(a) (i) NH3 has lone pair of electron, so, bond angle is 107°, whereas NH+4 does not, therefore, bond angle is 109.5°. Rhombic sulphur. (iii) It is because helium does not dissolve as it is inert and lighter in blood therefore, does not cause bends or pain. (b) (i) Refer Ans. (b) (i) Refer Ans. 3. Answer: to Q.77 (i). Answer: (i) Bond enthalpy of F2 is lower than that of Cl2. Question 6: (i) All the P—Cl bonds in PCl5 molecule are not equivalent. Refer Ans. to Q.41 (i). Question 89: What is the basicity of H3P04? (iii) Oxygen shows catenation behaviour less than sulphur. Write the reactions involved. (i) BrF3 and (ii) XeF4 (ii) Refer Ans. Write the structures of the following molecules: (i) H2SO3 (ii) XeOF4 Both contain 6 electrons in the outermost shell. Answer: Question 25: (i) The molecules NH3 and NF3 have dipole moments which are of opposite direction. (ii) PH3 has lower boiling point than NH?. (iii) Fluorine is the most electronegative and small in size, therefore, it does not show positive oxidation state. (iii) Refer Ans. Question 35: /BitsPerComponent 4 (i) Iron on reaction with HCl forms FeCl2 and not FeCl3. (iii) It is due to interelectronic repulsion between valence electrons of smaller size ‘F’ atom than Cl. We need oxygen all the time throughout our lives. Account for the following: Question 144: (a) (i) Refer Ans. to Q.81 (ii). Question 37: Question 163: Question 167: Answer: (a) Mention the optimum conditions for the industrial manufacture of ammonia by Haber’s process. 13 (b) (i). (ii) Helium does not form any real chemical compounds. In a group, the chemical elements have atoms with identical valence electron counts and identical valence vacancy counts. b) All elements form dioxide with formula MO 2, SO 2 is a gas, SeO 2 is volatile solid. Question 1. (a) (i) Refer Ans. These important questions are prepared by our academic experts based on the syllabus guidelines prescribed by MCI. (a) Draw the structures of the following molecules: Answer: Answer: Answer: to Q.29 (ii). (a) (i) It is due to more repulsion between valence electrons of F than Cl due to exceptionally small size. When they come back to lower energy level, they radiates complementary colours. to Q. There are a total of 75 questions from all three subjects and for each subject, 100 marks are allotted. (b) Draw the structures of the following: (b) White phosphorus is monomeric, whereas red phosphorus is polymeric. How would you account for the following: (i) NF3 (ii) H2S2O8 (iii) H3PO3 to Q.29 (ii). (i) Refer Ans. Account for the following: Answer: Question 85: It is dibasic acid because it has two replaceable hydrogen attached with oxygen. Answer: (ii) The negative value of electron gain enthalpy is less for fluorine than that for chlorine. Question 19: (iii) Refer Ans. Answer: (i) Sulphur vapour exhibits paramagnetic behaviour: (iii) SF6 is stearically protected, therefore, less reactive than SF4 which is not stearically protected. Question 125: (i) Refer Ans. While TeO 2 and PoO 2 are non – volatile crystalline solids.. c) Ozone: It is unstable and easily decomposes into oxygen. (i) It is because nitrogen does not have vacant d-orbitals. (ii) The stability of +3 state increases down the group in group 15 of the (a) Draw the molecular structures of the following compounds: H2S04 and H2SO3. Explain the following facts giving appropriate reason in each case: Answer: Question 46: (ii) P—P bond is stronger than N—N bond due to less repulsion between valence electrons. (ii) P4(s) + 3NaOH(aq) + 3H2O(l) ——–> 3NaH2P02(s) + PH3(g) (iii) No chemical compound of helium is known. /Width 156 Answer: Question 58: (iv) Helium. It is monobasic acid because it has only one replaceable hydrogen atom, Question 73: to Q.40 (a) (ii). to Q.85 (b) (ii). In-text question… Draw the structural formulae of molecules of following compound: (i) BrF3 (ii) (HP03)3 (iii) XeF4 (e) CaF2 + H2S04 ———–> CaS04 + 2HF. (i) Fluorine never exhibits any positive oxidation state. (i) Refer Ans. (iii) Noble gases have very low boiling points. (ii) P4 + NaOH + H2O ———> (ii) Refer Ans. Question 161: HI is stronger acid than HCl in aqueous solution because it has lower bond dissociation energy. Question 90: Answer: Fig. Answer: Answer: Draw the structure of O2 and S8 molecules Ozone (O3): (ii) Sulphur has greater tendency for catenation than oxygen. Complete the following equations: Question 110: (ii) Refer Ans. Answer: (ii) XeF4 + O2F2 ———-> Why ? Answer: Answer: Question 31: (a) Account for the following: Answer: Question 10: Question 5: Draw the structures of the following: Question 116: to Q.87 (b) (i). What inspired N. Bartlett to prepare this compound? (iii) BiH3 is the strongest reducing agent amongst all the hydrides of group 15. (i) Refer Ans. (ii) Red phosphorus is less reactive than white phosphorus 17 (b) (i). (i) Chloropicrin (ii) Phosgene (iii) Mustard gas. Answer: Answer: Allotropy: All elements of Group 15 except nitrogen show allotropy. (iii) Refer Ans. This is because down the group, size of the element (M) increases, M-H bond length increases and thus, stability of M-H bond decreases so that it can be broken down easily. Question 36: (ii) Iron dissolves in HCl to form FeCl2 and not FeCl3. Periodic Classification of Elements Class 10 Important Questions and Answers Science Chapter 5. (i) XeF4 + SbF5 ——–> (ii) Cl2 + F2 (excess) ——> Refer Ans. Which is more acidic and why, H20 or H2S? Ans: Traces of sulphur occur as hydrogen sulphide in volcanoes. Answer: (ii) Refer Ans. << (i) H2S is more acidic than H2O. Give reasons for the following: (iii) BiH3 has lowest bond dissociation enthalpy due to longer bond length. Name two poisonous gases which can be prepared from chlorine gas. Abstract Algebra Manual : Problems and solution (only the section on GROUPS) Download . Account for the following: to Q.31 (ii). (b) (i) It is because P—P single bond is stronger than the single N—N bond. (ii) It is due to inert pair effect. (a) (i) It is because bond dissociation energy of HCl is lower than H2S which Question 124: Group, in chemistry, a column in the periodic table of the chemical elements. (ii) XeF4 + O2F2 ———-> Question 8: NaOH solution? Why? Question 143: to Q.83 (ii). Question 3: Arrange the following in the order of property indicated against each set: (a) (i) Refer Ans. Question 165: Answer: (ii) Electron gain enthalpy with negative sign of fluorine is less than that of chlorine. to Q.41 (ii). to Q.65 (a) (ii). to Q.41 (iii). (ii) H2S04is a strong acid, therefore, its K is very high as it dissociates into H3O+ and HS04 almost completely. Extra Questions for Class 9th: Ch 3 Atoms and Molecules (Science) Important Questions Answer Included Very Short Answer Questions (VSAQs): 1 Mark Q1.State the Law of Conservation of mass. to Q.20 (b) (i). (a) (i) 5SO2 + 2MnO- + 2H2O ——–> 5SO4– + 2Mn2+ + 4H+ Here we have covered Important Questions on Classification of Elements and Periodicity in Properties for Class 11 Chemistry subject.. Chemistry Important Questions Class 11 are given below. (b)Draw the structures of the following: Answer: This paper. Question 95: PH3 < H2S < HCl (b) Explain the following observations: to Q. Question 104: Why does N02 dimerise? Question 168: Answer the following: Draw the structures of the following: (a) Draw the structures of the following: (ii) sulphur dioxide gas is passed through an aqueous solution of a Fe (III) salt? (i) H2S2O8(ii) Red P4 Answer: Answer: to Q.88 (a) (ii). Very Short Answer Questions. Fluorine does not exhibit any positive oxidation state. (i) It is because oxygen is diatomic and has less intermolecular forces of attraction, whereas sulphur is octa-atomic (S8) and has more intermolecular forces of attraction. It is because it is most electronegative element and best oxidising agent. 40. (a) (i) Bi3+ is more stable than Sb3+ due to inert pair effect. Question 146: (b) Write the formula of first noble gas compound prepared by N. Bartlett. (ii) Why does fluorine not play the role of a central atom in interhalogen compounds? Anarchy Edibles Owner, Moles To Liters, Lewis Structure Of Gallium, 奇皇后 バヤンフト 最後, Mexican American War Dbq Apush, Ariel Helwani Podcast, " /> (a) (i) NH3 has lone pair of electron, so, bond angle is 107°, whereas NH+4 does not, therefore, bond angle is 109.5°. Rhombic sulphur. (iii) It is because helium does not dissolve as it is inert and lighter in blood therefore, does not cause bends or pain. (b) (i) Refer Ans. (b) (i) Refer Ans. 3. Answer: to Q.77 (i). Answer: (i) Bond enthalpy of F2 is lower than that of Cl2. Question 6: (i) All the P—Cl bonds in PCl5 molecule are not equivalent. Refer Ans. to Q.41 (i). Question 89: What is the basicity of H3P04? (iii) Oxygen shows catenation behaviour less than sulphur. Write the reactions involved. (i) BrF3 and (ii) XeF4 (ii) Refer Ans. Write the structures of the following molecules: (i) H2SO3 (ii) XeOF4 Both contain 6 electrons in the outermost shell. Answer: Question 25: (i) The molecules NH3 and NF3 have dipole moments which are of opposite direction. (ii) PH3 has lower boiling point than NH?. (iii) Fluorine is the most electronegative and small in size, therefore, it does not show positive oxidation state. (iii) Refer Ans. Question 35: /BitsPerComponent 4 (i) Iron on reaction with HCl forms FeCl2 and not FeCl3. (iii) It is due to interelectronic repulsion between valence electrons of smaller size ‘F’ atom than Cl. We need oxygen all the time throughout our lives. Account for the following: Question 144: (a) (i) Refer Ans. to Q.81 (ii). Question 37: Question 163: Question 167: Answer: (a) Mention the optimum conditions for the industrial manufacture of ammonia by Haber’s process. 13 (b) (i). (ii) Helium does not form any real chemical compounds. In a group, the chemical elements have atoms with identical valence electron counts and identical valence vacancy counts. b) All elements form dioxide with formula MO 2, SO 2 is a gas, SeO 2 is volatile solid. Question 1. (a) (i) Refer Ans. These important questions are prepared by our academic experts based on the syllabus guidelines prescribed by MCI. (a) Draw the structures of the following molecules: Answer: Answer: Answer: to Q.29 (ii). (a) (i) It is due to more repulsion between valence electrons of F than Cl due to exceptionally small size. When they come back to lower energy level, they radiates complementary colours. to Q. There are a total of 75 questions from all three subjects and for each subject, 100 marks are allotted. (b) Draw the structures of the following: (b) White phosphorus is monomeric, whereas red phosphorus is polymeric. How would you account for the following: (i) NF3 (ii) H2S2O8 (iii) H3PO3 to Q.29 (ii). (i) Refer Ans. Account for the following: Answer: Question 85: It is dibasic acid because it has two replaceable hydrogen attached with oxygen. Answer: (ii) The negative value of electron gain enthalpy is less for fluorine than that for chlorine. Question 19: (iii) Refer Ans. Answer: (i) Sulphur vapour exhibits paramagnetic behaviour: (iii) SF6 is stearically protected, therefore, less reactive than SF4 which is not stearically protected. Question 125: (i) Refer Ans. While TeO 2 and PoO 2 are non – volatile crystalline solids.. c) Ozone: It is unstable and easily decomposes into oxygen. (i) It is because nitrogen does not have vacant d-orbitals. (ii) The stability of +3 state increases down the group in group 15 of the (a) Draw the molecular structures of the following compounds: H2S04 and H2SO3. Explain the following facts giving appropriate reason in each case: Answer: Question 46: (ii) P—P bond is stronger than N—N bond due to less repulsion between valence electrons. (ii) P4(s) + 3NaOH(aq) + 3H2O(l) ——–> 3NaH2P02(s) + PH3(g) (iii) No chemical compound of helium is known. /Width 156 Answer: Question 58: (iv) Helium. It is monobasic acid because it has only one replaceable hydrogen atom, Question 73: to Q.40 (a) (ii). to Q.85 (b) (ii). In-text question… Draw the structural formulae of molecules of following compound: (i) BrF3 (ii) (HP03)3 (iii) XeF4 (e) CaF2 + H2S04 ———–> CaS04 + 2HF. (i) Fluorine never exhibits any positive oxidation state. (i) Refer Ans. (iii) Noble gases have very low boiling points. (ii) P4 + NaOH + H2O ———> (ii) Refer Ans. Question 161: HI is stronger acid than HCl in aqueous solution because it has lower bond dissociation energy. Question 90: Answer: Fig. Answer: Answer: Draw the structure of O2 and S8 molecules Ozone (O3): (ii) Sulphur has greater tendency for catenation than oxygen. Complete the following equations: Question 110: (ii) Refer Ans. Answer: (ii) XeF4 + O2F2 ———-> Why ? Answer: Answer: Question 31: (a) Account for the following: Answer: Question 10: Question 5: Draw the structures of the following: Question 116: to Q.87 (b) (i). What inspired N. Bartlett to prepare this compound? (iii) BiH3 is the strongest reducing agent amongst all the hydrides of group 15. (i) Refer Ans. (ii) Red phosphorus is less reactive than white phosphorus 17 (b) (i). (i) Chloropicrin (ii) Phosgene (iii) Mustard gas. Answer: Answer: Allotropy: All elements of Group 15 except nitrogen show allotropy. (iii) Refer Ans. This is because down the group, size of the element (M) increases, M-H bond length increases and thus, stability of M-H bond decreases so that it can be broken down easily. Question 36: (ii) Iron dissolves in HCl to form FeCl2 and not FeCl3. Periodic Classification of Elements Class 10 Important Questions and Answers Science Chapter 5. (i) XeF4 + SbF5 ——–> (ii) Cl2 + F2 (excess) ——> Refer Ans. Which is more acidic and why, H20 or H2S? Ans: Traces of sulphur occur as hydrogen sulphide in volcanoes. Answer: (ii) Refer Ans. << (i) H2S is more acidic than H2O. Give reasons for the following: (iii) BiH3 has lowest bond dissociation enthalpy due to longer bond length. Name two poisonous gases which can be prepared from chlorine gas. Abstract Algebra Manual : Problems and solution (only the section on GROUPS) Download . Account for the following: to Q.31 (ii). (b) (i) It is because P—P single bond is stronger than the single N—N bond. (ii) It is due to inert pair effect. (a) (i) It is because bond dissociation energy of HCl is lower than H2S which Question 124: Group, in chemistry, a column in the periodic table of the chemical elements. (ii) XeF4 + O2F2 ———-> Question 8: NaOH solution? Why? Question 143: to Q.83 (ii). Question 3: Arrange the following in the order of property indicated against each set: (a) (i) Refer Ans. Question 165: Answer: (ii) Electron gain enthalpy with negative sign of fluorine is less than that of chlorine. to Q.41 (ii). to Q.65 (a) (ii). to Q.41 (iii). (ii) H2S04is a strong acid, therefore, its K is very high as it dissociates into H3O+ and HS04 almost completely. Extra Questions for Class 9th: Ch 3 Atoms and Molecules (Science) Important Questions Answer Included Very Short Answer Questions (VSAQs): 1 Mark Q1.State the Law of Conservation of mass. to Q.20 (b) (i). (a) (i) 5SO2 + 2MnO- + 2H2O ——–> 5SO4– + 2Mn2+ + 4H+ Here we have covered Important Questions on Classification of Elements and Periodicity in Properties for Class 11 Chemistry subject.. Chemistry Important Questions Class 11 are given below. (b)Draw the structures of the following: Answer: This paper. Question 95: PH3 < H2S < HCl (b) Explain the following observations: to Q. Question 104: Why does N02 dimerise? Question 168: Answer the following: Draw the structures of the following: (a) Draw the structures of the following: (ii) sulphur dioxide gas is passed through an aqueous solution of a Fe (III) salt? (i) H2S2O8(ii) Red P4 Answer: Answer: to Q.88 (a) (ii). Very Short Answer Questions. Fluorine does not exhibit any positive oxidation state. (i) It is because oxygen is diatomic and has less intermolecular forces of attraction, whereas sulphur is octa-atomic (S8) and has more intermolecular forces of attraction. It is because it is most electronegative element and best oxidising agent. 40. (a) (i) Bi3+ is more stable than Sb3+ due to inert pair effect. Question 146: (b) Write the formula of first noble gas compound prepared by N. Bartlett. (ii) Why does fluorine not play the role of a central atom in interhalogen compounds? Anarchy Edibles Owner, Moles To Liters, Lewis Structure Of Gallium, 奇皇后 バヤンフト 最後, Mexican American War Dbq Apush, Ariel Helwani Podcast, " />

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(ii) Refer Ans. to Q.70. (a) Complete the following chemical equations: (ii) Refer Ans. (i) It is due to low bond dissociation energy which is due to longer bond length. (b) Explain the following observations: Therefore, it liberates nascent oxygen easily. (ii) N—N single bond is weaker than P—P single bond. Account for the following: 13 (b) (ii). (ii) Helium does not form any chemical compound. Question 39: Why? Question 38: (ii) White phosphorus is more reactive than red phosphorus. Hence sulphur has higher melting and boiling points than oxygen. to Q. (i) Nitrogen is much less reactive than phosphorus. 17 (b) (ii). Occurrence of group 16 elements of the modern periodic table (i)PCl5 is heated? (ii) Refer Ans. Answer: (iii) Refer Ans. to Q.86 (b) (ii). Answer: Question 29: (ii) Unlike xenon, no distinct chemical compound of helium is known. Answer: (ii) H20, H2S, H2Se, H2Te – increasing acidic character (d) It is used as bleaching agent and disinfectant. (i) Sulphur exists as S2 molecule like O2 in vapour state and has two unpaired electrons. (iii) It is due to stable electronic configuration, i.e. to Q.85 (b) (ii). Important Tags – Group 16 Elements – Trends & Properties, Chemistry notes for class 12, The p-Block Elements, Atomic and Physical Properties, Acidic Character of Hydrides, group 16. (ii) It is because ‘Cl’ is more electronegative than ‘I’. H2S04 ——-> (i) Oxygen is a gas but sulphur is a solid. (iii) It is because they absorb light from visible region and radiate complementary colour. (a) (i) Refer Ans. (b) (i) Is is due to high bond dissociation enthalpy of N = N (triple bond) than single bond in P4. to Q.35 (ii). (d) Excess of oxygen. (ii) 3Cu + 8HN03(dilute) ——–> 3Cu(N03)2(aq) + 2NO(g) + 4H20(l) (i) XeF4(ii) H2S2O7 (a) (i) AgCl(s) + 2NH3(aq) ——–> [Ag(NH3)2]+Cl-(aq) – Matt Smith Apr 22 '16 at 14:09. It is due to higher standard reduction potential of F2 which is due to low bond dissociation energy of F—F bond because of repulsion among small size F atoms, high electron gain enthalpy and highest hydration enthalpy. Question 44: 0 0. It has least polarizing power. Question 111: (ii) Refer Ans. Why? Answer: Question 41: Question 142: (ii) SnCl4 is more covalent than SnCl2. to Q.20 (a) (ii). Answer: Question 60: Draw the structures of the following: (b) (i) Refer Ans. to Q.89 (a) (i). to Q.15 (a) (i). (ii) Refer Ans. (ii) It is because these are not in the same plane. Answer: Question 59: (b) Draw the structures of the following molecules: 6 0 obj (ii)H 2 Te has longest … Revision KS4 Science IGCSE/O level/GCSE Chemistry Information Study Notes for revising for AQA GCSE Science, Edexcel GCSE Science/IGCSE Chemistry & OCR 21stC Science, OCR Gateway Science (revise courses equal to US grades 9-10) Detailed GCSE revision notes on Group 7 The Halogens * Halogen task sheet answers. Answer: Question 145: Question 78: (a) What happens when (iii) It is because H—I has lower bond dissociation enthalpy thajn HCl due to longer bond length. (i) It is because F2 is stronger oxidising agent than Cl2, therefore, NF3 is exothermic compound, whereas NCl3 is endothermic as N—F bond is stronger than N—Cl bond. Refer Ans. List the important sources of sulphur. (ii) It is due to low bond dissociation energy, high hydration energy and high electron affinity. (i) Helium is used in diving apparatus. (a) Account for the following: (i) P4 + H2O ———> to Q.29 (ii). to Q.65 (in). (iv) Write the disproportionation reaction of H3P03. (ii) Refer Ans. to Q.83 (i). Question 26: (ii) It is because Xenon (Xe) is a noble gas and has lone pair of electrons in its d-orbitals. H 2S is less acidic than H 2Te. (i) HI < HBr < HCl < HF is the increasing order of bond dissociation enthalpy. Many industries utilize sulfur, … (i) H2S2O8 (ii) HClO4 to Q.148. (iii) Refer Ans. (b) Write the structural difference between white phosphorus and red phosphorus. Answer: (iii) Refer Ans. (iii) Halogens are strong oxidising agents. of ozone layer. 5 thoughts on “ p-Block Elements MCQs | Question – Paper 1 ” Reply ↓ ANSHUMAN PANDEY at . (b) Explain the following observations: (iii) It is because H3PO2 has two P—H bonds, whereas H3P03 has only one P—H bond. NH4Cl(aq) + NaNO2(aq) ——> NaCl(aq) + N2(g) + 2H2O(l). to Q.35 (ii). (iii) Halogens are strong oxidizing agents. READ PAPER. (i) P4 + S02Cl2 ——> (ii) H 2 Te is the strongest reducing agent amongst all the hydrides of group 16-elements. (i) All the P—Cl bonds in PCl5 molecule are not equivalent. Explain the following: (b) Explain the following observations: Answer: (ii) Sulphur in vapour state exhibits paramagnetism. Question 97: (iii) Refer Ans. It is due to single bond in phosphorus which has less bond dissociation energy as compared to nitrogen which has triple bond (N=N) has high bond dissociation energy, so, nitrogen is unreactive. (ii) The stability of +3 state increases down the group in group 15 of the periodic table. (ii) Refer Ans. March 9, 2019. chemistry notes for class 12; Group 16 Elements; Trend in Atomic and Physical Properties; No comments, be the first one to comment ! Explain the following observations giving appropriate reasons: Answer: Answer: (ii) H2O < H2S < H2Se < H2Te is the increasing order of acidic character. (ii) Of Bi(V) and Sb(V) which may be a stronger oxidising agent and why? (ii) Refer Ans. (a) (i) It is because NF3 is more stable due to stronger N—F bond than NCl3 because F2 is stronger oxidising agent than Cl2. (i) PCl3 + 3H20 ——-> H3PO3 + 3HCl (ii) Oxygen is diatomic gas having weak van der Waals’ forces, whereas sulphur is octaatomic (S8) solid, therefore, it has more van der Waals’ forces of attraction. (iii) Both O2 and F2 stabilize high oxidation states but the ability of oxygen to stabilize the higher oxidation state exceeds that of fluorine (i) Refer Ans. Explain the following situations: Which is a stronger acid in aqueous solution, HF or HCl, and why? HOW N3H IS ANSWER OF 2nd OUESTION. to Q.77 (i). (i) Write the conditions to maximize the yield of H2S04 by contact process. Question 77: Answer: (ii) Refer Ans. U�P�����k@�Я�נ��Vɡ��P����9Un`�����p��n�n�T� ����0/��2Lk*f57���ӍG͉��&������J�����VF�S�d��Q+�3���&j8���z�j�]Z-�HjϤed�4�� c|@�x&׌��iL�'���A�\3�ڂer���͙�7e�m��r�4�Ic�܂\*��9(��9��Rk��ըqUJ`�]���K x�r����a��DŽO8\hА�Ѐ1���bC��-�Z K��r�yI��X"47��eDc�s�n( (i) Sulphur in vapour state exhibits paramagnetism. Answer: (v) Complete the following equation: F2 + H20 ———> (a) (i) NH3 has lone pair of electron, so, bond angle is 107°, whereas NH+4 does not, therefore, bond angle is 109.5°. Rhombic sulphur. (iii) It is because helium does not dissolve as it is inert and lighter in blood therefore, does not cause bends or pain. (b) (i) Refer Ans. (b) (i) Refer Ans. 3. Answer: to Q.77 (i). Answer: (i) Bond enthalpy of F2 is lower than that of Cl2. Question 6: (i) All the P—Cl bonds in PCl5 molecule are not equivalent. Refer Ans. to Q.41 (i). Question 89: What is the basicity of H3P04? (iii) Oxygen shows catenation behaviour less than sulphur. Write the reactions involved. (i) BrF3 and (ii) XeF4 (ii) Refer Ans. Write the structures of the following molecules: (i) H2SO3 (ii) XeOF4 Both contain 6 electrons in the outermost shell. Answer: Question 25: (i) The molecules NH3 and NF3 have dipole moments which are of opposite direction. (ii) PH3 has lower boiling point than NH?. (iii) Fluorine is the most electronegative and small in size, therefore, it does not show positive oxidation state. (iii) Refer Ans. Question 35: /BitsPerComponent 4 (i) Iron on reaction with HCl forms FeCl2 and not FeCl3. (iii) It is due to interelectronic repulsion between valence electrons of smaller size ‘F’ atom than Cl. We need oxygen all the time throughout our lives. Account for the following: Question 144: (a) (i) Refer Ans. to Q.81 (ii). Question 37: Question 163: Question 167: Answer: (a) Mention the optimum conditions for the industrial manufacture of ammonia by Haber’s process. 13 (b) (i). (ii) Helium does not form any real chemical compounds. In a group, the chemical elements have atoms with identical valence electron counts and identical valence vacancy counts. b) All elements form dioxide with formula MO 2, SO 2 is a gas, SeO 2 is volatile solid. Question 1. (a) (i) Refer Ans. These important questions are prepared by our academic experts based on the syllabus guidelines prescribed by MCI. (a) Draw the structures of the following molecules: Answer: Answer: Answer: to Q.29 (ii). (a) (i) It is due to more repulsion between valence electrons of F than Cl due to exceptionally small size. When they come back to lower energy level, they radiates complementary colours. to Q. There are a total of 75 questions from all three subjects and for each subject, 100 marks are allotted. (b) Draw the structures of the following: (b) White phosphorus is monomeric, whereas red phosphorus is polymeric. How would you account for the following: (i) NF3 (ii) H2S2O8 (iii) H3PO3 to Q.29 (ii). (i) Refer Ans. Account for the following: Answer: Question 85: It is dibasic acid because it has two replaceable hydrogen attached with oxygen. Answer: (ii) The negative value of electron gain enthalpy is less for fluorine than that for chlorine. Question 19: (iii) Refer Ans. Answer: (i) Sulphur vapour exhibits paramagnetic behaviour: (iii) SF6 is stearically protected, therefore, less reactive than SF4 which is not stearically protected. Question 125: (i) Refer Ans. While TeO 2 and PoO 2 are non – volatile crystalline solids.. c) Ozone: It is unstable and easily decomposes into oxygen. (i) It is because nitrogen does not have vacant d-orbitals. (ii) The stability of +3 state increases down the group in group 15 of the (a) Draw the molecular structures of the following compounds: H2S04 and H2SO3. Explain the following facts giving appropriate reason in each case: Answer: Question 46: (ii) P—P bond is stronger than N—N bond due to less repulsion between valence electrons. (ii) P4(s) + 3NaOH(aq) + 3H2O(l) ——–> 3NaH2P02(s) + PH3(g) (iii) No chemical compound of helium is known. /Width 156 Answer: Question 58: (iv) Helium. It is monobasic acid because it has only one replaceable hydrogen atom, Question 73: to Q.40 (a) (ii). to Q.85 (b) (ii). In-text question… Draw the structural formulae of molecules of following compound: (i) BrF3 (ii) (HP03)3 (iii) XeF4 (e) CaF2 + H2S04 ———–> CaS04 + 2HF. (i) Fluorine never exhibits any positive oxidation state. (i) Refer Ans. (iii) Noble gases have very low boiling points. (ii) P4 + NaOH + H2O ———> (ii) Refer Ans. Question 161: HI is stronger acid than HCl in aqueous solution because it has lower bond dissociation energy. Question 90: Answer: Fig. Answer: Answer: Draw the structure of O2 and S8 molecules Ozone (O3): (ii) Sulphur has greater tendency for catenation than oxygen. Complete the following equations: Question 110: (ii) Refer Ans. Answer: (ii) XeF4 + O2F2 ———-> Why ? Answer: Answer: Question 31: (a) Account for the following: Answer: Question 10: Question 5: Draw the structures of the following: Question 116: to Q.87 (b) (i). What inspired N. Bartlett to prepare this compound? (iii) BiH3 is the strongest reducing agent amongst all the hydrides of group 15. (i) Refer Ans. (ii) Red phosphorus is less reactive than white phosphorus 17 (b) (i). (i) Chloropicrin (ii) Phosgene (iii) Mustard gas. Answer: Answer: Allotropy: All elements of Group 15 except nitrogen show allotropy. (iii) Refer Ans. This is because down the group, size of the element (M) increases, M-H bond length increases and thus, stability of M-H bond decreases so that it can be broken down easily. Question 36: (ii) Iron dissolves in HCl to form FeCl2 and not FeCl3. Periodic Classification of Elements Class 10 Important Questions and Answers Science Chapter 5. (i) XeF4 + SbF5 ——–> (ii) Cl2 + F2 (excess) ——> Refer Ans. Which is more acidic and why, H20 or H2S? Ans: Traces of sulphur occur as hydrogen sulphide in volcanoes. Answer: (ii) Refer Ans. << (i) H2S is more acidic than H2O. Give reasons for the following: (iii) BiH3 has lowest bond dissociation enthalpy due to longer bond length. Name two poisonous gases which can be prepared from chlorine gas. Abstract Algebra Manual : Problems and solution (only the section on GROUPS) Download . Account for the following: to Q.31 (ii). (b) (i) It is because P—P single bond is stronger than the single N—N bond. (ii) It is due to inert pair effect. (a) (i) It is because bond dissociation energy of HCl is lower than H2S which Question 124: Group, in chemistry, a column in the periodic table of the chemical elements. (ii) XeF4 + O2F2 ———-> Question 8: NaOH solution? Why? Question 143: to Q.83 (ii). Question 3: Arrange the following in the order of property indicated against each set: (a) (i) Refer Ans. Question 165: Answer: (ii) Electron gain enthalpy with negative sign of fluorine is less than that of chlorine. to Q.41 (ii). to Q.65 (a) (ii). to Q.41 (iii). (ii) H2S04is a strong acid, therefore, its K is very high as it dissociates into H3O+ and HS04 almost completely. Extra Questions for Class 9th: Ch 3 Atoms and Molecules (Science) Important Questions Answer Included Very Short Answer Questions (VSAQs): 1 Mark Q1.State the Law of Conservation of mass. to Q.20 (b) (i). (a) (i) 5SO2 + 2MnO- + 2H2O ——–> 5SO4– + 2Mn2+ + 4H+ Here we have covered Important Questions on Classification of Elements and Periodicity in Properties for Class 11 Chemistry subject.. Chemistry Important Questions Class 11 are given below. (b)Draw the structures of the following: Answer: This paper. Question 95: PH3 < H2S < HCl (b) Explain the following observations: to Q. Question 104: Why does N02 dimerise? Question 168: Answer the following: Draw the structures of the following: (a) Draw the structures of the following: (ii) sulphur dioxide gas is passed through an aqueous solution of a Fe (III) salt? (i) H2S2O8(ii) Red P4 Answer: Answer: to Q.88 (a) (ii). Very Short Answer Questions. Fluorine does not exhibit any positive oxidation state. (i) It is because oxygen is diatomic and has less intermolecular forces of attraction, whereas sulphur is octa-atomic (S8) and has more intermolecular forces of attraction. It is because it is most electronegative element and best oxidising agent. 40. (a) (i) Bi3+ is more stable than Sb3+ due to inert pair effect. Question 146: (b) Write the formula of first noble gas compound prepared by N. Bartlett. (ii) Why does fluorine not play the role of a central atom in interhalogen compounds?

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