The correct answer is $\boxed{\left[ {\mathop I\limits^{ + \delta } …\mathop O\limits^{ + \delta } {H_2}} \right]^ + }$.
Benzene cannot be iodinated with $${I_2}$$ directly because the aromatic ring is very stable and does not react with electrophiles. However, in presence of oxidants such as HNO3, iodination is possible. The oxidant abstracts a hydrogen atom from benzene, creating a radical cation. The radical cation is then attacked by the electrophilic iodine molecule, forming the iodonium ion. The iodonium ion is a good leaving group, and it is displaced by a nucleophile, such as water, to form the iodobenzene product.
The electrophile in this reaction is the iodonium ion, $\left[ {\mathop I\limits^{ + \delta } …\mathop O\limits^{ + \delta } {H_2}} \right]^ + $. The iodonium ion is formed when the oxidant abstracts a hydrogen atom from benzene, creating a radical cation. The radical cation is then attacked by the electrophilic iodine molecule, forming the iodonium ion. The iodonium ion is a good leaving group, and it is displaced by a nucleophile, such as water, to form the iodobenzene product.
The other options are not correct because they are not electrophiles. Option A, $\left[ {{I^ + }} \right]$, is a cation, but it is not an electrophile. Option B, $\left[ {{I^ – }} \right]$, is an anion, and it is not an electrophile. Option C, ${\left[ {\mathop I\limits^{ + \delta } …\mathop O\limits^{ – \delta } {H_2}} \right]^ + }$, is a zwitterion, and it is not an electrophile.