I have learnt that elements on the left-hand side of the periodic table such as sodium and magnesium prefer to lose electrons to form a cation because this requires less energy to obtain a stable octet, and vice-versa for the right-hand side of the periodic table e.g. fluorine. However, using this reasoning I am not sure why all transition metals tend to lose electrons rather than gain them.
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$\begingroup$ chemistry.stackexchange.com/a/34134/9961 - there is also tendency for anions. $\endgroup$– MithoronCommented Jul 23, 2015 at 15:55
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2 Answers
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See the outer configuration of some transition elements:
- Sc: $\mathrm{[Ar]~3d^1 4s^2}$
- Cr: $\mathrm{[Ar]~3d^5 4s^\color{red}{1}}$
- Co: $\mathrm{[Ar]~3d^7 4s^2}$
- Cu: $\mathrm{[Ar]~3d^{10} 4s^\color{red}{1}}$
- Zn: $\mathrm{[Ar]~3d^{10} 4s^2}$
If you are wondering about configuration of Cu and Cr read why this happens!
As you may notice, they can form ions by either losing or gaining electron in 4s orbital. In the chemistry of the transition elements, the 4s orbital behaves as the outermost, highest energy orbital. When these metals form ions, the 4s electrons are always lost first, leading to a positive charge on ion. Gaining any electron in 4s orbital would decrease the stability of anion thus formed. So, they form only cations (positive ions).
In general, the outer electronic configuration of transition elements is $n~\mathrm{s}^2~(n-1)\mathrm{d}^{1-10}$. Because of reasons above, the electron removes from the $\ce{ns}$ orbital, where n = principal quantum number of atom.
See $\ce{Zn^{2+}}$ configuration: $\mathrm{[Ar] (3d)^{10}}$ as an example. The electron exits from 4s orbital, leading to the formation of a positively charged ion.
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1$\begingroup$ Please tell me if I understand this correctly: the 4s orbital is higher in energy than the 3d orbital, so the metal will prefer to lose electrons from this orbital to lower its energy. These transition metals are not close to having a full outer shell unlike elements such as fluorine which only require 1 electron to gain a full outer shell, so fluorine gains an electron to become more stable. I think my problem was that I got orbitals and shells confused, thinking 4s and 3d were shells rather than orbitals in a shell. $\endgroup$– k--Commented Jul 24, 2015 at 3:52
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$\begingroup$ Yes you get it right. Transition elements don't form ions with complete outer shell. For ex. $Co^{2+}$ has the configuration $[Ar] 3d^7$. $\endgroup$– anshabhiCommented Jul 24, 2015 at 11:51
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$\begingroup$ In simple words, s,p,d,f are orbitals and 1,2,3,4.. are shells $\endgroup$– anshabhiCommented Jul 24, 2015 at 11:51
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$\begingroup$ I know this will probably start a confusion, just to add to your answer, lower TM's like Pt can form anions (Pt- and Pt2-) In Pt's case the relativistic stabilisation of the 6s electrons allows the negative oxidation states $\endgroup$ Commented Apr 26, 2018 at 11:12
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As you may notice, they can form ions by either losing or gaining electron in 4s orbital. In the chemistry of the transition elements, the 4s orbital behaves as the outermost, highest energy orbital. When these metals form ions, the 4s electrons are always lost first, leading to a positive charge on ion.
