JEE Main 2025 29th January Morning Shift Question Paper with Solutions

$$\begin{aligned} & \mathrm{n}_1 \sin \left(90-\theta_1\right)=\mathrm{n}_2 \sin \theta_3 \\ & \mathrm{n}_1 \cos \theta_1=\mathrm{n}_2 \sin \theta_3 \\ & \mathrm{n}_1 \frac{\mathrm{n}_2}{2 \mathrm{n}_1}=\mathrm{n}_2 \sin \theta_3 \\ & \frac{1}{2}=\sin \theta_3, \theta_3=30 \\ & \tan 30=\frac{\mathrm{d}}{2(\mathrm{t})} \\ & \mathrm{t}=\frac{\mathrm{d} \sqrt{3}}{2}=\frac{4 \sqrt{3} \times \sqrt{3}}{2} \mathrm{~cm}=6 \mathrm{~cm} \end{aligned}$$

$$\begin{aligned} & \mathrm{V}_{\text {Top }}=\sqrt{\mathrm{n}^2 \mathrm{gR}} \\ & \mathrm{~V}_{\text {Botom }}=\sqrt{\mathrm{n}^2 \mathrm{gR}+4 \mathrm{gR}} \\ & \text { Ratio }=\frac{\mathrm{n}^2+4}{\mathrm{n}^2} \end{aligned}$$

$$\begin{aligned} & \mathrm{B}=\frac{\rho \mathrm{gh}}{\left(\frac{\Delta \mathrm{v}}{\mathrm{v}}\right)} \\ & \frac{\Delta \mathrm{v}}{\mathrm{v}} \times 100=\frac{\rho \mathrm{gh}}{\mathrm{~B}} \times 100 \\ & \frac{1000 \times 10 \times 2.5 \times 10^3}{2 \times 10^9} \times 100 \% \\ & =1.25 \% \end{aligned}$$

Maximum possible magnetic field is at the surface

$$\begin{aligned} & \mathrm{B}_{\max }=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{a}} \\ & \frac{\mathrm{~B}_{\max }}{2}=\frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{a}} \end{aligned}$$

It can be obtained inside as well as outside the wire

For inside,

$$\begin{aligned} & \frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{a}}=\frac{\mu_0 \mathrm{Ir}}{2 \pi \mathrm{a}^2} \\ & \Rightarrow \mathrm{r}=\frac{\mathrm{a}}{2} \end{aligned}$$

For outside

$$\begin{aligned} & \frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{a}}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}} \\ & \Rightarrow \mathrm{r}=2 \mathrm{a} \end{aligned}$$

Correct answer $\left[\frac{\mathrm{a}}{2}, 2 \mathrm{a}\right]$

Assertion (A) is true because a sufficiently negative potential can repel and stop the emitted electrons from leaving the surface.

Reason (R) is also true because the stopping potential $V_{\text{stop}}$ varies linearly with the frequency $\nu$ of the incident light ($V_{\text{stop}} = \frac{h}{e}\,\nu - \frac{\phi}{e}$).

However, (R) does not directly explain why a negative potential suppresses electron emission; it only shows how much potential is needed for different frequencies.

Therefore, both (A) and (R) are true, but (R) is not the correct explanation of (A).

$$\begin{aligned} & \mathrm{I} \omega 2 \theta=\mathrm{q} \ell \mathrm{E}_0 \theta \\ & 2 \mathrm{~m}\left(\frac{\ell}{2}\right)^2 \omega^2=\mathrm{q} \ell \mathrm{E}_0 \\ & \omega^2=\frac{2 \mathrm{qE}_0}{\mathrm{~m} \ell} \\ & \mathrm{~T}=2 \pi \sqrt{\frac{\mathrm{~m} \ell}{2 \mathrm{qE}_0}} \end{aligned}$$

$(u+f)(v-f)=f^2$

$$\begin{aligned} &\begin{aligned} \phi & =\mathrm{BAN} \cdot \cos (\omega \mathrm{t}) \\ \varepsilon & =\frac{-\mathrm{d} \phi}{\mathrm{dt}}=\mathrm{BA} \omega \mathrm{~N} \cdot \sin (\omega \mathrm{t}) \end{aligned}\\ &\text { When } B \text { is parallel to plane, } \underline{\underline{\omega}} \mathrm{t}=\frac{\pi}{2}\\ &\Rightarrow \phi=0, \varepsilon=\mathrm{BA} \omega \mathrm{~N} \end{aligned}$$

The torque $\vec{\tau}$ acting on the particle with respect to the origin can be calculated using the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$:

$\vec{\tau} = \vec{r} \times \vec{F}$

Given the position vector $\vec{r} = (1, 1, 1) \, \text{m}$ and the force vector $\vec{F} = \hat{i} - \hat{j} + \hat{k}$, we need to calculate the cross product:

$$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix}$$

Calculating the determinant, we have:

$\vec{\tau} = \hat{i} \left(1 \cdot 1 - 1 \cdot (-1)\right) - \hat{j} \left(1 \cdot 1 - 1 \cdot 1\right) + \hat{k} \left(1 \cdot (-1) - 1 \cdot 1\right)$

This simplifies to:

$\vec{\tau} = \hat{i}(1 + 1) - \hat{j}(1 - 1) + \hat{k}(-1 - 1)$

$\vec{\tau} = 2\hat{i} - 0\hat{j} - 2\hat{k}$

The torque vector is $\vec{\tau} = 2\hat{i} - 2\hat{k}$.

To find the magnitude of the torque in the z-direction, we look at the $\hat{k}$ component:

$\tau_z = -2$

The magnitude of torque in the z-direction is:

$|\tau_z| = 2 \, \text{Nm}$

Thus, the magnitude of the torque in the z-direction is 2 Newton-meters.

To solve this problem, we can use the combined gas law in terms of pressure ($P$) and temperature ($T$). For a gas with a fixed volume, the relationship is given as:

$\frac{P_1}{T_1} = \frac{P_2}{T_2}$

where:

$P_1$ is the initial pressure,

$T_1$ is the initial temperature in Kelvin,

$P_2$ is the final pressure,

$T_2$ is the final temperature in Kelvin.

Given that the initial temperature $T_1 = 27°C$, we need to convert it to Kelvin:

$T_1 = 27°C + 273.15 = 300.15 \, \text{K}$

The final pressure $P_2$ is twice the initial pressure ($P_2 = 2P_1$). Substituting these values into the gas law equation gives:

$\frac{P_1}{300.15} = \frac{2P_1}{T_2}$

Solving for $T_2$:

Cancel $P_1$ from both sides:

$\frac{1}{300.15} = \frac{2}{T_2}$

Rearrange to solve for $T_2$:

$T_2 = 2 \times 300.15 = 600.3 \, \text{K}$

Convert the final temperature back to Celsius:

$T_2 = 600.3 \, \text{K} - 273.15 = 327.15°C$

Therefore, to double the pressure of the gas, the temperature should be raised to approximately 327°C.

$\Delta \mathrm{W}=-\Delta \mathrm{U}=-\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$

$$\begin{aligned} & \phi_1=\mathrm{L}_1 \mathrm{I}_1+\mathrm{M}_{12} \mathrm{I}_2 \\ & \varepsilon_1=-\frac{\mathrm{d} \phi_1}{\mathrm{dt}}=-\mathrm{L}_1 \frac{\mathrm{dI}_1}{\mathrm{dt}}-\mathrm{M}_{12} \frac{\mathrm{dI}}{\mathrm{dt}} \end{aligned}$$

$$\begin{aligned} & \sin \theta_{1 C}=\frac{n_1}{n_2} \\ & \sin \theta_{2 C}=\frac{n_1}{n_3} \\ & \sin \theta_{2 C}-\sin \theta_{1 C}=\frac{1}{2} \end{aligned}$$

$$\begin{aligned} & \mathrm{n}_1 \frac{\mathrm{n}_2}{\mathrm{n}_3}-\frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{1}{2} \\ & \mathrm{n}_1 \frac{\mathrm{n}_2}{\mathrm{n}_3}-\mathrm{n}_1=\frac{\mathrm{n}_2}{2} \\ & \mathrm{n}_1\left(\frac{2}{5}-1\right)=\frac{\mathrm{n}_2}{2} \\ & \frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{-5}{6} \\ & =\sin ^{-1}\left(-\frac{5}{6}\right) \end{aligned}$$

$$\begin{aligned} & \mathrm{H}_{\mathrm{Max}}=\frac{(\mathrm{usin} \theta)^2}{2 \mathrm{~g}} \\ & \frac{\left(\mathrm{H}_{\max }\right)_1}{\left(\mathrm{H}_{\max }\right)_2}=\frac{\mathrm{u}^2 \sin ^2(45-\alpha)}{\mathrm{u}^2 \sin ^2(45+\alpha)} \\ & =\frac{\left(\frac{1}{\sqrt{2}} \cos \alpha-\frac{1}{\sqrt{2}} \sin \alpha\right)^2}{\left(\frac{1}{\sqrt{2}} \cos \alpha+\frac{1}{\sqrt{2}} \sin \alpha\right)^2} \\ & =\frac{1-\sin 2 \alpha}{1+\sin 2 \alpha} \end{aligned}$$

$$\begin{aligned} &\text { As } h \text { increases, } g \text { decreases, } \mathrm{T} \text { increases }\\ &\begin{aligned} & \mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{~g}}} \\ & \mathrm{~g}=\frac{\mathrm{g}_0 \mathrm{R}^2}{(\mathrm{R}+\mathrm{h})^2} \end{aligned} \end{aligned}$$

(A) $\rightarrow 0$ (III)

(B) $\rightarrow \frac{\sigma}{2 \varepsilon_0}$

(C) $\rightarrow \frac{\sigma \mathrm{R}^2}{\varepsilon_0 \mathrm{r}^2}$ (No row matching)

$(\mathrm{D}) \rightarrow \frac{\sigma}{\varepsilon_0}(\mathrm{I})$

Velocity of a just before hitting :

$\mathrm{u}=\sqrt{2 \mathrm{~g} \frac{\mathrm{R}}{2}}=\sqrt{\mathrm{gR}}$

Just after collision, let velocity of A and B are $\mathrm{v}_1$ and $v_2$ respectively

$\therefore \text { by COM: }$

$$\begin{aligned} & \mathrm{mu}=\mathrm{mv}_1+\frac{\mathrm{m}}{2} \mathrm{v}_2 \\ & 2 \mathrm{v}_1+\mathrm{v}_2=2 \mathrm{u} \quad\text{...... (i)}\\ & \mathrm{e}=1=\frac{\mathrm{v}_2-\mathrm{v}_1}{\mathrm{u}} \\ & \Rightarrow \mathrm{v}_2-\mathrm{v}_1=\mathrm{u} \quad\text{...... (ii)} \end{aligned}$$

From (i) -(ii)

$\Rightarrow 3 \mathrm{v}_1=\mathrm{u} \Rightarrow \mathrm{v}_1=\frac{\mathrm{u}}{3}=\frac{1}{3} \sqrt{\mathrm{gR}}$

Assertion (A):

A choke coil is simply a coil having large inductance but small resistance.

Choke coils are used with fluorescent mercury-tube fittings.

If household electric power is directly connected to a mercury tube, the tube will be damaged.

A choke (or ballast) does have high inductance and low resistance.

In a fluorescent lamp (or mercury‐vapor lamp) circuit, the choke limits the current through the tube.

If we connect the tube directly to the full line voltage without any current-limiting device, excessive current will flow and damage the tube.

Therefore, (A) is true.

Reason (R):

By using the choke coil, the voltage across the tube is reduced by a factor

$\displaystyle \frac{R}{\sqrt{R^2 + (\omega L)^2}}$,

where $\omega$ is the supply frequency, and $R$ and $L$ are the resistance and inductance in series.

If the choke coil were not used, the voltage across the resistor (tube) would be the same as the applied voltage.

In a simple series R–L circuit (where $R$ represents the lamp’s effective resistance once conducting, and $L$ is the choke’s inductance), the total impedance is

$Z = \sqrt{\,R^2 + (\omega L)^2\,}.$

The current is the same through $R$ and $L$. The fraction of the total supply voltage appearing across $R$ alone is

$\frac{V_R}{V_{\text{total}}} \;=\; \frac{I\,R}{I\,Z} \;=\; \frac{R}{\sqrt{R^2 + (\omega L)^2}}.$

This shows that using a large inductance $L$ (the choke) can substantially reduce (or limit) the voltage—and hence the current—through the lamp.

Without the choke coil, the tube (treated here as a resistor once it starts conducting) would indeed see nearly the full line voltage, causing a large current that could destroy it.

Therefore, (R) is also true.

Does (R) correctly explain (A)?

(A) states why a choke coil is needed (to prevent the lamp from being damaged by excessive current).

(R) shows how the choke coil limits the voltage/current through the lamp—by giving the fraction of the supply voltage that appears across the lamp in a series R–L circuit.

Thus, (R) does provide the correct (physics-based) explanation: the choke’s inductive reactance drops a significant portion of the supply voltage, thereby protecting the lamp.

Hence, the correct option is:

$\boxed{\text{Option B: Both (A) and (R) are true and (R) is the correct explanation of (A).}}$

The first step is to determine the horizontal velocity of the ball as seen by an observer on the bank.

The velocity of the boat in still water is $27 \, \text{km/h}$. Since the boat is moving downstream and the river flows at $9 \, \text{km/h}$, the actual speed of the boat relative to the bank becomes:

$v_{\text{boat}} = 27 + 9 = 36 \, \text{km/h}$

Convert this speed from km/h to m/s by using the conversion factor $1 \, \text{km/h} = \frac{5}{18} \, \text{m/s}$:

$v_{\text{boat}} = 36 \times \frac{5}{18} = 10 \, \text{m/s}$

This is also the horizontal velocity of the ball as seen by the observer on the bank.

Next, calculate the time the ball spends in the air. The initial vertical velocity of the ball is $10 \, \text{m/s}$. The time to reach the maximum height $t_{\text{up}}$ is given by:

$t_{\text{up}} = \frac{v_{\text{initial}}}{g} = \frac{10}{10} = 1 \, \text{s}$

The total time of flight $t_{\text{total}}$ is twice the time to reach the maximum height:

$t_{\text{total}} = 2 \times 1 = 2 \, \text{s}$

The horizontal range $R$ of the ball is the horizontal velocity multiplied by the total time of flight:

$R = v_{\text{horizontal}} \times t_{\text{total}} = 10 \times 2 = 20 \, \text{m}$

Finally, convert this range into centimeters:

$R = 20 \times 100 = 2000 \, \text{cm}$

Thus, the range of the ball as observed by an observer at rest on the bank is $2000 \, \text{cm}$.

$$\begin{aligned} & \frac{\mathrm{F}_1}{\mathrm{~A}_1}=\frac{\mathrm{F}_2}{A_2}, \frac{100}{6}=\frac{\mathrm{F}}{1500}, \mathrm{~F}=\frac{50}{3} \times 1500 \\ & \mathrm{~F}=50 \times 500=25 \times 10^3 \mathrm{~N} \\ & \omega=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{~S}}=25 \times 10^3 \times \frac{20}{100} \\ & =5 \times 10^3=5 \mathrm{~kJ} \end{aligned}$$

Assertion is false because em waves have momentum.

$$\begin{aligned} &\begin{aligned} & \lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}} \\ & \lambda^2=\frac{\mathrm{h}^2}{2 \mathrm{~m}}\left(\frac{1}{\mathrm{k}}\right) \\ & \mathrm{Y}=\mathrm{cx}^2 \end{aligned}\\ &\text { Upward facing parabola passing through origin. } \end{aligned}$$

$\Rightarrow \text { OR Gate }$

$$\begin{aligned} & {[\tau]=[\mathrm{E}]} \\ & {[\sigma] \neq[\mathrm{I}]} \\ & {[\mathrm{L}]=[\mathrm{h}]} \\ & {[\mathrm{P}]=[\mathrm{Y}]} \end{aligned}$$

$$\begin{aligned} & {\left[\mathrm{LT}^{-1}\right]=[\mathrm{A}]\left[\mathrm{T}^2\right]=\frac{[\mathrm{B}][\mathrm{T}]}{[\mathrm{C}]+[\mathrm{T}]}} \\ & {[\mathrm{C}]=[\mathrm{T}]} \\ & {[\mathrm{A}]=\left[\mathrm{LT}^{-3}\right]} \\ & {[\mathrm{B}]=\left[\mathrm{LT}^{-1}\right]} \\ & {[\mathrm{ABC}]=\left[\mathrm{L}^2 \mathrm{~T}^{-3}\right]} \end{aligned}$$

$A B_2(g) \rightleftharpoons A B_{(g)}+\frac{1}{2} B_2({g})$

Degree of dissociation $\rightarrow x$ (small compared to unity)

Equilibrium constant interms of pressure $\rightarrow k_p$

Partial pressure of each gas $\rightarrow P_{A B_2}, P_{A B}$ and $P_{B_2}$

Total pressure $\rightarrow P$

ICE Table:

$\begin{aligned} & \text { partial pressure } \\ & \text { at equilibrium }\end{aligned}=\frac{\text { moles at equilibrium }}{\begin{array}{l}\text { total moles at } \\ \text { equilibrium }\end{array}} \times$ total pressure

Partial pressure of $A B_2(g)$

$P_{A B_2}=\left(\frac{1-x}{1+\frac{x}{2}}\right) P$

Partial pressure of $A B(g)$

$P_{A B}=\left(\frac{x}{1+\frac{x}{2}}\right) P$

Partial pressure of $B_2(g)$

$P_{B_2}=\left(\frac{\frac{x}{2}}{1+\frac{x}{2}}\right)^P$

$K_p$ for the equation $A_2(g) \rightleftharpoons A B_{(g)}+\frac{1}{2} B_2(g)$ can be written as

$k_p=\frac{P_{A B} P_{B_2}^{1 / 2}}{P_{A B_2}}$

$\begin{aligned} & =\frac{\frac{x}{1+\frac{x}{2}} P 1 \times\left(\frac{\frac{x}{2}}{1+\frac{x}{2}} P\right)^{1 / 2}}{\frac{1-x}{1+\frac{x}{2}} P} \\ & =\frac{x \times\left(\frac{x}{1+\frac{x}{2}} P\right)^{1 / 2}}{1-x}\end{aligned}$

$$\begin{aligned} &\text { given, } x \lll 1\\ &\begin{aligned} & \text { So, } 1-x \approx 1 \\ & K_p=\frac{x \times\left(\frac{\frac{x}{2}}{1+\frac{x}{2}} p\right)^{1 / 2}}{1} \end{aligned} \end{aligned}$$

$=x \times \frac{\left(\frac{x}{2}\right)^{1 / 2}}{\left(\frac{2+x}{2}\right)^{1 / 2}} p^{1 / 2}$

$=\frac{x \times \frac{x^{1 / 2}}{2^{1 / 2}} \times p^{1 / 2}}{\frac{(2+x)^{1 / 2}}{2^{1 / 2}}}$

$=\frac{x^{3 / 2} \times p^{1 / 2}}{2^{1 / 2}} \times \frac{2^{1 / 2}}{(2+x)^{1 / 2}}$

$\begin{gathered} x \lll 1 \\ 2+x \approx 2 \end{gathered}$

$$\begin{aligned} & =\frac{x^{3 / 2} \times p^{1 / 2}}{2^{1 / 2}} \\ & =\left(\frac{x^3 p}{2}\right)^{1 / 2} \end{aligned}$$

$$\begin{aligned} &k_p=\left(\frac{x^3 p}{2}\right)^{1 / 2}\\ &\text { Square on both sides, } \end{aligned}$$

$$\begin{aligned} k_p^2 & =\frac{x^3 p}{2} \\ x^3 p & =2 k_p^2 \\ x^3 & =\frac{2 k_p^2}{p} \\ x & =\sqrt[3]{\frac{2 k_p^2}{p}} \end{aligned}$$

Correct answer option (2) $\sqrt[3]{\frac{2 K_p^2}{p}}$

(A)

The substituents are written in the alphabetical order. So, ethyl is first and then methyl. Sok, lowest number is given for ethyl. 3-ethyl. And methyl has position 5. 5-methyl.

The parent chain has seven carbon atoms. So, the base name heptane.

IUPAC name : 3-ethyl-5-methylheptane

(A) Match with (II) 3-ethyl-5-methylheptane

(B)

Substituents get the same position when numbering is done from both ends. So, the substituent name (prefix) is 4,4-dimethyl. The parent chain has seven carbon atoms. So, the base name is heptane.

IUPAC name : 4,4-dimethylheptane

(B) Match with (III) 4,4-dimethylheptane.

(C)

Double bond has higher priority than the methyl group. So, the numbering is given as

Smallest possible position is 1 and 3 for the double bond present. So, 1, 3-diene is correct.

Methyl substituent is at position 2.

So, the prefix is 2-methyl-1,3

The parent chain has five carbon atoms and the base name is penta.

Include diene as suffix to penta- as pentadien.

So, IUPAC name is

2-methyl-1,3-pentadiene

(C) Match with (IV) 2-methyl-1,3-pentadiene.

(D)

Double bond has higher priority than methyl substituent. So, position 1 is for double bond and position 4 is for $-CH_3$ substituent. Prefix is 4-methyl.

Parent has five carbon atoms. So, the base name penta. The suffix used is ene for the double bond. So 'a' eliminated and written as pentane.

IUPAC name : 4-methylpent-1-ene

(D) match with (I) 4-methylpent-1-ene

The matching is written as

(A) - (II)

(B) - (III)

(C) - (IV)

(D) - (I)

The molar conductivity of a weak electrolyte is plotted against the square root of its concentration:

$c$-concentration of weak electrolyte

Weak electrolytes give curved decreasing graph.

1) Molar conductivity increases sharply with increase in concentration.

Incorrect.

With increase in concentration, molar conductivity decreases and it is lean from the graph of molar conductivity vs $\sqrt{c}$.

2) Molar conductivity decreases sharply with increase in concentration.

Correct

From the graph it is clean that, molar conductivity decreases sharply with increase in concentration.

3) A small increase in molar conductivity is observed at infinite dilution.

Incorrect

There is no small increase in molas conductivity with increase in concentration for a weak electrolyte. And there is no small increase in mola conductivity. Significant increase occurs.

4) A small decrease in molar conductivity is observed at infinite dilution.

Incorrect

For a weak electrolyte, the molar conductivity significantly increases with dilution, especially at infinite dilution.

Correct answer:

Option 2) Molen conductivity decreases sharply with increase in concentration.

Nucleophiles are electron rich species and can provide electrons for making new bonds.

Nucleophiles are nucleus loving species

* typically have a $-$ve (negative) charge or non-bonding electron pair available for making a new bond.

Nucleophiles are Lewis bases.

$\mathrm{NH}_3$ - It is a nucleophile.

$\mathrm{NH}_3$ has a lone pair of electrons on the nitrogen atom which allows it to be a nucleophile since it can provide electrons to form a new bond.

PhSH - It is a nucleophile

Lone pair of electrons are present on the sulphone atom and hence it is a nucleophile. The lone pair of electrons are readily donate to form new bonds. with electrophiles.

$\left(\mathrm{CH}_3\right)_2 \mathrm{~S}$ - It is a nucleophile

The lone pair of electrons on the sulphur atom makes this molecule nucleophile.

$\mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2-$ It is a nuclophile

Alkenes are nulleophiles because they have a high electron' density in their conbon-conbon double bond. The $\pi$ bondin $\mathrm{CH}_2=\mathrm{CH}_2$ is a region of high election density.

$\mathop O\limits^\Theta H$ - It is a nucleophile

Hydroxide ion is a nucleophile, not an electrophile. It is negatively changed and election rich. The electrons are easily donated.

$\mathrm{H}_3 \mathrm{O}^{\oplus}$ - It is not a nucleophile

Hychonium ion is an electrophile. It can gain a pair of electrons from a nucleophile. Also, it is positively changed.

$\left(\mathrm{CH}_3\right)_2 \mathrm{CO}$ - It is not a nucleophile.

Acetone primarily acts as an electrophile in reactions. It can readily accepts electron pairs from other molecules due to the partial positive charge on the carbonyl carbon atom.

- It is not a nucleophile.

It is an imine compound and generally act as electrophile in reactions. The carbon in the $C=N$ double bond is electrophile and attacked by nucleophiles.

So, the nucleophiles are

$\mathrm{NH}_3, \mathrm{PhSH}, \quad\left(\mathrm{CH}_3\right)_2 \mathrm{~S}, \mathrm{CH}_2=\mathrm{CH}_2, \mathrm{OH}^{-}$

Total number of nucleophiles is 5

$Fe{C_2}{O_4} + N{a_2}C{O_3}\buildrel {{O_2}} \over \longrightarrow \,?$

The chemical reaction can be written as

The insoluble product is $Fe_2O_3$.

Molar mass of $Fe_2O_3$

Fe (Atomic mass) $\to$ 55.845 amu

Molar mass = 55.845 g mol$^{-1}$

O (Atomic mass) $\to$ 15.999 amu

Molar mass = 15.999 g mol$^{-1}$

Molar mass of $Fe_2O_3$ = 55.845 g mol$^{-1}\times2$ + 15.999 g mol$^{-1}\times3$

$=(111.69+47.997)$ g mol$^{-1}$

$=159.687$ g mol$^{-1}\approx 160$ g mol$^{-1}$

Oxidizing capacity is an element's tendency to donate electrons, or the atmosphere's ability to oxidize compounds.

A higher standard reduction potential indicates a stronger oxidizing capacity. A substance with a higher standard reduction potential is move likely to gain electrons and act as an oxidizing agent in a redon reaction. The more positive the reduction potential, the stronger the oxidizing power of a species.

From the given options, the higher standard reduction potential is for lead $(\mathrm{Pb}) .(+1.67)$.

so, the p-block ion with strongest oxidizing capacity is Pb (lead).

Answer: Option 3) $E_{p_b{ }^{4+} / \mathrm{Pb}^{2+}}^0=+1.67 \mathrm{~V}$

(A) $\left[\mathrm{MnBr}_4\right]^{2-}$

$\mathrm{Mn}^{+2} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^5$

In presence of ligand field

$\Rightarrow \mathrm{sp}^3$ hybridization, paramagnetic in nature

(B) $\left[\mathrm{FeF}_6\right]^{3-}$

$\mathrm{Fe}^{+3} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^5$

In presence of ligand field

$\Rightarrow \mathrm{sp}^3 \mathrm{~d}^2$ hybridization, paramagnetic in nature

(C) $\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$

$\mathrm{Co}^{+3} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^6$

In presence of ligand field

$\Rightarrow \mathrm{d}^2 \mathrm{sp}^3$ hybridization, diamagnetic in nature

$\text { (D) }\left[\mathrm{Ni}(\mathrm{CO})_4\right]$

$\mathrm{Ni}^0 \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^2$

In presence of ligand field

$\Rightarrow \mathrm{sp}^3$ hybridization, diamagnetic in nature

$$\begin{aligned} &0.1 \text { mole of compound ( } \mathrm{S} \text { ) weight in gm }\\ &\begin{aligned} & =0.1 \times \text { molar mass of compound }(\mathrm{S}) \\ & =0.1 \times 130=13 \mathrm{gm} \end{aligned} \end{aligned}$$