QUESTION ONE Given below is some chemistry of a main group element, A. C+D E "HO Me, NF CIF n F₂ In H₂ A F G+ 2HCI BF3 N

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Question One Given Below Is Some Chemistry Of A Main Group Element A C D E Ho Me Nf Cif N F In H A F G 2hci Bf3 N 1 (57.24 KiB) Viewed 38 times

Question One Given Below Is Some Chemistry Of A Main Group Element A C D E Ho Me Nf Cif N F In H A F G 2hci Bf3 N 2 (49.25 KiB) Viewed 38 times

Question One Given Below Is Some Chemistry Of A Main Group Element A C D E Ho Me Nf Cif N F In H A F G 2hci Bf3 N 3 (49.49 KiB) Viewed 38 times

QUESTION ONE Given below is some chemistry of a main group element, A. C+D E "HO Me, NF CIF n F₂ In H₂ A F G+ 2HCI BF3 N₂F4 H All reactions involve a 1:1 ratio of reagents, unless indicated otherwise by the letter "n". Note that n is a variable and may vary in the different reactions. B has one lone pair on the central atom. C is a solid and D a gas at room temperature & pressure. C contains the element A; D does not. E contains 19.36% C. 4.88% H. 5.65 % N and 38.29% F. The element in its aximum oxidation state in compound F. A sample of F (0.3141 g) was decomposed in water and reacted with an excess of silver nitrate solution (20 mL., 0.84 M). After drying, 0.2150 g of a solid was recovered. G is a binary compound and contains two atoms of clement A. Two moles of HCI are generated for each mole of H; reagent. H contains one (nitrogen to element A) single bond. (i) Identify element A, and the compounds B 1. showing all your working and accounting for all the data provided.. Write balanced equations for all the reactions, except the reaction with silver nitrate, Draw the structures of E, G, H and I. (iii) (iv) For each reaction, suggest a reason why the reaction occurs based on your knowledge of chemistry. Note that explanations that only state the reaction is energetically favourable will score zero 120 marks] B
Given below is some chemistry of a main group halide, ACI, Cl₂ TÌNH C ACI, -B C + nHCI nE(MgC) F+G D+E Note that n is a variable and will vary in the different reactions. C has a molar mass of 347.64 g mol and contains 12.09% N. D and F contain no chlorine. E and G are ionic compounds that do not contain element A. Microanalysis data of F: C-60.97%, H = 12.82%. Ch TNH ACL CHO EtMgCl F+G D+E E and G are ionic compounds that do not contain element A E must be NaCl because C has to have some Cl in it, the other reagent is sodium ethoxide, and E is an ionic compound Similarly, G must be MgCl, because it's an ionic compound formed from a Grignard reaction with a main group chloride Starting with F F has no chiorine, so complete conversion of Cl to ethyl groups in F Fhas C-60.97, H-12.82, combined-73.29% The method below is not the quickest, but maybe the most reliable 73.79% is ethyl group, CH,, which has a molar mass of 29.07 g mol If one Et group total mass of F 29.07/0.7379-39.40 a mot 6 NaOEt 6 NaOE1 a) Identify element A, the halide ACI, and the compounds B - G, showing all your working and accounting for all the data provided. b) Write balanced equations for all the reactions. c) Draw the structure of C and discuss the bonding in C. [10 marks] Where to start? C has microanalysis and molar mass. F has microanalysis. To determine what element A is, I would start with either of these. But first... Starting with C: -3 Natoms 12.09% N of 347.64 g mol is 42.03 g mol Reaction of C with 6NaOEt suggests 6 Cl in C. 347.64 (3x14.01) (6x 35.45) 92.91 g molt This is same as Nb, but Nb not a main group element. 92.91/2 46.46 g mot). This is not the molar mass of an element. 92.91/3-30.97 g mol This is the molar mass of P and C is C,N,P, so A is P Check, 3 x [14.01+30.97+2(35.45)]- 347.64 g mot) If 2 Et groups: total mass of F is (2 x 29.07)/0.7379-78.79 g mor so A has mass of 20.65 g mot 178.79-(2x 29.07)) 20.65 g mol is not the molar mass of an element so it is not two Et groups. If 3 Et groups total mass of F is (3 29.07)/0.7379-118.19 g mol
If one Et group: total mass of F is 29.07/0.7379-39.40 g mol, so A has mass of 10.33 g mol! (39.40-29.07) 10.33 o mots not the molar mass of an element so it is not one Et group Advice: The microanalysis will always be close enough to determine the element. For example, for compound F, just saw 10.33 g mol was not an element (boron is 10.81 g mol-¹) Use the periodic table to consider what element it is even if you think you know what it is - double check (e.g. last year the element was Se, not Br) F is PET, - reason already given, see earlier calculation Gis MoC, since it is an ionic compound with no element A present - and only sensible product from the Grignard reaction (as said on slide 3) PCI, +3EtMgCl PET,+ 3MgCl, Structure of C K It can be described as with 6 m aromatic; with a significant contribution) from a charged canonical form no marks, (just a reminder of charged canonical form) so A has mass of 30.98 g mol¹ [118.19-(3x 29.07)) 30.98 g mol-¹ is the molar mass of PA so it is three Et groups. so F is PET, The halide ACT, is PCI, (reacts with more Cl, to form B) B is PCI, since has to be in higher 0/S than in ACI, (PCI) PCI₂ + Cl₂ PCI, Cis (NPCI₂), reason already given, see earlier calculation 3PCI, + 3NH CI (NPC₂), +12HCI D is (NP(OET),), since no CI in D, must substitute all the Ci by OE E is NaCl since it is an ionic compound with no element A present (and this is the only sensible product) (already mentioned, slide 3) (NPC₂), + 6NaOEt (NP(OET),), + 6NaCl Marking Determining the element A is phosphorus using data from C or F without assumptions Showing that (NPCI), or PEt, match the data for C or F ya mark each for PCI PCI (NP(OEt),), NaCl and MgCl, if correct with reason mark for each equation correctly balanced Diagram of (NP(OET)), Bonding in (NPYOEL))) Total Note N mark each for P & the seven compounds if correct with no reason 3 marks 1 mark each. 21 marks 2 marks 1 mark 15 mark 10 marks 2 marks