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Ira Lisetskai [31]
3 years ago
12

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

Chemistry
1 answer:
Stells [14]3 years ago
3 0

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

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Explanation:

Net ionic equation:

In net ionic equation we only write the ions that are involved in reaction. If the system have same moles of ions in initial and final stages we cancel them as they have the same amount and are present in ionic form in the reaction medium. To formulate an ionic equation we just cancel the ions which have the same moles in initial and final stages.

Chemical equation:

Pb(NO₃)₂ (aq) + Na₂CO₃(aq) → PbCO₃ (s) + NaNO₃ (aq)

Balanced chemical equation:

In a balanced chemical equation we write the reactants and products in molecular form with number of moles.

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Pb⁺² +2NO₃⁻¹ + CO₃⁻² + 2Na⁺¹  → PbCO3 (s) + 2NO₃⁻¹ (aq) + 2Na⁺¹ (aq)

Net ionic equation

In net ionic equation we cancel the ions that have equal moles on both sides. As we can see in the above ionic equation that Na⁺¹ and NO₃⁻¹ have equal moles on both sides so we canceled them.

Pb⁺²(aq) + CO₃⁻²(aq)  → PbCO₃ (s)

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