77
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Why is it wrong to use the concept of hybridization for transition metal complexes?
Tetrahedral complexes
Let's consider, for example, a tetrahedral $\ce{Ni(II)}$ complex ($\mathrm{d^8}$), like $\ce{[NiCl4]^2-}$. According to hybridisation theory, the central nickel ion has $\mathrm{...
- 71.1k
57
votes
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Why isn't the American nickel magnetic?
There are many types of magnetic properties, including ferromagnetism, paramagnetism, diamagnetism, antiferromagnetism, ferrimagnetism, superparamagnetism, metamagnetism, spin glasses, and ...
- 12.1k
42
votes
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Why is [PdCl4]2- square planar whereas [NiCl4]2- is tetrahedral?
The geometry of the complex changes going from $\ce{[NiCl4]^2-}$ to $\ce{[PdCl4]^2-}$. Clearly this cannot be due to any change in the ligand since it is the same in both cases. It is the other factor,...
- 71.1k
25
votes
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What is the oxidation state of Mn in HMn(CO)5?
On negative oxidation states, in general
Although it's usually a topic that's covered relatively late in a chemistry education, negative oxidation states for transition metals[1] are actually quite ...
- 71.1k
25
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Why isn't the American nickel magnetic?
None of the US coins are magnetic (ferromagnetic), except for the 1943 Lincoln penny (Steel Cents, made in steel and zinc to save copper for ammunition during wartime), which are considered magnetic. ...
- 39.5k
24
votes
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Why do transition elements make colored compounds?
You are absolutely correct, it all about the metal's electrons and also about their d orbitals.
Transition elements are usually characterised by having d orbitals. Now when the metal is not bonded ...
- 7,342
24
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Why is WF6 stable whereas CrF6 is unknown?
The answer has to do with two things. Note that HSAB theory is dubious at best and doesn't have very useful predictive power, so I am going to avoid talking about it.
(1) The accessibility of the high ...
- 71.1k
23
votes
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Why is manganese(II) coloured although the transition should be spin-forbidden?
Selection rules
The intensity of the transition from a state $\mathrm{i}$ to a state $\mathrm{f}$ is governed by the transition dipole moment $\mu_{\mathrm{fi}}$ (strictly, it is proportional to $|\...
- 71.1k
22
votes
Why isn't the orbital angular momentum also considered while calculating the magnetic moments 3d transition elements?
Simplistically speaking orbital angular momentum is present when some conditions are satisfied:
A set of orbitals are degenerate;
These orbitals can be "interconverted" by rotation about a certain ...
- 71.1k
21
votes
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Why does the Co³⁺/Co²⁺ couple have such a high reduction potential?
The electronic configuration has nothing to do with it. The reduction potentials of $\ce{Ni^3+}/\ce{Ni^2+}$, $\ce{Cu^3+}/\ce{Cu^2+}$ and $\ce{Zn^3+}/\ce{Zn^2+}$, if they have been/could be measured, ...
- 71.1k
19
votes
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How can the intense color of potassium permanganate be explained with molecular orbital theory?
Let’s take a look at a qualitative MO scheme for a tetrahedric transition metal complex whose ligands have three p-type orbitals each. On the left of figure 1 you have the metal orbitals ($\mathrm{3d}$...
- 67.3k
18
votes
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Is iron the most stable element in the periodic table?
Yes, $^{56}\ce{Fe}$ has the most stable nucleus, and $\ce{He}$ is the most chemically inert element. These are different and unrelated qualities, pretty much like physical fitness and intelligence in ...
- 31k
17
votes
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Why copper(I) is unstable in aqueous medium?
You are quite correct in that it appears at first sight that $\ce{Cu+}$ should be more stable than $\ce{Cu^2+}$, but in aqueous media it isn’t.
Stability in aqueous conditions depends on the ...
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16
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Cr(II) and Mn(III) - their oxidizing and reducing properties?
Related question with same answer but in a different context of the 4f block: Why don't we see these lanthanide species?
You have a misconception regarding the stability of oxidation states. The ...
- 71.1k
16
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Why is anhydrous copper(II) sulfate white while the pentahydrate is blue, even though both have one unpaired electron?
Let's compare the two compounds, here I plotted the $\ce{Cu(II)}$ centers of $\ce{CuSO4.5H2O}$ and $\ce{CuSO4}$ from their crystal structure data.
As you can see, it changes from a $\ce{[Cu(H2O)4[SO4]...
- 4,232
16
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Why is Ni[(PPh₃)₂Cl₂] tetrahedral?
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 ...
- 4,232
16
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Exchange energy of d6 configuration
As @orthocresol points out, the key is that you need to compare the exchange energy before vs after the ionization process. Anything that is unchanged by ionization cannot affect ionization energy. ...
- 817
15
votes
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While filling electrons, we follow Aufbau principle, but not while removing them. Why is this so?
Usually when adding electrons based on the Aufbau principle, you go from one element to the next highest one, e.g. from $\ce{Ti}: \ce{[Ar] 4s^2 3d^2}$ to $\ce{V: [Ar] 4s^2 3d^3}$. Thus you add not ...
- 4,709
14
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Why can mercury(I) exist, but not zinc(I)?
On the contrary, zinc(I) compounds do exist, though they are rare, and relatively unstable. Most zinc(I) compounds contain a $\ce{[Zn2]^{2+}}$ core, which is analogous to the $\ce{[Hg2]^{2+}}$ cation. ...
- 23.9k
13
votes
In crystal field theory, why do all the metal electrons enter the d orbitals and not the s orbital?
Your example is a little questionable. I would think you are talking about $\ce{[Fe(H2O)6]^2+}$, in which case you are actually supposed to have an $\ce{Fe^2+}$ ion with a $[\ce{Ar}]\mathrm{(3d)^6}$ ...
- 71.1k
13
votes
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What is the most common oxidation state of gold?
It's not obvious, but common oxidation state for gold is +3. It caused by destabilization of the $5d^{10}$ orbital. Detailed explanation you can find in The Chemistry of Gold, in Chapter 1.1.3.
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13
votes
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Why is MnO2 not a peroxide?
Peroxides contain two oxygens connected by a single bond. X-ray or neutron diffraction will show that the oxygens in $\ce{MnO2}$ are too far apart to be bonded, and therefore it is not a peroxide.
...
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13
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Why is MnO2 not a peroxide?
The question’s logical premise is skewed. $\ce{CO2}$ is carbon(IV) oxide or carbon dioxide. But carbon(II) also exists and so does carbon(II) oxide (carbon monoxide, $\ce{CO}$). Just because there are ...
- 67.3k
13
votes
Accepted
How do I find the ground state term symbol for transition metal complexes?
To find the ground state term symbol, you should be using symmetry and group theory arguments, you shouldn't have to resort to searching Tanabe-Sugano diagrams to get the answer.
We'll start with ...
- 71.1k
13
votes
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Derivation of the Orgel diagram for octahedral d2 complexes
1. Weak-field and strong-field limits
I will adopt the description used in Figgis and Hitchman's Ligand Field Theory and Its Applications (p 5), because I cannot really phrase it better:
It is ...
- 71.1k
13
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Why is Ni[(PPh₃)₂Cl₂] tetrahedral?
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 ...
- 4,312
12
votes
Is iron the most stable element in the periodic table?
No, nickel-62 is the most stable on a binding energy per nucleon basis. Fe-58 is second and Fe-56 is third.
See Fewell, M. P., "The Atomic Nuclide with the Highest Mean Binding Energy", Am. J. Phys....
- 40.2k
12
votes
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How does aqua regia dissolve gold?
The wikipedia article of aqua regia has beautifully mentioned the mechanism of gold dissolving in aqua regia:
Aqua regia dissolves gold, though neither constituent acid will do so alone, because, in ...
- 25k
12
votes
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Can aqua regia/royal water be produced with sources of chloride and nitrate other than hydrochloric acid and nitric acid?
Not even zinc would react with neutral nitrate + chloride, why should gold ?
Aqua regia must be strongly acidic for nitrates to have oxidative properties for oxidation of chlorides to chlorine and ...
- 39.1k
11
votes
Why is tetraamminecopper(II) a square planar and not a tetrahedral species?
The short answer is that $\ce{[Cu(NH3)4]^2+}$ does not exist; the compound you are observing is $\ce{[Cu(NH3)4(H2O)2]^2+}$. It is not square planar but a Jahn-Teller distorted octahedron. You can ...
- 67.3k
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