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I was studying about the periodic table recently, and was reading a topic associated with oxides of halogens, and came across the following line

The bromine oxides, $\ce{Br2O}$, $\ce{BrO2}$, $\ce{BrO3}$ are the least stable halogen oxides (Middle row anomaly) and exist only at low temperatures. They are very powerful oxidizing agents.

So, I went to search this on Google, and found these lines

It refers to the instability of oxides of bromine as compared to relative stability of oxides of chlorine and Iodine at room temperature, the former being stable only at low temperatures.

But the line above is simply restating what the book had already told. So, is their any specific reason why this happens, or is this only due to the experimental data we have gathered?

Gaurang Tandon
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Rajat Jain
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1 Answers1

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This is due to the transition metal contraction.

Bromine has the electron configuration $\ce{[Ar] 4s^{2} 3d^{10} 4p^{5}}. $The 3d orbital has no radial nodes and is therefore quite contracted (close to the nucleus), so there is relatively little repulsion between the 3d electrons and the 4p electrons. This makes it much harder to acheive high oxidation states of bromine (I, IV, VI for the examples you give) because the ionization energies are higher than you might expect from a simplistic approach to periodic trends.

J. LS
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    Why doesn't that argument apply to iodine as well? – ron May 09 '15 at 17:56
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    @ron he is correct, it does not apply for iodine as 4d orbitals have 1 radial node, while 3d does not have them. – Rajat Jain May 09 '15 at 23:38
  • It probably applies for astatine too due to the lanthanide contraction but I'm not sure if the redox chemistry has been investigated yet. – J. LS May 10 '15 at 10:10
  • So this argument should rather be made stating the high-than-expected effective nuclear charge in the $\rm{4p}$ and $\rm{5p}$ rows due to the presence of the d block in the fourth and fifth periods, which means 10 additional protons without perfect shielding in each case? The anomalous size of the $\rm{3d}$ orbitals might explain why bromine has the least stable oxides, but I'm never very comfortable with electron subshell repulsion arguments. – Nicolau Saker Neto May 10 '15 at 11:12
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    Your explanation to me seems equivalent. What is shielding other than electron-electron repulsion ? – J. LS May 10 '15 at 11:28
  • Fair enough, I actually hadn't made that connection in my mind before. Shielding is just a way to variationally compensate for the addition of the electron-electron repulsion term in the hamiltonian, after all. – Nicolau Saker Neto May 11 '15 at 12:02
  • can anyone cite any other examples of this middle row anomally? – Nilay Ghosh Jul 16 '15 at 09:39
  • @NilayGhosh Instability of AsCl5. Reduction potentials of H2SeO4. Next time, I highly recommend you ask a new question instead of asking in comments though. – orthocresol Oct 30 '15 at 13:52
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    @orthocresol, If I ask this as a new question, it would be regarded as duplicate. – Nilay Ghosh Oct 30 '15 at 16:32
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    @NilayGhosh It would only be marked as a duplicate if there is another question that specifically lists manifestations of the alternation effect, or middle row anomaly, or whatever you want to call it. This question does not deal with examples of the phenomenon, but rather the reason behind it. – orthocresol Oct 30 '15 at 17:10