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svetlana [45]
3 years ago
7

When metals combine with nonmetals, the metallic atoms tend to 1. lose electrons and become positive ions 2. lose electrons and

become negative ions 3. gain electrons and become positive ions 4. gain electrons and become negative ions?
Chemistry
2 answers:
Vika [28.1K]3 years ago
5 0
The correct answer is 1. Lose electrons and become positive ions.


I hope my answer was beneficial to you! c:
Andrews [41]3 years ago
4 0

Answer:

1) lose electrons and become positive ions

Explanation:

To be able to solve this question, it is crucial to understand generally the chemistry of metals and non metals.

Generally metals are found in groups 1,2,3 in the periodic table. What these  tells us is that they have 1,2,3 electrons in their outermost shells. whenever metals want to go into reactions to complete their octet, it is much more feasible energy wise to lose those few electrons than accept about 5,6,7 more electrons.

When a compound loses its electron(s), it becomes a positive ion due to the net increase of protons compared to electrons.

For instance, calcium has 20 electrons and protons. However when it loses 2 electrons to complete its octet and become Ca2+, it still retains its 20 protons but the total number of electrons left is 20 -2 = 1 8. The difference between the protons and the electrons is usually what is represented by the charge on the ion.

However, the reverse is the case when it comes to nonmetals. Non metals would rather accept the few electrons as it is much feasible energy wise to complete its octet. Also, when  non metals gain electrons they would become negative ions due to the excess in number of electrons compared to protons.

As such, the only logical answer would be 1) lose electrons and become positive ions.

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4 years ago
Consider the halogenation of ethene, where x is a generic halogen: h2c=ch2(g)+x2(g)→h2xc−ch2x(g) you may want to reference (page
KengaRu [80]
Consider the halogenation of ethene is as follows:
CH₂=CH₂(g) + X₂(g) → H₂CX-CH₂X(g)
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Part a) 
C=C break = +611 kJ
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2 C-Br formed = (2 * -280) = -560 kJ
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Part b)
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8 0
4 years ago
In the reaction 2H2O (1)+ 2Cl^- (aq)= H2(g)+Cl2 (g)+ 2OH^-(aq), which substance is reduced?
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As per the rules, oxidation number of hydrogen in its compounds is +1(except metal hydrides) and the oxidation number of oxygen in its compounds is -2.

The oxidation number in elemental form is zero.

In H_2O , the oxidation number of H is +1 and oxidation number of O is -2. Oxidation number of Cl in Cl^- is -1. On product side, the oxidation number of hydrogen in H_2 is zero and in OH^- the oxidation number of H is +1 and that of O is -2. Oxidation number of Cl in Cl_2 is 0.

From above data, Oxidation number of O is -2 on both sides so it is not reduced.

Oxidation number of Cl is changing from -1 to 0 which is oxidation.

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8 0
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