13. Write main differences between the properties of white phosphorus and red phosphorus.
Answer:
| White phosphorus |
Red Phosphorus |
| It is a soft and waxy solid. It possesses a garlic smell. |
It is a hard and crystalline solid, without any smell. |
| It is poisonous. |
It is non-poisonous. |
| It is insoluble in water but soluble in carbon disulphide. |
It is insoluble in both water and carbon disulphide. |
| It undergoes spontaneous combustion in air. |
It is relatively less reactive. |
In both solid and vapour states, it exists as a P4 molecule.
 |
It exists as a chain of tetrahedral P4 units.
 |
14. Why does nitrogen show catenation properties less than phosphorus?
Answer:
Catenation is much more common in phosphorous compounds than in nitrogen compounds. This is because of the relative weakness of the N−N single bond as compared to the P−P single bond. Since nitrogen atom is smaller, there is greater repulsion of electron density of two nitrogen atoms, thereby weakening the N−N single bond.
15. Give the disproportionation reaction of H3PO3.
Answer:
On heating, orthophosphorus acid (H3PO3) disproportionates to give orthophosphoric acid (H3PO4) and phosphine (PH3). The oxidation states of P in various species involved in the reaction are mentioned below.
16. Can PCl5 act as an oxidising as well as a reducing agent? Justify.
Answer:
PCl5 can only act as an oxidizing agent. The highest oxidation state that P can show is +5. In PCl5, phosphorus is in its highest oxidation state (+5). However, it can decrease its oxidation state and act as an oxidizing agent.
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:
The elements of group 16 are collectively called chalcogens.
(i) Elements of group 16 have six valence electrons each. The general electronic configuration of these elements is ns2 np4 , where n varies from 2 to 6.
(ii) Oxidation state: As these elements have six valence electrons (ns2 np4), they should display an oxidation state of −2. However, only oxygen predominantly shows the oxidation state of −2 owing to its high electronegativity. It also exhibits the oxidation state of −1 (H2O2), zero (O2), and +2 (OF2). However, the stability of the −2 oxidation state decreases on moving down a group due to a decrease in the electronegativity of the elements. The heavier elements of the group show an oxidation state of +2, +4, and +6 due to the availability of d-orbitals.
(iii) Formation of hydrides: These elements form hydrides of formula H2E, where E = O, S, Se, Te, PO. Oxygen and sulphur also form hydrides of type H2E2. These hydrides are quite volatile in nature.
18. Why is dioxygen a gas but sulphur a solid?
Answer:
Oxygen is smaller in size as compared to sulphur. Due to its smaller size, it can effectively form pπ−pπ bonds and form O2 (O==O) molecule. Also, the intermolecular forces in oxygen are weak van der Wall’s, which cause it to exist as gas. On the other hand, sulphur does not form M2 molecule but exists as a puckered structure held together by strong covalent bonds. Hence, it is a solid.
19. Knowing the electron gain enthalpy values for O → O− and O → O2− as −141 and 702 kJ mol−1 respectively, how can you account for the formation of a large number of oxides having O2− species and not O−?
(Consider lattice energy factor in the formation of compounds)
Answer:
Stability of an ionic compound depends on its lattice energy. More the lattice energy of a compound, more stable it will be.
Lattice energy is directly proportional to the charge carried by an ion. When a metal combines with oxygen, the lattice energy of the oxide involving O2− ion is much more than the oxide involving O− ion. Hence, the oxide having O2− ions are more stable than oxides having O−. Hence, we can say that formation of O2− is energetically more favourable than formation of O−.
20. Which aerosols deplete ozone?
Answer:
Freons or chlorofluorocarbons (CFCs) are aerosols that accelerate the depletion of ozone. In the presence of ultraviolet radiations, molecules of CFCs break down to form chlorinefree radicals that combine with ozone to form oxygen.
21. Describe the manufacture of H2SO4 by contact process?
Answer:
Sulphuric acid is manufactured by the contact process. It involves the following steps:
Step (i):
Sulphur or sulphide ores are burnt in air to form SO2.
Step (ii):
By a reaction with oxygen, SO2 is converted into SO3 in the presence of V2O5 as a catalyst.
\(2SO_{2(g)} + O_{2(g)}\xrightarrow{V_2O_5} 2SO_{3(g)}\)
Step (iii):
SO3 produced is absorbed on H2SO4 to give H2S2O7 (oleum).
\(SO_3+H_2SO_4\longrightarrow H_2S_2O_7\)
This oleum is then diluted to obtain H2SO4 of the desired concentration.
In practice, the plant is operated at 2 bar (pressure) and 720 K (temperature). The sulphuric acid thus obtained is 96-98% pure.
22. How is SO2 an air pollutant?
Answer:
Sulphur dioxide causes harm to the environment in many ways:
1. It combines with water vapour present in the atmosphere to form sulphuric acid. This causes acid rain. Acid rain damages soil, plants, and buildings, especially those made of marble.
2. Even in very low concentrations, SO2 causes irritation in the respiratory tract. It causes throat and eye irritation and can also affect the larynx to cause breathlessness.
3. It is extremely harmful to plants. Plants exposed to sulphur dioxide for a long time lose colour from their leaves. This condition is known as chlorosis. This happens because the formation of chlorophyll is affected by the presence of sulphur dioxide.
23. Why are halogens strong oxidising agents?
Answer:
The general electronic configuration of halogens is np5 , where n = 2-6. Thus, halogens need only one more electron to complete their octet and to attain the stable noble gas configuration. Also, halogens are highly electronegative with low dissociation energies and high negative electron gain enthalpies. Therefore, they have a high tendency to gain an electron. Hence, they act as strong oxidizing agents.
24. Explain why fluorine forms only one oxoacid, HOF.
Answer:
Fluorine forms only one oxoacid i.e., HOF because of its high electronegativity and small size.
25. Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not.
Answer:
Both chlorine and oxygen have almost the same electronegativity values, but chlorine rarely forms hydrogen bonding. This is because in comparison to chlorine, oxygen has a smaller size and as a result, a higher electron density per unit volume.
26. Write two uses of ClO2.
Answer:
Uses of ClO2:
(i) It is used for purifying water.
(ii) It is used as a bleaching agent.
