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Why NH4 plus ion does not form complexes?
Proton spin lattice relaxation by NH4+ ion motions and spin diffusion in some ammonium compounds
What is octahedral and tetrahedral voids?
Octahedral vs. tetrahedralSo far, we've seen the Crystal Field Theory in action in octahedral, tetrahedral and square planar complexes. But how do we tell whether a particular complex is octahedral, tetrahedral, or square planar? Obviously if we know the formula, we can make an educated guess: something of the type ML6will almost always be octahedral (there is an alternative geometry for 6-coordinate complexes, called trigonal prismatic, but it's pretty rare), whereas something of formula ML4will usually be tetrahedral unless the metal atom has the d8electron configuration, in which case it will probably be square planar.But what if we take a particular metal ion and a particular ligand? Can we predict whether it will form an octahedral or a tetrahedral complex, for example? To an extent, the answer is yes... we can certainly say what factors will encourage the formation of tetrahedral complexes instead of the more usual octahedral.Generally speaking, octahedral complexes will be favoured over tetrahedral ones because:It is more favourable to form six bonds rather than fourThe crystal field stabilisation energy is usually greater for octahedral than tetrahedral complexes.Remember that Δois bigger than Δtet(in fact, Δtetis approximately 4/9 Δo). If we make the assumption that Δtet= 4/9 Δo, we can calculate the difference in stabilisation energy between octahedral and tetrahedral geometries by putting everything in terms of Δo.For example: for a d3octahedral configuration, the CFSE is -1.2 Δo(refer back to theTableif you like).For a d3tetrahedral configuration (assuming high spin), the CFSE = -0.8 ΔtetRemember that because Δtetis less than half the size of Δo, tetrahedral complexes are often high spin.We can now put this in terms of Δo(we can make this comparison because we're considering the same metal ion and the same ligand: all that's changing is the geometry)So for tetrahedral d3, CFSE = -0.8 x 4/9 Δo= -0.355 Δo.And the difference in CFSE between the two geometries will be 1.2 - 0.355 = 0.845 Δo.If we do a similar calculation for the other configurations, we can construct aTableof Δo, Δtetand the difference between them (we'll ignore their signs since we're looking for the difference between them).We can then plot these values on a graph.Notice that:The CFSE favours octahedral over tetrahedral in most cases, but the degree of favourability varies with the electronic configuration. In other words, for d1there's only a small gap between the oct and tet lines, whereas at d3and d8there's a big gapFor d0, d5high spin and d10, there is no CFSE difference between octahedral and tetrahedral.The ordering of favourability of octahedral over tetrahedral is:d3, d8> d4, d9> d2, d7> d1, d6> d0, d5, d10The units of the graph are Δo. So if we have strong field ligands present, Δowill be bigger anyway (according to the spectrochemical series), and any energy difference between the oct and tet lines will be all the greater for it. A bigger Δomight also push the complexes over to low spin. Similarly, as we saw previously, high oxidation states and metals from the 2nd and 3rd rows of the transition series will also push up Δo.On the other hand, if large or highly charged ligands are present, they may suffer large interligand repulsions and thus prefer a lower coordination number (4 instead of 6). Consequently if you set out to make something that would have a tetrahedral geometry, you would use large, negatively charged, weak field ligands, and use a metal atom with a d0, d5or d10configuration from the first row of the transition series (though of course having weak field ligands doesn't matter in these three configurations because the difference between oct and tet is 0 Δo).As the following Table shows, you can find tetrahedral complexes for most configurations, but there are very few for d3and d8.d0MnO4-d5MnCl42-d1TiCl4-d6FeCl42-d2Cr(OR)4d7CoCl42-That's about it for the crystal field theory.Lecture 4starts by discussing how we actually make transition metal complexes, and this leads on to a section about stability constants.Back toteaching resourcespageThis page was written by Dr Mike Morris, March 2001Last updated March 2012
Does spinning water always spin in the same direction?
Does spinning water always spin in the same direction?
What is non zero nuclear spin?
It means that the spin has a value that is different from zero, such as 1/2, 1, 3/2, -1/2, etc. (spin is always a multiple of 1/2).It means that the spin has a value that is different from zero, such as 1/2, 1, 3/2, -1/2, etc. (spin is always a multiple of 1/2).It means that the spin has a value that is different from zero, such as 1/2, 1, 3/2, -1/2, etc. (spin is always a multiple of 1/2).It means that the spin has a value that is different from zero, such as 1/2, 1, 3/2, -1/2, etc. (spin is always a multiple of 1/2).
Will earth always spin at this speed if no how will it change?
No. Tidal interactions with the moon are gradually slowing the rate of Earth's spin
Give the name of outer orbital complexes?
There are two different types of outer orbital complexes. These two type of complexes are called low-spin or spin-paired complexes.
Why NH4 plus ion does not form complexes?
Proton spin lattice relaxation by NH4+ ion motions and spin diffusion in some ammonium compounds
What is high spin and low spin complexes of transition metal ion?
When metals are coordinated to ligands to form a complex, its "d" orbital splits into high and low energy groups of suborbitals. Depending on the nature of the ligands, the energy difference separating these groups can be large or small. In the first case, electrons of the d orbital tend to pair in the low energy suborbitals, a configuration known as "low spin". If the energy difference is low, electrons tend to distribute unpaired, giving rise to a "high spin" configuration. High spin is associated with paramagnetism (the property of being attracted to magnetic fields), while low spin is associated to diamagnetism (inert or repelled by magnets).
What is octahedral and tetrahedral voids?
Octahedral vs. tetrahedralSo far, we've seen the Crystal Field Theory in action in octahedral, tetrahedral and square planar complexes. But how do we tell whether a particular complex is octahedral, tetrahedral, or square planar? Obviously if we know the formula, we can make an educated guess: something of the type ML6will almost always be octahedral (there is an alternative geometry for 6-coordinate complexes, called trigonal prismatic, but it's pretty rare), whereas something of formula ML4will usually be tetrahedral unless the metal atom has the d8electron configuration, in which case it will probably be square planar.But what if we take a particular metal ion and a particular ligand? Can we predict whether it will form an octahedral or a tetrahedral complex, for example? To an extent, the answer is yes... we can certainly say what factors will encourage the formation of tetrahedral complexes instead of the more usual octahedral.Generally speaking, octahedral complexes will be favoured over tetrahedral ones because:It is more favourable to form six bonds rather than fourThe crystal field stabilisation energy is usually greater for octahedral than tetrahedral complexes.Remember that Δois bigger than Δtet(in fact, Δtetis approximately 4/9 Δo). If we make the assumption that Δtet= 4/9 Δo, we can calculate the difference in stabilisation energy between octahedral and tetrahedral geometries by putting everything in terms of Δo.For example: for a d3octahedral configuration, the CFSE is -1.2 Δo(refer back to theTableif you like).For a d3tetrahedral configuration (assuming high spin), the CFSE = -0.8 ΔtetRemember that because Δtetis less than half the size of Δo, tetrahedral complexes are often high spin.We can now put this in terms of Δo(we can make this comparison because we're considering the same metal ion and the same ligand: all that's changing is the geometry)So for tetrahedral d3, CFSE = -0.8 x 4/9 Δo= -0.355 Δo.And the difference in CFSE between the two geometries will be 1.2 - 0.355 = 0.845 Δo.If we do a similar calculation for the other configurations, we can construct aTableof Δo, Δtetand the difference between them (we'll ignore their signs since we're looking for the difference between them).We can then plot these values on a graph.Notice that:The CFSE favours octahedral over tetrahedral in most cases, but the degree of favourability varies with the electronic configuration. In other words, for d1there's only a small gap between the oct and tet lines, whereas at d3and d8there's a big gapFor d0, d5high spin and d10, there is no CFSE difference between octahedral and tetrahedral.The ordering of favourability of octahedral over tetrahedral is:d3, d8> d4, d9> d2, d7> d1, d6> d0, d5, d10The units of the graph are Δo. So if we have strong field ligands present, Δowill be bigger anyway (according to the spectrochemical series), and any energy difference between the oct and tet lines will be all the greater for it. A bigger Δomight also push the complexes over to low spin. Similarly, as we saw previously, high oxidation states and metals from the 2nd and 3rd rows of the transition series will also push up Δo.On the other hand, if large or highly charged ligands are present, they may suffer large interligand repulsions and thus prefer a lower coordination number (4 instead of 6). Consequently if you set out to make something that would have a tetrahedral geometry, you would use large, negatively charged, weak field ligands, and use a metal atom with a d0, d5or d10configuration from the first row of the transition series (though of course having weak field ligands doesn't matter in these three configurations because the difference between oct and tet is 0 Δo).As the following Table shows, you can find tetrahedral complexes for most configurations, but there are very few for d3and d8.d0MnO4-d5MnCl42-d1TiCl4-d6FeCl42-d2Cr(OR)4d7CoCl42-That's about it for the crystal field theory.Lecture 4starts by discussing how we actually make transition metal complexes, and this leads on to a section about stability constants.Back toteaching resourcespageThis page was written by Dr Mike Morris, March 2001Last updated March 2012
Does spinning water always spin in the same direction?
Does spinning water always spin in the same direction?
Which directrion did katriena spin?
Northern Hemisphere hurricanes always spin counterclockwise.
Is the spin quantum always negative?
No.
Which directions do tornadoes spin?
Tornadoes nearly always spin counterclockwise if they are in the northern hemisphere and clockwise if they are in the southern hemisphere.
What is non zero nuclear spin?
It means that the spin has a value that is different from zero, such as 1/2, 1, 3/2, -1/2, etc. (spin is always a multiple of 1/2).It means that the spin has a value that is different from zero, such as 1/2, 1, 3/2, -1/2, etc. (spin is always a multiple of 1/2).It means that the spin has a value that is different from zero, such as 1/2, 1, 3/2, -1/2, etc. (spin is always a multiple of 1/2).It means that the spin has a value that is different from zero, such as 1/2, 1, 3/2, -1/2, etc. (spin is always a multiple of 1/2).
What is better for a golf club high spin or low spin?
it is better to to a high spin The above answer is wrong. The answer is variable. One must decide what how they want to ball to behave. There are three main factors that affect flight: initial velocity, launch angle, spin rate.
Is there any connection between the wave length and spin of a photon?
No, the spin of different particles is a constant in each case; for example the spin of a photon is always 1.
Will earth always spin at this speed if no how will it change?
No. Tidal interactions with the moon are gradually slowing the rate of Earth's spin
