Redox Reactions

The reaction of oxygen gas with platinum hexafluoride can be written as:

$\mathrm{O_2 + PtF_6 \rightarrow X^{+}Y^{-}}$

The product formed is:

$\mathrm{O_2^{+}[PtF_6]^{-}}$

Hence, we identify the ions as:

$\mathrm{X^{+} \rightarrow O_2^{+}}$

$\mathrm{Y^{-} \rightarrow [PtF_6]^{-}}$

(A) For $\mathrm{O_2^{+}}$, Bond Order (BO) = 2.5

The molecular orbital electronic configuration of $\mathrm{O_2^{+}}$ is:

$\sigma 1s^2\\ \sigma^{*}1s^2\\ \sigma 2s^2\\ \sigma^{*}2s^2\\ \sigma 2p_z^2\\ \pi 2p_x^2 = \pi 2p_y^2\\ \pi^{*}2p_x^1 = \pi^{*}2p_y^0$

From this configuration, the bond order is calculated as:

$\text{BO} = 2.5$

Therefore, statement A is incorrect as given.

(B) In $\mathrm{O_2^{+}[PtF_6]^{-}}$, oxidation state of Pt = +5

We have $\mathrm{Pt^{+5}}$ with the electronic configuration $5d^5$. The atomic configuration of Pt is $5d^9 6s^1$. Hence, the oxidation state of +5 is correct.

So, statement B is correct.

(C) $\mathrm{PtF_6}$ acts as an oxidising agent

As $\mathrm{O_2}$ is oxidised to $\mathrm{O_2^{+}}$, $\mathrm{PtF_6}$ works as the oxidant that accepts electrons.

Hence, statement C is correct.

(D) No fluorine atom transfer occurs from $\mathrm{PtF_6}$

Since there is no fluorine transfer during the reaction, this statement is incorrect in this context.

Sn²⁺ of stannous chloride dihydrate (SnCl₂ . 2H₂O) tends to convert into Sn⁴⁺.

Hence, statement (a) is correct.

(b) SnO₂ reacts with KOH and gives K₂SnO₃ . 3H₂O or K₂[Sn(OH)₆] because it is amphoteric in nature.

$Sn{O_2} + KOH\buildrel {} \over \longrightarrow {K_2}Sn{O_3} + {H_2}O$

or ${K_2}[Sn{(OH)_6}]$

Hence, statement (b) is correct.

(c) In conc. HCl, PbCl₂ exists as chloroplumbous acid, H₂[PbCl₄]

$\mathop {PbC{l_2}}\limits^{II} + 2HCl\buildrel {} \over \longrightarrow \mathop {{H_2}[PbC{l_4}]}\limits^{II} $

Hence, statement (c) is incorrect.

(d) Pb₃O₄ is a mixture of $(2\mathop {PbO}\limits^{ + 2} + {\mathop {PbO}\limits^{ + 4} _2})$

$P{b_3}{O_4} + 4HN{O_3}\buildrel {} \over \longrightarrow \mathop {2Pb{{(N{O_3})}_2}}\limits^{ + 2} + \mathop {Pb{O_2}}\limits^{ + 4} + 2{H_2}O$

It is not a redox reaction. Thus, the statement (d) is incorrect.

Fe³⁺ is reduced to Fe²⁺ by H₂O₂/NaOH and Na₂O₂/H₂O.

$2F{e^{3 + }} + {H_2}{O_2} + O{H^ - } \to 2F{e^{2 + }} + 2{H_2}O + {O_2}$

$N{a_2}{O_2} + {H_2}O \to {H_2}{O_2} + NaOH$

Oxidation half-reaction,

$2{I^ - } \to {I_2} + 2{e^ - }$ .... (i)

Here, I^$-$ is converted into I₂. Oxidation number of I is increasing from $-$1 to 0 hence, this is a type of oxidation reaction.

Reduction half-reaction

$6{H^ + } + ClO_3^ - + 6{e^ - } \to C{l^ - } + 3{H_2}O$ ..... (ii)

Here, H₂O releases as a product. Hence, option (d) is correct.

Multiplying equation (i) by 3 and adding in equation (ii)


Stoichiometric coefficient of $HSO_4^ - $ is 6.

Hence, option (a), (b) and (d) are correct.

In an aqueous solution, the reduction of the metal center in permanganate ions ($MnO_4^-$) varies depending on the medium :

Acidic Medium : Permanganate acts as a strong oxidizing agent. The reaction can be represented as :

$8H^+ + 5e^- + MnO_4^- \to Mn^{2+} + 4H_2O$

This indicates that 5 electrons are involved in the reduction process in an acidic medium.

Neutral Medium : Permanganate acts as a moderate oxidizing agent. The reaction in a neutral medium is :

$2H_2O + 3e^- + MnO_4^- \to MnO_2 + 4OH^-$

Here, 3 electrons are involved in the reduction process in a neutral medium.

Alkaline Medium : In alkaline conditions, the reaction is similar to that in a neutral medium :

$2H_2O + 3e^- + MnO_4^- \to MnO_2 + 4OH^-$

Again, 3 electrons are involved in the reduction process in an alkaline medium.