6

@M.L has already provided the solution as per VBT, but that answer seems to have some flaws when it comes to CFT (looks like he has corrected the mistakes now). So, I would like to take some time to answer this question using CFT. According to crystal field theory, the degeneracy of the orbitals of a same orbital is lost in presence of a ligand and the ...


4

First iron(II) hydroxide $\ce{Fe(OH)2}$ is a precipitate made in aqueous solution, and it does not react with ammonia. Second, the iron(II) hydroxide $\ce{Fe(OH)2}$ is a green substance which is extremely sensitive to the oxygen of the air. In a couple of minutes, it gets brown, due to the formation of iron(III) hydroxide $\ce{Fe(OH)3}$ according to $$\ce{4 ...


3

When polydentate ligands binds to a metal in a ring form, it is called chelate ring and the corresponding complex is called chelate complex. Experimentally, it is observed that metal complexes of polydentate ligands with more rings are significantly more stable than the corresponding complexes of similar ligands but with less rings. Because, the larger the ...


3

Linkage isomerism is of major interest in the field of inorganic chemistry. Linkage isomerism occurs when an ambidentate ligand such as $\ce{NO2, SCN,}$ etc. binds to the metal center. Nitro-nitrito metal complexes can be regarded as good examples of linkage isomerism. It must be noted that the $\ce{NO2}$ group can be coordinated to the metal atom in ...


2

The spectrochemical series only gives info on the relative strength of the ligand field. Whether or not a metal complex is high or low spin (assuming there is an appropriate number of d electrons that both are possible) is a function of both the ligand and the metal center. Thus, it is possible for "weak field" ligand to be in a low spin complex if ...


2

As you would expect from increasing occupancy of antibonding orbitals, the result is weaker metal-ligand bonding. This is one reason why the octahedral geometry is not as strongly favored for metals with greater numbers of d electrons. For a d8 metal such as the Ni(II) in the example, square planar and octahedral complexes are both common, and planar ...


2

According to VBT, for $\ce{[Ni(CN)_4]^{2-}}$, because it is a square planar geometry, it will experience $\ce{dsp^2}$ hybridization. The following process and result will be the following: The diagram clearly shows that using VBT, we get that $\ce{[Ni(CN)_4]^{2-}}$ is diamagnetic. However, in general, VBT is not used often to describe these bonds, and more ...


1

$\ce{Co·6H2O}$ does not exist. But, if it would exist, il would be a special sort of metallic cobalt, surrounded by 6 water molecules fixed around each $\ce{Co}$ atom. This substance would be electrically neutral. It would exist in a sample containing no other substance. $\ce{[Co(H2O)6[^{2+}}$ does exist. It is a charged species (cation) that exists, but ...


1

The answer to your title question is: Yes, they are monodentate. But I think you have a misconception about when to use which type of counting: It does not depend on whether the ligand is polydentate or not. The IUPAC guide states that you use the bis-tris-tetrakis-etc. prefix when you want to avoid confusion: The prefixes are ‘di’, ‘tri’, ‘tetra’,etc., ...


1

For the first example, you are given that there are $6\, \ce{CN}$ ligands which are -1. That means that there is a total of 6- from the ligands. We also know the overall charge of the complex is -3. That means the charge of the central metal must be +3. In the second example, we can look at the charge again. Because the $\ce{CH3-}$ has a charge of -1, the ...


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