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# Complex chemistry

The complex chemistry ( lat. Complexum , "surround", "hugging", "clutching") is a field of inorganic chemistry . The term coordination chemistry is generally used synonymously. Complex chemistry deals with complexes or coordination compounds that are built up from one or more central particles and one or more ligands . The central particles are mostly atoms or ions of transition metals , which can be charged or uncharged.

In contrast to conventional covalent bonds, in complexes the ligands usually contribute all electrons to the bond, instead of each reactant contributing one electron to an electron pair bond; nevertheless there are also complexes of a more covalent nature. The formation of complexes can thus be understood as an acid-base reaction in the sense of the Lewis definition , in which the ligands (bases) act as electron pair donors and the central particle (acid) as an acceptor due to gaps in its electronic configuration.

Complex compounds play an important role in various areas: in technology (for example as a catalyst, see illustration of the Grubbs catalyst on the right), in biology ( hemoglobin and chlorophyll ), in medicine ( chelation therapy ) or in analytical chemistry . Compounds with organic ligands are also the subject of organometallic chemistry . Since complexes of transition metals are sometimes very colored, these are also used as dyes or pigments ( Berlin blue ). The charge-transfer complexes show particularly intense colors , such as B. Permanganates .

For a long time chemists had no idea of the structure of coordinative compounds known as “higher order compounds”. In addition, many of the behavior of complexes could not be explained with the theories of the time, such as the stability of cobalt (III) chloride in aqueous solution when ammonia is added. For the correct interpretation of the structure and bonding relationships in complexes, the Swiss Alfred Werner received the Nobel Prize for Chemistry in 1913 as the first and for decades the only inorganic chemist .

## history

Example of a typical complex: The hexacyanidoferrate (II) ion. The fact that cyanide is much less toxic than usual in this compound shows the special nature of the bond

Systematic research into the structure of complex compounds began in the late 19th century and is mainly characterized by the names Sophus Mads Jørgensen and Alfred Werner . Jørgensen made a name for himself with the synthesis of numerous new complexes, but was a supporter of the "chain theory" introduced by Christian W. Blomstrand . According to this theory, the ligands are lined up one behind the other and thus form chains, which makes little sense nowadays, but at least takes into account the value of the metal.

Werner, who was considered Jørgensen's opponent for decades, however, formulated a theory in 1892 that is basically still valid today and was first published the following year. The statements that can be derived from this, for example about possible stereoisomerism, could be confirmed experimentally and solidified the theory over the years. A few years after being awarded the Nobel Prize in Chemistry, Werner died in 1919.

In the decades that followed, the development of new theories about chemical bonding and new technologies such as X-ray structure analysis led to great advances in coordination chemistry. In addition, since then there have been other Nobel prizes in chemistry in the field of complex chemistry, for example in 1973 for the research of so-called sandwich complexes by Ernst Otto Fischer and Geoffrey Wilkinson , or in 2005 for the description of olefin metathesis and the development of suitable (complex) catalysts by Yves Chauvin , Robert Grubbs and Richard R. Schrock .

## nomenclature

### Complex formulas

The oxidation number is determined by looking at the original charge of the central atom, as if all ligands were removed while taking along the shared electron pairs. The sum of the charge contributions of the ligands and the oxidation number of the central particle (s) must give the charge of the complex.

1. The coordination unit is shown in square brackets. Any charge that may be present is written as an exponent behind the square brackets. In this case the coordination unit is also referred to as a complex ion.
2. The central particle is named before the ligand.
3. Ligands are listed alphabetically (as opposed to previous recommendations to avoid problems with so-called non-innocent ligands ) regardless of their charge (abbreviations for ligands are also given in alphabetical order, i.e. CH 3 CN before MeCN before NCMe).
4. Round brackets are used for polyatomic ligands (as well as abbreviations for ligands). Furthermore, the donor atom should be mentioned first if possible (for example, NCS and SCN are bond isomers ).
5. The oxidation number can be specified as an exponent (Roman numeral) on the central atom (the positive sign is neglected). This information is optional.
6. In polynuclear complexes, the type of binding of bridging ligands is denoted by the Greek letter μ . An index n indicates the number of linked central atoms. In the case of ligands with several donor atoms, the hapticity is indicated with the Greek letter η. Behind it is the number of donor atoms.

### Complex names

In the systematic naming of complex salts, the cation is given first and then the anion , regardless of whether they are complex or not. The components of a coordination unit are named in the following order:

1. Number of ligands : Multiple occurrences of ligands are indicated by using Greek numerals ( mono , di etc.) as a prefix. For ligands with more complex names or to avoid ambiguity (e.g. dithiosulphate ), their multiplicatives ( bis , tris etc.) are used. The part multiplied by this comes in brackets.
2. Type of ligands : The various ligands are listed in alphabetical order, regardless of their number or charge. Anionic ligands are given the ending “o” in their anion names (e.g. chlorido ) and radicals have the ending “-yl” (e.g. nitrosyl ). The names of neutral or cationic ligands are not changed. Exceptions to this rule are the names of water ( aqua ), ammonia ( ammin ), CO ( carbonyl ) and NO ( nitrosyl ).
3. Coordination unit: In the case of an anionic coordination unit, the central ion is given the Latin notation followed by the ending “at” (e.g. cuprate from Latin cuprum ). If the coordination unit is a cation or neutral, the unchanged German name is used.
4. Charge of the central ion: The oxidation number of the central ion (atom) is indicated by a Roman number in round brackets and placed after the name of the coordination unit. A plus sign is not written, the Arabic number is used for zero.

The full name of the coordination unit is written in one word. Except for the names of the ligands aqua , ammin and nitrosyl , the names of all neutral ligands are put in brackets. The names of inorganic anionic ligands are put in round brackets if they already contain numerical prefixes or if this avoids ambiguity. In the name of complex salts, a hyphen is written between the names of the cation and the anion.

## Building complexes

### Central particles

Mainly transition metals come into consideration as central particles , which have correspondingly free d orbitals with which the ligands can connect. Complexes are just as common with lanthanides and actinides . The number of valence electrons has a decisive influence on the type, stability and structure of the complexes that are formed. Explanations for this can be found in corresponding theories, which will be discussed later. Examples of central particles are:

Normally the metal particles in a complex have a positive oxidation number , but compounds with metal atoms in the oxidation state zero can also be produced by the reaction of metals or metal vapors with the corresponding ligands. An example is the reaction of nickel with carbon monoxide to form tetracarbonyl nickel ( Mond process ) or of iron to form pentacarbonyl iron . Such reactions can also be used to purify the corresponding metals ( chemical transport ). By reacting metals with ligands from the gas phase, for example, carbonyl , phosphine , olefin , aromatic and cyclopentadienyl complexes can be produced, although the formation of the latter is associated with a redox reaction.

### Ligands

Ethylenediamine as a ligand to M

Complexes can consist of the same or different ligands. If a complex contains only ligands of the same type, it is called homoleptic , otherwise heteroleptic . Since in complexes the number of partners in a bond with the central particle is often independent of the oxidation state of the central particle, there is also the term “higher order compounds” from the 19th century for complexes.

Ligands can bring different numbers of electrons into the bond, for example Cl - and PPh 3 bring two electrons, η 5 -Cp and η 6 -C 6 H 6 six electrons, unbridged μ 1 -CO (see bridge (chemistry) ) two electrons, bridged μ 2 -CO one electron (see hapticity ). Ligands can not only bring in different numbers of electrons, but also coordinate with several points at the same central particle if they are of the appropriate size. The number of bonds possible here is called tenacity . The common simple ligands such as aqua (H 2 O), ammine ( NH 3 ), chlorido (Cl - ) or cyanido (CN - ) are all monodentate or monodentate , and bind e.g. E.g .: H 3 N-M.

Ligands that have several coordination sites for the same metal particle are called chelating ligands (Greek chelé , "crab claw "). The chelate complexes formed have a higher stability both thermodynamically and kinetically. The high thermodynamic stability is based on the increase in the entropy of the system, since the following reaction takes place in aqueous solution to form an octahedral complex with a bidentate ligand (ligand with two coordination sites), for example:

${\ displaystyle \ mathrm {[M (H_ {2} O) _ {6}] + 3 \ X \ longrightarrow [MX_ {3}] + 6 \ H_ {2} O}}$

Here four free particles (on the left) become seven free particles (on the right). The kinetic stability is based on the fact that fewer particles have to meet in order to form the complex (according to the kinetic gas theory ) and all bonds of a ligand to the central particle have to be opened simultaneously during dissociation.

Common chelating ligands include bidentate ethylenediamine (s), tetradentate nitrilotriacetic acid (NTA) and hexadentate ethylenediaminetetraacetate (EDTA). The latter is used medically, for example, to bind toxic heavy metals such as lead or mercury in the body and then excrete them ( chelation therapy ).

#### Anionic ligands

Anionic ligands are formed from the name of the anion by adding the ending - o : -id becomes -ido , -it becomes -ito and -at becomes -ato . Some traditional names (such as chloro, cyano or oxo) were also allowed until the revision of the IUPAC nomenclature in 2005. As a result of the revision, these exceptions have now been eliminated and are no longer permitted.

## Spatial shape of complexes (molecular geometry)

### Coordination number and coordination polyhedron

Coordination number = 4:
tetrahedral complex
Coordination number = 6:
octahedral complex

The coordination number (KZ) indicates with how many donor atoms of the ligand a central particle surrounds itself. Depending on the coordination number, the ligands arrange themselves in certain arrangements around the center, which often, but not always, agree with the predictions of the VSEPR model . If you think of the lines connecting the ligands, you get the coordination polyhedra, which are usually used to describe the structure of complexes. Coordination numbers from 2 to 9 are common, and numbers beyond this can only be achieved with particularly large central particles and chelating ligands. The most common, however, are the coordination numbers 4 and 6. According to the VSEPR model, the following polyhedra can be assumed:

• KZ 2 : a linear complex, e.g. B. [AuCl 2 ] - or angled complex z. B. [Au (tBuXanthPhos)] [AuBr 2 ]
• KZ 3 : a trigonal-planar or a trigonal-aplanar structure (the central particle is not exactly in the middle of the triangle, but slightly above)
• KZ 4 : a tetrahedron or a square-planar structure
• KZ 5 : a square-pyramidal or trigonal-bipyramidal structure; Both can be converted into one another by the Berry pseudorotation and are in equilibrium at the appropriate temperature
• KZ 6 : mostly an octahedron , sometimes a trigonal antiprism or (more rarely) a trigonal prism
• KZ 7 (very rare): a pentagonal bipyramid, a simply capped octahedron or a simply capped trigonal prism
• KZ 8 : a square antiprism, a trigonal dodecahedron , a doubly capped trigonal prism or, more rarely, a hex vein ( cube )
• KZ 9 : a triple capped trigonal prism, e.g. B. [ReH 9 ] 2− or capped square antiprism z. B. [Ln (thf) [N (CH 2 CH 2 OH) 3 ]] 3+
• KZ 12 : icosahedron or a cuboctahedron gives z. B. [Ce (NO 3 ) 6 ] 2- or [Zr (η 3 −BH 4 ) 4 ]

### symmetry

Since the central particles and ligands of a stable complex, just like the ions, adopt geometrically ordered structures within crystal lattices , they are assigned to specific point groups . The labeling is usually based on the Schoenflies symbolism .

## Isomerism

If there are several distinguishable compounds for one and the same empirical formula, one speaks (as with organic compounds) of isomerism . For complex compounds, the geometric isomerism , optical isomerism , ionization isomerism and bond isomerism are relevant.

### Binding isomerism

Bond isomerism can occur when a ligand with different atoms can coordinate to the metal center. For example, the ligand SCN - can bond with the sulfur atom (thiocyanato) as well as with the nitrogen atom (isothiocyanato).

### Geometric isomerism

Depending on the coordination number and composition of a complex compound, cis - trans isomerism or facial or meridional arrangement occurs.

#### cis - trans isomerism

cis - trans isomerism on the complex

The cis - trans isomerism occurs in square planar (coordination number KZ = 4) and octahedral (KZ = 6) complexes. The figure on the right shows the cis and trans isomers of an octahedrally coordinated compound of the general formula MA 4 B L , where L can represent both a further ligand of type B and a ligand of the third type C. For the designation, it is only important whether the two “special” ligands (twice B or once B and once C) are arranged towards one another ( cis ) or away from one another ( trans ). Accordingly, the four ligands of type A form a rocker ( cis ) or a square plane ( trans ).

If you mentally remove the two A ligands that are not on the equatorial plane in the sketch, you get a square planar complex with the general formula MA 2 BL. As with the octahedral complex, it is important whether B and L point towards each other ( cis ) or away from each other ( trans ). As soon as there are four different types of ligands in the square planar complex, the number of possible isomers increases to three.

#### fac and mer arrangement

The facial (fac) or the meridional (mer) arrangement occurs for octahedral complexes of the general formula MA 3 B 3 . Can be the ligands of the type A and clearly separate from type B by a plane containing the central particles M, each have three ligands of the fac-assembly in one direction, such as two opposite faces (engl. Faces ). If the ligands of the two types cannot be separated from each other, but can only be divided into orthogonal planes of the pure type A or B, one speaks of the mer arrangement , since there must always be three similar ligands on the meridian of the sphere, its surface includes all six ligands.

### Optical isomerism

The prerequisite for optical isomerism is chirality , whereby a metal complex can also have a chiral metal center without a ligand molecule itself being chiral. Depending on the type of coordination, there are various requirements that must be met for optical isomerism to occur:

• tetrahedral (KZ = 4) metal complexes are chiral if four different ligands are bound to the metal center.
• Square planar (KZ = 4) metal complexes are chiral as soon as sterically demanding ligands interfere with one another and thus prevent rotation around a ligand-metal bond.
• Octahedral (KZ = 6) metal complexes often have chiral centers if they contain chelating ligands .

## Formation of complexes

The classic complex formation reaction is an acid-base reaction according to the theory of Gilbert Newton Lewis . Here the central particle represents the Lewis acid (electron pair acceptor); the ligand is the Lewis base, i.e. a molecule or an ion which can provide at least one free electron pair (electron pair donor) to form a bond.

An example of a typical complex formation is the addition of water to cupric sulfate . The colorless salt reacts with the water to form a blue complex:

${\ displaystyle \ mathrm {CuSO_ {4} +6 \ H_ {2} O \ longrightarrow [Cu (H_ {2} O) _ {6}] ^ {2 +} + SO_ {4} ^ {2-}} }$

The Cu 2+ reacts as a Lewis acid and the water with its free electron pairs reacts as a Lewis base and a hexaqua complex is formed. This reaction is often used as evidence of water in chemistry classes at school because of its highly visible effect.

Coordinate bond from ammonia and boron trifluoride

The type of chemical bond that results from the formation reaction is called coordinative bond (also - out of date - dative bond or donor-acceptor bond ), and thus from the other forms of chemical bond ( covalent bond , ionic bond , metal bond ) distinguished. This (controversial) distinction is justified by the fact that the binding pair of electrons originally mostly comes from the ligand and not (as in a covalent bond) one electron from each binding partner. In old textbooks, the bond is sometimes still indicated by an arrow in the direction of the acceptor, but these representations are out of date. Nowadays, a coordinative bond is drawn as a line in analogy to a covalent bond (see, for example, the sketch on the right), because although a complex is generally to be regarded as a Lewis acid-base adduct , the coordination can, as in In the area of homogeneous catalysis , also by oxidative addition , with part of the binding electrons being contributed by the metallic central atom, which can remain on this again in the course of a reductive elimination.

A typical example of a coordinative bond is the case shown in the adjacent sketch. Ammonia (NH 3 ) has a lone pair of electrons that are available for a coordinative bond to form the H 3 N-BF 3 molecule . Formally, the nitrogen atom transfers an electron to the boron atom, whereby the former (generally: the donor) receives a formal positive, the latter (generally: the acceptor) receives a formal negative charge. Note that these formal charges have nothing to do with the actual charge distribution: Since nitrogen has a much higher electronegativity than boron (3.0 versus 2.0), the bond to nitrogen is polarized (and the oxidation states remain unchanged).

In addition, there are complexes whose bonding relationships can only really adequately be described by more sophisticated concepts (such as molecular orbital theory ), such as metal clusters , sandwich complexes (e.g. bis (benzene) chromium , manganocene and ferrocene ) , but also the related half-sandwich compounds Olefin complexes (such as Zeise's salt ).

## Binding relationships

The shapes of the d orbitals involved in complex formation

The bond between central particles and ligands and the stability of complexes can be described in more detail using different models, which also enable statements to be made about properties such as color or magnetism.

### VB theory

The earliest explanation was provided by the valence bond theory (VB theory). This assumes that occupied ligand orbitals overlap with unoccupied orbitals of the central particle and thus form a bond. To explain the spatial structure of complexes, it is assumed that hybrid orbitals are formed in metal. The VB theory explains the geometry and magnetic properties, but not, for example, the color. The 18-electron rule can also be used to estimate the stability of transition metal complexes in certain cases, although the scope of the rule is very limited.

### Crystal and ligand field theory

The crystal field theory, which is based on pure electrostatic interactions between the ligands and the central particle and can also explain the color of the complexes, is a further development . The model is often mixed with the ligand field theory, which extends the crystal field theory and examines the influence of the point-like ligands on the energies of the d orbitals of the central metal. This approach is still very common today due to its ability to explain many properties despite its simplicity.

Energy level diagram of the d orbitals of a complex center in the octahedral field

The energies of the d orbitals of a metal are initially degenerate (equal in energy). If a spherically symmetrical ligand field approaches, the energy of all d orbitals increases to the same extent. The ligand field theory now considers the spatial shape of these orbitals and the geometry of the ligand field in the form of point charges. If you look at an octahedral complex, for example, you will see that in an octahedral ligand field, due to the geometry, the d z 2 and d x 2 −y 2 orbitals are energetically less favorable, and the orbitals d xy , d yz and d xz in turn cheaper. The sum of these energy differences is called the ligand field stabilization energy (LFSE) and in an octahedral complex it is denoted by Δ O. The absolute value of this splitting can be determined experimentally by spectroscopy and is dependent on both the center and the ligands, but is generally given as 10 Dq. The influence of the centers and ligands on the splitting can be read off from the spectrochemical series .

The energy increase makes up 3/5 of the LFSE and the decrease accordingly 2/5. If all orbitals are occupied, the sum total is 0 Dq, since the energy center of the d orbitals must not change (dashed line in the figure). The energetically increased orbitals are referred to as e g orbitals ( e for twice “degenerate”) and the lowered orbitals as t 2 g orbitals ( t for “triple degenerate”). In a complex, these orbitals are now filled with electrons according to Hund's rule . In contrast to this, the formation of electron pairs instead of being occupied by unpaired electrons is also observed at high field splittings, provided the LFSE is higher than the spin pairing energy . Such complexes are called low-spin complexes, in contrast to the usual high-spin complexes.

Similar considerations can also be made for coordination polyhedra other than the octahedron. Ligand field theory can thus provide simple explanations, among other things, for the magnetic behavior of complexes by considering the pairing of electrons, or for the color, which can be explained with electron transitions between the orbitals. The geometric distortion caused by the Jahn-Teller effect can also be explained here using the electron configuration.

#### Spectrochemical series

The experimentally established spectrochemical series arranges ligands and metal particles according to the strength of the ligand field splitting they cause. The following sequence results for the ligands:

I - <Br - <Cl - <F - <OH - <H 2 O <NC - <NH 3 <CN - <CO

In this series, the iodido ligand causes the smallest splitting and the carbonyl ligand the largest. This list corresponds in principle to the base strength according to the Lewis concept. You can sort the metal particles in a similar way:

Mn 2+ <Ni 2+ <Co 2+ <Fe 2+ <V 2+ <Fe 3+ <Cr 3+ <V 3+ <Co 3+ <Mn 4+ <Mo 3+ <Rh 3+ <Pd 4 + <Ir 3+ <Re 4+ <Pt 4+

From this the rule of thumb becomes clear that higher ion charges also lead to higher splitting. The further to the right a particle is, the more likely it will be a low-spin configuration in a corresponding complex.

### MO theory

However, the best results are provided by the molecular orbital theory, of which the ligand field theory is only a part. It treats both the central particle and the ligands quantum mechanically and is therefore the most precise, but also the most demanding.

## stability

Several considerations can be taken into account in order to estimate or explain the stability of complex compounds.

### HSAB concept

The principle of hard and soft acids and bases according to Ralph G. Pearson states that hard acids react preferentially with hard bases and soft acids correspondingly with soft bases. Particles with a high charge density are called hard and those with a low charge density that are easily polarized are called soft. This concept can also be applied to the stability of complex compounds.

### Tie back and forth

Back and forth binding in carbonyl complexes

The MO theory provides further information about the stability of, for example, metal carbonyls . Accordingly, all ligands in complexes serve as σ donors, but strong ligands such as CO are also strong π acceptors. The σ-binding takes place via the HOMO of the CO, which overlaps with an empty d orbital of the metal. In addition, however, the LUMO of the CO can overlap with an occupied d orbital of the metal of suitable symmetry and thus brings about a π backbonding, which helps the complex to be particularly strong. In addition, this back-connection strengthens the σ-connection, which is why one speaks of a synergy here .

### Application of the law of mass action

Equilibrium constants can be established for the quantitative description of the stability of complexes , since the Lewis acid-base reactions for complex formation are equilibrium reactions for which the law of mass action can be established. The overall reaction can be divided into individual steps (so-called elementary reactions), i. H. each for the addition of a ligand. The product of the equilibrium constants of the individual elementary reactions for complex formation then gives the equilibrium constant for the overall reaction.

{\ displaystyle {\ begin {aligned} \ mathrm {Ni ^ {2 +} + CN ^ {-}} & \ rightleftharpoons \ mathrm {[NiCN] ^ {+}} & \ quad K_ {1} & = \ mathrm {\ frac {[[Ni (CN)] ^ {+}]} {[Ni ^ {2 +}] \ cdot [CN ^ {-}]}} \\ [1em] \ mathrm {[Ni (CN) ] ^ {+} + CN ^ {-}} & \ rightleftharpoons \ mathrm {[Ni (CN) _ {2}]} & \ quad K_ {2} & = \ mathrm {\ frac {[[Ni (CN)) _ {2}]]} {[Ni (CN)] ^ {+} \ cdot [CN ^ {-}]}} \\ [1em] \ mathrm {[Ni (CN) _ {2}] + CN ^ {-}} & \ rightleftharpoons \ mathrm {[Ni (CN) _ {3}] ^ {-}} & \ quad K_ {3} & = \ mathrm {\ frac {[[Ni (CN) _ {3} ] ^ {-}]} {[Ni (CN) _ {2}] \ cdot [CN ^ {-}]}} \\ [1em] \ mathrm {[Ni (CN) _ {3}] ^ {- } + CN ^ {-}} & \ rightleftharpoons \ mathrm {[Ni (CN) _ {4}] ^ {2-}} & \ quad K_ {4} & = \ mathrm {\ frac {[[Ni (CN ) _ {4}] ^ {2 -}]} {[Ni (CN) _ {3} ^ {-}] \ cdot [CN ^ {-}]}} \\ [1em] \ mathrm {Ni ^ { 2 +} + 4CN ^ {-}} & \ rightleftharpoons \ mathrm {[Ni (CN) _ {4}] ^ {2-}} & \ quad K_ {A} & = K_ {1} * K_ {2} * K_ {3} * K_ {4} = \ mathrm {\ frac {[[Ni (CN) _ {4}] ^ {2 -}]} {[Ni ^ {2 +}] \ cdot [CN ^ { -}] ^ {4}}} = 1.0E + 29 \ end {aligned}}}

The resulting constant is called the complex formation constant . This constant also indicates how stable the complex is or whether it tends to dissociate. Therefore, the complexation constant is also complex stability constant or complex association constant K A called. Their reciprocal value is called the complex dissociation constant K D , i.e. K D = K A −1 . The higher the complex formation constant K A , the more stable the complex, the smaller, the easier the dissociation.

### Complex formation constants

According to Martell & Smith, 1982; Hyvönen & Aksela, 2010; Vasilev, et al. 1996; Vasilev, et al. 1998

Complexing agents abbreviation Complex formation constant against Ca 2+
Diethylenetriaminepentaacetic acid DTPA 10.8
Ethylenediaminetetraacetic acid EDTA 10.7
Methyl glycine diacetic acid MGDA 7th
Nitrilotriacetic acid NTA 6.4
Nitrilotrimethylene phosphonate NTMP 5.75
Tetrasodium iminodisuccinate IDS 5.2
Tetrasodium N, N-bis (carboxylatomethyl) -L-glutamate GLDA 5.2
Ethylenediamine disuccinic acid EDDS 4.6

## Special complex compounds

### Sandwich complexes

In sandwich complexes, the metal centers are enclosed by two planar and cyclic organic ligands like two halves of a bun, which is why this type of compound was given its name. The sandwich complexes include the metallocenes and the “piano chair complexes” . The most common ligand is the cyclopentadienyl anion (Cp), which has six π electrons and is therefore aromatic (see Hückel rule ). Corresponding complexes with benzene as a ligand are also possible, for example bis (benzene) chromium , or uranocene with cyclooctatetraenyl ligands.

The first sandwich complex to be synthesized was ferrocene in 1951 , but its possible structure puzzled for some time. Ernst Otto Fischer , Geoffrey Wilkinson and Robert B. Woodward finally clarified the correct structure independently of one another, for which the first two received the 1973 Nobel Prize in Chemistry. Ferrocene was discovered by chance while researching catalysts and was noticeable for the unusual stability of its orange-colored crystals. By fulfilling the 18-electron rule, it is more stable than similar compounds with other metals, such as cobaltocene or nickelocene .

In this type of compound, the organic ligands bind to the metal center with their π electrons. Since this does not necessarily have to happen with the entire ring, but other bonding states are also possible, the term hapticity is used to describe these relationships . The hapticity η in ferrocene is, for example, 5, since each ligand binds with five atoms. This is also reflected in the notation [Fe (η 5 -C 5 H 5 ) 2 ] for the complex.

### Multinuclear complexes

Bis (cyclooctadiene) di (µ-dichlorido) dirhodium ; a dimer with chloride bridging ligands

Polynuclear complexes contain more than one central particle. They are linked via a bridging ligand such as oxygen (O 2− , OH - , H 2 O, OR) or chlorine. However, there are also complex compounds with (partly non-integer) metal-metal multiple bonds, e.g. B. [Tc 2 X 9 ] 3− , X = Cl, Br.

### Macrocyclic Metal Complexes

Cryptate with a potassium ion and an azo polyether as ligand
macro-bicyclic cryptand (azo polyether)

Certain natural antibiotics of the cyclopeptide type (e.g. valinomycin ) are able to selectively bind and transport potassium ions. In 1967 Charles Pedersen first synthesized crown ethers , which belong to the type of macrocyclic polyethers and which are able to complex and transport especially alkali and alkaline earth ions . Starting from these macrocyclic polyethers, Jean-Marie Lehn's group in 1969 synthesized a macrobicyclic ligand (azo polyether ) for the first time, which was known as a cryptand . This also complexed alkali and alkaline earth ions in its cavity, the corresponding complexes were called cryptates . As a result, various cryptands were produced which have cavities of different sizes and are therefore adapted to the size of alkali or alkaline earth ions. The stability constants of the corresponding cryptates are relatively high, the complexes have a good selectivity for the ions and are therefore suitable for the selective separation of the ions from solutions. It was also possible to produce macrotricyclic cryptands and those with other heteroatoms. Many macrocyclic metal complexes also have biological significance . Further examples are complexes with phthalocyanine as a ligand, such as in the dye copper phthalocyanine .

### Lattice-shaped metal complexes

Lattice-shaped metal complexes are supramolecular complex compounds made up of several metal atoms and coordinating chelate ligands that form a lattice-shaped structural motif. The structure is usually formed through thermodynamic self-organization . They have properties that make them interesting for information technology as future storage materials.

## application

### Biological importance

In biology complexes play an important role. These can be catalytically active proteins ( enzymes ) or catalytically inactive proteins. Numerous enzymes contain complexes in their active centers. This topic is one of the main areas of bioinorganic chemistry . In general, a complexing metal atom is present here which is not completely complexed by amino acid side chains as ligands. A ligand site acts as an active center for the conversion or temporary binding of the substrate. The most common complex centers are iron, copper, zinc, calcium, magnesium and manganese. But there are also more unusual elements such as vanadium. Calcium and zinc complexes in particular are of structural importance (e.g. zinc fingers in DNA sequence recognition).

The catalytically inactive proteins include z. B. Porphyrin complexes such as the heme in hemoglobin and in cytochromes , or chlorophyll (each chelate complex). Coordination compounds are therefore responsible for blood to appear red and a leaf of a plant to appear green.

### technology

Complexes are used as catalysts in many chemical reactions . For example, in the already mentioned olefin metathesis , for which there was a Nobel Prize in 2005, carbene complexes with ruthenium or molybdenum are used (see Grubbs catalyst ). The Wilkinson catalyst is a square-planar rhodium (I) complex that is suitable for various applications such as the hydrogenation of olefins . It is also worth mentioning the industrial production of acetic acid from methanol and carbon monoxide with a rhodium catalyst in the Monsanto process .

Various complexing agents serve as food additives , as additives in the detergent and cleaning agent industry , in the electroplating and circuit board industry as well as in chemical analysis .

Phthalocyanine complexes are used in CDs as a storage medium.

In analog photography , the remaining, unexposed, hardly water-soluble silver bromide is removed from the layer with a fixing salt solution (ammonium or sodium thiosulphate ) after development : see fixing (photography) .

### research

It is fundamentally a problem to fix short-lived and unstable molecules that occur as intermediate products in reactions. One method is fixation through complex formation. The fixed molecules have different chemical properties, however, in this way bonding and structural relationships can be investigated. Examples are complexes with carbenes , cyclobutadiene , diimines and carbene-analogous silylenes. After being released from the complexes, the molecules are again highly reactive. Metal complexes with chromium, nickel, iron and manganese are used. A metal carbonyl complex is often used as the starting complex . Examples: tricarbonyl cyclobutadiene iron, methoxyphenyl carbene pentacarbonylchromium, tetrachlorobis (tetramethylcyclobutadiene) nickel.

## literature

Wikibooks: General and Inorganic Chemistry / Complexes  - Learning and Teaching Materials

## Individual evidence

1. ^ AF Holleman , E. Wiberg , N. Wiberg : Textbook of Inorganic Chemistry . 102nd edition. Walter de Gruyter, Berlin 2007, ISBN 978-3-11-017770-1 , p. 1316.
2. ^ William H. Brock: Viewegs Geschichte der Chemie , Vieweg, Braunschweig 1997, p. 364.
3. Lutz H. Gade: coordination chemistry . 1st edition, Wiley-VCH, Weinheim 1998, p. 6.
4. ^ Alfred Werner: Contribution to the constitution of inorganic compounds . in: Zeitschrift für Anorganische Chemie 1893 , 3, pp. 267-330, doi : 10.1002 / zaac.18930030136 .
5. Christoph Janiak: Complex / Coordination Chemistry , in: Erwin Riedel (Hrsg.): Moderne Anorganische Chemie . 3rd edition, de Gruyter, Berlin 2007.
6. ^ A b c Nomenclature of Inorganic Chemistry , IUPAC Recommendations 2005 , RSC Publishing, Cambridge, UK.
7. ^ IUPAC Red Book 2005, s. Section IR-9.2.2.3 for complexes (PDF file; 4.1 MB).
8. Max Herberhold: Complex chemistry with naked metal atoms . In: Chemistry in Our Time . tape 10 , no. 4 , 1976, p. 120–129 , doi : 10.1002 / ciuz.19760100405 .
9. Wolfgang Liebscher, Ekkehard Fluck: The systematic nomenclature of inorganic chemistry . Springer-Verlag, 1998, ISBN 978-3-540-63097-5 , pp. 127–150 ( limited preview in Google Book search).
10. ^ Karl-Heinz Hellwich: More systematics: Nomenclature of Inorganic Chemistry. Edited by the International Union of Pure and Applied Chemistry. RSC Publishing, Cambridge / UK, 2005. XII + 366 pp., Hardcover, 49.95. ISBN 0-85404-438-8 . In: News from chemistry. 54, 2006, pp. 807-808, doi : 10.1002 / nadc.20060540725 .
11. Entry on coordination . In: IUPAC Compendium of Chemical Terminology (the “Gold Book”) . doi : 10.1351 / goldbook.C01329 .
12. ^ Erwin Riedel: Allgemeine und Anorganische Chemie , Walter de Gruyter Verlag (1999), Section 2.2.3, ISBN 3-11-016415-9 .
13. ^ Charles E. Mortimer and Ulrich Müller: Chemistry. 10th edition, Thieme, Stuttgart 2010, p. 523.
14. ^ Bernard Dietrich, Jean-Marie Lehn , Jean-Marie Sauvage: Kryptate: macrocyclic metal complexes . In: Chemistry in Our Time . tape 7 , no. 4 , 1973, p. 120–128 , doi : 10.1002 / ciuz.19730070405 .
15. J.-M. Lehn et al., Angew. Chem., 2004, 116, pp. 3728-3747.
16. Günter Schmid : The fixation of short-lived molecules through complex formation . In: Chemistry in Our Time . tape 8 , no. 1 , 1974, p. 26–30 , doi : 10.1002 / ciuz.19740080105 .


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# Surfaces are homeomorphic iff are diffeomorphic.

I have read this statement in several places: "Two surfaces are homeomorphic iff are diffeomorphic". I think the nontrivial implication follows in this manner: First, we triangulate the surface and then we smooth the resulting piecewise linear manifold, but I haven't found and well-written proof. Could you possibly give me a reference for the proof? Thanks in advance.

-
What about a square and a sphere? Or you assume that both surfaces are smooth? – Ilya Jan 31 '13 at 19:04
Could you give a reference where you've read it? This page has some references for something similar to your question. – anon271828 Jan 31 '13 at 19:11
See this MathOverflow question, which gives a list of references. – David Moews Jan 31 '13 at 19:23
Ilya: Sorry, you're right, I assume smooth surfaces :) – math_failure Jan 31 '13 at 23:49
anon271828, I just read that in lecture notes on the net and google books without proof, because of that I need an strong reference – math_failure Jan 31 '13 at 23:51

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# Beginning mathematical proof

1. Mar 9, 2010

### Testify

1. The problem statement, all variables and given/known data

I originally made this thread for something else, but I have another problem that I need help with.

Suppose a and b are nonzero real numbers. Prove that if a < 1/a < b < 1/b then a < -1.

2. Relevant equations

A hint was given for the problem: Assume a < 1/a < b < 1/b. Now prove that a < 0, and then use this fact to prove that a < -1.

Thanks

Last edited: Mar 9, 2010
2. Mar 9, 2010

### Staff: Mentor

For b, you're missing the inequality in the triangle inequality.

3. Mar 9, 2010

### Testify

new problem bump

4. Mar 9, 2010

### Tinyboss

It's not like threads go in the landfill and pollute kindergarten playgrounds after we're done with them. No need to recycle. Make a new thread when you have a new problem.

For this problem, first think about what it means when $$a<\frac1a$$. What values could $$a$$ have? You should identify two possibilities (two open intervals where $$a$$ could be). The second part of the inequality will let you narrow it down to one.


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What is the greatest number of solid cubes with 0.2 meter edges that

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What is the greatest number of solid cubes with 0.2 meter edges that can be packed in a box whose inside dimensions are 1 meter by 1 meter by 2 meters?

(A) 10
(B) 25
(C) 50
(D) 125
(E) 250
[Reveal] Spoiler: OA

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Re: What is the greatest number of solid cubes with 0.2 meter edges that [#permalink]

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01 Dec 2017, 00:17
E.

The dimension of the smaller cube is 0.2.

Along the side of length 1m we can place 5 cubes.
These 5 cubes can be placed along the side of length 1m 5 times. Hence 5*5=25 cubes in one row.
These 25 cubes can be placed 2/0.2 times along the height. Hence total number is cubes are 25*10=250

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Re: What is the greatest number of solid cubes with 0.2 meter edges that [#permalink]

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01 Dec 2017, 10:21
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KUDOS
Bunuel wrote:
What is the greatest number of solid cubes with 0.2 meter edges that can be packed in a box whose inside dimensions are 1 meter by 1 meter by 2 meters?

(A) 10
(B) 25
(C) 50
(D) 125
(E) 250

1/0.2 = 5 & 2/0.2 = 10
Dimensions are 1 * 1 * 2
number of cubes will be = 5 * 5 * 10 = 250
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What is the greatest number of solid cubes with 0.2 meter edges that [#permalink]

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02 Dec 2017, 10:53
Bunuel wrote:
What is the greatest number of solid cubes with 0.2 meter edges that can be packed in a box whose inside dimensions are 1 meter by 1 meter by 2 meters?

(A) 10
(B) 25
(C) 50
(D) 125
(E) 250

Box's total volume / one cube's volume = number of cubes

Box's total volume: 1 * 1 * 2 = 2 $$m^3$$

Each cube's volume: .2 * .2 * .2 = .008 $$m^3$$

$$\frac{2}{.008}=\frac{2,000}{8}=250$$ cubes

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What is the greatest number of solid cubes with 0.2 meter edges that   [#permalink] 02 Dec 2017, 10:53
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# Prove/disprove $||A^2|| \leq ||A||^2$ for a nxn matrix

The question is: Prove/disprove $||A^2|| \leq ||A||^2$ where A is some nxn matrix.

I've played around with a while few matrices and I'm pretty sure that this is correct but I can't quite figure out how to prove it. My first guess is just define some arbitrary matrix A, and show that for any vector $x$ of size 1, $||A^2 x|| \leq ||Ax||^2$ however this method seems rather involved.

Any tips would be great! :)

Edit: $||A||= \max ||Ax||$, where $|x| = 1$

Edit: Wow this was rather fast. Thanks everyone.

• Knowing about eigenvalues maybe provides a shortcut. Have you done eigenvalues yet? – mathreadler Jan 21 '16 at 1:25

Recall that by the definition of $\|A\|$ we have $$\|Ay\| \leq \|A\|\|y\|.$$
For any $x \in \mathbb{R}^n$, using this inequality twice (on $y = Ax$ and on $y = x$), we have $$\|A^2x\| = \|A(Ax)\| \leq \|A\| \|Ax\| \leq \|A\| \left( \|A\| \|x\| \right) = \|A\|^2 \|x\|.$$
Therefore $$\|A^2\| = \sup_{x \neq 0}\dfrac{\|A^2x\|}{\|x\|} \leq \|A\|^2.$$
It is true. In fact, for all $x \in \mathbb{R}^n$ with $\|x\| = 1$. If $Ax \neq 0$, then by definition, $$\|A^2x\| = \|A(Ax)\| \leq \|Ax\| \|A\|\frac{\|x\|}{\|Ax\|} = \|A\|\|x\| = \|A\|.$$ The above inequality also holds for $x$ such that $Ax = 0$. Thus $$\|A^2\| = \sup_{\|x\| = 1} \|A^2 x \| \leq \|A\|.$$


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# Trig question

xXOfNiRXx
How do you go about finding the arctangent of an unfamiliar number. Example, arctan (-2)? I think it's in the direction of half-angles and double angels, but how do I get the angle to start with the formulas in the first place?

Bohrok
If x = arctan(-2), then tan(x) = -2. I don't think double or half-angle will help you for this problem. Sometimes you can figure out what x is, sometimes you can't without a calculator.

Mentor
How do you go about finding the arctangent of an unfamiliar number. Example, arctan (-2)? I think it's in the direction of half-angles and double angels, but how do I get the angle to start with the formulas in the first place?


text: … An acre is defined as a unit of the land measurement that gives an area commonly used in the US customary and the imperial systems. Definition: An acre (symbol: ac) is a unit of land area used in the imperial and US customary systems. An acre of land is further described as the area of one furlong by one chain measuring 660 feet by 66 feet that are precisely equivalent to 1/640 of a square mile, approximately 40% of an hectare, 43,560 square feet (or 4,047 square meters). We assume you are converting between acre and square foot. My current garden of both fruits and vegetables is roughly 4500 square feet. A township is measured … An area acre is an area of land one furlong (40 rods or 10 chains) long by 4 rods (or 1 chain) wide. A cent is one-hundredth (1/100) of an acre. 30 Layers( Eggs) 150 sq ft; 40 Broilers 50-100 sq ft; 2 Dairy cows (Milk) 500 sq ft; 4 Beef cows 600 sq ft ; 4 Sheep (Mutton) 300 sq ft; VEGETABLES. 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Half acre is 21,780. To help visualize it, a quarter acre is a little more than a 100 ft by 100 ft rectangle, and a half acre is a little less than a 150 by 150 ft rectangle. Here is the area conversion factors list from acres to other common area units: 1 Acre = 0.404686 hectare; 160 perches; 4 rood; 43560 square feet; 6272640 square inches; … T That's equal to 43,560 square feet for those unfamiliar with those units of measure. The answer is 43560. Type in unit A quarter acre is 10890 square feet. Sign Up For My Newsletter: http://theurbanfarmer.co/signup? Acres to Square Feet. But, just using whole numbers, there are 18 other possibilities for the shape of a 1/4 acre plot. 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Measurements in various units of area measurement equalling 1 acre. Definition: An acre (symbol: ac) is a unit of land area used in the imperial and US customary systems. One Acre consists of 43,560 square feet A quarter of an acre would thus consist of 10890 square feet Acres. It is about the size of an American football field. Did I translate "How Many Square Feet In An Acre" correctly in your language? An acre is a unit of area equal to 1/640 th of a square mile. In word, decimal point two five Acre (ac) is equal to ten thousands eight hundred and ninety Square Feet (sq ft) in Survey System. One mile is 5280 feet (1760 yards). 1.00 acres = 43,560 ft 2: 1.25 acres = 54,450 ft 2: 1.50 acres = 65,340 ft 2: 1.75 acres = 76,230 ft 2: 1.01 acres = 43,996 ft 2: … That means 5,280 x 5,280 is 27,878,400 and that means that that all sections are 27,878,400 square feet. To get the answer we do: 43,560 ÷ 4 = 10,890 square feet which is the answer. 43560. Did I translate "How Many Square Feet In An Acre" correctly in your language? I am administrator of the site is requesting to you that if any wrong information is found please let me know. It is the size of a square that is one foot on a side. Cent is written as "ct". These measurements are extremely important when buying or selling a piece of property. 240 square feet for the pigs. But your land could be shaped like a square, or more like a rectangle. Type in your own numbers in the form to convert the units! Cent: Cent is a term used in the measurement of land in Southern India. Unit Conversion Calculator Cooking Conversion Calculator Mass Conversion Calculator Acceleration Conversion Calculator Area Conversion Calculator Density Conversion … A quarter is double this size (Double the surface area). That is to say, there are 640 acres in a square mile or a "section". Ok, so the next step. How many feet are in a roll of barbed wire? A half acre equals to 21,780 square feet. All In One Unit Converter. The median lot size of a new single-family detached home sold in 2016 was 8,562 square feet, or just under one-fifth of an acre. 1 acre. How many square foot in 1 quarter acre? An acre is about 40% of a hectare (1 acre equals 0.404686 hectares). A quarter acre is 10890 square feet. ConvertUnits.com provides an online Type in 1.5 for 1 and a half, 1.25 for 1 and a quarter, 1.75 for 1 and three quarters, etc... How many? Use the calculator above to convert between acres and square feet. The precise meaning of this depends on the exact definition adopted for a foot: the international acre is 4 046.856 422 4 m² (for the UK, see). In this case, 0.5 acres is multiplied by 43,560 square feet. – AR15.COM – I have a rectangular shaped 1/2 acre sized lot. How Big Is An Acre In Length And Width. If you’re wondering how big an acre is in length and width, it’s not quite that simple. That is a square approximately 104.36 feet on a side. The acre is a US customary and imperial unit of area. The answer is 10889.99999462. The ideal, well-rounded, quarter acre farm should have fruits, vegetables and livestock that are easy to grow. It is roughly the area of 16 tennis courts (78 ft. by 36 ft.) or about 2.5 soccer field (393 ft. x 295 ft. as a regulation of the maximum size). We know how to calculate square footage which is length times width. quarter acre to square foot, or enter any two units below: A square foot is by definition the area enclosed by a square with sides each 1 foot long. The acre is related to the square mile, with 640 acres making up one square mile. How Many Square Feet in An Acre? um quarto acre equals 1012 square meters because 0.25 times 4047 (the conversion factor) = 1012 All In One Unit Converter Please, choose a physical quantity, two units, then type a … 1 Kattha Equal to How Many Decimal in Bihar, Gunta (guntha) to Square Feet (sq ft) Calculator. In mathematical expression, 0.25 Acre (ac) = 10890 Square Feet (sq ft). This measurement is roughly equivalent to 40 per cent of a hectare. Physics Chemistry Recipes ⇌ Please, choose a physical quantity, two units, then type a value in … Anonymous. How many acres in 29000 square feet? The land was frequently put to use with vegetable gardens, fruit trees, or lawns for family recreation. We multiply the value in acres by the conversion factor to find out how many square feet are in an acre. You can do the reverse unit conversion from 400 square feet for 15 egg layers. There are 43560 square feet in an acre so one quarter of an acre is square feet. Well, you'll have to start with square footage to get an idea. 3/4 of an acre is 32,670 square feet. (remember we only produce 10% of our fruit right now) 200 square feet for 50 meat chickens. One acre is 43,560 feet. 1 acre of land is approximately equal to 43,560 sq. 1 square meter is equal to 0.00024710538146717 acre, or 10.76391041671 square feet. How to convert acres to square feet? How much is 0.29 acre in square feet? For example, to calculate how many square feet is 2 acres, multiply 2 by 43560, that makes 87120 sq ft is 2 acres. area, mass, pressure, and other types. Please enable Javascript ›› It is defined as the area of 1 chain (66 feet) by 1 furlong (660 feet), which is exactly equal to 1/640 of a square mile, 43,560 square feet, or exactly 4046.8564224 square meters. When you compare an acre with sq km, it is nearly 1/1247.11 of a Square Kilometre, 4046.86 Square Metre. Acres. The global symbol used to denote an acre is ac, and the international acre is the commonly used measurement. How much should it cost to fence 1/2 acre??? How to transform acres in square feet? From the dimensions you have given, it's a quadrilateral, i.e. From. The acre is a unit of land commonly employed within North America and the Caribbean. Based upon the International yard and pound agreement of 1959, an acre may be declared as exactly 4,046.8564224 … If one acre is equal to 43,560 ft2, a quarter acre is equal to 43560/4 = 10890 sq ft. Once an acre is subdivided, the measurements reduce slightly because of the access roads for instance the 1/8=50 by 100 feet=15 by 30 metres =450/10,000 which becomes 0.045Ha I want to fence the entire thing, including an two gates large enough for vehicles to drive through…. You can find metric conversion tables for SI units, as well To convert any value in square feet to acres, just multiply the value in square feet by the conversion factor 2.29568411386595. Type in your own numbers in the form to convert the units! It is 6% of an acre. Quarter acre is 10,890. A quarter acre = 10,890 square feet. As we said above, an acre has no prescribed or definite shape. To convert acres into square feet, the number of acres is multiplied by the conversion factor. To convert acres into square feet, the number of acres is multiplied by the conversion factor. Did you know that there are 1,742,400 square feet in a standard 40 acre parcel. What is the conversion factor to convert from acres to square feet? A quarter acre is 10890 square feet. a polygon with 4 … The SI derived unit for area is the square meter. Date Time: 2020-12-18 16:04:52, My IP: 71.178.255.4. Deciding what you should do with your quarter acre can be a tricky decision. The acre is a unit of land area used in the imperial and US customary systems. Note that rounding errors may occur, so always check the results. It is equal to 43 560 square feet, 4840 square yards, or 160 square rods. acres to square feet formula. How many square foot in 1 quarter acre? That isn't even close to a quarter of an acre. There are 43,560 square feet per acre. Definition: An acre (symbol: ac) is a unit of land area used in the imperial and US customary systems. Your email address will not be published. A quarter of an acre is slightly larger than a house lot. This site is associated with unit conversion metrics that meet the daily needs of the public. The conversion says that it would be 2,740 square feet. Use this page to learn how to convert between square feet and quarter acre. About 43,560 square feet make one acre. It is approximately 1/847.55 of a Square Nautical Mile. It is about the size of an American football field. How many acres are in 0.29 square feet? It is defined as the area of 1 chain (66 feet) by 1 furlong (660 feet), which is exactly equal to 1/640 of a square mile, 43,560 square feet, or exactly 4046.8564224 square meters. Yahoo fait partie de Verizon Media. There are exactly 640 acres in a square mile. An acre is measured … Quarter acre is 10,890. As land is no different than a plane itself with its own length and breadth. If one acre is equal to 43,560 ft2, a quarter acre is equal to 43560/4 = 10890 sq ft. This helps convert viewer from one unit to other unit. Yes ALL of them Now we also know that a square acre is 43,560 feet because of our study tips in class. 1 square meter is equal to 10.76391041671 square feet, or 0.00024710538146717 acre. 1 acres to sq ft conversion. 1)For a rectangular plot, 50 by 100 refers to 50 feet by 100 feet which is equivalent to 15 meters by 30 meters and is also equal to 450 square meters. Acre. Required fields are marked *. Note that rounding errors may occur, so always check the results. One acre is defined as the area equal to a space that is one chain (66 ft) by one furlong (660 ft), or 10 square chains. Half acre is 21,780. 2 1. If each side of the square is F feet long then the area is F^2 square feet. How many square feet in an acre? What is an acre? This is a video about How Many Square Feet In An Acre. 0.25 acre = 10,890 square feet. Note: results are rounded. If each side of the square is F feet long then the area is F^2 square feet. It is equal to 4,047 square metres or 43,560 square feet. An acre is a unit of area that historically has been used to measure tracts of land. So, if the 1/4 acre was a square plot, it would be about 104.36 feet square, so perimeter would be 417.42 feet. 1 Acre is equal to 43560 square feet. An acre contains a little more than 43,000 square feet. The answer is 10889.99999462. For the best answers, search on this site https://shorturl.im/avdLh. When surveying land in the United States, the primary, modern definition of an acre is 1/640th of a square mile. 0.29 acres is equal to how many square feet? An acre of land is further described as the area of one furlong by one chain measuring 660 feet by 66 feet that are precisely equivalent to 1/640 of a square … Planning what exactly to grow or raise on this 10,000 square foot plot of land and in what quantities is a job unto itself. The area would be the same, but the lengths of the sides would be different and that would affect the amount of fencing you would need. 100 cents equals an acre. You can view more details on each measurement unit: square feet or acre The SI derived unit for area is the square meter. 1 acre. So no sort of different grade calculations or integration techniques. Thus an acre is 10 square chains, 160 square rods, 43,560 square feet, 4840 square yards, 1.184 arpent or 0.4047 hectares. Since there are 4,840 square yards in an acre, a quarter-acre is also defined as 1,210 square yards of land. That's equal to 43,560 square feet for those unfamiliar with those units of measure. 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Troels Ronnow , ETH Zürich, Zürich
ABSTRACT
In this article, we investigate an integral result presented by Herrick and Stillinger in 1975 [Phys. Rev. A 11, 42 (1975)] numerically and analytically. Based on their result we implement a custom integration routine. We demonstrate that custom made integration routines can be several times faster than the native routines. This may be beneficial in cases where many integrals need to be carried out. We compared the results of custom integration routine against those of the native integration routine. For cases, where the native routine could not perform the integration, we compared the custom integration routine against a Monte Carlo estimate. In all cases, we found excellent agreement. As an application of the implemented integration routine, we compute the $N$-body matrix elements of a Hamiltonian with two-body potentials.
CITATION
Troels Ronnow, "Evaluation of a special class of two-body integrals through extension of the integral routine in Maple", Computing in Science & Engineering, , no. 1, pp. 1, PrePrints PrePrints, doi:10.1109/MCSE.2013.40
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# Theoretical and numerical study of a few stochastic models of statistical physics

Abstract : In this thesis, we are mainly interested in three topics : functional inequalities and their probabilistic aspects, hydrodynamic limits for interacting continuous spin systems and discretizations of stochastic differential equations. This document, in addition to a general introduction (written in French), contains three parts. The first part deals with functional inequalities, especially logarithmic Sobolev inequalities, for canonical ensembles, and with hydrodynamic limits for continuous spin systems. We prove convergence to the hydrodynamic limit for several variants of the Ginzburg--Landau model endowed with Kawasaki dynamics, with quantitative bounds in the number of spins. We also study convergence of the microscopic entropy to its hydrodynamic counterpart. In the second part, we study links between gradient flows in spaces of probability measures and large deviations for sequences of laws of solutions to stochastic differential equations. We show that the large deviations principle is equivalent to the Gamma--Convergence of a sequence of functionals that appear in the gradient flow formulation of the flow of marginals of the laws of the diffusion processes. As an application of this principle, we obtain large deviations from the hydrodynamic limit for two variants of the Ginzburg-Landau model. The third part deals with the discretization of stochastic differential equations. We prove a transport-Entropy inequality for the law of the explicit Euler scheme. This inequality implies bounds on the confidence intervals for quantities of the form $\mathbb{E}[f(X_T)]$. We also study the discretization error for the evaluation of transport coefficients with the Metropolis-Adjusted Langevin algorithm (which is a combination of the explicit Euler scheme and the Metropolis algorithm).
Keywords :
Document type :
Theses

Cited literature [130 references]

https://tel.archives-ouvertes.fr/tel-01096886
Contributor : Max Fathi <>
Submitted on : Thursday, December 18, 2014 - 1:43:53 PM
Last modification on : Wednesday, May 15, 2019 - 3:36:48 AM
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• HAL Id : tel-01096886, version 1

### Citation

Max Fathi. Theoretical and numerical study of a few stochastic models of statistical physics. General Mathematics [math.GM]. Université Pierre et Marie Curie - Paris VI, 2014. English. ⟨NNT : 2014PA066349⟩. ⟨tel-01096886⟩

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# Vector Displacement| Vector Addition and Subtraction: Graphical Method| Two-Dimensional Kinematics| College Physics Problem 3.6

### Repeat the problem above, but reverse the order of the two legs of the walk; show that you get the same final result. That is, you first walk leg B , which is 20.0 m in a direction exactly 40º south of west, and then leg A , which is 12.0 m in a direction exactly 20º west of north. (This problem shows that A+B=B+A.)

SOLUTION:

First, we solve for the x and y components of vectors A and B.

For vector A, the components are:

$A_x=-\left(12.0\:m\right)sin\:\left(20^{\circ} \right)=-4.1042\:m$

$A_y=\left(12.0\:m\right)cos\left(20^{\circ} \right)=11.2763\:m$

For vector B, the components are:

$B_x=-\left(20\:m\right)cos\left(40^{\circ} \right)=-15.3209\:m$

$B_y=-\left(20\:m\right)sin\:\left(40^{\circ} \right)=-12.8557\:m$

Since we know the components, we can now solve for the resultant vector R by adding the components together.

The resultant of the x-components is:

$R_x=A_x+B_x=-4.1042+\left(-15.3209\right)=-19.4251\:m$

The resultant of the y-components is:

$R_y=A_y+B_y=11.2763+\left(-12.8557\right)=-1.5794\:m$

The magnitude of the resultant can be solved using the Pythagorean Theorem. That is:

$R=\sqrt{\left(R_x\right)^2+\left(R_y\right)^2}=\sqrt{\left(-19.4251\right)^2+\left(-1.5794\right)^2}=19.4892\:m$

The compass direction is

$\theta =tan^{-1}\left|\frac{R_y}{R_x}\right|=tan^{-1}\left(\frac{1.5794}{19.4251}\right)=4.6483^{\circ}$

The compass direction is $4.6483^{\circ} \:South\:of\:West$.

No matter what is the order of the vectors we are adding, the result is still the same. This proves that $\vec{A}+\vec{B}=\vec{B}+\vec{A}$.


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# Pacejka Formula - natural values

For my racing game I am using a simplified Pacejka Magical Formula:

$\mu_{lateral} = \mu_{0} * \mu_{1} * D * \sin(C * \arctan(B * x - E * (B * x - \arctan(B * x))))$

Where:

• $\mu_{lateral}$ is sideways coefficient of friction
• $\mu_{0}$ is wheel friction coefficient
• $\mu_{1}$ is surface coefficient friction
• $B$ is wheel lateral stiffness
• $C$ is wheel lateral peak factor
• $D$ is wheel lateral shape factor
• $E$ is wheel lateral curvature

However, I don't know what are realistic values for $B, C, D, E$. I know well the first ones and I also assume that $C$ may be around 1.1. However, what's realistic stiffness, shape factor and curvature in sideways directions for different types of real-life tires?

• Thanks for information. So $F_{z}$ is the load on wheel, right? Than I have it, with all the suspension stuff etc. That's great:) So I must try plugging up all the things. However, I'll still need to find out theese values for several tire types. – Adrians Netlis Mar 13 '16 at 21:59


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# Properties

 Label 2268.2.l.c Level $2268$ Weight $2$ Character orbit 2268.l Analytic conductor $18.110$ Analytic rank $0$ Dimension $2$ CM discriminant -3 Inner twists $4$

# Related objects

## Newspace parameters

 Level: $$N$$ $$=$$ $$2268 = 2^{2} \cdot 3^{4} \cdot 7$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 2268.l (of order $$3$$, degree $$2$$, not minimal)

## Newform invariants

 Self dual: no Analytic conductor: $$18.1100711784$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{-3})$$ Defining polynomial: $$x^{2} - x + 1$$ Coefficient ring: $$\Z[a_1, \ldots, a_{19}]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 756) Sato-Tate group: $\mathrm{U}(1)[D_{3}]$

## $q$-expansion

Coefficients of the $$q$$-expansion are expressed in terms of a primitive root of unity $$\zeta_{6}$$. We also show the integral $$q$$-expansion of the trace form.

 $$f(q)$$ $$=$$ $$q + ( -1 - 2 \zeta_{6} ) q^{7} +O(q^{10})$$ $$q + ( -1 - 2 \zeta_{6} ) q^{7} + ( -2 + 2 \zeta_{6} ) q^{13} + \zeta_{6} q^{19} -5 q^{25} + 7 \zeta_{6} q^{31} + 10 \zeta_{6} q^{37} -5 \zeta_{6} q^{43} + ( -3 + 8 \zeta_{6} ) q^{49} + ( 1 - \zeta_{6} ) q^{61} + 16 \zeta_{6} q^{67} + ( -17 + 17 \zeta_{6} ) q^{73} + ( 4 - 4 \zeta_{6} ) q^{79} + ( 6 - 2 \zeta_{6} ) q^{91} + 19 \zeta_{6} q^{97} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2q - 4q^{7} + O(q^{10})$$ $$2q - 4q^{7} - 2q^{13} + q^{19} - 10q^{25} + 7q^{31} + 10q^{37} - 5q^{43} + 2q^{49} + q^{61} + 16q^{67} - 17q^{73} + 4q^{79} + 10q^{91} + 19q^{97} + O(q^{100})$$

## Character values

We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/2268\mathbb{Z}\right)^\times$$.

 $$n$$ $$325$$ $$1135$$ $$1541$$ $$\chi(n)$$ $$-\zeta_{6}$$ $$1$$ $$-1 + \zeta_{6}$$

## Embeddings

For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.

For more information on an embedded modular form you can click on its label.

Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
109.1
 0.5 + 0.866025i 0.5 − 0.866025i
0 0 0 0 0 −2.00000 1.73205i 0 0 0
541.1 0 0 0 0 0 −2.00000 + 1.73205i 0 0 0
 $$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles

## Inner twists

Char Parity Ord Mult Type
1.a even 1 1 trivial
3.b odd 2 1 CM by $$\Q(\sqrt{-3})$$
63.g even 3 1 inner
63.n odd 6 1 inner

## Twists

By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 2268.2.l.c 2
3.b odd 2 1 CM 2268.2.l.c 2
7.c even 3 1 2268.2.i.d 2
9.c even 3 1 756.2.k.c 2
9.c even 3 1 2268.2.i.d 2
9.d odd 6 1 756.2.k.c 2
9.d odd 6 1 2268.2.i.d 2
21.h odd 6 1 2268.2.i.d 2
63.g even 3 1 inner 2268.2.l.c 2
63.g even 3 1 5292.2.a.g 1
63.h even 3 1 756.2.k.c 2
63.j odd 6 1 756.2.k.c 2
63.k odd 6 1 5292.2.a.f 1
63.n odd 6 1 inner 2268.2.l.c 2
63.n odd 6 1 5292.2.a.g 1
63.s even 6 1 5292.2.a.f 1

By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
756.2.k.c 2 9.c even 3 1
756.2.k.c 2 9.d odd 6 1
756.2.k.c 2 63.h even 3 1
756.2.k.c 2 63.j odd 6 1
2268.2.i.d 2 7.c even 3 1
2268.2.i.d 2 9.c even 3 1
2268.2.i.d 2 9.d odd 6 1
2268.2.i.d 2 21.h odd 6 1
2268.2.l.c 2 1.a even 1 1 trivial
2268.2.l.c 2 3.b odd 2 1 CM
2268.2.l.c 2 63.g even 3 1 inner
2268.2.l.c 2 63.n odd 6 1 inner
5292.2.a.f 1 63.k odd 6 1
5292.2.a.f 1 63.s even 6 1
5292.2.a.g 1 63.g even 3 1
5292.2.a.g 1 63.n odd 6 1

## Hecke kernels

This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(2268, [\chi])$$:

 $$T_{5}$$ $$T_{13}^{2} + 2 T_{13} + 4$$ $$T_{19}^{2} - T_{19} + 1$$

## Hecke characteristic polynomials

$p$ $F_p(T)$
$2$ $$T^{2}$$
$3$ $$T^{2}$$
$5$ $$T^{2}$$
$7$ $$7 + 4 T + T^{2}$$
$11$ $$T^{2}$$
$13$ $$4 + 2 T + T^{2}$$
$17$ $$T^{2}$$
$19$ $$1 - T + T^{2}$$
$23$ $$T^{2}$$
$29$ $$T^{2}$$
$31$ $$49 - 7 T + T^{2}$$
$37$ $$100 - 10 T + T^{2}$$
$41$ $$T^{2}$$
$43$ $$25 + 5 T + T^{2}$$
$47$ $$T^{2}$$
$53$ $$T^{2}$$
$59$ $$T^{2}$$
$61$ $$1 - T + T^{2}$$
$67$ $$256 - 16 T + T^{2}$$
$71$ $$T^{2}$$
$73$ $$289 + 17 T + T^{2}$$
$79$ $$16 - 4 T + T^{2}$$
$83$ $$T^{2}$$
$89$ $$T^{2}$$
$97$ $$361 - 19 T + T^{2}$$