Text Book Back Questions and Answers - Chapter 4 Hydrogen 11th Chemistry Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Hydrogen
Textual Evaluation Solved
I. Choose The Correct Answers
Question 1.
Which of the following statements about hydrogen is incorrect ? (NEET - 2016)
(a) Hydrogen ion, $\mathrm{H}_3 \mathrm{O}^{+}$exists freely in solution.
(b) Dihydrogen acts a,s a reducing agent.
(c) Hydrogen has three isotopes of which tritium is the most common.
(d) Hydrogen never acts as cation in ionic salts.
Answer:
(c) Hydrogen has three isotopes of which tritium is the most common.
Hint:
Correct statement:
Hydrogen has three isotopes of which protium is the most common.
Question 2.
Water gas is
(a) $\mathrm{H}_2 \mathrm{O}_{(\mathrm{g})}$
(b) $\mathrm{CO}+\mathrm{H}_2 \mathrm{O}$
(C) $\mathrm{CO}+\mathrm{H}_2$
(d) $\mathrm{CO}+\mathrm{N}_2$
Answer:
(c) $\mathrm{CO}+\mathrm{H}_2$
Question 3.
Which one of the following statements is incorrect with regard to ortho and para dihydrogen?
(a) They are nuclear spin isomers
(b) Ortho isomer has zero nuclear spin whereas the para isomer has one nuclear spin
(c) The para isomer is favoured at low temperatures
(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
Hints:
Correct statement:
Ortho isomer - one nuclear spin Para isomer - zero nuclear spin
Question 4.
Ionic hydrides are formed by
(a) halogens
(b) chalogens

(c) inert gases
(d) group one elements
Answer:
(d) group one elements
e.g., Sodium hydride $\left(\mathrm{Na}^{+} \mathrm{H}^{-}\right)$
Question 5.
Tritium nucleus contains
(a) $1 p+0 n$
(b) $2 p+1 n$
(c) $1 \mathrm{p}+2 \mathrm{n}$
(d) none of these
Answer:
(c) $1 \mathrm{p}+2 \mathrm{n}$
${ }_1 \mathrm{~T}^3\left(1 \mathrm{e}^{-}, 1 \mathrm{p}, 2 \mathrm{n}\right)$
Question 6.

Non-stoichiometric hydrides are formed by.
(a) palladium, vanadium
(b) carbon, nickel
(c) manganese, lithium
(d) nitrogen, chlorine
Answer:
(a) palladium, vanadium
Question 7.
Assertion : 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 false
Answer:
(a) Both assertion and reason are true and reason is the correct explanation of assertion. $\mathrm{Ca}^{2+}+\mathrm{Na}_2 \mathrm{CO}_3 \rightarrow \mathrm{CaCO}_3 \downarrow+2 \mathrm{Na}^{+}$

Question 8.
If a body of a fish contains $1.2 \mathrm{~g}$ 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) $1.2 \mathrm{~g}$
(b) $2.4 \mathrm{~g}$
(c) $3.6 \mathrm{~g}$
(d) $\sqrt{4.8} \mathrm{~g}$
Answer:
(a) $1.2 \mathrm{~g}$
Hints:
Mass of deuterium $=2 \times$ mass of protium
If all the $1.2 \mathrm{~g}$ hydrogen is replaced with deuterium, the weight will become $2.4 \mathrm{~g}$. Hence the increase in body weight is $(2.4-1.2=1.2 \mathrm{~g})$.
Question 9.
The hardness of water can be determined by volumetrically using the reagent
(a) sodium thio sulphate
(b) potassium permanganate
(c) hydrogen peroxide
(d) EDTA
Answer:
(d) EDTA
Question 10.
The cause of permanent hardness of water is due to
(a) $\mathrm{Ca}\left(\mathrm{HCO}_3\right)_2$
(b) $\mathrm{Mg}\left(\mathrm{HCO}_{3 \mathrm{k}}\right)_3$
(c) $\mathrm{CaCl}_2$
(d) $\mathrm{MgCO}_3$
Answer:
(c) $\mathrm{CaCl}_2$

Hints:
Permanent hardness if water is due to the presence of the chlorides, nitrates and sulphates of $\mathrm{Ca}^{2+}$ and $\mathrm{Mg}^{2+}$ ions.
Question 11.
Zeolite used to soften hardness of water is, hydrated
(a) Sodium aluminium silicate
(b) Calcium aluminium silicate
(c) Zinc aluminium borate
(d) Lithium aluminium hydride
Answer:
(a) Sodium aluminium silicate
Zeolite is sodium aluminium silicate.
$\left(\mathrm{NaAlSi}_2 \mathrm{O}_6 \cdot \mathrm{H}_2 \mathrm{O}\right)$
Question 12.
A commercial sample of hydrogen peroxide marked as 100 volume $\mathrm{H}_2 \mathrm{O}_2$, it means that
(a) $1 \mathrm{ml}$ of $\mathrm{H}_2 \mathrm{O}_2$ will give $100 \mathrm{ml} \mathrm{O}$ at $\mathrm{STP}$
(b) $1 \mathrm{~L}$ of $\mathrm{H}_2 \mathrm{O}_2$ will give $100 \mathrm{ml} \mathrm{O}_2$ at STP
(c) $1 \mathrm{~L}$ of $\mathrm{H}_2 \mathrm{O}_2$ will give $22.4 \mathrm{~L} \mathrm{O}_2$
(d) $1 \mathrm{ml}$ of $\mathrm{H}_2 \mathrm{O}_2$ will give 1 mole of $\mathrm{O}_2$ at STP
Answer:
(a) $1 \mathrm{ml}$ of $\mathrm{H}_2 \mathrm{O}_2$ will give $100 \mathrm{ml} \mathrm{O}_2$ at $\mathrm{STP}$

Question 13.
When hydrogen peroxide is shaken with an acidified solution of potassium dichromate in presence of ether, the ethereal layer turns blue due to the formation of
(a) $\mathrm{Cr}_2 \mathrm{O}_3$
(b) $\mathrm{CrO}_4^{2-}$
(c) $\mathrm{CrO}\left(\mathrm{O}_2\right)_2$
(d) none of these
Answer:
(c) $\mathrm{CrO}\left(\mathrm{O}_2\right)_2 \mathrm{CrO}\left(\mathrm{O}_2\right)_2$
Hints:
$
\mathrm{Cr}_2 \mathrm{O}_7^{2-}+2 \mathrm{H}^{+}+4 \mathrm{H}_2 \mathrm{O}_2 \rightarrow 2 \mathrm{CrO}\left(\mathrm{O}_2\right)_2+5 \mathrm{H}_2 \mathrm{O}
$
Question 14.
For de colorization of 1 mole of acidified $\mathrm{KMnO}_4$, the moles of $\mathrm{H}_2 \mathrm{O}_2$ required is
(a) $\frac{1}{2}$
(b) $\frac{3}{2}$
(c) $\frac{5}{2}$
(d) $\frac{7}{2}$
Answer:
(c) $\frac{5}{2}$
Hints:
$
2 \mathrm{MnO}_4^{-}+5 \mathrm{H}_2 \mathrm{O}_{2(\mathrm{aq})}+6 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+}+5 \mathrm{O}_2+8 \mathrm{H}_2 \mathrm{O}
$
Question 15.
Volume strength of $1.5 \mathrm{~N} \mathrm{H}_2 \mathrm{O}_2$ is .............
(a) 1.5
(b) 4.5
(c) 16.8
(d) 8.4
Answer:
(d) 8.4
Hints:
Volume strength of hydrogen peroxide $=$ Normality of hydrogen peroxide $\times 5.6=1.5 \times 5.6=8.4$
$
2 \mathrm{H}_2 \mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2 \uparrow
$
$
(2 \times 34 \mathrm{~g}) \quad(22.4 \text { litres) }
$
Volume strength of hydrogen peroxide
$
=\frac{\text { Normality } \times \text { Equivalent weight of } \mathrm{H}_2 \mathrm{O}_2 \times 2.24}{68}
$
$=$ Normality $\times \frac{17 \times 22.4}{68}$
Volume strength of hydrogen peroxide $=$ Normality x 5.6

Question 16.
The hybridization of oxygen atom is $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{H}_2 \mathrm{O}_2$ are respectively
(a) $s p$ and $\mathrm{sp}^3$
(b) $\mathrm{sp}$ and $\mathrm{sp}$
(c) $\mathrm{sp}$ and $\mathrm{sp}^2$
(d) $\mathrm{sp} 3$ and $\mathrm{sp}^3$
Answer:
(d) $\mathrm{sp}^3$ and $\mathrm{sp}^3$
Question 17.
The reaction $\mathrm{H}_3 \mathrm{PO}_2+\mathrm{D}_2 \mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{DPO}_2+\mathrm{HDO}$ indicates that hypo-phosphorus acid is
(a) tri basic acid
(b) di basic acid
(c) mono basic acid
(d) none of these

Answer:
(c) mono basic acid
Hints:
Hypophosphorus acid on reaction with $\mathrm{D}_2 \mathrm{O}$, only one hydrogen is replaced $\mathrm{P}$ by deuterium and hence it is mono basic.
Question 18.
In solid ice, oxygen atom is surrounded
(a) tetrahedrally by 4 hydrogen atoms
(b) octahedrally by 2 oxygen and 4 hydrogen atoms
(c) tetrahedrally by 2 hydrogen and 2 oxygen atoms
(d) octahedrally by 6 hydrogen atoms
Answer:
(a) tetrahedrally by 4 hydrogen atoms
Question 19.
The type of H-bonding present in ortho nitro phenol and p-nitro phenol are respectively
(a) inter molecular $\mathrm{H}$-bonding and intra molecular $\mathrm{H}$-bonding
(b) intra molecular $\mathrm{H}$-bonding and inter molecular $\mathrm{H}$-bonding
(c) intra molecular $\mathrm{H}$ - bonding and no $\mathrm{H}$ - bonding
(d) intra molecular $\mathrm{H}$ - bonding and intra molecular $\mathrm{H}$ - bonding

Answer:
(A) intra molecular $\mathrm{H}$-bonding and inter molecular $\mathrm{H}$-bonding
Question 20.
Heavy water is used as .............
(a) modulator in nuclear reactions
(b) coolant in nuclear reactions
(c) both (a) and (b)
(d) none of these
Answer:
(c) both (a) and (A)
Hints:
Heavy water is used as moderator as well as coolant in nuclear reactions.
Question 21.
Water is a
(a) basic oxide
(b) acidic oxide
(c) amphoteric oxide
(d) none of these
Answer:
(c) amphoteric oxide
II. Write brief answer to the following questions
Question 22.
Explain why hydrogen is not placed with the halogen in the periodic table.
Answer:
- Hydrogen resembles alkali metals as well as halogens.
- Hydrogen resembles more alkali metals than halogens.

- Electron affinity of hydrogen is much less than that of halogen atom. Hence the tendency to form hydride ion is low compared to that of halogens.
- In most of its compounds hydrogen exists in +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.
Question 23.
the cube at $0^{\circ} \mathrm{C}$ is placed in some liquid water at $0^{\circ} \mathrm{C}$, the ice cube sinks - Why ?
Answer:
- In an ice cube, each atom is surrounded tetrahedrally by four water molecules through hydrogen bond and its density is low.
- Liquid water at $0^{\circ} \mathrm{C}$ has the density as $999.82 \mathrm{~kg} / \mathrm{cm}^3$. Maximum density is attained by water only at $4^{\circ} \mathrm{C}$ as $1000 \mathrm{~kg} / \mathrm{cm}^3$.
- When the temperature changed from $4^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$, the density of water decreases rather than increases. This is called anomalous expansion of water.
- The reason for this phenomenon lies in the structure of ice lattice and hydrogen bonding in Water.
- At $0^{\circ} \mathrm{C}$, ice cube sinks in liquid water at $0^{\circ} \mathrm{C}$ because of the lesser density and greater volume of water.
Question 24.
Discuss the three types of Covalent hydrides.
Answer:
1. They are the compounds in which hydrogen is attached to another element by sharing of electrons.
2. The most common examples of covalent hydrides are methane, ammonia, water and hydrogen chloride.
3. Molecular hydrides of hydrogen are further classified into three categories as,
- Electron precise $\left(\mathrm{CH}_4, \mathrm{C}_2 \mathrm{H}_6, \mathrm{SiH}_4, \mathrm{GeH}_4\right)$
- Electron-deficient $\left(\mathrm{B}_2 \mathrm{H}_6\right)$ and
- Electron-rich hydrides $\left(\mathrm{NH}_3, \mathrm{H}_2 \mathrm{O}\right)$
4. Since most of the covalent hydrides consist of discrete, small molecules that have relatively weak intermolecular forces, they are generally gases or volatile liquids.
Question 25.
Predict which of the following hydrides is a gas on a solid (a) $\mathrm{HCl}$ (b) $\mathrm{NaH}$. Give your reason.

Answer:

- At room temperature, $\mathrm{HCl}$ is a colourless gas and the solution of $\mathrm{HCl}$ in water is called hydrochloric acid and it is in liquid state.
- Sodium hydride $\mathrm{NaH}$ is an ionic compound and it is made of sodium cations $\left(\mathrm{Na}^{+}\right)$and hydride $\left(\mathrm{H}^{-}\right)$anions. It has the octahedral crystal structure. It is an alkali metal hydride.
Question 26.
Write 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?
Answer:
The expected formulas for the hydrides of 4th period elements $\mathrm{MH}_4$ (electron precise). $\mathrm{M}_2 \mathrm{H}_6$ (electron deficient) and $\mathrm{MH}_3$ (electron rich).
The trend in formula is -
- Electron precise hydrides $-\mathrm{CH}_4 \mathrm{C}_2 \mathrm{H}_6, \mathrm{SiH}_4, \mathrm{GeH}_4$
- Electron deficient hydrides $-\mathrm{B}_2 \mathrm{H}_6$
- Electron rich hydrides $-\mathrm{NH}_3, \mathrm{H}_2 \mathrm{O}$
The first two members of the series $\mathrm{KH}, \mathrm{CaH}_2$ are ionic hydrides whereas the other members of the series $\mathrm{CH}_4, \mathrm{C}_2 \mathrm{H}_6, \mathrm{SiH}_4, \mathrm{~B}_2 \mathrm{H}_6, \mathrm{NH}_3$ are covalent hydrides.

Question 27.
Write chemical equation for the following reactions.
1. reaction of hydrogen with tungsten (VI) oxide $\mathrm{NO}_3$ on heating.
2. hydrogen gas and chlorine gas.
Answer:
1. $3 \mathrm{H}_2+\mathrm{WO}_2 \rightarrow \mathrm{W}+3 \mathrm{H}_2 \mathrm{O}$
Hydrogen reduces tungsten (VI) oxide. $\mathrm{WO}_3$ to tungsten at high temperature.
2. $\mathrm{H}_2+\mathrm{Cl}_2 \rightarrow 2 \mathrm{HCl}$ (Hydrogen Chloride)
Hydrogen reacts with chlorine at room temperature under light to give hydrogen chloride.
Question 28.
Complete the following chemical reactions and classify them in to (a) hydrolysis (b) redox (c) hydration reactions.
1. $\mathrm{KMnO}_4+\mathrm{H}_2 \mathrm{O}_2 \rightarrow$
2. $\mathrm{CrCl}_3+\mathrm{H}_4 \mathrm{O} \rightarrow$
3. $\mathrm{CaO}+\mathrm{H}_2 \mathrm{O} \rightarrow$
Answer:
1. $2 \mathrm{KMnO}_4+3 \mathrm{H}_2 \mathrm{O}_2 \rightarrow 2 \mathrm{MnO}_2+2 \mathrm{KOH}+3 \mathrm{H}_2 \mathrm{O}+3 \mathrm{O}_2(\mathrm{~g})$
This reaction is a redox reaction.
2. $\left.\mathrm{CrCl}_3+6 \mathrm{H}_2 \mathrm{O}_2 \rightarrow\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right)\right] \mathrm{Cl}_3$
This reaction is a hydration reaction.
3. $\mathrm{CaO}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Ca}(\mathrm{OH})_2$
This reaction is a hydrolysis reaction.
Question 29.
Hydrogen peroxide can function as an oxidizing agent as well as reducing agent. Substantiate this statement with suitable examples.
Answer:
Hydrogen peroxide can function as an oxidizing agent as well as reducing agent.
- $\mathrm{H}_2 \mathrm{O}_2$ act as oxidizing agent in acidic medium. For example,

$\text { - } \mathrm{H}_2 \mathrm{O}_2 \text { act as reducing agent in basic medium. For example, }$

Question 30.
Do you think that heavy water can be used for drinking purposes?
Answer:
- Heavy water $\left(\mathrm{D}_2 \mathrm{O}\right)$ 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 are 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 fluid in your inner ear. So it is unlikely to drink heavy water.
Question 31.
What is water-gas shift reaction?
Answer:
The carbon monoxide of water gas can be converted to carbon dioxide by mixing the gas mixture with more steam at $400^{\circ} \mathrm{C}$ and passed over a shift converter containing iron/copper catalysts. This reaction is called water-gas shift reaction.
$
\mathrm{CO}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{CO}_2+\mathrm{H}_2 \uparrow
$
Question 32.
Justify the position of hydrogen in the periodic table?
Answer:
Hydrogen resembles alkali metals in the following aspects.
1. Electronic configuration $\mathrm{Is}^1$ as alkali metals have $\mathrm{ns}^1$.
2. Hydrogen forms unipositive $\mathrm{H}+$ ion like alkali metals $\mathrm{Na}^{+}, \mathrm{K}^{+}$.
3. Hydrogen form halides $(\mathrm{HX})$, oxides $\left(\mathrm{H}_2 \mathrm{O}\right)$ peroxide $\left(\mathrm{H}_2 \mathrm{O}_2\right)$ like alkali metals $(\mathrm{NaX}$. $\mathrm{Na}_2 \mathrm{O}, \mathrm{Na}_2 \mathrm{O}_2$ ).
4. Hydrogen also acts as reducing agent like alkali metals. Hydrogen resembles halogens in the following aspects.
5. Hydrogen has a tendency to gain one electron to form hydride ion $\left(\mathrm{H}^{-}\right)$as halogens to form halide ion. (X).

6. 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 exist in +1 oxidation state.
7. 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.
Question 33.
What are isotopes? Write the names of isotopes of hydrogen.
Answer:
1. Isotopes are atoms of the same element that have the same atomic number but having different mass numbers (or) Isotopes are atoms with the same number of protons and electrons but differ in number of neutrons.
2. Hydrogen has three naturally occuring isotopes namely Protium $\left({ }_1 \mathrm{H}^1\right)$, Deuterium $\left({ }_1 \mathrm{H}^2\right)$ and Tritium $\left({ }_1 \mathrm{H}^3\right)$

Question 34.
Give the uses of heavy water.
Answer:
1. Heavy water is used as moderator in nuclear reactors as it can lower the energies of fast moving neutrons.
2. $\mathrm{D}_2 \mathrm{O}$ is commonly used as an tracer to study organic reaction mechanisms and mechanism of metabolic reactions.
3. It is also used as a coolant in nuclear reactors as it absorbs the heat generated.
Question 35 .
Explain the exchange reactions of deuterium.
Answer:
Deuterium can replace reversibly hydrogen in compounds either partially or completely depending upon the reaction conditions. These reactions occur in the presence of deuterium.

Question 36.
How do you convert para hydrogen into ortho hydrogen?
Answer:
Para hydrogen can be converted into ortho hydrogen by the following ways:
- By treating with catalysts platinum or iron.
- By passing an electric discharge
- By heating $>800^{\circ} \mathrm{C}$.
- By mixing with paramagnetic molecules such as $\mathrm{O}_2, \mathrm{NO}, \mathrm{NO}_2$.
- By treating with nascent/atomic hydrogen.
Question 37.
Mention the uses of deuterium.
Answer:
- Deuterium is used as a tracer element.
- Deuterium is used to study the movement of ground water by isotopic effect.
Question 38.
Explain preparation of hydrogen using electrolysis.
Answer:
High purity of hydrogen $(>99.9 \%)$ is obtained by the electrolysis of water containing traces of acid or alkali or electrolysis of 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 : $2 \mathrm{OH}^{-} \rightarrow \mathrm{H}_2 \mathrm{O}+1 / 2 \mathrm{O}_2+2 \mathrm{e}^{-}$
At cathode : $2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{OH}^{-}+\mathrm{H}_2$
Overall reaction : $\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2+14 \mathrm{O}_2$
Question 39.
A groups metal (A) which is present in common salt reacts with (B) to give compound (C) in which hydrogen is present in -1 oxidation state. (B) on reaction with a gas $(C)$ to give universal solvent (D). The compound (D) on reacts with (A) to give (B), a strong base. Identify A, B, C, D and E. Explain the reactions.
Answer:
1.Group (1) metal (A) is present in common salt $\mathrm{NaCl}$. So, (A) is sodium $-\mathrm{Na}$.
2. Sodium reacts with hydrogen (B) to give sodium hydride - $\mathrm{NaH}$ (C) in which hydrogen is in -1 oxidation state.

3. Hydrogen on reaction with oxygen $\left(\mathrm{O}_2\right)$ gas which is $(\mathrm{C})$ to give a universal solvent water (D).

4. Water (D) reacts with sodium metal (A) to give a strong base sodium hydroxide $\mathrm{NaOH}$ which is (E).

Question 40.
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 modulator in nuclear reaction. (A) adds on to a compound (C), which has the molecular formula $\mathrm{C}_3 \mathrm{H}_6$ to give (D). Identify $\mathrm{A}, \mathrm{B}$, $\mathrm{C}$ and $\mathrm{D}$.
Answer:
1. An isotope of hydrogen Deuterium (A) reacts with diatomic molecule of element belongs to group number 16 and period number 2 oxygen $\mathrm{O}_2$ to give a compound (B) which is heavy water $\mathrm{D}_2 \mathrm{O} . \mathrm{D}_2 \mathrm{O}$ is used as a moderator in nuclear reaction.

$\text { 2. Deuterium reacts with } \mathrm{C}_3 \mathrm{H}_6 \text { propane (C) to give Deutero propane } \mathrm{C}_2 \mathrm{D}_6 \text { (D). }$

Question 41.
$\mathrm{NH}_3$ has exceptionally high melting point and boiling point as compared to those of the hydrides of the remaining element of group 15- Explain.

Answer:
1. $\mathrm{NH}_3$ has exceptionally high melting point and boiling point due to hydrogen bonding between $\mathrm{NH}_3$ molecules.
2. Each molecule can form a maximum of 4 hydrogen bonds but on average 1 hydrogen bond per molecule as there is only one lone pair on $\mathrm{NH}_3$ available for hydrogen bonding.
3. Hydrogen bonding is strong intermolecular attraction as $\mathrm{H}$ on $\mathrm{NH}_3$ acts like a proton due to partial positive on it whole $\mathrm{N}$ has the partial negative charge. Thus when the very polarized
$\mathrm{H}$ comes close to a $\mathrm{N}$ atom in another $\mathrm{NH}_3$ molecule, a very strong hydrogen bond is formed.
4. Due to much strong intermolecular interactions compared to weaker permanent dipoledipole interactions between other $\mathrm{XH}_3$ molecules in group 15, large amount of energy are required to overcome the forces, giving it the highest boiling point and highest melting point.

Question 42 .
Why interstitial hydrides have a lower density than the parent metal.
Answer:
- d block elements form metallic or interstitial hydrides, on heating with dihydrogen under pressure.
- Hydrogen atom 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.
Question 43.
How do you expect the metallic hydrides to be useful for hydrogen storage?
Answer:
In metallic hydrides, hydrogen is adsorbed as $\mathrm{H}$-atoms. Due to the adsorption of $\mathrm{H}$ atoms the metal lattice expands and become unstable. Thus, when metallic hydride is heated, it decomposes to form hydrogen and finely divided metal. The hydrogen evolved can be used as fuel.
Question 44.
Arrange $\mathrm{NH}_3, \mathrm{H}_2 \mathrm{O}$ and $\mathrm{HF}$ in the order of increasing magnitude of hydrogen bonding and explain the basis for your arrangement.
Answer:
- Increasing magnitude of hydrogen bonding among $\mathrm{NH}_3, \mathrm{H}_2 \mathrm{O}$ and $\mathrm{HF}$ is $\mathrm{HF}>\mathrm{H}_2 \mathrm{O}>\mathrm{NH}_3$
- The extent of hydrogen bonding depends upon electronegativity and the number of hydrogne atoms available for bonding.
- Among N, F nd $O$ the increasing order of their electronegativities are $\mathrm{N}<\mathrm{O} \mathrm{H}_2 \mathrm{O}>\mathrm{NH}_3$.
Question 45.
Compare the structures of $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{H}_2 \mathrm{O}_2$.
Answer:
In water, $\mathrm{O}$ is $\mathrm{sp}^3$ hybridized. Due to stronger lone pair-lone pair repulsions than bond pair-bond pair repulsions, the $\mathrm{HOH}$ bond angle decreases from $109.5^{\circ}$ to $104.5^{\circ}$. Thus water molecule has a bent structure.

$\mathrm{H}_2 \mathrm{O}_2$ has a non-planar structure. The $0-\mathrm{H}$ bonds are in different planes. Thus, the structure of $\mathrm{H}_2 \mathrm{O}_2$ is like an open book.