how are normal metals bonded?

[albertlee]

We know that Potassium is very reactive...

 

so, is pure potasium bonded metalically? if so, then, how come it is very reactive? if not, what type of bonding?

[RyanJ]

We know that Potassium is very reactive...

 

so' date=' is pure potasium bonded metalically? if so, then, how come it is very reactive? if not, what type of bonding?[/quote']

 

I'm not shure about the bonding part (I'm pretty shure it is metallic as it is a metallic element) but I can start to explain why its so reactive

 

If you look at Potassium's electron configuration you'll find that ends in [ce]4s^1[/ce]. This meas that it has one electron in its outer shell and this is important becasue one is easier to get rid of than two

 

Also, you may notice that as you go down group I and group II the elements get more reactive. This has two reasons as far as I can emember:

 

1) Shielding from the other full electron shells reduces the nucleuses pull on the electron making it easier to remove and thus its more reactive.

2) its further from the nucleus anyway so that also reduces its pull on the electron.

 

If you combine that you'll see why its so damn reactive!

 

 

More Information:

• Potassium @ Wikipedia
  
• Potassium @ WebElements
  
• Potassium information
  

 

Cheers,

 

Ryan Jones

[albertlee]

but Ryan, Silver also has 1 electron in its outer shell, but it's so unreactive....

[RyanJ]

but Ryan, Silver also has 1 electron in its outer shell, but it's so unreactive....

 

You are quie correct - Silver does have one electron in its outer shell

 

Silvers electron configuration is: [ce][Kr]4d^{10}5s^1[/ce] and Gold also has one electron in its outer shell: [ce][Xe]4f^{14}5d^{10}6s^1[/ce]

 

But they can both be reactive at times too. For example gold does react with Hydrochloric acid buts its equilibrium of reaction is shifted more twards the redeposition of the gold due to the lack of the formation of stable gold ions - the same is true for silver.

 

Maybe one of the chemistry experts can give a better explination but I know that those are the reason(s) for the group I & II being reactive

 

Cheers,

 

Ryan Jones

[albertlee]

Ryan, I cant understand the notation you use for both silver and gold...which contain Krypon and Xenon...

 

maybe you can direct me to a brief easy website on this, so I can delve a bit myself?

 

thx

[RyanJ]

Ryan' date=' I cant understand the notation you use for both silver and gold...which contain Krypon and Xenon...

 

maybe you can direct me to a brief easy website on this, so I can delve a bit myself?

 

thx[/quote']

 

Ah, right - I see how that can be consusing. Its just shortand. Writing:

 

[math][Xe]4f^{14}5d^{10}6s^1[/math]

 

is easier than writing:

 

[math]1s^{2}2s^{2}2p^{6}3s^{2 }3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}5p^{6}4f^{14}5d^{10}6s^{1}[/math]

 

Here are two links you may find interesting on this

 

1) http://en.wikipedia.org/wiki/Electron_configuration

2) http://www.chemguide.co.uk/atoms/properties/elstructs.html#top

 

Cheers,

 

Ryan Jones

[woelen]

You are quie correct - Silver does have one electron in its outer shell

 

Silvers electron configuration is: [ce][Kr]4d^{10}5s^1[/ce] and Gold also has one electron in its outer shell: [ce][Xe]4f^{14}5d^{10}6s^1[/ce]

 

But they can both be reactive at times too. For example gold does react with Hydrochloric acid buts its equilibrium of reaction is shifted more twards the redeposition of the gold due to the lack of the formation of stable gold ions - the same is true for silver.

 

Maybe one of the chemistry experts can give a better explination but I know that those are the reason(s) for the group I & II being reactive

 

Cheers' date='

 

Ryan Jones[/quote']

A more important reason of the reactivity of the group I metals is that the energy difference for loosing an electron is much lower for these metals than for e.g. gold or silver.

 

E.g. if you mix K-metal and Cl2-gas, then the situation that you have K(+)/Cl(-) in a lattice is energetically MUCH more favorable than having K-metal and Cl2 around. If you look at the same thing with gold and chlorine, then the energy difference is very small.

 

The same holds for formation of K(+)(aq), when K-metal is added to water. In fact, formation of K(+) ions and subsequent solvation / crystallization releases a lot of energy and hence, the system K/H2O or K/Cl2 or whatever with K/... has a lot of potential energy.

 

So, you have to look at the total reaction (ionization costs energy, but formation of lattice or solvation releases energy).

[jdurg]

but Ryan, Silver also has 1 electron in its outer shell, but it's so unreactive....

 

Silver and Gold, and Platinum as well, both have 1 electron in their 6s shell, but the 6s shell is NOT their outer shell. Because of the energy levels of the 6s, 4f, and 5d levels, the 4f and 5d subshells are actually HIGHER in energy than the the 6s shell is. As a result, you would say that the 4f and 5d shells are the outer shells of gold, while 6s is actually an inner shell. As a result, it is not very reactive because the 6s shell is held closer to the nucleus than the 4f and 5 d shells are. The 4f and 5d shells are completely filled, so they are not going to be reactive at all.

 

If you take a look at this link, http://www.webelements.com/webelements/elements/text/Au/econ.html, you'll see the relativel levels of these shells and you'll see that gold's lone electron is closer to the nucleus than the filled 4f and 5d shells. This is even more pronounced in silver where the 5s shell is closer to the nucleus than the 4d shell is.

 

Cesium, like gold, has a lone 6s electron. The difference is, Cesium has ZERO electrons in its 4f and 5d subshell. As a result, the 6s electron IS the outermost electron for cesium. In gold, that's simply not the case.

[RyanJ]

@jdurg: Oh... I'm still learning about electron configurations so please forgive my ignorance

 

Cheers,

 

Ryan Jones

[jdurg]

No problem. Electron configuration, as boring and tedious as it may be to learn, is probably the most vital piece of knowledge in chemistry. By knowing how the electron shells fill up, you'll be able to see how reactive something is, how easily it is to pull those electrons away, whether or not the element forms a positive or negative ion more easily, etc. etc. I too was shocked that gold and silver were so unreactive, yet they had a lone 6s1 electron. Then I saw that the lone electron is pushed pretty deeply into the atom and isn't an outer shell one like it is in the alkali metals. It also makes me think that the elements in row 7 and 8 are probably INCREDIBLY unreactive. That is, if those elements' electrons fill up in the same manner. I state this because they would then have the g level filled up and further protecting that lone electron from reacting. So the analog of gold that has a full g-shell probalby wouldn't react at all if it was stable enough to exist.

[5614]

And the fact that all this electron config is QM and just generally physics, ahh, I love it!

[woelen]

....I too was shocked that gold and silver were so unreactive, yet they had a lone 6s1 electron. Then I saw that the lone electron is pushed pretty deeply into the atom and isn't an outer shell one like it is in the alkali metals....

I agree with you that in gold the 6s1 electron is not the electron at the highest energy level, and so, from an energetic point of view it is not the outermost one, but is this also true from a geometric point of view? Is the 6s1 orbital deeper (i.e. closer to the nucleus on average) than the 4f and 5d levels. I always thought that the principal quantum number (the shell) is a kind of measure of the size of an orbital, albeit a fairly fuzzy one.

 

And the fact that 4f and 5d levels are very close from an energetic point of view, compared to 6s1 only is true as far as I remember, due to the presence of the other electrons.

 

Suppose you have an imaginary atom, with just one electron. For such an electron, the 2s orbital and the 2p orbitals have the same energy level. In a real atom, such as carbon, however, the 2s orbital has a somewhat lower level of energy than the three 2p orbitals, and this is due to some repulsive forces of the electrons, already present in the 2s orbital. Please correct me (or refine my ideas) if you think that is appropriate.

[jdurg]

Well the energy levels define the amount of energy that an electron has, correct? So if energy level A is 'higher' than energy level B, the electrons in shell A would have more energy than B, correct? If an electron has more energy, it would be further from the nucleus because the energy it has is what keeps it away from the nucleus. (1s electrons have very low energy which is why they are so close to the nucleus. They simply don't have the 'oomph' to move away). Remember that the 6s1 designation is simply a solution to a mathematical equation. It doesn't denote that it's a higher energy level than the 5d10 electron is.

[builgate]

Well the energy levels define the amount of energy that an electron has, correct? So if energy level A is 'higher' than energy level B, the electrons in shell A would have more energy than B, correct? If an electron has more energy, it would be further from the nucleus because the energy it has is what keeps it away from the nucleus. (1s electrons have very low energy which is why they are so close to the nucleus. They simply don't have the 'oomph' to move away). Remember that the 6s1 designation is simply a solution to a mathematical equation. It doesn't denote that it's a higher energy level than the 5d10 electron is.

 

 

Since the 6s1 does not show the relative energy level, what is the significance of the principal number? How energy level n=5 is different from n=6 ? I can understand that 6s1 is not the outer shell so gold and silver are not reactive.

[jdurg]

The signifigance of the principal number just defines what solution of the equation you have. All of the electron 'shells' are really just solutions to a mathematical equation. 6s1 and 5s1 are just ways to differentiate between the different solutions. Think of a quadratic equation where you have two possible solutions. 6s1 is the term given for one answer, and 5s1 is the term given for a different answer. As to which one has more energy, that is determined experimentally.

[woelen]

The signifigance of the principal number just defines what solution of the equation you have. All of the electron 'shells' are really just solutions to a mathematical equation. 6s1 and 5s1 are just ways to differentiate between the different solutions. Think of a quadratic equation where you have two possible solutions. 6s1 is the term given for one answer, and 5s1 is the term given for a different answer. As to which one has more energy, that is determined experimentally.

This is a little bit too much simplifying to my opinion. The quantum numbers certainly do have a meaning.

Principal --> energy level

angular --> symmetry

magnetic --> instance corresponding to a symmetry plane

 

Also keep in mind, that the solutions here are solutions to differential equations and not solutions to algebraic equations. This is a huge difference. The symbols we assign to these solutions are just things to make handling of them more easy. Each orbital is a single possible solution to the set of partial differential equations, describing the electrons around an atom. Such an orbital is described as a function of space (x, y, z) and it is a measure for the chance-density for finding a particle at a certain point. Computing the chance of finding a particle in a certain volume V of space requires the computation of a three-dimensional space integral [math]K\int\int\int_V\psi^T\psi dx dy dz[/math], where K is some factor and [math]\psi[/math] represents the solution of the wave partial differential equation.

 

Frequently, orbitals are drawn as surfaces around the most compact regions of space, which cover 90% of the total chance of presence for a particle.

 

Under some assumptions about the shape of the solutions, the differential equations can be transformed to an eigenvalue problem, which is of the form [math]Hx = \lambda x[/math]

[5614]

Frequently, orbitals are drawn as surfaces around the most compact regions of space, which cover 90% of the total chance of presence for a particle.

I've always been taught standard definition of an orbital is the space in which you are 95% likely to find an electron.

[woelen]

I've always been taught standard definition of an orbital is the space in which you are 95%[/b'] likely to find an electron.

Apparently there are different standard definitions . I personally think it does not really matter. More important is that you know that this is just a way to visualize orbitals. In reality they are very abstract mathematical concepts. I think that the shape of the volumes of space with the 90% and 95% definitions do not differ that much.