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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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Answer:

The correct answer is 10.939 mol ≅ 10.94 mol

Explanation:

According to Avogadro's gases law, the number of moles of an ideal gas (n) at constant pressure and temperature, is directly proportional to the volume (V).

For the initial gas (1), we have:

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For the final gas (2), we have:

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We calculate n₂ as follows:

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