Class 11 Chemistry · Chapter 4

Samacheer Class 11 Chemistry - Hydrogen

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Chapter-wise textbook exercise answers for Hydrogen with validation-aware solutions.

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Sections in this chapter
I. Choose the best answer: 21II. Write brief answer to the following questions: 24
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1I. Choose the best answer:21 questions
Q.1Which of the following statements about hydrogen is incorrect?v
  1. (a) Hydrogen ion, H 3 O + exists freely in solution.
  2. (b) Dihydrogen acts as a reducing agent.
  3. (c) Hydrogen has three isotopes of which tritium is the most common.
  4. (d) Hydrogen never acts as cation in ionic salts.
Answer:

(c) Hydrogen has three isotopes of which tritium is the most common.

Q.2Water gas isv
  1. (a) H 2 O(g)
  2. (b) CO + H 2 O
  3. (c) CO + H 2
  4. (d) CO + N 2
Answer:

(c) CO + H 2

Q.3Which one of the following statements is incorrect with regard to ortho and para dihydrogen ?v
  1. (a) They are nuclear spin isomers
  2. (b) Ortho isomer has zero nuclear spin whereas the para isomer has one nuclear spin
  3. (c) The para isomer is favoured at low temperatures
  4. (d) The thermal conductivity of the para isomer is 50% greater than that of the ortho isomer.
Answer:

(b) Ortho isomer has zero nuclear spin whereas the para isomer has one nuclear spin

Q.4Ionic hydrides are formed byv
  1. (a) halogens
  2. (b) chalogens
  3. (c) inert gases
  4. (d) group one elements
Answer:

(d) group one elements

Q.5Tritium nucleus containsv
  1. (a) 1p + 0n
  2. (b) 2p + 1n
  3. (c) 1p + 2n
  4. (d) none of these
Answer:

(c) 1p + 2n

Q.6Non-stoichiometric hydrides are formed byv
  1. (a) palladium, vanadium
  2. (b) carbon, nickel
  3. (c) manganese, lithium
  4. (d) nitrogen, chlorine
Answer:

(b) carbon, nickel

Q.7Assertion: Permanent hardness of water is removed by treatment with washing soda. Reason: Washing soda reacts with soluble calcium and magnesium chlorides and sulphates in hard water to form insoluble carbonates (a) Both assertion and reason are true and reason is the correct explanation of assertion. (b) Both assertion and reason are true but reason is not the correct explanation of assertion. (c) Assertion is true but reason is false (d) Both assertion and reason are falsev
Answer:

(a) Both assertion and reason are true and reason is the correct explanation of assertion.

Q.8If a body of a fish contains 1.2 g hydrogen in its total body mass, if all the hydrogen is replaced with deuterium then the increase in body weight of the fish will be (a) 1.2 g (b) 2.4 g (c) 3.6 g (d) \(\sqrt{4.8}\) gv
Answer:

(a) 1.2 g

Q.9The hardness of water can be determined by volumetrically using the reagentv
  1. (a) sodium thiosulphate
  2. (b) potassium permanganate
  3. (c) hydrogen peroxide
  4. (d) EDTA
Answer:

(d) EDTA

Q.10The cause of permanent hardness of water is due tov
  1. (a) Ca(HCO 3 ) 2
  2. (b) Mg(HCO 3 ) 2
  3. (c) CaCl 2
  4. (d) MgCO 3
Answer:

(c) CaCl 2

Q.11Zeolite used to soften hardness of water is, hydratedv
  1. (a) Sodium aluminium silicate
  2. (b) Calcium aluminium silicate
  3. (c) Zinc aluminium borate
  4. (d) Lithium aluminium hydride
Answer:

(a) Sodium aluminium silicate

Q.12A commercial sample of hydrogen peroxide marked as 100 volume H 2 O 2, it means thatv
  1. (a) 1 ml of H 2 O 2 will give 100 ml O 2 at STP
  2. (b) 1 L of H 2 O 2 will give 100 ml O 2 at STP
  3. (c) 1 L of H 2 O 2 will give 22.4 L O 2
  4. (d) 1 ml of H 2 O2 will give 1 mole of O 2 at STP
Answer:

(a) 1 ml of H 2 O 2 will give 100 ml O 2 at STP

Q.13When hydrogen peroxide is shaken with an acidified solution of potassium dichromate in presence of ether, the ethereal layer turns blue due to the formation ofv
  1. (a) Cr 2 O 3
  2. (b) CrO 4 2-
  3. (c) CrO(O 2 ) 2
  4. (d) none of these
Answer:

(c) CrO(O 2 ) 2

Q.14For decolourisation of 1mole of acidified KMnO 4, the moles of H 2 O 2 required is (a) \(\frac{1}{2}\) (b) \(\frac{3}{2}\) (c) \(\frac{5}{2}\) (d) \(\frac{7}{2}\)v
Answer:

(c) \(\frac{5}{2}\)

Q.15Volume strength of 1.5 NH 2 O 2 isv
  1. (a) 1.5
  2. (b) 4.5
  3. (c) 16.8
  4. (d) 8.4
Answer:

(d) 8.4

Q.16The hybridisation of oxygen atom is H 2 O and H 2 O 2 are, respectivelyv
  1. (a) sp and sp 3
  2. (b) sp and sp
  3. (c) sp and sp 2
  4. (d) sp 3 and sp 3
Answer:

(d) sp 3 and sp 3

Q.17The reaction H 3 PO 2 + D 2 O → H 2 DPO 2 + HDO indicates that hypo-phosphorus acid isv
  1. (a) tribasic acid
  2. (b) dibasic acid
  3. (c) monobasic acid
  4. (d) none of these
Answer:

(c) monobasic acid
Hypophosphorous acid on reaction with D 2 O, only one hydrogen is replaced by deuterium and hence it is monobasic.

Q.18In solid ice, the oxygen atom is surrounded byv
  1. (a) tetrahedrally by 4 hydrogen atoms
  2. (b) octahedrally by 2 oxygen and 4 hydrogen atoms
  3. (c) tetrahedrally by 2 hydrogen and 2 oxygen atoms
  4. (d) octahedrally by 6 hydrogen atoms
Answer:

(a) tetrahedrally by 4 hydrogen atoms

Q.19The type of H-bonding present in ortho nitrophenol and p-nitrophenol are respectively (a) intermolecular H-bonding and intra molecular f H-bonding (b) intramolecular H-bonding and intermolecular H-bonding (c) intramolecular H – bonding and no H – bonding (d) intramolecular H -bonding and intramolecular H-bondingv
Answer:

(b) intramolecular H-bonding and intermolecular H-bonding

Q.20Heavy water is used as (a) the modulator in nuclear reactions (b) coolant in nuclear reactions (c) both (a) and (b) (d) none of thesev
Answer:

(c) both (a) and (b)
(c) Heavy water is used as a moderator as well as coolant in nuclear reactions.

Q.21Water is av
  1. (a) basic oxide
  2. (b) acidic oxide
  3. (c) amphoteric oxide
  4. (d) none of these
Answer:

(c) amphoteric oxide
Water is an amphoteric oxide.
II. Write brief answer to the following questions:

2II. Write brief answer to the following questions:24 questions
Q.22Explain why hydrogen is not placed with the halogen in the periodic table.v
Answer:
  • Hydrogen resembles alkali metals as well as halogens.
  • Hydrogen resembles more alkali metals than halogens.
  • The electron affinity of hydrogen is much less than that of halogen atoms. Hence the tendency to form hydride ion is low compared to that of halogens.
  • In most of its compounds hydrogen exists in a +1 oxidation state. Therefore it is reasonable to place the hydrogen in group 1 along with alkali metals as shown in the latest periodic table published by IUPAC.
Q.23A cube at 0°C is placed in some liquid water at 0°C, the ice cube sinks – Why?v
Answer:

In ice, each atom is surrounded tetrahedrally by four water molecules through hydrogen bonds. That is, the presence of two hydrogen atoms and two lone pairs of electrons on oxygen atoms in each water molecule allows the formation of a three-dimensional structure.
This arrangement creates an open structure, which accounts for the lower density of ice compared with water at 0°C. While in liquid water, unlike ice where hydrogen bonding occurs over a long-range, the strong hydrogen bonding prevails only in a short-range and therefore the denser packing. Hence, the ice cube sinks in water.

Q.24Discuss the three types of Covalent hydrides.v
Answer:
  • They are the compounds in which hydrogen is attached to another element by sharing electrons.
  • The most common examples of covalent hydrides are methane, ammonia, water, and hydrogen chloride.
  • Molecular hydrides of hydrogen are further classified into three categories,
  • Electron precise (CH 4, C 2 H 6, SiH 4, GeH 4 )
  • Electron-deficient (B 2 H 6 ) and
  • Electron-rich hydrides (NH 3, H 2 O)
  • Since most of the covalent hydrides consist of discrete, small molecules that have relatively weak intermolecular forces, they are generally gases or volatile liquids.
Q.25Predict which of the following hydrides is gas on a solid (a) HCl (b) NaH. Give your reason.v
Answer:

Sodium hydride (NaH) is gas on a solid. NaH is prepared by direct reaction of hydrogen gas with liquid sodium and it has a NaCl crystal structure. It is a tendency of deprotonating in many organic reactions. NaH is used in fuel cells.

Q.26Write the expected formulas for the hydrides of 4th-period elements. What is the trend in the formulas? In what way the first two numbers of the series different from the others?v
Answer:

The expected formulas for the hydrides of 4th-period elements MH 4 (electron precise). M 2 H 6 (electron-deficient) and MH 3 (electron-rich).
The trend in the formula is
* Electron precise hydrides – CH 4 C 2 H 6, SiH 4, GeH 4
* Electron deficient hydrides – B 2 H 6
* Electron rich hydrides – NH 3, H 2 O
The first two members of the series KH, CaH 2 are ionic hydrides whereas the other members of the series CH 4, C 2 H 6, SiH 4, B 2 H 6, NH 3 are covalent hydrides.

Q.27Write the chemical equation for the following reactions.v
Answer:

(i) the reaction of hydrogen with tungsten (VI) oxide NO 3 on heating.
(ii) hydrogen gas and chlorine gas.
(i) Hydrogen can be used to reduce metal oxide into metal. Hydrogen reduces tungsten(VI) oxide into tungsten.
WO 3 + 3H 2 → W + 3H 2 O
(ii) Hydrogen reacts with chlorine at room temperature under light gives hydrogen chloride.
H 2(g) + Cl 2(g) → 2HCl (g)

Q.28Complete the following chemical reactions and classify them into (a) hydrolysis (b) redox (c) hydration reactions. * KMnO 4 + H 2 O 2 → * CrCl 3 + H 2 O → * CaO + H 2 O →v
Answer:

* 2KMnO 4 + 3H 2 O 2 → 2MnO 2 + 2KOH + 3H 2 O + 3O 2(g)
This reaction is a redox reaction.
* CrCl 3 + 6H 2 O 2 → [Cr(H 2 O) 6 )] Cl 3
This reaction is a hydration reaction.
* CaO + H 2 O → Ca(OH) 2
This reaction is a hydrolysis reaction.

Q.29Hydrogen peroxide can function as an oxidizing agent as well as a reducing agent. Substantiate this statement with suitable examples.v
Answer:

Hydrogen peroxide can act both as an oxidizing agent and a reducing agent. Oxidation is usually performed in an acidic medium while the reduction reactions are performed in a basic medium.
In acidic conditions:
H 2 O 2 + 2H + + 2e – → 2H 2 O (E° = +1.77 V)
For example
2FeSO 4 + H 2 SO 4 + H 2 O 2 → Fe 2 (SO 4 ) 3 + 2H 2 O
In basic conditions:
HO 2 – + OH – → O 2 + H 2 O + 2e – (E° = + 0.08V)
For Example,
2 KMnO 4(aq) + 3H 2 O 2(aq) → 2MnO 2 + 2KOH + 2H 2 O + 3O 2 (g)

Q.30Do you think that heavy water can be used for drinking purposes?v
Answer:
  • Heavy water (D 2 O) contains a proton and a neutron. This makes deuterium about twice as heavy as protium, but it is not radioactive. So heavy water is not radioactive.
  • If you drink heavy water, you don’t need to worry about radiation poisoning. But it is not completely safe to drink, because the biochemical reaction in our cells is affected by the difference in the mass of hydrogen atoms.
  • If you drink an appreciable volume of heavy water, you might feel dizzy because of the density difference. It would change the density of the fluid in your inner ear. So it is unlikely to drink heavy water.
Q.31What is the water-gas shift reaction?v
Answer:

The carbon monoxide of the water gas can be converted to carbon dioxide by mixing the gas mixture with more steam at 400°C and passed over a shift converter containing iron/copper catalyst. This reaction is called as water-gas shift reaction.
CO + H 2 O → CO 2 + H 2
The CO 2 formed in the above process is absorbed in a solution of potassium carbonate.
CO 2 + K 2 CO 3 + H 2 O → 2KHCO 3

Q.32Justify the position of hydrogen in the periodic table?v
Answer:

Hydrogen resembles alkali metals in the following aspects.
* Electronic configuration 1s 1 as alkali metals have ns 1.
* Hydrogen forms unipositive H+ ion like alkali metals Na +, K +.
* Hydrogen form halides (HX), oxides (H 2 O) peroxide (H 2 O 2 ) like alkali metals (NaX. Na 2 O, Na 2 O 2 ).
* Hydrogen also acts as a reducing agent like alkali metals. Hydrogen resembles halogens in the following aspects.
* Hydrogen has a tendency to gain one electron to form hydride ion (H – ) as halogens to form halide ion. (X – ).
* Comparing the properties of hydrogen with alkali metals and with halogens, we can conclude that hydrogen resembles more alkali metals. In most of the compounds, hydrogen exists in a +1 oxidation state.
* Therefore, it is reasonable to place the hydrogen in group 1 along with alkali metals as shown in the latest periodic table published by IUPAC.

Q.33What are isotopes? Write the names of isotopes of hydrogen.v
Answer:

Atoms of the same element having same atomic number and different mass number are called isotopes.
Hydrogen has three naturally occurring isotopes, viz., protium ( 1 H 1 or H), deuterium ( 1 H 2 or D) and tritium ( 1 H 3 or T). Protium 1 H 1 is the predominant form (99.985 %) and it is the only isotope that does not contain a neutron. Deuterium, also known as heavy hydrogen, constitutes about 0.015 %. The third isotope, tritium is a radioactive isotope of hydrogen which occurs only in traces (~1 atom per 1018 hydrogen atoms). Due to the existence of these isotopes naturally occurring hydrogen exists as H 2, HD, D 2, HT, T 2, and DT.

Q.34Give the uses of heavy water.v
Answer:
  • Heavy water is used as a moderator in nuclear reactors as it can lower the energies of fast-moving neutrons.
  • D 2 O is commonly used as a tracer to study organic reaction mechanisms and the mechanism of metabolic reactions.
  • It is also used as a coolant in nuclear reactors as it absorbs the heat generated.
Q.35Explain the exchange reactions of deuterium.v
Answer:

When compounds containing hydrogen are treated with D 2 O, hydrogen undergoes an exchange for deuterium. This reaction is known as the exchange reaction of deuterium.
2NaOH + D 2 O → 2NaOD + HOD
HCl + D 2 O → DCl + HOD
NH 4 Cl + 4D 2 O → ND 4 Cl + 4HOD
These exchange reactions are useful in determining the number of ionic hydrogens present in a given compound. For example, when D 2 O is treated with hypo-phosphorus acid only one hydrogen atom is exchanged with deuterium. It indicates that it is a monobasic acid.
H 3 PO 2 + D 2 O → H 2 DPO 2 + HDO
It is also used to prepare some deuterium compounds:
Al 4 C 3 + 12D 2 O → 4Al(OD) 3 + 3CD 4
CaC 2 + 2D 2 O → Ca(OD) 2 + C 2 D 2
Mg 3 N 2 + 6D 2 O → 3Mg(OD) 2 + 2ND 3
Ca 3 P 2 + 6D 2 O → 3Ca(OD) 2 + 2PD 3

Q.36How do you convert parahydrogen into ortho hydrogen?v
Answer:

Para hydrogen can be converted into ortho hydrogen in the following ways:
* By treating with catalysts platinum or iron.
* Bypassing an electric discharge
* By heating > 800°C.
* By mixing with paramagnetic molecules such as O 2, NO, NO 2.
* By treating with nascent/atomic hydrogen.

Q.37Mention the uses of deuterium.v
Answer:
  • Deuterium is used to prepare heavy water which is used as a moderator in nuclear reactors.
  • Deuterium exchange reactions are useful in determining the number of ionic hydrogens present in a given compound.
  • It is also used to prepare some deuterium compounds.
Q.38Explain the preparation of hydrogen using electrolysis.v
Answer:

High purity of hydrogen (>99.9%) is obtained by the electrolysis of water containing traces of acid or alkali or electrolysis of an aqueous solution of sodium hydroxide or potassium hydroxide using a nickel anode and iron cathode. This process is not economical for large-scale production.
At anode: 2OH – → H 2 O + ½ O 2 + 2e –
At cathode: 2H 2 O + 2e – → 2OH – + H 2
Overall reaction: H 2 O → H 2 + 14 O 2

Q.39A group-1 metal (A) which is present ¡n common salt reacts with (B) to give compound (C) in which hydrogen is present in -1 oxidation state. (B) on reaction with gas to give universal solvent (D). The compound (D) reacts with (A) to give (B), a strong base. Identify A, B, C, D, and E. Explain the reactions.v
Answer:

The metal that belongs to Group-1 which is present in common salt is Sodium(A).
The metal reacts with (B) to give compound (C) in which hydrogen is present in a -1 oxidation state.
2Na + H 2 → 2NaH
Hence, the (B) is hydrogen, and (C) is sodium hydride. (B) on reaction with gas to give universal solvent (D).
H 2 + O 2 → 2NaOH
The universal solvent (D) is water. The compound (D) on reaction with (A) to give (E) which is a strong base.
H 2 O + 2Na → 2NaOH
The Compound (E) is Sodium hydroxide.
A – Sodium (Na)
B – Hydrogen (H 2 )
C – Sodium Hydride (NaH)
D – Water (H 2 O)
E – Sodium hydroxide (NaOH)

Q.40An isotope of hydrogen (A) reacts with diatomic molecule of element which occupies group number 16 and period number 2 to give compound (B) is used as a moderatorin nuclear reaction. (A) adds on to a compound (C), which has the molecular formula C 3 H 6 to give (D). Identify A, B, C and D.v
Answer:

An isotope of hydrogen (A) reacts with diatomic molecule of element which occupies group number 16 and period number 2 to give compound (B) is used as a moderator in nuclear reaction.
2D 2 + O 2 → 2D 2 O
The isotope of hydrogen is deuterium(A), diatomic element is oxygen and the compound (B) is heavy water.
(A) adds on to a compound (C), which has the molecular formula C 3 H 6 to give (D).
CH 3 – CH = CH 2 + D 2 → CH 3 – CHD – CH 2 D
(C) (D)
The compound (C) is propene and (D) is 1, 2 deutropropane.
A – Deuterium (D 2 )
B – Heavy water (D 2 O)
C – Propene (CH 3 – CH = CH 2 )
D – 1, 2 deutero propene (CH 3 – CHD – CH 2 D)

Q.41NH 3 has an exceptionally high melting point and boiling point as compared to those of the hydrides of the remaining element of group 15- Explain.v
Answer:

When a hydrogen atom is covalently bonded to a highly electronegative atom such as nitrogen, the bond is polarized. Due to this effect, the polarized hydrogen atom is able to form a weak electrostatic interaction with another electronegative atom present in the vicinity.
This interaction is called hydrogen bonding. Hence, NH 3 has an exceptionally high melting point and boiling point as compared to those of the hydrides of the remaining element of group 15 due to intermolecular hydrogen bonding.

Q.42Why interstitial hydrides have a lower density than the parent metal?v
Answer:
  • d block elements form metallic or interstitial hydrides, on heating with dihydrogen under pressure.
  • Hydrogen atoms being small in size occupy some in the metallic lattice producing distortion without any change in its type.
  • The densities of these hydrides are lower than those of metals from which they are formed since the crystal lattice expands due to the inclusion of dihydrogen.
Q.43How do you expect the metallic hydrides to be useful for hydrogen storage?v
Answer:

Metallic hydrides are usually obtained by hydrogenation of metals and alloys in which hydrogen occupies the interstitial sites (voids). Most of the hydrides are non-stoichiometric with variable composition (TiH 1.5 – 1.8 and PdH 0.6 – 0.8 ), some are relatively light, inexpensive, and thermally unstable which makes them useful for hydrogen storage applications.

Q.44Arrange NH 3, H 2 O, and HF in the order of increasing magnitude of hydrogen bonding and explain the basis for your arrangement.v
Answer:
  • The increasing magnitude of hydrogen bonding among NH 3, H 2 O, and HF is HF > H 2 O > NH 3
  • The extent of hydrogen bonding depends upon electronegativity and the number of hydrogen atoms available for bonding.
  • Among N, F, and O the increasing order of their electronegativities are N < O, H 2 O > NH 3.
Q.45Compare the structures of HO and HO.v
Answer:

Both in gas-phase and liquid-phase, the molecule adopts a skew confirmation due to repulsive interaction of the OH bonds with lone-pairs of electrons on each oxygen atom. Indeed, it is the smallest molecule known to show hindered rotation about a single bond.
H 2 O 2 has a non-polar structure. The molecular dimensions in the gas phase and solid phase differ as shown in the figure. Structurally, H 2 O 2 is represented by the dihydroxyl formula in which the two OH-groups do not lie in the same plane.
One way of explaining the shape of hydrogen peroxide is that the hydrogen atoms would lie on the pages of a partly opened book, and the oxygen atoms along the spine. In the solid phase of the molecule, the dihedral angle reduces to 90.2° due to hydrogen bonding and the O-O-H angle expands from 94.8° to 101.9°. Water has a bent structure and the H-O-H bond angle is 104.5°.