d and f Block Elements

The third ionisation energy of Mn is too high due to stable half filled 3d-orbital. Hence, it cannot further get oxidised to liberate hydrogen.

Complex	Metal	Oxidation state
${[Fe{(CN)_6}]^{2 - }}$ and ${[Cu{(CN)_4}]^{2 - }}$	Fe and Cu	+4 and +2
${[FeC{l_4}]^ - }$ and $C{o_2}{O_3}$	Fe and Co	+3 and +3
${[NiC{l_4}]^{2 - }}$ and ${[CoC{l_4}]^{2 - }}$	Ni and Co	+2 and +2
$Mn{O_2}$ and $Cr{O_2}C{l_2}$	Mn and Cr	+4 and +6

In order to determine which pair of compounds have metals in their highest oxidation states, we need to consider the common oxidation states of the metals involved, as well as the rules for determining oxidation states in compounds and complex ions. Let's evaluate each option:

Option A: MnO₂ and CrO₂Cl₂

In MnO₂, manganese (Mn) has an oxidation state of +4. This is not the highest oxidation state manganese can achieve; it can go up to +7, as seen in KMnO₄ (potassium permanganate).

In CrO₂Cl₂, chromium (Cr) has an oxidation state of +6, which is indeed its highest oxidation state.

Thus, this option contains one metal in its highest oxidation state (Cr) but not the other (Mn).

Option B: [NiCl₄]^2- and [CoCl₄]^2-

For both [NiCl₄]^2- and [CoCl₄]^2-, the nickel (Ni) and cobalt (Co) are in a +2 oxidation state. Neither of these represents the highest oxidation state for these metals. Nickel can have a +3 state, and cobalt can go up to +4 in very rare cases.

Option C: [Fe(CN)₆]^3- and [Cu(CN)₄]^2-

In [Fe(CN)₆]^3-, the iron (Fe) has an oxidation state of +3, which is a common high oxidation state for iron, though iron can also exist in a +2 state commonly and +6 in very rare conditions. Therefore, +3 is considered a high oxidation state but not the absolute highest.

In [Cu(CN)₄]^2-, copper (Cu) has an oxidation state of +2, which is the highest stable oxidation state for copper in most of its compounds.

Option D: [FeCl₄]^- and Co₂O₃

In [FeCl₄]^-, iron (Fe) has an oxidation state of +3. As mentioned, +3 is a high state for iron, but not the absolute highest oxidation state it can achieve.

In Co₂O₃, cobalt (Co) has an oxidation state of +3, as evidenced by the formula, where two Co atoms interact with three O atoms (each oxygen providing a -2 charge, for a total of -6, requiring each Co to be +3 to balance the charge). +3 is indeed among the highest common oxidation states for cobalt, although, as mentioned, Co can technically reach +4.

Conclusion: Based on above analysis, while none of the options perfectly fits the criteria of both elements being in their absolute highest oxidation states (especially considering the rare or less common states), Option A and Option D come closest, with chromium and cobalt being in their common high oxidation states in the provided compounds. However, the question asks for the highest oxidation state, and strictly speaking, none of the options perfectly satisfy the condition for both elements included. Among the choices, Option A is the closest because chromium is in its highest oxidation state of +6 in CrO₂Cl₂, and manganese in MnO₂ is in a +4 state, which while not its absolute highest, represents a commonly encountered high oxidation state. Given the context and conventional exams' focus on common oxidation states, Option A would be the best fit despite the slight discrepancy with the question's phrasing.

Zinc oxide (ZnO) when react with Na₂O it act as acid while with CO₂ it act as base. Therefore, it is an amphoteric oxide.

(a) ZnO + Na₂O $\to$ Na₂ZnO₂
     Acid      Base          Salt

(b) ZnO + CO₂ $\to$ ZnCO₃
     Base      Acid       Salt

Compounds which dissolve in both acids and bases are known as amphoteric oxides.

Amphoteric oxides are :

Cr₂O₃, BeO, SnO, SnO₂, ZnO, Al₂O₃ , PbO and PbO₂ Whereas, NO is a neutral oxide, B₂O₃ is an acidic oxide and CrO is a basic oxide.

(a) Conc. HNO₃ makes iron passive. Cold relatively concentrated HNO₃ will react with Fe.

$\mathop {Fe}\limits_{Iron} + \mathop {6HN{O_3}}\limits_{Nitric\,acid} \to \mathop {Fe{{(N{O_3})}_3}}\limits_{Iron\,nitrate} + \mathop {3N{O_2}}\limits_{Nitrogen\,dioxide} + \mathop {3{H_2}O}\limits_{Water} $

(b) $\mathop {Cu}\limits_{Copper} + \mathop {4HN{O_3}(conc.)}\limits_{Nitric\,acid} \to \mathop {Cu{{(N{O_3})}_2}}\limits_{Copper\,nitrate} + \mathop {2N{O_2}}\limits_{Nitrogen\,dioxide} + \mathop {2{H_2}O}\limits_{Water} $

(c) $\mathop {Zn}\limits_{Zinc} + \mathop {2NaO{H_{(aq)}}}\limits_{Sodium\,hydroxide} \to \mathop {N{a_2}Zn{O_2}}\limits_{Sodium\,zincate} + \mathop {{H_2}}\limits_{Hydrogen} $

(d) $\mathop {4Au}\limits_{Gold} + \mathop {8NaCN}\limits_{Sodium\,cyanide} + \mathop {{O_2}}\limits_{Oxygen} + \mathop {2{H_2}O}\limits_{Water} \to \mathop {4Na[Au{{(CN)}_2}]}\limits_{Sodium\,dicyanoaurate} + \mathop {4NaOH}\limits_{Sodium\,hydroxide} $

On addition of impurities to the colourless gemstones, brilliant colours are produced. On addition of chromium to colourless corundum (Al₂O₃), a red ruby is formed. Red colour ruby is obtained when Cr³⁺ ions replace Al³⁺ ions in the octahedral sites of corundum. A green emerald emerges when chromium is added to colourless beryl (Be₃Al₂(SiO₃)₆). Green colour emerald is obtained when Cr³⁺ ions replace Al³⁺ in the octahedral site of beryl.

MnO₄²⁻ disproportionate as (in neutral or acidic solution)

3MnO₄²⁻ + 4H⁺ → 2MnO₄²⁻ + MnO₂ + 2H₂O

Out of all the four given metallic oxides $Cr{O_2}$ is attracted by magnetic field very strongly. The effect persists even when the magnetic field is removed. Thus $Cr{O_2}$ is metallic and ferromagnetic in nature.

Consider the reactions of the given compounds with $\mathrm{S}^{2-}$ and $\mathrm{SO}_4^{2-}$.

(i) With $\mathrm{CuCl}_2$ :

$ \begin{aligned} &\mathrm{CuCl}_2+\mathrm{S}^{2-} \rightarrow \underset{\text { black ppt }}{\mathrm{CuS}}+2 \mathrm{Cl}^{-} \\\\ &\mathrm{CuCl}_2+\mathrm{SO}_4{ }^{2-} \rightarrow \text { No precipitate } \end{aligned} $

(ii) With $\mathrm{BaCl}_2$ : Alkaline earth metal sulphides are sparingly soluble in water but pass into solution with time and hence do not give precipitates from their aqueous solutions.

$ \begin{aligned} &\mathrm{BaCl}_2+\mathrm{S}^{2-} \rightarrow \underset{\text { No ppt }}{\mathrm{BaS}}+2 \mathrm{Cl}^{-} \\\\ &\mathrm{BaS}+2 \mathrm{H}_2 \mathrm{O} \rightleftharpoons \underset{\text { water soluble }}{\mathrm{Ba}(\mathrm{OH})_2+{\mathrm{Ba}(\mathrm{SH})_2}} \end{aligned} $

Alkaline earth metal sulphates, $\mathrm{BaSO}_4, \mathrm{CaSO}_4$ and $\mathrm{SrSO}_4$ are insoluble in water.

$\mathrm{BaCl}_2+\mathrm{SO}_4^{2-} \rightarrow \underset{\text { white ppt }}{\mathrm{BaSO}_4}+2 \mathrm{Cl}^{-}$

(iii) With $\mathrm{Pb}(\mathrm{OAc})_2$ :

$ \begin{aligned} &\mathrm{Pb}(\mathrm{OAc})_2+\mathrm{S}^{2-} \rightarrow \underset{\text { black ppt }}{\mathrm{PbS}}+2 \mathrm{CH}_3 \mathrm{COO}^{-} \\\\ &\mathrm{Pb}(\mathrm{OAc})_2+\mathrm{SO}_4^{2-} \rightarrow \underset{\text { white ppt }}{\mathrm{PbSO}_4}+2 \mathrm{CH}_3 \mathrm{COO}^{-} \end{aligned} $

The $K_{\mathrm{sp}}$ of $\mathrm{PbS}$ is $3 \times 10^{-28}$ and of $\mathrm{PbSO}_4$ is $25 \times$ $10^{-28}$. The difference is large, thus only $\mathrm{PbS}$ is selectively precipitated.

(iv) With $\mathrm{Na}_2\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]$ : When sodium nitroprusside is added to solution of sulphide ions, purple colouration is observed. This is a confirmative test for sulphides.

$ \begin{aligned} &{\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]^{2-}+\mathrm{S}^{2-} \rightarrow \underset{\text { Purple colour }}{\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]}} \\\\ &\mathrm{Na}_2\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]+\mathrm{SO}_4{ }^{2-} \rightarrow \text { No precipitate } \end{aligned} $

$A{g^ + }\mathop {2{S_2}O_3^{2 - }}\limits_{(Excess)} \to \mathop {{{[Ag{{({S_2}{O_3})}_2}]}^{3 - }}}\limits_{Argentothiosulphate\,(X)\,(Soluble)} \buildrel {A{g^ + }} \over \longrightarrow \mathop {A{g_2}{S_2}{O_3}}\limits_{(Y)\,(White\,ppt.)} \mathrel{\mathop{\kern0pt\longrightarrow} \limits_{(due\,to\,hydrolysis)}^{With\,time}} \mathop {A{g_2}S}\limits_{(Z)\,(Black\,ppt.)} $

$L \to M$ charge transfer spectra. $KMn{O_4}$ is colored because it absorbs light in the visible range of electromagnetic radiation. The permanganate ion is the source of color, as a ligand to metal, $\left( {L \to M} \right)$ charge transfer takes place between oxygen's $p$ orbitals and the empty $d$-orbitals on the metal. This charge transfer takes place when a photon of light is absorbed, which leads to the purple color of the compound.

To match the catalysts to the correct processes :

1. $TiCl_3$ is well-known as a catalyst in Ziegler-Natta polymerization which is used for the production of polymers of olefins. So, $(A) - (ii)$.




2. $PdCl_2$ is used in the Wacker process, which is used to convert ethene to acetaldehyde. Hence, $(B) - (i)$.




3. $V_2O_5$ is used in the Contact process to manufacture sulfuric acid by catalyzing the oxidation of sulfur dioxide to sulfur trioxide. Therefore, $(D) - (iii)$.




4. $CuCl_2$ is used in Deacon's process for the oxidation of hydrochloric acid to chlorine. Hence, $(C) - (iv)$.

Comparing these pairings with the options given, we find that the correct option is :

$ \left( A \right) - \left( {ii} \right),\,\,\left( B \right) - \left( {i} \right),\,\,\left( C \right) - \left( {iv} \right),\,\,\left( D \right) - \left( {iii} \right) $

The pairs Cu²⁺, Pb²⁺ and Hg²⁺, Bi³⁺ are precipitated as sulphides upon passing H₂S in acidic medium.

The correct statements are as follows:

(A) Cr³⁺ is more stable than Cr²⁺, thus, Cr²⁺ act as reducing agent.

(B) Mn²⁺ is more stable than Mn³⁺, thus, Mn³⁺ act as an oxidizing agent.

(C) Both Cr²⁺ and Mn³⁺ exhibit d⁴ electronic configuration.

In equation

(i) $\,\,\,$ $F{e_2}{\left( {S{O_4}} \right)_3}\,\,\,$

and in equation

(ii) $\,\,\,$ $F{e_2}{\left( {S{O_4}} \right)_3}$

on decomposing will form oxide instead of $Fe.$

The correct sequence of reactions is

$Fe\buildrel {{O_2},heat} \over \longrightarrow \,\,F{e_3}{O_4}\buildrel {Co,{{600}^ \circ }C} \over \longrightarrow $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,F{e_2}{\left( {S{O_4}} \right)_3}\buildrel \Delta \over \longrightarrow Fe$

To understand which of the given reagents will give copper metal upon heating with $Cu_2S$, let's discuss each one based on their chemical reactions during the process:

Option A: CuFeS₂

$CuFeS_2$ (Chalcopyrite) when heated with $Cu_2S$, it can react to form copper metal as part of the copper extraction process. However, the direct reaction between $CuFeS_2$ and $Cu_2S$ is more involved and typically requires a series of steps, including smelting and conversion to get to the copper metal. Therefore, while $CuFeS_2$ is a source of copper, it's not directly because of its reaction with $Cu_2S$, but through a complex metallurgical process.

Option B: CuO

$ \begin{aligned} & 4 \mathrm{CuO} \xrightarrow{\Delta} 2 \mathrm{Cu}_2 \mathrm{O}+\mathrm{O}_2 \\\\ & \mathrm{Cu}_2 \mathrm{~S}+2 \mathrm{Cu}_2 \mathrm{O} \xrightarrow{\Delta} 6 \mathrm{Cu}+\mathrm{SO}_2 \end{aligned} $

Option C: Cu₂O

Copper(I) oxide ($Cu_2O$) can also react with $Cu_2S$ to produce copper metal directly. The equation for this reaction is:

$2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$

Here, $Cu_2O$ serves as the oxidizing agent, reducing $Cu_2S$ to copper metal, while itself being reduced to copper, and sulfur is oxidized to sulfur dioxide. This reaction is similar in principle to the reaction with $CuO$, but involves copper(I) oxide instead.

Option D: CuSO₄

$\begin{array}{r}\mathrm{CuSO}_4 \xrightarrow{720^{\circ} \mathrm{C}} \mathrm{CuO}+\mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2 \\\\ 4 \mathrm{CuO} \xrightarrow{1100^{\circ} \mathrm{C}} 2 \mathrm{Cu}_2 \mathrm{O}+\mathrm{O}_2 \\\\ 2 \mathrm{Cu}_2 \mathrm{O}+\mathrm{Cu}_2 \mathrm{~S} \xrightarrow{\Delta} 6 \mathrm{Cu}+\mathrm{SO}_2\end{array}$

Reaction of alkali metals with ammonia depends upon the physical state of ammonia whether it is in gaseous state or liquid state. If ammonia is considered as a gas then reaction will be

(a) $Na + \mathop {N{H_3}}\limits_{(Excess)} \to NaN{H_2} + {1 \over 2}{H_2}$

(NaNH₂ + 1/2H₂ are diamagnetic) If ammonia is considered as a liquid then reaction will be

$M + (x + y)N{H_3} \to {[M{(N{H_3})_x}]^ + } + {[e{(N{H_3})_y}]^ - }$

$\bullet$ ammoniated electron

$\bullet$ blue colour

$\bullet$ paramagnetic

$\bullet$ very strong reducing agent

(b) $K + \mathop {{O_2}}\limits_{(Excess)} \to \mathop {K{O_2}({K^ + },O_2^ - )}\limits_{Potassium\,\sup eroxide\,Paramagnetic} $

(c) $3Cu + 8HN{O_3} \to \mathop {3Cu{{(N{O_3})}_2}}\limits_{Paramagnetic} + \mathop {2NO}\limits_{Paramagnetic} + 4{H_2}O$

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

(1)

V = 3d34s2	V2+ = 3d3	= 3 unpaired electron
Cr = 3d54s1	Cr2+ = 3d4	= 4 unpaired electron
Mn = 3d54s2	Mn2+ = 3d5	= 5 unpaired electron
Fe = 3d64s2	Fe2+ = 3d6	= 4 unpaired electron

hence the correct order of paramagnetic behaviour

${V^{2 + }} < Cr{}^{2 + } = F{e^{2 + }} < M{n^{2 + }}$

(2) $\,\,\,\,$

For the same oxidation state, the-ionic radii generally $dc$-creases as the atomic number increases in a particular transition series.

Hence the order is

$M{n^{ + + }} > F{e^{ + + }} > C{o^{ + + }} > N{i^{ + + }}$

(3)

In solution, the stability of the compounds depends upon electrode potentials, $SEP$ of the transitions metal ions are given as

$C{o^{3 + }}/Co = + 1.97,\,\,\,F{e^{3 + }}/Fe = + 0.77;$

$C{r^{3 + }}/C{r^{2 + }} = - 0.41,\,\,S{c^{3 + }}$ is highly stable as it does not show $+2$ $O.$ $S.$

(4)

$Sc - \left( { + 2} \right),\left( { + 3} \right)$

$Ti - \left( { + 2} \right),\left( { + 3} \right),\left( { + 4} \right)$

$Cr - \left( { + 1} \right),\left( { + 2} \right),\left( { + 3} \right),\left( { + 4} \right),\left( { + 5} \right),\left( { + 6} \right)$

$Mn - \left( { + 2} \right),\left( { + 3} \right),\left( { + 4} \right),\left( { + 5} \right),\left( { + 6} \right),\left( { + 7} \right)$

$i.e.\,Sc < Ti < Cr = Mn$

${E_a} = 53598.6J/mol$

$ = 53.6\,kJ/mol.$

On treatment with H₂S, Zn(II) gets precipitated as ZnS. Mg(II) does not get precipitated, while Fe³⁺ and Al³⁺ get precipitated as hydroxides.

$F{e^{3 + }}$ is easily hydrolysed than $F{e^{2 + }}$ due to more positive charge.

The first reaction is

${K_3}[Fe{(CN)_6}] + \mathop {KI}\limits_{excess} \to {K_4}[Fe{(CN)_6}] + \mathop {K{I_3}}\limits_{brownish\,yellow\,solution} $

This is a redox reaction as both oxidation and reduction are taking place.

$\mathop {{K_4}[Fe{{(CN)}_6}]}\limits_{soluble} \, + ZnS{O_4} \to \mathop {{K_2}Z{n_3}{{[Fe{{(CN)}_6}]}_3}}\limits_{white\,ppt} + NaOH \to \mathop {N{a_2}{{[Zn{{(OH)}_4}]}^{2 - }}}\limits_{soluble} $

$2I_3^ - + 2N{a_2}{S_2}{O_3} \to \mathop {N{a_2}{S_4}{O_6}}\limits_{clear\,solution} + 2NaI + \mathop {2{I_2}}\limits_{(blue\,colour)} $

When we consider the color of light absorbed by a substance, we are referring to the complementary color of the light that is transmitted or reflected by the substance. An aqueous solution of copper(II) sulfate (CuSO₄) appears blue to our eyes. This color results from the absorption of light in the complementary color range.

The visible spectrum of light can be divided into several regions, each corresponding to a different color. The complementary color of blue is orange-red. This means that the aqueous solution of CuSO₄ absorbs light in the orange-red region of the spectrum.

Therefore, the correct answer is:

Option A: orange-red

The number stable oxidation states of lanthanides is $+3$.

The configuration of $Gd$ is $4{f^7}\,5{d^1}\,6{s^2}.$

Lower oxidation state of an element forms more basic oxide and hydroxide, while the higher oxidation state will form more acidic oxide/hydroxide. For example,

Most of the $L{n^{3 + }}\,\,$ compounds except $L{a^{3 + }}$ and $L{u^{3 + }}$ are coloured due to the presence of $f$-electrons.

The titration of oxalic acid with $KMnO{}_4$ in presence of $HCl$ gives unsatisfactory result because of the fact that $KMn{O_4}$ can also oxidise $HCl$ along with oxalic acid. $HCl$ on oxidation gives $C{l_2}$ and $HCl$ reduces $KMnO{}_4$ to $M{n^{2 + }}$ thus the correct answer is $(c).$

NOTE : The main reason for exhibiting larger number of oxidation states by actinoids as compared to lanthanoids is lesser energy difference between $5f$ and $6d$ orbitals as compared to that between $4f$ and $5d$ orbitals.

NOTE : More the distance between nucleus and outer orbitals, lesser will be force of attraction on them. Distance between nucleus and $5f$ orbitals is more as compared to distance between $4f$ orbital and nucleus. So actinoids exhibit more number of oxidation states in general than the lanthanoids.

$4f$ orbital is nearer to nucleus as compared to $5f$ orbital therefore, shielding of $4f$ is more than $5f.$

The main reason of lanthanide contraction is, poor shielding of 4f electrons on the outer most shell electrons. As f orbital is defused nature so shielding of 4f electron is imperfect that is why nucleus can attract outer most shell electron easily and the size of atom decreases.

In lanthanides, there is poorer shielding of $5d$ electrons by $4f$ electrons resulting in greater attraction of the nucleus over $5d$ electrons and reduction of the atomic radii.

Metal atom in the lower oxidation state forms the ionic bond and in the higher oxidation state the covalent bond. Because higher oxidation state means small size and great polarizing power and hence greater the covalent character. Hence $MC{l_2}$ is more ionic than $MC{l_4}.$

The number of unpaired electrons in $N{i^{2 + }}\left( {aq} \right) = 2$

Water is weak ligand hence no pairing will take place spin magnetic moment

$\eqalign{ & = \sqrt {n\left( {n + 2} \right)} = \sqrt {2\left( {2 + 2} \right)} \cr & = \sqrt 8 = 2.82 \cr} $

$2C{u_2}O + C{u_2}S\buildrel \, \over \longrightarrow 6Cu + S{O_2}$ self reduction

$H{g_2}C{l_2} + 2N{H_4}OH\buildrel \, \over \longrightarrow HgN{H_2}Cl + N{H_4}Cl + 2{H_2}O$

$C{r_2}O_7^{2 - } + 6{{\rm I}^ - } + 14{H^ + }\buildrel \, \over \longrightarrow 3{{\rm I}_2} + 7{H_2}O + 2C{r^{3 + }}$

oxidation state of $Cr$ is $3+.$

The main reason of lanthanide contraction is, poor shielding of 4f electrons on the outer most shell electrons.

In lanthanides, there is poorer shielding of $5d$ electrons by $4f$ electrons resulting in greater attraction of the nucleus over $5d$ electrons and reduction of the atomic radii.

In the entire d-block,

Only in group 3, the size increase from top to bottom due to increase of no of shells in an atom. That is why size of Sc < Y < La. And shell size in group 3 is 3d < 4d < 5d.

But from group 4 to 12, the size of 3d < 4d but 4d $ \cong $ 5d due to lanthanide contraction. That is why Zr and Hf have about the same radius.

$\left( {n - 1} \right){d^5}n{s^2}\,\,$ attains the maximum $O.S.$ of $+7$

The $+4$ oxidation state of cerium is also known in solution.

$4\mathop {KI}\limits^{ - 1} + 2CuS{O_4} \to \mathop {{I_2}}\limits^0 + C{u_2}{I_2} + 2{K_2}S{O_4}$

$\mathop {{I_2}}\limits^0 + 2N{a_2}\mathop {{S_2}}\limits^{2 + } {O_3} \to N{a_2}\mathop {{S_4}}\limits^{ + 2.5} {O_6} + \mathop {2NaI}\limits^{ - 1} $

In this $Cu{I_2}$ is not formed

When a solution of potassium chromate is treated with an excess of dilute nitric acid. Potassium dichromate and ${H_2}O$ are formed.

$2{K_2}Cr{O_4} + 2HN{O_3} \to {K_2}C{r_2}{O_7} + 2KN{O_3} + {H_2}O$

Hence $C{r_2}{O_7}^ - $ and ${H_2}O$ are formed.

Fe (Z = 26) $\to$ e^- = 26

Fe²⁺ (Z = 26) $\to$ e^- = 24

Electronic configuration:

Fe (Z = 26) $\to$ 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s²

Fe²⁺ (Z = 26) $\to$ 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s⁰

as 4s is the outermost shell of Fe that's why e^- will always removed first from 4s, so retained e^- in d-shell is 6

When $K{\rm I}$ is added to mercuric iodide it dissolve in it and form complex.

$\mathop {Hg{{\rm I}_2}}\limits_{red,solid(inso{\mathop{\rm lub}} le)} \,\, + \,\,\,K{\rm I} \to \mathop {{K_2}\left[ {Hg{{\rm I}_4}} \right]}\limits_{(soluble)} $

On heating $Hg{{\rm I}_2}$ decomposes as

$Hg{{\rm I}_2}\,\,\,\,Hg\,\,\rightleftharpoons\, + \mathop {{{\rm I}_2}}\limits_{(violetvapours)} $