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Physics Chemistry Biology

Class 12th (Chemistry) Chapters
1. The Solid State 2. Solutions 3. Electrochemistry
4. Chemical Kinetics 5. Surface Chemistry 6. General Principles And Processes Of Isolation Of Elements
7. The P-Block Elements 8. The D-And F-Block Elements 9. Coordination Compounds
10. Haloalkanes And Haloarenes 11. Alcohols, Phenols And Ethers 12. Aldehydes, Ketones And Carboxylic Acids
13. Amines 14. Biomolecules 15. Polymers
16. Chemistry In Everyday Life



Chapter 7 The P – Block Elements



Group 15 Elements

The p-block elements are located in Groups 13 to 18 of the periodic table. Their general valence shell electronic configuration is ns$^2$np$^{1-6}$ (except Helium with 1s$^2$). Properties are influenced by atomic size, ionisation enthalpy, electron gain enthalpy, electronegativity, and the presence/absence of d or f orbitals.

Group 15 includes Nitrogen (N), Phosphorus (P), Arsenic (As), Antimony (Sb), Bismuth (Bi), and Moscovium (Mc).


Occurrence

Table 7.1 summarises properties of N, P, As, Sb, Bi.

Property N P As Sb Bi
Atomic number715335183
Atomic mass/g mol$^{-1}$14.0130.9774.92121.75208.98
Electronic configuration[He]2s$^2$2p$^3$[Ne]3s$^2$3p$^3$[Ar]3d$^{10}$4s$^2$4p$^3$[Kr]4d$^{10}$5s$^2$5p$^3$[Xe]4f$^{14}$5d$^{10}$6s$^2$6p$^3$
Ionisation enthalpy
( $\Delta_i H$/(kJ mol$^{-1}$)
I14021012947834703
II28561903179815951610
III45772910273624432466
Electronegativity3.02.12.01.91.9
Covalent radius/pma70110121141148
Ionic radius/pm (E$^{3-}$)171212222
Ionic radius/pm (E$^{3+}$)76103
Melting point/K63*317d1089e904544
Boiling point/K77.2*554d888f18601837
Density/[g cm$^{-3}$(298 K)]0.879g1.8235.778h6.6979.808

(a E$^{III}$ single bond (E = element); b E$^{3–}$; c E$^{3+}$; d White phosphorus; e Grey $\alpha$-form at 38.6 atm; f Sublimation temperature; g At 63 K; h Grey $\alpha$-form; * Molecular N$_2$).


Electronic Configuration

General valence shell configuration: ns$^2$np$^3$. The completely filled s-orbital and half-filled p-orbitals contribute to extra stability of the electronic configuration.


Atomic And Ionic Radii

Covalent and ionic radii increase down the group due to increased number of electron shells. Significant increase from N to P. Smaller increase from As to Bi due to presence of filled d and f orbitals in heavier elements.


Ionisation Enthalpy

Ionisation enthalpy decreases down the group due to increasing atomic size. Group 15 elements have higher ionisation enthalpy values compared to Group 14 elements in the same period due to their extra stable half-filled p-orbitals. Order of successive ionisation enthalpies: $\Delta_i H_1 < \Delta_i H_2 < \Delta_i H_3$.


Electronegativity

Electronegativity generally decreases down the group with increasing atomic size. Difference is less pronounced among heavier elements.


Physical Properties

All elements are polyatomic. Nitrogen is a diatomic gas (N$_2$). Others are solids. Metallic character increases down the group (N, P are non-metals; As, Sb are metalloids; Bi is a metal), due to decreasing ionisation enthalpy and increasing atomic size. Boiling points generally increase down the group. Melting point increases up to As, then decreases to Bi. All elements except N show allotropy.


Chemical Properties

Common oxidation states: –3, +3, +5. Stability of –3 state decreases down the group (Bi rarely shows –3). Stability of +5 state decreases down the group (due to inert pair effect), while +3 state stability increases (Bi(V) is rare). Nitrogen also shows +1, +2, +4 with oxygen. Nitrogen's maximum covalency is 4 (limited by s and p orbitals). Heavier elements can expand covalency beyond 4 due to vacant d orbitals (e.g., PF$_6^{-}$).

Anomalous properties of nitrogen:

(i) Reactivity towards hydrogen: Form hydrides of type $\textsf{EH}_3$ ($\textsf{E} = \textsf{N}, \textsf{P}, \textsf{As}, \textsf{Sb}, \textsf{Bi}$). Properties of hydrides (Table 7.2):

Property NH$_3$ PH$_3$ AsH$_3$ SbH$_3$ BiH$_3$
Melting point/K195.2139.5156.7185
Boiling point/K238.5185.5210.6254.6290
(E–H) Distance/pm101.7141.9151.9170.7
HEH angle ($^\circ$)107.893.691.891.3
$\Delta_f H$/kJ mol$^{-1}$–46.113.466.4145.1278
$\Delta_{diss}H$(E–H)/kJ mol$^{-1}$389322297255

(ii) Reactivity towards oxygen: Form $\textsf{E}_2\text{O}_3$ and $\textsf{E}_2\text{O}_5$ oxides. Acidity decreases down the group. $\textsf{N}_2\text{O}_3$, $\textsf{P}_2\text{O}_3$ are acidic; $\textsf{As}_2\text{O}_3$, $\textsf{Sb}_2\text{O}_3$ are amphoteric; $\textsf{Bi}_2\text{O}_3$ is basic. Higher oxidation state oxide is more acidic than lower one.

(iii) Reactivity towards halogens: Form $\textsf{EX}_3$ and $\textsf{EX}_5$. N forms only $\textsf{NF}_3$ (no d orbitals). $\textsf{EX}_5$ are more covalent than $\textsf{EX}_3$ (due to higher +5 ox. state). Trihalides except $\textsf{NF}_3$ and $\textsf{BiF}_3$ are covalent and stable. $\textsf{BiF}_3$ is ionic.

(iv) Reactivity towards metals: React to form binary compounds with –3 oxidation state (e.g., $\textsf{Ca}_3\text{N}_2$, $\textsf{Ca}_3\text{P}_2$).

Intext Questions

7.1 Why are pentahalides of P, As, Sb and Bi more covalent than their trihalides?

7.2 Why is BiH$_3$ the strongest reducing agent amongst all the hydrides of Group 15 elements ?

Answer:

7.1 In pentahalides of P, As, Sb, and Bi (e.g., PCl$_5$), the central atom is in the +5 oxidation state, while in trihalides (e.g., PCl$_3$), the central atom is in the +3 oxidation state. According to Fajan's rules, a higher positive charge (or oxidation state) on the cation increases its polarising power. When the central atom is in a higher positive oxidation state (+5), it can polarise the electron cloud of the halide anion more effectively. This increased polarisation leads to a greater degree of covalent character in the bond. Therefore, pentahalides are generally more covalent than the corresponding trihalides for these elements.

7.2 The reducing character of the hydrides of Group 15 elements ($\textsf{EH}_3$) increases down the group from $\textsf{NH}_3$ to $\textsf{BiH}_3$. This trend is directly related to the decrease in the stability of the E-H bond down the group. As the atomic size of the central element (E) increases down the group (from N to Bi), the E-H bond length increases, and the bond dissociation enthalpy decreases. This means the E-H bond becomes weaker and is easier to break down the group. A weaker E-H bond leads to a greater tendency to donate hydrogen atoms or hydride ions, which is characteristic of reducing agents. Since the Bi-H bond in $\textsf{BiH}_3$ is the weakest among all the Group 15 hydrides, $\textsf{BiH}_3$ is the most easily decomposed and thus acts as the strongest reducing agent.


Example 7.1 Though nitrogen exhibits +5 oxidation state, it does not form pentahalide. Give reason.

Answer:

Nitrogen is the first element in Group 15 and is in the second period (n=2). The valence shell of nitrogen has only s and p orbitals (2s and 2p). It does not possess vacant d orbitals in its valence shell. The maximum covalency an element can exhibit is related to the number of valence orbitals available for bonding. With only one 2s and three 2p orbitals, nitrogen has a total of four valence orbitals available for bonding. Therefore, nitrogen can form a maximum of four covalent bonds, limiting its covalency to four. In pentahalides (like PCl$_5$), the central atom forms five covalent bonds, requiring five valence orbitals and thus expanding its octet (covalency > 4). Since nitrogen cannot expand its octet due to the absence of d orbitals, it cannot form pentahalides with five covalent bonds.


Example 7.2 PH$_3$ has lower boiling point than NH$_3$. Why?

Answer:

The boiling point of a substance depends on the strength of the intermolecular forces between its molecules. In the liquid and solid states, ammonia ($\textsf{NH}_3$) molecules are associated through strong hydrogen bonding. This occurs because nitrogen is highly electronegative and small, allowing hydrogen atoms bonded to nitrogen to form hydrogen bonds with nitrogen atoms of neighboring molecules.

Phosphorus, on the other hand, is less electronegative and larger than nitrogen. As a result, phosphine ($\textsf{PH}_3$) molecules do not form significant hydrogen bonds with each other in the liquid state. The intermolecular forces in liquid $\textsf{PH}_3$ are primarily weak van der Waals forces (specifically dipole-dipole and dispersion forces, as $\textsf{PH}_3$ is polar, but the dipole moment is small). Since hydrogen bonding is much stronger than van der Waals forces, more energy (higher temperature) is required to overcome the intermolecular forces in liquid $\textsf{NH}_3$ compared to liquid $\textsf{PH}_3$ to cause boiling. Therefore, $\textsf{PH}_3$ has a lower boiling point than $\textsf{NH}_3$.


Dinitrogen

Dinitrogen ($\textsf{N}_2$) is the elemental form of nitrogen, a diatomic gas with a triple bond ($\textsf{N}\equiv\textsf{N}$).


Preparation


Properties

Uses: Manufacture of ammonia and other nitrogen compounds. Provides inert atmosphere (iron, steel industry, chemical reactions). Liquid $\textsf{N}_2$ as refrigerant (preserving biological materials, food, cryosurgery).

Example 7.3 Write the reaction of thermal decomposition of sodium azide.

Answer:

Thermal decomposition of sodium azide ($\textsf{NaN}_3$) is used to obtain very pure dinitrogen gas.

$2\textsf{NaN}_3\text{(s)} \xrightarrow{\text{Heat}} 2\textsf{Na(s)} + 3\textsf{N}_2\text{(g)}$


Intext Question

7.3 Why is N$_2$ less reactive at room temperature?

Answer:

Nitrogen ($\textsf{N}_2$) is less reactive at room temperature primarily because of the very strong triple bond ($\textsf{N}\equiv\textsf{N}$) between the two nitrogen atoms. This triple bond consists of one sigma ($\sigma$) bond and two pi ($\pi$) bonds. The bond dissociation enthalpy of the $\textsf{N}\equiv\textsf{N}$ bond is very high (941.4 kJ mol$^{-1}$). Breaking this strong bond requires a large amount of energy. At room temperature, most collisions between $\textsf{N}_2$ molecules and other substances do not provide sufficient energy to break this bond, making $\textsf{N}_2$ kinetically inert under normal conditions. Reactivity increases significantly at higher temperatures where molecules possess more energy.



Ammonia

Ammonia ($\textsf{NH}_3$) is an important compound of nitrogen.


Preparation

Flow chart for the manufacture of ammonia by Haber's process

Properties

Structure of ammonia molecule showing trigonal pyramidal shape and lone pair on nitrogen

Uses: Manufacture of nitrogenous fertilisers ($\textsf{NH}_4\text{NO}_3$, urea), nitric acid. Liquid $\textsf{NH}_3$ as refrigerant.

Example 7.4 Why does NH$_3$ act as a Lewis base ?

Answer:

A Lewis base is a substance that can donate a lone pair of electrons. The nitrogen atom in the ammonia ($\textsf{NH}_3$) molecule has one lone pair of electrons (in its sp$^3$ hybrid orbital) that is not involved in bonding. This lone pair is available for donation to electron-deficient species (Lewis acids), such as metal ions or protons ($\textsf{H}^{+}$). By donating its lone pair, ammonia forms a coordinate covalent bond. Therefore, $\textsf{NH}_3$ acts as a Lewis base.


Intext Questions

7.4 Mention the conditions required to maximise the yield of ammonia.

7.5 How does ammonia react with a solution of Cu$^{2+}$?

Answer:

7.4 Ammonia is manufactured by the Haber's process ($\textsf{N}_2\text{(g)} + 3\textsf{H}_2\text{(g)} \rightleftharpoons 2\textsf{NH}_3\text{(g)}; \Delta_f H^0 = -46.1$ kJ mol$^{-1}$). This is an exothermic and reversible reaction with a decrease in the number of moles of gas. According to Le Chatelier's principle, the following conditions favour the forward reaction (formation of ammonia) and maximise the yield:

  1. High pressure: The forward reaction involves a decrease in volume (4 moles of gas reactants $\to$ 2 moles of gas products). High pressure shifts the equilibrium towards the side with fewer moles of gas, which is the product side (ammonia). Optimum pressure is about 200 atm.
  2. Moderate temperature: The forward reaction is exothermic ($\Delta H^0 < 0$). Low temperature favours exothermic reactions. However, the reaction rate is slow at very low temperatures. A compromise is reached, and a moderate temperature of about 700 K is used to achieve a reasonable reaction rate and equilibrium yield.
  3. Use of a catalyst: A catalyst (e.g., iron oxide with $\textsf{K}_2\text{O}$ and $\textsf{Al}_2\text{O}_3$) increases the rate of both forward and backward reactions equally, helping the system reach equilibrium faster without changing the equilibrium position. This is essential for commercial production to achieve high yields in a reasonable time.
  4. Continuous removal of ammonia: Removing ammonia as it is formed shifts the equilibrium to the right, increasing the yield.

So, the conditions required to maximise the yield of ammonia are high pressure, moderate temperature, and the use of a catalyst.

7.5 Ammonia ($\textsf{NH}_3$) acts as a Lewis base due to the presence of a lone pair of electrons on the nitrogen atom. When ammonia reacts with a solution of $\textsf{Cu}^{2+}$ ions (e.g., copper sulfate solution), it forms a deep blue coloured complex ion, tetraamminecopper(II) ion, $[\textsf{Cu(NH}_3)_4]^{2+}$. The reaction is:

$\textsf{Cu}^{2+}\text{(aq)} + 4 \textsf{NH}_3\text{(aq)} \rightleftharpoons [\textsf{Cu(NH}_3)_4]^{2+}\text{(aq)}$

Initially, adding dilute ammonia solution to $\textsf{Cu}^{2+}$ solution may cause the precipitation of copper(II) hydroxide, which is a pale blue precipitate: $\textsf{Cu}^{2+}\text{(aq)} + 2\textsf{OH}^{-}\text{(aq)} \to \textsf{Cu(OH)}_2\text{(s)}$. However, adding excess ammonia causes the precipitate to dissolve, forming the soluble deep blue complex ion:

$\textsf{Cu(OH)}_2\text{(s)} + 4\textsf{NH}_3\text{(aq)} \rightleftharpoons [\textsf{Cu(NH}_3)_4]^{2+}\text{(aq)} + 2\textsf{OH}^{-}\text{(aq)}$



Oxides Of Nitrogen

Nitrogen forms a variety of oxides with different oxidation states (+1 to +5). Their names, formulas, methods of preparation, and properties are summarised in Table 7.3.

Oxidation state of nitrogen Name Formula Common methods of preparation Physical appearance and chemical nature
+1Dinitrogen oxide
[Nitrogen(I) oxide]
N$_2$ONH$_4$NO$_3$ $\xrightarrow{\text{Heat}}$ N$_2$O + 2H$_2$Ocolourless gas, neutral
+2Nitrogen monoxide
[Nitrogen(II) oxide]
NO2NaNO$_2$ + 2FeSO$_4$ + 3H$_2$SO$_4$ $\to$ Fe$_2$(SO$_4$)$_3$ + 2NaHSO$_4$ + 2H$_2$O + 2NOcolourless gas, neutral
+3Dinitrogen trioxide
[Nitrogen(III) oxide]
N$_2$O$_3$2NO + N$_2$O$_4$ $\xrightarrow{\text{250K}}$ 2N$_2$O$_3$blue solid, acidic
+4Nitrogen dioxide
[Nitrogen(IV) oxide]
NO$_2$2Pb(NO$_3$)$_2$ $\xrightarrow{\text{Heat}}$ 4NO$_2$ + 2PbO + O$_2$brown gas, acidic
+4Dinitrogen tetroxide
[Nitrogen(IV) oxide]
N$_2$O$_4$2NO$_2$ $\xrightarrow{\text{Cool}}$ N$_2$O$_4$ $\xrightarrow{\text{Heat}}$ 2NO$_2$colourless solid/liquid, acidic
+5Dinitrogen pentoxide
[Nitrogen(V) oxide]
N$_2$O$_5$4HNO$_3$ + P$_4$O$_{10}$ $\to$ 4HPO$_3$ + 2N$_2$O$_5$colourless solid, acidic

Nitrogen oxides often exhibit resonance structures and multiple bonds. Their Lewis structures and bond parameters are provided (Table 7.4).

Name Lewis dot structure Main resonance structures Bond parameters
Dinitrogen oxide
Lewis structure of N2O
Resonance structures of N2O
N-N bond length 112 pm
N-O bond length 119 pm
Linear shape
Nitrogen monoxide
Lewis structure of NO
Resonance structures of NO
N-O bond length 115 pm
Linear shape
Dinitrogen trioxide
Lewis structure of N2O3
Resonance structures of N2O3
N-N bond length 186 pm
N-O bond lengths 114, 121 pm
Planar shape
Nitrogen dioxide
Lewis structure of NO2
Resonance structures of NO2
N-O bond length 120 pm
Bond angle 134$^\circ$
Bent shape
Dinitrogen tetroxide
Lewis structure of N2O4
Resonance structures of N2O4
N-N bond length 175 pm
N-O bond length 121 pm
Planar shape
Dinitrogen pentoxide
Lewis structure of N2O5
Resonance structures of N2O5
N-O bond lengths 118, 132 pm
Bond angles 114$^\circ$, 134$^\circ$
Non-planar shape

NO$_2$ contains an odd number of valence electrons (17). It can dimerise to form $\textsf{N}_2\text{O}_4$ which has an even number of valence electrons (34), making it more stable.

Intext Question

7.6 What is the covalence of nitrogen in N$_2$O$_5$ ?

Answer:

Dinitrogen pentoxide ($\textsf{N}_2\text{O}_5$) has the structure shown in Table 7.4. Each nitrogen atom is bonded to three oxygen atoms and one other nitrogen atom. Counting the covalent bonds formed by each nitrogen atom, including the double bond with one oxygen atom and single bonds with two oxygen atoms and the other nitrogen atom, the total number of covalent bonds formed by each nitrogen atom is 4. Therefore, the covalence of nitrogen in $\textsf{N}_2\text{O}_5$ is four.



Nitric Acid

Nitric acid ($\textsf{HNO}_3$) is the most important oxoacid of nitrogen.


Preparation


Properties

Brown Ring Test for nitrates: Based on reduction of nitrates to NO by $\textsf{Fe}^{2+}$, and NO reacting with $\textsf{Fe}^{2+}$ to form a brown complex $[\textsf{Fe(H}_2\text{O)}_5\text{(NO)}]^{2+}$.

Uses: Manufacture of fertilisers ($\textsf{NH}_4\text{NO}_3$), explosives (nitroglycerin, TNT), pigments, paints, dyes, etching metals, oxidiser in rocket fuels.

Example 7.5 Why does NO$_2$ dimerise ?

Answer:

Nitrogen dioxide ($\textsf{NO}_2$) is a molecule with an unpaired electron (it is an odd-electron molecule). The Lewis structure of $\textsf{NO}_2$ shows that the nitrogen atom has 5 valence electrons, and each oxygen atom has 6, totaling $5 + 2(6) = 17$ valence electrons. An unpaired electron makes the molecule paramagnetic and highly reactive.

$\textsf{NO}_2$ dimerises to form dinitrogen tetroxide ($\textsf{N}_2\text{O}_4$). In $\textsf{N}_2\text{O}_4$, the two $\textsf{NO}_2$ units join together via a covalent bond between the two nitrogen atoms. This process involves the pairing up of the unpaired electrons on the nitrogen atoms of two $\textsf{NO}_2$ molecules, forming a stable N-N bond. The resulting $\textsf{N}_2\text{O}_4$ molecule has an even number of electrons (34) and is diamagnetic and significantly more stable than the $\textsf{NO}_2$ monomer, especially at lower temperatures. This increase in stability drives the dimerisation process.

Reaction: $2\textsf{NO}_2\text{(g)} \rightleftharpoons \textsf{N}_2\text{O}_4\text{(g)}$


Phosphorus — Allotropic Forms

Phosphorus exists in several allotropic forms, the most important being white, red, and black phosphorus.

At elevated temperatures ($\sim 1000$ K), sulphur exists as $\textsf{S}_2$ molecules which are paramagnetic like $\textsf{O}_2$ (due to unpaired electrons in $\pi^*$ antibonding orbitals).

Intext Questions

7.7 (a) Bond angle in PH$_4^+$ is higher than that in PH$_3$. Why?

(b) What is formed when PH$_3$ reacts with an acid?

7.8 What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO$_2$ ?

Answer:

7.7 (a) Both $\textsf{PH}_3$ and $\textsf{PH}_4^{+}$ involve sp$^3$ hybridisation of the phosphorus atom. In $\textsf{PH}_3$, phosphorus has three bond pairs and one lone pair of electrons. According to VSEPR theory, the lone pair-bond pair repulsions are greater than bond pair-bond pair repulsions. This increased repulsion caused by the lone pair compresses the bond angles between the P-H bonds, making the HPH bond angle in $\textsf{PH}_3$ (93.6$^\circ$) less than the ideal tetrahedral angle (109.5$^\circ$). In $\textsf{PH}_4^{+}$, phosphorus has four bond pairs and no lone pairs. The four bond pairs are arranged tetrahedrally around the phosphorus atom to minimize repulsions, resulting in the ideal tetrahedral bond angle of 109.5$^\circ$. Since there are no lone pairs in $\textsf{PH}_4^{+}$ to cause extra repulsion, the bond angle in $\textsf{PH}_4^{+}$ is higher than that in $\textsf{PH}_3$.

(b) Phosphine ($\textsf{PH}_3$) is weakly basic due to the presence of a lone pair of electrons on the phosphorus atom. Like ammonia, it can accept a proton ($\textsf{H}^{+}$) from an acid, forming a phosphonium ion ($\textsf{PH}_4^{+}$). When $\textsf{PH}_3$ reacts with an acid like HBr or HI, it forms the corresponding phosphonium salt, e.g., phosphonium bromide ($\textsf{PH}_4\text{Br}$) or phosphonium iodide ($\textsf{PH}_4\text{I}$).

$\textsf{PH}_3 + \textsf{HBr} \to \textsf{PH}_4\text{Br}$

$\textsf{PH}_3 + \textsf{HI} \to \textsf{PH}_4\text{I}$

So, when $\textsf{PH}_3$ reacts with an acid, a phosphonium compound (salt) is formed.

7.8 When white phosphorus ($\textsf{P}_4$) is heated with a concentrated solution of $\textsf{NaOH}$ in an inert atmosphere of $\textsf{CO}_2$, it undergoes a disproportionation reaction. Phosphorus (oxidation state 0) is simultaneously oxidised and reduced. It is oxidised to phosphite (in hypophosphite, $\textsf{NaH}_2\text{PO}_2$, P has oxidation state +1) and reduced to phosphine ($\textsf{PH}_3$, P has oxidation state -3).

$\textsf{P}_4\text{(s)} + 3\textsf{NaOH(aq)} + 3\textsf{H}_2\text{O(l)} \to \textsf{PH}_3\text{(g)} + 3\textsf{NaH}_2\text{PO}_2\text{(aq)}$

Phosphine is a highly poisonous gas with a rotten fish smell.



Phosphine

Phosphine ($\textsf{PH}_3$) is a hydride of phosphorus.


Preparation

Pure phosphine is non-inflammable, but it becomes inflammable due to impurities like $\textsf{P}_2\text{H}_4$ or $\textsf{P}_4$ vapors. Purification involves absorbing it in HI to form phosphonium iodide ($\textsf{PH}_4\text{I}$), which on treatment with KOH liberates pure $\textsf{PH}_3$.

Uses: Spontaneous combustion used in Holme's signals (containers with $\textsf{CaC}_2$ and $\textsf{Ca}_3\text{P}_2$ thrown in sea produce $\textsf{C}_2\text{H}_2$ and $\textsf{PH}_3$, which ignite). Also used in smoke screens.


Properties

Example 7.6 In what way can it be proved that PH$_3$ is basic in nature?

Answer:

Phosphine ($\textsf{PH}_3$) is a weak Lewis base due to the presence of a lone pair of electrons on the phosphorus atom. It accepts a proton ($\textsf{H}^{+}$) from strong acids to form phosphonium ions ($\textsf{PH}_4^{+}$). The reaction with hydroiodic acid ($\textsf{HI}$) is a good example:

$\textsf{PH}_3\text{(g)} + \textsf{HI(aq)} \to \textsf{PH}_4\text{I(s)}$ (Phosphonium iodide)

The formation of the salt $\textsf{PH}_4\text{I}$ demonstrates that $\textsf{PH}_3$ has accepted a proton from $\textsf{HI}$, which is characteristic of a basic substance (Lewis base). This reaction proves the basic nature of phosphine.

Intext Questions

7.7 (a) Bond angle in PH$_4^+$ is higher than that in PH$_3$. Why?

(b) What is formed when PH$_3$ reacts with an acid?

7.8 What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO$_2$ ?

Answer:

7.7 (a) Both $\textsf{PH}_3$ and $\textsf{PH}_4^{+}$ involve sp$^3$ hybridisation of the phosphorus atom. In $\textsf{PH}_3$, phosphorus has three bond pairs and one lone pair of electrons. According to VSEPR theory, the lone pair-bond pair repulsions are greater than bond pair-bond pair repulsions. This increased repulsion caused by the lone pair compresses the bond angles between the P-H bonds, making the HPH bond angle in $\textsf{PH}_3$ (93.6$^\circ$) less than the ideal tetrahedral angle (109.5$^\circ$). In $\textsf{PH}_4^{+}$, phosphorus has four bond pairs and no lone pairs. The four bond pairs are arranged tetrahedrally around the phosphorus atom to minimize repulsions, resulting in the ideal tetrahedral bond angle of 109.5$^\circ$. Since there are no lone pairs in $\textsf{PH}_4^{+}$ to cause extra repulsion, the bond angle in $\textsf{PH}_4^{+}$ is higher than that in $\textsf{PH}_3$.

(b) Phosphine ($\textsf{PH}_3$) is weakly basic due to the presence of a lone pair of electrons on the phosphorus atom. Like ammonia, it can accept a proton ($\textsf{H}^{+}$) from an acid, forming a phosphonium ion ($\textsf{PH}_4^{+}$). When $\textsf{PH}_3$ reacts with an acid like HBr or HI, it forms the corresponding phosphonium salt, e.g., phosphonium bromide ($\textsf{PH}_4\text{Br}$) or phosphonium iodide ($\textsf{PH}_4\text{I}$).

$\textsf{PH}_3 + \textsf{HBr} \to \textsf{PH}_4\text{Br}$

$\textsf{PH}_3 + \textsf{HI} \to \textsf{PH}_4\text{I}$

So, when $\textsf{PH}_3$ reacts with an acid, a phosphonium compound (salt) is formed.

7.8 When white phosphorus ($\textsf{P}_4$) is heated with a concentrated solution of $\textsf{NaOH}$ in an inert atmosphere of $\textsf{CO}_2$, it undergoes a disproportionation reaction. Phosphorus (oxidation state 0) is simultaneously oxidised and reduced. It is oxidised to phosphite (in hypophosphite, $\textsf{NaH}_2\text{PO}_2$, P has oxidation state +1) and reduced to phosphine ($\textsf{PH}_3$, P has oxidation state -3).

$\textsf{P}_4\text{(s)} + 3\textsf{NaOH(aq)} + 3\textsf{H}_2\text{O(l)} \to \textsf{PH}_3\text{(g)} + 3\textsf{NaH}_2\text{PO}_2\text{(aq)}$

Phosphine is a highly poisonous gas with a rotten fish smell.



Phosphorus Halides

Phosphorus forms two types of halides: $\textsf{PX}_3$ ($\textsf{X} = \textsf{F}, \textsf{Cl}, \textsf{Br}, \textsf{I}$) and $\textsf{PX}_5$ ($\textsf{X} = \textsf{F}, \textsf{Cl}, \textsf{Br}$).


Phosphorus Trichloride

Phosphorus trichloride ($\textsf{PCl}_3$) is obtained by passing dry chlorine over heated white phosphorus or by action of thionyl chloride ($\textsf{SO}_2\text{Cl}_2$) on phosphorus.

Properties: Colourless oily liquid. Fumes in moist air due to hydrolysis to phosphorous acid and $\textsf{HCl}$ fumes.

Reacts with compounds having –OH groups (alcohols, carboxylic acids) to replace –OH with –Cl.

Structure: Pyramidal shape (sp$^3$ hybridisation, 3 bond pairs, 1 lone pair).

Pyramidal structure of PCl3 molecule

Example 7.7 Why does PCl$_3$ fume in moisture ?

Answer:

Phosphorus trichloride ($\textsf{PCl}_3$) reacts readily with moisture in the air (hydrolyses) to form phosphorous acid ($\textsf{H}_3\text{PO}_3$) and hydrogen chloride ($\textsf{HCl}$) gas:

$\textsf{PCl}_3\text{(l)} + 3\textsf{H}_2\text{O(l)} \to \textsf{H}_3\text{PO}_3\text{(aq)} + 3\textsf{HCl(g)}$

The $\textsf{HCl}$ gas produced is highly soluble in water and forms tiny droplets of hydrochloric acid with atmospheric moisture. These tiny acid droplets are visible as dense fumes in the air. Therefore, $\textsf{PCl}_3$ fumes in moisture due to the formation of $\textsf{HCl}$ gas upon hydrolysis.


Phosphorus Pentachloride

Phosphorus pentachloride ($\textsf{PCl}_5$) is prepared by the reaction of white phosphorus with excess dry chlorine or by action of $\textsf{SO}_2\text{Cl}_2$ on phosphorus.

Properties: Yellowish white powder. Hydrolyses in moist air to $\textsf{POCl}_3$ and finally to phosphoric acid ($\textsf{H}_3\text{PO}_4$).

Sublimes on heating, decomposes on stronger heating to $\textsf{PCl}_3$ and $\textsf{Cl}_2$.

Reacts with compounds containing –OH group (alcohols, carboxylic acids) to convert them to chloro derivatives.

Reacts with finely divided metals on heating to give chlorides.

Uses: Synthesis of organic chloro compounds.

Structure: In gaseous and liquid phases, it has a trigonal bipyramidal structure (sp$^3$d hybridisation). Three equatorial P-Cl bonds are equivalent, two axial P-Cl bonds are longer and weaker due to greater repulsion from equatorial bond pairs.

Trigonal bipyramidal structure of PCl5 molecule

Example 7.8 Are all the five bonds in PCl$_5$ molecule equivalent? Justify your answer.

Answer:

No, not all five bonds in a $\textsf{PCl}_5$ molecule are equivalent. In the gaseous and liquid phases, $\textsf{PCl}_5$ has a trigonal bipyramidal structure. This structure has two types of bonds:

  1. Three equatorial bonds lying in a plane.
  2. Two axial bonds located above and below the equatorial plane.

The axial bonds are slightly longer and weaker than the equatorial bonds. This is due to greater repulsion between the axial bond pairs and the electron pairs of the equatorial bonds compared to the repulsions between the equatorial bond pairs themselves. These greater axial repulsions push the axial chlorine atoms slightly further away from the phosphorus atom.


Intext Questions

7.9 What happens when PCl$_5$ is heated?

7.10 Write a balanced equation for the reaction of PCl$_5$ with water.

Answer:

7.9 When phosphorus pentachloride ($\textsf{PCl}_5$) is heated, it undergoes thermal decomposition into phosphorus trichloride ($\textsf{PCl}_3$) and chlorine gas ($\textsf{Cl}_2$). This decomposition is reversible, meaning $\textsf{PCl}_3$ and $\textsf{Cl}_2$ can recombine to form $\textsf{PCl}_5$ upon cooling.

$\textsf{PCl}_5\text{(s)} \xrightarrow{\text{Heat}} \textsf{PCl}_3\text{(g)} + \textsf{Cl}_2\text{(g)}$

7.10 Phosphorus pentachloride ($\textsf{PCl}_5$) reacts vigorously with water (hydrolyses). The reaction with limited water yields phosphorus oxychloride ($\textsf{POCl}_3$) and hydrogen chloride ($\textsf{HCl}$). With excess water, the hydrolysis proceeds further to produce phosphoric acid ($\textsf{H}_3\text{PO}_4$) and $\textsf{HCl}$.

Reaction with limited water:

$\textsf{PCl}_5\text{(s)} + \textsf{H}_2\text{O(l)} \to \textsf{POCl}_3\text{(l)} + 2\textsf{HCl(g)}$

Reaction with excess water:

$\textsf{PCl}_5\text{(s)} + 4\textsf{H}_2\text{O(l)} \to \textsf{H}_3\text{PO}_4\text{(aq)} + 5\textsf{HCl(aq/g)}$

A balanced equation for the reaction of $\textsf{PCl}_5$ with water (assuming complete hydrolysis) is:

$\textsf{PCl}_5\text{(s)} + 4\textsf{H}_2\text{O(l)} \to \textsf{H}_3\text{PO}_4\text{(aq)} + 5\textsf{HCl(aq)}$



Oxoacids Of Phosphorus

Phosphorus forms a variety of oxoacids, which contain phosphorus, oxygen, and hydrogen. The composition of these acids is often related by loss or gain of $\textsf{H}_2\text{O}$ or O atoms. Important oxoacids are listed in Table 7.5.

Name Formula Oxidation state of phosphorus Characteristic bonds and their number Preparation
Hypophosphorous acid (Phosphinic)H$_3$PO$_2$+1One P–OH, Two P–H, One P=Owhite P$_4$ + alkali
Orthophosphorous acid (Phosphonic)H$_3$PO$_3$+3Two P–OH, One P–H, One P=OP$_2$O$_3$ + H$_2$O
Pyrophosphorous acidH$_4$P$_2$O$_5$+3Four P–OH, One P–O–P, Two P=OPCl$_3$ + H$_3$PO$_3$
Hypophosphoric acidH$_4$P$_2$O$_6$+4Four P–OH, Two P=O, One P–Pred P$_4$ + alkali
Orthophosphoric acid (Phosphoric acid)H$_3$PO$_4$+5Three P–OH, One P=OP$_4$O$_{10}$+H$_2$O
Pyrophosphoric acidH$_4$P$_2$O$_7$+5Four P–OH, Two P=O, One P–O–Pheat phosphoric acid
Metaphosphoric*(HPO$_3$)$_n$+5One P–OH, Two P=O, One P–O–P
(in trimer (HPO$_3$)$_3$)
phosphorus acid + Br$_2$, heat in a sealed tube

* Exists in polymeric forms only. Characteristic bonds of (HPO$_3$)$_3$ have been given in the Table.

Key structural features of phosphorus oxoacids:

Acids with P in +3 oxidation state disproportionate on heating to higher (+5) and lower (–3) oxidation states. Example: Orthophosphorous acid ($\textsf{H}_3\text{PO}_3$) disproportionates to orthophosphoric acid ($\textsf{H}_3\text{PO}_4$) and phosphine ($\textsf{PH}_3$).

$4\textsf{H}_3\text{PO}_3 \xrightarrow{\text{Heat}} 3\textsf{H}_3\text{PO}_4 + \textsf{PH}_3$

The acids with $\textsf{P–H}$ bonds (e.g., $\textsf{H}_3\text{PO}_2$, $\textsf{H}_3\text{PO}_3$) have strong reducing properties. $\textsf{H}_3\text{PO}_2$ is a strong reducing agent (two $\textsf{P–H}$ bonds). $\textsf{H}_3\text{PO}_3$ has one $\textsf{P–H}$ bond.

Only the hydrogen atoms attached to oxygen in $\textsf{P–OH}$ bonds are ionisable and contribute to basicity. $\textsf{H}$ atoms directly bonded to P are not acidic. $\textsf{H}_3\text{PO}_2$ is monobasic (one $\textsf{P–OH}$). $\textsf{H}_3\text{PO}_3$ is dibasic (two $\textsf{P–OH}$). $\textsf{H}_3\text{PO}_4$ is tribasic (three $\textsf{P–OH}$).

Structures of important oxoacids of phosphorus: Hypophosphorous acid, Orthophosphorous acid, Orthophosphoric acid, Pyrophosphoric acid, Pyrophosphorous acid, Metaphosphoric acid (Cyclic and Polymeric forms)

Example 7.9 How do you account for the reducing behaviour of H$_3$PO$_2$ on the basis of its structure ?

Answer:

The structure of hypophosphorous acid ($\textsf{H}_3\text{PO}_2$) is shown in Figure 7.4. It contains one P=O bond, one P–OH bond, and two P–H bonds. The hydrogen atoms directly bonded to the phosphorus atom (P–H bonds) are not acidic and do not dissociate to form $\textsf{H}^{+}$ ions. However, these P–H bonds are responsible for the reducing properties of the acid. When $\textsf{H}_3\text{PO}_2$ acts as a reducing agent, the hydrogen atoms attached to phosphorus are oxidised (their oxidation state changes from -1 in P-H bonds to +1 or higher as they eventually end up in P-O-H or P=O bonds as phosphorus's oxidation state increases). The presence of two such P–H bonds makes $\textsf{H}_3\text{PO}_2$ a strong reducing agent, capable of reducing substances like $\textsf{AgNO}_3$ to metallic silver.

Intext Questions

7.11 What is the basicity of H$_3$PO$_4$?

7.12 What happens when H$_3$PO$_3$ is heated?

Answer:

7.11 The basicity of an oxoacid is the number of ionisable $\textsf{H}^{+}$ ions it can donate in aqueous solution. These are the hydrogen atoms bonded to oxygen atoms ($\textsf{P–OH}$ bonds). The structure of orthophosphoric acid ($\textsf{H}_3\text{PO}_4$) has one $\textsf{P=O}$ bond and three $\textsf{P–OH}$ bonds (Figure 7.4). The three hydrogen atoms bonded to oxygen are ionisable.

$\textsf{H}_3\text{PO}_4\text{(aq)} \rightleftharpoons \textsf{H}^{+}\text{(aq)} + \textsf{H}_2\text{PO}_4^{-}\text{(aq)}$

$\textsf{H}_2\text{PO}_4^{-}\text{(aq)} \rightleftharpoons \textsf{H}^{+}\text{(aq)} + \textsf{HPO}_4^{2-}\text{(aq)}$

$\textsf{HPO}_4^{2-}\text{(aq)} \rightleftharpoons \textsf{H}^{+}\text{(aq)} + \textsf{PO}_4^{3-}\text{(aq)}$

Since $\textsf{H}_3\text{PO}_4$ can donate three $\textsf{H}^{+}$ ions, its basicity is three.

7.12 When orthophosphorous acid ($\textsf{H}_3\text{PO}_3$, phosphorus in +3 oxidation state) is heated, it undergoes a disproportionation reaction. Phosphorus in the +3 oxidation state is simultaneously oxidised to the +5 oxidation state (in orthophosphoric acid, $\textsf{H}_3\text{PO}_4$) and reduced to the –3 oxidation state (in phosphine, $\textsf{PH}_3$).

The reaction is:

$4\textsf{H}_3\text{PO}_3\text{(s)} \xrightarrow{\text{Heat}} 3\textsf{H}_3\text{PO}_4\text{(s)} + \textsf{PH}_3\text{(g)}$

So, when $\textsf{H}_3\text{PO}_3$ is heated, it forms orthophosphoric acid and phosphine gas.



Group 16 Elements

Group 16 includes Oxygen (O), Sulphur (S), Selenium (Se), Tellurium (Te), Polonium (Po), and Livermorium (Lv). These are collectively called chalcogens (meaning 'ore producers').


Occurrence

Table 7.6 summarises properties of O, S, Se, Te, Po.

Property O S Se Te Po
Atomic number816345284
Atomic mass/g mol$^{-1}$16.0032.0678.96127.60210.00
Electronic configuration[He]2s$^2$2p$^4$[Ne]3s$^2$3p$^4$[Ar]3d$^{10}$4s$^2$4p$^4$[Kr]4d$^{10}$5s$^2$5p$^4$[Xe]4f$^{14}$5d$^{10}$6s$^2$6p$^4$
Covalent radius/(pm)a66104117137146
Ionic radius, E$^{2-}$/pm140184198221230b
Electron gain enthalpy/
$\Delta_{eg}H$ kJ mol$^{-1}$
–141–200–195–190–174
Ionisation enthalpy ($\Delta_i H_1$)/kJ mol$^{-1}$13141000941869813
Electronegativity3.502.582.552.011.76
Density /g cm$^{-3}$ (298 K)1.32c2.06d4.19e6.25
Melting point/K55393f490725520
Boiling point/K9071895812601235
Oxidation states*–2,–1,1,2–2,2,4,6–2,2,4,6–2,2,4,62,4

(aSingle bond; bApproximate value; cAt the melting point; d Rhombic sulphur; eHexagonal grey; fMonoclinic form, 673 K. * Oxygen shows oxidation states of +2 and +1 in oxygen fluorides OF$_2$ and O$_2$F$_2$ respectively).


Electronic Configuration

General configuration: ns$^2$np$^4$. Six valence electrons.


Atomic And Ionic Radii

Increase down the group due to more shells. Oxygen is exceptionally small. Ionic radii (E$^{2-}$) also increase down the group.


Ionisation Enthalpy

Decreases down the group (increasing size). Lower than corresponding Group 15 elements due to less stable np$^4$ configuration vs Group 15's half-filled np$^3$.


Electron Gain Enthalpy

Negative electron gain enthalpy (tendency to accept electron). Oxygen's value is less negative than sulfur's (due to small size and interelectronic repulsions). Becomes less negative from S to Po.


Electronegativity

Decreases down the group (increasing size). Oxygen is highly electronegative (next to Fluorine). Metallic character increases down the group (O, S non-metals; Se, Te metalloids; Po metal).


Physical Properties

O, S are non-metals. Se, Te are metalloids. Po is a metal (radioactive, short-lived). All show allotropy. Melting and boiling points increase down the group. Large difference between O and S is due to atomicity (O$_2$ diatomic, S$_8$ polyatomic).


Chemical Properties

Exhibits various oxidation states (Table 7.6). Stability of –2 state decreases down the group (Po rarely shows –2). Oxygen shows –2 (except in OF$_2$ (+2) and $\textsf{O}_2\textsf{F}_2$ (+1)). Other elements show +2, +4, +6 (common). Stability of +6 decreases, +4 increases down the group (inert pair effect). Bonding in +4, +6 states is primarily covalent.

Anomalous behaviour of oxygen: Small size, high electronegativity, absence of d orbitals. Forms strong hydrogen bonding in $\textsf{H}_2\textsf{O}$ (unlike $\textsf{H}_2\textsf{S}$). Covalency limited to 4 (rarely exceeds 2 in practice).

(i) Reactivity with hydrogen: Form hydrides $\textsf{H}_2\textsf{E}$ ($\textsf{E} = \textsf{O}, \textsf{S}, \textsf{Se}, \textsf{Te}, \textsf{Po}$). Properties in Table 7.7.

Property H$_2$O H$_2$S H$_2$Se H$_2$Te
m.p/K273188208222
b.p/K373213232269
H–E distance/pm96134146169
HEH angle ($^\circ$)104929190
$\Delta_f H$/kJ mol$^{-1}$–286–2073100
$\Delta_{diss}H$(H–E)/kJ mol$^{-1}$463347276238
Dissociation constanta1.8$\times$10$^{-16}$1.3$\times$10$^{-7}$1.3$\times$10$^{-4}$2.3$\times$10$^{-3}$

(ii) Reactivity with oxygen: Form $\textsf{EO}_2$ and $\textsf{EO}_3$ (E = S, Se, Te, Po). Acidic character. $\textsf{SO}_2$ reducing, $\textsf{TeO}_2$ oxidising. Reducing property of dioxide decreases $\textsf{SO}_2$ to $\textsf{TeO}_2$.

(iii) Reactivity towards halogens: Form $\textsf{EX}_6$, $\textsf{EX}_4$, $\textsf{EX}_2$. Stability decreases $\textsf{F}^{-}> \textsf{Cl}^{-} > \textsf{Br}^{-} > \textsf{I}^{-}$. Hexafluorides are stable (octahedral, e.g., $\textsf{SF}_6$ exceptionally stable). Tetrafluorides (e.g., $\textsf{SF}_4$) have see-saw geometry (sp$^3$d, one lone pair). Dihalides (except O) are tetrahedral (sp$^3$). Monohalides are dimeric (e.g., $\textsf{S}_2\text{Cl}_2$).

Example 7.10 Elements of Group 16 generally show lower value of first ionisation enthalpy compared to the corresponding periods of group 15. Why?

Answer:

Elements of Group 16 (ns$^2$np$^4$) have six valence electrons, while corresponding elements in Group 15 (ns$^2$np$^3$) have five. In Group 15, the p-orbitals are exactly half-filled (np$^3$), which is an extra stable electronic configuration. Removing an electron from this stable half-filled configuration requires a relatively higher amount of energy. In contrast, Group 16 elements have one extra electron in the p-orbitals compared to the half-filled configuration (np$^4$). Removing this one electron from the np$^4$ configuration (which is less stable than the half-filled np$^3$) requires less energy compared to removing an electron from the stable np$^3$ configuration of Group 15 elements in the same period. Therefore, Group 16 elements generally have lower first ionisation enthalpy values than Group 15 elements in the same period.

Example 7.11 H$_2$S is less acidic than H$_2$Te. Why?

Answer:

The acidic character of the hydrides of Group 16 elements ($\textsf{H}_2\textsf{E}$) increases down the group from $\textsf{H}_2\textsf{O}$ to $\textsf{H}_2\textsf{Te}$. This trend is explained by the decrease in the strength of the H–E bond down the group. As we move from Sulphur to Tellurium, the atomic size of the central element (E) increases. This leads to an increase in the H–E bond length (Table 7.7) and a decrease in the bond dissociation enthalpy. A weaker H–Te bond compared to an H–S bond means it is easier for the $\textsf{H}_2\textsf{Te}$ molecule to release a proton ($\textsf{H}^{+}$) in solution, making it a stronger acid than $\textsf{H}_2\textsf{S}$.

Intext Questions

7.13 List the important sources of sulphur.

7.14 Write the order of thermal stability of the hydrides of Group 16 elements.

7.15 Why is H$_2$O a liquid and H$_2$S a gas ?

Answer:

7.13 Important sources of sulphur include:

  1. Sulphates: Gypsum ($\textsf{CaSO}_4.\textsf{2H}_2\text{O}$), Epsom salt ($\textsf{MgSO}_4.\textsf{7H}_2\text{O}$), Baryte ($\textsf{BaSO}_4$).
  2. Sulphides: Galena ($\textsf{PbS}$), Zinc blende ($\textsf{ZnS}$), Copper pyrites ($\textsf{CuFeS}_2$).
  3. Traces of hydrogen sulphide ($\textsf{H}_2\textsf{S}$) in volcanoes.
  4. Sulphur is also found in organic materials like eggs, proteins, garlic, onion, mustard, hair, and wool.

7.14 The thermal stability of the hydrides of Group 16 elements ($\textsf{H}_2\textsf{E}$) decreases down the group from $\textsf{H}_2\textsf{O}$ to $\textsf{H}_2\textsf{Po}$. This is due to the decrease in the H–E bond dissociation enthalpy as the size of element E increases from O to Po, making the bond weaker and easier to break thermally.

Order of thermal stability: $\textsf{H}_2\textsf{O} > \textsf{H}_2\textsf{S} > \textsf{H}_2\textsf{Se} > \textsf{H}_2\textsf{Te} > \textsf{H}_2\textsf{Po}$.

7.15 Water ($\textsf{H}_2\textsf{O}$) is a liquid at room temperature, while hydrogen sulphide ($\textsf{H}_2\textsf{S}$) is a gas. This significant difference in physical state is due to the presence of strong hydrogen bonding between water molecules in the liquid state. Oxygen is highly electronegative and small, allowing hydrogen atoms bonded to oxygen in one water molecule to form hydrogen bonds with the oxygen atoms of neighboring water molecules. These strong intermolecular forces require more energy (higher temperature) to overcome, resulting in a relatively high boiling point for water (373 K).

Sulphur is less electronegative and larger than oxygen. Hydrogen atoms bonded to sulphur in $\textsf{H}_2\textsf{S}$ do not form significant hydrogen bonds. The intermolecular forces in liquid $\textsf{H}_2\textsf{S}$ are primarily weak van der Waals forces (dipole-dipole and dispersion forces). These weaker forces are easily overcome at room temperature, so $\textsf{H}_2\textsf{S}$ exists as a gas with a much lower boiling point (213 K).



Dioxygen

Dioxygen ($\textsf{O}_2$) is the elemental form of oxygen, a diatomic molecule.


Preparation


Properties

Uses: Respiration, combustion, oxyacetylene welding, manufacture of steel, in hospitals, high altitude flying, mountaineering. In rocket fuels.

Intext Questions

7.16 Which of the following does not react with oxygen directly?

Zn, Ti, Pt, Fe

7.17 Complete the following reactions:

(i) C$_2$H$_4$ + O$_2$ $\to$

(ii) 4Al + 3 O$_2$ $\to$

Answer:

7.16 Among the given metals (Zn, Ti, Pt, Fe), Platinum (Pt) is a noble metal and does not react with oxygen directly under ordinary conditions. Zinc, Titanium, and Iron are reactive metals that readily react with oxygen to form oxides.

7.17 Complete the reactions:

(i) $\textsf{C}_2\text{H}_4 + 3\textsf{O}_2 \to 2\textsf{CO}_2 + 2\textsf{H}_2\text{O}$ (Combustion of ethene) - Assuming complete combustion.

(ii) $4\textsf{Al} + 3\textsf{O}_2 \to 2\textsf{Al}_2\text{O}_3$ (Formation of aluminium oxide) - This reaction is already complete and balanced as provided in the question.



Simple Oxides

A binary compound of oxygen with another element is called an oxide. Elements form various oxides.

Simple oxides (containing one element besides oxygen) are classified by their acid-base character:

Mixed oxides (e.g., $\textsf{Pb}_3\text{O}_4$, $\textsf{Fe}_3\text{O}_4$) contain more than one oxidation state of the metal.



Ozone

Ozone ($\textsf{O}_3$) is an allotropic form of oxygen. It is formed in the stratosphere by UV radiation acting on atmospheric oxygen. The ozone layer protects Earth from harmful UV radiation.


Preparation

By passing a slow dry stream of oxygen through a silent electrical discharge. This converts about 10% of $\textsf{O}_2$ to $\textsf{O}_3$ (ozonised oxygen). Silent discharge is used to prevent decomposition of $\textsf{O}_3$ (formation is endothermic).

$3\textsf{O}_2 \to 2\textsf{O}_3; \Delta H = +142$ kJ mol$^{-1}$.


Properties

Structure of ozone molecule and its resonance structures

Nitrogen oxides and freons (CFCs) contribute to ozone depletion in the stratosphere.

Uses: Germicide, disinfectant, sterilizing water, bleaching (oils, ivory, flour), oxidising agent (in $\textsf{KMnO}_4$ manufacture).

Intext Questions

7.18 Why does O$_3$ act as a powerful oxidising agent?

7.19 How is O$_3$ estimated quantitatively?

Answer:

7.18 Ozone ($\textsf{O}_3$) acts as a powerful oxidising agent because it is thermodynamically unstable and readily decomposes to oxygen ($\textsf{O}_2$) and releases an atom of nascent oxygen (O): $\textsf{O}_3 \to \textsf{O}_2 + \text{O}$. Nascent oxygen is extremely reactive and acts as a strong oxidising agent, reacting with other substances to oxidise them. The release of this highly reactive nascent oxygen atom is the reason for ozone's powerful oxidising ability.

7.19 Ozone ($\textsf{O}_3$) can be estimated quantitatively using its reaction with potassium iodide ($\textsf{KI}$) solution. When ozone is passed through an excess of potassium iodide solution buffered with a borate buffer (pH $\approx$ 9.2), it oxidises iodide ions ($\textsf{I}^{-}$) to iodine ($\textsf{I}_2$):

$2\textsf{I}^{-}\text{(aq)} + \textsf{H}_2\text{O(l)} + \textsf{O}_3\text{(g)} \to 2\textsf{OH}^{-}\text{(aq)} + \textsf{I}_2\text{(s)} + \textsf{O}_2\text{(g)}$

The amount of iodine ($\textsf{I}_2$) liberated in the solution is directly proportional to the amount of ozone that reacted. The liberated iodine can then be titrated against a standard solution of sodium thiosulphate ($\textsf{Na}_2\text{S}_2\text{O}_3$), using starch as an indicator:

$\textsf{I}_2\text{(s)} + 2\textsf{Na}_2\textsf{S}_2\text{O}_3\text{(aq)} \to \textsf{Na}_2\textsf{S}_4\text{O}_6\text{(aq)} + 2\textsf{NaI(aq)}$

By determining the amount of sodium thiosulphate consumed in the titration, the amount of iodine liberated can be calculated, which in turn allows for the calculation of the initial amount of ozone.



Sulphur — Allotropic Forms

Sulphur exhibits numerous allotropic forms, with rhombic ($\alpha$-sulphur) and monoclinic ($\beta$-sulphur) being the most significant.

The temperature 369 K is the transition temperature where both forms are stable in equilibrium.

Both forms consist of $\textsf{S}_8$ molecules, which are puckered rings with a crown shape. The packing of these $\textsf{S}_8$ rings differs in the two allotropes.

Structures of the S8 ring in rhombic sulfur and the S6 ring

Other forms with different ring sizes ($\textsf{S}_6$ to $\textsf{S}_{20}$) exist. At elevated temperatures ($\sim 1000$ K), $\textsf{S}_2$ is the dominant species, which is paramagnetic like $\textsf{O}_2$.

Example 7.12 Which form of sulphur shows paramagnetic behaviour ?

Answer:

Sulphur exhibits paramagnetic behaviour in its $\textsf{S}_2$ form, which exists predominantly in the vapour state at high temperatures (around 1000 K). Similar to the $\textsf{O}_2$ molecule, the $\textsf{S}_2$ molecule has two unpaired electrons in its antibonding $\pi^*$ molecular orbitals. Substances with unpaired electrons are paramagnetic, meaning they are weakly attracted by a magnetic field. Solid sulphur (like $\textsf{S}_8$ in rhombic or monoclinic forms) and $\textsf{S}_6$ are diamagnetic because all electrons are paired in these molecules.



Sulphur Dioxide

Sulphur dioxide ($\textsf{SO}_2$) is a significant oxide of sulphur.


Preparation


Properties

Angular structure of SO2 molecule and its resonance structures

Uses: Refining petroleum and sugar, bleaching wool and silk, anti-chlor, disinfectant, preservative. Manufacture of $\textsf{H}_2\text{SO}_4$, sodium hydrogen sulphite. Liquid $\textsf{SO}_2$ as solvent.

Intext Questions

7.20 What happens when sulphur dioxide is passed through an aqueous solution of Fe(III) salt?

7.21 Comment on the nature of two S–O bonds formed in SO$_2$ molecule. Are the two S–O bonds in this molecule equal ?

7.22 How is the presence of SO$_2$ detected ?

Answer:

7.20 Sulphur dioxide ($\textsf{SO}_2$) acts as a reducing agent when moist. When passed through an aqueous solution of an Iron(III) salt (e.g., Iron(III) sulfate, $\textsf{Fe}_2\text{(SO}_4)_3$), it reduces Iron(III) ions ($\textsf{Fe}^{3+}$) to Iron(II) ions ($\textsf{Fe}^{2+}$). In this reaction, $\textsf{SO}_2$ is oxidised to sulphate ions ($\textsf{SO}_4^{2-}$).

The reaction is:

$2\textsf{Fe}^{3+}\text{(aq)} + \textsf{SO}_2\text{(aq)} + 2\textsf{H}_2\text{O(l)} \to 2\textsf{Fe}^{2+}\text{(aq)} + \textsf{SO}_4^{2-}\text{(aq)} + 4\textsf{H}^{+}\text{(aq)}$

So, when $\textsf{SO}_2$ is passed through an aqueous solution of an Iron(III) salt, the Iron(III) ions are reduced to Iron(II) ions, and the solution may change colour (e.g., from yellow/brown for $\textsf{Fe}^{3+}$ to pale green for $\textsf{Fe}^{2+}$).

7.21 The $\textsf{SO}_2$ molecule has one sulphur atom double bonded to one oxygen atom and single bonded to another oxygen atom, with a lone pair of electrons on the sulphur atom. However, due to resonance, the electron density is delocalised over the molecule, resulting in two equivalent S–O bonds that have partial double bond character. The two S–O bonds are indeed equal in length and strength, intermediate between a single bond and a double bond. This is represented by two resonance structures.

7.22 The presence of $\textsf{SO}_2$ gas can be detected by its pungent, suffocating smell. A more specific chemical test is to pass the gas through an acidified potassium permanganate(VII) solution ($\textsf{KMnO}_4$ acidified with dilute $\textsf{H}_2\textsf{SO}_4$). Sulphur dioxide acts as a reducing agent and decolourises the characteristic purple colour of the permanganate ion ($\textsf{MnO}_4^{-}$). The reaction is:

$5\textsf{SO}_2\text{(g)} + 2\textsf{MnO}_4^{-}\text{(aq)} + 2\textsf{H}_2\text{O(l)} \to 5\textsf{SO}_4^{2-}\text{(aq)} + 2\textsf{Mn}^{2+}\text{(aq)} + 4\textsf{H}^{+}\text{(aq)}$

The colour change from purple ($\textsf{MnO}_4^{-}$) to colourless ($\textsf{Mn}^{2+}$) indicates the presence of $\textsf{SO}_2$.



Oxoacids Of Sulphur

Sulphur forms several oxoacids, many of which are unstable and known only in solution or as salts. Their structures involve sulphur atoms bonded to oxygen atoms (P=O), hydroxyl groups (P–OH), and sometimes other sulphur atoms (P–P) or peroxide linkages (O–O).

Structures of some important oxoacids are shown in Fig. 7.6.

Structures of some important oxoacids of sulphur: Sulfurous acid, Sulfuric acid, Thiosulfuric acid, Peroxodisulfuric acid, Oleum, Pyrosulfuric acid, Disulfuric acid

All sulphur oxoacids contain at least one $\textsf{S=O}$ bond and one $\textsf{S–OH}$ bond. Acidity is due to ionisation of $\textsf{H}$ from $\textsf{S–OH}$ bonds. Number of $\textsf{S–OH}$ groups determines basicity.



Sulphuric Acid

Sulphuric acid ($\textsf{H}_2\text{SO}_4$) is one of the most important industrial chemicals.


Manufacture

Manufactured by the Contact Process, involving three main steps:

  1. Burning sulphur or sulfide ores to produce $\textsf{SO}_2$.
  2. Catalytic oxidation of $\textsf{SO}_2$ to $\textsf{SO}_3$ using $\textsf{O}_2$ in the presence of $\textsf{V}_2\text{O}_5$ catalyst.
    • $2\textsf{SO}_2\text{(g)} + \textsf{O}_2\text{(g)} \xrightarrow{\text{V}_2\text{O}_5} 2\textsf{SO}_3\text{(g)}; \Delta_r H^0 = -196.6$ kJ mol$^{-1}$
    This step is exothermic and reversible. Favourable conditions (Le Chatelier's principle): low temperature and high pressure. Plant operates at $\sim 720$ K and 2 bar pressure (compromise for rate).
  3. Absorption of $\textsf{SO}_3$ in concentrated $\textsf{H}_2\text{SO}_4$ to form Oleum ($\textsf{H}_2\text{S}_2\text{O}_7$).
    • $\textsf{SO}_3\text{(g)} + \textsf{H}_2\text{SO}_4\text{(aq)} \to \textsf{H}_2\text{S}_2\text{O}_7\text{(l)}$
    Oleum is then diluted with water to get $\textsf{H}_2\text{SO}_4$ of desired concentration.
    • $\textsf{H}_2\text{S}_2\text{O}_7\text{(l)} + \textsf{H}_2\text{O(l)} \to 2\textsf{H}_2\text{SO}_4\text{(aq)}$

$\textsf{SO}_2$ is purified before catalytic oxidation. Process is continuous to reduce cost.

Flow diagram for the manufacture of sulfuric acid by Contact process

Properties

Uses: Manufacture of fertilisers, pigments, paints, dyes, detergents, petroleum refining, pickling metals, storage batteries, manufacture of nitrocellulose products, laboratory reagent.

Example 7.13 What happens when (i) Concentrated H$_2$SO$_4$ is added to calcium fluoride (ii) SO$_3$ is passed through water?

Answer:

(i) When concentrated sulphuric acid ($\textsf{H}_2\text{SO}_4$) is added to calcium fluoride ($\textsf{CaF}_2$), it reacts to form calcium sulphate ($\textsf{CaSO}_4$) and hydrogen fluoride ($\textsf{HF}$) gas. This is a reaction where a less volatile acid ($\textsf{H}_2\text{SO}_4$) is used to prepare a more volatile acid ($\textsf{HF}$) from its salt.

$\textsf{CaF}_2\text{(s)} + \textsf{H}_2\text{SO}_4\text{(conc)} \to \textsf{CaSO}_4\text{(s)} + 2\textsf{HF(g)}$

(ii) When sulphur trioxide ($\textsf{SO}_3$) is passed through water, it dissolves readily to form sulphuric acid ($\textsf{H}_2\text{SO}_4$). However, in the Contact process, $\textsf{SO}_3$ is absorbed in concentrated $\textsf{H}_2\text{SO}_4$ to form oleum ($\textsf{H}_2\text{S}_2\text{O}_7$), which is then diluted with water to form $\textsf{H}_2\text{SO}_4$. This is because dissolving $\textsf{SO}_3$ directly in water is highly exothermic and produces a fine mist of $\textsf{H}_2\text{SO}_4$ that is difficult to condense.

$\textsf{SO}_3\text{(g)} + \textsf{H}_2\text{O(l)} \to \textsf{H}_2\text{SO}_4\text{(aq)}$

Intext Questions

7.23 Mention three areas in which H$_2$SO$_4$ plays an important role.

7.24 Write the conditions to maximise the yield of H$_2$SO$_4$ by Contact process.

7.25 Why is K$_{a2}$ > K$_{a1}$ for H$_2$SO$_4$ in water ?

Answer:

7.23 Three areas in which $\textsf{H}_2\text{SO}_4$ plays an important role are:

  1. Fertiliser Industry: Used in the manufacture of ammonium sulphate and superphosphate fertilisers.
  2. Chemical Industry: Used as a raw material in the production of hundreds of other chemicals, pigments, paints, dyes, detergents, and explosives.
  3. Metallurgical Applications: Used for pickling (cleaning) metal surfaces before processes like enameling, electroplating, and galvanizing.
  4. Petroleum Refining: Used in the purification of petroleum.
  5. Storage Batteries: Used as the electrolyte in lead-acid batteries.

(Mention any three from this list).

7.24 In the Contact process for the manufacture of $\textsf{H}_2\text{SO}_4$, the key step is the catalytic oxidation of $\textsf{SO}_2$ to $\textsf{SO}_3$: $2\textsf{SO}_2\text{(g)} + \textsf{O}_2\text{(g)} \rightleftharpoons 2\textsf{SO}_3\text{(g)}; \Delta_r H^0 = -196.6$ kJ mol$^{-1}$. This reaction is exothermic and reversible, with a decrease in the number of moles of gas. According to Le Chatelier's principle, the conditions to maximise the yield of $\textsf{SO}_3$ (and thus $\textsf{H}_2\text{SO}_4$) are:

  1. Low temperature: Exothermic reaction is favored by low temperature. However, very low temperature makes the reaction rate too slow. A compromise temperature of around 720 K is used to balance yield and rate.
  2. High pressure: The forward reaction reduces the number of moles of gas (3 moles $\to$ 2 moles). High pressure shifts the equilibrium towards the side with fewer moles, favouring $\textsf{SO}_3$ formation. Pressure of 2 bar is used in practice.
  3. Use of a catalyst: $\textsf{V}_2\text{O}_5$ (or platinized asbestos previously) increases the rate of attainment of equilibrium, allowing the reaction to reach maximum yield faster.
  4. Excess of oxygen: Using excess oxygen shifts the equilibrium to the right, favouring product formation.
  5. Continuous removal of $\textsf{SO}_3$: Removing $\textsf{SO}_3$ as it is formed shifts the equilibrium to the right, increasing the yield.

(Mention the key conditions like low temperature, high pressure, catalyst).

7.25 Sulphuric acid ($\textsf{H}_2\text{SO}_4$) is a strong diprotic acid. Its ionisation in water occurs in two steps:

Step 1: $\textsf{H}_2\text{SO}_4\text{(aq)} + \textsf{H}_2\text{O(l)} \to \textsf{H}_3\text{O}^{+}\text{(aq)} + \textsf{HSO}_4^{-}\text{(aq)}$ ; $\textsf{K}_{\text{a}1} > 10$ (very large)

Step 2: $\textsf{HSO}_4^{-}\text{(aq)} + \textsf{H}_2\text{O(l)} \rightleftharpoons \textsf{H}_3\text{O}^{+}\text{(aq)} + \textsf{SO}_4^{2-}\text{(aq)}$ ; $\textsf{K}_{\text{a}2} = 1.2 \times 10^{-2}$

The first ionisation ($\textsf{K}_{\text{a}1}$) involves removing a proton from a neutral $\textsf{H}_2\text{SO}_4$ molecule to form a negatively charged $\textsf{HSO}_4^{-}$ ion. This process is very easy, and $\textsf{H}_2\text{SO}_4$ is almost completely dissociated in the first step in water, making it a very strong acid.

The second ionisation ($\textsf{K}_{\text{a}2}$) involves removing a proton from the already negatively charged $\textsf{HSO}_4^{-}$ ion to form a doubly charged $\textsf{SO}_4^{2-}$ ion. Removing a proton from a negatively charged species is more difficult than removing it from a neutral species because of the increased electrostatic attraction between the proton and the negative charge of the $\textsf{HSO}_4^{-}$ ion. Therefore, the second ionisation of $\textsf{H}_2\text{SO}_4$ is significantly weaker than the first, and $\textsf{K}_{\text{a}2}$ is much smaller than $\textsf{K}_{\text{a}1}$. It's not that $\textsf{K}_{\text{a}2}$ is greater than $\textsf{K}_{\text{a}1}$; it's actually the reverse, $\textsf{K}_{\text{a}1}$ is much greater than $\textsf{K}_{\text{a}2}$.



Group 17 Elements

Group 17 includes Fluorine (F), Chlorine (Cl), Bromine (Br), Iodine (I), Astatine (At), and Tennessine (Ts). These are called halogens (salt producers). They are highly reactive non-metallic elements. Astatine and Tennessine are radioactive.


Occurrence

Table 7.8 provides properties of F, Cl, Br, I, At.

Property F Cl Br I At$^\text{a}$
Atomic number917355385
Atomic mass/g mol$^{-1}$19.0035.4579.90126.90210
Electronic configuration[He]2s$^2$2p$^5$[Ne]3s$^2$3p$^5$[Ar]3d$^{10}$4s$^2$4p$^5$[Kr]4d$^{10}$5s$^2$5p$^5$[Xe]4f$^{14}$5d$^{10}$6s$^2$6p$^5$
Covalent radius/pm6499114133
Ionic radius X$^{-}$/pm133184196220
Ionisation enthalpy/kJ mol$^{-1}$1680125611421008
Electron gain enthalpy/kJ mol$^{-1}$–333–349–325–296
Electronegativityb43.23.02.72.2
$\Delta_{hyd}H$(X$^{-}$)/kJ mol$^{-1}$515381347305
F$_2$Cl$_2$Br$_2$I$_2$
Melting point/K54.4172.0265.8386.6
Boiling point/K84.9239.0332.5458.2
Density/g cm$^{-3}$1.5 (85K)c1.66 (203K)c3.19(273K)c4.94(293K)d
Distance X – X/pm143199228266
Bond dissociation enthalpy/(kJ mol$^{-1}$)158.8242.6192.8151.1
E$^\circ$/Ve2.871.361.090.54

(a Radioactive; b Pauling scale; c For the liquid at temperatures (K) given in the parentheses; d solid; e The half-cell reaction is X$_2$(g) + 2e$^-$ $\to$ 2X$^-$(aq)).


Electronic Configuration

General configuration: ns$^2$np$^5$. One electron short of noble gas configuration.


Atomic And Ionic Radii

Smallest atomic radii in their periods (max effective nuclear charge). Radii increase down the group (more shells). Fluorine has exceptionally small atomic radius.


Ionisation Enthalpy

Very high ionisation enthalpy (tendency to lose electron). Decreases down the group (increasing size).


Electron Gain Enthalpy

Maximum negative electron gain enthalpy in their periods (readily accept electron to achieve noble gas configuration). Becomes less negative down the group. Fluorine's value is less negative than chlorine's (small size of F atom, interelectronic repulsions in 2p orbitals).

Example 7.14 Halogens have maximum negative electron gain enthalpy in the respective periods of the periodic table. Why?

Answer:

Halogens have a general electronic configuration of ns$^2$np$^5$ in their valence shell, which is just one electron short of the stable noble gas configuration (ns$^2$np$^6$). Due to their small size and high effective nuclear charge, they have a strong attraction for an additional electron to achieve this stable configuration. When a halogen atom gains one electron, a significant amount of energy is released, resulting in a large negative electron gain enthalpy. This tendency to readily accept an electron is highest for halogens in their respective periods, giving them the maximum negative electron gain enthalpy values.


Example 7.15 Although electron gain enthalpy of fluorine is less negative as compared to chlorine, fluorine is a stronger oxidising agent than chlorine. Why?

Answer:

The oxidising power of a halogen ($\textsf{X}_2$) in aqueous solution depends on several factors related to the energy changes involved in the process:

$1/2 \textsf{X}_2\text{(g)} \xrightarrow{\text{Dissociation}} \textsf{X(g)} \xrightarrow{\text{Electron Gain}} \textsf{X}^{-}\text{(g)} \xrightarrow{\text{Hydration}} \textsf{X}^{-}\text{(aq)}$

The overall enthalpy change for this process contributes to the standard electrode potential ($\textsf{E}^0$) of the $\textsf{X}_2$/X$^-$ couple, which is a measure of oxidising power. Fluorine has a lower (less positive) bond dissociation enthalpy for the F-F bond (158.8 kJ/mol) compared to the Cl-Cl bond (242.6 kJ/mol). This makes it easier to break the F-F bond. Also, the hydration enthalpy of the small $\textsf{F}^{-}$ ion (515 kJ/mol) is much higher (more negative) than that of the larger $\textsf{Cl}^{-}$ ion (381 kJ/mol). Although the electron gain enthalpy of fluorine is less negative than chlorine's, the overall process for fluorine (including bond dissociation and very high hydration of $\textsf{F}^{-}$) releases more energy than for chlorine. This makes the standard electrode potential for $\textsf{F}_2$/F$^-$ (+2.87 V) significantly higher than for $\textsf{Cl}_2$/Cl$^-$ (+1.36 V), indicating that fluorine is a much stronger oxidising agent than chlorine.


Electronegativity

Very high electronegativity, decreases down the group. Fluorine is the most electronegative element.


Physical Properties

Gradation in properties: F$_2$ (yellow gas), Cl$_2$ (greenish yellow gas), Br$_2$ (red liquid), I$_2$ (violet solid). M.p. and b.p. increase with atomic number. All are coloured due to light absorption. Soluble in organic solvents. Anomalously low bond dissociation enthalpy of F$_2$ compared to Cl$_2$ is due to electron repulsion in small F atom.


Chemical Properties

Anomalous behavior of fluorine: Small size, highest electronegativity, low F-F bond dissociation enthalpy, no d orbitals. Leads to higher ionization enthalpy, electronegativity, electrode potential, and lower ionic/covalent radii, m.p./b.p., bond dissociation enthalpy than expected. Forms only one oxoacid (HOF). HF is liquid due to strong hydrogen bonding (others are gases).


Chlorine

Chlorine ($\textsf{Cl}_2$) is a greenish yellow gas.


Preparation


Properties

Uses: Bleaching (woodpulp, cotton), extraction of gold/platinum, manufacture of dyes, drugs, organic compounds ($\textsf{CCl}_4$, $\textsf{CHCl}_3$, DDT, refrigerants), sterilizing water, preparing poisonous gases (phosgene, tear gas, mustard gas).

Example 7.17 Write the balanced chemical equation for the reaction of Cl$_2$ with hot and concentrated NaOH. Is this reaction a disproportionation reaction? Justify.

Answer:

When chlorine gas ($\textsf{Cl}_2$) reacts with hot and concentrated sodium hydroxide ($\textsf{NaOH}$) solution, it forms sodium chloride ($\textsf{NaCl}$), sodium chlorate ($\textsf{NaClO}_3$), and water ($\textsf{H}_2\textsf{O}$).

The balanced chemical equation is:

$3\textsf{Cl}_2\text{(g)} + 6\textsf{NaOH(hot, conc.)} \to 5\textsf{NaCl(aq)} + \textsf{NaClO}_3\text{(aq)} + 3\textsf{H}_2\text{O(l)}$

This reaction is a disproportionation reaction. In a disproportionation reaction, an element in a particular oxidation state is simultaneously oxidised to a higher oxidation state and reduced to a lower oxidation state. In this reaction, chlorine ($\textsf{Cl}_2$) in its elemental form has an oxidation state of 0. In the product sodium chloride ($\textsf{NaCl}$), chlorine has an oxidation state of –1 (reduction from 0 to –1). In the product sodium chlorate ($\textsf{NaClO}_3$), chlorine has an oxidation state of +5 (oxidation from 0 to +5). Since chlorine (0) is both reduced to chlorine (–1) and oxidised to chlorine (+5), it is a disproportionation reaction.


Manufacture Of Chlorine

Intext Questions

7.29 Give the reason for bleaching action of Cl$_2$.

7.30 Name two poisonous gases which can be prepared from chlorine gas.

Answer:

7.29 Chlorine acts as a powerful bleaching agent in the presence of moisture. When chlorine dissolves in water, it forms hydrochloric acid ($\textsf{HCl}$) and hypochlorous acid ($\textsf{HOCl}$):

$\textsf{Cl}_2\text{(g)} + \textsf{H}_2\text{O(l)} \to \textsf{HCl(aq)} + \textsf{HOCl(aq)}$

Hypochlorous acid is unstable and decomposes to give hydrochloric acid and a highly reactive atom of nascent oxygen (O):

$\textsf{HOCl(aq)} \to \textsf{HCl(aq)} + \text{O}$

This nascent oxygen is a powerful oxidising agent. It reacts with coloured substances (vegetable or organic matter), oxidising them to colourless substances. The bleaching action of chlorine is therefore due to the oxidation caused by the nascent oxygen released from hypochlorous acid in the presence of moisture.

7.30 Two poisonous gases that can be prepared from chlorine gas are:

  1. Phosgene (Carbonyl chloride), $\textsf{COCl}_2$: Prepared by reacting carbon monoxide ($\textsf{CO}$) with chlorine ($\textsf{Cl}_2$) in the presence of a catalyst (e.g., activated charcoal).
    $\textsf{CO(g)} + \textsf{Cl}_2\text{(g)} \to \textsf{COCl}_2\text{(g)}$
  2. Tear gas (Chloropicrin), $\textsf{CCl}_3\text{NO}_2$: Prepared by the reaction of chloroform ($\textsf{CHCl}_3$) with nitric acid ($\textsf{HNO}_3$) or by reaction of picric acid with a chlorinating agent. A simpler synthesis involves reacting nitromethane with sodium hypochlorite and chlorine.
    $\textsf{CH}_3\text{NO}_2 + 3\textsf{NaClO} \to \textsf{CCl}_3\text{NO}_2 + 3\textsf{NaOH}$ (Simplified, overall involves reactions with chlorine and hypochlorite)
  3. Mustard gas (2,2'-Dichlorodiethyl sulfide), $\textsf{ClCH}_2\text{CH}_2\text{SCH}_2\text{CH}_2\text{Cl}$: Prepared by the reaction of ethene ($\textsf{C}_2\text{H}_4$) with sulphur monochloride ($\textsf{S}_2\text{Cl}_2$).
    $2\textsf{C}_2\text{H}_4 + \textsf{S}_2\text{Cl}_2 \to (\textsf{ClCH}_2\text{CH}_2)_2\textsf{S} + \textsf{S}$

(Name any two from the list provided in the text: phosgene, tear gas (chloropicrin), mustard gas).



Hydrogen Chloride

Hydrogen chloride ($\textsf{HCl}$) is a pungent smelling gas, highly soluble in water forming hydrochloric acid.


Preparation

$\textsf{HCl}$ gas can be dried by passing through concentrated $\textsf{H}_2\text{SO}_4$.


Properties

Uses: Manufacture of $\textsf{Cl}_2$, $\textsf{NH}_4\text{Cl}$, glucose. Extracting glue, purifying bone black, medicine, laboratory reagent.

Example 7.18 When HCl reacts with finely powdered iron, it forms ferrous chloride and not ferric chloride. Why?

Answer:

When hydrochloric acid ($\textsf{HCl}$) reacts with finely powdered iron ($\textsf{Fe}$), it undergoes a single displacement reaction where iron displaces hydrogen from the acid to form hydrogen gas ($\textsf{H}_2$) and iron(II) chloride (ferrous chloride, $\textsf{FeCl}_2$).

The balanced chemical equation is:

$\textsf{Fe(s)} + 2\textsf{HCl(aq)} \to \textsf{FeCl}_2\text{(aq)} + \textsf{H}_2\text{(g)}$

In this reaction, iron is oxidised from the 0 oxidation state to +2 in $\textsf{FeCl}_2$. Hydrogen from $\textsf{HCl}$ (+1 oxidation state) is reduced to $\textsf{H}_2$ (0 oxidation state).

The presence of nascent hydrogen gas ($\textsf{H}_2$) produced during the reaction creates a reducing atmosphere around the reaction mixture. This reducing atmosphere prevents the further oxidation of $\textsf{Fe}^{2+}$ ions to $\textsf{Fe}^{3+}$ ions, which would be required to form ferric chloride ($\textsf{FeCl}_3$). Therefore, under these conditions, ferrous chloride ($\textsf{FeCl}_2$) is formed instead of ferric chloride ($\textsf{FeCl}_3$).



Oxoacids Of Halogens

Halogens form several oxoacids, where the halogen is the central atom bonded to oxygen and hydrogen (as –OH groups). Most are unstable and exist only in solution or as salts. Fluorine forms only one oxoacid (HOF or HFO) due to its high electronegativity and small size.

Structures of oxoacids of halogens are based on VSEPR theory (Fig. 7.8).

Structures of oxoacids of chlorine: Hypochlorous acid (HOCl), Chlorous acid (HClO2), Chloric acid (HClO3), Perchloric acid (HClO4)

Oxidation states of halogens (except F) in oxoacids range from +1 to +7.

Oxidation State Halic (I) acid (Hypohalous acid) Halic (III) acid (Halous acid) Halic (V) acid (Halic acid) Halic (VII) acid (Perhalic acid)
+1HOF (Hypofluorous acid)
HOCl (Hypochlorous acid)
HOBr (Hypobromous acid)
HOI (Hypoiodous acid)
+3HOClO (Chlorous acid)
+5HOClO$_2$ (Chloric acid)
HOBrO$_2$ (Bromic acid)
HOIO$_2$ (Iodic acid)
+7HOClO$_3$ (Perchloric acid)
HOBrO$_3$ (Perbromic acid)
HOIO$_3$ (Periodic acid)

Acidic strength of oxoacids increases with increasing oxidation state of the halogen. For oxoacids of the same halogen, acidic strength increases with number of non-hydroxylated oxygen atoms.

Intext Question

7.26 Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F$_2$ and Cl$_2$.

Answer:

The oxidising power of a halogen ($\textsf{X}_2$) in aqueous solution depends on the overall energy change involved in the process of converting gaseous molecular halogen ($\textsf{X}_2$) into aqueous halide ion ($\textsf{X}^{-}$). This process involves three steps: dissociation of the $\textsf{X-X}$ bond ($\Delta_{diss}H$), electron gain by the halogen atom ($\Delta_{eg}H$), and hydration of the halide ion ($\Delta_{hyd}H$).

$1/2 \textsf{X}_2\text{(g)} \xrightarrow{1/2 \Delta_{diss}H} \textsf{X(g)} \xrightarrow{\Delta_{eg}H} \textsf{X}^{-}\text{(g)} \xrightarrow{\Delta_{hyd}H} \textsf{X}^{-}\text{(aq)}$

The net energy change (which influences the standard electrode potential $\textsf{E}^0_{\text{X}_2/\text{X}^-}$) is approximately related to these enthalpies. A more positive $\textsf{E}^0$ indicates stronger oxidising power.

Comparing $\textsf{F}_2$ and $\textsf{Cl}_2$:

  • Bond dissociation enthalpy: F$_2$ has a lower bond dissociation enthalpy (158.8 kJ/mol) than Cl$_2$ (242.6 kJ/mol). This is favourable for fluorine.
  • Electron gain enthalpy: Chlorine has a more negative electron gain enthalpy (–349 kJ/mol) than fluorine (–333 kJ/mol). This is favourable for chlorine.
  • Hydration enthalpy: The hydration enthalpy of $\textsf{F}^{-}$ ion (–515 kJ/mol) is much higher (more negative) than that of $\textsf{Cl}^{-}$ ion (–381 kJ/mol) due to the smaller size of $\textsf{F}^{-}$. This is highly favourable for fluorine.

Although chlorine has a more negative electron gain enthalpy, the lower bond dissociation enthalpy of F$_2$ and the significantly higher hydration enthalpy of the small $\textsf{F}^{-}$ ion more than compensate for this difference. The overall energy released in the hydration of $\textsf{F}^{-}$ ions is a major driving force that makes the reduction of $\textsf{F}_2$ (oxidation of other species) very favorable. Consequently, the standard electrode potential for $\textsf{F}_2$/F$^-$ (+2.87 V) is much higher than for $\textsf{Cl}_2$/Cl$^-$ (+1.36 V), indicating that $\textsf{F}_2$ is a stronger oxidising agent than $\textsf{Cl}_2$.



Interhalogen Compounds

Interhalogen compounds are formed when two different halogens react with each other. General formulas are $\textsf{XX}'$, $\textsf{XX}'_3$, $\textsf{XX}'_5$, and $\textsf{XX}'_7$, where X is the larger, more electropositive halogen, and X' is the smaller, more electronegative halogen. The number of X' atoms increases with the ratio of atomic radii between X and X'. Example: $\textsf{IF}_7$.


Preparation

By direct combination of halogens or by reaction of a halogen with a lower interhalogen compound under specific conditions.


Properties

Type Formula Physical state and colour Structure
XX′ClF
BrF
IFa
BrClb
ICl
IBr
colourless gas
pale brown gas
detected spectroscopically
gas
ruby red solid (α-form)
brown red solid (β-form)
black solid
Linear
XX′$_3$ClF$_3$
BrF$_3$
IF$_3$
ICl$_3$c
colourless gas
yellow green liquid
yellow powder
orange solid
Bent T-shaped
XX′$_5$IF$_5$
BrF$_5$
ClF$_5$
colourless gas but solid below 77 K
colourless liquid
colourless liquid
Square pyramidal
XX′$_7$IF$_7$colourless gasPentagonal bipyramidal

(aVery unstable; bThe pure solid is known at room temperature; cDimerises as Cl–bridged dimer (I$_2$Cl$_6$)).

Uses: Non-aqueous solvents, fluorinating agents (e.g., $\textsf{ClF}_3$, $\textsf{BrF}_3$ for $\textsf{UF}_6$ production in uranium enrichment).

Example 7.19 Discuss the molecular shape of BrF$_3$ on the basis of VSEPR theory.

Answer:

In the $\textsf{BrF}_3$ molecule, Bromine (Br) is the central atom. Bromine is in Group 17, so it has 7 valence electrons. It forms three single covalent bonds with three fluorine (F) atoms. The number of electrons involved in these bonds is 3 $\times$ 1 = 3. The remaining valence electrons on bromine are 7 - 3 = 4. These 4 electrons form two lone pairs ($4/2 = 2$).

So, the central atom Br has 3 bond pairs and 2 lone pairs of electrons.

According to VSEPR theory, the electron pairs (both bonding and non-bonding) arrange themselves around the central atom to minimize repulsions. The arrangement that minimizes repulsion for 5 electron pairs is trigonal bipyramidal geometry. In a trigonal bipyramidal arrangement, lone pairs preferentially occupy equatorial positions to minimize repulsions (especially lone pair-lone pair and lone pair-bond pair repulsions).

In $\textsf{BrF}_3$, the two lone pairs occupy two of the three equatorial positions, and the three bond pairs (with fluorine atoms) occupy one equatorial and two axial positions. Due to the presence of the lone pairs, the molecular shape is not trigonal bipyramidal, but rather derived from the arrangement of the atoms. The arrangement of the three fluorine atoms and the bromine atom results in a bent T-shape.

Diagram showing the VSEPR structure of BrF3 with 3 bond pairs and 2 lone pairs arranged in a trigonal bipyramid, resulting in a bent T-shape

The bond angle between the two axial Br-F bonds and the equatorial Br-F bond is ideally 90$^\circ$, but it is slightly less due to the repulsion from the lone pairs. The shape is described as bent T-shaped.


Intext Question

7.31 Why is ICl more reactive than I$_2$?

Answer:

The reactivity of halogens and interhalogens is related to the strength of the bond between the atoms. Chemical reactions often involve breaking existing bonds and forming new ones. Substances with weaker bonds tend to be more reactive.

Iodine monochloride ($\textsf{ICl}$) is an interhalogen compound, while iodine ($\textsf{I}_2$) is a halogen molecule. In $\textsf{ICl}$, the bond is between two different halogen atoms (I and Cl). In $\textsf{I}_2$, the bond is between two identical iodine atoms (I-I).

In general, the bond between two different atoms (like I-Cl) is weaker than the bond between two identical atoms (like I-I) if the bond in the parent halogen molecule is particularly strong (like Cl-Cl). However, when considering halogens beyond chlorine, the bond dissociation enthalpy decreases down the group (Cl-Cl > Br-Br > I-I). For interhalogens, the X-X' bond is usually weaker than the average of the X-X and X'-X' bonds.

More specifically, the I-Cl bond is weaker than the I-I bond in $\textsf{I}_2$. This is because the I-Cl bond is polar due to the electronegativity difference between I and Cl, but the overall strength is less than the nonpolar I-I bond in $\textsf{I}_2$. The bond dissociation enthalpy of I-Cl is approximately 208 kJ/mol, while that of I-I in $\textsf{I}_2$ is approximately 151 kJ/mol. Wait, this suggests I-Cl is stronger than I-I. Let's reconsider the general statement from the textbook: "In general, interhalogen compounds are more reactive than halogens (except fluorine). This is because X–X¢ bond in interhalogens is weaker than X–X bond in halogens except F–F bond." The F-F bond is anomalously weak due to electron repulsion. For others, the trend might hold. Let's recheck bond energies. I-Cl bond energy is listed around 208 kJ/mol. I-I bond energy is listed around 151 kJ/mol. So the bond energy of I-Cl is actually HIGHER than I-I. This contradicts the provided reasoning in the textbook.

However, the general reactivity trend is that interhalogens like ICl ARE more reactive than I$_2$. Why? It could be due to the polarity of the I-Cl bond. The I$^{\delta+}$-Cl$^{\delta-}$ bond is polar. This polarity can make the molecule more susceptible to attack by other molecules, particularly polar molecules or ions (like water or nucleophiles/electrophiles), compared to the nonpolar $\textsf{I}_2$ molecule. The polar bond can also facilitate the breaking of the I-Cl bond in certain reaction environments.

While the bond energy reason given in the text seems incorrect for ICl vs I$_2$, the observation is that ICl is indeed more reactive than I$_2$. The reason is likely related to the polarity of the I-Cl bond making it more susceptible to reaction pathways compared to the nonpolar I-I bond.



Group 18 Elements

Group 18 includes Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe), Radon (Rn), and Oganesson (Og). They are called noble gases due to their low reactivity.


Occurrence

Table 7.12 provides properties of noble gases (except Og).

Propery He Ne Ar Kr Xe Rn*
Atomic number21018365486
Atomic mass/ g mol$^{-1}$4.0020.1839.9583.80131.30222.00
Electronic configuration1s$^2$[He]2s$^2$2p$^6$[Ne] 3s$^2$3p$^6$[Ar]3d$^{10}$4s$^2$4p$^6$[Kr]4d$^{10}$5s$^2$5p$^6$[Xe]4f$^{14}$5d$^{10}$6s$^2$6p$^6$
Atomic radius/pm120160190200220
Ionisation enthalpy/kJmol$^{-1}$237220801520135111701037
Electron gain enthalpy/kJmol$^{-1}$4811696967768
Density (at STP)/gcm$^{-3}$1.8$\times$10$^{-4}$9.0$\times$10$^{-4}$1.8$\times$10$^{-3}$3.7$\times$10$^{-3}$5.9$\times$10$^{-3}$9.7$\times$10$^{-3}$
Melting point/K24.683.8115.9161.3202
Boiling point/K4.227.187.2119.7165.0211
Atmospheric content (% by volume)5.24$\times$10$^{-4}$1.82$\times$10$^{-3}$0.9341.14$\times$10$^{-4}$8.7$\times$10$^{-6}$

* radioactive


Electronic Configuration

General valence configuration: ns$^2$np$^6$ (He is 1s$^2$). Full outermost shells contribute to stability.


Ionisation Enthalpy

Very high due to stable electron configurations. Decreases down the group (increasing size).


Atomic Radii

Increase down the group (increasing atomic number and number of shells).


Electron Gain Enthalpy

Large positive values (or close to zero). No tendency to accept electron due to stable configuration.


Physical Properties

Example 7.21 Noble gases have very low boiling points. Why?

Answer:

Noble gases exist as monoatomic molecules. They do not form chemical bonds with each other. The only type of intermolecular forces present between noble gas atoms are very weak dispersion forces (also known as London forces). These forces arise from temporary fluctuations in electron distribution. Since these forces are extremely weak, very little energy is required to overcome them. Therefore, noble gases condense into liquids and freeze into solids only at very low temperatures. This results in their characteristic very low melting and boiling points.


Chemical Properties

Generally least reactive (inertness) due to:

Reactivity increases down the group due to decreasing ionisation enthalpy and increasing size (electron cloud is more easily distorted). Xenon is the most reactive among stable noble gases.

Compounds of noble gases: Reactivity was discovered by Neil Bartlett (1962) by reacting Xe with $\textsf{PtF}_6$. Compounds are formed mainly with highly electronegative elements like fluorine and oxygen. Compounds of Kr are few (only $\textsf{KrF}_2$ studied). No true compounds of Ar, Ne, He known.

Structures (VSEPR theory): $\textsf{XeF}_2$ (linear), $\textsf{XeF}_4$ (square planar), $\textsf{XeF}_6$ (distorted octahedral), $\textsf{XeO}_3$ (pyramidal), $\textsf{XeOF}_4$ (square pyramidal).

Structures of XeF2, XeF4, XeF6, XeO3, XeOF4 based on VSEPR theory

Example 7.22 Does the hydrolysis of XeF$_6$ lead to a redox reaction?

Answer:

Let's consider the hydrolysis reactions of $\textsf{XeF}_6$ with different amounts of water:

Partial hydrolysis: $\textsf{XeF}_6 + \textsf{H}_2\text{O} \to \textsf{XeOF}_4 + 2\textsf{HF}$

Partial hydrolysis: $\textsf{XeF}_6 + 2\textsf{H}_2\text{O} \to \textsf{XeO}_2\textsf{F}_2 + 4\textsf{HF}$

Complete hydrolysis: $\textsf{XeF}_6 + 3\textsf{H}_2\text{O} \to \textsf{XeO}_3 + 6\textsf{HF}$

In these reactions, let's check the oxidation states of the elements:

  • In $\textsf{XeF}_6$: Xe is +6 (assuming F is -1). F is -1.
  • In $\textsf{H}_2\text{O}$: H is +1, O is -2.
  • In $\textsf{XeOF}_4$: Xe is +6 (O is -2, F is -1), O is -2, F is -1.
  • In $\textsf{XeO}_2\text{F}_2$: Xe is +6 (O is -2, F is -1), O is -2, F is -1.
  • In $\textsf{XeO}_3$: Xe is +6 (O is -2), O is -2.
  • In $\textsf{HF}$: H is +1, F is -1.

In all these products (and reactants), the oxidation state of Xenon remains +6, Oxygen remains -2, Fluorine remains -1, and Hydrogen remains +1. Since there is no change in the oxidation states of any of the elements involved in these hydrolysis reactions, these reactions are not redox reactions.

Uses: Helium (non-inflammable, light) in balloons (meteorological), gas-cooled nuclear reactors, cryogenic agent (liquid He), diluent for $\textsf{O}_2$ in diving apparatus (low solubility in blood). Neon in discharge tubes, fluorescent bulbs (advertisements). Argon in inert atmosphere (welding, electric bulbs), handling air-sensitive substances. Xe and Kr in special light bulbs. Radon (radioactive) in cancer therapy.

Intext Questions

7.32 Why is helium used in diving apparatus?

7.33 Balance the following equation: XeF$_6$ + H$_2$O $\to$ XeO$_2$F$_2$ + HF

7.34 Why has it been difficult to study the chemistry of radon?

Answer:

7.32 Helium is used in diving apparatus (in mixtures with oxygen and nitrogen, e.g., in Trimix or Heliox) for deep-sea diving for two main reasons:

  1. Low solubility in blood: Nitrogen from air becomes significantly soluble in blood and body tissues at high pressures experienced during deep dives. As the diver ascends, the pressure decreases, and the dissolved nitrogen bubbles out of the blood, causing a painful and dangerous condition called the "bends" or decompression sickness. Helium is much less soluble in blood than nitrogen at high pressures, so substituting helium for nitrogen in the breathing mixture significantly reduces the risk of bends.
  2. Lower density: Helium is much less dense than nitrogen. Breathing dense gases at high pressure is difficult. The lower density of helium makes breathing easier at depth, reducing the work of breathing.

7.33 Balance the equation: $\textsf{XeF}_6 + \textsf{H}_2\text{O} \to \textsf{XeO}_2\text{F}_2 + \textsf{HF}$

Reactants: 1 Xe, 6 F, 2 H, 1 O

Products: 1 Xe, 2 O, 2 F, 1 H, 1 F

Number of Xe atoms is balanced (1 on each side). Number of O atoms is unbalanced (1 on left, 2 on right in XeO$_2$F$_2$). Let's put a coefficient of 2 in front of $\textsf{H}_2\text{O}$ to get 2 oxygen atoms on the left.

$\textsf{XeF}_6 + 2\textsf{H}_2\text{O} \to \textsf{XeO}_2\text{F}_2 + \textsf{HF}$

Reactants: 1 Xe, 6 F, 4 H, 2 O

Products: 1 Xe, 2 O, 2 F (in XeO$_2$F$_2$), 1 H (in HF), 1 F (in HF)

Now oxygen is balanced. Hydrogen is unbalanced (4 on left, 1 on right). Let's put a coefficient of 4 in front of $\textsf{HF}$.

$\textsf{XeF}_6 + 2\textsf{H}_2\text{O} \to \textsf{XeO}_2\text{F}_2 + 4\textsf{HF}$

Reactants: 1 Xe, 6 F, 4 H, 2 O

Products: 1 Xe, 2 O, 2 F (in XeO$_2$F$_2$), 4 H (in 4HF), 4 F (in 4HF). Total F in products = 2 + 4 = 6.

Now all atoms are balanced.

The balanced equation is: $\textsf{XeF}_6\text{(g)} + 2\textsf{H}_2\text{O(l)} \to \textsf{XeO}_2\text{F}_2\text{(l)} + 4\textsf{HF(aq)}$

7.34 Radon (Rn) is a radioactive element with a relatively short half-life (the most stable isotope, $^{222}\text{Rn}$, has a half-life of 3.8 days). This means that any sample of radon rapidly decays into other elements. Working with radioactive substances requires special handling precautions. The short half-life also limits the time available to perform chemical experiments and characterise its compounds before the sample decays away. While some compounds like RnF$_2$ have been identified using radiotracer techniques, isolating and studying stable compounds of radon has been difficult due to its inherent radioactivity and limited lifespan.



Intext Questions



Question 7.1. Why are pentahalides of P, As, Sb and Bi more covalent than their trihalides?

Answer:

Question 7.2. Why is $BiH_3$ the strongest reducing agent amongst all the hydrides of Group 15 elements ?

Answer:

Question 7.3. Why is $N_2$ less reactive at room temperature?

Answer:

Question 7.4. Mention the conditions required to maximise the yield of ammonia.

Answer:

Question 7.5. How does ammonia react with a solution of $Cu^{2+}$?

Answer:

Question 7.6. What is the covalence of nitrogen in $N_2O_5$ ?

Answer:

Question 7.7. (a) Bond angle in $PH_4^+$ is higher than that in $PH_3$. Why?

(b) What is formed when $PH_3$ reacts with an acid?

Answer:

Question 7.8. What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of $CO_2$ ?

Answer:

Question 7.9. What happens when $PCl_5$ is heated?

Answer:

Question 7.10. Write a balanced equation for the reaction of $PCl_5$ with water.

Answer:

Question 7.11. What is the basicity of $H_3PO_4$?

Answer:

Question 7.12. What happens when $H_3PO_3$ is heated?

Answer:

Question 7.13. List the important sources of sulphur.

Answer:

Question 7.14. Write the order of thermal stability of the hydrides of Group 16 elements.

Answer:

Question 7.15. Why is $H_2O$ a liquid and $H_2S$ a gas ?

Answer:

Question 7.16. Which of the following does not react with oxygen directly?

Zn, Ti, Pt, Fe

Answer:

Question 7.17. Complete the following reactions:

(i) $C_2H_4 + O_2 \rightarrow$

(ii) $4Al + 3O_2 \rightarrow$

Answer:

Question 7.18. Why does $O_3$ act as a powerful oxidising agent?

Answer:

Question 7.19. How is $O_3$ estimated quantitatively?

Answer:

Question 7.20. What happens when sulphur dioxide is passed through an aqueous solution of Fe(III) salt?

Answer:

Question 7.21. Comment on the nature of two S–O bonds formed in $SO_2$ molecule. Are the two S–O bonds in this molecule equal ?

Answer:

Question 7.22. How is the presence of $SO_2$ detected ?

Answer:

Question 7.23. Mention three areas in which $H_2SO_4$ plays an important role.

Answer:

Question 7.24. Write the conditions to maximise the yield of $H_2SO_4$ by Contact process.

Answer:

Question 7.25. Why is $K_{a_2} > K_{a_1}$ for $H_2SO_4$ in water ?

Answer:

Question 7.26. Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of $F_2$ and $Cl_2$.

Answer:

Question 7.27. Give two examples to show the anomalous behaviour of fluorine.

Answer:

Question 7.28. Sea is the greatest source of some halogens. Comment.

Answer:

Question 7.29. Give the reason for bleaching action of $Cl_2$.

Answer:

Question 7.30. Name two poisonous gases which can be prepared from chlorine gas.

Answer:

Question 7.31. Why is $ICl$ more reactive than $I_2$?

Answer:

Question 7.32. Why is helium used in diving apparatus?

Answer:

Question 7.33. Balance the following equation:

$XeF_6 + H_2O \rightarrow XeO_2F_2 + HF$

Answer:

Question 7.34. Why has it been difficult to study the chemistry of radon?

Answer:



Exercises



Question 7.1. Discuss the general characteristics of Group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity.

Answer:

Question 7.2. Why does the reactivity of nitrogen differ from phosphorus?

Answer:

Question 7.3. Discuss the trends in chemical reactivity of group 15 elements.

Answer:

Question 7.4. Why does $NH_3$ form hydrogen bond but $PH_3$ does not?

Answer:

Question 7.5. How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions involved.

Answer:

Question 7.6. How is ammonia manufactured industrially?

Answer:

Question 7.7. Illustrate how copper metal can give different products on reaction with $HNO_3$.

Answer:

Question 7.8. Give the resonating structures of $NO_2$ and $N_2O_5$.

Answer:

Question 7.9. The HNH angle value is higher than HPH, HAsH and HSbH angles. Why? [Hint: Can be explained on the basis of $sp^3$ hybridisation in $NH_3$ and only s–p bonding between hydrogen and other elements of the group].

Answer:

Question 7.10. Why does $R_3P = O$ exist but $R_3N = O$ does not (R = alkyl group)?

Answer:

Question 7.11. Explain why $NH_3$ is basic while $BiH_3$ is only feebly basic.

Answer:

Question 7.12. Nitrogen exists as diatomic molecule and phosphorus as $P_4$. Why?

Answer:

Question 7.13. Write main differences between the properties of white phosphorus and red phosphorus.

Answer:

Question 7.14. Why does nitrogen show catenation properties less than phosphorus?

Answer:

Question 7.15. Give the disproportionation reaction of $H_3PO_3$.

Answer:

Question 7.16. Can $PCl_5$ act as an oxidising as well as a reducing agent? Justify.

Answer:

Question 7.17. Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation state and hydride formation.

Answer:

Question 7.18. Why is dioxygen a gas but sulphur a solid?

Answer:

Question 7.19. Knowing the electron gain enthalpy values for $O \rightarrow O^-$ and $O \rightarrow O^{2-}$ as –141 and 702 kJ mol$^{-1}$ respectively, how can you account for the formation of a large number of oxides having $O^{2-}$ species and not $O^-$? (Hint: Consider lattice energy factor in the formation of compounds).

Answer:

Question 7.20. Which aerosols deplete ozone?

Answer:

Question 7.21. Describe the manufacture of $H_2SO_4$ by contact process?

Answer:

Question 7.22. How is $SO_2$ an air pollutant?

Answer:

Question 7.23. Why are halogens strong oxidising agents?

Answer:

Question 7.24. Explain why fluorine forms only one oxoacid, HOF.

Answer:

Question 7.25. Explain why inspite of nearly the same electronegativity, nitrogen forms hydrogen bonding while chlorine does not.

Answer:

Question 7.26. Write two uses of $ClO_2$.

Answer:

Question 7.27. Why are halogens coloured?

Answer:

Question 7.28. Write the reactions of $F_2$ and $Cl_2$ with water.

Answer:

Question 7.29. How can you prepare $Cl_2$ from HCl and HCl from $Cl_2$? Write reactions only.

Answer:

Question 7.30. What inspired N. Bartlett for carrying out reaction between Xe and $PtF_6$?

Answer:

Question 7.31. What are the oxidation states of phosphorus in the following:

(i) $H_3PO_3$

(ii) $PCl_3$

(iii) $Ca_3P_2$

(iv) $Na_3PO_4$

(v) $POF_3$?

Answer:

Question 7.32. Write balanced equations for the following:

(i) NaCl is heated with sulphuric acid in the presence of $MnO_2$.

(ii) Chlorine gas is passed into a solution of NaI in water.

Answer:

Question 7.33. How are xenon fluorides $XeF_2$, $XeF_4$ and $XeF_6$ obtained?

Answer:

Question 7.34. With what neutral molecule is $ClO^–$ isoelectronic? Is that molecule a Lewis base?

Answer:

Question 7.35. How are $XeO_3$ and $XeOF_4$ prepared?

Answer:

Question 7.36. Arrange the following in the order of property indicated for each set:

(i) $F_2$, $Cl_2$, $Br_2$, $I_2$ - increasing bond dissociation enthalpy.

(ii) HF, HCl, HBr, HI - increasing acid strength.

(iii) $NH_3$, $PH_3$, $AsH_3$, $SbH_3$, $BiH_3$ – increasing base strength.

Answer:

Question 7.37. Which one of the following does not exist?

(i) $XeOF_4$

(ii) $NeF_2$

(iii) $XeF_2$

(iv) $XeF_6$

Answer:

Question 7.38. Give the formula and describe the structure of a noble gas species which is isostructural with:

(i) $ICl_4^-$

(ii) $IBr_2^-$

(iii) $BrO_3^-$

Answer:

Question 7.39. Why do noble gases have comparatively large atomic sizes?

Answer:

Question 7.40. List the uses of neon and argon gases.

Answer: