Question

In the Lewis structure for ICl2–, how many lone pairs of electrons are around the central iodine atom.

Answer 1

Answer :  The number of lone pairs of electrons around the central iodine atom are, 6 electrons

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The representation of valence electrons of an atom by dots around the symbol of an element is said to be the Lewis-dot symbol.

In the representation of Lewis-symbol, one electron dot must put on each side of the element symbol then paired the electrons.

The given molecule is, $ICl_2^-$

As we know that iodine ans chlorine has '7' valence electrons.

Therefore, the total number of valence electrons in $ICl_2^-$ = 7 + 2(7) + 1 = 22

According to Lewis-dot structure, there are 4 number of bonding electrons and 18 number of non-bonding electrons.

Hence, the number of lone pairs of electrons around the central iodine atom are, 6 electrons

Answer 2

There are three lone pairs of electrons around the central iodine atom in ICl2–.

In the ion, ICl2–, the central atom is iodine. It is necessary to note that in every chemical specie, the central atom is always the least electronegative atom. In the ion shown, the least electronegative atom is iodine.

The Lewis structure of the ICl2– ion is shown in the image attached. We can see that there are three lone pairs of electrons around the central iodine atom in ICl2–.

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