ONS. Problem. Lewis structures: 2. a diamond (sp. carbon. very strong hammer). Each question was. graded with max. 0.5 points.

Size: px

Start display at page:

Download "ONS. Problem. Lewis structures: 2. a diamond (sp. carbon. very strong hammer). Each question was. graded with max. 0.5 points."

Baldric Chapman
5 years ago
Views:

www.biosan.lv 11 ND ROUND, OLUTIO ON Problem 1 (Lithuania) If you do not knw ask google! (5 points) 1. sp, sp, sp hybridization http:/ //en.wikipedia.

Heating organi materials, they deomposes forming amorphous arbon and water or other gaseous ompounds. 4. Amorphous arbon.

The remaining eletrons are deloalized over the moleule (or a part of it) in pi it bonding orbitals. These eletrons are the valene shell of the moleule and thereforee attempts to fill the shell.

http:/ //en.wikipedia.org/wiki/aggregated_diamond_nanorod 8. Diamonds are thermodynamially unstable but kinetially stable (metastablee modifiation of arbon).

1 11 ND ROUND, OLUTIO ON Problem 1 (Lithuania) If you do not knw ask google! (5 points) 1. sp, sp, sp hybridization //en.wikipedia.org/wiki/carbon_allotropes sp ethane; sp methane; sp ethyne Lewis strutures: More information:. a diamond (sp ), b graphite (sp ), lonsdaleitee (sp ). Heating organi materials, they deomposes forming amorphous arbon and water or other gaseous ompounds. 4. Amorphous arbon. Amorphous arbon materials may also be stabilized by terminating dangling π bonds with hydrogen. These materials are then alled hydrogenated amorphous arbon. 5. Fullerene. The remaining eletrons are deloalized over the moleule (or a part of it) in pi it bonding orbitals. These eletrons are the valene shell of the moleule and thereforee attempts to fill the shell. Usually, fullerenes reat as eletrophiles; also, they an be hydrogenated plus fullerenes also have superondutor properties. 6. Carbon nanotube. 7. Agregated diamond nanoroads (ADNR) or fullerite. //en.wikipedia.org/wiki/aggregated_diamond_nanorod 8. Diamonds are thermodynamially unstable but kinetially stable (metastablee modifiation of arbon). Transformation is thermodynamially possible but transformation rate is to slow (not enough ativation energy). Heating of diamond at anaerobi onditions it transforms to amorphous arbon (at temperatures aprox. 1 o C) ). At aerobi onditions it oxidizes and forms arbon dioxide. Liquid arbon annot exist at normal pressure, sublimation is observed. 9. It is graphene. Hybridization is sp. 1. Carbon fibre / modified graphite. The arbon form is graphene (one-sheet thik graphite) and it forms arbon fiber. Carbon fiber is very strong while bending, ompared to its weight, but will easily break if ompressed or exposed to shok (hit with hammer). Eah question was graded with max.5 points.

Problem (Estonia) Compounds of sulfur (5 points) Google translated english: Väävli ühendid O HO O Cl O HO Cl H H a).

Therefore, oxygen aounts for twoand three membered moleules, but sulphur forms long hains ontaining sigma bonds.

In O moleulee there are four bonding MOs, whih are formed from AOs: p z p z p z (π ), p y p y p y (π 1 ), p x p x p x (σ ), p x s p x (σσ 1 ); In O only

2 Problem (Estonia) Compounds of sulfur (5 points) Google translated english: Väävli ühendid O HO O Cl O HO Cl H H a). O O Polar polar polar b) Compared to O OO pi bonds, pi bonds are muh weaker. Therefore, oxygen aounts for twoand three membered moleules, but sulphur forms long hains ontaining sigma bonds. ) O moleule ontains double bonds, O has 1,5 bonds. Common Lewis struturess are: Lewis struture provide a few information and atually are wrong. In O moleulee there are four bonding MOs, whih are formed from AOs: p z p z p z (π ), p y p y p y (π 1 ), p x p x p x (σ ), p x s p x (σσ 1 ); In O only three: p z p z pp z (π ), p y p y pp y (π 1 ), p x p x pp x (σ ), π and π 1 has almost the same energy. In O moleule p x s p x MO is non bonding due to the low energy of oxygen s AO and shielding effet of s s s MO. O O d) + 4 planar, is quasiaromati ompound; 4 does not exist boat; 4 a hain. e) F 6 is strongly shielded.

3 Problem (Lithuania) Quantum observations due to laser ooling (1 points) Rb atom has 87 nuleons and 7 eletrons, hene: subatomi partiles in total, and 14 is even. Na has nuleons and 11 eletrons 4 subatomi partiles. The number of eletrons is equal to the number of protons in an atom, so it atually only depends on the number of neutrons if it is even, then the atom is a boson.. Two lasers in every possible dimension, therefore 6 lasers in D.. The general equation is: kt, where f is the number of degrees of freedom and for monatomi gas it is. As we an see, the kineti energy only depends on the type of the gas. Beause both sodium and rubidium gasses are monatomi their average kineti energies are equal: kt J/K J 4. T.^..^ λ. 6. This energy orresponds to the energy needed for the eletron of rubidium to go from the s orbital of the fifth layer to the p orbital of the same layer: 5s 5 7. The general Doppler effet is expressed in the following way for the moving detetor (atom) and stationary soure (laser): v d v Where f is the frequeny observed by the detetor and f is the frequeny of the soure. If we put f f(atom), f f(laser), v, v(d) v(atom) then: 1 Beause we want for the atoms to slow down when they are moving towards the laser, so the frequeny deteted by the atom inreases. o we take v/ positive. If we then divide by1 we obtain the desired equation 1 Due to the fat that the speed of the atoms is really small, ompared to the speed of light, we do not have to take the relativity into aount. 8. λ T λ Compared to the Hz frequeny of thelaser it is not a lot, so during the experiments the sientists usually detune the lasers using the trial and error method.

4 Problem 4 (Latvia) Chemistry of green ompounds (5 points) 4

5 5

6 6 olutions for questions 8 1 oppied from Jurgis Kuliesius (Lithuania). Eah question was graded with max.5 points. Questions 1 5 were awarded with maximum points also without omplete solution (due to errors in problem text). Problem 5 (Chemial rebuss) Chemial rebuss (5 points) 1.. Δ H A NOCl B HCl C HNO D NO E Cl f ( NOCl( g ) ) 51.71kJ / mol Δ H f ( NO( g ) ) 9.5kJ / mol NOCl g ) 61.69J ( ( ) NO g ) 1.761J ( ( ) Cl ).66J g ( ( ) NOCl ( g ) NO( g ) + Cl ( g ) K e F HNO G H O 4 H NOHO 4 I Cu J Cu(NO ) H Δ r 77.8kJ / mol Δ 11.1J ΔH r TΔ RT r e K K C ( RT) NOCl.5M r 778J / mol 4K 11.1J /( mol K ) 8.14J /( mol K ) 4K 4 p 5.5 Δ n [ ] [ Cl ] [ NO].M K HCN L CNCl M (CN) N NO 4

7 K 4 x [ NO] [ Cl ] [ NOCl] -.1 x x.145 [ NOCl].5 [ NO].145 [ Cl ].145M 4 x (.5 x) x M.9M. n( n( χ( n( + n( NO) + n( Cl ) 1.5n(.5n( For example 1bar, then:.149bar NO).985bar Cl ).49bar K K p (.149) ΔH 778J / mol T 14K Δ R ln K 11.1J /( molk) 8.14J /( molk) ln(.15 1 ) 7

General Equilibrium. What happens to cause a reaction to come to equilibrium?

General Equilibrium. What happens to cause a reaction to come to equilibrium? General Equilibrium Chemial Equilibrium Most hemial reations that are enountered are reversible. In other words, they go fairly easily in either the forward or reverse diretions. The thing to remember