d and f Block Elements

(A) Complementarty color of yellow is violet. As [CoCl(NH₃)₅]²⁺ absorbs yellow so it will show complementary color violet.

(B)The building block of ruby is aluminium oxide (Al₂O₃) containing about 0.5 – 1% Cr³⁺ ions which are randomly distributed in the position normally occupied by Al³⁺ ions.

(C) Total number of unpaired electons in [Fe(H₂O)₆]²⁺ = 4
and total number of unpaired electons in [Cr(H₂O)₆]²⁺ = 4
So, the spin-only magnetic moments are same in both cases.

(D) Total number of unpaired electons in [Ni(NH₃)₄(H₂O)₂]²⁺ = 2
So, $\mu $ = $\sqrt {2\left( {2 + 2} \right)} $ BM = 2.83 BM

The highest possible oxidation states of uranium and plutonium can be identified by considering the electronic configurations of these elements and noting the number of valence electrons available for bonding.

Uranium (U, atomic number 92) has the electronic configuration: $ [Rn] 5f^3 6d^1 7s^2 $

Plutonium (Pu, atomic number 94) has the electronic configuration: $ [Rn] 5f^6 7s^2 $

To find the highest oxidation state, we sum the number of $d$, $f$, and $s$ electrons in the valence shell.

For uranium :

$ 3 (f) + 1 (d) + 2 (s) = 6 $

So the highest oxidation state of uranium is +6.

For plutonium :

$ 6 (f) + 2 (s) = 8 $

So the highest oxidation state of plutonium is +8. However, in practice, plutonium is known to exhibit a highest oxidation state of +7.

Comparing with the given options, the correct answer is :

Option D : 6 and 7.

As we know that

$\mu = \sqrt {n\left( {n + 2} \right)} $ where n = no. of unpaired electrons i.e. greater the no. of unpaired electron more will be the spin-only magnetic moments.

. Electronic configuration of the given transition metal ions are

Sc³⁺ (Z = 21) = 1s²2s²2p⁶3s²3p⁶

$ \therefore $ n = 0

Ti²⁺ (Z = 22) = 1s²2s²2p⁶3s²3p⁶3d²

$ \therefore $ n = 2

Ti³⁺ (Z = 22) = 1s²2s²2p⁶3s²3p⁶3d¹

$ \therefore $ n = 1

V²⁺ (Z = 23) = 1s²2s²2p⁶3s²3p⁶3d³

$ \therefore $ n = 3

So the correct increasing order of magnetic moment is

Sc³⁺ < Ti³⁺ < Ti²⁺ < V²⁺

Actinoid 'Th' shows oxidation state = +4

Actinoid 'Ac' shows oxidation state = +3

Actinoid 'Pu' shows oxidation states = +3, +4, +5, +6, +7

Actinoid 'Np' shows oxidation states = +3, +4, +5, +6, +7

Actinoid 'Bk' shows oxidation states = +3, +4

Actinoid 'Cm' shows oxidation states = +3, +4

Actinoid 'Lr' shows oxidation states = +3

$ \therefore $ Maximum oxidation state is shown by Np and Pu.

Interstitial compound are –

(i) very hard

(ii) chemically inert

(iv) high melting point.

As interstitial compounds are chamically inert.

In lanthanide series from left to right atomic radius decreases but the exception is Europiun (Eu) and Ytterbium (Yb).

So the correct order is : Eu > Ce > Ho > N

$Mn{O_2}(A) \,\,\buildrel {4KOH,{O_2}} \over \longrightarrow \,\,\mathop {2K_2MnO_4(B) }\limits_{\left( {Green} \right)} \,\, + \,\,2{H_2}O$

$3K_2MnO_4(B) \,\,\buildrel {4HCl} \over \longrightarrow \,\,\mathop {2K_2MnO_4(C) }\limits_{\left( {Purple} \right)} \,\, + \,\,Mn{O_2} + 2{H_2}O$

$2K_2MnO_4(C) \,\,\buildrel {{H_2}O.KI} \over \longrightarrow \,\,\mathop {2 Mn{O_2}(A) }\limits_{\left( {Purple} \right)} \,\, + \,\,2KOH\,\, + \,\,KIO_3(D) $

$ \therefore $ A $ \to $ MnO₂

D $ \to $ KIO₃

Sc³⁺ has noble gas configuration hence only +3 exists.

Most common oxidation states of :

(i) Sc : +3

(ii) V : +2, +3, +4, +5

(iii) Ti : +2, +3, +4

(iv) Cu : +1, +2

In lutetium (⁷¹Lu) the 71^st electron enters in 5d-orbital.

Electron configuration is [Xe]4f¹⁴5d¹6s²

Wilkinsion catalyst is [(Ph₃P)₃RhCl]

Due to Lanthanoid contraction both atomic radii and ionic radii decreases gradually in the lanthanoid series.

Since Zn is not a transition element so transition element having lowest atomisation energy out of Cu, V, Fe is Cu.

The explanation of given statements are as follows :

(a) Aqua-regia is prepared by mixing conc. HCl and conc. HNO₃ in 3 : 1 (v/v) ratio and is used in oxidation of gold and platinum. Hence, option (a) is correct.

(b) Yellow colour of aqua-regia is due to its decomposition into NOCl (orange yellow) and Cl₂ (greenish yellow). Hence, option (b) is correct.

(c) When gold reacts with aqua-regia then it produces $AuCl_4^ - $ anion complex in which Au has +3 oxidation state.



Hence, option (c) is correct.

(d) Reaction of gold with aqua-regia produces NO gas in absence of air. Hence, option (d) is incorrect.

$MnO_4^{2 - }$ ion has one unpaired electrons, therefore it gives d-d transition to form green colour. Y complex has paramagnetic nature due to presence of one unpaired electron.

In aqueous solution,



$MnO_4^{2 - }$ ions gives charge transfer spectrum in which a fraction of electronic charge is transferred between the molecular entities.

$ \because $ $\mathop {MnO_4^{2 - }}\limits_{(Y)} \mathrel{\mathop{\kern0pt\longrightarrow} \limits_{oxidation}^{Electrolytic}} \mathop {MnO_4^ - }\limits_{(Z)} + {e^ - }$

In acidic medium, Y undergoes disproportionation reaction.



$\mathop {MnO_4^{2 - }}\limits_{(Y)} $ and $\mathop {MnO_4^ - }\limits_{(Z)} $ both ions form $\pi $-bonding between p-orbitals of oxygen and d-orbitals of manganese.

Thus, options (a, c, d) are correct.

Borax bead test is performed only for coloured salt. Borax (sodium pyroborate), Na₂B₄O₇.10H₂O on heating gets fused and lose water of crystallisation. It swells up into fluffy white porous mass which melts into a colourless liquid which later form a clear transparent glassy bead consisting of boric anhydride and sodium metaborate.



Boric anhydride is non-volatile. When it react with Cr(III) salt then deep green complex is formed.



Hence, option (c) is correct.

Cu²⁺ and Ni²⁺ ions belong to Group II and Group IV respectively.

Group II sulphides precipitated out even under low concentration of sulphides, while Group IV sulphides require higher concentration of sulphides. To fulfill this reaction condition, dil. HCl is chosen in Group II reagents and NH₄OH is chosen in Group IV reagents.

X is Mn.

MnO₂  +  2KOH  +  KNO₃  $ \to $  K₂MnO₄  +  KNO₂  +  H₂O

Here   K₂MnO₄  is dark green.

3MnO₄^$-$2  +  4H⁺  $ \to $  2MnO₄^$-$  +  MnO₂  +  2H₂O

In acidic solution, K₂MnO₄ changes to dark purple solution of KMnO₄.

Statement wise explanation is

Statement (a) Mn²⁺ produces yellow-green colour in flame test while Cu²⁺ produces bluish-green colour in flame test. Thus, due to the presence of green colour in both the cases, flame test is not the suitable method to distinguish between nitrate salts of Cu²⁺ and Mn²⁺. Hence this statement is wrong.

Statement (b) Cu²⁺ belong to group II of cationic or basic radicals. It gives black ppt. of CuS if H₂S is passed through it in the presence of acid (e.g. HCl). Mn₂₊ does not show this property hence this can be considered as a suitable method to distinguish between Mn²⁺ and Cu²⁺. Hence, this statement is correct.

Statement (c) In faintly basic medium when H₂S is passed both Cu²⁺ and Mn²⁺ forms precipitates. Thus, it is not suitable method to distinguish between them. Hence, this statement is incorrect.

Statement (d) The standard reduction potential of Cu²⁺ /Cu is +0.34 V while that of Mn²⁺/Mn is $-$1.18 V. This can be used to distinguish between Cu²⁺ and Mn²⁺. In general less electropositive metals have higher SRP. Hence, this statement is correct.

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 $(c).$