻𝑒+ = 1

𝜆𝐻𝑒+ = RZ2 (

1

22−

1

42)

= R x 4 (1

4−

1

16)

= R x (4

4−

4

16) = R x (1 −

1

4) .......... (ii)

Comparing equation (i) and (ii)

∴ 1

𝑛12 = 1, n1 = 1

1

𝑛22 =

1

4, n2 = 2

Q12.

Solution: (c)

E ∝ 𝑍2

𝑛2

rn = 𝑟1 𝑓𝑜𝑟 𝐻 𝑥 𝑛

2

𝑍

rn ∝ n2

rn ∝ (2)2 a0 (rn = a0)

rn ∝ 4a0

mvr ∝ 𝑛ℎ

2𝜋 =

2

2

ℎ

𝜋

∴ 𝜈 = ℎ

𝜋𝑚𝑟 (r = a0)

= ℎ

𝜋𝑚𝑎0 =

ℎ

𝜋𝑚 𝑥 4𝑎0

So, KE = 1

2𝑚𝑣2

= 1

2𝑚(

ℎ

4𝜋 𝑥 𝑚𝑎0)2 =

ℎ2

32 𝜋2𝑚𝑎02

Q13.

Ans. (a)

[Mn(H2O)6]2+ = d5 (H.S)

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As all the levels are electrically non-degenerate. Hence, No Jahn-Teller distribution. Therefore all the

Mn-O bond length will be equal.

Q.14

2P3/2

ANSWER: The electronic configuration of the given element is: 1s22s22p5 As 2s orbital is completely filled, it will not contribute towards ground term. Hence

ground term will be determined entirely from 2P5 configuration. The electron occupancy of 2P orbital is as follows: +1 0 -1

Hence, spin S = 3(+1/2) + 2(-1/2) = 3/2 – 2/2 = ½ So, the spin multiplicity, 2S+1 = 2. (1/2) + 1 = 2

Now, L = 2(+1) +2(0)+1(-1) = 1 ⇒ P Lastly, J = |L + S| to |L - S| = |1 + 1/2| to |1 – 1/2| = 3/2, ½ According to Hund for the more than half-filled electronic configuration ground-state will

be determined by highest J value. Hence, the ground state term symbol of the element having configuration 1s22s22p5: (2S+1)LJ = 2P3/2

Hence, the correct answer is: b Q15.

d According to Soddy-Fajan group displacement law:

1. When an 𝛼-particle is emitted by an element, it mass and atomic no. will be decreased by four and two units respectively. 2. When an β-particle is emitted by an element, it mass no. will remain same but atomic

no. will be increased by one unit. When Ac will emit one beta particle, it will be converted into an element which has mass no. 90 i.e. Th. When Th will emit one alpha particle, it will convert to an element which

has mass no. 88 i.e. Ra.

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89𝐴𝑐227 −𝛽 → 90𝑇ℎ227

−𝛼 → 88𝑅𝑎223

−𝛼 → Rn

[A] [B] Hence, A and B are: Th and Ra respectively.

Q16.

(a)

(a) Cu, Ag, Au are called coinage metals because in the early times, they were used in making coins.

Q17.

Ans. 2.5 x 10-5

b) pH = 2.85

[H3O+] = 10-2.85 = 10-3 100.15 = 1.41 x 10-3

Also in a weak acid, by Ostwald’s dilution law,

[H3O+] = CKa

1.41 x 10-3 = 0.08Ka

Ka = 2.48 x 10-5 2.5 x 10-5

MULTIPLE SELECT QUESTION (MSQs)

Q1.

Solution: (b, c) Heisenberg principle is only for microscopic particles which are moving with very high speed. Q2. Solution: (a, b, c)

(a) Moles of O2 = 1.6

32= 0.05

(b) Molecules of O2 = 0.05 x 6.023 x 1023 = 3.011 x 1022 (c) Volume of O2 at STP = 0.05 x 22.4 = 1.12 L Q3.

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(b) (c) Ans. (b) Mn2+ and Fe3+ ; (c) Ti2+ and Ni2+ Conceptual approach: Mn2+ and Fe3+ have 5 unpaired electrons each; Ti2+ and Ni2+ have 2 unpaired electrons each.

Q4.

(a,b,c) Eu2+, Gd3+ and Tb4+ have 7 unpaired electrons. Lu3+ have no unpaired electron.

NUMERICAL ANSWER TYPE (NAT) Q. 1 – Q. 10 carry ONE mark each

Q1.

1.6

The value of the Ka from the appendix: Ka = 6.3 x 10-8 (We are using Ka2 since we dealing with the

equilibrium in which the second hydrogen ion is being lost).

Determine the pKa using pKa = -log 6.3 x 10-8 = 7.200659451 (unrounded)

Use the Henderson-Hasselbalch equation:

pH = pKa + log

]PO[H

][HPO

42

24

7.40 = 7.200659451 + log

]PO[H

][HPO

42

24

0.19934055 = log

]PO[H

][HPO

42

24

]PO[H

][HPO

42

24

= 1.582486 = 1.6

Q2.

2.8 x 10-11 The equation and ion-product expression for silver dichromate, Ag2Cr2O7, is:

Ag2Cr2O7 (s) ⇌ 2 Ag+ (aq) + 2

72OCr (aq) Ksp = [Ag+]2[2

72OCr ]

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The solubility of Ag2Cr2O7, converted from g / 100 mL to M is:

Molar solubility = S =

722

7223

7223

OCrAgg431.8

OCrAgmol1

10

1

mL100

OCrAgg108.3

L

mL

= 0.0001922186 1 M (unrounded)

Since 1 mole of Ag2Cr2O7 dissociates to form 2 moles of Ag+, the concentration of Ag+ is

2S = 2(0.00019221861 Al) = 0.00038443723 M(unrounded). The concentration of 2

72OCr is

S = 0.0001922 1861 M because I mole of Ag2Cr2O7 dissociates to form 1 mole of 2

72OCr .

Ksp = [Ag+]2[2

72OCr ] = (2 S)2(S) = (0.00038443723)2(0.00019221861) = 2.8408 x 10-11

= 2.8 x 10-11.

Q3.

2.4 L The mass of hydrochloric acid is obtained from the density and the volume.

Mass = density x volume = 1.096 g/mL x 50.0 mL = 54.80 g

Next, from the law of conservation of mass, -.

Mass of marble + mass of acid = mass of solution + mass of carbon dioxide gas

Plugging in gives

10.0 g + 54.80 g = 60.4 g + mass of carbon dioxide gas

Mass of carbon dioxide gas = 10.0 g + 54.80 g – 60.4 g = 4.40 g

Finally, use the density to convert the mass of carbon dioxide gas to volume.

Volume = g/L1.798

0g44.

density

mass = 2.447 L = 2.4 L

Q4.

6 30ml of 0.01 M [Cr(H2O)5Cl]Cl2 => 2Cl- ions ∴ Number of Cl- ions = 30 x 0.01 x 2 = 0.6 mmol

∴ mmoles of Ag+ ions required = 0.6 mmol

∴ Volume = ? V x 0.1 = 0.6 => V = 6 ml

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Q11.

40 ANSWER: We know that for a radioactive element

kt = 2.3031og(N0/N) where, N0 = no. of radioactive atoms present initially, N = no. of radioactive atoms present after time t, k = rate constant.

For 20% decay; kt20 = 2.3031og(100/80) = 2.3031og(5/4) (1) For 80% decay : kt80 = 2.3031og(100/20) = 2.303log(5/1) (2)

From equations (1) and (2), we have, k(t20 - t80) = 2.303[log(5/4) - log(5/1)] = 2.303log[(5 x 1)/(5 x 4)] = 2.3031og(1/4) or, (t80 - t20) = [2.303log(4)]/k = [2.303log(4)].to 5/0.693 = [2.303 x0.6202 x 20]/[0.693] =

40 Hence, the time between 20% and 80% decay will be: 40 minutes. Hence, the correct answer is: 40 min. Q12.

1 ANSWER: Mn(CO)3 : The total no. of electrons present in it = 7 + 3(2) = 13.

Fe(CO)4 : The total no. of electrons present in it = 8 + 4(2) = 16. V(CO)6 : The total no. of electrons present in it = 5 + 6(2) = 17.

Cr(CO)6 : The total no. of electrons present in it = 6 + 6(2) = 18. Hence, the complex that obeys the 18-electron rule is Cr(CO)6.

Q13.

3

ANSWER: When the solution is allowed to pass through a cation exchanger, the chloride ions present outside the co-ordination sphere will become free. Then, these chloride ions will react with AgNO3 to give the white precipitate of AgCl. Let x no. of Cl- ligands are

present outside the co-ordination sphere. Now, we can write that

CoC13.6NH3 → x Cl- → x AgC1 ↓ From the equation, it is evident that

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x × no. of moles of CoC13.6NH3 = no. of moles of AgC1

x[2.675/267.5] = 4.78/143.5 or, x = 3. Hence, three chloride ions are present outside the co-ordination sphere. Hence, the correct answer is: 3.

Q14.

47 ANSWER: In case of a first order reaction the rate equation is: 2.303 log{a/(a-x)} = kt....(1)

Where, a = initial conc. of the reactant; (a-x) = final conc. of the reactant ; k = rate constant ; t = time. Here, a = 100, (a-x) = 40 and t = 60 min. Putting these values into the equation (1), we have

2.3031og (100/40) = k x 60 or, k = (2.303/60)[log10 - log4] = (2.303/60) x (1 - 0.6) = (2.303 x 0.4)/(60) = 0.01535 min-1

Now, k = 0.01535 min-1, a = 100 and (a-x) = 50 Putting these values into the equation (1), we have 2.303 log (100/50) = 0.01535 x t or, t = (2.303/0.01535) [log10-log5] = 150 [1-0.69] = 47

min. Hence, the correct answer is: 47 minutes.

Physical Solution for 3rd Year Exam

held on 21st Jan, 2018.

SECTION A 5. Ans: (c)

Conceptual Approach: Please note that in case of problems of this kind, you cannot

offer a direct solution. Under such circumstance, you must follow a tricky method of hit and trial followed by logical elimination of the incorrect options so as to focus on the perfect option.

Technique of solution:-

For 0 order, [A] = [A]0 – 𝑘𝑡. Thus, [A]0 –[A]

𝑡= 𝑘 must be constant.

For 1st order, ln ([𝐴]0

[A]) = 𝑘𝑡. Thus,

1

𝑡𝑙𝑛(

[𝐴]0

[A]) = 𝑘 must be constant.

For 2nd order, 1

[𝐴] =

1

[𝐴]0 + kt. Thus,

1

𝑡(1

[𝐴] =

1

[𝐴]0) = 𝑘 must be constant.

If the data is not consistent with these three then, it is pseudo 1st order.

Now, [A]0 –[A]1

𝑡1= 0.17mM 𝑚𝑖𝑛−1 and

[A]0 –[A]2

𝑡2= 0.14mM 𝑚𝑖𝑛−1. So it is not 0 order.

Again, 1

𝑡1𝑙𝑛 (

[A]0

[𝐴]1) = 0.19𝑚𝑖𝑛−1 and

1

𝑡2𝑙𝑛 (

[A]0

[𝐴]2) = 0.16𝑚𝑖𝑛−1. So it is not 1st order.

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Again, 1

𝑡1(1

[A]1 −

1

[𝐴]0) = 0.2 mM−1 𝑚𝑖𝑛−1and

1

𝑡2(1

[A]2 −

1

[𝐴]0) = 0.19mM−1 𝑚𝑖𝑛−1. So, it must be a

2nd order reaction. 6. Ans: (C)

𝐂𝐨𝐧𝐜𝐞𝐩𝐭𝐮𝐚𝐥 𝐚𝐩𝐩𝐫𝐨𝐚𝐜𝐡: – The basic equation is given by:-

Now, according to the given problem,

Thus, comparing the equations we have,

Thus, we have and

7. (b)

PV =1

3U

Or, PdV + VdP =1

3dU

Or, PdV + VdP = −1

3PdV (Since, the process is adiabatic in nature)

Thus, 4

3PdV + VdP = 0

Again, PVα = constant Or, 𝛼PVα−1dV + VαdP = 0 Thus, 𝛼PdV + VdP = 0

Hence, α =4

3

8. (a)

Since, U = 0 involving an ideal gas, the process must be Isothermal in nature.

Thus, H = 0 Again, considering reversible isothermal pathway (as S is a state function);

S = R ln 2 = 0.693R

Thus, G = −298 × 0.693R = −206.514R 18. (a)

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19. Ans: 2 or B According to Debye Huckel limiting law

log 𝛾𝑛 = – 𝐴𝑍𝑛2√𝐼

and A = (𝑒2𝑁𝐴

4.606∈𝑅𝑇) (

8𝜋𝑒2𝑁𝐴

∈𝑘𝐵𝑇)1

2 = (const.) (1

∈𝑇)3

2

20. Ans: 1 or A

This problem is based on Walden’s rule. ‘When any ion flows through an aqueous medium in the presence of electric field, the viscous force is neutralized by the electrical force.

Hence, 6𝜋𝜂0𝑣0𝑟 = 𝑍𝑒𝑋

Therefore, 6𝜋𝜂0𝜇0𝑟 = 𝑍𝑒 (𝑠𝑖𝑛𝑐𝑒, 𝜇0 =𝑣0

𝑋)

Now the molar conductance is enhanced by the increase in 𝜇0 and it is maximized by the

minimization of 𝜂0which is co-related to the frictional force between the liquid layers. 21. Ans: (C)

In the standard reduction form, O2 + 4H+ + 4e = 2H2O, E0 = 1.23 V

Now, E = E0 – 2.303RT pH/F Now the required EMF for the given reaction is – E

22. Ans: (4 or D) 23. Ans: (A)

𝐄 = (𝐄𝐑𝟎 − 𝐄𝐋

𝟎) +𝐑𝐓

𝟐𝐅𝐥𝐧[𝐀𝐠+]

[𝐂𝐮𝟐+]

Or, 𝐄 = (𝐄𝐑𝟎 − 𝐄𝐋

𝟎) +𝐑𝐓

𝟐𝐅𝐥𝐧

[𝐀𝐠+]

[𝐂𝐮𝟐+]

Or, 𝐄 = [(𝟎. 𝟖 − 𝟎. 𝟑𝟒) +𝟎.𝟎𝟓𝟗𝟏

𝟐𝐥𝐨𝐠

𝟎.𝟎𝟓

𝟎.𝟑] 𝐕 = 𝟎. 𝟒𝟒𝐕

24. Ans: A

SECTION B 5. Ans: (a), (b), (c)

∆𝐻 = 𝐸𝑃 − 𝐸𝑅 = (E* − 𝐸1) – (E* − 𝐸−1)

∴ ∆𝐻 = E* − 𝐸−1 Now, for the given problem

𝐸1 ≡ 𝐸𝑎 Now, from the basic energy profile diagram it is evident that ∆𝐻 ≯ 𝐸𝑎 i.e., 𝐸𝑎 ≥ ∆𝐻

But, 𝐸𝑎 = ∆𝐻 is practically meaningless since under such circumstance the product and transition state becomes energetically equal.

Hence, (𝐸𝑎)𝑚𝑎𝑥 > ∆𝐻

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6. Ans: (a), (b), (c) Adiabatic free expansion involving ideal gas.

According to 1st law of thermodynamics, dQ + dW =dU…………… (1)

dW = −PdV …………. (2)

dU = 𝐶𝑣 𝑑𝑇 + (𝜕𝑈

𝜕𝑉)𝑇𝑑𝑉 …………. (3)

Now for adiabatic process, dQ = 0 𝑃𝑜𝑝 = 0 (constant) → Irreverible

∴ dW = 0 (Using equation 2)

∴ dU = 0 (Using equation 1)

Hence, 𝐶𝑣 𝑑𝑇 + (𝜕𝑈

𝜕𝑉)𝑇𝑑𝑉 = 0 (using equation 3)

Now, for an ideal gas, (𝜕𝑈

𝜕𝑉)𝑇 = 0

∴ 𝐶𝑣 𝑑𝑇 = 0

Now, 𝐶𝑣 ≠ 0

dT = 0

∴ Q = W = ∆𝑇 = 0 7. Ans: (a), (b), (c), (d)

8. Ans: (a), (b)

SECTION C 5. 16

According to the given data v1 = k(nA

V)2 and v2 = k(

nAV

4

)2 = 16v1

6. 3

P = (180xA + 90)units The general equation is given by P = PB

∗ + (PA∗ − PB

∗)xA Comparing the equations, PB

∗ = 90 units and (PA∗ − PB

∗) = 180 units; so PA∗ = 270 units

Therefore, PA∗

PB∗ = 3

7. 0.11 According to the given data,

pOH = 14 − 12 = 2; Thus,[OH−] = 10−2M = 0.01M; [BOH] = (0.1 − 0.01)M and [B+] = 0.1M

Now, i =0.1+0.1+(0.1−0.01)

0.1= 1.1

Therefore, π = iCRT = (1.1 × 0.1)RT = 0.11RT

8. 0.0585

The basic equation is given by 𝛥𝑇𝑓

𝐾𝑓= 𝑖𝑚

Where, m is the molality of NaCl which is almost equivalent to molarity in aqueous solution since the density of water is 1 kg.L-1

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Now, considering α = 1 for NaCl, i = 2

Thus, 2m =1

500

Or, m =1

1000

Or, Mass = (1

1000× 58.5) g = 0.0585g

15. 0.5

16. 2

N2O has two possible orientations w.r.to its Principal axis of symmetry (𝐶∞).

Now at absolute zero the residual entropy is given by S =kB ln(No. of orientations)

(No.of molecules)

Now, No. of molecules = NAvo and No. of orientations = 2 Thus, S̅ = R ln 2 17. 0.5

𝑄 = 1 + 𝑒−𝜀𝑘𝐵𝑇 , 𝑠𝑖𝑛𝑐𝑒 𝐸0 = 0 𝑎𝑛𝑑 𝐸1 = 𝜀

𝑁𝑜𝑤,< 𝜀 >=𝑁0𝐸0 +𝑁1𝐸1

𝑁=𝜀𝑒−𝜀𝑘𝐵𝑇

𝑄, 𝑠𝑖𝑛𝑐𝑒, 𝐸0 = 0, 𝐸1 = 𝜀 𝑎𝑛𝑑

𝑁1𝑁=𝑒−𝜀𝑘𝐵𝑇

𝑄

𝑁𝑜𝑤, 𝑎𝑡 𝑇 → ∞, 𝑒−𝜀𝑘𝐵𝑇 → 1,

𝑇ℎ𝑢𝑠, 𝑄 → 2 𝑎𝑛𝑑 ℎ𝑒𝑛𝑐𝑒, < 𝜀 >≈𝜀

2

18. 283

Applying Lambert Beer’s law,

A = 2 − log%T = λcl =λPl

RT

Thus, 2−log%T1

2−log%T2=P1

P2

Therefore, P2 = P1 ×2−log%T2

2−log%T1

Now, substituting the values, P2 = {100 ×2−log2

2−log25.1}mm of Hg = 283 mm of Hg

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Organic Solution for 3rd Year Exam

held on 21st Jan, 2018.

MCQ

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MSQ

7. Ans. (A), (C), (D)

In these molecules all the C are not sp2 hybridised. 8. Ans. (A), (D)

9. Ans. (A), (C)

O

O

COOH

COOH

COO

CN

COOH

OH

NH

(A) (C)

-CN H+, H2O H+, H2O

10. Ans. (C), (D)

O

Br

*

O

MeO *

NaOMe/ MeOH

MeO-

O-

MeO *

Br

NAT

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inorganic solution for 3 year m.sc. entrance exam held on ... · m.sc. entrance test series 2017...

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