Question

In the Lewis structure of $IC{l}_2^{-}$, how many lone pairs of electrons are around the iodine atom?

Answer 1

Hint: In $IC{l}_2^{-}$, the central atom is iodine which is surrounded by two chlorine atoms. The electronic configuration of iodine is $[Kr]4d^{10}5s^{2}5p^5$and the electronic configuration of chlorine is $[Ne]3s^{2}3p^5$.

Complete step by step answer:
Lewis structure is defined as the representation for the arrangement of valence electrons surrounding each atom present in the molecule.
The compound $IC{l}_2^{-}$ is an anion, where the central iodine atom is attached with two chlorine atoms by a single bond and a negative charge is present on the overall molecule. The atomic number of iodine is 53 and the electronic configuration of iodine is $[Kr]4d^{10}5s^{2}5p^5$. The valence electrons in iodine is 7 where two electrons take part in bonding and six electrons are present as three lone pairs. The atomic number of chlorine is 17 and the electronic configuration of chlorine is $[Ne]3s^{2}3p^5$. The valence electrons in chlorine is 7. As two chlorine atoms are present one electron from each chlorine takes part in bonding and each chlorine atom contains 6 electrons as three lone pairs. The negative charge shows that one electron is gained by iodine atom.
The Lewis structure of $IC{l}_2^{-}$ is shown below.

Therefore, In the Lewis structure of $IC{l}_2^{-}$, 3 lone pairs of electrons are around the iodine atom.

Note:
The relative atomic mass has no units. The calculation of relative formula mass is done the same as the calculation of molecular weight where atomic mass of the atoms are added but the unit of molecular weight is g/mol.