VITEEE 2025 Question Paper with Solutions

Let’s look at the pattern formed by the letters according to their positions in the English alphabet.

For the first three clusters, we can see this sequence of operations on the positions of letters:

$\begin{array}{cc}& I\stackrel{-3}{\to \leftarrow }F\stackrel{+7}{\to \leftarrow }M\\ & T\stackrel{-3}{\to \leftarrow }Q\stackrel{+7}{\to \leftarrow }X\\ & G\stackrel{-3}{\to \leftarrow }D\stackrel{+7}{\to \leftarrow }K\end{array}$

In each case above, the first change is by $-3$ (three letters backward), and the second change is by $+7$ (seven letters forward).

Now, for the last cluster:

$O\stackrel{-3}{\to \leftarrow }L\stackrel{+6}{\to \leftarrow }R$

Here, the second change is $+6$ instead of $+7$. So, this letter-cluster does not follow the same pattern and therefore does not belong to the group.

$\text{ The pattern of the given series is as follows, }$

As,

And

Similarly,

$\text{ The pattern of the given series is as follows, }$

$\text{ Logic : Opposite letter place value. }$

Total export during 2013, 2014 and 2018

$=140+240+270=650$

Total production during 2015, 2016 and 2017

$=300+290+340=930$

∴ Required percentage

$\begin{array}{cc}& =\frac{(930-650)}{930}\times 100=\frac{280}{930}\times 100\\ & =30.1\%\end{array}$

Total production of computers in 2013, 2014 and 2015

$=180+300+320=800$

Total export of computers

$=140+240+200+350+300+270=1500$

Now, $𝑥=\frac{800}{1500}\times 100\%=53\frac{1}{3}\%$

According to the statements,

Conclusions

I. ( $✓)$II. ( $𝑋$ )III. ( $✓$ )

$\text{ Hence, only Conclusions I and III follow. }$

As,

Similarly,

Statement I alone is not sufficient as wedding anniversary month is not known. Statement II alone is not sufficient as wedding anniversary date in February is not known. From both the statements together Yuvi's wedding anniversary is on 7th of February. Thus, both Statements I and II together are sufficient to answer the question.

Oxidation state changes as follows Cr goes from $+6\mathrm{in}{\mathrm{Cr}}_2{O}_{2-}^7$ to $+3\mathrm{in}{\mathrm{Cr}}^{3+}$ that is 2 Cr atoms gain 6 electrons in total.

S goes from +4 in ${\mathrm{SO}}_2$ to +6 in ${\mathrm{SO}}_{2-}^4$ that is 2 electrons lost per sulphur.

So, total balanced equation:

$\begin{array}{cc}{\mathrm{Cr}}_2{O}_{2-}^7\left(𝑎𝑞\right)+& 3{\mathrm{SO}}_{2-}^3\left(𝑎𝑞\right)+8H^{+}\left(𝑎𝑞\right)⟶2{\mathrm{Cr}}^{3+}\left(𝑎𝑞\right)+3{\mathrm{SO}}_{2-}^4\left(𝑎𝑞\right)+4H_{2}O\left(𝑙\right)\end{array}$

Hence, $𝑥=1,𝑦=3,𝑧=8$

$\Rightarrow \text{ sum }=1+3+8=12$

$\begin{array}{cc}& \text{ We know that }\\ & \begin{array}{cc}{𝑊}_{\mathrm{irr}}& =-{𝑝}_{\mathrm{ext}}\Delta V\\ & =-2\mathrm{bar}(0.25-0.1)L\\ & =-0.30L-\mathrm{bar}\\ & =-0.30\times 100J\\ & =-30J\end{array}\end{array}$

Depression in freezing point,

$\Delta {𝑇}_{𝑓}\propto 𝑖\times \text{ concentration }$

where, $𝑖=$ van't Hoff factor (number of particles):

$\Delta {𝑇}_{𝑓}={𝑇}_{\circ }^{𝑓}-{𝑇}_{𝑓}$

Glucose (non-electrolyte),   $𝑖=1$

$\begin{array}{c}{\mathrm{NaNO}}_3⟶{\mathrm{Na}}^{+}+{\mathrm{NO}}_{-}^3,𝑖=2\\ {\mathrm{AlCl}}_3⟶{\mathrm{Al}}^{3+}+3{\mathrm{Cl}}^{-},𝑖=4\end{array}$

Hence, freezing point is lowest for ${\mathrm{AlCl}}_3$ due to maximum ionisation.

Nucleophilicity increases with negative charge and less solvation.

• ${\mathrm{CN}}^{-}$has strong negative charge, less hindered, and not heavily solvated.

• ${\mathrm{OH}}^{-}$is more solvated due to H -bonding.

• ${\mathrm{NH}}_3$ and $H_{2}O$ are neutral and weaker nucleophiles, therefore the order will be ${\mathrm{CN}}^{-}>{\mathrm{OH}}^{-}>{\mathrm{NH}}_3>H_{2}O$.

Thus, ${\mathrm{CN}}^{-}$is the strongest nucleophile in aqueous medium.

In Williamson synthesis, a primary alkyl halide reacts with alkoxide. However, tertiary alkyl halides undergo elimination instead of substitution.

tert-butyl bromide + methoxide → alkene $($ via ${𝐸}_2)$ Thus, this method fails for the preparation of methyl tert-butyl ether.

Bond order = (number of bonding electrons antibonding electrons)

${\mathrm{NO}}^{+}$(14 electrons): B.O. $=3$

${\mathrm{CN}}^{-}$(14 electrons): B.O. $=3$

They both have same molecular orbital configuration. Thus, have same bond order.

The correct sequence is

${𝐸}_{2𝑠}\left(O\right)<{𝐸}_{2𝑠}\left(N\right)<{𝐸}_{2𝑠}\left(C\right)<{𝐸}_{2𝑠}\left(B\right)$

With increasing atomic number, nuclear charge increases, pulling electrons closer, so orbital energy decreases.

The complete reaction is as follows

Step $1{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}+$ alc. $\mathrm{KOH}⟶{\mathrm{CH}}_2={\mathrm{CH}}_2$

( $𝐴$, via $𝛽$-elimination)

$\begin{array}{cc}& \text{ Step }2{\mathrm{CH}}_2={\mathrm{CH}}_2+{\mathrm{Br}}_2/\mathrm{hv}\to {\mathrm{CH}}_2\mathrm{BrCH}2\mathrm{Br}\\ & \text{ (B) }\end{array}$

Step $3{\mathrm{CH}}_2{\mathrm{BrCH}}_2\mathrm{Br}+\mathrm{Alc}.\mathrm{KCN}⟶{\mathrm{CH}}_2=\mathrm{CHCN}$ ( C , via double $S_N$ 2 and elimination)

Final product $𝐶$ is acrylonitrile.

$\begin{array}{cc}& \text{ Moles of }{\mathrm{MgCl}}_2=\frac{0.048}{95g/\mathrm{mol}}\\ & =5.05\times {10}^{-4}\mathrm{mol}\\ & \begin{array}{cc}& \text{ van’t Hoff factor }\left(𝑖\right)=3({\mathrm{Mg}}^{2+}+2{\mathrm{Cl}}^{-})\\ & 𝜋=𝑖\text{ CRT }=3\times (5.05\times {10}^{-4}\mathrm{mol}/0.5L)\times 0.0821\times 298\\ & 𝜋\approx 4.98\times {10}^{-2}\mathrm{atm}\end{array}\end{array}$

The complete reaction is as follow

$\begin{array}{cc}& {\mathrm{CH}}_3{\mathrm{COCH}}_3+{\mathrm{CH}}_3\mathrm{MgBr}⟶\\ & {\left({\mathrm{CH}}_3\right)}_3{\mathrm{CO}}^{-}\mathrm{MgBr}\stackrel{H_3{O}^{+}}{\to \leftarrow }{\left({\mathrm{CH}}_3\right)}_3\mathrm{COH}\end{array}$

Grignard reagents react with ketones to form tertiary alcohols. Acetone $\left({\mathrm{CH}}_3{\mathrm{COCH}}_3\right)$ is a ketone, and upon reaction with ${\mathrm{CH}}_3\mathrm{MgBr}$ followed by $H_3{O}^{+}$, yields tert-butanol.

The correct match is A-I, B-II, C-III, D-IV. F has highest electronegativity (I), Cs has lowest I.E. (II), Na is a strong reducing agent (III), O forms hydrogen bonds due to high electronegativity and lone pairs (IV).

Given:

• Molarity of HCl $=2\text{M}$

• Density of solution $=1.08{\text{g mL}}^{-1}$

We have to find the molality.

Molality is defined as:

$\text{Molality}=\frac{\text{moles of solute}}{\text{mass of solvent in kg}}$

Since the molarity is $2\text{M}$, this means:

$1\text{L solution contains }2\text{ moles of HCl}$

Now find the mass of $1\text{L}$ solution.

Since density $=1.08{\text{g mL}}^{-1}$,

$\text{Mass of }1000\text{mL solution}=1.08\times 1000=1080\text{g}$

Mass of HCl present in $1\text{L}$ solution:

Molar mass of HCl $=36.5{\text{g mol}}^{-1}$

$\text{Mass of HCl}=2\times 36.5=73\text{g}$

Now mass of solvent (water):

$\text{Mass of solvent}=\text{mass of solution}-\text{mass of solute}$

$=1080-73=1007\text{g}$

Convert into kg:

$1007\text{g}=1.007\text{kg}$

Now molality:

$𝑚=\frac{2}{1.007}$

$𝑚\approx 1.98𝑚$

So, the correct answer is:

$\overline{)1.98𝑚}$

Hence, option A is correct.

Sodium bisulphite reacts with aldehydes and simple methyl ketones to form white crystalline adducts. Benzophenone is a hindered ketone (diphenyl), so it does not react appreciably with ${\mathrm{NaHSO}}_3$.

Among the given options, ${\mathrm{NH}}_3$ has the highest dipole moment. This is because in ${\mathrm{NH}}_3,N$ is more electronegative than H . So, it will attract the shared pair of electrons towards itself. Because of this the dipole due to N is in same

direction as the resultant dipole moment (i.e. due to $N-H$ bond).

Radial nodes $=𝑛-𝐼-1$

For $4𝑠:𝑛=4,1=0\to 4-0-1=3$

For $3𝑝:𝑛=3,𝑙=1\to 3-1-1=1$

We use the Arrhenius equation in logarithmic form,

$\ln ({𝑘}_2/{𝑘}_1)=({𝐸}_{𝑎}/𝑅)\times [({𝑇}_2-{𝑇}_1)/({𝑇}_1\times {𝑇}_2)]$

Given,

$\begin{array}{cc}& {𝐸}_{𝑎}=9.0\mathrm{kcal}/\mathrm{mol}=9000\mathrm{cal}/\mathrm{mol}\\ & 𝑅=1.987\mathrm{cal}/\mathrm{mol}\cdot K\\ & {𝑇}_1=298K\\ & {𝑇}_2=308K\end{array}$

Now substituting values,

$\begin{array}{cc}& \ln \left(\frac{{𝑘}_2}{{𝑘}_1}\right)=\left(\frac{9000}{1.987}\right)\times \left(\frac{10}{298\times 308}\right)\\ & \ln \left(\frac{{𝑘}_2}{{𝑘}_1}\right)\approx 4531.2\times \left(\frac{10}{91784}\right)\\ & \ln \left(\frac{{𝑘}_2}{{𝑘}_1}\right)=4531.2\times 0.000109\\ & \ln \left(\frac{{𝑘}_2}{{𝑘}_1}\right)=0.494\end{array}$

Now, $\frac{{𝑘}_2}{{𝑘}_1}={𝑒}^{0.497}\approx 1.64$

Percentage increase in rate constant

$=(1.64-1)\times 100=64\%$

So, the rate constant increases by approximately $64\%$.

Moles of $\mathrm{Cu}=6.35/63.5=0.1\mathrm{mol}$

${\mathrm{Cu}}^{2+}+2{𝑒}^{-}⟶\mathrm{Cu}$

$\begin{array}{cc}\text{ Electrons needed }& =0.1\times 2=02\mathrm{mol}\\ & =02\text{ Faraday }\end{array}$

Reaction at anode:

$2H_{2}O⟶O_2+4H^{+}+4e^{-}$

So, $4\mathrm{mol}{𝑒}^{-}\to 1{\mathrm{molO}}_2$

$0.2{\mathrm{mole}}^{-}⟶\left(\frac{0.2}{4}\right)=0.05\mathrm{mol}{O}_2$

Volume at STP $=0.05\times 22.4=1.12L$

In acidic medium, $H_2{O}_2$

Following reaction will take place.

$\begin{array}{cc}{\mathrm{Na}}_2{\stackrel{+6}{\mathrm{Cr}}}_2{O}_7+& 4H_2{O}_2++H_2{\mathrm{SO}}_4\stackrel{\text{ Ether }}{\to \leftarrow }2{\mathrm{CrO}}_{+6}^5+{\mathrm{Na}}_2{\mathrm{SO}}_4+5H_{2}O\end{array}$

Thus, oxidation state of Cr remains same i.e. +6 .

$\mathrm{CuO}+H_2⟶\mathrm{Cu}+H_{2}O$

Moles of $\mathrm{CuO}=\frac{8}{79.5}=0.1006\mathrm{mol}$

So, $0.1006\mathrm{mol}{H}_2$ needed

$\begin{array}{cc}& 0.1006\mathrm{mol}{H}_2=0.1006\times 2\times 96500=19391C\\ & 𝑡=\frac{Q}{1}=\frac{19391}{50}=387.8s\approx 1930s\end{array}$

The correct match is A-I, B-II, C-III, D-IV. Ethanal and propanone → 2,4-DNP (Brady's) Benzaldehyde → Tollen's reagent (silver mirror) Phenol → Violet colour with neutral ${\mathrm{FeCl}}_3$ Cyclohexene → Decolourises Baeyer's reagent (alkene test)

Redox reaction in acidic medium:

$\begin{array}{cc}2{\mathrm{MnO}}_{-}^4+5C_2{O}_{2-}^4+16H^{+}⟶2{\mathrm{Mn}}^{2+}+& 10{\mathrm{CO}}_2+8H_{2}O\end{array}$

So, $2\mathrm{mol}{\mathrm{KMnO}}_4$ oxidises 5 mol oxalic acid.

For 1 mol oxalic acid $\to \left(\frac{2}{5}\right)=0.4\mathrm{mol}{\mathrm{KMnO}}_4$

$\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_3{\mathrm{Cl}}_3\right]$ is an octahedral complex of the type ${\mathrm{MA}}_3{B}_3$, which can exhibit fac and mer geometrical isomers.

To suppress the concentration of free amine otherwise it will couple with the diazonium salt to give the corresponding azo dye.

The complete reaction sequence is as follows.

Zinc has a fully filled ${𝑑}^{10}$ configuration in ${\left[\mathrm{Zn}{\left(H_{2}O\right)}_6\right]}^{2+}$, so nod-d transition occurs, making it colourless. Other options involve ions with unpaired $𝑑$-electrons, which absorb visible light due to $𝑑-𝑑$ transition.

From $\mathrm{pH}=3.85$, we get:

$\left[{𝐻}^{+}\right]={10}^{-3.85}=1.41\times {10}^{-4}M$

For conjugate acid $\left(C_6{H}_5{\mathrm{NH}}_{+}^3\right)$, hydrolysis happens:

$C_6{H}_5{\mathrm{NH}}_{+}^3+H_{2}O\rightleftharpoons C_6{H}_5{\mathrm{NH}}_2+H_3{O}^{+}$

Using the formula,

$\begin{array}{cc}{𝐾}_{𝑏}& =({𝐾}_{𝑤}\times 𝐶)/{\left[H^{+}\right]}^2\\ & =(1\times {10}^{-14}\times 0.05)/{(1.41\times {10}^{-4})}^2\\ & \approx 2.51\times {10}^{-8}\end{array}$

Thus, ${𝐾}_{𝑏}\approx 2.5\times {10}^{-8}$

NO has an odd number of electrons and one unpaired electron, making it paramagnetic. It also has a lone pair on nitrogen.

The complete reaction sequence will yield 2, 4, 6-tribromoaniline.

As we move down the group in halogens, the number of shells increases, so atomic size increases as $F<\mathrm{Cl}<\mathrm{Br}<I$.

$\begin{array}{cc}\Delta {𝐺}^{\circ }& =-𝑅𝑇\ln 𝑘\\ 𝑘& ={𝑒}^{(-\Delta 𝐺/𝑅𝑇)}\\ & ={𝑒}^{(-(-50200)/(8.314\times 298\left)\right)}\\ & \approx {𝑒}^{\left(20.2\right)}\\ & \approx 7.9\times {10}^8\end{array}$

So, ${𝐾}_{\mathrm{sp}}=7.9\times {10}^{-8}$ (reverse sign of exponent because dissolution is written forward).

Vitamin C and B complex are water-soluble. $𝐴,𝐷,𝐸,𝐾$ are fat-soluble.

$\begin{array}{cc}& \text{ For first-order reaction, }\\ & 𝑘=\left(\frac{2.303}{𝑡}\right)\times \log \left(\frac{[𝐴{]}_0}{\left[𝐴\right]}\right)\end{array}$

$\begin{array}{cc}& =\left(\frac{2.303}{20}\right)\times \log \left(\frac{0.5}{0.4}\right)\\ & =0.11515\times 0.0969\approx 0.0112{\min }^{-1}\end{array}$

If hydrolysis of an ester yields ethanol, it indicates the original ester was likely ethyl acetate or a similar derivative.

To confirm the presence of ethanol, we can use the iodoform test, which is specific for compounds containing the ${\mathrm{CH}}_3-\mathrm{CH}\left(\mathrm{OH}\right)$ group (ethanol) or ${\mathrm{CH}}_3-\mathrm{CO}-$ group.

$\begin{array}{cc}& \text{ Ethanol }+I_2+\mathrm{NaOH}⟶{\mathrm{CHI}}_3\text{ (yellow ppt.) }\\ & +\mathrm{HCOONa}+\mathrm{NaI}+H_{2}O\end{array}$

This test gives a yellow precipitate of iodoform, confirming ethanol is present.

Osazone formation occurs by reacting a sugar with excess phenylhydrazine

$\left(C_6{H}_5{\mathrm{NHNH}}_2\right)$. The reaction occurs at $C-1$ and $C-2$ carbons of the sugar.

Glucose:

$\mathrm{CHO}-(\mathrm{CHOH}{)}_4-{\mathrm{CH}}_2\mathrm{OH}+3C_6{H}_5{\mathrm{NHNH}}_2$

⟶ Glucosazone (yellow crystals)

$+2{\mathrm{NH}}_3+H_{2}O$

Fructose:

${\mathrm{CH}}_2\mathrm{OH}-\mathrm{CO}-(\mathrm{CHOH}{)}_3-{\mathrm{CH}}_2\mathrm{OH}+3C_6{H}_5{\mathrm{NHNH}}_2⟶$ Same glucosazone

$+2{\mathrm{NH}}_3+H_{2}O$

The reaction involved are:

$\begin{array}{cc}& O_3+2\mathrm{KI}+H_{2}O⟶I_2+O_2+2\mathrm{KOH}\\ & I_2+2{\mathrm{Na}}_2{S}_2{O}_3⟶2\mathrm{Nal}+{\mathrm{Na}}_2{S}_4{O}_6\\ & \text{ Moles }=\text{ Molarity }\text{ × }\text{ Volume }\text{ (in litres) }\\ & =0.01\mathrm{mol}/L\times (20.0\mathrm{mL}/1000)=2.00\times {10}^{-4}\mathrm{mol}\\ & I_2+2{\mathrm{Na}}_2{S}_2{O}_3⟶2\mathrm{Nal}+{\mathrm{Na}}_2{S}_4{O}_6\end{array}$

The reaction, tells us that 1 mole of $I_2$ reacts with 2 moles of ${\mathrm{Na}}_2{S}_2{O}_3$.

$\begin{array}{cc}& \text{ So, moles of }{I}_2=\frac{2\times {10}^{-4}}{2}=1.00\times {10}^{-4}\mathrm{mol}\\ & O_3+2\mathrm{KI}+H_{2}O⟶I_2+O_2+2\mathrm{KOH}\end{array}$

The reaction, shows that 1 mole of $O_3$ gives 1 mole of $I_2$.

Hence, moles of $O_3=$ moles of $I_2$

$=1.00\times {10}^{-4}\mathrm{mol}\text{. }$

The correct option is Option A: thinks outside the box.

Explanation:

• The idiom “think outside the box” means to think creatively or in an innovative way, going beyond conventional ideas.

• The sentence—“The team always _ to come up with unique advertisements for all its clients”—suggests creativity and originality.

• Therefore, the most appropriate idiom is “thinks outside the box.”

Correct sentence:

The team always thinks outside the box to come up with unique advertisements for all its clients.

The word "Prudent" means careful and wise in handling practical matters or showing good judgment.

Let us understand each option:

• Option A: Precious – means valuable or dear. (Not related in meaning)

• Option B: Diffident – means shy or lacking confidence. (Not related)

• Option C: Judicious – means showing good judgment or sense. (Same meaning)

• Option D: Doubtful – means uncertain. (Not related)

Hence, the correct synonym of "Prudent" is:

Option C: Judicious

The underlined word is "hysterics", which means a state of extreme emotion or uncontrolled laughter or crying.

The antonym would therefore describe a state of composure, peace, or self-control.

• Option A: Aghast — means shocked or filled with horror (not opposite).

• Option B: Cray — informal and not a proper antonym.

• Option C: Calm — means peaceful, composed, and not emotionally disturbed.

• Option D: Insanity — means mental instability (not opposite).

Hence, the most appropriate antonym is:

Option C: Calm