Vapour pressure of water at Kis Calculate the vapour pressure of water at Kwhen 25 g of glucose is dissolved in g of water. What should be the molarity of such a sample of the acid if the density of the solution is 1.
You are given benzene, conc. Write the equations for the preparation of phenol using these reagents. Q3 Explain with two examples each of the following: coordination entity, ligand, coordination number, coordination p Q4 What is meant by unidentate, didentate and ambidentate ligands?
Give two examples for each. Q8 List various types of isomerism possible for coordination compounds, giving an example of each Q9 How many geometrical isomers are possible in the following coordination entities? Q13 Aqueous copper sulphate solution blue in colour gives: i a green precipitate with aqueous potassium fluori Q14 What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulph Q15 Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory: i Book Store Download books and chapters from book store.
Currently only available for. Class 10 Class Coordination Compounds. Aqueous copper sulphate solution blue in colour gives: i a green precipitate with aqueous potassium fluoride and ii a bright green solution with aqueous potassium chloride. Explain these experimental results. Switch Flag Bookmark. What is crystal field splitting energy?
Geometrical isomerism. Complexes with coordination number six, having bidentate ligands provide examples of optical isomerism, e. These ligands differ quite a lot in their affinity for a particular metal ion, but the rules governing this situation are not simple.
In other words, a ligand which is a strong Lewis base with respect to one metal ion is not necessarily a strong base with respect to another. There are some ions, however, which almost always function as very weak Lewis bases. The perchlorate ion, ClO 4 — in particular, forms almost no complexes.
The nitrate ion, NO 3 — , and sulfate ion, SO 4 2— , only occasionally form complexes. The addition of ligands to a solution in order to form a highly colored complex is often used to detect the presence or absence of a given metal in solution. This can be seen in the following video, where a aqueous solution of ammonia is added to a copper sulfate solution:.
When ammonia is added, a precipitate of Cu OH 2 s is formed. If you are working towards a UK-based exam and haven't got a copy of your syllabus , follow this link to find out how to get one. Use the BACK button on your browser to return quickly to this page. Zinc with the electronic structure [Ar] 3d 10 4s 2 doesn't count as a transition metal whichever definition you use. In the metal, it has a full 3d level.
When it forms an ion, the 4s electrons are lost - again leaving a completely full 3d level. At the other end of the row, scandium [Ar] 3d 1 4s 2 doesn't really counts as a transition metal either.
Although there is a partially filled d level in the metal, when it forms its ion, it loses all three outer electrons. Note: If you aren't happy about naming complex ions , you might find it useful to follow this link. The corresponding transition metal ions are coloured. Some, like the hexaaquamanganese II ion not shown and the hexaaquairon II ion, are quite faintly coloured - but they are coloured. And why does the colour vary so much from ion to ion? Complex ions containing transition metals are usually coloured, whereas the similar ions from non-transition metals aren't.
That suggests that the partly filled d orbitals must be involved in generating the colour in some way. Remember that transition metals are defined as having partly filled d orbitals. For simplicity we are going to look at the octahedral complexes which have six simple ligands arranged around the central metal ion.
The argument isn't really any different if you have multidentate ligands - it's just slightly more difficult to imagine! Note: If you aren't sure about the shapes of complex ions , you might find it useful to follow this link before you go on. You only need to read the beginning of that page. If you don't know what a ligand is, you should read the introduction to complex ions as a matter of urgency! When the ligands bond with the transition metal ion, there is repulsion between the electrons in the ligands and the electrons in the d orbitals of the metal ion.
That raises the energy of the d orbitals. However, because of the way the d orbitals are arranged in space, it doesn't raise all their energies by the same amount. Instead, it splits them into two groups. Whenever 6 ligands are arranged around a transition metal ion, the d orbitals are always split into 2 groups in this way - 2 with a higher energy than the other 3.
When white light is passed through a solution of this ion, some of the energy in the light is used to promote an electron from the lower set of orbitals into a space in the upper set. Each wavelength of light has a particular energy associated with it. Red light has the lowest energy in the visible region. Violet light has the greatest energy. Suppose that the energy gap in the d orbitals of the complex ion corresponded to the energy of yellow light.
The yellow light would be absorbed because its energy would be used in promoting the electron. That leaves the other colours. Your eye would see the light passing through as a dark blue, because blue is the complementary colour of yellow. Warning: This is a major simplification, but is adequate for this level UK A level or the equivalent. It doesn't, for example, account for absorption happening over a broad range of wavelengths rather than a single one, or for cases where there is more than one colour absorbed from different parts of the spectrum.
If your syllabus wants you to know about the way the shapes of the d orbitals determine how the energies split, then follow this link for a brief explanation. Non-transition metals don't have partly filled d orbitals. Visible light is only absorbed if some energy from the light is used to promote an electron over exactly the right energy gap.
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