When the paramagnetic [co(cn)6] 4– ion is oxidized to [co(cn)6] 3– , the ion becomes diamagnetic. however, when the paramagnetic
[co(ox)3] 4– is oxidized to [co(ox)3] 3– , the ion remains paramagnetic. here ‘ox' denotes a bidentate chelating ligand. explain these observations in detail using crystal field theory and crystal field splitting diagrams?
Answer: Basically, paramagnetic and diamagnetic refer to the way a chemical species interacts with a magnetic field. More specifically, it refers to whether or not a chemical species has any unpaired electrons or not. A diamagnetic species has no unpaired electrons, while a paramagnetic species has one or more unpaired electrons. Now, I won't go into too much detail about crystal field theory in general, since I assume that you're familiar with it. So, you're dealing with the hexafluorocobaltate(III) ion, [CoF6]3â’, and the hexacyanocobaltate(III) ion, [Co(CN)6]3â’. You know that [CoF6]3â’ is paramagnetic and that [Co(CN)6]3â’ is diamagnetic, which means that you're going to have to determine why the former ion has unpaired electrons and the latter does not. Both complex ions contain the cobalt(III) cation, Co3+, which has the following electron configuration Co3+:1s22s22p63s23p63d6 For an isolated cobalt(III) cation, all these five 3d-orbitals are degenerate. The thing to remember now is that the position of the ligand on the spectrochemical series will determine how these d-orbtals will split. More specifically, you can say that a strong field ligand will produce a more significant splitting energy, Δ a weak field ligand will produce a less significant splitting energy, Δ Now, the spectrochemical series looks like this http://chemedu.pu.edu.tw/genchem/delement/9.htmhttp://chemedu.pu.edu.tw/genchem/delement/9.htm Notice that the cyanide ion, CNâ’, is higher on the spectrochemical series than the fluoride ion, Fâ’. This means that the cyanide ion ligands will cause a more significant energy gap between the eg and t2g orbitals when compared with the fluoride ion ligands. http://wps.prenhall.com/wps/media/objects/3313/3393071/blb2405.htmlhttp://wps.prenhall.com/wps/media... In the case of the hexafluorocobaltate(III) ion, the splitting energy is smaller than the electron pairing energy, and so it is energetically favorable to promote two electrons from the t2g orbitals to the eg orbitals → a high spin complex will be formed. This will ensure that the hexafluorocobaltate(III) ion will have unpaired electrons, and thus be paramagnetic. On the other hand, in the case of the hexacyanocobaltate(III) ion, the splitting energy is higher than the electron pairing energy, and so it is energetically favorable to pair up those four electrons in the t2g orbitals → a low spin complex is formed. Since it has no unpaired electrons, the hexacyanocobaltate(III) ion will be diamagnetic.
Energy required to raise the temperature from 35°C - 45 °C= 25116 J.
specific heat, the quantity of warmth required to raise the temperature of one gram of a substance by means of one Celsius degree. The units of precise warmth are generally energy or joules consistent with gram according to Celsius diploma. for instance, the unique warmth of water is 1 calorie (or 4.186 joules) according to gram in step with Celsius degree.
solving,
Sample of liquid = 400. 0 g
temperature = 30. 0 ºc
joules of energy are required to raise the temperature of the water to 45. 0 ºc
therefore rise in temperature 45 - 30 = 15°C
Specific heat capacity = 4.186 J/g m °C
In kelvin = 273 + 15 = 288
= ∴ energy required = Q = m s ( t final - t initial)
Calcium forms an ion with a positive 2 charge and chlorine forms an ion with a negative one charg, so the formula is <span>CaC<span>l2</span></span>
Group 1 metals and group 2 metals form positive ions by losing 1 and 2 electrons respectively. Non-metals in group 17 gain 1, group 16 gain 2 and group 15 gain 3. Elements which lose electrons form positive ions while elements that gain electrons form negative ions.
To write a formula, you must balance charges so the overall charge is zero. A simple way to do this is to swap the # of the ion's charge and make it the subscript of the other ion. However, leave off the number 1 and reduce to lowest whole number ratio.
