Block- P Important Question Answers with Explaination / In-organic Chemistry

Question 1: Why are pentahalides more covalent than trihalides? Answer: In pentahalides, the oxidation state is +5 and in trihalides, the oxidation state is +3. Since the metal ion with a high charge has more polarizing power, pentahalides are more covalent than trihalides.

Question 2: Why is BiH3 the strongest reducing agent amongst all the hydrides of Group 15 elements? Answer: As we move down a group, the atomic size increases and the stability of the hydrides of group 15 elements decreases. Since the stability of hydrides decreases on moving from NH3 to BiH3, the reducing character of the hydrides increases on moving from NH3 to BiH3.

Question 3: Why is N2 less reactive at room temperature? Answer: The two N atoms in N2 are bonded to each other by very strong triple covalent bonds. The bond dissociation energy of this bond is very high. As a result, N2 is less reactive at room temperature.

Question 4: Mention the conditions required to maximise the yield of ammonia. Answer: Ammonia is prepared using the Haber’s process. The yield of ammonia can be maximized under the following conditions: (i) High pressure (~ 200 atm) (ii) A temperature of ~700 K (iii) Use of a catalyst such as iron oxide mixed with small amounts of K2O and Al2O3

Question 5: How does ammonia react with a solution of Cu2+? Answer: NH3 acts as a Lewis base. It donates its electron pair and forms a linkage with metal ion.

Question 6: What is the covalence of nitrogen in N2O5? Answer: From the structure of N2O5, it is evident that the covalence of nitrogen is 4.

Question 7: What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO2? Answer: White phosphorous dissolves in boiling NaOH solution (in a CO2 atmosphere) to give phosphine, PH3.

Question 8: List the important sources of sulphur. Answer: Sulphur mainly exists in combined form in the earth’s crust primarily as sulphates [gypsum (CaSO4.2H2O), Epsom salt (MgSO4.7H2O), baryte (BaSO4)] and sulphides [(galena (PbS), zinc blends (ZnS), copper pyrites (CuFeS2)].

Question 9: What is the basicity of H3PO4? Answer:  H3PO4 Since there are three OH groups present in H3PO4, its basicity is three i.e., it is a tribasic acid.

Question 10: Why is H2O a liquid and H2S a gas? Answer: H2O has oxygen as the central atom. Oxygen has smaller size and higher electronegativity as compared to sulphur. Therefore, there is extensive hydrogen bonding in H2O, which is absent in H2S. Molecules of H2S are held together only by weak van der Waal’s forces of attraction. Hence, H2O exists as a liquid while H2S as a solid.

Question 11: Which of the following does not react with oxygen directly? Zn, Ti, Pt, Fe Answer: Pt is a noble metal and does not react very easily. All other elements, Zn, Ti, Fe, are quite reactive. Hence, oxygen does not react with platinum (Pt) directly.

Question 12: Why does O3 act as a powerful oxidising agent? Answer: Ozone is not a very stable compound under normal conditions and decomposes readily on heating to give a molecule of oxygen and nascent oxygen. Nascent oxygen, being a free radical, is very reactive. Therefore, ozone acts as a powerful oxidising agent

Question 13: How is O3 estimated quantitatively? Answer: Quantitatively, ozone can be estimated with the help of potassium iodide. When ozone is made to react with potassium iodide solution buffered with a borate buffer (pH 9.2), iodine is liberated. This liberated iodine can be titrated against a standard solution of sodium thiosulphate using starch as an indicator.

Question 14: What happens when sulphur dioxide is passed through an aqueous solution of Fe(III) salt? Answer: SO2 acts as a reducing agent when passed through an aqueous solution containing Fe(III) salt. It reduces Fe(III) to Fe(II) i.e., ferric ions to ferrous ions.

Question 14: Comment on the nature of two S−O bonds formed in SO2 molecule. Are the two S−O bonds in this molecule equal? Answer: The electronic configuration of S is 1s2 2s2 2p6 3s2 3p4. During the formation of SO2, one electron from 3p orbital goes to the 3d orbital and S undergoes sp2 hybridization. Two of these orbitals form sigma bonds with two oxygen atoms and the third contains a lone pair. p-orbital and d-orbital contain an unpaired electron each. One of these electrons forms pπ- pπ bond with one oxygen atom and the other forms pπ- dπ bond with the other oxygen. This is the reason SO2 has a bent structure. Also, it is a resonance hybrid of structures I and II. Both S−O bonds are equal in length (143 pm) and have a multiple bond character.

Question 15: How is the presence of SO2 detected? Answer: SO2 is a colourless and pungent smelling gas. It can be detected with the help of potassium permanganate solution. When SO2 is passed through an acidified potassium permanganate solution, it decolonizes the solution as it reduces.

Question 16: Mention three areas in which H2SO4 plays an important role. Answer: Sulphuric acid is an important industrial chemical and is used for a lot of purposes. Some important uses of sulphuric acid are given below. (i) It is used in fertilizer industry. It is used to make various fertilizers such as ammonium sulphate and calcium super phosphate. (ii) It is used in the manufacture of pigments, paints, and detergents. (iii) It is used in the manufacture of storage batteries.

Question 17: Write the conditions to maximize the yield of H2SO4 by Contact process. Answer: Manufacture of sulphuric acid by Contact process involves three steps. 1. Burning of ores to form SO2 2. Conversion of SO2 to SO3 by the reaction of the former with O2 (V2O5 is used in this process as a catalyst.) 3. Absorption of SO3 in H2SO4 to give oleum (H2S2O7) The key step in this process is the second step. In this step, two moles of gaseous reactants combine to give one mole of gaseous product. Also, this reaction is exothermic. Thus, in accordance with Le Chatelier’s principle, to obtain the maximum amount of SO3 gas, temperature should be low and pressure should be high.

Question 18: Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F2 and Cl2. Answer: Fluorine is a much stronger oxidizing agent than chlorine. The oxidizing power depends on three factors. 1. Bond dissociation energy 2. Electron gain enthalpy 3. Hydration enthalpy The electron gain enthalpy of chlorine is more negative than that of fluorine. However, the bond dissociation energy of fluorine is much lesser than that of chlorine. Also, because of its small size, the hydration energy of fluorine is much higher than that of chlorine. Therefore, the latter two factors more than compensate for the less negative electron gain enthalpy of fluorine. Thus, fluorine is a much stronger oxidizing agent than chlorine.

Question 19: Give two examples to show the anomalous behaviour of fluorine. Answer: Anomalous behaviour of fluorine (i) It forms only one oxoacid as compared to other halogens that form a number of oxoacids. (ii) Ionisation enthalpy, electronegativity, and electrode potential of fluorine are much higher than expected.

Question 20: Sea is the greatest source of some halogens. Comment Answer: Sea water contains chlorides, bromides, and iodides of Na, K, Mg, and Ca. However, it primarily contains NaCl. The deposits of dried up sea beds contain sodium chloride and carnallite, KCl.MgCl2.6H2O. Marine life also contains iodine in their systems. For example, sea weeds contain upto 0.5% iodine as sodium iodide. Thus, sea is the greatest source of halogens.

Question 21: Give the reason for bleaching action of Cl2. Answer: When chlorine reacts with water, it produces nascent oxygen. This nascent oxygen then combines with the coloured substances present in the organic matter to oxide them into colourless substances Coloured substances + [O] → Oxidized colourless substance

Question 22: Name two poisonous gases which can be prepared from chlorine gas. Answer: Two poisonous gases that can be prepared from chlorine gas are (i) Phosgene (COCl2) (ii) Mustard gas (ClCH2CH2SCH2CH2Cl)

Question 23: Why is ICl more reactive than I2? Answer: ICl is more reactive than I2 because I−Cl bond in ICl is weaker than I−I bond in I2.

Question 24: Why is helium used in diving apparatus? Answer: Air contains a large amount of nitrogen and the solubility of gases in liquids increases with increase in pressure. When sea divers dive deep into the sea, large amount of nitrogen dissolves in their blood. When they come back to the surface, solubility of nitrogen decreases and it separates from the blood and forms small air bubbles. This leads to a dangerous medical condition called bends. Therefore, air in oxygen cylinders used for diving is diluted with helium gas. This is done as He is sparingly less soluble in blood.

Question 25: Balance the following equation: XeF6 + H2O → XeO2F2 + HF Answer: Balanced equation XeF6 + 2 H2O → XeO2F2 + 4 HF

Question 26: Why has it been difficult to study the chemistry of radon? Answer: It is difficult to study the chemistry of radon because it is a radioactive substance having a half-life of only 3.82 days. Also, compounds of radon such as RnF2 have not been isolated. They have only been identified.