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I just noticed that on the periodic table it doesn't say how many valence electrons there are for each column in the d block.

How do I find out how many valence electrons elements in the d block have?

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  • $\begingroup$ Firstly, it depends on what you count as "valence electrons". If you say that, for all the d-block metals, the ns and (n-1)d electrons count as "valence electrons", then the answer is to just look at the group number. However, that obviously doesn't work for Zn, which effectively only has 2 valence electrons. That's really an extreme case though. It is sometimes said that the d electrons transition from being valence electrons in the early d-block (hence Sc only forms $\ce{Sc^3+}$), to being core electrons in the late d-block (hence Zn only forms $\ce{Zn^2+}$). $\endgroup$ – orthocresol Oct 20 '15 at 16:28
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The simplest answer is: d-block elements have number of valence electrons equal to their group number, which is equal to the number of electrons in the "valence shell". This works if you are using the definition of valence shell to be the outermost shell.

But it is wrong to apply the concept to determine the valency of the element, because that varies with the compound it is present in. For example, iron can show two valencies, $2$ and $3$ in the compunds iron(II) sulphate ($\ce{FeSO4}$) and iron(III) sulphate ($\ce{Fe2(SO4)3}$) respectively. But from the above definition, iron has $8$ valence electrons.

As a matter of fact, this applies even to many elements outside d-block too, for example chlorine typically shows a valency of ($-1$) in compunds like $\ce{NaCl}$, but it can also show a different valency (more appropriately, oxidation state) in ions like perchlorate anion ($\ce{ClO4^-}$), where its valency is $\mathbf 7$ (while the oxidation state is $\mathbf{+7}$, notice the difference that there is no plus or minus sign in valency).

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d- block elements have 1-10 electrons in the d- shell. but the (n+1)th S outer shell has either 2 or 1 electron so as to complete half filled or full filled states of the elements which have one electron less than completion of that state.

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    $\begingroup$ This also isn't really true. There are multiple exceptions to your rule; $\ce{Nb, W, Ru, Sg}$; to name but a few. The real rule is the electrons occupy the orbitals that give the lowest energy configuration. Calculating what this configuration is is a complex topic. $\endgroup$ – bon Oct 20 '15 at 15:49
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Here is the list of Transition Metals and the generalized information about their valence electrons. The valence electrons (VE) are the electrons in the outer shell of an atom. In the above image valence electrons column imparts about the different orbits of an atom (But again this is a generalized rule there are some exceptions). So in order to find the valence electron you have to check the last digit of the series. at Comparisonofmetals.com you can compare and explore more about these metals with the help of various parameters like their various properties, facts, uses, periodic table details, etc. I came across this website while i was working on my science project. They have some unique and interesting features like comparison of metals, Metals Quiz, plot graphs, compare statistics and much more! I hope it will be helpful. enter image description here

For example, Scandium metal's electrons per shell are 2, 8, 9, 2 and its outer shell has only 2 electrons. Hence, the valence electron count of scandium is 2.

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    $\begingroup$ Hi Pravin, as is, the response here doesn't really answer the question. The table just lists numbers without saying which orbital the electrons are in. $\endgroup$ – John Snow May 17 '16 at 7:05
  • $\begingroup$ Hi John Snow, to answer your query; the valence electrons (VE) are the electrons in the outer shell of an atom. In the above image valence electrons column imparts about the different orbits of an atom. So in order to find the valence electron you have to check the last digit of the series. For example, Scandium metal's electrons per shell are 2, 8, 9, 2 and its outer shell has only 2 electrons. Hence, the valence electron count of scandium is 2. $\endgroup$ – Pravin Mastud May 17 '16 at 9:37
  • $\begingroup$ This is not true, because the transition metals can easily lose electrons from their d orbitals as well as the outer s orbitals. This question cannot be answered without considering the orbital electronic configurations of the elements. $\endgroup$ – bon May 17 '16 at 14:05
  • $\begingroup$ Hi Bon, your are right that we can not generalize this rule because there are some exceptions. The image attached above gives generalize idea about the valence electrons of d block element. But if want it the precise information, please refer this site (comparisonofmetals.com) Just for an example, the valence elctrons of Gold is 1 according to the table. Gold (Au) is a d-block element and has 1 valence electron (sometimes 3 or even 11 depending on how “valence electron is defined.) $\endgroup$ – Pravin Mastud May 18 '16 at 10:52
  • $\begingroup$ Gold (Au) is a d-block element and has 1 valence electron (sometimes 3 or even 11 depending on how “valence electron is defined.) However Gold most often has chemical reactions with 1 or 3 valence electrons. Reactions with three valence electrons would be marked as gold(III.) If the formula doesn’t connote the usage of gold(III), then gold with one valence electron could be considered. $\endgroup$ – Pravin Mastud May 18 '16 at 10:57
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Elements in the d block are transition elements, and each posses one or two valence electrons in their respective s orbitals. Most have two, with several notable exceptions: elements may "steal" an electron from the outermost s block and relocate it to the d block in order to reach a filled or half-filled ($5$ of $10$ electrons) state in the d block. These elements must be taken on a case-by-case basis.

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    $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the help center. $\endgroup$ – Martin - マーチン Oct 20 '15 at 5:23

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