To determine the relative stability of isomers, it is helpful to consider the molecular structure, bond types, and electronegativity of the atoms involved.
First, let's analyze the three isomers of NOBr, OBrN, and ONBr:
1. NOBr: In this isomer, nitrogen (N) is bonded to oxygen (O) via a double bond, and bromine (Br) is attached to nitrogen. The molecule can be represented as O=N-Br.
2. OBrN: In this isomer, oxygen is bonded to bromine, and nitrogen is attached to bromine. The molecule can be represented as O-Br=N.
3. ONBr: In this isomer, oxygen is bonded to nitrogen, and bromine is attached to nitrogen. The molecule can be represented as O-N=Br.
To determine the most stable isomer, consider the electronegativity difference between the atoms:
1. Nitrogen is more electronegative than bromine, so the double bond in NOBr is polar, with the oxygen having a partial positive charge and the bromine having a partial negative charge.
2. Oxygen is more electronegative than bromine, so the bond in OBrN is polar, with the nitrogen having a partial positive charge and the bromine having a partial negative charge.
3. Since bromine is more electronegative than nitrogen, the bond in ONBr is polar, with the oxygen having a partial negative charge and the nitrogen having a partial positive charge.
Now, let's consider the stability of these isomers based on the electronegativity difference and the overall electron arrangement.
Since oxygen is more electronegative than nitrogen, the oxygen atom in ONBr will stabilize the partial positive charge on nitrogen through electron donation. This results in a more stable arrangement compared to the other two isomers.
Therefore, based on the analysis, ONBr is the most stable isomer among NOBr, OBrN, and ONBr due to the electron-donating effect of oxygen stabilizing the partial positive charge on nitrogen.