$H$ 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}$
Explanation:
Among the given species only K2CrO4 is diamagnetic as central metal atom Cr in it has [Ar]3d0 electronic configuration i.e., all paired electrons. The structure and oxidation state of central metal atom of this compound are as follows
Structure
Oxidation state Cr6+
Rest all the compounds are paramagnetic. Reasons for their paramagnetism are given below.
(i) H-atom have 1s1 electronic configuration i.e., 1 unpaired electron.
(ii) NO2 i.e.
in itself is an odd electron species.
(iii) O$_2^ - $ (Superoxide) has one unpaired electron in $\pi$* molecular orbital.
(iv) S2 in vapour phase has O2 like electronic configuration i.e., have 2 unpaired electrons in $\pi$* molecular orbitals.
(v) Mn3O4 has following structure
Thus, Mn is showing +2 and +4 oxidation states. The outermost electronic configuration of elemental Mn is 3d54s2. Hence, in both the above oxidation states it has unpaired electron as
(vii) (NH4)2NiCl4 has Ni as central metal atom with +2 oxidation state. The electronic configuration of Ni2+ in the complex is
(viii) In K2MnO4 central metal atom Mn has +6 oxidation state with following structure
Electronic configuration of Mn6+ is
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.
The oxidation number of Mn in the product of alkaline oxidative fusion of MnO$_2$ is ___________.
Explanation:
The reaction for alkaline oxidative fusion is
2MnO$_2$ + 4KOH + O$_2$ $\to$ 2K$_2$MnO$_4$ + 2H$_2$O
In potassium manganite formed as product :
K$_2$MnO$_4$
2(+1) + $x$ + 4($-$2) = 0
$2+x-8=0$
$x-6=0\Rightarrow x=+6$
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.
Explanation:
In the development of photographic film, the black and white photographic film is developed as follows:
The silver bromide reacts with hydroquinone producing black coloured silver particles along with HBr and quinone. During this process, silver bromide gets reduced to silver and hydroquinone is oxidised to quinone.
The hydroquinone acts as a developer during this process.
The unreacted silver bromide then reacts with sodium thiosulphate to produce soluble complex and sodium bromide.
$\mathrm{AgBr + \mathop {2N{a_2}{S_2}{O_3}}\limits_{hypo\,solution} \to \mathop {N{a_3}[Ag{{({S_2}{O_3})}_2}]}\limits_{so{\mathop{\rm lub}} le} + NaBr}$
The reaction occurs in acidic medium. In acidic medium, sulphur from sodium thiosulphate will precipitate out. The colloidal sulphur is obtained which gives milky white turbidity.
$\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3+2 \mathrm{H}^{+}(a q) \longrightarrow 2 \mathrm{Na}^{+}+\mathrm{H}_2 \mathrm{SO}_3+\underset{\substack{\text { colloidal } \\ \text { sulphur }}}{\mathrm{S} \downarrow}$
Final Answer :
$\mathrm{AgBr + \mathop {2N{a_2}{S_2}{O_3}}\limits_{hypo\,solution} \to \mathop {N{a_3}[Ag{{({S_2}{O_3})}_2}]}\limits_{so{\mathop{\rm lub}} le} + NaBr}$