3<\/sub>) is
\n(a) 3
\n(b) 2
\n(c) 1
\n(d) 0
\nAnswer:
\n(b) 2<\/p>\nQuestion 3.
\nThe stable allotrope of sulphur at room temperature is ___________ .
\nAnswer:
\nRhombic sulphur<\/p>\n
Question 4.
\nWhich element would readily replace oxygen from an oxide?
\n(a) N
\n(b) S
\n(c) F
\n(d) Cl
\n(Answer:
\n(c) F<\/p>\n
Question 5.
\nThe geometry of XeF4<\/sub> is.
\nAnswer:
\nSquare planar<\/p>\nQuestion 6.
\nThe shape and hybridisation of XeF4<\/sub> molecule
\nAnswer:
\nSquare planar and sp3<\/sup>d2<\/sup><\/p>\nQuestion 7.
\nThe least stable hydride of 15th<\/sup> group elements is.
\nAnswer:
\n(BiH3<\/sub>)<\/p>\nQuestion 8.
\nChoose the weak monobasic acid among the following.
\n(a) H3<\/sub>BO3<\/sub>
\n(b) H3<\/sub>PO2<\/sub>
\n(c) H3<\/sub>PO4<\/sub>
\n(d) HNO3<\/sub>
\nAnswer:
\n(a) H3<\/sub>BO3<\/sub><\/p>\nQuestion 9.
\nThe bond enthalpy is highest for ________
\nAnswer:
\nH2<\/sub><\/p>\nQuestion 10.
\nMagnetic moment of an atom with atomic no. 24 in aqeous solution is ____________.
\nAnswer:
\n4.90 B.M.<\/p>\n
Plus Two Chemistry The p Block Elements Two Mark Questions and Answers<\/h3>\n
Question 1.
\nIn the class, a student argued that \u201cheterogenous\u201d\u00a0catalysis and Le-Chatlier\u2019s principles are applied in contact process.”<\/p>\n
\n- Do you agree with this statement?<\/li>\n
- Justify.<\/li>\n<\/ol>\n
Answer:
\n1. Yes
\n
\nThe reaction is exothermic, reversible and the forward reaction leads to a decrease in volume. Therefore, low temperature and high pressure are the favourable conditions for maximum yield according to Le Chatelier\u2019s principle.<\/p>\n
2. Contact process is an example of heterogeneous catalysis since the catalyst V2<\/sub>O5<\/sub> is solid while the reactants are gases.<\/p>\nQuestion 2.
\nClF3<\/sub> exists, but FCl3<\/sub> doesn\u2019t, why?
\nAnswer:
\nIn the valence shell of Cl vacant 3d orbitals are available. Hence it can form ClF3<\/sub>. But, F has no vacant d orbital to show higher oxidation state and FCl3<\/sub> is not possible.<\/p>\nQuestion 3.
\nIn a classroom discussion, a student argued that carbon has the maximum catenation property.<\/p>\n
\n- Do you agree?<\/li>\n
- What do you understand by the term catenation?<\/li>\n
- List out some other elements which can show catenation?<\/li>\n<\/ol>\n
Answer:<\/p>\n
\n- Yes.<\/li>\n
- Catenation is the self linking property of an atom. It is maximum for carbon, because C-C bond is very strong.<\/li>\n
- Sulphur and Phosphorus.<\/li>\n<\/ol>\n
Question 4.
\nSome elements and their ores are given in table. Arrange them correctly.<\/p>\n
\n\n\nA<\/td>\n | B<\/td>\n<\/tr>\n |
\na) Calcium<\/td>\n | Galena<\/td>\n<\/tr>\n |
\nb) Potassium<\/td>\n | Magnetite<\/td>\n<\/tr>\n |
\nc) Lead<\/td>\n | Fluorapatite<\/td>\n<\/tr>\n |
\nd) Tin<\/td>\n | Gypsum<\/td>\n<\/tr>\n |
\ne) Phosphorus<\/td>\n | Cassiterite<\/td>\n<\/tr>\n |
\n<\/td>\n | Carnallite<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Answer:<\/p>\n \n\n\nA<\/td>\n | B<\/td>\n<\/tr>\n | \na) Calcium<\/td>\n | Gypsum<\/td>\n<\/tr>\n | \nb) Potassium<\/td>\n | Carnallite<\/td>\n<\/tr>\n | \nc) Lead<\/td>\n | Galena<\/td>\n<\/tr>\n | \nd) Tin<\/td>\n | Cassiterite<\/td>\n<\/tr>\n | \ne) Phosphorus<\/td>\n | Fluorapatite<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Question 5. \nAnalyse the statement: \u2018Ionisation enthalpy of halogens decreases with a decrease in atomic size\u2019. Is it true? Justify your answer. \nAnswer: \nNo. On moving from top to bottom in a given group, the size of the atom increases and hence ionisation enthalpy decreases.<\/p>\n Question 6. \nHydrides of group 15 are Lewis bases.<\/p>\n \n- Give reason.<\/li>\n
- Arrange the group 15 hydrides in the decreasing order of basic strength.<\/li>\n<\/ol>\n
Answer:<\/p>\n \n- Due to the presence of lone pair on the central atom which is available for donation.<\/li>\n
- NH3<\/sub> > PH3<\/sub> > ASH3<\/sub> > SbH3<\/sub> \u2265 BiH3<\/sub>.<\/li>\n<\/ol>\n
Question 7. \nAccording to VSEPR theory assign structure to XeOF4<\/sub>. \nAnswer: \nXeOF4<\/sub> possess square pyramidal shape. \n<\/p>\nQuestion 8. \nWhy does the reactivity of nitrogen differ from phosphorus? \nAnswer:<\/p>\n \n- Nitrogen has a small size, high electronegativity, high ionisation enthalpy as compared to phosphorus.<\/li>\n
- Nitrogen does not contain vacant d-orbitals in its valence shell whereas phosphorus contains vacant d-orbitals in its valence shell.<\/li>\n
- Nitrogen has the ability to form triple bond ( N \u2261 N) as a result of which its bond enthalpy (941.4 kJ mol-1<\/sup>) is very high making it less reactive.<\/li>\n<\/ol>\n
Question 9. \nWhy does NH3<\/sub> form hydrogen bond but PH3<\/sub> does not? \nAnswer: \nIn NH3<\/sub>, the nitrogen atom forms hydrogen bond because of the following reasons:<\/p>\n\n- Small size of nitrogen<\/li>\n
- High electronegativity (3.0) of nitrogen<\/li>\n<\/ul>\n
N-H bond is polar forming hydrogen bond. \nP-H bond is almost purely covalent due to larger size and lesser electronegativity.<\/p>\n Question 10. \nThe HNH angle is higher than HPH, HAsH and HSbH angles. Why? \nAnswer: \nBecause in NH3<\/sub> is sp3<\/sup> hybridised. Due to lone pair of electrons the bond angle contracts from 109\u00b0 28\u2019 to 106.5\u00b0. The sp3<\/sup> hybridisation becomes less and less distinct with increasing size of the central atom. Thus, the bond angle of the hydrides of group 15 decreases as. \n<\/p>\nQuestion 11. \nCan PCl3<\/sub> act as an oxidising as well as a reducing agent? Justify. \nAnswer: \nThis is because in PCl3<\/sub> phosphorus is in the intermediate oxidising state of -3. \n1. As reducing agent: \nThe following reactions support the reducing behaviour of PCl3<\/sub>.<\/p>\n\n- PCl3<\/sub> + SO2<\/sub>Cl2<\/sub> \u2192 PCl5<\/sub> + SO2<\/sub><\/li>\n
- PCl3<\/sub> + SO3<\/sub> \u2192 POCl3<\/sub> + SO2<\/sub><\/li>\n<\/ul>\n
2. As an oxidising agent: \nIt oxidises metals to their respective chlorides.<\/p>\n \n- 12Ag + 4PCl3<\/sub> \u2192 12AgCl + P4<\/sub><\/li>\n
- 6Na + PCl3<\/sub> \u2192 3NaCl + Na3<\/sub>P<\/li>\n<\/ul>\n
Question 12. \nExplain why inspite of nearly the same electronegativity, nitrogen forms hydrogen bonding while chlorine does not. \nAnswer: \nDue to its larger size (99 pm) as compared to oxygen (66 pm).<\/p>\n Question 13. \nOzone is a strong oxidising agent. Why? \nAnswer: \nO3<\/sub> undergoes dissociation to form nascent oxygen. This nascent oxygen is responsible for oxidising property of ozone.<\/p>\nO3<\/sub> \u2192 O2<\/sub> + [O]<\/p>\nQuestion 14. \nWhite phosphorus is more reactive than other solid phases of phosphorus. Give reason. \nAnswer: \nThis is beacuse of angular strain in the P4<\/sub> molecule where the angles are only 60\u00b0.<\/p>\nQuestion 15. \nWhat happens when<\/p>\n \n- Concentrated H2<\/sub>SO4<\/sub> is added to CaF2<\/sub>?<\/li>\n
- SO3<\/sub> is passed through water?<\/li>\n<\/ol>\n
Answer: \n1. It form hydrogen fluoride. \nCaF2<\/sub> + H2<\/sub>SO4<\/sub> \u2192 CaSO4<\/sub> + 2HF<\/p>\n2. It dissolves SO3<\/sub> to give H2<\/sub>SO4<\/sub>. \nSO3<\/sub> + H2<\/sub>O \u2192 H2<\/sub>SO4<\/sub><\/p>\nQuestion 16. \nGive two important fluorides of Xenone. Predict their structure. \nAnswer:<\/p>\n \n- XeF2<\/sub> – linear<\/li>\n
- XeF4<\/sub> – Square planar<\/li>\n
- XeF6<\/sub> – Distorted octahedral<\/li>\n<\/ul>\n
Question 17. \nWhy is N2<\/sub> less reactive at room temperature? \nAnswer: \nDue to the presence of triple bond between N atoms N2<\/sub> has high bond dissociation energy and is less reactive.<\/p>\nQuestion 18. \nArrange the following in the increasing order of thermal stability. \nASH3<\/sub>, NH3<\/sub>, PH3<\/sub> \nAnswer: \nThermal stability of hydrides, decreases from nitrogen to bismuth: NH3<\/sub> > PH3<\/sub> > AsH3<\/sub>.<\/p>\nQuestion 19. \nThe 3 allotropic forms of Pare white, red and black.<\/p>\n \n- Name the thermodynamically most stable allotrope.<\/li>\n
- Which allotrope of P is stored in water?<\/li>\n<\/ol>\n
Answer:<\/p>\n \n- Black phosphorus<\/li>\n
- White phosphorus<\/li>\n<\/ol>\n
Question 20. \nNitrogen and Phosphorus are in the same group. PCI5<\/sub> is known but NCI5<\/sub> is not known. Why? \nAnswer: \nIn phosphorus vacant d-orbitals are present. It can form pentahalides. But due to absence of d-orbitals nitrogen cannot form NCI5<\/sub>.<\/p>\nQuestion 21. \nWhy is helium used in diving apparatus? \nAnswer: \nHelium is used in diving apparatus because of its very low solubility in blood.<\/p>\n Question 22. \nWhy has it been difficult to study the chemistry of radon? \nAnswer: \nRadon is radioactive with very short half-life which makes the study difficult.<\/p>\n Question 23. \nAccount for the following: \nSO3<\/sub> is more covalent than SO2<\/sub>. \nAnswer: \nDue to high charge and small size of sulphur in +6 oxidation state in SO3<\/sub> it is more covalent than SO2<\/sub>, in which sulphur is in +4 oxidation state.<\/p>\nQuestion 24. \nI2<\/sub> is more soluble in KI than in water. \nAnswer: \nI2<\/sub> combines with KI to form the soluble complex, KI3<\/sub>. \nKI + I2<\/sub> \u2192 KI3<\/sub><\/p>\nQuestion 25. \nH3<\/sub>PO3<\/sub> is dibasic while H3<\/sub>PO4<\/sub> is tribasic. \nAnswer: \nThere are three \u2018P-OH\u2019 bonds that are ionisable in H3<\/sub>PO4<\/sub>. So, it is tribasic. In H3<\/sub>PO3<\/sub> there are only 2 ionisable \u2018P-OH\u2019 bonds. Hence it is dibasic. \n<\/p>\nPlus Two Chemistry The p Block Elements Three Mark Questions and Answers<\/h3>\nQuestion 1. \nConsider the given reaction: \n2A + 6SiO2<\/sub> \u2192 6CaSiO3<\/sub> + B \nB + 10C \u2192 D + 10CO<\/p>\n\n- Identify A, B and D.<\/li>\n
- D is stored underwater. Why?<\/li>\n
- Give two examples of oxoacids of D and compounds of D.<\/li>\n<\/ol>\n
Answer: \n1. A – Ca3<\/sub>(PO4<\/sub>)2<\/sub> \nB – P4<\/sub>O10<\/sub> \nD – P4<\/sub><\/p>\n2. Because P4<\/sub> readily catches fire in air.<\/p>\n3. Two examples of oxoacids of D and compounds of D<\/p>\n \n- Oxoacids of D:
\nOxoacids – Phosphorous acid(H3<\/sub>PO3<\/sub>), Phosphoric acid (H3<\/sub>PO4<\/sub>).<\/li>\n- Compounds of D:
\nphosphine (PH3<\/sub>), PCI5<\/sub>, PCI4<\/sub>, P4<\/sub>O6<\/sub><\/li>\n<\/ul>\nQuestion 2.<\/p>\n \n- Suggest a method for the preparation of PCI3<\/sub>.<\/li>\n
- Why does PCI3<\/sub> fume in moist air?<\/li>\n<\/ol>\n
Answer: \n1. By passing dry chlorine overheated white phosphorus. \nP4<\/sub> + 6CI2<\/sub> \u2192 4PCI3<\/sub><\/p>\n2. PCI3<\/sub> hydrolyses in the presence of moisture giving fumes of HCI. \nPCI3<\/sub> + 3H2<\/sub>O \u2192 H3<\/sub>PO3<\/sub> + 3HCI<\/p>\nQuestion 3. \nA student argued that electronegativity of p-block elements decrease along period and increases down the group.<\/p>\n \n- Do you agree with this? Explain.<\/li>\n
- Write about the metallic character of p-block elements.<\/li>\n
- Arrange the following of p-block elements in the order of decreasing oxidising power.<\/li>\n<\/ol>\n
F2<\/sub>\/F-( E\u00b0 =+2.85 V), Br\/Br–<\/sup>( E\u00b0 =+1.07 V), CI2<\/sub>\/CI–<\/sup> (E\u00b0 =+1.36V), I2<\/sub>\/I–<\/sup>( E\u00b0 =+0.57V). \nAnswer:<\/p>\n\n- No. electronegativity of p-block elements increases along a period and decreases down a group.<\/li>\n
- Most of the p-block elements are non-metallic in nature. Among p-block elements the metallic character decreases along a period and increases down a group.<\/li>\n
- F2<\/sub> > CI2<\/sub> > Br2<\/sub> > I2<\/sub>, because the standard electrode potential values decreases in the same order.<\/li>\n<\/ol>\n
Question 4. \nThe shape and hybridisation of some interhalogen compounds are given below in wrong order. Match them correctly. \n \nAnswer: \n<\/p>\n Question 5. \nAccount for the following:<\/p>\n \n- H2<\/sub>O\u00a0is a liquid while H2<\/sub>S is a gas.<\/li>\n
- H2<\/sub>S is more acidic than H2<\/sub>O<\/li>\n
- SF6<\/sub> is known while SH6<\/sub> is not known.<\/li>\n<\/ol>\n
Answer:<\/p>\n \n- There is intermolecular hydrogen bonding in H2<\/sub>O\u00a0molecule but there is no hydrogen bonding in H2<\/sub>S .<\/li>\n
- The S-H bond is weaker than O-H bond because size of S atom is greater than that of O atom. Hence H2<\/sub>S can dissociate to give H+<\/sup> ions in aqueous solution.<\/li>\n
- In the higher oxidation state, S can combine only with highly electronegative elements like F.<\/li>\n<\/ol>\n
Question 6. \nSulphur forms many allotropes such as a – sulphur, \u03b2 -sulphur etc.<\/p>\n \n- What do you mean by allotropy?<\/li>\n
- How can you convert a – sulphur to b-sulphur?<\/li>\n<\/ol>\n
Answer:<\/p>\n \n- Certain elements can exist in different forms with different physical property and same chemical properties.<\/li>\n
- b -sulphur is prepared by melting a -sulphur in a dish and cooling, till crust is obtained. Holes are then pierced into the crust, and the liquid is taken out. On removing the crust, needle shaped \u03b2 -sulphur is obtained.<\/li>\n<\/ol>\n
Question 7. \nAccount for the following:<\/p>\n \n- PH3<\/sub> has lower boiling point than NH3<\/sub>.<\/li>\n
- Pentahalides are more covalent than trihalides.<\/li>\n
- ICI is more reactive than I2<\/sub>.<\/li>\n<\/ol>\n
Answer:<\/p>\n \n- NH3<\/sub>, PH
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