Question

1. Give reasons for the following:
a. $M{n^{3 + }}$is a good oxidizing agent.
b. $E_{{M^{2 + }}/M}^ \circ$ values are not regular first-row transition metals (3d series).
c. Although F is more electronegative than O the highest Mn fluoride is $Mn{F_4}$, whereas the highest oxides is $M{n_2}{O_7}$
2. Complete the following equations:
a. $2C{r_{}}O_4^{2 - } + 2{H^ + } \to$
b. $KMn{O_4}\xrightarrow{\Delta }$

Answer 1

1 a. Let’s recall the definition of an oxidizing agent, an oxidizing agent is an electron acceptor, meaning if there is a chemical reaction taking place, an oxidizing agent would take up electrons from the other reactant, resulting in its oxidation number to decrease.
b. $E_{{M^{2 + }}/M}^ \circ$ value is called the standard electrode potential of metal undergoing reduction(${M^{2 + }}$ 2{e^\\_} \to M). Now understand that metals are electropositive, meaning they would prefer to get a positive charge on them(${M^{n + }}$). Hence, expect negative $E_{{M^{2 + }}/M}^ \circ$ values for metals.
c. Fluorine and oxygen both are forming bonds with manganese, there could be something related to the bond formation property, which is the governing factor in the above statement.

2 a. Try to visualize the standard chemical reaction for the above, which is given in the ionic form, the above reaction is in between a chromate ($Cr$) compound and ${H^ + }$ion, the source of this ${H^ + }$ion is probably an acid.
b. The above reaction involves a chemical reaction of potassium permanganate ($KMn{O_4}$) with heat($\Delta$), therefore, it is a thermal decomposition reaction, recall what happens in a thermal decomposition reaction.

Complete step by step solution:
1 a. To understand why$M{n^{3 + }}$is a good reducing agent, first, write down the electronic configuration of $Mn$ and $M{n^{3 + }}$.
Electronic configuration of $Mn$ $\to \left[ {Ar} \right]4{s^2}3{s^5}$
Electronic configuration of $M{n^{3 + }}$ $\to \left[ {Ar} \right]4{s^0}3{d^4}$
Now, we can clearly understand that if Manganese accepts an electron it will attain the${d^5}$ configuration which is a half-filled electronic configuration resulting in more stability, remember in chemistry everything tries to attain stability of some form or the other.
By accepting an electron the $M{n^{3 + }}$species converts to $M{n^{2 + }}$which has greater stability as it is a half-filled configuration. Therefore, by definition of the oxidizing agent, we can conclude that $M{n^{3 + }}$is an electron acceptor, hence, a good oxidizing agent.

Additional information:
i. An oxidizing agent itself undergoes reduction(accepting electrons).
ii. A reducing agent itself undergoes oxidation(giving away electrons).

b. First, look at the definition of the standard electrode potential, it is the measured potential of an electrode under the standard conditions(temperature-273k, pressure-1atm) in a half cell redox reaction.
$E_{{M^{2 + }}/M}^ \circ$ value gives us the total potential needed for the conversion of ${M^{2 + }} + 2{e^ - } \to M$ the metal atom. Various factors determine this value, such as 1st and 2nd ionization enthalpies of the atom in gaseous state(${\Delta _i}{H_1} + {\Delta _i}{H_2}$)and the enthalpy of sublimation.
Due to the presence of irregularity in the ionization enthalpies(${\Delta _i}{H_1} + {\Delta _i}{H_2}$) values, and relatively lower enthalpy of sublimation for Mn and V atoms. One can observe the irregularity in the standard electrode potential trend in the first-row transition metal series(3d).

Additional information:
i. Ionization enthalpy for an isolated atom at a gaseous state is the amount of energy required to remove an electron from its outermost shell. Energy is released during this reaction, hence, it is an exothermic reaction.
ii. Enthalpy of sublimation is the energy required to convert one mole of a substance in a solid-state to its gaseous state. It involves enthalpy of fusion (solid to liquid) and enthalpy of vaporization(liquid to gas).

c. Here, both fluorine and oxygen tend to stabilize the higher oxidation state of Manganese(Mn), but oxygen exceeds because of its ability to form multiple bonds ($\pi$bonds) with Mn. The ability to form double bonds will give the $M{n_2}{O_7}$ molecule a relatively smaller and spatial structure.
Whereas, fluorine can only form sigma($\sigma$) type bonds with Mn. Even if fluorine tries to form $Mn{F_7}$, there will be an increase in electron-electron repulsion between fluorine atoms, resulting in the formation of highly unstable $Mn{F_7}$, hence, it does not exists.

Additional information:
i. Electronegativity is the tendency of an element to attract the shared paired electron towards itself. The three most electronegative elements are Fluorine, Oxygen, and Nitrogen (F>O>N).
ii. Manganese has the maximum number of oxidation states (+2 to +7).

2 a. The given reaction is in ionic form, meaning reactants are converted into their corresponding ions.
Two molecules of Chromite($CrO_4^{2 - }$) and two ${H^ + }$ions are reacting to form dichromate($C{r_2}O_7^{2 - }$ )and water (${H_2}O$)
$2CrO_4^{2 - }$+$2{H^ + }$ $\to C{r_2}O_7^{2 - } + {H_2}O$
Remember it is not a redox reaction as the oxidation number of Cr is not changing.

Additional information:
i. Potassium dichromate is a good oxidizing agent.
ii. The ${H^ + }$ used in the above reaction is from acids like ${H_2}S{O_4}$,$HN{O_3}$, etc.

b. The above reaction shows the thermal decomposition of potassium permanganate($KMn{O_4}$), which results in the formation of potassium manganate(${K_2}Mn{O_4}$) and manganese dioxide($Mn{O_2}$) along with the release of oxygen gas.
The balanced chemical equation is written below;
$2KMn{O_4}\xrightarrow{\Delta }{K_2}Mn{O_4} + Mn{O_2} + {O_2}$

Additional information:
i. Thermal decomposition or thermolysis(thermo meaning heat, lysis meaning to break down) is a chemical reaction in which heat is used to break down the compound. The reaction is usually endothermic as heat is being provided to the reaction mixture.
ii. Potassium permanganate is a strong oxidizing agent.

Note:
1a. The definition of oxidizing agent must not be confused with oxidation, these two are separate concepts. Also, while filling/removing the electrons from the Mn atom the electrons will be filled/removed first from 4s as it has a lower energy state than 3d.
b. Understand the difference between reduction potential and standard electrode potential properly, to avoid confusion.
c. Manganese can accommodate 7 atoms, but understand that stability is what the elements are looking for.

2 a. If the chemical reaction is given in ionic form, always write its answer in ionic form unless it is explicitly mentioned in the question to write it in standard form. Otherwise don’t write it in standard form.
b. If you are not comfortable with your balanced chemical reaction do not write it, unless it is asked in the question. Writing a correct balanced chemical reaction can leave a good impression on the examiner.