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lyudmila [28]
3 years ago
5

Ionic bonds are formed between metal and nonmetal elements. Potassium has only one valence electron and it is stripped away in t

he process of forming an ionic bond. Nonmetals, such as chlorine, tend to gain valence electrons in the formation of ionic bonds. Determine which element(s) are likely to have their electrons stripped away in the formation of ionic bonds. I) aluminum II) sulfur III) oxygen IV) silver V) neon
Chemistry
2 answers:
lisabon 2012 [21]3 years ago
6 0

Answer:

The answer is alimunum and silver.

Explanation:

An ionic bond is produced between metallic and non-metallic atoms, where electrons are completely transferred from one atom to another. During this process, one atom loses electrons and another one gains them, forming ions. Usually, the metal gives up its electrons forming a cation to the nonmetal element, which forms an anion.

As mentioned, metallic elements are those that tend to give up electrons, which bind with atoms that have the ease of accepting them, which are non-metallic elements. Then, <u><em>aluminum and silver, being metals, are the most likely elements to eliminate their electrons in the formation of ionic bonds. </em></u>

Sulfur, oxygen, being nonmetals, will tend to accept electrons. Neon is a noble gas, also called inert, and having its last layer of electrons complete, tends not to react with other elements.

Nonamiya [84]3 years ago
4 0
Aluminum and Silver, as they are metals. Metals are more likely to lose valence electrons.
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Although both compounds are weak electrolytes, NH₃ is somewhat stronger base than HCN is a strong acid, so the solution reacts alkaline. We can prove this using Ka and Kb values:

Ka(HCN) = 4.9 x × 10⁻¹⁰

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Let's first calculate Ka for NH₄⁺: 
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</span>Ka(NH₄⁺) = Kw/Kb(NH₃) = 5.6 x 10⁻¹⁰

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Kb(CN⁻) x Ka(HCN) = pKw

Kb(CN⁻) = Kw/Ka(HCN) = 2 x 10⁻⁵

 
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