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Q1 Why are pentahalides more covalent than trihalides? Ans: The group 15 elements (nitrogen, phosphorus,arsenic,antimony and bismuth) when reacted with halogen atom ,tend to form halides. The halides formed are of two types:
1) Trihalides (MX3)
2) Pentahalides (MX5) where M -15TH group element X- halogen atom
The trihalides are formed by all the elements of group 15 while pentahalides are formed by all the elements except nitrogen because there is absence of vacant d- orbital in its outermost shell.
The oxidation state of +5 in pentahalides is more as compared to +3 oxidation state in trihalides. Due to the higher positive oxidation state of central atom in pentahalide state, these atoms will have larger polarizing power than the halogen atom attached to them. The central atom in pentahalide state will tend to polarize more the halide ion to which it is attached.
But In case of trihalides due to +3 oxidation state the central atom will polarize the halogen atom to a lesser extent as compared to pentahalide state. Therefore, more the polarization, larger will be the covalent character of the bond.
Hence due larger polarization of bond in pentahalide state as compared to trihalide state, the pentahalides are more covalent than trihalides.
Q2 Why is BiH3 the strongest reducing agent amongst all the hydrides of Group 15 elements? Ans: The reducing character of the hydrides of Group 15 elements increases from NH3 to BiH3(Bismuthine) because the reducing character depends upon the stability of the hydride.The greater the unstability of anhydride,the greater is the its reducing character.Since,the BiH3 is the least stable (because the size of central atom is greatest & therefore its tendency to form stable covalent bond with small hydrogen atom decreases,as a result the bond strength decreases) in this series,hence the reducing character increases.
Q3 Why is N2 less reactive at room temperature? Ans: Dinitrogen (N2) IS formed by sharing three electron pairs between two nitrogen atoms.The two nitrogen atoms are joined by triple bond(N≡N). The nitrogen atom is very small in size ,therefore the bond length is also quite small(109.8 pm) & as a result the bond dissociation energy is quite high(946Kj / mol).This reason leads N2 to be very less reactive at room temperature.
Q4 Mention the conditions required to maximise the yield of ammonia. Ans: Ammonia is prepared using the Haber’s process.
The conditions required to maximize the yield of ammonia are as follows:
1 In accordance to Le Chatelier’s principle ,low temperature will shift the equilibrium to the right because the reaction is exothermic.This gives greater yield of ammonia.Therefore a temperature of about 450°C will be optimal for the preparation of ammonia.
2 High pressure on the reaction at equilibrium favours the shift of the equilibrium to the right because the forward reaction proceeds with a decrease in number of moles.Hence a pressure of about 200 atm will be optimal for the higher yield of ammonia.
3 A catalyst should be used to increase the rate of reaction & to quickly attain equilibrium.For eg as iron oxide mixed with small amounts of K2O and Al2O3 can be used as catalyst.
4 N2 & H2 gases should be pure in nature to increase the yield of ammonia.
Q5 How does ammonia react with a solution of Cu2+? Ans: Ammonia acts as a Lewis base due to the presence of lone pair of electrons on the nitrogen atom.Therefore,it can form coordinate bond with electron deficient molecules or a number of transition metal cations to from complex compounds.Ammonia reacts with a solution of Cu2+ to from a deep blue coloured complex.The reaction for the above complex can be written as follows:
Cu2+(aq) + 4 NH3(aq) ⇔ [Cu(NH3)4]2+(aq)
Blue Deep Blue
This property of ammonia to form complex compounds is useful in detection of metal ions.
Q6 What is the covalence of nitrogen in N2O5? Ans: Covalency is defined as the tendency of an atom to form number of covalent bond with other molecule. N2O5 (Dinitrogen pentoxide) is one of the form of the nitrogen atom which belongs to group 15 in the periodic table. Nitrogen combines with oxygen to form a number of oxides having different oxidation states. Nitrogen has tendency to form pπ-pπ multiple bonds, which decides the structure of oxides. N2O5 has following structure:
From the above structure, it is clear that Nitrogen atom is sharing its electrons with oxygen atom. Nitrogen shares it four pair of electrons with oxygen, therefore nitrogen covalency is four (4).
Q7 Bond angle in PH+4 is higher than that in PH3. Why? Ans: PH3 (Phosphine) is a hydride of Phosphorus. Phosphine is a gas and has pyramidal structure. Phosphorus involved is sp3 hybridized.
Now both PH4+ and PH3 have sp3 hybridization state for phosphorous .In PH4+,all the four orbitals are bonded, whereas in PH3,there is a lone pair of electrons on Phosphorous .In PH4+, the HPH bond angle is tetrahedral angle of 109.5°.But in PH3 ,lone pair- bond pair repulsion is more than bond pair-bond pair repulsion so that bond angles become less than normal tetrahedral angle .The bond angle in PH3 is about 93.6°.
Q8 What happens when white phosphorus is heated with concentrated NaO Hsolution in an inert atmosphere of CO2? Ans: Phosphorus exist in many allotropic forms. The three main forms are:
1) White Phosphorus
2) Red Phosphorus
3) Black Phosphorus
White Phosphorus is obtained from phosphorite rock with coke and sand. It consist of P4 units.The four P- atoms lie at the corners of a regular tetrahedron. It is less stable and therefore more reactive than other allotropic forms of Phosphorus. On heating with concentrated NaOH in an inert environment of CO2 following reaction takes place:
Phosphine (PH3) is formed.This reaction is an example of disproportionation reaction in which oxidation state of phosphorus decreases from 0 in P4 to -3 in PH3,while it increases from 0 in P4 to +1 in NaH2PO2.
Q9 What happens when PCl5 is heated? Ans: Phosphorus forms two type of halides i.e phosphorus trihalides (PCl3) and phosphorus pentahalide PCl5. In PCl5 phosphorus undergoes sp3 d hybridization and has trigonal bipyramidal structure .It has three equatorial P-Cl bonds and two axial P-Cl bonds which are different. The axial bonds(219pm) are larger than equatorial bonds(204pm). PCl5 is thermally less stable than PCl3.On heating it sublimes but decomposes on stronger heating into trichloride and chlorine.The reaction involved is as under:
heat
PCL5 → PCL3 + CL2
It is a reversible reaction,in which PCl3 combines with Cl2 to form PCl5.
Q10 Write a balanced equation for the hydrolytic reaction of PCl5 in heavy water. Ans: In moist air ,PCl5 hydrolysis to POCl3 and finally gets converted to phosphoric acid.The reaction is as follows:
PCL5 + H2O → POCl3 + 2HCl
phosphorus oxychloride
POCl3 + H2O → H3PO4 + 3HCl
phosphoric acid
But on treatment of PCl5 with heavy water (D2O) only phosphorus oxychloride if formed with the liberation of 2 molecules of deuterium chloride as a side product. The reaction involved is:
PCl5 + D2O → POCl3 + 2DCl
phosphorus oxychloride
Q11 What is the basicity of H3PO4? Ans: H3PO4
Since there are three OH groups present in H3PO4, its basicity is three i.e., it is a tribasic acid.
Q12 What happens when H3PO3 is heated? Ans: H3PO3,on heating, undergoes disproportionation reaction to form PH3 and H3PO4. The oxidation numbers of P in H3PO3,PH3, and H3PO4 are +3, - 3, and +5 respectively. As the oxidation number of the same element is decreasing and increasing during a particular reaction, the reaction is a disproportionation reaction.
Q13 List the important sources of sulphur. Ans: 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)].
Q14 Write the order of thermal stability of the hydrides of Group 16 elements. Ans: The thermal stability of hydrides decreases on moving down the group. This is due to a decrease in the bond dissociation enthalpy (H-E) of hydrides on moving down the group.
Therefore,
Q15 Why is H2O a liquid and H2S a gas? Ans: 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 gas.
Q16 Which of the following does not react with oxygen directly? Zn, Ti, Pt, Fe Ans: 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.
Q17 Complete the following reactions: (i) C2H4 + O2 → (ii) 4Al + 3O2 → Ans: (i) C2H4 + 3O2 → 2CO2 + 2H2O
Ethene Oxygen Carbon dyoxide Water
(ii) 4Al + 3O2 → 2Al2O3
Aluminium Oxygen Alumina
Q18 Why does O3 act as a powerful oxidising agent? Ans: 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.
O3 O2 + [O]
Ozone Oxygen Nascent Oxygen
Therefore, ozone acts as a powerful oxidising agent.
Q19 How is O3 estimated quantitatively? Ans: 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. The reactions involved in the process are given below.
2I- + H2O + O3 → 2OH- + I2 + O2
Iodide Ozone Iodine
I2 + 2Na2S2O3 → Na2S4O6 + NaI
Sodium Sodium
thiosulphate tetrathionate
Q20 What happens when sulphur dioxide is passed through an aqueous solution of Fe(III) salt? Ans: 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.
2Fe3+ + SO2 + 2H2O → 2Fe2+ + SO42- + 4H+
Q21 Comment on the nature of two S-O bonds formed in SO2 molecule. Are the two S-O bonds in this molecule equal? Ans: 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 p π- p π bond with one oxygen atom and the other forms p π- d π 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.
Q22 How is the presence of SO2 detected? Ans: 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 MnO4- ions to Mn2+ ions
5SO2 + 2MnO4- + 2H2O → 5SO42- + 4H+ + 2Mn2+
Q23 Mention three areas in which H2SO4 plays an important role. Ans: Sulphuric acid is an important industrial chemicaland 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.
Q24 Write the conditions to maximize the yield of H2SO4 by Contact process. Ans: 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.
Q25 Why is Ka2 << Ka1 for H2SO4 in water? Ans: It can be noticed that Ka1 >> Ka2
This is because a neutral H2SO4 has a much higher tendency to lose a proton than the negatively charged HSO4- . Thus, the former is a much stronger acid than the latter.
Q26 Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F2 and Cl2. Ans: 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.
Q27 Give two examples to show the anomalous behaviour of fluorine. Ans: 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.
Q28 Sea is the greatest source of some halogens. Comment. Ans: 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.
Q29 Give the reason for bleaching action of Cl2. Ans: 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.
Cl2 + H2O → 2HCl + [O]
Coloured substances + [O] → Oxidized colourless substance
Q30 Name two poisonous gases which can be prepared from chlorine gas. Ans: Two poisonous gases that can be prepared from chlorine gas are
(i) Phosgene (COCl2)
(ii) Mustard gas (ClCH2CH2SCH2CH2Cl)
Q31 Why is ICl more reactive than I2? Ans: ICl is more reactive than I2 because I-Cl bond in ICl is weaker than I-I bond in I2.
Q32 Why is helium used in diving apparatus? Ans: 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.
Q33 Balance the following equation: XeF6 + H2O → XeO2F2 + HF Ans: Balanced equation:
XeF6 + 2 H2O → XeO2F2 + 4 HF
Q34 Why has it been difficult to study the chemistry of radon? Ans: 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.