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7

Dichlorobis(triphenylphosphine)nickel(II), or $\ce{NiCl2[P(C6H5)3]2}$ in square planar form is red and diamagnetic. The blue form is paramagnetic and features tetrahedral Ni(II) centers. Both tetrahedral and square planar isomers coexist in solutions. Weak field ligands, favor tetrahedral geometry and strong field ligands favor the square planar isomer. Both ...


11

We sometimes call this type of complex 'pseudotetrahedral' since there is an isomerism from a tetrahedral to a square planar complex possible. I was unable to find the original work here but this link gives some information. As you already mentioned there are two strong and two weak ligands so it's hard to tell how strong the ligand field splitting will be. ...


0

The point is not really whether chloride or ammonia is a strong or weak field ligand, the point is $\ce{Co^3+}$ is $\ce{d^6}$, and virtually all "octahedral" $\ce{d^6}$ complexes are low spin - essentially some complexes of $\ce{Fe^2+}$ and a very small number of fluoro complexes of $\ce{Co^3+}$ are the only exceptions to the rule that all $\ce{d^6}$ ...


13

Hybridisation won't explain anything in transition metal complexes, so please stop using it, at least to the extent where it is possible to avoid using it. Quite literally everything about transition metal complexes is better rationalised using MO theory, and I am not exaggerating. The reason why low-spin $T_\mathrm d$ complexes are rare is because the ...


1

Tetraamminecopper(II) Sulfate or cuprammonium sulfate can be synthesized as its monohydrate (Ref.1): A solution of $\pu{50 g}$ of finely divided $\ce{CuSO4 . 5 H2O}$ in $\pu{75 mL}$ of conc. ammonia and $\pu{50 mL}$ of water is filtered and precipitated by slow addition of $\pu{75 mL}$ of alcohol. After standing for several hours in the cold, the crystals ...


0

Name of this complex is. Mercury tetrathiocyanato cobaltate (2) . First take the stable oxidation state one . Another factor is Hg +2 a soft acid and SCN- is a soft ligand their is soft -soft interactions... Cobalt is in +2 oxidation state which is a borderline acid but due to symbiosis it behaves more like soft acid than hard .


2

Conventionally, first complex ion is positively charged and the latter is negatively charged. Suppose thess charges are $p+$ and $q-$, respectively. Since two ion ratio is 1:1, $|p+|= |q-|$. Suppose oxidation number of $\ce{Co}$ is $n$ and that of $\ce{Cr}$ is $m$. Hence, for cation complex (Note that the net charges of $\ce{NH3}$ and $\ce{NO2-}$ is $0$ and ...


0

Alright; let me try. In Cl-, the “HOMO” of the ion itself are the p orbitals; the two pi-bonding p orbitals act as a pi bonding (weak field) ligand. In NH3, the HOMO is the 3a1 bonding MO (sigma donor) and the LUMO is the 4a1 antibonding MO. The LUMO+1 are the antibonding 2e orbitals (which involve the p orbitals). As the LUMO can’t act as a pi acceptor, ...


0

$\ce{PR3}$ is π-acceptor (from metal d orbital to phosphorous d orbital electron cloud is transferred). $\ce{Cl-}$ is π-donor (minus charge form σ bond with the metal orbitals, $\mathrm{e_g}$, $\mathrm{t_{1u}}$ and $\mathrm{a_{1g}}$ orbitals in $O_\mathrm{h}$ field. The lone pair of electron on $\ce{Cl}$ will try to form π-bonding with $\mathrm{t_{2g}}$ ...


0

Even for weak ligands, the electrons will eventually have to pair up, once the number becomes more than $5$. So, $d^7$ is indeed possible with a weak field ligand. Consider : $[Co(Cl)_6]^{-4}$ for an example.


0

This is a diagram of the d orbitals of a generic d8 complex in a tetrahedral and square planar configuration. The tetrahedral complex would be expected for pi donor ligands (Cl-, OH-, etc.) where the pairing energy is greater than delta t, while the square planar complex would be expected for strong field pi* acceptor ligands (think CO, NO+, and CN-, which ...


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