IIT JEE 1980 Chemistry Question Paper with Answer and Solution

(A) Since $x$ is strongly electropositive and univalent,it forms a cation $x^+$.
Since $y$ is strongly electronegative and univalent,it forms an anion $y^-$.
The combination of $x^+$ and $y^-$ results in the formation of an ionic compound with the formula $xy$.

(C) In the $N_2$ molecule,the two nitrogen atoms are connected by a triple bond.
Each nitrogen atom contributes $3$ electrons to the bond formation.
Therefore,the total number of electrons involved in the bond formation is $3 + 3 = 6$ electrons.

(A) covalent bond is formed by the sharing of electrons between atoms,typically between non-metals.
$H_2$ consists of two hydrogen atoms sharing a pair of electrons,making it a covalent molecule.
$CaO$,$KCl$,and $Na_2S$ are ionic compounds formed by the electrostatic attraction between metal cations and non-metal anions.
Therefore,the correct option is $A$.

(C) In benzene $(C_6H_6)$,the carbon atoms are arranged in a hexagonal ring with alternating single and double bonds.
Due to the phenomenon of resonance,the $\pi$-electrons are delocalized over the entire ring.
As a result,all carbon-carbon bonds are equivalent,and the bond order is the average of a single bond $(1)$ and a double bond $(2)$,which is $1.5$.
Therefore,the bond order of individual carbon-carbon bonds in benzene is between $1$ and $2$.

(D) $C_2H_5OH$ (ethanol) is a polar molecule that contains an $-OH$ group.
It can form intermolecular hydrogen bonds with water molecules,which makes it soluble in water.
In contrast,$CCl_4$,$CS_2$,and $CHCl_3$ are non-polar or weakly polar organic solvents that do not form hydrogen bonds with water and are therefore insoluble.

(C) $NH_4OH$ is a weak base because it undergoes partial dissociation in aqueous solution,producing a low concentration of $OH^{-}$ ions compared to strong bases like $NaOH$,$KOH$,and $Ca(OH)_2$,which dissociate completely. Therefore,$NH_4OH$ is the weakest base among the given options.

(C) The reaction is: $P_4 + 3NaOH + 3H_2O \to 3NaH_2PO_2 + PH_3$.
In this reaction,the oxidation state of phosphorus $(P)$ changes from $0$ in $P_4$ to $-3$ in $PH_3$ (reduction) and to $+1$ in $NaH_2PO_2$ (oxidation).
Since both oxidation and reduction occur simultaneously,it is a disproportionation reaction,which is a type of redox reaction.

(D) The strength of a base depends on its ability to release $OH^-$ ions in an aqueous solution.
$NaOH$,$KOH$,and $Ca(OH)_2$ are strong bases as they are alkali or alkaline earth metal hydroxides that dissociate completely in water.
$Zn(OH)_2$ is an amphoteric hydroxide,meaning it acts as a very weak base and also exhibits acidic properties.
Therefore,$Zn(OH)_2$ is the weakest base among the given options.

(D) Normal hydrogen gas is colourless at $STP$ conditions.
If an electric discharge is passed through hydrogen gas at low pressure,it emits a characteristic light blue glow.

(C) The reaction of hydrated barium peroxide with dilute hydrochloric acid is a standard laboratory method for the preparation of hydrogen peroxide.
$BaO_2 \cdot 8H_2O(s) + 2HCl(aq) \to BaCl_2(aq) + H_2O_2(aq) + 8H_2O(l)$
Among the given options,only $BaO_2$ reacts with $HCl$ to produce $H_2O_2$.

(D) $MnO_2$,$PbO_2$,and $BaO$ do not produce $H_2O_2$ when reacted with $HCl$.
$MnO_2$ reacts with $HCl$ to produce $MnCl_2$,$Cl_2$,and $H_2O$.
$PbO_2$ reacts with $HCl$ to produce $PbCl_2$,$Cl_2$,and $H_2O$.
$BaO$ reacts with $HCl$ to produce $BaCl_2$ and $H_2O$.
$H_2O_2$ is typically prepared by the action of acids on hydrated barium peroxide $(BaO_2 \cdot 8H_2O)$,not on $BaO$ or other oxides listed.

(D) Hydrogen is a colourless,odourless,and tasteless gas.

(D) The correct answer is $D$.
Calcium is an alkaline earth metal,which is highly reactive and cannot be obtained by carbon reduction.
It is extracted by the electrolysis of molten $CaCl_2$ (fused salt) because electrolysis of aqueous $CaCl_2$ would result in the evolution of hydrogen gas at the cathode instead of calcium metal.
At the cathode: $Ca^{+2} + 2e^{-} \to Ca$
At the anode: $2Cl^{-} \to Cl_2 + 2e^{-}$

(C) The chemical reaction is: $P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2$.
In this reaction,the oxidation state of phosphorus changes from $0$ in $P_4$ to $-3$ in $PH_3$ (reduction) and to $+1$ in $NaH_2PO_2$ (oxidation).
Since the same element is simultaneously oxidized and reduced,this is a disproportionation reaction,which is a type of redox reaction involving both oxidation and reduction.

(B) Marsh gas is formed by the anaerobic decomposition of organic matter in wetlands and swamps.
It primarily consists of methane $(CH_4)$,which is the simplest alkane.
Therefore,the correct option is $(B)$.

(C) The bond order of individual carbon-carbon bonds in benzene is $1.5$,which is between $1$ and $2$.
This is due to the resonance structure of $C_6H_6$.
In the benzene molecule,the $\pi$-electrons are delocalized over the entire ring.
Each carbon-carbon bond in benzene is equivalent,and the bond order is calculated as the average of the single and double bonds in the resonance structures,which is $(1+2)/2 = 1.5$.

(B) $Zn$ is an amphoteric metal. It reacts with hot concentrated $NaOH$ solution to form sodium zincate and hydrogen gas.
$Zn + 2NaOH \xrightarrow{\text{heat}} Na_2ZnO_2 + H_2$

(B) Let the thickness of each layer be $x$ and the thermal conductivity of layer $B$ be $K$. Then the thermal conductivity of layer $A$ is $2K$.
In steady state,the rate of heat flow $H$ through both layers is the same.
$H = \frac{Q}{t} = \frac{K_A A \Delta T_A}{x} = \frac{K_B A \Delta T_B}{x}$
Since the cross-sectional area $A$ and thickness $x$ are the same for both layers,we have:
$K_A \Delta T_A = K_B \Delta T_B$
$(2K) \Delta T_A = (K) \Delta T_B$
$\Delta T_B = 2 \Delta T_A$
The total temperature difference is given as $\Delta T_A + \Delta T_B = 36\,^{\circ}C$.
Substituting $\Delta T_B = 2 \Delta T_A$ into the equation:
$\Delta T_A + 2 \Delta T_A = 36\,^{\circ}C$
$3 \Delta T_A = 36\,^{\circ}C$
$\Delta T_A = 12\,^{\circ}C$
Therefore,the temperature difference across layer $A$ is $12\,^{\circ}C$.

(C) The strength of a base depends on its ability to dissociate into $OH^-$ ions in an aqueous solution.
$NaOH$,$KOH$,and $Ca(OH)_2$ are strong bases because they dissociate almost completely in water.
$NH_4OH$ (ammonium hydroxide) is a weak base because it undergoes partial dissociation in water,represented by the equilibrium: $NH_4OH \rightleftharpoons NH_4^+ + OH^-$.
Therefore,$NH_4OH$ is the weakest base among the given options.

(B) German silver is an alloy that consists of copper $(Cu)$,zinc $(Zn)$,and nickel $(Ni)$.
Specifically,its composition is approximately $Cu = 60\%$,$Zn = 20\%$,and $Ni = 20\%$.
Therefore,$Cu$ is one of the constituents of German silver.

(B) The correct answer is $(B)$.
$C_2H_5OH$ (ethanol) is soluble in water because it can form hydrogen bonds with water molecules.
$CS_2$,$CCl_4$,and $CHCl_3$ are non-polar or weakly polar organic solvents that do not form hydrogen bonds with water,making them insoluble.