iCON Education - d and f Block Elements

2025 JEE Advanced MSQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Advanced 2025 Paper 1 Online

The pair(s) of diamagnetic ions is(are)

A.

La³⁺, Ce⁴⁺

B.

Yb²⁺, Lu³⁺

C.

La²⁺, Ce³⁺

D.

Yb³⁺, Lu²⁺

Correct Answer: A,B

Explanation:

(6) Diamagnetic - No unpaired electrons. All the electrons are paired.

A) $\mathrm{La}^{3+}, \mathrm{Ce}^{4+}$.

Element $L a$ election configuration is $[\mathrm{Xe}] 5 \mathrm{~d}^{\prime} 6 \mathrm{~s}^2$

For $\mathrm{La}^{3+}$, three valence electrons are removed So, the electron configuration of $\mathrm{La}^{3+}$ is $\left[\mathrm{Xe}\right]$

The configuration [ $x_e$ ] is the noble gas configuration and all the elections are paired. So, $\mathrm{La}^{3+}$ is diamagnetic.

Element $\mathrm{Ce}_{\text { }}$ election configuration is $\left[x_e\right] 4 f^{\prime} 5 d^{\prime} 6 s^2$

For $\mathrm{Ce} ^{4+}$, for valence elections are removed. so, the electron configuration of $\mathrm{Ce}^{4+}$ is $\left[\mathrm{Xe}\right]$.

The origination $\left[X_e\right]$ is the noble gas configuration and all the electrons are paired.

So, $Ce^{4+}$ is diamagnetic.

So, It is a pair of a diamagnetic ions.

B) $\mathrm{Yb}^{2+}, \mathrm{Lu}^{3+}$

Element Yb electron configuration is : $\left[x_e\right] 4 f^{14} 6 s^2$

For $y b^{2+}$, two valence electrons are removed from $y b$ - So, the election configuration of ${y} b^{2+}$ is $[x e] \& f^{14}$.

The configuration $\left[X_e\right] 4 f^{14}$ has no unpaired electrons. All the electrons are paired and hence $yb^{2+}$ is diamagnetic.

Element Lu electron configuration is $\left[x_e\right] 4 f^{\prime 4} 5 d^1 6 s^2$

For $\mathrm{Lu}^{3+}$, three valence electrons we removed from. Lu . So, the electron configuration of $\mathrm{Lu}^{3+}$ is $\left[x_e\right] 4 f^{14}$.

The configuration $\left[\mathrm{Xe}\right] 4 f^{14}$ has no unpaired electrons. All the electrons are paired and hence $L u^3 t$ is diamagnetic.

So, It is a pair of diamagnetic ions:

c) $\mathrm{La}^{2+}, \mathrm{Ce}^{3+}$

Element la electron configuration is [ $x e$ ] $5 d^{\prime} 6 s^2$.

For $\mathrm{La}^{2+}$, two valence electrons are removed so, the electron configuration of $\mathrm{La}^{2+}$ is $[\mathrm{Xe}] 5 d^{\prime}$.

The configuration $\left[X_e\right] 5 d^{\prime}$ has one unpaired election and hence it is paramagnetic, not diamagnetic.

Element $C_e$ electron configuration is $[x e] 4 f^{\prime} 5 d^{\prime} 6 s^2$

For $\mathrm{Ce}^{3+}$, three valence electrons we removed. So, the electron configuration of $\mathrm{Ce}^{3+}$ is $[x e] 4 f^{\prime}$

The configuration $[\mathrm{Xe}]$. $4f^{\prime}$ has one unpaired electron and hence it is paramagnetic, not diamagnetic.

So, it is not a pair of diamagnetic ions.

D) $\mathrm{Yb}^{3+}, \mathrm{Lu}^{2+}$

yb electron configuration is $[x e] 4 f^{14} 6 s^2$.

For $y b^{3+}$, three valence electrons are removed from $y b$. So, the electron configuration of $\mathrm{Yb}^{3+}$ is $\left[x_e\right] 4 f^{13}$

The configuration $\left[x_e\right] 4 f^{13}$ has unpaired election and it is paramagnetic, not diamagnetic.

Lu electron configuration is $\left[X_e\right] 4 f^{14} 5 d^1 6 s^2$.

For $L u^{2+}$, two valence electrons are removed from LU. So, the electron configuration of $\mathrm{Lu}^{2+}$ is $\left[\mathrm{Xe}_e\right] 4 f^{14} 5 d^1$.

The configuration $[x]$. $4 f^{\prime 4} 5 d^{\prime}$ has unpaired electron and it is paramagnetic, not diamagnetic.

So, it is not a pair of diamagnetic ions.

The pairs of diamagnetic ions are

A) $\mathrm{La}^{3+}, \mathrm{Cl}^{4+}$

B) $\mathrm{Yb}^{2+}, \mathrm{Lu}^{3+}$

2022 JEE Advanced MSQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Advanced 2022 Paper 1 Online

The treatment of galena with $\mathrm{HNO}_{3}$ produces a gas that is

A.

paramagnetic

B.

bent in geometry

C.

an acidic oxide

D.

colorless

2019 JEE Advanced MSQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Advanced 2019 Paper 2 Offline

With reference to aqua-regia, choose the correct option(s).

A.

Aqua-regia is prepared by mixing conc. HCl and conc. HNO₃ in 3 : 1 (v / v) ratio

B.

The yellow colour of aqua-regia is due to the presence of NOCl and Cl₂

C.

Reaction of gold with aqua-regia produces an anion having Au in +3 oxidation state

D.

Reaction of gold with aqua regia produces NO₂ in the absence of air

2019 JEE Advanced MSQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Advanced 2019 Paper 1 Offline

Fusion of MnO₂ with KOH in presence of O₂ produces a salt W. Alkaline solution of W upon electrolytic oxidation yields another salt X. The manganese containing ions present in W and X, respectively, are Y and Z. Correct statement(s) is (are)

A.

Both Y and Z are coloured and have tetrahedral shape

B.

Y is diamagnetic in nature while Z is paramagnetic

C.

In both Y and Z, $\pi $-bonding occurs between p-orbitals of oxygen and d-orbitals of manganese

D.

In aqueous acidic solution, Y undergoes disproportionation reaction to give Z and MnO₂

2018 JEE Advanced MSQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Advanced 2018 Paper 2 Offline

The correct option(s) to distinguish nitrate salts of $M{n^{2 + }}$ and $C{u^{2 + }}$ taken separately is (are)

A.

$M{n^{2 + }}$ shows the characteristic green color in the flame test

B.

Only $C{u^{2 + }}$ shows the formation of precipitate by passing ${H_2}S$ in acidic medium

C.

Only $M{n^{2 + }}$ shows the formation of precipitate by passing ${H_2}S$ in faintly basic medium

D.

$C{u^{2 + }}/Cu$ has higher reduction potential than $M{n^{2 + }}/Mn$ (measured under similar conditions)

2017 JEE Advanced MSQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Advanced 2017 Paper 2 Offline

The options(s) with only amphoteric oxides is (are)

A.

$C{r_2}{O_3},BeO,SnO,Sn{O_2}$

B.

$C{r_2}{O_3},CrO,SnO,PbO$

C.

$NO,{B_2}{O_3},PbO,Sn{O_2}$

D.

$ZnO,A{l_2}{O_3},PbO,Pb{O_2}$

2016 JEE Advanced MSQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Advanced 2016 Paper 1 Offline

The reagent(s) that can selectively precipitate S^2- from a mixture of S^2- and SO₄^2- in aqueous solution is (are) :

A.

CuCl₂

B.

BaCl₂

C.

Pb(OOCCH₃)₂

D.

Na₂[Fe(CN)₅NO]

Correct Answer: A,C

Explanation:

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} $

2015 JEE Advanced MSQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Advanced 2015 Paper 2 Offline

The pairs of ions where BOTH the ions are precipitated upon passing H₂S gas in presence of dilute HCl, is(are)

A.

Ba²⁺, Zn²⁺

B.

Bi³⁺, Fe³⁺

C.

Cu²⁺, Pb²⁺

D.

Hg²⁺, Bi³⁺

2015 JEE Advanced MSQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Advanced 2015 Paper 1 Offline

The correct statements about Cr²⁺ and Mn³⁺ is(are) (Atomic numbers of Cr = 24 and Mn = 25)

A.

Cr²⁺ is a reducing agent.

B.

Mn³⁺ is an oxidizing agent.

C.

both Cr²⁺ and Mn³⁺ exhibit d⁴ electronic configuration.

D.

when Cr²⁺ is used as a reducing agent, the chromium ion attains d⁵ electronic configuration.

2014 JEE Advanced MSQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Advanced 2014 Paper 1 Offline

Upon heating with Cu₂S, the reagent(s) that give copper metal is/are

A.

CuFeS₂

B.

CuO

C.

Cu₂O

D.

CuSO₄

Correct Answer: B,C,D

Explanation:

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

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

2014 JEE Advanced MSQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Advanced 2014 Paper 1 Offline

The pair(s) of reagents that yield paramagnetic species is/are

A.

Na and excess of NH₃

B.

K and excess of O₂

C.

Cu and dilute HNO₃

D.

O₂ and 2-ethylanthraquinol

2012 JEE Advanced MSQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2012 Paper 2 Offline

For the given aqueous reactions, which of the statement(s) is(are) true?

IIT-JEE 2012 Paper 2 Offline Chemistry - d and f Block Elements Question 8 English

A.

The first reaction is a redox reaction.

B.

White precipitate is Zn₃[Fe(CN)₆]₂.

C.

Addition of filtrate to starch solution gives blue colour.

D.

White precipitate is soluble in NaOH solution.

2023 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Advanced 2023 Paper 1 Online

In the scheme given below, $\mathbf{X}$ and $\mathbf{Y}$, respectively, are

JEE Advanced 2023 Paper 1 Online Chemistry - d and f Block Elements Question 2 English

JEE Advanced 2023 Paper 1 Online Chemistry - d and f Block Elements Question 2 English

A.

$\mathrm{CrO}_4{ }^{2-}$ and $\mathrm{Br}_2$

B.

$\mathrm{MnO}_4{ }^{2-}$ and $\mathrm{Cl}_2$

C.

$\mathrm{MnO}_4{ }^{-}$and $\mathrm{Cl}_2$

D.

$\mathrm{MnSO}_4$ and $\mathrm{HOCl}$

2019 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Advanced 2019 Paper 1 Offline

The green colour produced in the borax bead test of a chromium (III) salt is due to

A.

Cr₂O₃

B.

CrB

C.

Cr(BO₂)₃

D.

Cr₂(B₄O₇)₃

2017 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Advanced 2017 Paper 2 Offline

Which of the following combination will produce ${H_2}$ gas ?

A.

Fe metal and conc. $HN{O_3}$

B.

$Cu$ metal and conc. $HN{O_3}$

C.

$Zn$ metal and $NaOH(aq)$

D.

Au metal and $NaCN\left( {aq} \right)$ in the presence of air

2016 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Advanced 2016 Paper 2 Offline

In the following reaction, sequence in aqueous solution, the species X, Y and Z, respectively, are

${S_2}O_3^{2 - }\buildrel {A{g^ + }} \over \longrightarrow \mathop X\limits_{Clear\,solution} \buildrel {A{g^ + }} \over \longrightarrow \mathop Y\limits_{White\,precipitate} \buildrel {With\,time} \over \longrightarrow \mathop Z\limits_{Black\,precipitate} $

A.

[Ag(S₂O₃)₂]^3$-$, Ag₂S₂O₃, Ag₂S

B.

[Ag(S₂O₃)₂]^5$-$, Ag₂SO₃, Ag₂S

C.

[Ag(SO₃)₂]^3$-$, Ag₂S₂O₃, Ag

D.

[Ag(SO₃)₃]^3$-$, Ag₂SO₄, Ag

2013 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Advanced 2013 Paper 1 Offline

Upon treatment with ammoniacal H₂S, the metal ion that precipitates as a sulphide is

A.

Fe(III)

B.

Al(III)

C.

Mg(II)

D.

Zn(II)

2012 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2012 Paper 1 Offline

The colour of light absorbed by an aqueous solution of CuSO₄ is

A.

orange-red

B.

blue-green

C.

yellow

D.

violet

2022 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 JEE Advanced 2022 Paper 2 Online

The reaction of $\mathrm{Xe}$ and $\mathrm{O}_{2}{F}_{2}$ gives a $\mathrm{Xe}$ compound $\mathbf{P}$. The number of moles of $\mathrm{HF}$ produced by the complete hydrolysis of $1 \mathrm{~mol}$ of $\mathbf{P}$ is __________.

2022 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 JEE Advanced 2022 Paper 2 Online

Thermal decomposition of $\mathrm{AgNO}_{3}$ produces two paramagnetic gases. The total number of electrons present in the antibonding molecular orbitals of the gas that has the higher number of unpaired electrons is ____________.

2020 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 JEE Advanced 2020 Paper 2 Offline

An acidified solution of potassium chromate was layered with an equal volume of amyl alcohol. When it was shaken after the addition of 1 mL of 3% H₂O₂, a blue alcohol layer was obtained. The blue colour is due to the formation of a chromium (VI) compound 'X'. What is the number of oxygen atoms bonded to chromium through only single bond in a molecule of X?

2018 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 JEE Advanced 2018 Paper 1 Offline

Among the species given below, the total number of diamagnetic species is _____________.

$H$ atom, $N{O_2}$ monomer, ${O_2}^ - $ (superoxide), dimeric sulphur in vapor phase, $M{n_3}{O_4},\,\,$ ${\left( {N{H_4}} \right)_2}\left[ {FeC{l_4}} \right],$ ${\left( {N{H_4}} \right)_2}\left[ {NiC{l_4}} \right],\,$ $\,{K_2}Mn{O_4},\,$${K_2}Cr{O_4}$

2014 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 JEE Advanced 2014 Paper 1 Offline

Among PbS, CuS, HgS, MnS, Ag₂S, NiS, CoS, Bi₂S₃ and SnS₂ the total number of black coloured sulphides is

Correct Answer: 6

Explanation:

(i) Lead in +2 oxidation state as $\mathrm{Pb}^{2+}$ reacts with hydrogen sulphide gas $\left(\mathrm{H}_2 \mathrm{~S}\right)$ to form black coloured precipitate of lead sulphide $(\mathrm{PbS})$.

$\begin{aligned} & \mathrm{Pb}^{2+}(a q)+\mathrm{H}_2 \mathrm{~S}(g) \rightarrow \mathrm{PbS}(s) \downarrow\text { (Black } \text { ppt) }+2 \mathrm{H}^{+}(a q) \\\\ & \mathrm{PbS} \text { is black coloured precipitate. } \\ & \end{aligned}$

(ii) Silver in +1 oxidation state as $\mathrm{Ag}^{+}$reacts with hydrogen sulphide in neutral or acidic medium to form black coloured precipitate of silver sulphide $\left(\mathrm{Ag}_2 \mathrm{~S}\right)$.

$ 2 \mathrm{Ag}^{+}(a q)+\mathrm{H}_2 \mathrm{~S}(g) \xrightarrow[\text { or acidic medium }]{\text { Neutral }} \mathrm{Ag}_2 \mathrm{~S}(s) $(Black ppt)

$\mathrm{Ag}_2 \mathrm{~S}$ is black coloured precipitate.

(iii) Mercury in +2 oxidation state as $\mathrm{Hg}^{2+}$ reacts with hydrogen sulphide in dilute hydrochloric acid to form black coloured precipitate of mercury sulphide (HgS).

$ \mathrm{Hg}^{2+}(a q)+\underset{\begin{array}{c} \text { (saturated aq. } \\ \text { solution or gas) } \end{array}}{\mathrm{H}_2 \mathrm{~S}} \xrightarrow{\text { dil. } \mathrm{HCl}} \underset{\begin{array}{c} \text { Black } \\ \text { ppt } \end{array}}{\mathrm{HgS}(s)} $

HgS is black coloured precipitate.

(iv) Copper in +2 oxidation state as $\mathrm{Cu}^{2+}$ reacts with dilute hydrochloric acid to form black coloured precipitate of copper sulphide (CuS).

$ \begin{aligned} & \mathrm{Cu}^{2+}(a q)+\mathrm{H}_2 \mathrm{~S}(\mathrm{~g}) \text { (saturated aq. solution) } \xrightarrow[\text { HCl }]{\text { dil }} \mathrm{CuS}(\mathrm{s}) \text { (black ppt) } \\ & \end{aligned} $

CuS is black coloured precipitate.

(v) Cobalt in +2 oxidation state as $\mathrm{Co}^{2+}$ reacts with hydrogen sulphide in neutral or alkaline solution to form black coloured precipitate of cobalt sulphide.

$\mathrm{Co}^{2+}(a q) \text { (ammonical hydrogen sulphide) }+\mathrm{H}_2 \mathrm{~S} \longrightarrow \mathrm{CoS}(\mathrm{s})\text { (Black ppt.) }$

$(\mathrm{CoS})$ is black coloured precipitate.

(vi) Nickel in +2 oxidation state as $\mathrm{Ni}^{2+}$ reacts with hydrogen sulphide in neutral or slightly alkaline solution to form, black coloured precipitate of nickel sulphide.

$ \mathrm{Ni}^{2+}(a q)+\mathrm{H}_2 \mathrm{~S} \longrightarrow \underset{\substack{\text { Black } \\ \text { ppt. }}}{\mathrm{NiS}(\mathrm{s})} $

NiS is black coloured precipitate.

(vii) Bismuth in +3 oxidation state as $\mathrm{Bi}^{3+}$ reacts with hydrogen sulphide in cold dilute hydrochloric acid to form crystalline dark brown coloured precipitate, but appears black coloured solid of $\mathrm{Bi}_2 \mathrm{~S}_3$.

$\mathrm{Bi}^{3+}(a q) \text { (Saturated solution) }+\mathrm{H}_2 \mathrm{~S} \longrightarrow \mathrm{Bi}_2 \mathrm{~S}_3(\mathrm{~s})\text { (Dark brown or Black precipitate brownish) }$

$\mathrm{Bi}_2 \mathrm{~S}_3$ may appear black in colours.

Either compounds of sulphur are black in colour.

(viii) In mildly acidic, medium tin in +2 state, i.e., $\mathrm{Sn}^{2+}$ reacts with Hydrogen sulphide $\left(\mathrm{H}_2 \mathrm{~S}\right)$ to form brown coloured tin (II) sulphide which further reacts with excess of hydrogen sulphide to form light yellow coloured tin (IV) sulphide $\left(\mathrm{SnS}_2\right)$.

$\begin{aligned} \mathrm{Sn}^{2+}(a q)+\mathrm{H}_2 \mathrm{~S} \rightleftharpoons \underset{\text { Brown }}{\mathrm{SnS}(s)} \\ \mathrm{SnS}(\mathrm{s})+\mathrm{H}_2 \mathrm{~S} \rightleftharpoons \underset{\text { Yellow }}{\mathrm{SnS}_2(\mathrm{~s})}+2 \mathrm{H}^{+}\end{aligned}$

Hence, tin (IV) sulphide or $\mathrm{SnS}_2$ is yellow in colour.

(ix) MnS is known to be dirty pink coloured.

2009 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2009 Paper 2 Offline

The oxidation number of Mn in the product of alkaline oxidative fusion of MnO$_2$ is ___________.

2005 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2005 Mains

Write the balanced chemical equation for developing a back and white photographic film. Also explain why the solution of sodium thiosulphate on acidification turns milky white.

Chemistry Chapters

d and f Block Elements

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