1/28/2017

JAMB Sets Different TRAPS For Cheaters Including The Use Of CCTV.



Tech!!! UTME 2017 – JAMB Plans to use CCTV Camera Come 2017
    
JAMB says CCTV cameras will help to monitor the exams worldwide.

The Joint Admissions and Matriculation Board (JAMB) has disclosed plans to install Closed-Circuit Television (CCTV) cameras in all computer-based centres to be accredited for the registration of 2017 Unified Tertiary Matriculation Examination (UTME).

JAMB registrar, Professor Ishaq Oloyede, revealed this during an interactive meeting with operators of CBT centres from across the country, at the University of Lagos (UNILAG), Daily Independent reports.

Oloyede said the introduction of the CCTV’s will help to monitor whatever is going on at the centres. 

He also stated that the new process will exclude many cybercafe operators as the computer-based centres to be accredited must have 250 desktop or laptop systems in a single room, with provision for additional 25 systems as back-ups.

Oloyede said:“What we have said is that we are in a period of change, and what that means is that everything we have been doing must be reviewed. Whatever we have been doing rights will have to be reinforced while what were doing wrongly will be changed.

“For the purpose of the CBT centres, we have to introduce the CCTV so that whatever that is going on at our centres can be monitored anywhere in the world.

We are also standardizing the CBT centres so that all CBT centres will have 250 computers with 10 per cent of the figure as backups.”

In a related development, JAMB has urged the general public to disregard the speculations that it had started the sales of its application forms for 2017 UTME.

Dr. Fabian Benjamin, the board’s Head, Media and Information, debunked the speculations in a statement on Sunday, December 4, in Lagos.

According to the statement, the Registrar of the examination body, Prof. Ishaq Oloyede, expressed disappointment with the development and calls on Nigerians to ignore the claims.

JAMB STUDENTS WILL USE ATM TO PRODUCE THEIR REG. PIN (JAMB 2017 PAYMENT PLANS)


New Jamb 2017 payment plans.

The Joint Admissions and Matriculation Board says it will adopt “pin vending” for the 2017 UTME test, advising intending candidates to get familiar with the new approach.

Dr. Fabian Benjamin, the board’s Head of Media and Information, told the News Agency of Nigeria on Wednesday in Lagos that JAMB would no longer use scratch cards.

Benjamin said: “Candidates, wishing to sit for the 2017 examination, should start getting themselves familiar with the newly adopted process of pin vending by the board.

“We must make ourselves open to change like it is obtained in other climes.

“We are no longer going back to the use of scratch card; that method is outdated.

“Candidates wishing to register for the examination will just make online payment and get a pin with which they can upload their data.

“This new pin vending will be accessible through the options of web payment, ATM issued cards like Visa, Verve, and Master card, online Quick Teller, mobile application and Bank Branch case/card.”

Benjamin assured that the board was working hard to redress all challenges experienced by candidates during its 2016 UTME as it was preparing for the 2017 diet.

He said that all hands were on deck to ensure a hitch free conduct of the examination across the country.

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He said: “Preparations are on to ensure that all the technical hitches that manifested in the 2016 Unified Tertiary Matriculation Examination do not arise again.

“That is not to say that the examination will be completely hitch-free.

“But we are deploying resources to correct the ones identified already.

“In the course of the examination, should there be any other new challenges, we will act promptly.”

The spokesman also said that the board would begin validation of UTME centres across the country soon.

He stated: “We shall be going round to the proposed centres to check the state of their facilities and also to ensure that such facilities could accommodate a minimum of 250 candidates.”

NAN reports that some candidates, who sat for the 2016 examination, complained of various technical hitches.

Such challenges ranged from incomplete upload of questions from servers to poor backup facilities in some centres among others.

NAN also recalls that the board’s Registrar, Prof. Ishaq Oloyede, recently announced the scrapping of the use of scratch card, describing it as outdated.

He said the board decided to do away with the method because of its consistent subjection to fraudulent practices.

JAMB FINALLY SETS A DATE FOR 2017 JAMB EXAMS


See The Date JAMB Plans to fix 2017 UTME Date Here



JAMB has now concluded plans to fix the dates for the 2017 UTME. This was made known today by The Head, Media and Information of the board, Dr Fabian Benjamin.


Members of the JAMB board have been scheduled to meet with board members of WAEC, NECO and NABTEB on Tuesday, 17th January, to find suitable dates for the 2017 UTME.
Remember that the new JAMB CBT no longer holds in one day, and may require up to 2 weeks to fully test all candidates, who apply for the exam. This meeting will therefore help avoid clashes in exam dates for other exam bodies, a problem experienced in previous exams.

According to Mr. Fabian, this is the final preparation required before the Sale of 2017 UTME Registration Forms can commence.

We therefore hope that another press briefing will be conducted just after the meeting on 17th January, to reveal the possible 2017 UTME Exam Dates, and when sale of registration forms will commence

1/23/2017

THE GENERAL CHARACTERISTICS OF ALL TRANSITION METAL CHEMISTRY.


THE GENERAL FEATURES OF TRANSITION METAL CHEMISTRY

This page explains what a transition metal is in terms of its electronic structure, and then goes on to look at the general features of transition metal chemistry. These include variable oxidation state (oxidation number), complex ion formation, coloured ions, and catalytic activity.

You will find some of this covered quite briefly on this page with links to other parts of the site where the topics are covered in more detail.

The electronic structures of transition metals

What is a transition metal?

The terms transition metal (or element) and d block element are sometimes used as if they mean the same thing. They don't - there's a subtle difference between the two terms.

We'll explore d block elements first:

d block elements

You will remember that when you are building the Periodic Table and working out where to put the electrons using the Aufbau Principle, something odd happens after argon.

At argon, the 3s and 3p levels are full, but rather than fill up the 3d levels next, the 4s level fills instead to give potassium and then calcium.

Only after that do the 3d levels fill.

If you do follow the link, use the BACK button on your browser (or the History file or Go menu) to return quickly to this page.


The elements in the Periodic Table which correspond to the d levels filling are called d block elements. The first row of these is shown in the shortened form of the Periodic Table below.




The electronic structures of the d block elements shown are:

Sc[Ar] 3d14s2Ti[Ar] 3d24s2V[Ar] 3d34s2Cr[Ar] 3d54s1Mn[Ar] 3d54s2Fe[Ar] 3d64s2Co[Ar] 3d74s2Ni[Ar] 3d84s2Cu[Ar] 3d104s1Zn[Ar] 3d104s2

You will notice that the pattern of filling isn't entirely tidy! It is broken at both chromium and copper.

Note:  This is something that you are just going to have to accept. There is no simple explanation for it which is usable at this level. Any simple explanation which is given is faulty!

People sometimes say that a half-filled d level as in chromium (with one electron in each orbital) is stable, and so it is - sometimes! But you then have to look at why it is stable. The obvious explanation is that chromium takes up this structure because separating the electrons minimises the repulsions between them - otherwise it would take up some quite different structure.

But you only have to look at the electronic configuration of tungsten (W) to see that this apparently simple explanation doesn't always work. Tungsten has the same number of outer electrons as chromium, but its outer structure is different - 5d46s2. Again the electron repulsions must be minimised - otherwise it wouldn't take up this configuration. But in this case, it isn't true that the half-filled state is the most stable - it doesn't seem very reasonable, but it's a fact! The real explanation is going to be much more difficult than it seems at first sight.

Neither can you use the statement that a full d level (for example, in the copper case) is stable, unless you can come up with a proper explanation of why that is. You can't assume that looking nice and tidy is a good enough reason!

If you can't explain something properly, it is much better just to accept it than to make up faulty explanations which sound OK on the surface but don't stand up to scrutiny!


Transition metals

Not all d block elements count as transition metals! There are discrepancies between the various UK-based syllabuses, but the majority use the definition:

A transition metal is one which forms one or more stable ions which have incompletely filled d orbitals.

Note:  The most recent IUPAC definition includes the possibility of the element itself having incomplete d orbitals as well. This is unlikely to be a big problem (it only really arises with scandium), but it would pay you to learn the version your syllabus wants. Both versions of the definition are currently in use in various UK-based syllabuses.

If you are working towards a UK-based exam and haven't got a copy of your syllabus, follow this link to find out how to get one. Use the BACK button on your browser to return quickly to this page.


On the basis of the definition outlined above, scandium and zinc don't count as transition metals - even though they are members of the d block.

Scandium has the electronic structure [Ar] 3d14s2. When it forms ions, it always loses the 3 outer electrons and ends up with an argon structure. The Sc3+ ion has no d electrons and so doesn't meet the definition.

Zinc has the electronic structure [Ar] 3d104s2. When it forms ions, it always loses the two 4s electrons to give a 2+ ion with the electronic structure [Ar] 3d10. The zinc ion has full d levels and doesn't meet the definition either.

By contrast, copper, [Ar] 3d104s1, forms two ions. In the Cu+ ion the electronic structure is [Ar] 3d10. However, the more common Cu2+ ion has the structure [Ar] 3d9.

Copper is definitely a transition metal because the Cu2+ ion has an incomplete d level.

Transition metal ions

Here you are faced with one of the most irritating facts in chemistry at this level! When you work out the electronic structures of the first transition series (from scandium to zinc) using the Aufbau Principle, you do it on the basis that the 3d orbitals have a higher energy than the 4s orbital.

That means that you work on the assumption that the 3d electrons are added after the 4s ones.

However, in all the chemistry of the transition elements, the 4s orbital behaves as the outermost, highest energy orbital. When these metals form ions, the 4s electrons are always lost first.

You must remember this:

When d-block elements form ions, the 4s electrons are lost first.

Note:  The problem here is that the Aufbau Principle can only really be used as a way of working out the electronic structures of most atoms. It is a simple way of doing that, although it fails with some, like chromium or copper, of course, and you have to learn these.

There is, however, a flaw in the theory behind it which produces problems like this. Why are the apparently higher energy 3d electrons not the ones to get lost when the metal ionises?

I have written a detailed explanation of this on another page called the order of filling 3d and 4s orbitals. If you are a teacher or a very confident student then you might like to follow this link.

If you aren't so confident, I suggest that you ignore it. Make sure that you can work out the structures of these atoms using the Aufbau Principle on the assumption that the 3d orbitals fill after the 4s, and learn that when the atoms ionise, the 4s electrons are always lost first. Just ignore the contradictions between these two ideas!


To write the electronic structure for Co2+:

Co[Ar] 3d74s2Co2+[Ar] 3d7

The 2+ ion is formed by the loss of the two 4s electrons.

To write the electronic structure for V3+:

V[Ar] 3d34s2V3+[Ar] 3d2

The 4s electrons are lost first followed by one of the 3d electrons.

Note:  You will find more examples of writing the electronic structures for d block ions, by following this link.

Use the BACK button on your browser to return quickly to this page.


Variable oxidation state (number)

One of the key features of transition metal chemistry is the wide range of oxidation states (oxidation numbers) that the metals can show.


It would be wrong, though, to give the impression that only transition metals can have variable oxidation states. For example, elements like sulphur or nitrogen or chlorine have a very wide range of oxidation states in their compounds - and these obviously aren't transition metals.

However, this variability is less common in metals apart from the transition elements. Of the familiar metals from the main groups of the Periodic Table, only lead and tin show variable oxidation state to any extent.

Examples of variable oxidation states in the transition metals

Iron

Iron has two common oxidation states (+2 and +3) in, for example, Fe2+ and Fe3+. It also has a less common +6 oxidation state in the ferrate(VI) ion, FeO42-.

Manganese

Manganese has a very wide range of oxidation states in its compounds. For example:

+2in Mn2++3in Mn2O3+4in MnO2+6in MnO42-+7in MnO4-

Other examples

You will find the above examples and others looked at in detail if you explore the chemistry of individual metals from the transition metal menu. There is a link to this menu at the bottom of the page.

Explaining the variable oxidation states in the transition metals

We'll look at the formation of simple ions like Fe2+ and Fe3+.

When a metal forms an ionic compound, the formula of the compound produced depends on the energetics of the process. On the whole, the compound formed is the one in which most energy is released. The more energy released, the more stable the compound.

There are several energy terms to think about, but the key ones are:

The amount of energy needed to ionise the metal (the sum of the various ionisation energies)

The amount of energy released when the compound forms. This will either be lattice enthalpy if you are thinking about solids, or the hydration enthalpies of the ions if you are thinking about solutions.

The more highly charged the ion, the more electrons you have to remove and the more ionisation energy you will have to provide.

But off-setting this, the more highly charged the ion, the more energy is released either as lattice enthalpy or the hydration enthalpy of the metal ion.

Thinking about a typical non-transition metal (calcium)

Calcium chloride is CaCl2. Why is that?

If you tried to make CaCl, (containing a Ca+ ion), the overall process is slightly exothermic.

By making a Ca2+ ion instead, you have to supply more ionisation energy, but you get out lots more lattice energy. There is much more attraction between chloride ions and Ca2+ ions than there is if you only have a 1+ ion. The overall process is very exothermic.

Because the formation of CaCl2 releases much more energy than making CaCl, then CaCl2 is more stable - and so forms instead.

What about CaCl3? This time you have to remove yet another electron from calcium.

The first two come from the 4s level. The third one comes from the 3p. That is much closer to the nucleus and therefore much more difficult to remove. There is a large jump in ionisation energy between the second and third electron removed.

Although there will be a gain in lattice enthalpy, it isn't anything like enough to compensate for the extra ionisation energy, and the overall process is very endothermic.

It definitely isn't energetically sensible to make CaCl3!

Thinking about a typical transition metal (iron)

Here are the changes in the electronic structure of iron to make the 2+ or the 3+ ion.

Fe[Ar] 3d64s2Fe2+[Ar] 3d6Fe3+[Ar] 3d5

The 4s orbital and the 3d orbitals have very similar energies. There isn't a huge jump in the amount of energy you need to remove the third electron compared with the first and second.

The figures for the first three ionisation energies (in kJ mol-1) for iron compared with those of calcium are:

metal1st IE2nd IE3rd IECa59011504940Fe76215602960

There is an increase in ionisation energy as you take more electrons off an atom because you have the same number of protons attracting fewer electrons. However, there is much less increase when you take the third electron from iron than from calcium.

In the iron case, the extra ionisation energy is compensated more or less by the extra lattice enthalpy or hydration enthalpy evolved when the 3+ compound is made.

The net effect of all this is that the overall enthalpy change isn't vastly different whether you make, say, FeCl2 or FeCl3. That means that it isn't too difficult to convert between the two compounds.

The formation of complex ions

What is a complex ion?

A complex ion has a metal ion at its centre with a number of other molecules or ions surrounding it. These can be considered to be attached to the central ion by co-ordinate (dative covalent) bonds. (In some cases, the bonding is actually more complicated than that.)

The molecules or ions surrounding the central metal ion are called ligands.

Simple ligands include water, ammonia and chloride ions.




What all these have got in common is active lone pairs of electrons in the outer energy level. These are used to form co-ordinate bonds with the metal ion.

Some examples of complex ions formed by transition metals

[Fe(H2O)6]2+

[Co(NH3)6]2+

[Cr(OH)6]3-

[CuCl4]2-

Other metals also form complex ions - it isn't something that onlytransition metals do. Transition metals do, however, form a very wide range of complex ions.

Note:  You will find much more about complex ions by following this link. It will take you to a part of the site dealing exclusively with complex ions.

If you do follow the link, use the BACK button on your browser (or the History file or Go menu) if you want to return to this page again.


The formation of coloured compounds

Some common examples

The diagrams show aproximate colours for some common transition metal complex ions.




You will find these and others discussed if you follow links to individual metals from the transition metal menu (link at the bottom of the page).

Alternatively, you could explore the complex ions menu (follow the link in the help box which has just disappeared off the top of the screen).

The origin of colour in the transition metal ions

When white light passes through a solution of one of these ions, or is reflected off it, some colours in the light are absorbed. The colour you see is how your eye perceives what is left.

Attaching ligands to a metal ion has an effect on the energies of the d orbitals. Light is absorbed as electrons move between one d orbital and another. This is explained in detail on another page.

Note:  You will find a detailed explanation of the origin of colour in complex ions and the factors which cause it to change by following this link. That page is on the part of the site dealing with complex ions.

Use the BACK button on your browser if you want to return to this page again.


Catalytic activity

Transition metals and their compounds are often good catalysts. A few of the more obvious cases are mentioned below, but you will find catalysis explored in detail elsewhere on the site (follow the link after the examples).

Transition metals and their compounds function as catalysts either because of their ability to change oxidation state or, in the case of the metals, to adsorb other substances on to their surface and activate them in the process. All this is explored in the main catalysis section.

Transition metals as catalysts

Iron in the Haber Process

The Haber Process combines hydrogen and nitrogen to make ammonia using an iron catalyst.



Reasons Why You Fail Chemistry

Top Reasons Why Students Fail Chemistry


Avoiding Failure in Chemistry

Are you taking a chemistry class? Are you worried you might not pass? Chemistry is a subject many students prefer to avoid, even if they have an interest in science, because of its reputation for lowering grade point averages. However, it isn't as bad as it seems, especially if you avoid these common mistakes.

1.  Procrastinating


You can pass chemistry if you pace yourself studying.Arne Pastoor, Getty Images

Never do today what you can put off until tomorrow, right? Wrong! The first few days in a chemistry classmay be very easy and could lull you into a false sense of security. Don't put off doing homework or studying until halfway through the class. Mastering chemistry requires you to build concept upon concept. If you miss the basics, you'll get yourself into trouble. Pace yourself. Set aside a small segment of time each day for chemistry. It will help you to gain long-term mastery. Don't cram.



2.  Insufficient Math Preparation

Don't go into chemistry until you understand the basics of algebra. Geometry helps, too. You will need to be able to perform unit conversions. Expect to work chemistry problems on a daily basis. Don't rely too much on a calculator. Chemistry and physics use math as an essential tool.



3.  Not Getting or Reading the Text

Yes, there are classes in which the text is optional or completely useless. This isn't one of those classes. Get the text. Read it! Ditto for any required lab manuals. Even if the lectures are fantastic, you'll need the book for the homework assignments. A study guide may be of limited use, but the basic text is a must-have.


4.  Psyching Yourself Out

I think I can, I think I can... you have to have a positive attitude toward chemistry. If you truly believe you will fail you may be setting yourself up for a self-fulfilling prophecy. If you have prepared yourself for the class, you have to believe that you can be successful. Also, it's easier to study a topic you like than one you hate. Don't hate chemistry. Make your peace with it and master it.


5.  Not Doing Your Own Work

Study guides and books with worked answers in the back are great, right? Yes, but only if you use them for help and not as an easy way to get your homework done. Don't let a book or classmates do your work for you. They won't be available during the tests, which will count for a big portion of your grade.

Basic OXIDATION AND REDUCTION Lesson...

DEFINITIONS OF OXIDATION AND REDUCTION (REDOX)

This page looks at the various definitions of oxidation and reduction (redox) in terms of the transfer of oxygen, hydrogen and electrons. It also explains the terms oxidising agent and reducing agent.

Oxidation and reduction in terms of oxygen transfer

Definitions

Oxidation is gain of oxygen.

Reduction is loss of oxygen.

For example, in the extraction of iron from its ore:




Because both reduction and oxidation are going on side-by-side, this is known as a redox reaction.

Oxidising and reducing agents

An oxidising agent is substance which oxidises something else. In the above example, the iron(III) oxide is the oxidising agent.

A reducing agent reduces something else. In the equation, the carbon monoxide is the reducing agent.

Oxidising agents give oxygen to another substance.

Reducing agents remove oxygen from another substance.

Oxidation and reduction in terms of hydrogen transfer

These are old definitions which aren't used very much nowadays. The most likely place you will come across them is in organic chemistry.

Definitions

Oxidation is loss of hydrogen.

Reduction is gain of hydrogen.

Notice that these are exactly the opposite of the oxygen definitions.

For example, ethanol can be oxidised to ethanal:




You would need to use an oxidising agent to remove the hydrogen from the ethanol. A commonly used oxidising agent is potassium dichromate(VI) solution acidified with dilute sulphuric acid.

Note:  The equation for this is rather complicated for this introductory page. If you are interested, you will find a similar example (ethanol to ethanoic acid) on the page dealing with writing equations for redox reactions.

Ethanal can also be reduced back to ethanol again by adding hydrogen to it. A possible reducing agent is sodium tetrahydridoborate, NaBH4. Again the equation is too complicated to be worth bothering about at this point.




An update on oxidising and reducing agents

Oxidising agents give oxygen to another substance or remove hydrogen from it.

Reducing agents remove oxygen from another substance or give hydrogen to it.

Oxidation and reduction in terms of electron transfer

This is easily the most important use of the terms oxidation and reduction at A' level.

Definitions

Oxidation is loss of electrons.

Reduction is gain of electrons.

It is essential that you remember these definitions. There is a very easy way to do this. As long as you remember that you are talking about electron transfer:




A simple example

The equation shows a simple redox reaction which can obviously be described in terms of oxygen transfer.



Copper(II) oxide and magnesium oxide are both ionic. The metals obviously aren't. If you rewrite this as an ionic equation, it turns out that the oxide ions are spectator ions and you are left with:




A last comment on oxidising and reducing agents

If you look at the equation above, the magnesium is reducing the copper(II) ions by giving them electrons to neutralise the charge. Magnesium is a reducing agent.

Looking at it the other way round, the copper(II) ions are removing electrons from the magnesium to create the magnesium ions. The copper(II) ions are acting as an oxidising agent.

Warning!

This is potentially very confusing if you try to learn both what oxidation and reduction mean in terms of electron transfer, and also learn definitions of oxidising and reducing agents in the same terms.

Personally, I would recommend that you work it out if you need it. The argument (going on inside your head) would go like this if you wanted to know, for example, what an oxidising agent did in terms of electrons:

An oxidising agent oxidises something else.

Oxidation is loss of electrons (OIL RIG).

That means that an oxidising agent takes electrons from that other substance.

So an oxidising agent must gain electrons.

Or you could think it out like this:

An oxidising agent oxidises something else.

That means that the oxidising agent must be being reduced.

Reduction is gain of electrons (OIL RIG).

So an oxidising agent must gain electrons.

Understanding is a lot safer than thoughtless learning

Easiest Way To Name Any Compound In Chemistry


Naming Inorganic Compounds


With over 10 million known chemicals, and potentially dangerous results if chemicals are combined in an incorrect manner, imagine the problem if you are in the lab and say "mix 10 grams of that stuff in with this stuff". We need to be very clear on identification of chemicals.

Two early classifications of chemical compounds:

Organic compounds. These contain the element Carbon (C). "Life on earth is carbon based"Inorganic compounds. All other compounds

Organic compounds were associated with living organisms, however, a large number of organic compounds have been synthesized which do not occur in nature, so this distinction is no longer valid.

Ionic compounds: (an association of a cation and an anion)

The positive ion (cation) is always named first and listed first in writing the formula for the compound.

The vast majority of monatomic (composed of a single atom) cations are formed from metallic elements:

Na+ Sodium ionZn2+ Zinc ionAl3+ Aluminum ion

If an element can form more than one positive ion, the positive charge of the ion is indicated by a Roman numeral in parentheses following the name of the metal:

Fe2+ iron(II) ionFe3+ iron(III) ionCu+ copper(I) ionCu2+ copper(II) ion

Iron and copper are examples of transition metals. They occur in the block of elements from IIIB to IIB of the periodic table.

The transition metals often form two or more different monoatomic cations.



click on picture for larger image

 

An older nomenclature for distinguishing between the different ions of a metal is to use the suffixes -ous and -ic. The suffix -ic will indicate the ion of higher ionic charge:

Fe2+ ferrous ionFe3+ ferric ionCu+ cuprous ionCu2+ cupric ion

Note that the different ions of the same element often have quite different chemical properties (again, pointing to the importance of electrons in determining chemical reactivity).

Ionic compounds: Anions

Monatomic anions are usually formed from non-metallic elements. They are named by dropping the ending of the element name and adding -ide:

Cl- chloride ionF- flouride ionS2- sulfide ionO2- oxide ion

Some common polyatomic anions include:

OH- hydroxide ionCN- cyanide ion

Many polyatomic anions contain oxygen, and are referred to as oxyanions. When an element can form two different oxyanions the name of the one that contains more oxygen ends in -ate, the one with less ends in -ite:

NO2- nitrite ionNO3- nitrate ionSO32- sulfite ionSO42- sulfate ion

Note that unlike the -ous and -ic suffix nomenclature to distinguish the different cations of a metal, the -ite and -ate suffix is used to distinguish the relative amounts of the oxygen atoms in a (polyatomic) oxyanion (in the above examples the ionic charge is the same for the -ite and -ate ions of a specific oxyanion).

Now to get really perverse....

Some compounds can have multiple oxyanion forms (the oxyanions involving the halogens, for example):

ClO-ClO2-ClO3-ClO4-

Note again, that the number of Oxygens relative to the Chlorine is changing, but that the ionic charge is not.

How do we name these? The -ite and -ate suffixes are still used, but we have to add an additional modification to allow us to distinguish between the four forms:

ClO- hypochlorite ionClO2- chlorite ionClO3- chlorate ionClO4- perchlorate ion




It should be pointed out that some of the naming of ions is historical and is not necessarily systematic. It may be frustrating and confusing, but its all part of chemistry's rich history.

Many polyatomic anions that have high (negative) charges can add one or more hydrogen cations (H+) to form anions of lower effective charge. The naming of these anions reflects whether the H+ addition involves one or more hydrogen ions:

HSO4- hydrogen sulfate ionH2PO4- dihydrogen phosphate ion

Acids

An acid is a substance whose molecules yield hydrogen (H+) ions when dissolved in water.The formula of an acid consists of an anionic group whose charged is balanced by one or more H+ ions.The name of the acid is related to the name of the anionAnions whose names end in -ide have associated acids that have the hydro- prefix and an -ic suffix:

 

Cl- chloride anion

HCl hydrochloric acid 

S2- sulfide anion

H2S hydrosulfuric acid 

Using the -ic suffix here may seem a bit inconsistent since it was used in naming metal cations to indicate the form which had the higher positive charge. However, when you think about it, the acid compound has a higher net positive charge than the anion from which it is derived (the anion is negatively charge and the associated acid is neutral).

Again, things get complicated when we consider the acids of oxyanions:

If the anion has an -ate ending, the corresponding acid is given an -ic endingIf the anion has an -ite ending, the corresponding acid has an -ous ending.Prefixes in the name of the anion are kept in naming the acid

 

ClO- hypochlorite ion 

HClO hypochlorous acid 

ClO2- chlorite ion 

HClO2 chlorous acid 

ClO3- chlorate ion 

HClO3 chloric acid 

ClO4- perchlorate ion 

HClO4 perchloric acid 

This is confusing: we previously had used the -ous and -ic suffixes to indicate the ionic charge differences in metal cations (-ic had a higher positive charge). Although in comparison to the ionic form, the -ic and -ous acid forms have a higher net positive charge, the -ic suffix would indicate forms with a higher oxygen content, and not an apparent charge difference.



Molecular compounds

Although they may not be ionic compounds, chemically bonded compounds of two different elements can be thought of as being made up of an element with a more positive chemical nature, and one that has a more negative nature in comparison. Elements on the left hand side of the periodic table prefer to donate electrons (thus taking on a more positive chemical nature), and elements on the right hand side prefer to accept electrons (thus taking on a more negative chemical nature). The element with the more positive nature in a compound is named first. The second element is named with an -ide ending.

Often a pair of elements can form several different molecular compounds. For example, Carbon and Oxygen can form CO and CO2. Prefixes are used to identify the relative number of atoms in such compounds:

CO carbon monoxide (carbon mono oxide)CO2 carbon dioxide

Such prefixes can extend for quite a way for some organic and polymeric compounds (a common detergent in shampoos is sodium dodecylsulfate, or "SDS", also known as Sodium Laurel Sulfate because it sounds more benign and holistic). The list of such prefixes includes: 
 

Prefix & Meaning

Mono-1

Di-2

Tri-3

Tetra-4

Penta-5

Hexa-6

Hepta-7

Octa-8

Nona-9

Deca-10

Undeca-11

Dodeca-12

Easy Way To Calculate The Empirical And Molecular Formula Of Compounds...

 

 


Empirical and Molecular Formula Calculation

Empirical formula is the smallest whole number ratio of moles of each element in a compound.

CaCl2 --> there is 1 mole of calcium for every 2 moles of chlorine

 

Level 1 Simple Empirical formula questions

What is the empirical formula of the following compounds? (so reduce the formula if you can)

molecular formulaempirical formulaC2H4CH2C11H22O11CH2OH2OH2OC25H50CH2

Level 2  Empirical Formula Calculation Steps

Step 1 If you have masses go onto step 2.

If you have %.  Assume the mass to be 100g, so the % becomes grams.

Step 2 Determine the moles of each element.

Step 3 Determine the mole ratio by dividing each elements number of moles by the smallest value from step 2.

Step 4 Double, triple … to get an integer if they are not all whole numbers

 

Molecular Formula (additional steps)

The question should have included a molecular mass.

Step 5 Determine the mass of your empirical formula

Step 6 Divide the given molecular mass by your E.F. mass in step 5

Step 7 Multiply the atoms in the empirical formula by this number

 

Examples-Caffeine has an elemental analysis of 49.48% carbon, 5.190% hydrogen, 16.47% oxygen, and 28.85% nitrogen. It has a molar mass of 194.19 g/mol. What is the molecular formula of caffeine?

(Hint-Save the molar mass 194.19g/mol until the end)

 

49.48% C, 5.190%H,  16.47% O and 28.85% N

Step 1 Assume a mass of 100g so % becomes grams

49.48g C, 5.190gH,  16.47g O and 28.85g N

Step 2 determine the moles of each element

49.48g  C x ( 1 mole/12.0 g C) = 4.123moles C
5.190g H x (1 mole / 1.0 g H) = 5.190 moles H
16.47g  O x (1 mole /16.0 g O ) =1.029moles O
 28.85g N x ( 1 mole/ 14.0 g N) = 2.061 moles N

Step 3 determine the mole ratio by dividing each elements number of moles by the smallest

Dividing by the smallest (1.029) we get

C: 4.123 / 1.029 = 4.007
H: 5.190 / 1.029 = 5.044
 O: 1.029 / 1.029 =1.000
N: 2.061 / 1.029 = 2.002

Step 4 Double, triple .. to get an integer is they are not all whole numbers

The values are all really close to whole numbers.

Empirical Formula= C4H5ON2

Example- Molecular Formulas (Steps 5-7)

It has a molar mass of 194.19 g/mol.

Step 5 After you determine the empirical formula, determine its mass.

Empirical Formula= C4H5ON2

(4 carbon x 12.0) + (5 hydrogen x1.0) + (1 oxygen x 16.0) + (2 nitrogen x 14.0) =97.0g/mol

Step 6 Determine how many times greater the molecular mass is compared to the mass of the empirical formula.

molecular mass/ empirical formulas mass

194.19g/mol  / 97.0g/mol = 2

Step 7 Multiply the empirical formula by this number

2x C4H5ON2 =C8H10O2N4

better==> C8H10N4O2

***note

if step 6 does not work out to be a whole number your empirical formula is wrong or your teacher screwed up.

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1/22/2017

How To Write Chemical Formula With Valency.

How to write a Chemical Formula for a Compound?

A chemical formula is the representation of a substance by symbols. More importantly, it denotes the number of atoms of each element present in the compound. For example, the formula for Ferric oxide or Iron [III] oxide is Fe2O3, which implies that 2 atoms of Fe and 3 atoms of O are present in an electrically-neutral molecule of the compound. To write a chemical formula, one must know the symbols and valencies of the elements / radicals.
This is for CACO3

Example 1. Write the chemical formula for Calcium hydroxide. 
Valency of Calcium (Ca) = 2 ; Valency of Hydroxide (OH) = 1. 
Interchanging their valencies and writing as subscripts (ignoring 1), 
formula for Calcium hydroxide is Ca(OH)2. 
Note that 1 calcium ion [Ca2+] and 2 hydroxide ions [OH1-] are present in an electrically-neutral molecule of calcium hydroxide [Ca(OH)2].

Example 2. Write the chemical formula for Magnesium sulfate. 
Valency of Magnesium (Mg) = 2 ; Valency of Sulfate (SO4) = 2. 
Interchanging their valencies and simplifying (on dividing by 2), 
formula for Magnesium sulfate is MgSO4. 
Note that 1 magnesium ion [Mg2+] and 1 sulfate ion [SO42-] are present in an electrically-neutral molecule of magnesium sulfate [MgSO4].

Points to remember:

When the subscript is 1, it is ignored.The radical is written in parenthesis when the subscript is 2 or greater.Whenever possible, subscripts are simplified by dividing by the highest common factor (HCF)

Valency of Elements and radicals


Valency or Valence

When atoms of one element combine with the atoms of another element to form formula units, they do so in fixed numbers depending upon the capacities of the atoms to form bonds.

Valency of an element is a measure of the combining capacity of its atom to form chemical bonds.

Valency / Valence

Valency is defined as the number of hydrogen or chlorine atoms with which 1 atom of the element would combine.

As a general rule, if an atom participates in ionic bonding, the valency tells the charge on the ion formed. If the atom participates in covalent bonding, the valency tells the number of electrons the atom shares with its partner atom(s).

The following example will clarify the concept of valency:

 (a) 1 atom of chlorine combines with 1 atom of hydrogen to form HCl (hydrogen chloride). The valency of chlorine is thus 1 (from the definition of valency).

(b) 1 atom of oxygen combines with 2 atoms of hydrogen to form H2O (water or hydrogen monoxide). The valency of oxygen is thus 2 (from the definition of valency).

(c) 1 atom of carbon combines with 4 atoms of hydrogen to form CH4 (methane). The valency of carbonis thus 4.

(d) 1 atom of magnesium combines with 1 atom of oxygen to form MgO (magnesium oxide). Since we know that oxygen has a valency of 2, i.e. its combining capacity is twice that of hydrogen, so the valency ofmagnesium also is 2.

Easy formula by balancing valencies

It is clear that if we know the valencies of elements, then we can work out the chemical formulae of their compounds by balancing the valencies of the different atoms which occur in the compound, i.e. the total of the valencies of one set of atoms should balance the total of the valencies of the other set.

To illustrate:

 (a) Carbon dioxide is made up of carbon and oxygen. We know that the valency of carbon is 4 and that of oxygen is 2. In order for them to combine, we have to balance the valency of 4 of the carbon atom with 2 atoms ofoxygen, each of valency 2. Thus, one atom of carbon will combine with two atoms of oxygen to form the molecule CO2(carbon dioxide).

(b) From the chemical formula of the compound aluminium oxide Al2O3, the valency of aluminium can be determined. Since we know that the valency of oxygen is 2, and there are 3 oxygen atoms, this gives a total value 6 for the valencies of the oxygen atoms, which has to be matched by the 2 atoms of aluminium. Hence the valency ofaluminium is 3.

Valency of an element is a whole number and varies from 1 to 8, and is 0 for an element that does not combine with any other element.

Elements with multiple valencies

Most elements have a fixed value of valency, but some exhibit two or more different valenciesand hence can form two or more different compounds with another element.

There are two naming methods for distinguishing the names of the different compounds formed by the same multiple-valency element. In the older system, the latin name of the element was suffixed with -ous to indicate lower valence or -ic to indicate higher valence. In the newer Stock System, the element's name is followed by its valence in parentheses in Roman numerals.

(a) Iron forms two chlorides – FeCl2 (ferrous chloride or iron(II) chloride) and FeCl3(ferric chloride or iron(III) chloride. The valency of iron is 2 in the former compound and 3 in the latter compound.

(b) Similarly, tin is divalent in SnCl2 (stannous chloride or tin(II) chloride) and tetravalent inSnCl4 (stannic chloride or tin(IV) chloride).

(c) Valency of sulphur is 4 in SO2 (sulphur dioxide) and 6 in SO3 (sulphur trioxide).

Radicals

As we know, a radical is a group of atoms which can combine with another element as a single unit. Each radical takes part as a whole in chemical reactions and has its own valency.

Some examples of radicals are the sulphate radical SO42−, the carbonate radical CO32−, the nitrate radicalNO3−, the phosphate radical PO43− and the ammonium radical NH4+.

(a) In H2SO4(sulphuric acid), the radical SO4 combines with 2 atoms of hydrogen to exhibit a valency of 2.

(b) In H2CO3 (carbonic acid), the radical CO3 combines with 2 atoms of hydrogen to exhibit a valency of2.

(c) In HNO3 (nitric acid), the radical NO3 combines with 1 atom of hydrogen to exhibit a valency of 1.

(d) In H3PO4, (phosphoric acid) the radical PO4 combines with 3 atoms of hydrogen to exhibit a valency of3.

(e) In NH4Cl (ammonium chloride), the radical NH4 combines with 1 atom of chlorine to exhibit a valency of 1.

Easy formula by valency interchange

Notice how while writing the chemical formula for a compound, the subscripts of the constituent elements/radicals are obtained by interchanging the values of valencies (a subscript of 1 is not written by convention), and if there happens to be a common factor between the subscripts, it is factored out. The following exemplifies this "interchanging" of valencies.

(a) Na (valency 1) + Cl (valency 1) gives NaCl (interchange, but 1 need not be written).
(b) Al (valency 3) + O (valency 2) gives Al2O3 (interchange).
(c) Mg (valency 2) + O (valency 2) gives MgO (interchange; common factor 2 cancels out).
(d) Ba (valency 2) + CO3 (valency 2) gives BaCO3.
(e) Fe (valency 3) + SO4 (valency 2) gives Fe2(SO4)3.

Valencies of common elements & radicals

Table 1 gives a list of valencies of some common elements and radicals.

Table 1: Some commonly used valenciesElementsRadicalsValency 1: MonovalentbromineBracetateCH3COO−*chlorineClammoniumNH4+cuprous or copper(I)Cubicarbonate(hydrogencarbonate)HCO3−fluorineFbisulphate (hydrogensulphate)HSO4−hydrogenHbisulphite (hydrogensulphite)HSO3−iodineIchlorateClO3−mercurous or mercury(I)Hg2chloriteClO2−potassiumKcyanideCN−silverAghydroxide (hydroxyl)OH−sodiumNahypochloriteClO−nitrateNO3−nitriteNO2−perchlorateClO4−permanganateMn4−* CH3COO− is another way of writing C2H3CO2−, as it is easier to remember. This radical falls under organic chemistry, where complex formulae are often written in ways easy to memorise.Valency 2: DivalentbariumBacarbonateCO32−cadmiumCdchromateCrO42−calciumCadichromateCr2O72−cobaltCoperoxideO22−cupric or copper(II)CusilicateSi4O32−ferrous or iron(II)FesulphateSO42−leadPbsulphiteSO32−magnesiumMgmanganous or manganese(II)Mnmercuric or mercury(II)HgoxygenOplumbous or lead(II)Pbstannous or tin(II)SnsulphurSzincZnValency 3: TrivalentaluminiumAlferricyanideFe(CN)63−auric or gold(III)AuphosphatePO43−chromic or chromium(III)CrphosphitePO33−ferric or iron(III)FenitrogenNphosphorus(III)PValency 4: TetravalentcarbonCferrocyanideFe(CN)64−plumbic or lead(IV)PbsiliconSistannic or tin(IV)SnValency 5: PentavalentnitrogenNphosphoric or phosphorus(V)P

Rules for writing chemical formula

There are some basic rules that should be known for writing a chemical formula.

Ionic compounds

In general, the entity forming the cation is written first, and then the entity forming the anion. Once the entities are positioned correctly, the subscripts are obtained by interchanging their valencies.

For the case of an ionic compound consisting of a metal and non-metal, the metal is written first (since it forms cations), and then the non-metal (as it forms anions). Thus, we have the formula for common salt as NaCl, not ClNa.

When polyatomic ions, or radicals, are involved, they are treated the same way but as a single unit. The ammonium(NH4+) cation is one of the few polyatomic cations. Most other polyatomic ions are anions.

Binary covalent compounds

Except for binary compoynd of hydrogen, the formula is written with the first element being the one which is either farther to the left, or lower in the periodic table, unless that element isoxygen or fluorine. Oxygen is always named last, except in its compounds with fluorine. Also, remember thatthe position towards the left has a higher priority than a position lower in the periodic table.

Once the elements are positioned correctly, the subscripts are obtained by interchanging their valencies.

(a) For a compound of sulphur and chlorine, sulphur will be written first because it is to the left of chlorinein the periodic table.

(b) For a compound of sulphur and iodine, sulphur is written first even though it is lower in the periodic table, since it is to the left of iodine, and the left position has a higher priority than the lower position.

(c) For a compound of oxygen and chlorine, even though oxygen lies to the left of chlorine, chlorine is written first since oxygen is an exception.

Hydrogen-based covalent compounds

Binary compounds of hydrogen that are not acids have the hydrogen written last such as in NH3 (ammonia), except for the case of H2O (water) and H2O2 (hydrogen peroxide).

Acids are a special type of hydrogen containing compounds (some of which are binary) that will be covered later. In the formula for acids, hydrogen is written first, such as in the binary acid HCl (hydrochloric acid).