Preliminary Test Questions in English

(A) Starch paper is used to detect the presence of $I_{2}$ (iodine).
When an oxidizing agent reacts with an iodide ion $(I^{-})$,it produces iodine $(I_{2})$.
$2I^{-} + \text{oxidant} \rightarrow I_{2} + \text{reduced product}$
The produced $I_{2}$ reacts with starch to form a characteristic blue-black complex.
$I_{2} + \text{starch} \rightarrow \text{blue-black complex}$

(A) fusion mixture is an equimolar mixture of sodium carbonate $(Na_2CO_3)$ and potassium carbonate $(K_2CO_3)$.
It is commonly used in analytical chemistry to fuse insoluble substances like silicates or metal oxides to make them soluble in water or acids.

(C) The chemical formula for microcosmic salt is $Na(NH_4)HPO_4 \cdot 4H_2O$.
It is also known as sodium ammonium hydrogen phosphate tetrahydrate.

(D) The correct answer is $(D)$.
In the borax bead test,$Mn^{2+}$ ions form manganese metaborate,$Mn(BO_2)_2$,which exhibits a violet or amethyst colour in the oxidizing flame.
This colour is characteristic of the $Mn^{2+}$ ion.

(C) The charcoal cavity test with $Co(NO_3)_2$ is used for the identification of certain metals.
When aluminum salts are heated with $Na_2CO_3$ and $Co(NO_3)_2$ on a charcoal piece,they form cobalt aluminate $(CoAl_2O_4)$,which is known as Thenard's blue.
The reaction is: $2Co(NO_3)_2 \rightarrow 2CoO + 4NO_2 + O_2$ and $CoO + Al_2O_3 \rightarrow CoAl_2O_4$ (blue ash).

(A) The flame test is used to identify metal ions based on the characteristic colour they impart to a non-luminous flame.
$A$. Barium $(Ba^{2+})$ salts impart a characteristic apple-green colour to the flame.
$B$. Calcium $(Ca^{2+})$ salts impart a brick-red colour.
$C$. Borates (like borax) can produce a green flame,but the question specifically asks for a salt tested by $Pt$ wire in the context of metal cation identification,where Barium is the standard answer.
$D$. Lead $(Pb^{2+})$ salts typically do not give a characteristic flame colour.
Therefore,the correct option is $A$.

(C) The salt used for the bead test is known as Microcosmic salt,which is $Na(NH_4)HPO_4 \cdot 4H_2O$.
Upon heating,it decomposes to form a transparent bead of sodium metaphosphate:
$Na(NH_4)HPO_4 \cdot 4H_2O \xrightarrow{\Delta} NaPO_3 + NH_3 + 5H_2O$.
This $NaPO_3$ bead reacts with metal oxides to form colored beads,which helps in the identification of metal ions.

(C) The borax-bead test is used to identify transition metal ions.
$Cr^{3+}$ ions produce an emerald green bead in both the oxidising and reducing flames due to the formation of chromium$(III)$ metaborate,$Cr(BO_2)_3$.
Therefore,the correct option is $(C)$.

(A) The flame test is a qualitative analysis method used to identify the presence of certain metal ions based on the characteristic color they impart to a non-luminous Bunsen flame.
$Ba^{2+}$ ions produce a characteristic grassy green color in the flame.
$Ca^{2+}$ ions produce a brick-red color.
$K^+$ ions produce a lilac color.
$Sr^{2+}$ ions produce a crimson-red color.
Therefore,the correct metal is Barium $(Ba)$.

(A) The process described is known as $Tempering$.
In this process,steel is heated to a temperature well below red heat and then cooled slowly.
This treatment reduces the brittleness of the steel while retaining its hardness.

(C) The bead test (specifically the microcosmic salt bead test) uses microcosmic salt,which is $Na(NH_4)HPO_4 \cdot 4H_2O$.
Upon heating,it decomposes to form a transparent bead of sodium metaphosphate:
$Na(NH_4)HPO_4 \cdot 4H_2O \xrightarrow{\Delta} NaPO_3 + NH_3 + 5H_2O$.
This sodium metaphosphate bead reacts with metal oxides to form colored beads.

(D) In the borax bead test,$Mn^{2+}$ ions form manganese metaborate,$Mn(BO_2)_2$,which exhibits a violet color in the oxidizing flame.

(C) $Na(NH_4)HPO_4 \cdot 4H_2O$ is known as microcosmic salt.
It is commonly used in qualitative inorganic analysis,specifically for the bead test.

(C) The salt used in the bead test is known as microcosmic salt,which has the chemical formula $Na(NH_4)HPO_4 \cdot 4H_2O$.

(B) Microcosmic salt is a chemical compound known as sodium ammonium hydrogen phosphate tetrahydrate.
Its chemical formula is $NaNH_4HPO_4 \cdot 4H_2O$.
It is commonly used in qualitative analysis for the bead test.

(C) $Na(NH_4)HPO_4 \cdot 4H_2O$ (Microcosmic salt) is used for performing the bead test in qualitative inorganic analysis.
Upon heating,it forms a glassy bead of sodium metaphosphate $(NaPO_3)$,which reacts with metal oxides to form colored beads.
So,the correct answer is $C$.

(A) The chromyl chloride test is a qualitative test used to detect the presence of chloride ions $(Cl^-)$ in a sample.
In this test,a chloride salt is heated with potassium dichromate $(K_2Cr_2O_7)$ and concentrated sulfuric acid $(H_2SO_4)$.
If chloride ions are present,they react to form chromyl chloride $(CrO_2Cl_2)$,which is a deep red vapor.
This test is specifically applicable to ionic chlorides.
$KCl$ is an ionic chloride and readily gives this test.
$PbCl_2$ and $HgCl_2$ are covalent in nature and do not provide the necessary chloride ions in the reaction mixture to form chromyl chloride.
Therefore,only $KCl$ gives the chromyl chloride test.

(A) The borax bead test is used to identify transition metal ions that form colored metaborates upon heating with borax $(Na_2B_4O_7 \cdot 10H_2O)$.
Transition metals like $Co^{+2}$,$Cu^{+2}$,and $Ni^{+2}$ form colored beads (blue,green/blue,and brown/yellow respectively).
$Al^{+3}$ is a representative element (non-transition metal) and does not form colored metaborates,thus it does not give a positive borax bead test.

(A) The chromyl chloride test is a characteristic test for the presence of chloride ions $(Cl^-)$ in inorganic compounds.
It is specifically given by ionic chlorides.
Among the given options,$KCl$ is an ionic chloride,whereas $PbCl_2$ and $HgCl_2$ are covalent in nature and do not respond to this test.

(B) Microcosmic salt is a chemical compound with the formula $Na(NH_4)HPO_4 \cdot 4H_2O$.
It is historically known as sodium ammonium hydrogen phosphate tetrahydrate.
It is obtained from urine and is used in the laboratory for the microcosmic salt bead test to identify metallic radicals based on the characteristic colors they produce in an oxidizing or reducing flame.

(B) The reaction of a chloride salt with solid $K_2Cr_2O_7$ and concentrated $H_2SO_4$ is known as the $Chromyl$ $Chloride$ test.
$4NaCl_{(s)} + K_2Cr_2O_{7(s)} + 6H_2SO_4(conc.) \to 2CrO_2Cl_2(orange-red \ vapours) + 2KHSO_4 + 4NaHSO_4 + 3H_2O$
The orange-red vapours of $CrO_2Cl_2$ (chromyl chloride) dissolve in aqueous $NaOH$ to form a yellow solution of sodium chromate:
$CrO_2Cl_2 + 4NaOH_{(aq)} \to Na_2CrO_4(yellow \ solution) + 2NaCl + 2H_2O$
Thus,the correct salt is $NaCl$.

(B) The reaction of sodium chloride with potassium dichromate and concentrated sulfuric acid is known as the chromyl chloride test.
The balanced chemical equation is:
$4NaCl + K_2Cr_2O_7 + 6H_2SO_4 \to 2CrO_2Cl_2 + 4NaHSO_4 + 2KHSO_4 + 3H_2O$
In this reaction,$CrO_2Cl_2$ (Chromyl chloride) is formed as a deep red gas,where the oxidation state of $Cr$ is $+6$.

(C) The borax bead test is used to identify transition metal ions that form colored metaborates.
$Ag^{+}$ ion does not give a characteristic colored bead because the formed silver metaborate,$AgBO_2$,is white or colorless,making it indistinguishable from the borax bead itself.

(D) The reaction involves the interaction of metal chromate precipitates ($BaCrO_4$ or $PbCrO_4$) with acetic acid $(AcOH)$.
Since $AcOH$ is a weak acid,it is not strong enough to dissolve these specific chromate precipitates.
Therefore,the observation 'No dissolution' indicates that no significant chemical reaction occurs between the precipitate and the acid.
Thus,the correct classification is 'For no reaction'.

(C) The reaction $2KI + FeCl_2 \longrightarrow$ No reaction indicates that no chemical change occurs between potassium iodide $(KI)$ and ferrous chloride $(FeCl_2)$.
Since the reactants remain in the solution without forming any precipitate or changing color,the resulting mixture remains a clear/colourless solution.
Therefore,the correct assignment is $C$.

(C) The reaction between $Na_2CO_3$ and $KNO_3$ involves two soluble salts.
According to the solubility rules,all sodium and potassium salts are soluble in water.
Since no insoluble precipitate is formed and no redox reaction occurs to produce a coloured species,there is no observable reaction.
Therefore,the mixture results in a clear/colourless solution.
Thus,the correct assignment is $C$.

(D) The reaction is: $(NH_4)_2SO_4 + Ba(OH)_2 \longrightarrow BaSO_4 \downarrow + 2NH_3 \uparrow$.
In this reaction,$BaSO_4$ (Barium sulfate) is formed as a solid precipitate.
$BaSO_4$ is known to be a white coloured precipitate.

(B) Statement $I$ is false because cupric metaborate $(Cu(BO_2)_2)$ is blue in color,not colourless.
Statement $II$ is false because heating boric anhydride $(B_2O_3)$ and copper sulphate $(CuSO_4)$ in a non-luminous (oxidizing) flame produces cupric metaborate $(Cu(BO_2)_2)$,which is blue. Cuprous metaborate $(CuBO_2)$ is formed in a luminous (reducing) flame.

(A) The flame test is a diagnostic tool used to identify the presence of certain metal ions in a sample based on the characteristic color they impart to a non-luminous flame.

Ion	Colour of the flame
$A. Cu^{2+}$	Green flame with blue centre
$B. Sr^{2+}$	Crimson Red
$C. Ba^{2+}$	Apple green
$D. Ca^{2+}$	Brick red

Based on the observation,the presence of a green flame with a blue centre is characteristic of the $Cu^{2+}$ ion.

(A) $Fe^{+2}$ ions undergo hydrolysis in water.
To prevent this hydrolysis and to keep the solution clear,dilute $H_2SO_4$ is added to the mixture of ferrous sulphate and ammonium sulphate before dissolving it in water.

(A) The correct matches are as follows:
$A$. Borax bead test: $MSO_4$ $\xrightarrow[\Delta]{Na_2B_4O_7} M(BO_2)_2$ $\rightarrow MBO_2$ $\rightarrow M$ (Matches $III$)
$B$. Charcoal cavity test: $MSO_4$ $\xrightarrow[\Delta]{Na_2CO_3} MCO_3$ $\rightarrow MO$ $\rightarrow M$ (Matches $IV$)
$C$. Cobalt nitrate test: $MCO_3$ $\rightarrow MO$ $\xrightarrow[+\Delta]{Co(NO_3)_2} CoO \cdot MO$ (Matches $I$)
$D$. Flame test: $MCO_3$ $\rightarrow MCl_2$ $\rightarrow M^{2+}$ (Matches $II$)
Therefore,the correct sequence is $A-III, B-IV, C-I, D-II$.

(C) Mohr's salt is $FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$. It contains $Fe^{2+}$ ions,which impart a light green colour to the salt.
Alum (specifically Potash Alum) is $K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O$. It contains $Al^{3+}$ ions,which are colourless in aqueous solution and solid form.
Therefore,the correct order is Light Green and Colourless.

(A) The reaction of solid salts with dilute $H_2SO_4$ is a standard qualitative analysis test for anions:
$1$. $CO_3^{2-}$ reacts with dil. $H_2SO_4$ to release $CO_2$ gas,which causes effervescence $(A-II)$.
$2$. $S^{2-}$ reacts with dil. $H_2SO_4$ to release $H_2S$ gas,which has the smell of rotten eggs $(B-III)$.
$3$. $Cl^{-}$ (in some cases like $NaCl$ with conc. $H_2SO_4$ or specific conditions) can release $Cl_2$ gas (yellowish-green,pungent) if an oxidizing agent is present,but in standard qualitative analysis,$Cl^-$ is identified by the evolution of $HCl$ gas. However,based on the provided options,$C$ matches with $IV$ $(Cl^-)$ and $D$ matches with $I$ $(NO_2^-)$ which releases brown fumes of $NO_2$ gas.
Therefore,the correct matching is $A-II, B-III, C-IV, D-I$.

(A) The borax bead test is used to identify transition metal cations that form coloured metaborates when heated with borax $(Na_2B_4O_7 \cdot 10H_2O)$.
Transition metals like $Cu^{2+}$,$Co^{2+}$,$Ni^{2+}$,$Fe^{3+}$,and $Cr^{3+}$ form characteristic coloured beads.
$Zn^{2+}$,$Pb^{2+}$,and $Ag^{+}$ do not form coloured metaborates in the borax bead test.
Therefore,the pair $Cu^{2+}$ and $Co^{2+}$ can both be identified by this test.

(B) $1$. Morphology of nanomaterial: $FTIR$ does not provide information about the morphology (shape,structure) of nanomaterials. Techniques like $SEM$ or $TEM$ are used for this.
$2$. Absorption of functional group: $FTIR$ identifies functional groups by detecting characteristic absorption bands in the infrared region,such as those for $-OH$,$-CH$,$-NH$,and $-CO$ groups. This is the correct information collected by $FTIR$.
$3$. Geometry of particles: $FTIR$ does not give information about particle geometry (e.g.,spherical or cubic shapes).
$4$. Particle size: $FTIR$ cannot determine particle size. Techniques like $DLS$ (Dynamic Light Scattering) or electron microscopy are used for this.

(B) The wet chemical synthesis (Sol-Gel process) of nanomaterials typically follows these steps: hydrolysis,condensation,gelation (formation of oxide or alcohol-bridged network),aging,drying,and finally dehydration (or calcination) to remove residual organic groups and water to obtain the final nanomaterial.

(D) $FTIR$ (Fourier Transform Infrared Spectroscopy) is the technique used to identify the functional groups and the binding nature of nanomaterials by measuring the absorption of infrared radiation.
$SEM$ (Scanning Electron Microscopy) is used for surface morphology.
$TEM$ (Transmission Electron Microscopy) is used for internal structure and size.
$XRD$ ($X$-Ray Diffraction) is used for crystal structure analysis.

(A, B, D) When $NaCl$ and $K_{2}Cr_{2}O_{7}$ are warmed with conc. $H_{2}SO_{4}$,they produce deep red vapours of chromyl chloride $(CrO_{2}Cl_{2})$. Thus,statement $(A)$ and $(D)$ are correct.
When these vapours are passed through $NaOH$ solution,they form sodium chromate $(Na_{2}CrO_{4})$,which is yellow in color. Thus,statement $(B)$ is correct.
Chlorine gas is not evolved in this reaction. Thus,statement $(C)$ is incorrect.
The reactions are as follows:
$(I)$ $K_{2}Cr_{2}O_{7} + 6 H_{2}SO_{4} \longrightarrow 2 KHSO_{4} + 2 CrO_{3} + 3 H_{2}O$
$(II)$ $NaCl + H_{2}SO_{4} \longrightarrow NaHSO_{4} + HCl$
$(III)$ $CrO_{3} + 2 HCl \longrightarrow CrO_{2}Cl_{2} + H_{2}O$
$(IV)$ $CrO_{2}Cl_{2} + 4 NaOH \longrightarrow Na_{2}CrO_{4} + 2 NaCl + 2 H_{2}O$