KCET 2020 Chemistry Questions with Answers Key Solutions

Solution:

The formula of copper pyrites is $CuFeS _{2} .$ In auto reduction, the sulphide ores of the metals which are less electropositive like $Hg , Pb$ and $Cu$ are heated in the air so as to convert a part of it to its oxides. These oxides then react with the remaining sulphide ore in the absence of air, to give the metal and sulphur dioxide. The corresponding reaction is:

$Cu _{2} O +(1 / 2) Cu _{2} S \rightarrow 3 Cu +(1 / 2) SO _{2}$

Solution:

The function of potassium ethylxanthate in froth floatation process is to make the ore water repellant, i.e. hydrophobic.

Solution:

The corresponding reactions are;

$2 Cu _{2} S +3 O _{2} \rightarrow 2 Cu _{2} O +2 SO _{2}( s )$ (Reaction involved in roasting)

$SO _{2}+ Cl _{2}+$ activatedcharcoal $\rightarrow SO _{2} Cl _{2}$

So here $X$ is $SO _{2}$ (sulphur dioxide)

And $Y$ is $SO _{2} Cl _{2}$ (Sulfuryl chloride)

Solution:

The corresponding reactions are

$BaCl _{2}+ Na _{2} SO _{3} \rightarrow BaSO _{3}+2 NaCl$

$BaSO _{3}+2 HCl \rightarrow BaCl _{2}+ SO _{2}+ H _{2} O$

Salt $A$ is $Na _{2} SO _{3}$ and the dense white precipitate is $BaSO _{3}$ (Barium sulfite).

Gas $B$ is $SO _{2}$. It decolourises acidified $KMnO _{4}$ solution because of its reducing nature.

The corresponding reaction is:

$2 KMnO _{4}+5 SO _{2}+2 H _{2} O \rightarrow K _{2} SO _{4}+2 MnSO _{4}+2 H _{2} SO _{4}$

Solution:

The hydrides of group $15,16$ below the $3^{rd}$ period, follows Drago's rule. The rule states that due to a large energy difference between the atomic orbitals, these compounds do not exhibit hybridization. Thus, $PH _{3}$ will not exhibit hybridization and here the bond formation takes place due to the overlap of pure $p$-orbitals and $s$-orbitals. $PH _{3}$ has a lone pair on the central $P$ atom, which is absent in $PH _{4}{ }^{+}$. Thus in $PH _{3}$, there will be bond pair - lone pair repulsion and this is the reason why the bond angle in $PH _{3}$ is less than that of $PH _{4}{ }^{+}$.

Solution:

In $XeF _{6}$ there are $6$ bond pairs and one lone pair. Thus the shape of $XeF _{6}$ is distorted octahedral.

In $XeOF _{4}, Xe$ is $sp ^{3} d$ hybridized and the shape is square pyramidal.

In $XeO _{3}$, $Xe$ is $sp ^{3}$ hybridized and the shape is trigonal pyramidal.

In $XeF _{4}, Xe$ is $sp ^{3} d ^{2}$ hybridized. It has $4$ bond pairs and two lone pairs. Thus, the shape is square planar.

Solution:

(a) Phosphorous pentachloride $\left( PCl _{5}\right)$ has a trigonal bipyramidal structure. Here, the five bonds are not of equal lengths as the axial bonds are slightly longer than the equatorial bonds.

(b) $PCl _{5}$ solid exists as $\left[ PCl _{4}\right]^{+}$ and $\left[ PCl _{6}\right]^-$ and they have tetrahedral and octahedral structures respectively.

(c) On hydrolysis, $PCl _{5}$ gives $H _{3} PO _{4}$ which is tribasic.

$PCl _{5}+4 H _{2} O \rightarrow H _{3} PO _{4}+5 HCl$

In $H _{3} PO _{4}, P$ is in its $+5$ oxidation state. So it is in its highest oxidation state and hence is not a good reducing agent.

Solution:

The ions which contain unpaired electrons exhibit paramagnetic behaviour.

The electronic configuration of the ions are given below:

$Ti ^{3+}=1 s ^{2} 2 s ^{2} 2 p ^{6} 3 s ^{2} 3 p ^{5}$

$Cu ^{2+}=1 s ^{2} 2 s ^{2} 2 p ^{6} 3 s ^{2} 3 p ^{5} 3 d ^{9}$

$Mn ^{3+}=1 s ^{2} 2 s ^{2} 2 p ^{6} 3 s ^{2} 3 p ^{5} 3 d ^{4}$

Thus, all of them have unpaired electrons. Hence, they will show paramagnetic behaviour.

In option (B) $V ^{3+}$ does not possess any unpaired electrons.

In option (C) $Zn ^{2+}$ does not possess any unpaired electrons.

Similarly, in option (D) $Zn ^{2+}$ does not possess any unpaired electrons.

Solution:

The balanced redox reaction will be:

$K _{2} Cr _{2} O _{7}+6 KI +7 H _{2} SO _{4} \rightarrow $
$4 K _{2} SO _{4}+3 I_{2}+ Cr _{2}\left( SO _{4}\right)_{3}+7 H _{2} O$

$3$ moles of $I_{2}$ requires $1$ mole of $K _{2} Cr _{2} O _{7}$

Hence, $6$ moles of $I_{2}$ will require $2$ moles of $K _{2} Cr _{2} O _{7}$

Solution:

$CuCl _{2}$ is more stable than $Cu _{2} Cl _{2}$. The size of $Cu ^{2+}$ is smaller than $Cu ^{+}$. The hydration enthalpy

depends on the size of the cation. Smaller the size of the cation more will be the hydration enthalpy. Hence, more will be stability.

Solution:

$C _{2} O _{4}^{2-}$ is the oxalate ligand. It binds through two oxygen atoms and thus, is a bidentate ligand.

Hence the co-ordination number of $Fe$ in $\left[ Fe \left( C _{2} O _{4}\right)_{3}\right]^{3-}$ is $6$ .

$SCN^-$ is the thiocyanate ligand. It is a monodentate ligand. Hence the co-ordination number of

$Co$ in the complex ion $\left[ Co ( SCN )_{4}\right]^{2-}$ is $4$ .

Solution:

Solution:

$NH _{3}$ is neutral, making the first complex positively charged overall.
$Cl$ has a $-1$ charge, making the second complex the anion and so platinum ends with -ate.
Therefore, the complex with $NH _{3}$ comes first, followed by the one with $Cl$.
Option (A) is correct. The I.U.P.A.C name for $\left[ Pt \left( NH _{3}\right)_{4}\right]\left[ PtCl _{4}\right]$
is tetraammineplatinum(II) tetrachloridoplatinate(II)

Solution:

When chloroform is oxidised, it forms phosgene which is poisonous.

$CHCl _{3} \ce{->[(o)]}COCl _{2}+ HCl$

Solution:

The rate of an $S _{ N } 1$ reaction is determined by the stability of the intermediate formed. In this case, the most stable intermediate formed would be for $C _{6} H _{5} CH _{2} Cl$ and the intermediate formed would be $C _{6} H _{5} CH _{2}{ }^{+}$. This intermediate is stabilised via resonance.

Solution:

Solution:

Primary alkyl halides/ benzyl halides react with alkoxide/phenoxide to give ethers via $S _{N} 2$ mechanism. This reaction is known as Williamson's ether synthesis. Aryl halides are the least reactive as the $C-X$ bond has a partial double bond characteristic and hence, does not cleave easily. $3^{\circ}$ halides do not give ethers as they undergo elimination in the presence of a strong base like alkoxides.

Solution:

Solution:

To check the basicity of an ion, the conjugate acid must be weak. The conjugate acids are given below: $H _{2} O , CH _{3} OH , CH _{3} COOH$ and $HCl$.

We know that methanol is a stronger acid than water. Weaker the acid, stronger is the conjugate base, by that logic, $OH ^{-}$ is the strongest base among the given options.

Solution:

Ethers cannot form strong hydrogen bonds and hence they have the lowest boiling point among the given options.

Solution:

At least one $\alpha$-hydrogen should be present for an aldehyde or a ketone to undergo aldol reaction. The only compound that does not have even one $\alpha$-hydrogen is trichloroacetaldehyde, and hence, it does not undergo aldol reaction.

Solution:

Solution:

Hinsberg's reagent is benzene sulphonyl chloride $\left( C _{6} H _{5} SO _{2} Cl \right)$.

It is used to test for and distinguish between primary, secondary and tertiary amines.

Solution:

Adipose tissue stores those vitamins which are fat-soluble. Among the given vitamins, only $A , D$ and $E$ are fat-soluble. Vitamin $B _{6}$ is water-soluble and hence, is not stored in adipose tissue.

Solution:

Hypothyroidism is caused by the deficiency of the hormone thyroxine which is produced by the thyroid gland.

Solution:

Sucrose has $C_1 - C_2$ glycosidic cond and not $C_1 - C_4$ glycosidic bond

Solution:

The monomer of polythene is ethane $\left( C _{2} H _{4}\right)$

The monomer of polystyrene is styrene $\left( C _{6} H _{5} CHCH _{2}\right)$

The monomer of neoprene is chloroprene $\left( CH _{2}= C ( Cl )- CH = CH _{2}\right)$

The monomer of Terylene is $1,4$ -benzene dicarboxylic acid and $1,2$ -ethanediol. Thus, it possesses intermolecular hydrogen bonding and has the strongest intermolecular forces among the given options.

Solution:

Glycine can undergo condensation polymerisation. Given below is an example:

Solution:

BHA (Butylated hydroxyl anisole) is widely used as an anti-oxidant in foods like vegetable oils to stop them from getting oxidised and thereby, preventing them from going rancid.

Solution:

Antagonists impede the action of drugs by occupying the enzyme sites themselves and hence is the odd one out.

Solution:

The corresponding balanced reaction is:

$H _{2}+ Cl _{2} \rightarrow 2 HCl$

Again, $0.4 \,g H _{2}=0.4 / 2 \,mol =0.2 \,mol$

$7.1 \, g Cl _{2}=7.1 / 71 \,mol =0.1 \,mol$

So, $Cl _{2}$ is the limiting reagent here.

Hence $0.2$ moles of $HCl$ will be formed.

Again, we need to find out the volume of $HCl$ formed in the STP condition.

1 mol of gas in STP occupies $22.7 \, L$

Thus, $0.2$ mol of $HCl$ will occupy $=4.54 \, L$

Solution:

Photoelectric effect is the emission of electrons from a certain metal surface when it is irradiated with photons or light. The minimum frequency required to eject an electron from a metal surface is called threshold frequency. $\left(v_{0}\right)$

The equation for the photoelectric effect is:

$H U = h v _{0}+ KE$, where $v =$ frequency of the incident radiation

Here $hv _{0}$ is the work function. The number of electrons ejected does not depend on work function.

The number of electrons ejected increases with the increase in the intensity of incident light, an increase in the intensity of incident light means that the number of photons incident per unit surface area of the metal increases. (Provided the incident photons has its frequency more than the threshold frequency)

The energy of the ejected electrons increases with an increase in the frequency of incident light.

Solution:

The last element of the $p$-block in $6^{th}$ period is Radon $(Rn)$

The electronic configuration of $Rn$ is $[ Xe ] 4 f ^{14} 5 d ^{10} 6 s ^{2} 6 p ^{6}$

Solution:

A conjugate base has one proton less from the acid. If we remove one proton from $NH _{3}$, we get $NH _{2}^{-} .NH _{3} \rightarrow NH _{2}{ }^{-}+ H ^{+} .$

Solution:

Since temperature and pressure are constant, we can assume it to be in STP condition.

Now, let the total volume be $100\, L$

Moles of $H e$ present $=\frac{25}{22.4}=1.12\, mol$

Mass of $H e$ present $=1.12 \times 4=4.48 \,g$

Moles of $C H_{4}$ present $=\frac{75}{22.4}=3.35\, mol$

Mass of $C H_{4}$ present $=3.35 \times 16=53.60 \,g$

Hence, the mass percentage of $C H_{4}=\frac{53.60}{58.08} \times 100=92.28 \%$

Solution:

In $NO _{2}{ }^{+}$, the steric number of $N$ is two (two sigma bonds). Thus, it is sp hybridized. Hence, the percentage s-character is $=1 / 2 \times 100=50 \%$.

In $O _{3}^{-}$, the steric number of $N$ is three (three sigma bonds). Thus, it is $sp ^{2}$ hybridized. Hence the percentage s-character is $=1 / 3 \times 100=33.33 \%$.

In $NH _{4}^{+}$, the steric number of $N$ is four (four sigma bonds). Thus, it is $sp ^{3}$ hybridized.

Hence, the percentage s-character is $=1 / 4 \times 100=25 \%$

Solution:

(number of valence electron(s) in the neutral atom) $-1 / 2$ (number of valence electron(s) in the

covalent bond)-(number of electrons in lone pair)

$6-1 / 2 \times 6-2=+1$

Formal charge on the central oxygen i.e. oxygen $(1)$ is $+1$.

Solution:

In thermodynamics, we are concerned about entropy change due to the transfer of heat.

Which we can write in the form, $\Delta S = q / T$, here $q$ is constant.

So, $\Delta S \propto 1 / T$

We are given, $T _{1}> T _{2}$, so the actual relation between the entropies will be $\Delta S _{1}<\Delta S _{2}$

Solution:

$NH _{4} NO _{3}$ is a salt. The cationic part is $NH _{4}{ }^{+}$ and the anionic part is $NO _{3}^{-}$.

In $NH _{4}{ }^{+}$, let the oxidation number be $x$. So,

$x+4=1$

$\Rightarrow x =-3$

In $NO^- _{3}$, let the oxidation number be $y$. So,

$y-6=-1$

$\Rightarrow y=+5$

Solution:

Potassium generally forms three oxides: $K _{2} O , K _{2} O _{2}$ and $KO _{2}$.

$K _{2} O _{3}$ is not formed in normal conditions and hence, is the answer.

Solution:

$Zn$ produces $H _{2}$ with both dil. $HCl$ and $NaOH ( aq )$. The corresponding reactions are:

$Zn +2 HCl \rightarrow ZnCl _{2}+ H _{2} \uparrow$

$Zn +2 NaOH \rightarrow Na _{2} ZnO _{2}+ H _{2} \uparrow$

Solution:

Functional isomers are compounds having the same molecular formula but different functional groups. Both the compounds given in option c are ethers. $C _{2} H _{5} OC _{2} H _{5}$ and $C _{3} H _{7} OCH _{3}$ are metamers, while the rest of the options are examples of functional isomers.

Solution:

Solution:

Greenhouse gases are those compounds which increase the temperature of the atmosphere by absorbing the IR rays of the sun. They include compounds like water vapour, $CO _{2}, NO _{2}, O _{3}$, $CFCs$ etc. $O _{2}$ is not a greenhouse gas.

Solution:

Let, the number of $M ^{2+}$ ions $= x$

Then, the number of $M ^{3+}$ ions will be $0.96- x$

We know, the overall charge in the metal oxide is zero.

So, $x(2)+(0.96-x)(3)+1(-2)=0$

$\Rightarrow 2 x+2.88-3 x=2$

$\Rightarrow - x =-0.88$

$\Rightarrow x =0.88$

$\therefore $ Number of $M ^{3+}$ ions $=0.96-0.88=0.08$

$\therefore $ Percentage of $M ^{3+}$ ions $=0.08 / 0.96 \times 100=8.33 \%$

Solution:

For FCC,

Atomic radius $(r)=(\sqrt{2} a) / 4$

$\Rightarrow \sqrt{2} \times 10^{-10}=(\sqrt{2} a ) / 4$

$\Rightarrow a =\left(4 \times \sqrt{2} \times 10^{-10}\right) / \sqrt{2}=4 \times 10^{-10} m$

Volume of unit cell $= a ^{3}=\left(4 \times 10^{-10}\right)^{3}$

$=64 \times 10^{-30} m ^{3}=6.4 \times 10^{-29} m ^{3}$

Solution:

An intrinsic semiconductor is a semiconductor where we do not use any dopant, i.e. it is an undoped semiconductor.

Silicon has $4$ valence electrons whereas gallium has only $3$ valence electrons. Thus, the dopant has less number of valence electrons in this case. Hence, silicon doped with gallium forms a p-type semiconductor.

Solution:

The value of the constant $A$ in the Debye $-$ Huckel -Onsager equation, $\lambda_{ m }=$ $\lambda_{m}^{0}-A \sqrt{C}$ depends on temperature, charges of the ions, the dielectric constant of the solvent and also the viscosity of the solvent. Here, since we are not given the solvent, we will assume that the solvent is the same for each case. Here, the deciding factor is the charge on the ions. Again, in $NH _{4} Cl$ and $NaBr$, the charges on the ions are the same. Hence the pair of electrolytes that will possess the same value for the constant (A) will be $NH _{4} Cl , NaBr$.

Solution:

Isotonic solutions are those which have the same osmotic pressure $(\pi = iCRT)$. But here we have different concentration of the solutions and also they have different Van't Hoff factors (i). So the solutions for which the product of $i$ and $c$ will be the same and isotonic.

a. For, $0.001\, M \,CaCl _{2}, i =3$.
So $i \times C =3 \times 0.001=0.003$

For, $0.001 \,M\, Al _{2}\left( SO _{4}\right)_{3}, i =5 .$
So $i \times C =5 \times 0.001=0.005$

b. For, $0.01 \,M\, BaCl _{2}, I =3 .$
So $i \times C =3 \times 0.01=0.03$

For, $0.001\, M \,CaCl _{2}, i =3 .$
So $i \times C =3 \times 0.001=0.003$

c. For, $0.01 \,M \,BaCl _{2}, i =3$.
So $i \times C =3 \times 0.01=0.03$

For, $0.015 \,M \,NaCl , i =2 .$
So $i \times C =2 \times 0.015=0.03$

Thus $0.01 \,M \,BaCl _{2}$ and $0.015 M NaCl$ are isotonic in nature.

d. For, $0.001 \,M\, Al _{2}\left( SO _{4}\right)_{3}, i =5 .$
So $i\times C =5 \times 0.001=0.005$

For, $0.01 \,M \,aCl _{2}, i =3 .$
So $i \times C =3 \times 0.01=0.03$

Solution:

We are given,

$2 X \rightarrow X _{2}$

Therefore, $i =1-\alpha+\propto / 2=1-0.8+0.4=0.6$

(since $\alpha=0.8$, given $)$

We know, $\Delta T _{ b }= ik _{ b } \times W _{ B } / M \times 1000 /\left( W _{ A }( g )\right)$

Where, $W _{ B }=$ Mass of solute,

$W _{ A }=$ Mass of solvent,

$M =$ Molar mass

Given, $\Delta T _{ b }=0.3$

Therefore, $0.3=0.6 \times 0.52 \times 2.5 / M \times 1000 / 100$

$\Rightarrow M =2 \times 0.52 \times 2.5 \times 10=26 \,g$

Solution:

Given, $E_{F e^{+3} / F e^{+2}}^{o}=+0.76 V$ (Cathode)

$E_{I_{2 / I^{-}}}^{o}=+0.55 V$ (Anode)

$\therefore E_{C e l l}^{o}=E_{C}^{o}-E_{A}^{o}=0.76-0.55=0.21$

The complete cell reaction is

$2 Fe ^{3+}+2 I^{-} \rightarrow 2 Fe ^{2+}+I_{2}$

$\therefore E_{\text {cell }}^{o}=\frac{0.059}{2} \log K_{C}$

$\Rightarrow \log K_{C} \approx 7$

$\Rightarrow K_{C}=10^{7}$

Solution:

At the anode, oxidation takes place. Here water will be oxidized. The reaction at the anode will be:

$2 H _{2} O \rightarrow 4 H ^{+}+ O _{2}+4 e ^{-}$

Solution:

Platinum and gold lie at the bottom of the reactivity series. Thus it does not react with $HNO _{3}$ and $HCl$. Again, $Zn$ is more reactive than copper. Thus, copper will be displaced by $Zn$, if $ZnSO _{4}$ is placed in a copper vessel. But $Ag ^{+}$ is less reactive than copper. Thus, copper will displace silver. The corresponding reaction is:

$2 AgNO _{3}+ Cu \rightarrow Cu \left( NO _{3}\right)_{2}+2 Ag$

Solution:

For a first order reaction the, the rate constant is given by

$k=\frac{2.303}{t} \log \frac{\left[R_{0}\right]}{[R]}$

Given, at $50 \,min , 60 \%$ of the reaction is completed

$\therefore k=\frac{2.303}{t} \log \frac{\left[R_{0}\right]}{[R]}$

$=\frac{2.303}{50} \log \frac{100}{40}$

$=\frac{2.303}{50} \times 0.397$

So, when $93.6 \%$ of the reaction is completed,

$\Rightarrow \frac{2.303}{50} \times 0.397$

$=\frac{2.303}{t} \log \frac{100}{6.4}$

$\Rightarrow \frac{2.303}{50} \times 0.397$

$=\frac{2.303}{t} \times 1.19$

$\Rightarrow t \approx 150\, min$

Solution:

The given reaction is,

$2 A+3 B \rightarrow 4 C+D$

So, $-\frac{1}{3} \frac{d[B]}{d t}=\frac{1}{4} \frac{d[c]}{d t}$

$\Rightarrow -\frac{d[B]}{d t}=\frac{3}{4} \frac{d[C]}{d t}$

$=\frac{3}{4}\left(2.8 \times 10^{-3}\right) m o l\, L^{-1} S^{-1}$

Solution:

The expression of rate constant given in the question ( $\left.k = PZ e ^{(- Ea / RT )}\right)$ is according to Arrhenius theory. To speed up the reaction, we will have to decrease the value of $E_{a}$ i.e. activation energy and will have to increase the value of temperature $(T)$ and the number of collisions $(Z)$.

Solution:

Given reaction is,

$AgNO _{3}+ KI \rightarrow Agl ( Sol )+ KNO _{3}$

Here, the amount of $AgNO _{3}$ present is $0.1 \,M$.

Amount of $KI$ present is $0.2\, M$.

Since $KI$ is excess here, thus, sol obtained is a negative sol with $I$ adsorbed on $AgI$.

Solution:

When a gas is adsorbed on a solid surface, its movement is restricted leading to a decrease in the entropy of the gas i.e. $\Delta S$ is negative. Now, we know that, for a process to be spontaneous, $\Delta G$ should be negative.

We know, $\Delta G =\Delta H - T \Delta S$

Since $\Delta S$ is negative, $\Delta H$ has to be negative so as to make $\Delta G$ negative. Hence adsorption is always an exothermic process.