With Singapore having the highest incidence of diabetes in the world, every Singapore should take Berberine.
Berberine helps to prevent diabetes if you don't have diabetes, it helps to prevent diabetes if you're pre-diabetic, and helps control your diabetes if you already kena diabetes.
In other words, Berberine is always helpful, for everyone : non-diabetics, pre-diabetics and diabetics.
There are dozens of different brands of supplements that use Berberine, which is a herbal extract from several different plants of the same genus.
Berberine is a quaternary ammonium salt from the protoberberine group of benzylisoquinoline alkaloids found in such plants as Berberis (e.g. Berberis vulgaris - barberry, Berberis aristata - tree turmeric, Mahonia aquifolium - Oregon-grape, Hydrastis canadensis - goldenseal, Xanthorhiza simplicissima - yellowroot, Phellodendron amurense - Amur cork tree, Coptis chinensis - Chinese goldthread, Tinospora cordifolia, Argemone mexicana - prickly poppy, and Eschscholzia californica - Californian poppy. Berberine is usually found in the roots, rhizomes, stems, and bark. - https://en.wikipedia.org/wiki/Berberine
Berberine is supplemented for its anti-inflammatory and anti-diabetic effects. It can also improve intestinal health and lower cholesterol. Berberine is able to reduce glucose production in the liver. Human and animal research demonstrates that 1500mg of berberine, taken in three doses of 500mg each, is as effective as taking 1500mg of metformin or 4mg glibenclamide, two pharmaceuticals for treating type II diabetes, but with less side-effects. - https://examine.com/supplements/berberine/
Share this info with all your family and friends, and help reduce the pain & suffering that comes with diabetes (eg. blindness, amputation of limbs, kidney failure, etc).
On a related medical note, you should also be aware that Glutathione (the best supplemental form being Setria Glutathione, superior to N-acetylcysteine with less side-effects), or lack of it in diabetics, is the main reason for many of the pathological symptoms of diabetes. In other words, ensuring you've sufficient levels of Glutathione (eg. by Setria supplementation) will control your symptoms of diabetes, as well as help (via different biochemical pathways compared to Berberine) to prevent diabetes if you're non-diabetic or pre-diabetic.
Medical studies :
The central role of glutathione in the pathophysiology of human diseases.
Glutathione Synthesis Is Diminished in Patients With Uncontrolled Diabetes and Restored by Dietary Supplementation With Cysteine and Glycine
[China] - Female university student radioactively poisoned by roommate until she became mental retarded and paralyzed for life, but roommate escapes any legal consequences due to her rich family connections.
[Russia] - Assassination of ex-FSB ex-KGB intelligence officer via lethal radioactive poisoning in London.
'O' & 'A' Level Qn.
Copper can reduce nitric acid to NO2. Write a balanced equation for this reaction.
Is it :
1) Cu +2HNO3 +2H+ --> Cu (2+) + 2NO2 + 2H20 OR
2) Cu +4H(+) + 2NO3(-) --> Cu(2+) +2NO2+ 2H20
Write a balanced equation for the reduction of nitric acid to NH4+ by aluminium.
Suggest a reason why the reduction products of nitric acid are different for each of these two metals.
It depends on whether concentrated (ie. pure liquid, no water present) or aqueous acid is used. If concentrated, the HNO3 exists as a covalent, unionized molecule. If aqueous, it dissociates and ionizes to H+ and NO3-.
If concentrated, use equation #1. If aqueous, use equation #2. If the exam question is ambiguous, give both with qualifications (ie. explanations).
Regarding the QA test for nitrate(V) ions, the aluminium foil or dervada's alloy will reduce nitrate(V) to ammonium cation, which will undergo a Bronsted-Lowry acid-base proton transfer reaction with the hydroxide ion (from the alkali added), to generate aqueous ammonia, which when heated overcomes the hydrogen bonds with water and generates gaseous ammonia, which can be detected by the hydrolysis of ammonia on moist red litmus paper generating the hydroxide ion, turning the litmus paper blue.
Reduction half-equation :
NO3(-) + 10H(+) + 8e- ---> NH4(+) + 3H2O
Oxidation half-equation :
Al ---> Al(3+) + 3e-
Combine the two half-equations to get the balanced redox equation.
The reasons for the different redox products of nitric(V) acid with aluminium versus copper, has to do with the different oxidation potentials (ie. reducing strength) of Al versus Cu. Al is a lot more reactive, has a more positive oxidation potential, and is hence a more powerful reducing agent than Cu.
Hence Al is able to reduce the O.S. of N from +5 (in HNO3) all the way to -3 (in NH4+), while Cu is only able to reduce the O.S. of N from +5 (in HNO3) down slightly to +4 (in NO2).
'A' Level Qn
Draw the uninegative iodine dichloride ion. State all formal charges and oxidation states within the ion. State its electron and ionic geometry.
Write a balanced equation to represent the redox reaction (under acidic conditions) between the iodate(V) ion, the iodide ion, and the chloride ion, to generate the uninegative iodine dichloride ion.
In the iodine dichloride ion, [Cl-I-Cl]-, the 2 Cl atoms have no formal charge, and an O.S. of -1. The I atom has a 1- formal charge, and an O.S. of +1. The sum of formal charges, as well as the sum of oxidation states, adds up to the uninegative ionic charge.
Electron geometry = trigonal bipyramidal.
Ionic geometry = linear.
What is happening during the redox reaction : the iodate(V) ion is reduced, and the iodide ion is oxidized, both becoming the iodine atom in the iodine dichloride ion (O.S. of +1). Notice that the O.S. of Cl does not change (-1 before redox, -1 after redox).
[reduction] 4e- + 6H+ + 2Cl- + IO3(-) ---> [Cl-I-Cl]- + 3H2O
[oxidation] 2Cl- + I- ---> [Cl-I-Cl]- + 2e-
[balanced redox] 6Cl- + 6H+ + 2I- + IO3(-) ---> 3[Cl-I-Cl]- + 3H2O
'A' Levels Notes.
"The Thermodynamics of Ruling a Kingdom."
"Emperor Gibbs and his prime ministers, Mr Enthalpy and Mr Entropy."
You are Emperor Gibbs, ruler of a vast empire and kingdom. You have 2 prime ministers, Mr Enthalpy and Mr Entropy. Mr Entropy is married to a lovely lass by the name of Temperature. (the proof of marriage is in the Gibbs free energy formula G = H - T*S ; see? I told you they were married to each other!)
Recall that an exothermic reaction is usually a more favourable & feasible one, having products, eg. a H2 molecule, that are more stable (ie. less kinetic energy) than the reactants, eg. 2 H atoms. Where did the energy difference go to? lost to the environment, as heat! This is why bond-forming is exothermic.
Recall also, that a reaction that has a positive entropy increase is usually a more favourable & feasible one, eg. when two substances dissolve in each other, and are freely miscible even though the solute-solvent interactions are no stronger than the solute-solute and solvent-solvent interactions. In other words, enthalpy is approximately neutral (or perhaps even slightly endothermic), but entropy is positive.
For your kingdom to surely prosper (ie. reaction to be feasible & spontaneous), you will need the staunch undying support from both your prime ministers. Both Enthalpy and Entropy must be favourable (ie. enthalpy must be endothermic and entropy must be positive).
If both prime ministers do not support you, want to overthrow you and usurp your throne, then your rule of the kingdom is over (ie. reaction is non-feasible & non-spontaneous). It's a matter of time before you get assasinated by either of them; you don't stand a chance.
The more interesting scenario is, when one prime minister supports you, while the other wants to overthrow you.
You see, as Emperor, you have one power - to seduce the wife of your prime minister. The power of a married man lies in his marriage. If you please the wife and make her hot-hot, she will please her husband and he will be powerful (don't try this in rl, folks.) If you do not satisfy the wife, and make her cold-cold, her husband will also be left unsatisfied and his power will weaken.
Hence, if Mr Enthalpy supports you (ie. is exothermic) and Mr Entropy does not support you (ie. is negative), then you will of course want to weaken the Mr Entropy's power ("I always knew you would betray me some day, you bastard!"), so that he will have less power to assasinate you. So go make his wife cold-cold already! In other words, the reaction is feasible & spontaneous (your rule of the kingdom will prosper) only at low temperatures.
Conversely, if Mr Enthalpy does not supports you (ie. is endothermic) and Mr Entropy supports you (ie. is positive), then you will of course want to strengthen Mr Entropy's power ("I always knew you would stay loyal to me forever, my dear friend!"), so that he can help you to overcome the political thorn of Mr Enthalpy. So go make his wife hot-hot already! In other words, the reaction is feasible & spontaneous (your rule of the kingdom will prosper) only at high temperatures.
If enthalpy and entropy are favourable, the reaction is feasible & spontaneous at all temperatures.
If enthalpy and entropy are not favourable, the reaction is non-feasible & non-spontaneous at all temperatures.
If enthalpy is favourable (ie. exothermic) and entropy is not favourable (ie. negative), the reaction is feasible & spontaneous only at low temperatures (cold-cold).
If enthalpy is not favourable (ie. endothermic) and entropy is favourable (ie. positive), the reaction is feasible & spontaneous only at hot-hot high temperatures (hot-hot).
Real-life Everyday Example :
Condensation of gaseous water vapour into liquid water.
Because this involves the formation of intermolecular hydrogen bonds, it is exothermic and hence favourable in terms of enthalpy.
However, because the formation of intermolecular hydrogen bonds results in a decrease in entropy, hence it is unfavourable in terms of entropy.
As predicted by the Gibbs free energy formula (and/or by our "Kingdom of Gibbs" fable), the condensation reaction is only feasible & spontaneous at low temperatures.
(Hence water vapour only condenses on the outer surface of the cold-cold glass containing your iced drink, and not on the surface of the hot-hot cup containing your freshly brewed coffee).
'A' Level Qn.
Compound A C13H15O3Cl decolourises hot acidified KMnO4 forming B, C, and D with formulae C4H8O3, C3H4O3, C6H5OCl respectively. 1 mol B reacts with 0.5 moles of Na2CO3, and also reacts with conc H2SO4 to give sweet smelling E C8H12O4. C gives positive idoform test and reacts with SOCl2. D reacts with aq Br2 to form white ppt. Draw the structures of A to E.
A - HOC(CH3)2CH=C(CH3)COOC6H4Cl
B - HOC(CH3)2COOH
C - CH3COCOOH
D - ClC6H4OH
E - (cyclic diester) ~C(CH3)2COOC(CH3)2COO~
'O' & 'A' Level Qn
"When 20cm3 of a gaseous hydrocarbon was sparked with 150cm3 of oxygen and the residual gases cooled to rtp, a contraction of 60cm3 occurred. A further contraction of 80cm3 took place when the residual gases were subjected to aq NaOH. All volumes measured at 20 deg C. Determine the formula of the hydrocarbon."
This question is ambiguous because "a contraction of 60cm3" may refer either to this process ocurring on reactant gases (ie. 170cm3 - 60cm3 = 110cm3 residual gases at rtp; total gases produced exceeds total gases used up, in addition to water vapour condensing), or on residual gases (ie. 60cm3 refers *only* to water vapour condensing).
Carry out your calculations along both interpretations and arrive at the two correspondingly different identities for the hydrocarbon, and draw + name all 5 possible isomers for *each* of the two hydrocarbons.
Assuming reactant gases, C4H8 has the following isomers :
Assuming residual gases, C4H6 has the following isomers :
'A' Level Qn.
8.00g of potassium chromate is dissolved in acid to make a 100.0cm3 solution. A dynamic equilibrium occurs, according to the equation below:
2(CrO4)2-(aq) + 2H+(aq) -> (Cr2O7)2-(aq) + H2O(l)
Kc = 7.55x10^12 mol-3 dm9
Write the Kc expression for the above equilibrium and calculate the concentration of H+ at which one-fifth of the original amount of chromate ions remain.
Molarity = 0.4119 mol/dm3
2(CrO4)2-(aq) + 2H+(aq) ---> (Cr2O7)2-(aq) + H2O(l)
Initial | 0.4119 | ??? | ---> | 0 | ??? |
Change | - (4/5)0.4119 | -(4/5)0.4119 | ---> | +(1/2)(4/5)(0.4119) | +(1/2)(4/5)(0.4119)
Equilibrium | 0.08239 | ??? | ---> | 0.1648 | ??? |
Since Kc = 7.55x10^12 mol-3 dm9 = (0.1648) / [ (0.08239)^2 ([H+]^2 ]
[ (0.08239)^2 ([H+]^2 ] = 2.18278
[H+]^2 = 3.2156 x 10^-12
[H+] = 1.7932 x 10^-6
pH = 5.746 = 5.75 (3 sf)
Note that Le Chatelier's principle predicts that at acidic pH, the position of equilibrium shifts to the right, using up more chromate(VI) ions to generate more dichromate(VI) ions. If the pH was more alkaline, you'll notice more dichromate(VI) ions used up to generate more chromate(VI) ions, which is consistent with the idea of maintaining the equilibrium constant Kc value, for any given temperature.
'A' Level Qn.
An engine cylinder volume of 500cm3 contains air at 60 deg C and has a pressure of 101 kPa. If the average molar mass of petrol is 100g and undergoes combustion with O2 in a 1:12 molar ratio, what is the sample mass of petrol that should be injected into the cylinder to react with the O2 present? (Assume negligible change in pressure upon injection of the liquid petrol.)
'A' Level Qn.
For the reaction : 2A(g) + B(s) --> C(g) + 3D(g)
When 1.2 mol of A was reacted with 1.0 mol of B in a 2.0dm3 vessel to form C and D, it was found that 30% of A had been converted. Given that the total pressure at equilibrium was 200kPa, determine the Kp value for this temperature.
At a certain temperature the Kc for the formation of gaseous hydrogen iodide from gaseous hydrogen and gaseous iodine has a value of 0.191. A 1dm3 vessel contains 0.30 mol of I2, 0.40 mol of H2, and 0.10 mol of HI at initial. At this temperature, what is the amount of H2 that must be introduced into the vessel at initial such that the equilibrium molarity of HI is 0.150 mol/dm3?
Q1) 660 kPa2 (3 sf)
Q2) 0.0534 mol (3 sf)
'O' & 'A' Level Qns.
1) An aqueous solution solution contains 1 mol of S2O3 2- ions and it reduces 4 mol of Cl2 molecules. What is the sulfur containing product of this reaction?
B] SO3 2-
C] SO4 2-
D] S4O6 2-
2) 25.0cm3 of 0.05 mol/dm3 KClO4(aq) required 50.0cm3 of 0.20mol/dm3 TiCl3(aq) for complete reaction. Given that titanium(III) is oxidised to titanium(IV) in this reaction, which one of the following formulae correctly represents the reduction product of the ClO4 - ion?
B] ClO2 -
C] ClO3 -
D] OCl -
4 moles of Cl2 molecules accept 8e- to be reduced to 8Cl-. The OS of S in thiosulfate ion is +2. Hence, each of the 2 S atoms must donate 4 e- each. Hence, the OS of S in the oxidized species must be (+2) + (+4) = +6. Hence the answer is C.
1.25 x 10^-3 mols of perchlorate ions / chlorate(VII) ions, oxidize 0.01 mol of titanium(III) ions to titanium(IV) ions. This implies 1.25 x 10^-3 mols of perchlorate ions / chlorate(VII) ions accept 0.01 mol of e-. This implies 1 mol of perchlorate ions / chlorate(VII) ions accept 8 mol of e-. Hence the final OS of chlorine in the reduced species must be (+7) + (-8) = -1. Hence the answer is A.
'A' Level Qn
Using Kekule structures, draw the electron-flow mechanism to illustrate the formation of pyroglutamic acid (a lactam) from glutamic acid (2-aminopentanedioic acid).
Hint : this is an intramolecular nucleophilic acyl substitution involving an addition-elimination mechanism including an intramolecular proton transfer.
#1 - nucleophilic attack (or "addition") : lone pair on N atom becomes a dative sigma bond with the partially-positively charged C atom of the 'R' group carboxylic acid; simultaneously, the pi bond shifts up to give the O atom a -ve formal charge. In addition, the OH group of this carboxylic acid is protonated (from the solution).
#2 - A proton is lost to allow the N atom to lose its +ve formal charge. A lone pair on the -ve formal charged O atom shifts back to to become a pi bond, reforming the carbonyl group. The H2O+ leaving group is "eliminated".
Thus, the "addition" of the amino group nucleophile to the 'R' group carboxylic acid group, and the subsequent "elimination" of the H2O+ leaving group, constitutes the "addition-elimination" mechanism for this intramolecular nucleophilic acyl substitution reaction.
The resulting product generated is a lactam (ie. a cyclic amide).
Posted by JetGrey :
Oh my gosh.. H2 Chem is at a so so so so much higher level than A Level Chem previously.
UltimaOnline replied :
Uhhh no, it's just me.
I teach students through understanding of concepts, rather than blind memorizing of notes. The H2 Chem syllabus content is still the same level as the old (or current UK or international) A level Chem syllabus, just that the topics have been reorganized (eg. instead of core and optional topics in the old A level Chem, now in H2 Chem the topics are all already pre-determined by MOE and there's no choice on the part of the students).
Having said that (ie. even though the difficulty level of the syllabus content remains the same as the old A levels), the difficulty level of the exam qns are certainly increasing in recent years, especially in the 2010 paper. Curiously, the difficulty of the Physics and Biology papers remained unchanged.
It's just the Chemistry papers. One possible reason (direct or indirect) could be due to the increasing competitiveness of getting into Medicine. Unlike the triple science requirements of the old A levels, because the new H2/H1 system for JC students require a contrasting subject, so today's requirement for Medicine is a distinction in H2 Chemistry and either H2 Biology or H2 Physics. Notice that H2 Chemistry is the only non-negotiable subject required for Medicine.
'A' Level Qns.
BedokFunland JC's "Fun with Oxygen and Nitrogen" questions.
1.71g of barium reacts with oxygen to form 2.109g of a compound.
1.71g of calcium reacts with oxygen to form 4.439g of a compound.
1.71g of strontium reacts with oxygen to form 3.584g of a compound.
Deduce the formulae of, and name, each of these 3 compounds, and that of their constituent ions.
Draw the displayed structure (resonance contributor) of the 3 anions (indicating all formal charges).
Work out the oxidation state (O.S.) of the individual O atoms in the 3 anions, and hence the average O.S. of oxygen in each of these 3 anions.
Verify that “ionic charge = sum of formal charges” and “ionic charge = sum of oxidation states”.
Draw the displayed structures of the following nitrogen species, and label them by their formulae.
Indicate all formal charges, oxidation states (in parenthesis) of all atoms, and dative bonds.
Verify that “ionic charge = sum of formal charges” and “ionic charge = sum of oxidation states”.
Nitrite anion [latin name] or nitrate(III) anion [stock name], uninegative
Nitrate anion [latin name] or nitrate(V) anion [stock name], uninegative
Nitronium cation, unipositive (utilized in the nitration of benzene ring)
Nitrogen monoxide, molecule
Nitrogen dioxide, molecule
Dinitrogen monoxide, molecule
Dinitrogen trioxide, molecule
Dinitrogen tetroxide, molecule
Dinitrogen pentoxide, molecule
BaO2, barium peroxide, OS = -1 and -1, average OS = -1
CaO4 or Ca(O2)2, calcium superoxide, OS = 0 and -1, average OS = -1/2
SrO6 or Sr(O3)2, strontium ozonide, OS = -1, +1 and -1, average OS = -1/3
All structures refer to Wikipedia to check.
'A' Levels Organic Chemistry (OC) Synthesis & Mechanism Challenge
May be attempted by students from H1/H2/H3 'A' Levels, International Baccalaureate (IB), Advance Placement (AP).
You may refer to Wikipedia for relevant background information in attempting this question.
Suggest a sythesis pathway, and include all relevant mechanisms, for synthesizing "methyl 4-chlorobenzoate" aka "methyl parachlorobenzoate" (an ester) starting from benzene.
Hints : You may use Lewis acid catalysts (for electrophilic aromatic substitution of halogens onto the benzene ring), magnesium metal in diethyl ether, to generate a Grignard reagent (in turn to generate a carbon nucleophile on the benzene ring), carbon dioxide (which undergoes hydrolysis to generate a species which is electrophilic under strongly acidic conditions), and diazomethane (to generate the methyl ester, via proton transfer and nucleophilic aliphatic substitution).
The hints already give most of the answer away. Honestly.
But additional hints are :
#1 Note that halogens may be overall deactivating against electrophilic aromatic substitution (because their electron-withdrawing effects by induction slightly outweigh their electron-donating effects by resonance), but are nonetheless ortho-para directing (because they donante electrons by resonance, aka mesomeric effect).
#2 By looking at the mechanism by which Grignard reagents are generated, and the activation energies involved, do you expect chlorine or bromine to react more readily with magnesium to generate your Grignard reagent? Selective reactivity is possible.
#3 Under strongly acidic conditions, the species generated upon hydrolysis of carbon dioxide, may be protonated to generate an excellent leaving group of H2O+ (which is eliminated as water), and hence may function as an electrophile that may be readily attacked by a carbon nucleophile on the benzene ring.
#4 There are two resonance contributors for diazomethane, both of which have a positive formal charge on the central nitrogen. Accordingly, the carbon on diazomethane has basic properties, and proton transfer with a carboxylic acid is possible (consequently generating the carboxylate nucleophile).
#5 R-N(triple-bond)N: with a positive formal charge on the central N, includes an excellent leaving group, which is eliminated as molecular gaseous nitrogen.
You may search online for the relevant mechanisms; alternatively upload your mechanisms and PM me if you wish me to check them for you.
'A' Level Qns on Chemical Bonding
Originally posted by bonkysleuth:
Hello. Have a question on bond angles of H2S and SO2 in which I was asked to determine whether bond angle of H2S < that of SO2.
I found that H2S has 2 bonding pair of e- and 2 lone pair of e- while SO2 has 2 bond pair of e- and 1 lone pair of 2-. Which gives H2S a bent shape of 105deg and SO2 also a bent shape of 118deg. Thus H2S has smaller bond angle than SO2.
Just verifying, the answer my tutor has provided didnt indicate that this statement is true. Thank you!
SO2 has electron geometry of trigonal planar, molecular geometry of bent / v-shape. The S atom has 3 electron charge clouds : 1 lone pair, 2 double bonds. Bond angle approx 120 deg, modified to slightly less (eg. 118 deg) due to greater repulsion between lone pair - bond pair, compared to bond pair - bond pair repulsion (this is based on complete sp2 hybridization as taught in H1/H2/H3 'A' level Chemistry; in reality extent of hybridization is incomplete for some compounds of period 3 elements, due to less repulsion from a larger atomic radius; in SO2 the bond angle is actually 119 deg, so indeed sigma bonds have significant sp2-sp2 character).
H2S has electron geometry tetrahedral, molecular geometry of bent / v-shape. The S atom has 4 electron charge clouds : 2 lone pairs, 2 single bonds. Bond angle approx 109.5 deg, modified to somewhat less (eg. 105 deg) due to greater repulsion between lone pairs, compared to lone pair - bond pair repulsion, compared to bond pair - bond pair repulsion (this is based on complete sp3 hybridization as taught in H1/H2/H3 'A' level Chemistry; in reality extent of hybridization is minimal for some compounds of period 3 elements, due to minimal repulsion from a larger atomic radius; in H2S the bond angle is 92.1 deg, sigma bonds have significant s-p character, rather than s-sp3).
At H1/H2/H3 'A' levels, the actual Cambridge paper will presumably (though Cambride isn't always so reliable) set the question such that whether you use your limited (and not completely correct) 'A' level knowledge, or you use higher level (more correct) knowledge, either way you should get the same correct answer (eg. if MCQ) or still score the marks (eg. if P2 or P3 question).
Originally posted by bonkysleuth:
Thanks for explaining so clearly.
I have 2 more questions on this topic which I am unsure of.
1. The N2O5 molecule is present in the vapour state, but the structure is ionic in the solid state, existing as NO2+ NO3-.
Later on, I was asked to draw the displayed formula of NO2+ and NO3-, which was done easily.
Then the part I am unsure of is: Draw a possible displayed formula for the N2O5 molecule and suggest values for the bond angles in the molecule.
I realized I could not really make "complete" use of the 2 displayed formula I was asked to draw earlier on because the answer showed merely TWO of the NO3- oxoanion, repetitively. Why is NO2+ not included in helping me draw this N2O5 molecule? What are the steps I should take in deriving the displayed formula for N2O5?
2. Nitrogen and phosphurous are elements of Group V in the Periodic Table.
Nitrogen exists naturally as gaseous diatomic molecules, with triple bonds, whereas phosphorus is a solid and exists as P4 molecules comprising of P-P single bonds.
SUggest why phosphorus does not occur naturally as triple bonds.
My sch tutor said it is because P is a relatively big atom with diffused orbitals, so side-on overlap of p-orbitals to form pi bonds is much less effective than the head-on overlap to form sigma bond.
Well, dont really get why a big atom will be less effective in having side-on overlap...
NO2+ has a unipositive formal charge on the central N atom.
NO3- has a unipositive formal charge on the central N atom, and 2 uninegative formal charges on the 2 singly bonded O atoms. There are a total of 3 formal charges, and one dative bond.
NO2 (nitrogen dioxide) has an N atom with a unipositive formal charge as well as an unpaired electron on the central N atom, doubly bonded to an O atom with no formal charge, and singly bonded to a O atom with a uninegative formal charge. Hence NO2 is a free radical, with 2 formal charges, and one dative bond.
N2O4 is simply a dimer of two NO2 molecules joined together using their unpaired electrons. Hence N2O4 is a non-radical, with 4 formal charges, and two dative bonds.
N2O5 is similar to N2O4, except that instead of directly joining two NO2 free radical molecules together using their unpaired electrons, you join the two NO2 free radical molecules together using an O atom in the center, using the unpaired electrons of the two NO2 molecules. N2O5 has a total of 4 formal charges, with two dative bonds.
The bond geometries in N2O5 are hence :
V-shaped about the O atom (ie. 105 deg).
trigonal planar about the 2 N atoms (ie. 120 deg).
2. Yes, your JC teacher is pretty much right about the reason. But I'll elaborate a little here, to help you understand a little better.
Note that the inter-nuclei or inter-atomic distance is significantly larger between two P atoms, compared to two N atoms. Remember that single bonds are longer (sigma bonds are end-on or head-on overlap of any orbitals, usually hybridized), and double or triple bonds are much shorter (a sigma bond and one or two pi bonds, which are formed by the side-on overlap of unhybridized p orbitals).
Hence, P atoms find it easier (ie. feasibility under standard conditions) to form single bonds with each other, while N atoms find it easier (ie. feasibility under standard conditions) to form triple bonds with each other.
The specific reason for the greater ease for nitrogen, and a decreased ease for phosphorus, to achieve p electron sideways orbital overlap to form pi bonds, can be traced to their electron configuration :
Phosphorus, being in period 3, will have to utilize its significantly more diffuse 3p electron orbitals to form much weaker side-on overlaps, which means the consequent attempted triple bond will be so weak, that it is unable to stabilize the P atoms and molecule well.
Nitrogen, being in period 2, is able to utilize its significantly more concentrated 2p electron orbitals to form much stronger side-on overlaps, which means the consequent triple bond formed will be so strong, that it is able to stabilize the N atoms and molecule well.
And remember that in Chemistry, whichever possible path achieves greater stability, will be the more feasible one. (Under standard conditions) A triply bonded N2 molecule would be more stable than a singly bonded N4 molecule, and is hence favoured. (Under standard conditions) A singly bonded P4 molecule would be more stable than a triply bonded P2 molecule, and is hence favoured. (There are actually many allotropes of phosphorus, but P4 is the most common.)
'O' & 'A' Level Qns.
Have a few qns below
1. Why do alkali metals have low densities & low melting/boiling point
2. Why do the densities of alkali increase down the group
3. Why does melting/boiling point of alkali metal decrease down the group
4. Why does melting/boiling point of halogens increase down the grp
5. Why does the colour of halogens become darker down the grp
6. Why does reactivity/oxidation power of halogens decrease down the grp
All help is greatly appreciated[/QUOTE]
1. Because their unipositive cations have low charge densities, and hence experience relatively weak electrostatic attraction between their cations and delocalized sea of valence electrons.
2. Density = mass / volume. Going down the group, both atomic mass and volume (ie. atomic radius) increases, but the rate of increase in mass outweighs rate of increase in volume, hence density increases.
3. Because charge densities of cations decrease down the group (due to increasing ionic radius, in turn due to increasing number of electron shells), hence decreasing electrostatic attraction between their cations and delocalized sea of valence electrons.
4. Increasing number of electrons and increasing electron charge cloud polarizability (due to increasing atomic & molecular size), results in stronger and/or more extensive instantaneous dipole - induced dipole van der Waals interactions between molecules, requiring more energy to overcome to melt / boil the halogens, down the group.
5. There are two different reasons for this phenomenon. The first relates to the different colours, while the second relates to the intensity / darkness of the colours.
Firstly, absorption of visible light by a molecule depends on the magnitude of energy differences involved in the electron transitions between molecular orbital energy levels. Going down the group, the magnitude of these energy differences decrease; consequently the wavelength of visible light absorbed increases (from 'higher energy' colours to 'lower energy' colours), and accordingly the wavelengths of light not absorbed (ie. the visible colour of the halogens) decreases (from 'lower energy' colours to 'higher energy' colours).
Secondly, at any given temperature and pressure, (going down the group) the strength and/or extent of instantaneous dipole - induced dipole van der Waals interactions increase, resulting in decreasing volumes and hence increasing densities (from gaseous to liquid to solid), which in turn gives the halogens (down the group) more intense, darker colours.
6. Due to differences in the distance and hence strength of electrostatic attractions between the positively charged nucleus and the valence shell electrons,
The reduction potentials of halogens decrease (ie. become less positive, or more negative) down the group; accordingly the oxidizing power of halogens decrease down the group.
The oxidation potentials of halide ions increase (ie. become more positive, or less negative) down the group; accordingly the reducing power of the halide ions increase down the group.
Thanks for ur reply...a little difficult to understand for the colour part as I'm a sec4 student...
Anyway does highly reactive = Good oxidizing/reducing agent?[/QUOTE]
You're welcome. If you intend to continue studying Chemistry after your 'O' levels (eg. H2 Chem in JC, or Chem Engineering in Poly, etc), you will benefit from learning and understanding a little deeper about these matters now.
Yes, the more reactive the metal, the more positive its oxidation potential. The more reactive the halogen, the more positive its reduction potential.
Check out the standard redox potentials below (even though you're only taking 'O' levels, and do not need to know how to use the values in the image file below, but once you understand what these values mean, you can use it during your revision to determine which is a more reactive metal, which is a more powerful reducing agent, which is a more powerful oxidizing agent, etc.)
Notice that since the half-equations (in the image file below) are written as reduction equations, the values given are standard reduction potentials.
Although potential has a more technical definition in Physics, in the context of ElectroChemistry, you will do well to regard "potential" as "tendency or propensity or willingness".
As an example, notice that :
The standard reduction potential of Cu2+ to Cu is +0.34V.
Hence, the standard oxidation potential of Cu to Cu2+ is -0.34V.
The standard reduction potential of Ca2+ to Ca is -2.87V.
Hence, the standard oxidation potential of Ca to Ca2+ is +2.87V.
How to interpret these vaules :
Positive reduction potential means Cu2+ prefers to be reduced to Cu.
Negative oxidation potential means Cu does not like to be oxidized to Cu2+.
Hence copper is an unreactive metal, prefering to remain as Cu metal, and does not like to react (eg. with oxygen) to be oxidized to Cu2+ (eg. in copper(II) oxide).
Negative reduction potential means Ca2+ does not like to be reduced to Ca.
A very positive oxidation potential means that Ca strongly wants to be oxidized to Ca2+.
Hence calcium is a very reactive metal, eager to oxidize itself from Ca metal to Ca2+ cationic form (eg. by reacting with oxygen to form calcium oxide).
Similarly, comparing fluorine/fluoride, iodine/iodide, you'll notice :
The reduction potential of F2 to F- is more positive than the reduction potential of I2 to I-.
This means F2 is more eager than I2 to be reduced, and hence F2 is a more powerful oxidizing agent compared to I2.
The oxidation potential of I- to I2 is more positive (or less negative) than the oxidation potential of I2 to I-.
This means I- is more eager than F- to be oxidized, and hence I- is a more powerful reducing agent compared to F-.
BedokFunland JC's 'A' Levels H2 Chemistry Notes on :
Chemical Bonding (intermolecular interactions) and Proteins (tertiary structure)
Many Singapore JC teachers still subscribe to the outdated idea of regard van der Waals strictly as purely instantaneous dipole - induced dipole between non-polar molecules or parts of molecules. This is outdated, and the correct updated version that Cambridge uses today is : Van der Waals forces are a collective term for all 3 types : instantaneous dipole - induced dipole, permanent dipole - permanent dipole, and permanent dipole - induced dipole.
Increasing number of electrons, and/or increasing size of atom/ion/molecule, and/or increasing polarizability of electron clouds; any and all of these (which are intimately related) will result in greater magnitude of the dipoles (whether permanent or induced) and partial charges, and hence stronger or more extensive van der Waals interactions, stronger instantaneous dipole - induced dipole interactions, stronger permanent dipole - permanent dipole interactions, and stronger permanent dipole - induced dipole interactions, all of these.
For hydrophobic interactions of non-polar or weakly polar 'R' groups (ie. groups not capable of hydrogen bonding or ionic bonding) of amino acid residues of a polypeptide chain in a protein : as hydrogen bonds are significantly stronger, water would rather hydrogen bond amongst themselves (because the stronger the bonds or interactions formed, the more stable the resulting structure/product/species), hence pushing or repelling away non-polar 'R' groups. These non-polar 'R' groups subsequently get pushed until they come within close proximity of each other, whereupon the consequent van der Waals forces of attraction (whether they be permanent dipole - permanent dipole, or permanent dipole - induced dipole, or instantaneous dipole - induced dipole) will hold these non-polar or weakly polar 'R' groups together, contributing to the protein's tertiary structure.
The exam-smart student will include both key phrases "hydrophobic" and "van der Waals" in his answer on the nature of the interaction betwen non-polar or weakly polar 'R' groups in a protein.
Disulfide bridges, or disulfide bonds, are covalent bonds between two S atoms, usually present in proteins as occurring between cysteine amino acid residues of the polypeptide. These are the covalent bonds that are involved in the perming and rebonding of hair. (the term "rebond" means to recreate the disulfide covalent bonds that allow hair to hold its shape, whether it be curly or straight.)
If your hair is curly, the disulfide bonds are cleaved (via adding of hydrogen; a reduction process for sulfur since the OS of the sulfur atoms is decreased; since sulfur is more electronegative than hydrogen), then your hair is straightened, and the disulfide bonds are recreated (hence "rebond") (via removal of hydrogen; an oxidation process for sulfur since the OS of the sulfur atoms are increased; since sulfur and sulfur have equal electronegativity) these newly recreated disulfide bonds hold your hair straight after a rebonding session.
Hydrogen bonds, belong to a class on its own. They are stronger than permanent dipole - permanent dipole forces of attraction, but weaker than covalent bonds. They are therefore neither, but possess some characteristics of both.
Hydrogen bonds have similar strength to ion-dipole attractions (of which there are two types : ion - permanent dipoles are naturally usually stronger than ion - induced dipoles), though (depending on the charge densities of the ions involved) ion - dipole attractions are usually slightly stronger than hydrogen bonds.
An illustrative example of this, is that phenol precipitate or benzoic acid precipitate, dissolves when aqueous sodium hydroxide is added : deprotonation of phenol or benzoic acid, which only had limited hydrogen bonding with water, results in the phenoxide anion or benzoate anion which experiences the somewhat stronger (than hydrogen bonding) ion - (permanent) dipole interactions with water, allowing the precipitate to dissolve. An upgrade in the class of inter-species physical interactions, so to speak, resulting in greater solubility in water.
An interesting case in point, would be whether amino-acids in zwitterionic form have hydrogen bonds or ionic bonds with each other. The answer is : the forces of attraction that results between the R-NH3+ group on one zwitterionic amino acid, and the R-COO- group on the next zwitterionic amino acid, may be regarded as either. Again, the exam-smart candidate would write both, but with brief qualifications or explanations, so the examiner understands, "this candidate really knows his stuff!" and award the full marks.
The interaction between R-NH3+ and R-COO-, may be regarded as hydrogen bonding, because the interaction is between a partially positively charged hydrogen on R-NH3+ and a formally negatively charged oxygen on R-COO-. But because R-NH3+ may also be regarded as cationic (though the positive formal charge is on N, not H; more on that in a while) and the R-COO- group may be regarded as anionic; hence one may argue ionic bonding occurs between zwitterionic amino acids. Either way, this explains why amino acids are least soluble in water, at their isoelectric point.
Related to the above discussion, is whether ammonium cations and hydroxide anions experience hydrogen bonding or ion-dipole attractions with water. Both are acceptable (some chemists would label such as the former, while other chemists would label such as the latter), and the exam-smart student will again, write both, but with brief qualifications and explanations.
Technically, NH4+ experiences hydrogen bonding rather than ion-dipole interactions with water. This is because :
1) the tetrahedral geometry of the NH4+ ion will result in water molecule physically interacting with the partially positively charged H atoms, rather than the N atom.
2) the positively formally charged N atom would be even more strongly electron-withdrawing by induction, resulting in a larger magnitude of a partial positive charge on the H atoms, resulting in stronger hydrogen bonding with water (compared to say, ammonia with water).
The hydroxide ion, consists of a partially positively charged H atom, and a formally negatively charged O atom. The H atom therefore experiences hydrogen bonding with water molecules, while the O atom experiences what may be regarded as ion - permanent dipole interactions with water, or what may be regarded as especially strong hydrogen bonding with water.
For hydrogen bonding to occur, there must be :
#1 - a partially positively charged H atom; the magnitude of the partial positive charge must be significant enough, this occurs naturally when H is covalently bonded to electronegative F, O or N atoms.
#2 - a partially or formally negatively charged F, O or N atom with at least one available lone pair; because hydrogen bonding is mainly electrostatic in nature, hence there needs to be a partial or formal negative charge on the F, O or N atom for hydrogen bonding to occur with the partially positively charged H atom. Because hydrogen bonding is also somewhat covalent in nature, there needs to be at least one available lone pair for donation of the hydrogen bond (which may be viewed as almost dative in nature, but not quite, since H atoms only has one 1S orbital and cannot violate its duplet, ie. a H atom can have at most only one lone pair or one bond pair at any time).
Therefore, the negatively formally charged O atom in hydroxide ion experiences particularly strong hydrogen bonding, which accordingly may arguably also be labelled as an ion - permanent dipole interaction.
Being exam smart, means to write both possible answers if the question is ambiguous, or the context or concepts involved, allow for such. This will greatly increase the exam candidate's chances of successfully scoring the allocated marks. However, the exam smart student *must* qualify and explain (at least briefly) why both answers may be argued to be correct or relevant. If this qualification or explanation is not done, the examiner/marker may think the student is trying to smoke/bluff his way through, by giving multiple answers to a single question; which will result in the student losing the marks.
Example of an ambiguous question : "What is the product formed when the following compound is reacted with KMnO4(aq)?".
The exam smart candidate would write : "Dear Examiner Sir/Mdm, because the question failed to specify umambiguously the conditions (eg. temperature) at which the reaction is carried out, therefore I will explore both alternative possibilites as follows."
"If hot, concentrated, acidified KMnO4(aq) is used, oxidative cleavage of the alkene double bond would occur, and the structure of the product would be... If cold, dilute, alkaline KmNO4(aq) is used, then two OH groups would be added across the alkene double bond, and the structure of the product would be..."
'A' Level Qn.
From Wikipedia :
Snake wine (è›‡é…’, pinyin: shéjÇ�u; rÆ°á»£u ráº¯n in Vietnamese) is an alcoholic beverage produced by infusing whole snakes in rice wine or grain alcohol. The drink was first recorded to have been consumed in China during the Western Zhou dynasty and considered an important curative and believed to reinvigorate a person according to Traditional Chinese medicine. It can be found in China, Vietnam and throughout Southeast Asia.
The snakes, preferably venomous ones, are not usually preserved for their meat but to have their "essence" and snake poison dissolved in the liquor.
'A' Level Qn :
By the use of relevant diagrams and/or equations, suggest why drinking venomous snake wines will not usually poison the drinker.
'A' Level Qn - Kinetics.
Originally posted by Audi:
1. Do we include a catalyst when writing down the rate equation?
2.Explain the following observations as fully as you can.
When ammonia is passed over a platinum gauze, the rate of decomposition into nitrogen and hydrogen is independent of the partial pressure of ammonia. However, at very low pressure, the rate is directly proportional to the partial pressure of ammonia.
1. Yes. Since the rate of reaction is affected by the presence and molarity of a catalyst. For any given reaction, the numerical value of the rate constant k is said to be affected only by temperature. This is true. But when a catalyst is employed, the reaction pathway involves an alternate mechanism with a lower activation energy (at the rate determining elementary step), which means to say that this may be considered a different reaction altogether, and hence has a different (a much larger) rate constant k numerical value.
2. At sufficiently high pressure, the catalyst (ie. platinum) is saturated with its substrate (ie. ammonia, the excess reactant), and hence altering (eg. further increasing) the partial pressure (or molarity) of ammonia (ie. the excess reactant) does not alter the rate of reaction, ie. the reaction is zero order with respect to ammonia. At lower partial pressures (or molarities) of ammonia, the limiting reactant is the ammonia itself, and increasing partial pressure (or molarity) of ammonia (ie. the limiting reactant) will result in an increase in rate of reaction, ie. the reaction is first order with respect to ammonia.
'A' Level Qn
Can iron(III) iodide exist in aqueous solution? Explain.
aka "The Legend of the Mythical FeI3(aq)"
No, because Fe3+ would oxidize I- to I2, and be itself reduced to Fe2+. Or you can think of it as I- would reduce Fe3+ to Fe2+, and be itself oxidized to I2.
Always quote relevant Data Booklet values in your answer. Under standard conditions, ie. 1 mol/dm3 of Fe3+/Fe2+ and I2/2I- , the standard reduction potential of Fe3+ to Fe2+ is +0.77V; while the standard oxidation potential of 2I- to I2 is -0.54V.
Hence, under standard molarities, we have cell potential = reduction potential @ cathode + oxidation potential @ anode (alternatively, cell potential = reduction potential @ cathode - reduction potential @ anode) = +0.77V + -0.54V = +0.23V, which proves that our conjecture as stated in our opening paragraph above holds true.
Moreover, notice that the stoichiometry or mole ratio of I- to Fe3+ in a solution of FeI3(aq) is not the required (ie. for our calculations to hold true) 2 : 1 ratio, but is actually 3 : 1.
This means that as predicted by Le Chatelier's principle, having a greater molarity (ie. molar concentration) of I- will shift position of equilibrium to the left, meaning that the oxidation potential of our I- to I2 is now even more positive than before, which results in our cell potential being even positive than +0.23V, which in effect means all the more I- will react with Fe3+ in a redox reaction to generate I2 and Fe2+.
[Reduction] 2Fe3+ + 2e- ---> 2Fe2+
[Oxidation] 2I- --> I2 + 2e-
[Balanced Redox] 2Fe3+ + 2I- --> I2 + 2Fe2+
So we have 2 moles of FeI3 reacting to generate one mole of I2 and 2 moles of Fe2+ and there are 4 moles of I- left over, which is equivalent to having 2 moles of FeI2(aq), which can indeed exist happily ever after in aqueous solution, unlike the non-existent fairy-tale mythical FeI3(aq).
'A' Level Qn.
You have 2 solutions of Cu2+, both having equal molarities of 1mol/dm3 Cu2+(aq). In one solution, you add 2 mol/dm3 of methylamine. In the other solution, you add 1mol/dm3 of ethylenediamine (aka ethane-1,2-diamine). In which solution do you expect to have a larger Kc value (to be precise : stability constant aka formation constant) for the formation of the complexed Cu2+(aq) ions? Explain.
Although the enthalpy of Cu-N bond formation is approximately the same for both ligands (or put in another way, both ligands' donor N atoms have approximately equal donor power towards Cu2+), however there will be a significantly greater molarity of the ethylenediamine copper(II) complex ions, over the dimethylamine copper(II) complex ions.
There are primarily 2 reasons for this phenomenon, which is known as the Chelate Effect.
The 1st reason, is that of Proximity. Once one end of a bidentate ligand (eg. ethylenediamine) is coordinated, the other end is forced to be in close proximity to the central metal ion. Only a rotation of the ligand is required for further coordination, which makes it far more likely that this would occur. In contrast, just because a monodentate ligand (eg. methylamine) is coordinated, has no direct effect on the chance or rate of another monodentate ligand being coordinated as well.
The 2nd reason, is that of Thermodynamics, specifically Entropy. With the monodentate methylamine ligand, we have :
2CH3NH2 + [Cu(H2O)6]2+ --> [Cu(CH3NH2)2(H2O)4]2+ + 2H2O
Notice we have 3 species on the left, and 3 species on the right. Almost no change in entropy.
With the bidentate ethylenediamine ligand, we have :
NH2CH2CH2NH2 + [Cu(H2O)6]2+ --> [Cu(NH2CH2CH2NH2)(H2O)4]2+ + 2H2O
Notice we have 2 species on the left, and 3 species on the right. A significant, positive, favourable change in entropy.
Hence, with the bidentate (or in general, multidentate or polydentate ligands) ligands, compared to monodentate ligands, we observe a more positive (ie. favourable) change in entropy, which in effect resuts in a more negative (ie. favourable) change in Gibbs Free Energy.
The two reasons above, explain why the Kc (stability constant or formation constant) value for multidentate or polydentate ligands, are larger than the corresponding Kc value for monodentate ligands, ceteris paribus (ie. all else being equal).