Valence electrons

[shaq786]

You have the configuration 4s2 3d1.

How many valence electrons do you have? is it 2 or 3?

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[Mowgli]

Three.

 

[shaq786]

Yea I figured hehe.

What about 3s2 4d5???

isnt that 2? becuase you have half filled orbitals which then move closer to the nucleas allowing 3s2 to shine in the outside.

 

[abcxyz0123]

Three.

thats right

 

[shaq786]

What about 3s2 4d5???

isnt that 2? becuase you have half filled orbitals which then move closer to the nucleas allowing 3s2 to shine in the outside.

 

[sdnstud]

this is a trick question!!!

the two electrons in the 3s orbital are NOT valance electrons

the five electrons in the 4d orbital MAY be valance electrons.

To figure out how many total valance electrons, you need to know how many electrons are in 5s orbital. Also, you can't assume there are 2 electrons in the 5s orbital because the atom can be excited.

Yea I figured hehe.

What about 3s2 4d5???

isnt that 2? becuase you have half filled orbitals which then move closer to the nucleas allowing 3s2 to shine in the outside.

 

[shaq786]

Good analysis SDN, but I accidentally misphrased the question. But I'll keep your explanation in mind for the tricky question I accidentally posted...But heres what I was asking:

You have a 4s2 3d5....How many valence electrons? wouldnt it be 2? Becuase 3d5 gets pulled into the nucleas becuase of the phenomenon of half filled orbitals.

What about 4s2 3d10?? How many valences?

 

[QofQuimica]

Good analysis SDN, but I accidentally misphrased the question. But I'll keep your explanation in mind for the tricky question I accidentally posted...But heres what I was asking:

You have a 4s2 3d5....How many valence electrons? wouldnt it be 2? Becuase 3d5 gets pulled into the nucleas becuase of the phenomenon of half filled orbitals.

What about 4s2 3d10?? How many valences?

No, assuming the atom is the ground state, the 3d electrons are still valence electrons even though the 4s would be emptied first if you were ionizing the atom. So there are seven in the first case, 12 in the second.

 

[[email protected]]

No, assuming the atom is the ground state, the 3d electrons are still valence electrons even though the 4s would be emptied first if you were ionizing the atom. So there are seven in the first case, 12 in the second.

I don't think there are 12 valence electrons in the second case of 4s2 3d10. An atom's valence refers to the number of bonds it can form due to available electrons. Once a subshell becomes full then it becomes unlikely that those electrons will participate in the formation of bonds. So you only count d subshell electrons if the subshell is not full.

So 4s2 3d5 = 7 valence electrons BUT 4s2 3d10 = 2 valence electrons.

That's how remember it. I think it's a little abstract but maybe worth clarifying. Anybody else back me up?

 

[HippocratesX]

mit is correct. If you just think about it with logic for a moment, its impossible to have 12 VALENCE electrons.

 

[fpr85]

8 is max per el, except for the first one; 2, 🙂

 

[Medikit]

There is no such thing as valence electrons. Or orbitals 🙁

 

[HistoRocks]

You have the configuration 4s2 3d1.

How many valence electrons do you have? is it 2 or 3?

Three valence electrons. And you have a transition metal, scandium. s and p correspond to the orbitals of the representative elements, while f (14 electrons in the valence shell) is for cesium and beyond. When you see d, immediatly think "transition metal," and visualize that middle portion of the periodic table.

 

[QofQuimica]

I don't think there are 12 valence electrons in the second case of 4s2 3d10. An atom's valence refers to the number of bonds it can form due to available electrons. Once a subshell becomes full then it becomes unlikely that those electrons will participate in the formation of bonds. So you only count d subshell electrons if the subshell is not full.

So 4s2 3d5 = 7 valence electrons BUT 4s2 3d10 = 2 valence electrons.

That's how remember it. I think it's a little abstract but maybe worth clarifying. Anybody else back me up?

Well, maybe I misunderstood the OP; if s/he meant to ask about the valency of Zn (i.e., how many bonds it can form or how many electrons it tends to lose) then you are correct; it does usually only lose two (although Zn 3+ is known; the third electron is lost from the 3d orbitals). But if the question is about how many valence electrons there are versus how many core electrons there are, then all of the 4s and 3d electrons will be valence electrons for the 3d block of transition metals; their core is [Ar]. I interpreted the question in the second sense, which is why I answered 12.

Calling an electron a "valence electron" doesn't necessarily mean that it can be involved in bonding. To give a simpler example from the main group elements, consider fluorine. Its core is [He] and we say it has seven valence electrons (from the 2s and 2p orbitals) although those 2s electrons won't ever come off or form bonds, and generally F tends to want to gain an electron to have eight in the second quantum level. Moving on to neon, we'd say it has 8 valence electrons, although neon doesn't really form any compounds.

The transition metals are trickier, since the 3d and 4s orbitals are so close to each other in energy that these elements can lose electrons from either level, and most of them have multiple valencies. They often tend to lose their 4s electrons first, but again, they CAN also lose some of their 3d electrons as well, which is why they too are considered to be valence electrons.

 

[QofQuimica]

One other example I thought of as soon as I had already hit the "send" button on the last message is the 18-electron rule. It is kind of like the octet rule, but it applies to transition metals that are forming coordination complexes with various ligands. It states that stable complexes will be ones that surround the transition metal with 18 valence electrons, because that is the configuration of the noble gas for that period (Kr in this case). Of course, there are exceptions to the 18-electron rule just as there are to the octet rule, but I'm mentioning it because it also supports the idea of calling all of the 4s and 3d electrons "valence electrons".

 

[MedSchoolFool]

I just finished taking a chemistry test on this material. According to my teacher and my text the number of valence electrons would be 2. I think that for almost all the transition metals the valence electrons are 2. The coinage metals may be different.

but I'm sure this is a trick question as was alluded to earlier.