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sdas [7]
3 years ago
6

Write a Lewis structure for each atom or ion. Draw the particle by placing atoms on the grid and connecting them with bonds. Inc

lude all lone pairs of electrons and non-bonding electrons. Show the charge of the atom. Particles: S2-, Mg, Mg+2, P.
Chemistry
1 answer:
andrew11 [14]3 years ago
8 0

Answer:

The Lewis structure to this question can be described as follows:

Explanation:

Structure of Lewis for  S^{2-}:  

The maximum number of electrons from valence in S^{2-}  is 8 (6 from S as well as 2 from negative change).  

The valence electrons in the Lewis structure are placed on four sides of the atom.  

Thus the structure of Lewis for S^{2-} is as follows:

\left[\begin{array}{ccc} &. .&\\: &S&:\\&. .&\end{array}\right] ^{2-}

Lewis Mg Structure:  

Complete valence electrons are 2 in Mg.  

The Lewis structure for Mg, therefore, is as follows:

\ . \\ Mg\\ \ .

The Lewis structure for  Mg^{2+}

The maximum valence of electrons   Mg^{2+} in is=  0.

Thus, the structure for   Mg^{2+} is as follows:

 Mg^{2+}

Lewis structure for P :

The maximum number of valence electrons in P is = 5.

Thus, the structure for P is=

\ \ \ . \\ : P \ : \\

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<u>Answer:</u> The final equation has hydroxide ions which indicate that the reaction has occurred in a basic medium.

<u>Explanation:</u>

Redox reaction is defined as the reaction in which oxidation and reduction take place simultaneously.

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when the oxidation number of a species increases.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when the oxidation number of a species decreases.

The given redox reaction follows:

MnO_4^-(aq)+NO_2^-(aq)\rightarrow MnO_2(s)+NO_3^-(aq)

To balance the given redox reaction in basic medium, there are few steps to be followed:

  • Writing the given oxidation and reduction half-reactions for the given equation with the correct number of electrons

Oxidation half-reaction: NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-

Reduction half-reaction: MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-

  • Multiply each half-reaction by the correct number in order to balance charges for the two half-reactions

Oxidation half-reaction: NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-         ( × 3)

Reduction half-reaction: MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-             ( × 2)

The half-reactions now become:

Oxidation half-reaction: 3NO_2^-+6OH^-\rightarrow 3NO_3^-+3H_2O+6e^-

Reduction half-reaction: 2MnO_4^-+4H_2O+3e^-\rightarrow 2MnO_2+8OH^-

  • Add the equations and simplify to get a balanced equation

Overall redox reaction: 3NO_2^-+2MnO_4^-+H_2O\rightarrow 3NO_3^-+2MnO_2+2OH^-

As we can see that in the overall redox reaction, hydroxide ions are released in the solution. Thus, making it a basic solution

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