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?
Answer
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.
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