Chem Hydrogen

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

R. K. MALIK’S NEWTON CLASSES

14

Chapter

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Hydrogen

Solutions SECTION - A

Objective Type Questions (One option is correct)

1.

Water gas is a mixture of (1) CO and H2O

(2)

O2 and CO

(3)

CO and N2

(4)

CO and H2

Sol. 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.

An orange coloured solution of K2Cr2O7 acidified with H2SO4 and treated with a substance X gives a blue coloured solution of CrO5. The substance X is

(1) H2O

(2)

dil. HCl

(3)

H2O2

(4)

conc. HCl

(3)

1

(4)

4

(1) H2 is liberated

(2)

O2 is liberated

(3) 4 moles of H2O2 consumed

(4)

Sulphur is converted to sulphite

Sol. Answer (3)



H Cr2O7–2  H2O2   CrO5

3.

The number of H-bonds formed by a water molecule is

(1) 2

(2)

8

Sol. Answer (4) H

O

O

H

O

H

H H

4.

O

H

H

H

H

O

H

When 1 mole of PbS reacts completely with H2O2

Sol. Answer (3) PbS + 4H2O2  PbSO4 + 4H2O

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R. K. MALIK’S NEWTON CLASSES 2 5.

Hydrogen

Solution of Assignment

The term hydride gap refers to which region of periodic table? (1) Groups 6 to 9

(2)

Groups 7 to 9

(3)

Groups 7 to 10

(4)

Groups 5 to 7

Sol. Answer (2) The elements of 7, 8 and 9 group do not form hydrides. This is actually known as HYDRIDE GAP. Which one is true about nascent hydrogen?

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6.

(1) More reactive than molecular hydrogen (2) Can be produced in situ

(3) Show similarity exactly with hydrogen in reduction reactions (4) Both (1) & (2)

Sol. Answer (4)

Nascent hydrogen is more reactive than molecular hydrogen and can be produced in situ.

7.

2-ethyl anthraquinol when oxidised in air produces

(1) O3

(2)

H2O2

(3)

H2O

(4)

C2H5OH

Sol. Answer (2)

2 Ethyl anthraquinol produces H2O2 on oxidation.

8.

Which are the compounds, which can be oxidised and reduced by H2O2 in acidic and basic medium respectively?

(1) [K2O, NaAlO2]

(2)

[KMnO4, K2MnO4]

(3) {K4[Fe(CN)6], K3[Fe(CN)6]}

(4)

[K3[Fe(CN)6, K4Fe(CN)6]

Sol. 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.

In alkaline medium, which elements can produce hydrogen? (1) Zn, Si

(2) Cu, Ag

(3) Cu, N2

(4) Al, C

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

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R. K. MALIK’S NEWTON CLASSES Solution of Assignment

Hydrogen

3

10. Select the incorrect statement (1) Ortho and para hydrogen are different due to difference in their nuclear spins (2) Ortho and para hydrogen are different due to difference in their electron spins (3) Para hydrogen has a lower internal energy than that of ortho hydrogen

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(4) Para hydrogen is more stable at lower temperature Sol. Answer (1)

Ortho and Para hydrogen have different nuclear spins.

11. Hydride ion is a

(1) Strong conjugate acid of H2

(2)

Strong conjugate base of H2

(3) Strong conjugate acid of H+

(4)

Strong conjugate base of H–

(1) Give acidic solutions

(2)

Give basic solutions

(3) Produce hydride ions

(4)

Produce protons

(1) Covalent bond between H and O

(2)

Linear shape

(3) Hydrogen bonding

(4)

Non-linear shape

Sol. Answer (2) H

H

–H

H (Conjugate base of H)2 Hydride ion

12. Ionic hydrides react with water to

Sol. Answer (2)

Na H + H — OH

NaOH + H2

So the solution would be basic.

13. The boiling point of water is exceptionally high due to

Sol. Answer (3)

Boiling point of water is exceptionally high due to hydrogen bonding.

14. Metallic hydrides (1) Are also called interstitial hydrides

(2)

Are non-stoichiometric, being deficient in hydrogen

(3) Pt and Pd hydrides are the example of it

(4)

Have all properties given above

Sol. Answer (4) Metallic hydrides are interstitial and non-stoichiometric and poor conductor of electricity. 15. The volume strength of 3.57 M solution of hydrogen peroxide is (1) 30 volume

(2)

40 volume

(3)

20 volume

(4)

25 volume

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R. K. MALIK’S NEWTON CLASSES 4

Hydrogen

Solution of Assignment

Sol. Answer (2)

M

Volume strength 11.2

Volume strength= 3.57 × 11.2 = 40 volume

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16. In the following reaction using isotopic 18O in H2O2,

2MnO4– + 3H2O218  2MnO2 + 3O2 + 2H2O + 2OH– isotopic oxygen goes (1) Both with O2

(2)

Both with MnO2

(3) Both with OH–

(4)

One with O2 and one with MnO2

Sol. Answer (1)

oxidising 18 2H2O18 2  2H2 O  O2 agent

17. H2O2 can be obtained when following reacts with H2SO4 except with (1) PbO2

(2)

BaO2

(3)

Na2O2

(4)

SrO2

(1) 1 ml H2O2 solution gives 10 ml O2 at NTP

(2)

1 gm H2O2 solution gives 10 ml O2 at NTP

(3) 1 mol H2O2 solution gives 10 ml O2 at NTP

(4)

1 ml H2O2 solution gives 100 ml O2 at NTP

Sol. Answer (1)

PbO2 is not a peroxide so, it does not produce H2O2.

18. 10 volumes of H2O2 means

Sol. Answer (1)

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

19. Out of LiH, MgH2 and CuH (1) All are ionic hydrides

(2) LiH, MgH2 are ionic and CuH is covalent hydride

(3) All are covalent hydrides

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

Sol. Answer (4) LiH is ionic hydride, while MgH2 and CuH are intermediate hydrides. 20. Consider the following reactions I.

AlH3 + H–  AlH4–

II.

H2O + H–  H2 + OH–

Select the correct statements based on these reactions. (1) H– is a Lewis acid in I and Lewis base in II

(2)

H– is a Lewis base in I and Bronsted base in II

(3) H– is a Lewis acid in I and Bronsted acid in II

(4)

H– is a Lewis base in I and II

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R. K. MALIK’S NEWTON CLASSES Solution of Assignment

Hydrogen

5

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

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21. Which among the following is interstitial hydride? (2)

(3)

TiH1.7

(1) H2O with dissolved Mg(HCO3)2

(2)

D2O

(3) D2O with heavy metal impurities

(4)

H2O with CaCO3

(1) Two dipole moments are opposite but unequal

(2)

Two dipole moments are opposite and equal

(3) Two dipole moments are equal but not at 180°

(4)

Two dipole moments are equal but non-planar

(3)

[H+] = [OH–]

(4)

Amphoteric nature

(3)

K+

(4)

All of these

(1) ScH2

LaH2

(4)

All of these

Sol. Answer (4)

Transition elements form interstitial hydride.

22. Heavy water is

Sol. Answer (2)

It is deuterium oxide.

23. Dipole moment of H2O2 is non-zero as

Sol. Answer (4)

As both ‘O’ atoms are in sp3.

24. Neutrality of water is due to (1) pH = 7

(2)

Presence of salts

Sol. Answer (3)

This is the condition of neutrality.

25. Which of the following cation is stable in water? (1) H+

(2)

Na+

Sol. Answer (4)

SECTION - B Objective Type Questions (More than one options are correct) 1.

Which compound/s is/are saline hydride? (1) CaH2

(2)

HCl

(3)

ScH2

(4)

SrH2

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

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R. K. MALIK’S NEWTON CLASSES 6 2.

Hydrogen

Solution of Assignment

Which compound/s is/are metallic hydrides? (1) KH

(2)

VH

(3)

PH3

(4)

TiH3

Sol. Answer (2, 4) VH and TiH3 are metallic hydride. Which compound/s is/are covalent hydrides?

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3.

(1) CH4

(2)

CsH

(3)

HCl

(4)

NaH

(3)

NaH

(4)

CaH2

(3)

9

(4)

10

(1) Lighter than water

(2)

Denser than water

(3) More viscous than water

(4)

Less viscous than water

Sol. Answer (1, 3)

CH4 and HCl are covalent hydride.

4.

Examples of polymeric hydrides are (1) BH3

(2)

BeH2

Sol. 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

H

5.

H

B

H (BH3)2

Hs

Which group element of d-block do not form hydride at all? (1) 7

(2)

8

Sol. Answer (1, 2, 3)

7, 8, 9 group do not form hydride.

6.

Which reaction shows oxidising nature of H2O2? (1) Ag2O + H2O2  Ag + H2O + O2

(2) MnO2 + H2O2 + H2SO4  MnSO4 + 2H2O + O2 (3) PbS + 4H2O2  PbSO4 + 4H2O

(4) K2Cr2O7 + H2SO4 + 4H2O2  K2SO4 + 2CrO5 + 5H2O Sol. Answer (3) H2O2 act as oxidising agent for PbS. 7.

H2O2 is

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

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R. K. MALIK’S NEWTON CLASSES Solution of Assignment

8.

Hydrogen

7

Hardness of water may be due to (1) CaCO3

(2)

CaSO4

(3)

MgSO4

(4)

MgCl2

Sol. Answer (2, 3, 4) Hardness of water is due to CaSO4, MgCl2 and MgSO4. The soaps contain salts of higher fatty acids like

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9.

(1) Stearic acid

(2)

Oxalic acid

(3)

Palmitic acid

(4)

Oleic acid

Sol. 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. Chemical additive which can be used to remove water hardness is (1) Na2[Na4(PO3)6]

(2)

Ca(OH)2

(3)

Na2CO310H2O

(4)

CaCO3

(4)

MgCO3

Sol. 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. Ca(OH)2 removes temporary hardness by forming (1) CaCl2

(2)

CaSO4

(3)

CaCO3

(1) The two nuclei have same spin

(2)

The two nuclei have opposite spin

(3) The two electrons have same spin

(4)

The two electrons have opposite spin

(1) On increasing temperature

(2)

On increasing ortho concentration

(3) On decreasing temperature

(4)

On decreasing ortho concentration

Sol. Answer (3, 4)

Ca(OH)2 form carbonate of Ca & Mg.

12. In ortho H2 the correct statement/s is/are

Sol. Answer (1, 4) Two nucleus have same spin two e– have opposite spin. 13. Consider the following reversible conversion: Ortho (H2)

para (H2)

This equilibrium will shift in forward direction

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R. K. MALIK’S NEWTON CLASSES 8

Hydrogen

Solution of Assignment

Sol. 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. 14. Hydrogen can be prepared by (2)

Bosch’s process

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(1) Lane’s process (3) Electrolysis of acidulated water

(4)

Kolbe’s electrolysis

Sol. Answer (1, 2, 3, 4)

Hydrogen gas can be prepared by Lane’s process, Bosch’s process, electrolysis of acidulated water and Kolbes electrolysis.

15. H2O2 can act as

(1) Oxidising agent

(2)

Reducing agent

(3)

Bleaching agent

(4)

Acid

(4)

He

Sol. Answer (1, 2, 3, 4)

H2O2 acts as oxidising agent, reducing agent, bleaching agent and acid.

16

Radio-activity can be detected in hydrogen due to the presence of (3)

3 1H

(1) Same state of hybridisation

(2)

Both are protic solvents

(3) Both are acidic

(4)

Both are basic

(1)

1 1H

2 1H

(2)

Sol. Answer (3) n  1.5 p

17

H2O and Ph – OH resemble as

Sol. Answer (2, 3)

In aqueous solution both ionize.

SECTION - C Linked Comprehension Type Questions Comprehension-I Hydrogen peroxide is a hydride of oxygen. It can be obtained by sodium peroxide or barium peroxide in laboratory. Pure hydrogen peroxide is a colourless, syrupy liquid with bitter taste and odour like that of nitric acid. It is highly soluble in water, alcohol and ether. It is an unstable liquid and readily decomposes to water and oxygen. It has ability of both oxidising and reducing nature. It is used for restoring colour of oil painting. H2O2 on treatment with acidified solution of titanium salt, orange colour is produced due to formation of pertitanic acid. Its strength is measured as volume strength or percentage weight/volume strength.

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R. K. MALIK’S NEWTON CLASSES Solution of Assignment

1.

Hydrogen

9

Bleaching action of H2O2 is due to its ability to_______the colouring matter. (1) Decompose

(2)

Oxidise

(3)

Reduce

(4)

Destroy

Sol. Answer (2)

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Bleaching action of H2O2 is due to its ability to oxidize. 2.

A 30 volume solution of hydrogen peroxide means that 1 litre of the solution gives (1) 30 ml of O2

(2)

30 litres of O2

(3) 30 ml of O2 at STP

(4)

30 litres of O2 at STP

(3)

Weak acid

Sol. Answer (4)

Oxygen evolved = 30 × 1000

= 3 × 104 mol

= 30 litre

3.

Pure hydrogen peroxide is a (1) Weak base

(2)

Strong base

(4)

Strong acid

Sol. Answer (3)

Pure H2O2 is a weak acid.

Comprehension-II

Heavy water is the oxide of heavy hydrogen (deuterium) and is also called deuterium oxide. It is represented as D2O. Heavy water is chemically similar to ordinary water. Heavy water is used for the neutron moderator, as a tracer compound and for the preparation of deuterium. Reaction of heavy water with alkali metals liberates heavy hydrogen. Heavy water can also be used for exchanging labile hydrogen with deuterium completely or partially. Heavy water reacts slower than ordinary water but forms stronger bonds with other elements as compared to hydrogen. 1.

The products of the reaction Al4C3 + D2O  is

(1) DC  CD

(2)

CD4

(3)

Al(OD)3

(4)

Both (2) & (3)

(3)

NO

(4)

NO2

Sol. Answer (4) Al4C3 + D2O  CD4 + Al(OD)3 2.

Reaction of N2O5 and D2O produces (1) DNO3

(2)

DNO2

Sol. Answer (1) N2O5 + D2O  DNO3

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R. K. MALIK’S NEWTON CLASSES 10 3.

Hydrogen

Solution of Assignment

Which property of heavy water is lesser in magnitude as that compared with normal water? (1) Molecular mass

(2)

Density

(3)

Boiling point

(4)

Ionisation constant

Sol. Answer (4) Heavy water is weaker acid than normal water.

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Kw of D2O < Kw of H2O.

Comprehension-III

Size of nucleus increases from protium to tritium so in H2 D2 & T2 area of overlapping also increases in the same order. 1.

Which overlapping is responsible for bond formation in H2, D2, T2 respectively? (1) 1s – 1s in each

(2)

2s – 2s in each

(3) 1s – 2s in each

(4)

1s – 2s in T2, 1s – 1s in rest

(3)

H2 < D2 < T2

Sol. Answer (1)

At no. is same for all.

2.

H2, D2 & T2 show their bond-enthalpies as

(1) H2 = D2 = T2

(2)

H2 > D2 > T2

(4)

D2 < H2 < T2

Sol. Answer (3)

Bond-strength and area overlapping.

3.

Which of the following has highest bond dissociation energy?

(1) D2

(2) H2

(3) HD

(4) All of these have identical bond dissociation energy

Sol. Answer (1)

SECTION - D Assertion-Reason Type Questions 1.

STATEMENT-1 : H2O2 liberates O2 when it reacts with acidified KMnO4 solution. and STATEMENT-2 : KMnO4 oxidises H2O2 to O2.

Sol. Answer (1) H SO

2 4 KMnO4  H2O2   MnSO4  H2O  O2

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Solution of Assignment

2.

Hydrogen

11

STATEMENT-1 : The O — O bond length in H2O2 is longer than that of O2F2. and STATEMENT-2 : H2O2 is a polar covalent molecule.

Sol. Answer (2)

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O—O bond in O2F2 is shorter than that of H2O2. H2O2 is a polar covalent molecule. 3.

STATEMENT-1 : Complete hydrolysis of one mole of peroxydisulphuric acid gives one mole of H2O2 and two moles of H2SO4. and

STATEMENT-2 : Peroxydisulphuric acid has one peroxy linkage.

Sol. Answer (1)

Hydrolysis HO3 S— O — O — SO3H   H2O2  2HSO4  2H 1mole

4.

1mole

STATEMENT-1 : Zinc hydroxide dissolves in excess of NaOH solution to evolve dihydrogen gas.

and

STATEMENT-2 : Zinc hydroxide is amphoteric in nature.

Sol. Answer (4)

Zn + 2NaOH  Na2ZnO2 + H2

Zn(OH)2 is amphoteric in nature.

5.

STATEMENT-1 : Hydrogen gas is liberated by the action of aluminium with a concentrated solution of NaOH.

and

STATEMENT-2 : Aluminium with NaOH forms sodium metaaluminate.

Sol. Answer (2)

2Al + 2NaOH   2NaAlO2 + H2

6.

STATEMENT-1 : With non-metals covalent hydrides are formed. and

STATEMENT-2 : For covalent hydrides electronegativity difference should be less.

Sol. Answer (1) Statement-2 explains statement-1. 7.

STATEMENT-1 : Valence factor of H2O2 is always 2. and STATEMENT-2 : In redox reactions change of O.S./H2O2 molecule is 2.

Sol. Answer (4)

 H2 O  In thermal decomposition H2 O2 

1 O2 , it is “1”. 2

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R. K. MALIK’S NEWTON CLASSES 12

Hydrogen

Solution of Assignment

SECTION - E Matrix-Match Type Questions 1.

Match the following. Column-I

(p) Oxidising agent

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(A) H2O2

Column-II

(B) HNO2

(q) Reducing agent

(C) O3

(r) Bleaching agent

(D) H2S

(s) Acidic in nature

Sol. 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.

Match the following. Column-I

Column-II

(A) H2O

(p) Central atom is sp 3 hybridised

(B) H2O2

(q) Central atom is sp2 hybridised

(C) CH4

(r) Presence of H-bonding

(D) NH3

(s) Presence of lone pair of electron on central atom

Sol. 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.

Match the following. Column I

Column II

(A) H2O (solid)

(p) Hydrogen bonding is present

(B) H2O2 (solid)

(q) Cage like structure

(C) H2O2 (gas)

(r) O-H bond length 95 pm

(D) H2O (liquid)

(s) Non-linear structure (t) Polar molecule

Sol. 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.

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R. K. MALIK’S NEWTON CLASSES Solution of Assignment

4.

Hydrogen

13

Match the following Column I

Column II (p) Cause of permanent hardness in water

(B) MgCl2

(q) Used for removal of hardness of water

(C) CaSO4

(r) Cause temporary hardness

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(A) Ca(HCO3)2

(D) (NaPO3)6

(s) Ionised in water

(t) Can be removed from water by using washing soda

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

Ca(HCO3)2 + Na2CO3  CaCO3  + 2NaHCO3CaSO4 + Na2CO3  CaCO3  + Na2SO4

(NaPO3)6 is calgon.

5.

Match the following Column I

Column II

(A) H+

(p) Hydrogen molecule

(B) H

(q) Free radical

(C) H2

(r) Hydrogen atom

(D) H–

(s) Anion of hydrogen (t) Proton

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

SECTION - F

Integer Answer Type Questions

1.

In liquid water, the number of H2O molecules surrounded to one H2O molecule are ____.

Sol. Answer (4)

Each molecules is surrounded by 4 water. 2.

When H2O2 is decomposed to O2 gas its nfactor is _______________.

Sol. Answer (2) H2O2  O2 + 2e– 3.

Determine volume strength of 0.45 M H2O2.

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

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R. K. MALIK’S NEWTON CLASSES 14

Hydrogen

Solution of Assignment

SECTION - G Multiple True-False Type Questions 1.

STATEMENT-1 : H2O2 is more polar than H2O. STATEMENT-2 : D2O has higher boiling point than H2O. STATEMENT-3 : H2 bond energy is less than D2. (2)

TTF

(3)

TFF

(4)

FFT

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(1) T T T

Sol. Answer (1)

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

2.

STATEMENT-1 : Tritium is radioactive form of Hydrogen.

STATEMENT-2 : NaCl is more soluble in water as compared to D2O. STATEMENT-3 : pH of water depends on temperature. (1) T T T

(2)

TTF

(3)

TFF

(4)

FFF

Sol. 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.

STATEMENT-1 : Hydrogen is used for cutting and welding purpose. STATEMENT-2 : In H2, protium is 99% by mass.

STATEMENT-3 : H2 & O2 react only under vigorous conditions.

(1) T F T

(2)

FTT

(3)

TTF

(4)

FFT

Sol. Answer (1) Factual

SECTION - H

Aakash Challengers Questions

1.

Order of bond length can be given as (1) H2 = D2 = T2

(2) H2 < D2 < T2

(3) H2 > D2 > T2

(4) D2 < H2 < T2

Sol. Answer (3) Bond length  2.

1 bond strength

Incorrect among the following. (1) H+ exist in water as H+.(H2O)n

(2) H– exist in water as H–.(H2O)n

o o (3) hydH(H )  hydH(H ) (hyd  hydration)

(4) H exist in water as H.(H2O)n

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

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Solution of Assignment

3.

Hydrogen

15

Not incorrect among the following. (1) N2 & H2 do not react at room temperature (2) Cl2 & H2 react in presence of sunlight (3) N2 & D2 react at room temperature (4) D2 on combustion forms heavy water

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Sol. Answer (3)

N2  3D2  2ND3

4.

For which  f H○ is zero?

(1) H

(2) H–

(3) H+

(4) H+(aq)

Sol. Answer (4) Factual

5.

Correct regarding bond-strength is (1) C – H > C – D > C – T

(2) C – H < C – D < C – T

(3) C – T < C – D = C – H

(4) C – D > C – H = C – T

Sol. Answer (2)

1 Bond strength  bond length

6.

Which among the following is used as moderator in nuclear-reactors?

(1) Hard water

(2) Soft water

(3) Pure water

(4) Heavy water

Sol. Answer (4)

As it slows-down the neutron.

7.

How many grams of barium hydride must be treated with water to obtain 4.36 L of hydrogen at 20°C and 0.975 atm pressure (Ba = 137)?

Sol. 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

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R. K. MALIK’S NEWTON CLASSES 16 8.

Hydrogen

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX Solution of Assignment

When a substance A reacts with water, it produces a combustible gas B and a solution of substance C in water. When another substance D reacts with this solution of C, it produces the same gas B on warming but D can produce gas B on reaction with dilute sulphuric acid at room temperature. A imparts a deep golden yellow colour to a smokeless flame of Bunsen burner. A, B, C and D respectively are _____?

Sol. Na impart golden yellow colour to Bunsen burner flame. 2Na + 2H2O  2NaOH H 2 'B'

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C

Zn  2NaOH  Na2ZnO 2  H2 D

C

B

Zn + H2SO4  ZnSO4 + H2

9.

An element X reacts with H2 gas at 200°C to form compound Y. When Y is heated to a higher temperature, it decomposes to element X and H2 gas in the ratio of 559 ml of H2 (measured at STP) for 1.00 g of X reacted. X also combines with Cl2 to form a compound Z which contains 63.89 percent by mass of chlorine. Identify X, Y and Z.

 Sol. X + H2  XH2 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 10. Hydrogen peroxide acts both as an oxidising agent and as a reducing agent in alkaline solution towards certain first row transition metal ions. Illustrate both these properties of H2O2 using chemical equations. Sol. 2K4[Fe(CN)6] + H2O2 + H2SO4  2K3[Fe(CN)6] + H2SO4 + H2O 2K3[Fe(CN)6] + H2O2 + KOH  3K4[Fe(CN)6] + 2H2O + O2

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Solution of Assignment

Hydrogen

17

11. A mixture of hydrazine and H2O2 with Cu (II) catalyst is used as a rocket propellant, why? Sol. 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) NH2 – NH2(l) + 2H2O2(l)   N2(g) + 4H2O(g)

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12. To a 25 ml H2O2 solution, excess of acidified solution of potassium iodide was added. The iodine liberated required 20 ml of 0.3 N sodium thiosulphate solution. Calculate the volume strength of H2O2 solution. Sol. 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  1000 25

4 = 2.4  10  1000

= 0.24 N

So, volume strength = 0.24 × 5.6

= 1.34 volume

13. Ionic metal hydrides react with water to give hydrogen gas and an aqueous solution of the metal hydroxide. (i) On reaction of equal masses of LiH and CaH2 with water, which compound gives more H2?

(ii) How many kilograms of CaH2 is needed to fill a 100 L tank with compressed H2 gas at 150 atm pressure and at 25°C? (Li = 9, H = 1, Ca = 40)

LiH  H2O  LiOH 

1 H2 2

CaH2 + 2H2O  Ca(OH)2 + 2H2

Sol. (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 =

42  6.13  102 kg 1000  2

= 12.873 kg

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R. K. MALIK’S NEWTON CLASSES 18

Hydrogen

Solution of Assignment

14. H2O2 ionises as H+ + HO2–

H2O2

If pH of H2O2 at 20°C is found to be 5.91, calculate ionic product (auto-protolysis constant) of H2O2. Sol. H2O2 ionise as H2O2

H+ + HO2–

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pH = –log(H+) = 5.9

H+ = 10–5.9

Xionic = 10–5.9 × 10–5.9 = 10–11.8 = 1.58 × 10–12

15. 1 mega litre water (density 1 g/cc) needs 106 Kg of Na2CO3 for removal of its permanent hardness. Determine its hardness in the multiples of 200 ppm. Sol. Answer (5)







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