Question

Charge distribution in iodine chloride can be best represented as:
A. ${{I}^{+}}C{{l}^{-}}$
B. ${{I}^{\delta +}}-C{{l}^{\delta -}}$
C. ${{I}^{-}}C{{l}^{+}}$
D. ${{I}^{\delta -}}-C{{l}^{\delta +}}$

Answer 1

Hint: In ideal covalent bond the electrons shared stays right between the nucleuses of two atoms. But when there is electronegativity difference between two atoms these electrons slightly move towards the higher electronegativity atoms and that generates partial positive charge on less electronegativity atom and partial positive charge on high electronegativity atom and such covalent bond is called as polar covalent bond.

Complete step by step answer:
Iodine and chlorine both are the elements of group 17 called halogen. They both are non-metal with atomic numbers 17 and 53.
Valence shells of iodine and chlorine both contain 7 electrons and to have an octet like a noble gas both need one electron and they both are non-metals so sharing an electron will be a perfect option for them.
So they form a covalent sharing one electron between each-other and both the electrons are supposed to be right between their nucleuses.
But as we know that electronegativity of chlorine is higher than electronegativity of iodine, so these electrons slightly shift towards chlorine giving it a partial negative charge and iodine acquires partial positive charge.
Therefore, their charge distribution must be like
${{I}^{\delta +}}-C{{l}^{\delta -}}$

So the correct answer is Option (B).

Additional information:
Iodine is a lustrous purple-black solid which is found in seafoods like fish, seaweed, shrimp etc. Chlorine is a gas and its major source for humans is sodium chloride.

Note: Magnitude of partial charges on atoms depend upon electronegativity difference. Higher the electronegativity difference, higher will be the magnitude of partial charges on the atoms. In fact rough approximation suggests that when we have electronegativity differences of higher than $1.8$, bonds will become completely ionic. Generally for the electronegativity difference of $0.4$ to $1.8$ makes the bond polar covalent.