BITSAT 2025 Question Paper with Solutions

The energy of an electron in a hydrogen atom is given by the formula:

${𝐸}_{𝑛}=-\frac{{𝑅}_H}{{𝑛}^2}$

We are told that the energy of the atom is equal to $-\frac{{𝑅}_H}{16}$. To find the principal quantum number $𝑛$, we compare the two expressions:

$-\frac{{𝑅}_H}{{𝑛}^2}=-\frac{{𝑅}_H}{16}$

Canceling out $-{𝑅}_H$ from both sides gives:

$\frac{1}{{𝑛}^2}=\frac{1}{16}$

Therefore, ${𝑛}^2=16\Rightarrow 𝑛=4$.

In a hydrogen atom, all orbitals that share the same principal quantum number $𝑛$ have the same energy. This means they are called degenerate. The total number of these degenerate orbitals is given by ${𝑛}^2$.

So, for $𝑛=4$:

Degeneracy $={𝑛}^2=4^2=16$.

Magnesium nitrate separates into ions in solution as shown below:

$\mathrm{Mg}{\left({\mathrm{NO}}_3\right)}_2\to {\mathrm{Mg}}^{2+}+2{\mathrm{NO}}_{-}^3$

This means that one unit of magnesium nitrate produces a total of 3 particles after complete dissociation (1 magnesium ion and 2 nitrate ions).

If the salt is only 40% dissociated, it means $𝛼=0.4$.

The Van’t Hoff factor, $𝑖$, which shows how many times the number of effective particles increases due to dissociation, is given by:

$𝑖=1+𝛼(𝑛-1)$

Substituting the values, we get:

$𝑖=1+0.4(3-1)=1+0.8=1.8$

$\text{ The complete reaction sequence is as follows }$

In a cyclic process, net heat absorbed = work done $=$ area enclosed in $𝑝𝑉$ diagram.

Area $=(3-1)\mathrm{atm}\times (4-2)L$

$=2\mathrm{atm}\times 2L=4L\mathrm{atm}$.

Now, convert to energy units if needed,

$4\mathrm{Latm}\times 101.3J<text>=</text><text>\sim </text>405J$ (optional)

But in $L\cdot \mathrm{atm}$, answer is 4 L atm

First, separate into two half-reactions:

Oxidation $({\mathrm{Fe}}^{2+}$ to ${\mathrm{Fe}}^{3+})$,

${\mathrm{Fe}}^{2+}⟶{\mathrm{Fe}}^{3+}+1{𝑒}^{-}$

Reduction ( ${\mathrm{MnO}}_{-}^4$to ${\mathrm{Mn}}^{2+}$ ),

${\mathrm{MnO}}_{-}^4⟶{\mathrm{Mn}}^{2+}$

Balance O by adding $4H_{2}O$,

${\mathrm{MnO}}_{-}^4⟶{\mathrm{Mn}}^{2+}+4H_{2}O$

Balance H by adding $8H^{+}$,

$8H^{+}+{\mathrm{MnO}}_{-}^4⟶{\mathrm{Mn}}^{2+}+4H_{2}O$

Add $5{𝑒}^{-}$to balance charge,

$8H^{+}+{\mathrm{MnO}}_{-}^4+5{𝑒}^{-}⟶{\mathrm{Mn}}^{2+}+4H_{2}O$

Now, multiply oxidation half-reaction by 5 :

$5{\mathrm{Fe}}^{2+}⟶5{\mathrm{Fe}}^{3+}+5{𝑒}^{-}$

Now add both:

$8H^{+}+{\mathrm{MnO}}_{-}^4+5{\mathrm{Fe}}^{2+}⟶{\mathrm{Mn}}^{2+}+5{\mathrm{Fe}}^{3+}+4H_{2}O$

Thus, the stoichiometric coefficient of $H_{2}O=4$

The de-Broglie wavelength is given by the formula:

$𝜆=\frac{ℎ}{𝑚𝑉}$

According to the question, the wavelength is also given as:

$𝜆=10\sqrt{\surd \frac{ℎ}{𝑚}}$

Now, equating the two expressions for $𝜆$:

$\frac{ℎ}{𝑚𝑉}=10\sqrt{\surd \frac{ℎ}{𝑚}}$

Multiply both sides by $𝑚𝑉$:

$ℎ=10𝑚𝑉\sqrt{\surd \frac{ℎ}{𝑚}}$

Divide both sides by 10:

$\frac{ℎ}{10}=𝑚𝑉\sqrt{\surd \frac{ℎ}{𝑚}}$

Square both sides to remove the square root:

${\left(\frac{ℎ}{10}\right)}^2=(𝑚𝑉{)}^2\times \frac{ℎ}{𝑚}$

Simplify the equation:

$\frac{{ℎ}^2}{100}={𝑚}^2{𝑉}^2\frac{ℎ}{𝑚}$

$\frac{ℎ}{100}=𝑚{𝑉}^2$

Now, solve for $𝑉$:

$𝑉=\sqrt{\surd \frac{ℎ}{100𝑚}}$

To find the ratio of speed to mass, $\frac{𝑉}{𝑚}$:

$\frac{𝑉}{𝑚}=\frac{1}{𝑚}\sqrt{\surd \frac{ℎ}{100𝑚}}=\frac{\sqrt{\surd ℎ}}{10{𝑚}^{3/2}}$

Electron affinity is highest when addition of electron releases maximum energy.Chlorine (Cl) has highest EA due to optimal size and its electronic configuration.

Oxygen and fluorine have lower EA due to electron-electron repulsion in small orbitals.

Let, ${𝐴}_2=𝑥$, then ${𝐵}_2=4𝑥,𝐴𝐵=2𝑥$

$\begin{array}{cc}\Delta 𝐻& =[\frac{1}{2}{𝐴}_2+\frac{1}{2}{𝐵}_2⟶𝐴𝐵]\\ & =(\frac{1}{2}𝑥+2𝑥)-2𝑥=\frac{1}{2}𝑥=-100\to 𝑥=-200\end{array}$

So, ${𝐵}_2=4𝑥=-800\mathrm{kJ}/\mathrm{mol}$

$\mathrm{HCl}<{\mathrm{HNO}}_2<{\mathrm{CH}}_3\mathrm{COOH}$

HCl is a strong acid (complete ionisation), ${\mathrm{HNO}}_2$ is a stronger weak acid than ${\mathrm{CH}}_3\mathrm{COOH}$.

So, HCl has lowest $\mathrm{pH},{\mathrm{CH}}_3\mathrm{COOH}$ highest pH .

Order of acidity: $\mathrm{HCl}>{\mathrm{HNO}}_2>{\mathrm{CH}}_3\mathrm{COOH}$

Hence, the correct order of $\mathrm{pH}:\mathrm{HCl}<{\mathrm{HNO}}_2<{\mathrm{CH}}_3\mathrm{COOH}$

The correct order is $N^{3-}>O^{2-}>F^{-}$.

All are isoelectronic (10 electrons), and nuclear charge increases from $N^{3-}$ and $F^{-}$.

Greater nuclear charge pulls electrons closer, reducing size.

Hence, size order: $N^{3-}>O^{2-}>F^{-}$.

$\begin{array}{cc}& \text{ Henderson equation, }\mathrm{pH}=p{𝐾}_{𝑎}+\log \left(\frac{\left[\text{ Salt }\right]}{\left[\text{ Acid }\right]}\right)\\ & \begin{array}{cc}{\mathrm{pK}}_{𝑎}& =-\log (1.8\times {10}^{-5})=4.74\\ \text{ Ratio }& =\frac{(0.1\times 60)}{(0.1\times 30)}=2\\ \mathrm{pH}& =4.74+\log \left(2\right)\\ & \approx 4.74+0.3=5.04\end{array}\end{array}$

Elevation of boiling point depends on the vant' Hoff factor (i).

Urea is non-electrolyte, $𝑖=1$

${\mathrm{CaCl}}_2$ dissociates into 3 ions, $𝑖=3$

So, effective particles $=0.05\times 3=0.15\mathrm{mol}$ for ${\mathrm{CaCl}}_2$ is more than 0.1 mol for urea is equal amount of solvent.

Hence, ${\mathrm{CaCl}}_2$ solution will show greater elevation.

The compound has two double bonds:

${\mathrm{CH}}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}={\mathrm{CH}}_2$

Only the central $C=C$ (second one) can show geometrical isomerism, as the terminal alkene ( $\mathrm{CH}={\mathrm{CH}}_2$ ) lacks two different groups on one carbon.

So, only one C=C shows cis-trans isomerism $\to 2$ isomers

$\text{ The complete reaction occurs as follows }$

Electrolysis of NaCl produces NaOH at the cathode.

Charge passed $=4\times 1200=4800C$

Moles of electrons $=\frac{4800}{96500}\approx 0.04974\mathrm{mol}$

Since 2 moles of electron produce 2 moles NaOH , NaOH formed $\approx 0.04974\mathrm{mol}$

$\begin{array}{c}\text{ Concentration }=\frac{0.04974}{0.5}=0.0995M\\ \text{ pOH }=-\log 0.0995\\ \text{ pOH }\approx 1,\text{ so }\mathrm{pH}\approx 14-1=13\end{array}$

Salicylaldehyde has - OH and - CHO ortho to each other, enabling intramolecular H-bonding.

$𝑚$-hydroxy and benzoic acid form intermolecular H-bonds.

$\begin{array}{cc}& \\ & \begin{array}{cc}& {\mathrm{Ca}}^{2+}+2{𝑒}^{-}⟶\mathrm{Ca}\text{ (at cathode) }\\ & 1\text{ mole of }{\mathrm{CaCl}}_2=2F\\ & 𝑄=𝐼𝑡\Rightarrow 𝑛\cdot 2𝐹=𝐼𝑡\\ & \Rightarrow 𝑡=\frac{𝑛\cdot 2𝐹}{𝐼}=\frac{20}{40}\times \frac{2\times 96500}{4}\\ & =24125s=6.7\text{ hours. }\end{array}\end{array}$

${\mathrm{XeF}}_2$ has $𝑠{𝑝}^3𝑑$ hybridisation involving 5 orbitals (one of them $𝑑$ ). The rest are $𝑠{𝑝}^3$ (no $𝑑$-orbitals).

Aldol condensation requires the presence of at least one $𝛼$-hydrogen in the aldehyde. Among the options, acetaldehyde $\left({\mathrm{CH}}_3\mathrm{CHO}\right)$ has $𝛼$-hydrogens and undergoes the reaction.

$2{\mathrm{CH}}_3\mathrm{CHO}+\text{ dilute }\mathrm{NaOH}⟶{\mathrm{CH}}_3\mathrm{CH}\left(\mathrm{OH}\right){\mathrm{CH}}_2\mathrm{CHO}\text{ (3-hydroxy butanal) }$

This $𝛽$-hydroxy aldehyde may further dehydrate to form an $𝛼,𝛽$-unsaturated aldehyde:

${\mathrm{CH}}_3\mathrm{CH}\left(\mathrm{OH}\right){\mathrm{CH}}_2\mathrm{CHO}⟶{\mathrm{CH}}_3\mathrm{CH}=\mathrm{CHCHO}+H_{2}O$

The other compounds either lack $𝛼$-hydrogens or are deactivated (e.g., benzaldehyde, phenylacetylene derivative).

Balanced reaction,

$\begin{array}{cc}& 5{\mathrm{Fe}}^{2+}+{\mathrm{MnO}}_{-}^4+8H^{+}⟶5{\mathrm{Fe}}^{3+}+{\mathrm{Mn}}^{2+}+4H_{2}O\\ & \text{ Using, molarity }=\frac{\text{ Number of moles of solute }}{\text{ Volume of solution }\left(L\right)}\\ & \begin{array}{cc}\text{ or }\left({\mathrm{KMnO}}_4\right)& =\left({\mathrm{FeSO}}_4\right)\\ \frac{{𝑀}_1{𝑉}_1}{{𝑛}_1}& =\frac{{𝑀}_2{𝑉}_2}{{𝑛}_2}\\ {\mathrm{Fe}}^{2+}\text{ moles }& =0.02\times 0.5=0.01\mathrm{mol}\end{array}\end{array}$

5 moles of ${\mathrm{Fe}}^{2+}$ react with 1 mole of ${\mathrm{KMnO}}_4$. So,

${\mathrm{KMnO}}_4$ required $=0.01/5=0.002\mathrm{mol}$

Volume $=0.002/0.1=0.02L=20\mathrm{mL}$

Riboflavin ( $B_2$ ) deficiency leads to cheilosis (cracks at mouth corners), glossitis, and digestive issues. Vitamin A is related to vision; C for scurvy; Niacin for pellagra.

$\text{ The complete reaction sequence is as follow }$

From the balanced reaction,

$2{\mathrm{SO}}_2+O_2\rightleftharpoons 2{\mathrm{SO}}_3$

Using stoichiometry,

$\begin{array}{cc}& \text{ Rate of disapperance of }{O}_2=2\times {10}^{-1}{\mathrm{molL}}^{-1}{s}^{-1}\\ & \frac{-𝑑\left[O_2\right]}{𝑑𝑡}=\frac{1}{2}\frac{𝑑\left[{\mathrm{SO}}_3\right]}{𝑑𝑡}\\ & \frac{𝑑\left[{\mathrm{SO}}_3\right]}{𝑑𝑡}=2\times \left[\frac{-𝑑\left[O_2\right]}{𝑑𝑡}\right]\end{array}$

Rate of appearance of

${\mathrm{SO}}_3=2\times (2\times {10}^{-1})=4\times {10}^{-1}{\mathrm{molL}}^{-1}{s}^{-1}$

Half-life for first-order reaction,

${𝑡}_{1/2}=0.693/k$

Given, ${𝑡}_{1/2}=600s$

$𝑘=\frac{0.693}{600}\approx 1.155\times {10}^{-3}{s}^{-1}$

Using Arrhenius,

$𝑘=𝐴{𝑒}^{(-{𝐸}_0/𝑅𝑇)}$

Take $\log$ both sides and solve,

$\ln 𝑘=\ln \left(𝐴\right)-\frac{{𝐸}_{𝑎}}{𝑅𝑇}$

$\begin{array}{cc}& \text{ Substituting the values, }\\ & \begin{array}{cc}\ln (1.155\times {10}^{-3})& =\ln (4\times {10}^{13})-\left(\frac{98600}{𝑅𝑇}\right)\\ & \approx 311.49𝑘\end{array}\end{array}$

Given that, alanine is the C-terminal amino acid and N-terminal must be chiral. So, glycine is not used at the N-terminal because it is achiral. So, possible sequence of tetrapeptides are

Val-Gly-Ser-Ala

Val-Ser-Gly-Ala

Ser-Gly-Val-Ala

Ser-Val-Gly-Ala

We calculate the number of unpaired electrons in each ion,

(i) ${\mathrm{Cr}}^{3+}\to \left[\mathrm{Ar}\right]3{𝑑}^3\to 3$ unpaired electrons

(ii) ${\mathrm{Fe}}^{3+}\to \left[\mathrm{Ar}\right]3{𝑑}^5\to 5$ unpaired electrons

(iii) ${\mathrm{Mn}}^{2+}\to \left[\mathrm{Ar}\right]3{𝑑}^5\to 5$ unpaired electrons

(iv) ${\mathrm{Co}}^{2+}\to \left[\mathrm{Ar}\right]3{𝑑}^7\to 3$ unpaired electrons

(v) ${\mathrm{Ni}}^{2+}\to \left[\mathrm{Ar}\right]3{𝑑}^8\to 2$ unpaired electrons

Now compare:

• ${\mathrm{Cr}}^{3+}$ and ${\mathrm{Co}}^{2+}\to$ both have 3 unpaired electrons.

• ${\mathrm{Fe}}^{3+}$ and ${\mathrm{Mn}}^{2+}\to$ both have 5 unpaired electrons.

Two distinct pairs have same number of unpaired electrons.

${\mathrm{Fe}}^{3+}$ is $3{𝑑}^5$, all electrons unpaired (high-spin ${𝑑}^5$ ): Spin-only moment

$\begin{array}{cc}& =\sqrt{\surd \left[𝑛\right(𝑛+2\left)\right]}=\sqrt{\surd [5\times 7]}\\ & =\sqrt{\surd 35}\approx 5.92\mathrm{BM}\approx 6\mathrm{BM}\end{array}$

${\mathrm{CN}}^{-}$is a strong field ligand; it causes pairing in ${\mathrm{Ni}}^{2+}\left({𝑑}^8\right)$.

The $𝑑𝑠{𝑝}^2$ hybridisation forms a square planar geometry with no unpaired electrons.

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

A. $\mathrm{Ni}(\mathrm{CO}{)}_4$, all ligands neutral. So, oxidation state, $\mathrm{Ni}=0$

B. $[\mathrm{Fe}{\left(H_{2}O\right)}_5{\mathrm{NO}}^{2+},{\mathrm{NO}}^{+}=+1,H_{2}O=0$. So oxidation state, $\mathrm{Fe}=+1$

C. ${\left[\mathrm{Co}\right(\mathrm{CO}{)}_5]}^{2-},𝑥+5\left(0\right)=-2$, So, oxidation state, $\mathrm{Co}=-2$

D. ${\left[{\mathrm{Cr}}_2\right(\mathrm{CO}{)}_{10}]}^{2-}$, each $\mathrm{Cr}=-1($ since total $=-2)$

Williamson ether synthesis An $S_{N}2$ reaction between an alkoxide and an alkyl halide. Williamson synthesis work best with $2^{\circ }$ and $3^{\circ }$. If often gives E2 elimination product.

'eat' can be used in place of 'had eaten' in the given sentence. As the sentence expresses a habit, simple present, tense will be used in the sentence.

'fast' is most appropriate to fill the blank in the given sentence as per the context.

'The children opened the door for their father', best expresses the given sentence in active voice.

'Famous' means known about many people in many places. 'Obscure' from the given options is the most appropriate antonym of 'famous' as 'Obscure' means not well known and usually not very important.

'Austere' means strict and serious in manner. Hence, 'Stern' from the given options is the most appropriate synonym of 'Austere'. 'Stern' means serious and strict, and showing disapproval of someone's behaviour.

'Divisible' means able to be divided, for example by a number. Hence, it can be used as a one-word substitute for the given group of words.

'Altruism' can be used as a one-word substitute for the given group of words. 'Altruism' is when you care about or help other people, even though this brings no advantage to yourself.

The highlighted idiom 'A chip off the old block' means someone who is very similar in character or appearance to their father or mother. Hence,option (b) gives its most appropriate meaning.

$\text{ DCBA }$

Option (b) best expresses the given sentence in passive voice.

As per the options,

(a) $𝑆\stackrel{-1}{\to \leftarrow }𝑅\stackrel{+5}{\to \leftarrow }𝑊\stackrel{+1}{\to \leftarrow }𝑋$

(b) $H^{-1}G\stackrel{+5}{\to \leftarrow }L\stackrel{+1}{\to \leftarrow }M$

(c) $P\stackrel{-1}{\to \leftarrow }O\stackrel{+5}{\to \leftarrow }T\stackrel{+1}{\to \leftarrow }U$

From figure 1st to 2nd the element at top left corner shifted to the top right corner in the clockwise direction. Similarly, all the element at corner are shifted one place in clockwise direction, but arrow in the middle shifted ${45}^{\circ }$ in anti-clockwise direction. The same pattern follow in the next steps. Hence, option

(d) is the correct answer.

Pattern of the given series is as follows,

Hence, option (c) is the correct answer.

As, $25:18$

$25-(2+5)=25-7=18$

And $49:36$

$49-(4+9)=49-13=36$

Similarly, 37 : 27

$37-(3+7)=37-10=27$

Hence, option (d) is the correct answer.

As, barometer is an instrument which is used to measure pressure. Similarly, multimeter is used to measure electricity. Hence, option (d) is the correct answer.

The given figure is embedded in option figure (a) as its part.

Hence, option (a) is the correct answer.

As $(112,327,215)$

$112+215=327$

And, $(98,254,156)$

$98+156=254$

Similarly, $(126,268,142)$

$126+142=268$

Hence, option (c) is the correct answer.

In the series,

Hence, option (c) is the correct answer.

As per the question, given figure is hidden in option (c) figure the answer.

Hence, option (c) is the correct answer.

As per the question,

$\therefore D$ is sister of U .

Hence, option (a) is the correct answer.

As per the question,

A1 is opposite to U5.

Hence, option (a) is the correct answer.

As per the question,

Hence, on opposite of C, 2 will come. Hence, option (b) is the correct answer.

$\text{ As per the question, }$

From the above diagram,

$\begin{array}{cc}& 6+3=9=2𝑥+1\\ \Rightarrow & 9-1=2𝑥\\ \Rightarrow & 8=2𝑥\\ \Rightarrow & 𝑥=4\end{array}$

Hence, option (c) is the correct answer.