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3 years ago

35 Which element reacts with oxygen to form ionic bonds?(1) calcium (3) chlorine

Chemistry

2 answers:

QveST [7]3 years ago

8 0

Covalent bonds exist between two nonmetal elements while ionic bonds exist between a metal and a nonmetal element. In this case, oxygen is a nonmetal and in order to create an ionic bond, the other substance should be a metal. The answer therefore to this problem is calcium.

kow [346]3 years ago

3 0

$\boxed{\left( 1 \right){\text{ calcium}}}$ reacts with oxygen to form ionic bonds.

Further Explanation:

Covalent bonds are the types of bonds that involve the sharing of electrons between the bonded atoms. This bond is usually formed between two or more non-metals.

Ionic bonds are the chemical bonds that involve the transference of electrons from one atom to another. These bonds are generally formed between metals and non-metals. In these types of bonds, ions are formed by the atoms. Cations are formed by the loss of electrons and anions result from the gain of electrons.

Oxygen is a non-metal. In order to form an ionic bond with oxygen, the other element has to be a metal.

Hydrogen, chlorine, and nitrogen are non-metals so these can form covalent bonds with oxygen. But calcium is a metal and it forms ionic bonds with oxygen. Calcium loses two electrons to form ${\text{C}}{{\text{a}}^{2 + }}$ and these electrons are gained by oxygen, resulting in the formation of ${{\text{O}}^{2 - }}$ . The attraction between ${\text{C}}{{\text{a}}^{2 + }}$ and ${{\text{O}}^{2 - }}$ is an ionic bond and the resulting compound is CaO (For structure, refer to the attached image).

Learn more:

Identification of ionic bonding: brainly.com/question/1603987
What type of bond exists between phosphorus and chlorine? brainly.com/question/81715

Answer details:

Grade: High School

Chapter: Ionic and covalent compounds

Subject: Chemistry

Keywords: ionic bond, covalent bond, nitrogen, hydrogen, chlorine, calcium, oxygen, CaO, metal, non-metal.

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The number electrons transferred in the reaction are, 10 electron.

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Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Rules for the balanced chemical equation in basic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the more number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydroxide ion $(H^+)$ at that side where the less number of hydrogen are present.

Now balance the charge.

The given chemical reaction is,

$2BrO_3^-+5SnO_2^{2-}+H_2O\rightarrow 5SnO_3^{2-}+Br_2+2OH^-$

The oxidation-reduction half reaction will be :

Oxidation : $SnO_2^{2-}\rightarrow SnO_3^{2-}$

Reduction : $BrO_3^-\rightarrow Br_2$

First balance the main element in the reaction.

Oxidation : $SnO_2^{2-}\rightarrow SnO_3^{2-}$

Reduction : $2BrO_3^-\rightarrow Br_2$

Now balance oxygen atom on both side.

Oxidation : $SnO_2^{2-}\rightarrow SnO_3^{2-}+H_2O$

Reduction : $2BrO_3^-+H_2O\rightarrow Br_2$

Now balance hydrogen atom on both side.

Oxidation : $SnO_2^{2-}+OH^-\rightarrow SnO_3^{2-}+H_2O$

Reduction : $2BrO_3^-+3H_2O\rightarrow Br_2+12OH^-$

Now balance the charge.

Oxidation : $SnO_2^{2-}+OH^-\rightarrow SnO_3^{2-}+H_2O+2e^-$

Reduction : $2BrO_3^-+3H_2O+10e^-\rightarrow Br_2+12OH^-$

The charges are not balance on both side of the reaction. We are multiplying oxidation reaction by 5 and then added both equation, we get the balanced redox reaction.

Oxidation : $5SnO_2^{2-}+5OH^-\rightarrow 5SnO_3^{2-}+5H_2O+10e^-$

Reduction : $2BrO_3^-+3H_2O+10e^-\rightarrow Br_2+12OH^-$

The balanced chemical equation in basic medium will be,

$2BrO_3^-+5SnO_2^{2-}+H_2O\rightarrow 5SnO_3^{2-}+Br_2+2OH^-$

The number electrons transferred in the reaction are, 10 electron.

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