VITEEE 2023 Question Paper with Solutions

Hydrogen bonding is maximum in ethanol. This bonding is observed when $\mathrm{H}$ atom is attached to more electronegative atoms like N, F or O atom.

In ethanol, the hydrogen atom is attached directly to the electronegative element. Hence, hydrogen bonding occurs while in other given compound its not takes place.

To evaluate which statement is not correct about the lanthanoids, let's examine each option individually for accuracy:

Option A: This statement is correct. The lanthanoids consist of the 14 elements in the sixth period of the periodic table, with atomic numbers 58 (Cerium, Ce) through 71 (Lutetium, Lu). These elements are characterized by the filling of their $4f$ sublevel.

Option B: This statement is also correct. The most common oxidation state of lanthanoids is +3. This is because in this oxidation state, they tend to lose their three outer electrons (two from the $6s$ sublevel and one from the $4f$ or $5d$ sublevels) to achieve a more stable electronic configuration.

Option C: This statement is not correct. Although the availability of $4f$-electrons does indeed allow for a variety of oxidation states among the lanthanoids, not all elements in the series form compounds in a +4 oxidation state. While certain lanthanoids like Cerium (Ce) can exhibit a +4 oxidation state, it is not a general characteristic for all elements in the series. Different lanthanoids have different capacities to achieve higher oxidation states, and the +4 state is not universally applicable to all lanthanoids.

Option D: This statement is missing an important part of the information that seems to have been cut off. The statement likely refers to the "lanthanide contraction," which is the gradual decrease in atomic and ionic radii of the lanthanoid elements with increasing atomic number. This phenomenon is due to the imperfect shielding of nuclear charge by the $4f$ electrons, leading to a stronger attraction between the nucleus and the outer electrons, resulting in smaller atomic sizes. Assuming the statement intended to reference the lanthanide contraction, it would be correct assuming proper context was given.

Therefore, Option C is the statement that is not correct in the context of the lanthanoids.

$C{H_4}(g) + 2{O_2} \to C{O_2}(g) + 2{H_2}O$

$\Delta H = \Delta E + \Delta {n_g}RT$

$\Delta H = - 1770778 - 2 \times 8.314 \times 298$ [$\because \Delta ng=2$]

$-1770778-4955.144=-1775733.14$ Jmol$^{-1}$

Chromite $[\mathrm{FeCr}_2 \mathrm{O}_4]$ is used as a starting material for preparation of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$.

Due to inert pair effect, the stability of +2 oxidation state increases as we move down this group. Therefore, the order is

$\therefore \mathrm{PbX}_2>\mathrm{SnX} \mathrm{S}_2>\mathrm{GeX}_2>\mathrm{SiX}_2$

$$\begin{aligned} \text { Volume of unit cell } & =a^3=\left(8 \times 10^{-8} \mathrm{~cm}\right)^3 \\ =512 \times 10^{-24} & =5.12 \times 10^{-22} \mathrm{~cm}^3 \end{aligned}$$

To distinguish between aldehyde and ketone the Tollen's reagent is used. It is because the aldehydes gives positive test to this reagent whereas ketones do not react with this reagent.

The reaction is represented as

$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ and $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$, inner orbital complex ions are diamagnetic.

$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ oxidation state of $\mathrm{Co}$ in is +3.

Electronic configuration of $\mathrm{Co}^{3+}=[\mathrm{Ar}] 3 d^6$

As NH$_3$ is a strong field ligand thus, pairing of electrons takes place.

In $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$ oxidation state of $\mathrm{Fe}$ is +2. E.C of $\mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 d^6$

Combination of four amino acid units gives a linear tetrapeptide, in which three peptide bonds are present.

In the given compounds, only acetophenone is formed by substitution. Whereas, other compounds are prepared by addition reactions. Acetophenone is prepared when benzene reacts with acetyl chloride in presence of anhydrous $\mathrm{AlCl}_3$ as

Mole fraction and molality does not involve volume therefore, they are independent of temperature. Normality and molarity changes with temperature because they involve volumes.

For zero order reaction, rate of reaction is independent of concentration of reactants.

$\frac{d x}{d t}=k[A]$

Unit of rate constant is concentration time$^{-1}$. For a zero order reaction, the plot of concentration of reactant versus time is a straight line (linear) with a negative slope and non-zero intercept.

$\mathrm{XeOF}_4$ has square pyramidal structure. In $\mathrm{XeOF}_4, \mathrm{Xe}$ is $s p^3 d^2$-hybridised and has 5 bond-pairs and 1 lone-pair of electrons.

Conductance of anion in aqueous solution is dependent upon its hydrated size.

Ionic conductance $\propto \frac{1}{\text { ionic size of hydrated ion }}$

Thus, ionic conductance in aqueous solution increases in the order

$\mathrm{Li}^{+}<\mathrm{Na}^{+}<\mathrm{K}^{+}<\mathrm{Rb}^{+}<\mathrm{Cs}^{+}$. As the size of hydrated ion is minimum for $\mathrm{Cs}^{+}$ ion and maximum for Li ion among all the hydrated alkali metal ions.

Arrhenius equation is given as

$k=A e^{-E_a / R T}$

On taking In on both sides, we get

$\ln k=\ln A+\left(-\frac{E_a}{R T}\right) \text { or } \log A-\log k=\frac{E_a}{2.303 R T}$

or $\log A=\log k+E_a / 2.303 R T$

Hence, equation $\log k-\left(E_a / R T\right)=A$, is not represented as an Arrhenius equation.

The correct match is A-IV, B-III, C-II, D-I.

$T \propto i \times$ molality

Molality is assumed to be equal to molarity, then $T \propto i \times M$

Larger the value of $i$ and $M$, higher will be the boiling point

$$\begin{aligned} & \text { I. } 0.0001 \mathrm{~M} \mathrm{~NaCl}, i=2, M=0.0001 \\ & \therefore \quad T=2 \times 0.0001=0.0002 \end{aligned}$$

$\text { II. } 0.0001 \mathrm{~M} ~\text {urea, } i=1, \therefore T=1 \times 0.0001=0.0001$

III. $0.001 \mathrm{~M} \mathrm{~MgCl}_2, i=3$

$\therefore \quad T=0.001 \times 3=0.003$

IV. $0.01 \mathrm{~M} \mathrm{~NaCl}, i=2 \therefore T=2 \times 0.01=0.02$

But-1-yne on reaction with $\mathrm{H}_2 \mathrm{O}$ in presence of $\mathrm{Hg}^{2+}$ ions as a catalyst produces butan-2-one.

The half reaction are

$$\begin{aligned} & \left.\mathrm{Fe}(\mathrm{s}) \longrightarrow \mathrm{Fe}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-}\right] \times 2 \text { [at Anode] } \\ & \frac{\mathrm{O}_2(g)+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_2 \mathrm{O} \text { [at cathode] }}{2 \mathrm{Fe}(\mathrm{s})+\mathrm{O}_2(g)+4 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Fe}^{2+}(\mathrm{aq})+2 \mathrm{H}_2 \mathrm{O}(l)} \\ \end{aligned}$$

$$\begin{aligned} & E=E^{\circ}-\frac{0.0591}{4} \log \frac{\left(10^{-5}\right)^2}{\left(10^{-5}\right)^4 \times 3} \\ & =1.67+\frac{0.0591}{4} \log 3 \times 10^{-5} \times 10^{-5} \simeq 1.529 \mathrm{~V} \end{aligned}$$

The reaction is represented as

Reducing character is based upon higher negative value of reduction electrode potential The more negative is the standard reduction potential, stronger is the reducing agent.

Thus, order of reducing character is given as $B>C>A$

Statement II and III are incorrect. Acid rain is caused by the release of sulphur dioxide and nitrogen oxides into the atmosphere, where they react with water, oxygen and other chemicals to form sulphuric and nitric acid.

$\mathrm{NO}_2$ is more harmful than NO. Nitric oxide is a sharp sweet-smelling gas at room temperature. Whereas nitrogen dioxide has a strong, harsh odour and is a liquid at room temperature. The elevated levels of nitrogen dioxide can cause damage to the human respiratory tract and increase a person's vulnerability to respiratory infections and asthma.

Boiling point of ammonia $(\mathrm{NH}_3)$ is much higher than phosphine $(\mathrm{PH}_3)$. It is due to extensive inter molecular hydrogen bonding found in ammonia.

The reaction of alkyl halides with sodium alkoxide or sodium phenoxide to form ethers is called Williamson synthesis. Here, in this reaction alkyl halide should be primary and alkoxide, should be bulkier as shown below,

Glyptal is a condensation copolymer of ethylene glycol and phthalic acid. The reaction is represented as.

Freundlich adsorption isotherm equation is

$x / m=k p^{1 / n}$

On taking log both sides,

$$\begin{aligned} & \log \frac{x}{m}=\log k+\frac{1}{n} \log p, \log \frac{x}{m}=\log 10+\frac{1}{n} \log 2 \\ & \left(\because \text { Slope }=\frac{1}{n}=\tan \theta=\tan 45^{\circ}=1\right) \\ & \log x / m=1+1 / 1 \log 2, \log x / m=1+1 \times 0.3010 \\ & \log x / m=1.3010, x / m=20 \mathrm{~g} \end{aligned}$$

The correct match is A-III, B-II, C-IV, D-I.

$\text { A. } \underset{100 \mathrm{~g}}{\mathrm{CaCO}_3} \xrightarrow[\text { Decomposition }]{\Delta} \mathrm{CaO}+\underset{22.4 \mathrm{~L}}{\mathrm{CO}_2}$

$\because 100 \mathrm{~g}$ of $\mathrm{CaCO}_3$ on decomposition gives

$=22.4 \mathrm{~L}^2 \text { of } \mathrm{CO}_2$

$\therefore 20 \mathrm{~g}$ of $\mathrm{CaCO}_3$ on decomposition will give

$=\frac{22.4 \times 20}{100} \mathrm{Lof}^2 \mathrm{CO}_2=4.48 \mathrm{LCO}_2$

B. $\mathrm{\mathop C\limits_{12\,g} \mathrel{\mathop{\kern0pt\longrightarrow} \limits_{Combustion}^{Excess\,{O_2}}} \mathop {C{O_2}}\limits_{22.4\,L}}$

$12 \mathrm{~g}$ of carbon on combustion gives

$=22.4 \mathrm{~L} \text { of } \mathrm{CO}_2=8.96 \mathrm{~LCO}_2$

C. $\underset{106 \mathrm{~g}}{\mathrm{Na}_2 \mathrm{CO}_3} \xrightarrow{\text { Excess HCl}} 2 \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}+\underset{22.4 \mathrm{~L}}{\mathrm{CO}_2}$

$$\begin{aligned} & \quad 106 \mathrm{~g} \text { of } \mathrm{Na}_2 \mathrm{CO}_3 \text { gives }=22.4 \mathrm{~L} \text { of } \mathrm{CO}_2 \\ & \therefore \quad 2.12 \mathrm{~g} \text { of } \mathrm{Na}_2 \mathrm{CO}_3 \text { will give } \\ & \quad=\frac{22.4 \times 2.12}{106} \mathrm{~L} \text { of } \mathrm{CO}_2=0.448 \mathrm{~L} \mathrm{CO}_2 \end{aligned}$$

D. $2 \mathrm{CO} \xrightarrow[\text { Combustion }]{\text { Excess }} 2 \mathrm{CO}_2$

$$\begin{aligned} & 2(12+16) \mathrm{g} \quad 2 \times 22.4 \mathrm{~L} \\ & 56 \mathrm{~g} \end{aligned}$$

$56 \mathrm{~g}$ of carbon monoxide on combustion gives $=2 \times 22.4 \mathrm{~L}$ of $\mathrm{CO}_2$

$1.12 \mathrm{~g}$ of carbon monoxide on combustion will give $=\frac{2 \times 22.4 \times 1.12}{56} \mathrm{L~of~CO}_2=0.896 \mathrm{~L~CO}_2$

In a carboxylic acid group the presence of halogens on adjacent carbons increases the acidity of the carboxylic acid group by stabilising the carboxylate ion (conjugate base).

$-\mathrm{CH}_3$ group shows + $I$-effect and increases electron density and destabilise the carboxylate ion (conjugate base). So, decreases acidic strength.

Hence, the order of decreasing acidic strength

According to Heisenberg uncentainty principle,

$\Delta x \times \Delta p \geq \frac{h}{4 \pi}$

Where, $\Delta x=$ uncertainty in position

$\Delta p=$ uncertainty in momentum

$$\begin{aligned} & 2.0 \times 10^{-8} \mathrm{kgms}^{-1} \\ \therefore & \Delta x \times 2.0 \times 10^{-8} \geq \frac{6.62 \times 10^{-34}}{4 \times 3.14} \\ & \Delta x \geq \frac{6.62 \times 10^{-34}}{4 \times 3.14 \times 2.0 \times 10^{-8}} \geq 2.63 \times 10^{-27} \mathrm{~m} \end{aligned}$$

$\stackrel{+7}{\mathrm{MnO}_4^{-}} \rightarrow \mathrm{Mn}^{2+}$ change in oxidation number 5

$\longrightarrow \quad \stackrel{+6}{\mathrm{MnO}_4^{2-}}$ change in oxidation number 1

$\longrightarrow \quad \stackrel{+4}{\mathrm{MnO}_2}$ change in oxidation number 3

$\longrightarrow \quad \stackrel{+3}{\mathrm{Mn_2O}_3}$ change in oxidation number 4

Aniline reacts with bromine water at room temperature to give white precipitate of $2,4,6$ - tribromoaniline. The reaction involved is as follows.

$$\begin{aligned} K_C & =\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}=\frac{(1)^2}{(0.3)(0.3)} \\ & =\frac{1}{0.09}=11.11 \end{aligned}$$

The properties of the system whose value is independent of the amount of substance present in the system are called intensive properties e.g. temperature. Whereas the properties of the system whose value is depends upon the amount of substance present in the system is called extensive property. e.g. Mass, volume, enthalpy.

The ortho-silicate ion or group have $\mathrm{SiO}_4^{4-}$ formula. It has tetrahedral shape, with one silicon atom surrounded by four oxygen atoms. In the ion, each oxygen carries a unit negative charge. The $\mathrm{Si}-\mathrm{O}$ bond is 162 pm long.

Thus, the total number of metal-metal bond = 1.

The magnetic field due to two wires at $P$

$B_1=\frac{\mu_0 j}{2 \pi(2 d+x)} \Rightarrow B_2=\frac{\mu_0 i}{2 \pi(2 d-x)}$

Both the magnetic field act in opposite direction, $B=B_2-B_1=\frac{\mu_0 j}{2 \pi}\left(\frac{1}{2 d-x}-\frac{1}{2 d+x}\right)$

$=\frac{\mu_0 i}{2 \pi}\left(\frac{2 d+x-2 d+x}{4 d^2-x^2}\right)=\frac{\mu_0 i x}{\pi\left(4 d^2-x^2\right)}$

For solid sphere rolling without slipping on inclined plane, acceleration is given as,

$a_1=\frac{g \sin \theta}{1+k^2 / R^2}$

For solid sphere slipping on inclined plane without rolling, acceleration is as follows.

$a_2=g \sin \theta$

Therefore, required ratio $=a_1 / a_2$

$=\frac{1}{1+\frac{k^2}{R^2}}=\frac{1}{1+\frac{2}{5}}=\frac{5}{7}$

No. of electrons reducing the collector,

$n_c=\frac{94}{100} \times 10^{10}=0.94 \times 10^{10}$

Emitter current, $i_e=\frac{n_e \times e}{t}$

Collector current, $i_c=\frac{n_c \times e}{t}$

Current transfer ratio,

$\alpha=\frac{i_c}{i_e}=\frac{n_c}{n_e}=\frac{0.94 \times 10^{10}}{10^{10}}=0.94=\alpha$

Here, initial momentum of particle $=$ Final momentum of particles,

$m v \hat{\mathbf{i}}+m v \hat{\mathbf{j}}=2 m v$

Where, $\mathbf{v}=\frac{v}{2}(\hat{\mathbf{i}}+\hat{\mathbf{j}})$

Hence, $v=\frac{v}{2} \times \sqrt{2}=v / \sqrt{2} \quad\left(\because v=10 \mathrm{~ms}^{-1}\right)$

So, $v=\frac{10}{\sqrt{2}} \mathrm{~ms}^{-1}$

$\frac{\Delta Q}{\Delta t}=\frac{\Delta W}{\Delta t}=$ work done per unit time

$\frac{d W}{d t}=P \frac{d v}{d t}=\frac{k a Q}{L}$

as $P=\frac{n R T}{v} \Rightarrow v=\frac{k a}{R}\left(\frac{27}{300}\right)=\frac{k}{100 R}$

Using Newton's Law of cooling,

$\frac{\theta_1-\theta_2}{t}=-k\left(\frac{\theta_1+\theta_2}{2}-\theta_0\right)$

for cooling from $82^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$

$\frac{82-50}{10}=-k\left(\frac{82+50}{2}-\theta_0\right)\quad \text{..... (i)}$

$\frac{50-42}{10}=-k\left(\frac{50+40}{2}-\theta_0\right) \quad \text{..... (ii)}$

Dividing Eq. (i) by Eq. (ii), we get

$\frac{66-\theta_0}{46-\theta_0}=4 \Rightarrow 3 \theta_0=118 \text { or } \theta_0=39.3^{\circ} \mathrm{C}$

Given, $n=10$ turns

$R_{\text {coil }}=40 \Omega, R_G=-30 \Omega$

total resistance in the circuit $=40+30=70 \Omega$

$A=10^{-2} \mathrm{~m}^2, B=10^{-2} \mathrm{~T} \text {, and } \theta_1=0, \theta=60^{\circ}$

we know that, flux of magnetic field is $\phi$

$\phi=n B A$

By Faraday's law, $e=-\frac{d \phi}{d t} \quad(\because e=$ induced emf$)$

$$\begin{aligned} i= & \frac{e}{R} \Rightarrow \frac{d q}{d t}=\frac{d \phi}{R d t} \Rightarrow q=\frac{d \phi}{R} \\ q & =\frac{\phi_1-\phi_2}{R}=\frac{B n A \cos \theta_1-B n A \cos \theta_2}{R} \\ & =\frac{B n A\left(\cos 0^{\circ}-\cos 60^{\circ}\right)}{R} \\ q & =\frac{B n A(1-0.5)}{R}=\frac{0.5 \times 10^{-2} \times 10 \times 10^{-2}}{70} \\ & =\frac{50 \times 10^{-5}}{70} \\ \end{aligned}$$

$q=0.714 \times 10^{-5} \mathrm{C}$. (Charge induced in a coil)