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Common mistake in drawing hydrogen bond

Mr Khemistry

Common mistakes in drawing hydrogen bond

One of the most common mistakes is to write the dipole across the hydrogen bond, e.g.

A dipole exists as a pair, partial positive and partial negative. It also depicts the uneven electron density distribution in a covalent bond thus it first forms in a single molecule where
the hydrogen atom is bonded to a very electronegative element such as F/O/N. As the hydrogen is almost “stripped” bare of its electron, it is strongly attracted to the lone pair on a highly electronegative F/O/N atom. This electrostatic attraction is known as a hydrogen bond. e.g.

Another type of hydrogen bonding within the same molecule is known as Intramolecular hydrogen bonding. Which type is overcome during boiling?

More here

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My Last Class T-shirt

Mr Khemistry

Thankful that i had the chance to teach such a wonderful bunch of students before i left the college. Even though i told them i didn’t need one as i wouldn’t have
any occasion to wear it. Thank you 17S305, for my last class t-shirt 🙂

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Tips for organic deductive questions

Mr Khemistry

Organic deductive questions

If the organic deductive questions gives the molecular formula, see if the number of C and H is comparable. In addition, if the molecule contains more than 6 carbon atoms, the compound is likely to contain a benzene ring.

Even if C and H is not comparable, the degree of unsaturation can tell us a lot. (Every two H lesser than the alkane general formula is one degree of unsaturation)

For example, one degree of unsaturation means that the compound either has one C=C or is cyclic. Two degree of unsaturation means
the compound could have one C=C and is cyclic, has two C=C bonds or one triple bond. Degree of unsaturation is also known as index of hydrogen deficiency.

Another vital piece of information is how many products a strong oxidative cleavage produces. If only one single product is formed with no loss in C, it means the the original compound is likely to be cyclic. If there is effervescence, the gas is likely to be carbon dioxide and the original compound contains terminal alkene. Be careful though, ethanedioic acid can also oxidize to 2 moles of carbon dioxide under strong oxidation conditions e.g. hot acidic potassium manganate (VII)

More tips here

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Why do we need to acidify after heating with NaOH(aq) in test for halides?

Mr Khemistry

Halides test

Ans: The purpose of acidifying is to neutralise the excess OH- ions that will form a precipitate with Ag+ ions.

This is an often overlooked step in the halides test.

Another common question is why does the precipitate take varying time to appear? This has to do with the type of halide we have.

More common questions here.

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Simple tips for pattern recognition questions

Mr Khemistry

Tips for pattern recognition questions

Tip #1: Use your highlighter.

Complicated rearrangement of atoms can be followed simply by highlighting the example given.
Do the same for the question you are given and you will be able to predict the reaction product easily!

Tip #2: Number the carbon chain.

This tip is more for those questions with long chains, sometimes with numerous branches. It allows
you to keep track of the number of carbons involved so that you don’t miss out any in your answer.
Especially for those cyclic organic compounds.

Want more tips to help you master those pattern recognition questions quickly? Click here!

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Changes to H2 Chemistry Syllabus Part 2

Mr Khemistry

H2 Chemistry Syllabus 2017

Part 2 (part 1 here)

H2 chemistry syllabus 2017 updates:

Organic: Constitutional (structural) isomerism. Cis-trans isomerism, used to be geometric isomerism. Enantiomerism used to be optical isomerism.

Nitrogen Compounds (proteins): Not required to learn primary – quaternary structures, as well as list major functions of proteins in the body. Not required to recognize the twenty amino acids and describe α-helix and β-pleated sheet. Not required to describe protein components of haemoglobin. Not required to explain denaturation of proteins.

Electrochemistry: state and apply the relationship Δ = −nFEθ to electrochemical cells. spontaneous redox reaction (ΔG < 0) would mean that Ecell > 0. In other words, the extent of a redox reaction is governed by the sign and magnitude of cell, like in the case of Δ.

Since Eθcell is dependent on both temperature and concentration (Nernst equation), a non-spontaneous reaction under standard conditions can be made feasible under non-standard conditions (e.g. increasing temperature or concentration).

Similar to the use of ΔGθ to predict the feasibility of a reaction, Eθcell only gives an indication of the thermodynamic feasibility (spontaneity), i.e. based on energy consideration.

The other important aspect, the kinetic feasibility (looking at the activation energy, Ea) must also be considered in order to understand fully whether a reaction is likely to proceed on its own. 

Note that “standard conditions” for both Electrochemistry and Energetics is now 298K, 1 bar and 1 moldm-3 

Atomic structure: Describe the shapes of d orbitals. This ties in well with the explanation for splitting of degenerate (same energy level) d orbitals.

Transition elements: Describe, using the shape and orientation of the d orbitals, the splitting of degenerate d orbitals into two energy levels in octahedral complexes.

In an octahedral complex, a central metal atom is surrounded by 6 lone pairs of electrons (on the 6 ligands). dx2_y2 orbital and the dz2 orbitals have their lobes pointing at the ligands along the x and y axes, and z axis respectively, hence they experienced greater repulsion from the ligands. The dxy, dyz and dxz orbitals experienced less repulsion since their lobes do not point at the ligands along the axes. Thus the 5 d orbitals are split into 2 energy levels, with higher energy level dx2_y2 and dz2 orbitals and the dxy, dyz and dxz orbitals having the lower energy level

End of H2 Chemistry Syllabus 2017 update~

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4s vs 3d, which subshell has the higher energy level?

Mr Khemistry

4s vs 3d – Which has the higher energy?

Some of us may have learn this phrase “First In, First Out” or “FIFO” regarding 4s vs 3d. But why is the 4s orbital filled first?

Ans: The simple answer is because 4s orbital is lower in energy than 3d orbital when they are empty.

Qn: But isn’t principle quantum n=3 supposed to have lower energy than n=4?

Ans: Generally yes, but within the quantum shell, each subshell has slightly different energy levels due to their differing distance from the nucleus (when the subshell is further away from the nucleus, it has higher energy, as it experiences weaker attractive force) . It so happens that because of their shapes, there is an overlap between 3d and 4s orbitals energy levels, but they are very close in energy levels, which explains why transition elements can lose both 3d and 4s electrons to have variable oxidation states.

Qn: Woah. Wait, then why after filling the 3d orbitals, the 4s becomes higher in energy level?

Ans: Once 3d orbitals are occupied by electrons, like in the case of transition elements, because they are closer to the nucleus, they will repel the 4s electrons further away from the nucleus and cause it to have higher energy level. So this explains why even though we fill the 4s before 3d orbitals, we will still ionize 4s electrons before 3d electrons.

Qn: Ok ok, last question. Why is a half-filled/fully-filled d subshell more stable?

Ans: A symmetrical distribution (half/fully-filled subshell) of electron density results in a more stable atom which has lower energy.

Qn: I still don’t get it. Why flip-flop like that?

Ans: Sign up for our group tuition lessons, you will understand 4s vs 3d better after some guided practice 🙂

Call us @ 98537960 to book your place now!

For more commonly asked questions, click here.

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Changes in H2 Chemistry syllabus 2017 – part 1

Mr Khemistry

Theories of acids and bases *New!
Arrhenius theory – acid produce H+ in water, base produces OH in water.
Bronsted Lowry theory – acid produces H+, base accepts H+. Conjugate pair concept.
Lewis theory – acid as electron pair acceptor, base as electron pair donor.

Gases: Vm = 22.7 dm3 at stp (1 bar, 273 K), 24 dm3 rtp (1 atm, 293 K)

Group 2 & 17 (no longer Roman Numerals) now under Periodicity.

Group 17: Not required to describe the reactions of halide ions with sulfuric acid. Not required to describe the reaction of chlorine with cold and hot aqueous sodium hydroxide.

Group 2: learn thermal stability of carbonates instead of nitrates.

Equilibria: Show understanding that the position of equilibrium is dependent on the standard Gibbs free energy change of reaction, ΔG.
ΔG < 0 means forward reaction is spontaneous (equilibrium constant is >> 1, position of equilibrium lies to the right), > 0 means forward reaction is not spontaneous but backward reaction is (equilibrium constant is < 1, position of equilibrium lies to the left). ΔG = 0 means the system is at equilibrium.

Part 2 here

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Changes in weightage for H2 Chemistry syllabus 2017

Mr Khemistry

‘A’ Level Chemistry Syllabus

Overview of changes in weightage:

A level H2 Chemistry weightage

Table showing weightage

Table taken from SEAB website

All students required to take practical paper 4, including private candidates.

This means that candidates retaking their A level chemistry paper this year will have to sit for a 20{8445fa0408f68c331c03e03f10d6a7bff33fb7168cc52e9b2191ddf5ec3671a6} weightage practical paper.
Also, their previous SPA scores will not be counted.

So you are a private student retaking A level H2 chemistry this year, do try to enroll yourself in a private school which provides practical lessons.

‘A’ Level Chemistry Syllabus weightage

Paper 1 will be 30 MCQ (with 5-8 Multiple Completion Type) instead of 40 MCQ (10 MCT). Lower weightage.

Paper 2 data-based question will be 20-25 marks out of 75 instead of 15-20 marks out of 72, which gives it a higher weightage.

Planning question shifted from Paper 2 to Paper 4. Weightage is similar ~11/12 marks.

Paper 3 now has two sections, Section A 60 marks of 3-4 compulsory questions. Section B 20 marks, choose one of two questions.
Previously students choose 4 of 5 20 mark questions.
This means that there is lesser chances of avoiding a certain topic since bulk of the questions are compulsory.
Questions will include different topics.

Overall, 5{8445fa0408f68c331c03e03f10d6a7bff33fb7168cc52e9b2191ddf5ec3671a6} of weightage shifted over from MCQ to Paper 2.

There is a shift away from remembering large chunks of information and a shift toward application of knowledge.
Hence there is a real need to guide students to often and consciously apply their knowledge to unfamiliar contexts.

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Why do we use only sulfuric acid for permanganate titrations?

Mr Khemistry

Permanganate titration

Qn: Why can’t we use hydrochloric acid or nitric acid for manganate (VII) redox titrations?

Ans: For HCl, the chloride ion will be oxidised to chlorine gas by manganate (VII) ion.

HNO3 is also an oxidising agent and hence would compete with permanganate.

Lastly, permanganate cannot oxidize sulfate ion and sulfuric acid is not an oxidizing agent.

Therefore, sulfuric acid is the mineral acid of choice for manganate (VII) titrations.

More common questions here.

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