MOLECULAR ORBITAL THEORY SIMPLIFIED

TOTAL NUMBER OF ELECTRONS:

10 11 12 13 14 15 16 17 18

1 1.5 2 2.5 3 2.5 2 1.5 1 <--- BOND ORDER

D P P P D P P P D

D ----> Dimagnetic

P------> Paramagnetic

the bond order calculated by the formula will always give you the correct answer.

but as for the magnetic nature...90% will be correct....

their are some exceptions like B2

both are paramagnetic

Inorganic 4 ALL ORES from METALLURGY(SIMPLIFIED)

SOME IMPORTANT METALS AND THEIR ORES:

1. SODIUM (Na):

Sodium Chloride NaCl (rock salt, table salt, common salt)

Sodium Carbonate Na₂CO₃ (Soda Ash)

Na₂CO₃.10H₂O(Washing Soda)

Na₂CO₃.H₂O(Crystal Carbonate)

Sodium Nitrate NaNO₃(chlie salt petre or chile nitre)

Borax Na₂B₄O₇.10H₂O(Tincal)

Sodium Sulphate Na₂SO₄.10H₂O(Glauber's Salt)

Cryolite Na₃AlF₆

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2.Potassium:

Potassium Chloride KCl(Sylvine)

Potassium Carbonate K₂CO₃(Pearl Ash)

Potassium Nitrate KNO₃(Indian Salt Petre,nitre or salt petre)

Carnallite KCl.MgCl₂.6H₂O

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3. Copper:

Cuprite Cu₂O (Ruby Copper)

Copper glance Cu₂S (Chalcocite)

Chalcopyrite CuFeS₂ or Cu₂S.Fe₂S₃(Copper pyrites)

Malachite CuCO₃.Cu(OH)₂

Azurite 2CuCO₃.Cu(OH)₂

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4.SIVER:

Native Silver Ag

Argentite Ag₂S

Horn silver AgCl (Kerargyrite)

Pyragyrite Ag₂S.Sb₂S₃(Ruby Silver)

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5. GOLD:

Native Gold Au

Sylvanite A telluride of Gold and Silver

Bismuth Auride AuBi

Calaverite AuTe₂

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6.MAGNESIUM:

Magnesite MgCO₃

Dolomite MgCO₃.CaCO₃

Carnallite KCl.MgCl₂.6H₂O

Epsom salt MgSO₄.7H₂O

Brucite Mg(OH)₂

Kieserite MgSO₄.H₂O

Schonite K₂SO₄.MgSO₄.6H₂O

Asbestos CaSiO₃.3MgSiO₃

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7.CALCIUM:

Limestone CaCO₃(iceland spar, marble, chalk, calcite)

Gypsom CaSO₄.2H₂O

Anhydrite CaSO₄

Flouride CaF₂(Flourspar)

Phosphite Ca₃(PO₄)₂

Chlorapatite 3Ca₃(PO₄)₂ .CaCl₂

Fluor apatite 3Ca₃(PO₄)₂ .CaF₂

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8. STRONTIUM:

Strontianite SrCO₃

Celestine SrSO₄

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9.BARIUM:

Witherite BaCO₃

Heavy spar BaSO₄

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10.ZINC:

Zincite ZnO

Calamine ZnCO₃

Zinc Blende ZnS (Black Jack)

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11. CADMIUM:

Greenokite CdS

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12. MERCURY:

Cinnabar HgS

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13. ALUMINIUM:

Corundum Al₂O₃

Diaspore Al₂O_3.H₂O

Bauxite Al₂O_3.2H₂O

Alunite K₂SO₄.Al₂(SO₄)₃.4Al(OH)₃ (Alum stone)

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14. TIN:

Cassiterite SnO₂

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15. LEAD:

Galena PbS

Cerrusite PbCO₃

Matlockite PbCl₂

Anglesite PbSO₄

Lanarkite PbO.PbSO₄

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16. IRON:

Haemitite Fe₂O₃

Load Stone Fe₃O₄

Limonite 2Fe₂O₃.3H₂O

Siderite FeCO₃(Spathic iron ore)

Iron Pyrites FeS₂

Chalcopyrite CuFeS₂

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17.MANGANESE:

Pyrolusite MnO₂

Braunite Mn₂O₃

Hausmannite Mn₃O₄

________________________________________________

18. TITANIUM:

Rutile TiO₂

Illmenite FeTiO₃

important things usually left in inorganic.

SECONDARY FORCES:

These forces decide the physical properties of the molecule like boiling point, meltiing point, refractive index, viscoscity and solubility

1. HYDROGEN BOND:

it is a type of dipole dipole interaction...

it is only shown by N, O , F

to learn it...

hydrogen ne FON (phone) lagaya to H-bonding hui.

NOTE: boiling point of water is greater than HF and NH3 exceptionally.

Q. Enthalpy of vapourization of H2O is greater than HF. WHY?

A. water evaporates as a monomer so we have to break all the H bonds but HF evaporates in the form of a dimer so we have to break less bonds.

(draw the diagram...it'll be clear to you at once)

2.INTRA MOLECULAR HYDROGEN BOND:

It is the hydrogen bond formed within the molecule:

example o-nitrophenol

meta and para isomers don't form intermolecular H-bond.

APPLICATIONS:

1. Boiling Point inter molecular H-bond

1/ intramolecular H bond Solubility

3. VISCOSCITY No. of H bonds Molecular weight

4. ACIDIC NATURE less H-bonds

Q.Paranitrophenol is more acidic than orthonitrophenol?

A. in orthonitrophenol due to intramolecular H-bonding a strong cyclic ring caleed as chelate is formd which provides extra stability to the molecule.

To phir khud hi soch lo wo apna H kyun dega.

kyunki agar usne H de diya to chelate ring toot jayegi.

1st disassociation of maleic acid is grtr dn fumaric acid because it acquires stability due to the chelate formation but the 2nd disassociation of maleic acid is less dn the fumaric acid

ALCOHOLS ARE SOLUBLE IN WATER DUE TO H-BOND.

small important things in inorganic

APPLICATION OF DIPOLE DIPOLE FORCE:

diple- dipole boiling point

Molecule with 0 dipole moment shows minimum boiling point and with maximum boiling point shows maximum boiling point.

ex: Cis isomers have high boiling point than trans.

MELTING POINT SYMMETRY PACKING 1/size

ex: s-block have low Melting point while d-block have high MP

Trans isomers have higher melting point than cis due 2 symmetry

Para isomers have high MP due 2 symmetry.

WHY ICE FLOATS ON WATER?

Ice has tetrahedral arrangement of water in which one H2O moleule is surrounded by 4 other H2O molecules and leave large vacant spaces due to which its volume increases and density decreases.

WHAT ARE ELATHRATES OF CAGE COMPOUNDS?

when noble gases like Ar, Kr, Xe are passed into H2O and suddenly freezed then dese gases get trapped in the vacant spaces of ice. such compounds are called cage-compounds.

ION DIPOLE FORCE:

When ionic compound or polar covalent compound is dissolved in water or polar solvents then

1. cation is surrounded by negative part of the solvent by ion dipole force for Group 1 and Group 2 cations. Coordination No is 6 but 4 Li and Be it is 4.

due to absence of d-orbital

NaCl when hydrated gives [Na(H2O)6]

Anion is surrounded by positive part of the solvent by ion dipole force

Which is:

a)H(delta +ve) if solvent is polar protic (which can donate H) and here the ion dipole force is called as H bonding

b) other dn H if the solvent is polar protic solvent and the ion dipole force is not the H bond

CONCLUSION: solvation energy is more in case of polar protic solvents hence the stability for the formation of cation or anion is more.

Q. Why does Li doesn't form alum?

A. for the formation of alum the required coordination number for the cation is six but Li has coordination number 4.

TO CHECK SOLUBILITY:

1.HYDRATION ENERGY

2. ABILITY TO GET POLARIZED

3.PHYSICAL STATE

find the solubility order of iodine, bromine and Chlorine:

answer: Cl2I2

EXPLANATION: I2 is polarized to a greater extent but due to its large size and its solid state it is less soluble.

Br2 secondly polarizes to a greater extent ( less dn I) and due to itz liquid nature it is more soluble in water.

INSTANTANEOUS DIPOLE INDUCED DIPOLE FORCES:

In non polar gaseous molecules due to collison or friction momentarily dipole is produced which in turn produced the dipole produce the dipole in the neighbouring moleule.

Ex: all noble gases

this force only is called as VANDER WAAL FORCE.

order of Vander waal force:

He

more the vander waal force more easy is the liquification.

Boiling Point: NH3 > PH3

H2O > H2S

HF > HCl <>

such order is due to H-bonding

and the boiling point increases downward due to increase in vanderwaal force of attraction.

Vander Waal force exists in all states of matter.

Liquids/solids in which molecules are held together by weak V.W. forces are volatile/sublime

VOLATILE LIQUIDS ACT AS GOOD FUELS.

INORGANIC 5 HYBRIDIZATION and GEOMETRY SIMPLIFIED

PS:THIS FORMULA WILL ALWAYS GIVE YOU THE CORRECT ANSWER

Just get through wid all this u'll find inorganic petty easy....

SHORTEST FORMULAE FOR HYBRIDIZATION:

No of valence electrons + No. of atoms attached to central atom

2

CONDITIONS:

1. Don't count the multiple bonded attached atoms.

2. Don't coun't odd or unpaired electrons.

after putting the values in the formula if ya get:

2 then compound is sp hybridized

3 then compound is sp2 hybridized

4 then compound is sp3/dsp2 hybridized

5 then compound is sp3d/dsp3 hybridized

6 then compound is sp3d2/d2sp3 hybridized

TO DIFFERENTIATE b/w sp3 and dsp2 hybridization and so on.

1. Easiest way is to check whether d-orbital present or not.

2. If the compound undergoes reaction with strong ligand than dsp2, dsp3,d2sp3 hybridization occurs.

3.If the compound undergoes reaction with weak ligand than sp3, sp3d, sp3d2 hybridization occurs.

TO IDENTIFY STRONG LIGAND AND WEAK LIGAND:

ALL NITROGEN CONTAINING except nitrate ion are strong ligands.

CO is exceptionally strong ligand...

ALL STRONG LIGANDS CAUSE PAIRING OF UNPAIRED ELECTRON IN D-ORBITAL.

GEOMETRIES:

REQUIRED (R): the total no. of atoms which are necessary for perfect geometry of that hybridization or number of hybrid orbitals.

AVAILABLE(A): Atoms attached.

R-A=no. lone pair

1.sp hybridization:

R=2, A=2

R-A=0

linear geometry, 0 dipole moment.

R=2, A=1

R-A=1 lone pair

so dipole moment is not 0.

2.sp2 hybridization:

a. R=3

-A=3

0 = no. of lone pairs.

Geometry: trigonal planar and dipole moment =0

b. R=3

A=2

lone pair = 1

Geometry: V shape/angular

c. R=3

A=1

lone pair = 2

Geometry: Linear

3. sp3

a. R=4

A=4

lone pair=0

Geometry: Tetrahedral

b. R=4

A=3

lone pair = 1

Geometry: pyramidal

c. R=4

A=2

lone pair = 2

Geometry: V-shape

d. R=4

A=1

lone pair = 3

Geometry: Linear

4.sp3d

a. R=5

A=5

lone pair = 0

Geometry: Trigonal Bipyramidal

b. R=5

A=4

lone pair = 1

Geometry: see-saw

c. R=5

A=3

lone pair = 2

Geometry: T-shape

d. R=5

A=2

lone pair = 3

Geometry: Linear

LONE PAIR CAUSE MAXIMUM repulsion so they will be placed at equitorial postion (larger bond angle)

5. sp3d2:

a. R=6

A=6

lone pair = 0

Geometry: square bipyramidal

b. R=6

A=5

lone pair = 1

Geometry:square pyramidal

c. R=6

A=4

lone pair = 2

Geometry : square planar.

PRACTICE PROBLEMS:

1. BeCl2 = 2+2 = 2 =sp Linear

2

2. CO2 = 4/2 =2 =sp Linear

3.XeOF2 = 8/2 + 2/2 = sp3d T-shape

4. XeF6 = 8/2 + 6/2 = Sp3d3 Distorted octahedral or caped octahedrl

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, 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.
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] 3d14s2 Ti [Ar] 3d24s2 V [Ar] 3d34s2 Cr [Ar] 3d54s1 Mn [Ar] 3d54s2 Fe [Ar] 3d64s2 Co [Ar] 3d74s2 Ni [Ar] 3d84s2 Cu [Ar] 3d104s1 Zn [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.
On the basis of this definition, 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 You have already come across the fact that when the Periodic Table is being built, the 4s orbital is filled before the 3d orbitals. This is because in the empty atom, 4s orbitals have a lower energy than 3d orbitals. However, once the electrons are actually in their orbitals, the energy order changes - and in all the chemistry of the transition elements, the 4s orbital behaves as the outermost, highest energy orbital. The reversed order of the 3d and 4s orbitals only applies to building the atom up in the first place. In all other respects, you treat the 4s electrons as being the outer electrons.
	Note: This is another of those things that you just have to accept. The explanation again lies well beyond the level you are working at. Just remember that once you have the full electronic structure for one of these atoms, the 4s electrons are the outermost electrons.
Remember this: When d-block elements form ions, the 4s electrons are lost first. To write the electronic structure for Co2+: Co [Ar] 3d74s2 Co2+ [Ar] 3d7 The 2+ ion is formed by the loss of the two 4s electrons. To write the electronic structure for V3+: V [Ar] 3d34s2 V3+ [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.
	Note: If you aren't sure about oxidation states, you really need to follow this link before you go on. Use the BACK button on your browser to return quickly to this page.
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: +2 in Mn2+ +3 in Mn2O3 +4 in MnO2 +6 in MnO42- +7 in 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] 3d64s2 Fe2+ [Ar] 3d6 Fe3+ [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: metal 1st IE 2nd IE 3rd IE Ca 590 1150 4940 Fe 762 1560 2960 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 only transition metals do. Transition metals do, however, form a very wide range of complex ions.
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.
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 expored 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. Nickel in the hydrogenation of C=C bonds This reaction is at the heart of the manufacture of margarine from vegetable oils. However, the simplest example is the reaction between ethene and hydrogen in the presence of a nickel catalyst. Transition metal compounds as catalysts Vanadium(V) oxide in the Contact Process At the heart of the Contact Process is a reaction which converts sulphur dioxide into sulphur trioxide. Sulphur dioxide gas is passed together with air (as a source of oxygen) over a solid vanadium(V) oxide catalyst. Iron ions in the reaction between persulphate ions and iodide ions Persulphate ions (peroxodisulphate ions), S2O82-, are very powerful oxidising agents. Iodide ions are very easily oxidised to iodine. And yet the reaction between them in solution in water is very slow. The reaction is catalysed by the presence of either iron(II) or iron(III) ions.

Short note on halogens

The halogens or halogen elements are a series of nonmetal elements from Group 17 (old-style: VII or VIIA; Group 7 IUPAC Style) of the periodic table, comprising fluorine, F, chlorine, Cl, bromine, Br, iodine, I, astatine, At.

The group of halogens is the only group which contains elements in all three familiar states of matter at standard temperature and pressure.

Abundance

Owing to their high reactivity, the halogens are found in the environment only in compounds or as ions. Halide ions and oxoanions such as IO₃^? can be found in many minerals and in seawater. Halogenated organic compounds can also be found as natural products in living organisms. In their elemental forms, the halogens exist as diatomic molecules, but these only have a fleeting existence in nature and are much more common in the laboratory and in industry. At room temperature and pressure, fluorine and chlorine are gases, bromine is a liquid and iodine and astatine are solids; Group 7 is therefore the only periodic table group exhibiting all three states of matter.

Etymology

The term halogen originates from 18th century scientific French nomenclature based on erring adaptations of Greek roots; the Greek word halos meaning "salt", and genes meaning "production" ? referring to elements which produce a salt in union with a metal.

Properties

The halogens show a number of trends when moving down the group - for instance, decreasing electronegativity and reactivity, increasing melting and boiling point.

Halogen	Standard Atomic Weight (u)	Melting Point (K)	Boiling Point (K)	Electronegativity (Pauling)
Fluorine	18.998	53.53	85.03	3.98
Chlorine	35.453	171.6	239.11	3.16
Bromine	79.904	265.8	332.0	2.96
Iodine	126.904	386.85	457.4	2.66
Astatine	(210)	575	610 ?	2.2
Ununseptium	(291)*	*	*	*

* Ununseptium has not yet been discovered; values are either unknown if no value appears, or are estimates based on other similar chemicals.

Diatomic halogen molecules

halogen	molecule	structure	model	d(X?X) / pm (gas phase)	d(X?X) / pm (solid phase)
fluorine	F2			143	149
chlorine	Cl2			199	198
bromine	Br2			228	227
iodine	I2			266	272

Chemistry

Reactivity

Halogens are highly reactive, and as such can be harmful or lethal to biological organisms in sufficient quantities. Fluorine is the most reactive element in existence, attacking such inert materials as glass, and forming compounds with the heavier noble gases. It is a corrosive and highly toxic gas. The reactivity of fluorine is such that, if used or stored in laboratory glassware, it can react with glass in the presence of small amounts of water to form SiF₄. Thus flourine must be handled with substances such as Teflon, extremely dry glass, or metals such as copper or steel which form a protective layer of fluoride on their surface.

Both Chlorine and bromine are used as disinfectants for drinking water, swimming pools, fresh wounds, dishes, and surfaces. They kill bacteria and other potentially harmful microorganisms through a process known as sterilization. Their reactivity is also put to use in bleaching. Chlorine is the active ingredient of most fabric bleaches and is used in the production of most paper products.

Hydrogen halides

The halogens all form binary compounds with hydrogen, the hydrogen halides, HX (HF, HCl, HBr, HI), a series of particularly strong acids. When in aqueous solution, the hydrogen halides are known as hydrohalic acids. HAt, or "hydrastatic acid", should also qualify, but it is not typically included in discussions of hydrohalic acid due to astatine's extreme instability toward alpha decay.

Interhalogen compounds

The halogens react with each other to form interhalogen compounds. Diatomic interhalogen compounds (e.g. BrF, ICl, ClF) bear resemblance to the pure halogens in some respects. The properties and behaviour of a diatomic interhalogen compound tend to be intermediate between those of its parent halogens. Some properties, however, are found in neither parent halogen ? Cl₂ and I₂ are soluble in CCl₄ but ICl is not, since it is a polar molecule due to the relatively large electronegativity difference between I and Cl.

Organohalogen compounds

Many synthetic organic compounds such as plastic polymers, and a few natural ones, contain halogen atoms; these are known as halogenated compounds or organic halides. Chlorine is by far the most abundant of the halogens, and the only one needed in relatively large amounts (as chloride ions) by humans. For example, chloride ions play a key role in brain function by mediating the action of the inhibitory transmitter GABA and are also used by the body to produce stomach acid. Iodine is needed in trace amounts for the production of thyroid hormones such as thyroxine. On the other hand, neither fluorine nor bromine are believed to be essential for humans, although small amounts of fluoride can make tooth