Hydrogen (Package Solutions)

Solutions of Assignment (Set-2)

Hydrogen (Solutions)

9 Hydrogen

Chapter Section-A Q.No. 1.

Solutions Answer (4) Water gas is 1 : 1 mixture of CO and H2. As this mixture is used for the synthesis of methanol and a number of hydrocarbons, it is also known as synthesis gas.

2.

Answer (3) 

H Cr2O7– 2  H2O 2   CrO5

3.

Answer (4) O

O H

H

H

H O H

H O H

4.

H

O H

H

Answer (3) PbS + 4H2O2  PbSO4 + 4H2O

5.

Answer (2) The elements of 7, 8 and 9 group do not form hydrides. This is actually known as HYDRIDE GAP.

6.

Answer (4) Nascent hydrogen is more reactive than molecular hydrogen and can be produced in situ.

7.

Answer (2) 2 Ethyl anthraquinol produces H2O2 on oxidation

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Hydrogen (Solutions)

Solutions of Assignment (Set-2)

Q.No. 8.

Solutions Answer (3) H2O2 acting as oxidizing agent In acidic medium 2K4[Fe(CN)6] + H2O2 + H2SO4  2K3[Fe(CN)6] + K2SO4 + H2O H2O2 acting as reducing agent when it reacts with K3[Fe(CN)6] In Basic medium 2K3[Fe(CN)6] + 2KOH + H2O2  2K4[Fe(CN)6] + 2H2O + O2

9.

Answer (1) Zn and Sn react with NaOH and produce hydrogen Zn 2NaOH  Na2ZnO2 + H2

10.

Answer (1) Ortho and Para hydrogen have different nuclear spins

11.

Answer (2)

H

12.

H

–H

H (Conjugate base of H2) Hydrideion

Answer (2)

Na H + H — OH

NaOH + H2

So the solution would be basic. 13.

Answer (3) Boiling point of water is exceptionally high due to hydrogen bonding

14.

Answer (4) Metallic hydrides are interstitial and non-stoichiometric and poor conductor of electricity.

15.

Answer (2) M

Volume strength 11.2

volume strength = 3.57 × 11.2 = 40 volume

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Solutions of Assignment (Set-2)

Hydrogen (Solutions)

Q.No. 16.

Solutions Answer (1) oxidising

18 2H2O18 2   2H2O  O 2 agent

17.

Answer (1) PbO2 is not a peroxide so, it does not produce H2O2

18.

Answer (1) 10 volumes of H2O2 means, 10 ml of oxygen is produced by the decomposition of 1 ml of H2O2 at STP.

19.

Answer (4) LiH is ionic hydride, while MgH2 and CuH are intermediate hydrides

20.

Answer (2) AlH3 is a Lewis acid H– is a Lewis base in I In II — H– accept H+ so Bronsted base

21.

Answer (4) Transition elements form interstitial hydride.

22.

Answer (2) It is deuterium oxide.

23.

Answer (4) As both ‘O’ atoms are in sp3

24.

Answer (3) This is the condition of neutrality.

25.

Answer (4)

Section-B Q.No. 1.

Solutions Answer (1, 4) Saline hydride or ionic hydrides are CaH2 and SrH2

2.

Answer (2, 4) VH and TiH3 are metallic hydride

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Hydrogen (Solutions)

Solutions of Assignment (Set-2)

Q.No. 3.

Solutions Answer (1, 3) CH4 and HCl are covalent hydride

4.

Answer (1, 2) BH3 and BeH2 exist in polymeric form

Be

H

Be

H

H

Be

H

H H

Be

(BeH2)n H

H B

B H

5.

H

H (BH3)2

H

Answer (1, 2, 3) 7, 8, 9 group do not form hydride

6.

Answer (3) H2O2 act as oxidising agent for PbS

7.

Answer (2, 3) Density and viscosity of H2O2 is higher than water.

8.

Answer (2, 3, 4) Hardness of water is due to CaSO4, MgCl2 and MgSO4.

9.

Answer (1, 3, 4) Stearic acid, palmitic acid and oleic acids are long chain fatty acid and hence their sodiumsalts can participate in colloidal formation.

10.

Answer (1, 2, 3 ) Na2[Na4(PO3)6]  in Calgon method Ca(OH)2  Clark’s method Na2CO3 also used to convert soluble salt of Ca & Mg to insoluble bicarbonate

11.

Answer (3, 4) Ca(OH)2 form carbonate of Ca & Mg

12.

Answer (1, 4) Two nucleus have same spin two e– have opposite spin

13.

Answer (2, 3) On increasing ortho concentration equilibrium will shift towards right. Decreasing temp will also increase the concentration of parahydrogen and hence equilibrium will shift in forward direction.

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Solutions of Assignment (Set-2)

Hydrogen (Solutions)

Q.No. 14.

Solutions Answer (1, 2, 3, 4) Hydrogen gas can be prepared by Lene’s process, Bosh’s process, electrolysis of acidulated water and Kolbes electrolysis

15.

Answer (1, 2, 3, 4) H2O2 acts as oxidising agent, reducing agent, bleaching agent and acid

16.

Answer (3) n  1.5 p

17.

Answer (2, 3) In eq, solution both ionize

Section-C Q.No.

Solutions Comprehension-I

1.

Answer (2) Bleaching action of H2O2 is due to its ability to oxidise

2.

Answer (4) Oxygen evolved = 30 × 1000 = 3 × 104 mol = 30 litre

3.

Answer (3) Pure H2O2 is a weak acid Comprehension-II

1.

Answer (4) Al4C3 + D2O  CD4 + Al(OD)3

2.

Answer (1) N2O5 + D2O  DNO3

3.

Answer (4) Heavy water is weaker acid than normal water. Kw of D2O < Kw of H2O.

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Hydrogen (Solutions)

Solutions of Assignment (Set-2)

Q.No.

Solutions Comprehension-III

1.

Answer (1) At no. is same for all.

2.

Answer (3) Bond-strength and area overlapping.

3.

Answer (1)

Section-D Q.No. 1.

Solutions Answer (1) H SO

4 KMnO 4  H2O 2 2   MnSO 4  H2O  O 2

2.

Answer (2) O—O bond in O2F2 is shorter than that of H2O2. H2O2 is a polar covalent molecule.

3.

Answer (1) Hydrolysis HO 3 S — O — O — SO 3H    H2O 2  2HSO 4  2H 1 mole

4.

1 mole

Answer (4) Zn + 2NaOH  Na2ZnO2 + H2 Zn(OH)2 is amphoteric in nature

5.

Answer (2) 2Al + 2NaOH   2NaAlO2 + H2

6.

Answer (1) Statement-2 explains statement-1

7.

Answer (4) In thermal decomposition H2O2  H2 O 

1 O2 , it is “1” 2

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Solutions of Assignment (Set-2)

Hydrogen (Solutions)

Section-E Q.No. 1.

Solutions Answer : A(p, q, r, s), B(p, q, s), C(p, r), D(q, s) (A) H2O2 act as oxidizing agent, reducing agent, bleaching agent and acidic in nature (B) HNO2 act as both oxidising agent, reducing agent and acidic in nature. (C) O3 is an oxidising and bleaching agent (D) H2S  reducing agent and acidic in nature

2.

Answer : A(p, r, s), B(p, r, s), C(p), D(p, r, s) (A) sp3 hybridisation is present in H2O, H2O2, B2H6 and NH3. H2O, H2O2 and NH3 can form hydrogen bonding. H2O, H2O2 and NH3 also have lone pair in their structure.

3.

Answer A(p, q, s, t), B(p, r, s, t), C(p, s, t), D(p, s, t) Structures of H2O2 are same but with different bond lengths and bond angles. All are like half open book structure. H2O is bent molecule.

4.

Answer A(r, s, t), B(p, s, t), C(p, s, t), D(q, s) Ca(HCO3)2 + Na2CO3  CaCO3  + 2NaHCO3 CaSO4 + Na2CO3  CaCO3  + Na2SO4 (NaPO3)6 is calgon

5.

Answer A(t), B(q, r), C(p), D(s) Self explanatory.

Section-F Q.No. 1.

Solutions Answer (4) Each molecules is surrounded by 4 water.

2.

Answer (2) H2O2  O2 + 2e–

3.

Answer (5) Vol. Strength = M 11.2 = 0.45 11.2 = 5

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Hydrogen (Solutions)

Solutions of Assignment (Set-2)

Section-G Q.No. 1.

Solutions Answer (1) Dielectric constant of H2O2 is greater than H2O. Boiling point and freezing point of D2O are greater & H2 requires less energy.

2.

Answer (1) Tritium is radioactive. Solubility of NaCl is more in H2O due to high dielectric constant. pH value of H2O decreases by increasing temperature.

3.

Answer (1) Factual

Section-H Q.No. 1.

Solutions Answer (3) Bond length 

2.

1 bond strength

Answer (3, 4) For same atom, hydration enthalpies of cation anion

3.

Answer (3) N2  3D2   2ND3

4.

Answer (4) Factual

5.

Answer (2) Bond strength 

6.

1 bond length

Answer (4) As it slows-down the neutron

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Solutions of Assignment (Set-2)

Hydrogen (Solutions)

Q.No. 7.

Solutions BaH2 + 2H2O  Ba(OH)2 + 2H2 Moles of hydrogen produced =

4.36  0.975 = 0.1768 mole 293  1  0.0821

2 mole of hydrogen is produced from = 1 mole of BaH2 1 mole of hydrogen is produced =

1 mole of BaH2 2

0.176 mole of hydrogen is produced =

1  0.176 = 0.088 moles of BaH2 2

wt. of BaH2 require : 0.088 × 139 = 12.232 g 8.

Na impart golden yellow colour to Bunsen burner flame. 2Na + 2H2O  2NaOH H2 C

'B'

Zn 2NaOH  Na2ZnO2  H2 D

C

B

Zn + H2SO4  ZnSO4 + H2 9.

 XH2 X + H2  200 º C



XH2  X + H2 1 gm of X can produce = 559 ml of H2 X + Cl2  XCl2 One equivalent of hydrogen is produced by =

1  11200 = 20 gm 559

63.89 gm chlorine with metal = 36.11 gm 1 gm combine with metal = 35.5 gm combine with

36.11 63.89

36.11  35.5 = 20 g 63.89

n factor: 2 So atomic wt. = 40 So metal is Ca X = Ca Y = CaH2 Z = CaCl2 Aakash Educational Services Pvt. Ltd. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.011-47623456

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Hydrogen (Solutions)

Solutions of Assignment (Set-2)

Q.No.

10.

Solutions

2K4[Fe(CN)6] + H2O2 + H2SO4  2K3[Fe(CN)6] + H2SO4 + H2O 2K3[Fe(CN)6] + H2O2 + KOH  3K4[Fe(CN)6] + 2H2O + O2

11.

The reaction between hydrazine and H2O2 is highly exothermic and is accompanied by large increase in the volume of the products and hence, this mixture is used as a rocket propellant Cu(II)  N2(g) + 4H2O(g) NH2 – NH2(l) + 2H2O2(l) 

12.

Equivalent of I2 liberated = equivalent of hypo used = equivalent of H2O2 So,

20 25  0.3   N(H2O 2 ) 1000 1000

N(H2O2) =

20 0.3   2.4  10 4  1000 = 0.24 N 1000 25

So, volume strength = 0.24 × 5.6 = 1.34 volume 13.

(i)

LiH will give more amount of H2 Due to low molecular weight, in equal amount it has high no. of moles. 2LiH + H2O  LiOH + H2

(ii)

CaH2 + 2H2O  Ca(OH)2 + 2H2 Moles of H2 produced :

150  100 = 6.13×102 298  1.0821

2 mole of H2 is produced by = 42 g 6.13 × 102 mole of H2 will be produced by 

14.

2

42  6.13  10 kg = 12.873 kg 1000  2

H2O2 ionise as H2O2

H+ + HO2–

pH = –log(H+) = 5.9 H+ = 10–5.9 Xionic = 10–5.9 × 10–5.9 = 10–11.8 = 1.58 × 10–12 15.

5

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