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Scorpion4ik [409]
2 years ago
9

Which of the following is most likely to form multiple (double or triple) bonds? 1. Li 2. Cl 3. N 4. H 5. F

Chemistry
1 answer:
alukav5142 [94]2 years ago
3 0

Answer: Option (3) is the correct answer.

Explanation:

Atomic number of lithium is 3 and its electronic distribution is 2, 1. So, to attain stability it will loose an electron and hence, it forms a single bond.

Atomic number of chlorine is 17 and it has 7 valence electrons. Hence, in order to attain stability it will gain one electron and therefore, it forms a single bond only.

Atomic number of nitrogen is 7 and its electronic distribution is 2, 5. Therefore, to attain stability it needs to gain 3 more electrons. Hence, a nitrogen atom is able to form a triple bond and also it is able to form a double bond.

Hydrogen has atomic number 1 and it attains stability by gaining one electron. Therefore, a hydrogen atoms always forms a single bond.

Atomic number of fluorine is 9 and its electronic distribution is 2, 7. To complete its octet it needs to gain one electron. Hence, a fluorine atom always forms a single bond.

Thus, we can conclude that out of the given options nitrogen is most likely to form multiple (double or triple) bonds.

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C. 49.8 grams of ki is dissolved in enough water to make 1.00 l of solution. what is the molarity?
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Hey There!:


Molar Mass KI => 166.003 g/mol


* number of moles:


n = mass of solute / molar mass


n = 49.8 / 166.003


n = 0.3 moles KI


Therefore:


M = n / V


M = 0.3 / 1.00


M = 0.3 mol/L


hope this helps!

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