IIT-JEE 2007 Paper 2 Question Paper with Solutions

Young's modulus $\left(𝑌\right)=\frac{\text{ Stress }}{\text{ Strain }}=\frac{4𝑚𝑔𝑙}{𝜋{𝑑}^2\Delta 𝑙}$

$\begin{array}{c}Y=\frac{4𝑚𝑔𝑙}{𝜋{𝑑}^2𝑙}\\ \\ =\frac{4\times 1\times 9.8\times 2}{3.14\times (0.4{)}^2\times (0.8)\times {10}^{-6}\times {10}^{-3}}\\ \\ =2\times {10}^{11}N/m^2\end{array}$

Where, $$ Mass $\left(𝑚\right)=1\mathrm{kg}$

$\begin{array}{cc}𝑔& =9.8m/s^2\\ \\ \text{ Length of wire }& =2m\\ \\ 𝜋& =3.14\\ \\ \text{ Diameter }\left(𝑑\right)& =0.4\mathrm{mm}\times {10}^{-3}m\end{array}$

Extension in the length of wire

$\begin{array}{cc}\left(\Delta 𝑙\right)& =0.8\mathrm{mm}\\ \\ & =0.8\times {10}^{-3}m\\ \\ \frac{\Delta 𝑦}{𝑦}& =\frac{2\Delta 𝑑}{𝑑}+\frac{\Delta 𝑙}{𝑙}\\ \\ & =\frac{2\times \left(0.01\right)}{0.4}+\frac{\left(0.05\right)}{0.8}\\ \\ & =0.1125\end{array}$

$\begin{array}{cc}\Rightarrow \Delta Y& =\left(Y\right)\times 0.1125\\ \\ & =2\times {10}^{11}\times 0.1125\\ \\ & =0.225\times {10}^{11}\\ \\ & \simeq 0.2\times {10}^{11}\\ \\ \text{ (b) }(2\pm 0.2\times {10}^{11}N/& m^2)\end{array}$

Given that,

$\overrightarrow{P}\left(𝑡\right)=A[\widehat{𝑖}\cos (𝑘𝑡)-\widehat{𝑗}\sin (𝑘𝑡\left)\right]$

where, $A$ and $𝑘=$ constant

$\begin{array}{cc}\text{ Force }\overrightarrow{F})& =\frac{𝑑\overrightarrow{P}}{𝑑𝑡}(\text{ from Newton’s }{2}^{\text{nd }}\text{ law) }\\ \\ \Rightarrow \overrightarrow{F}& =\frac{𝑑}{𝑑𝑡}\left[A\right(\widehat{𝑖}\cos \left(𝑘𝑡\right)-\widehat{𝑗}\sin \left(𝑘𝑡\right)\left)\right]\\ \\ & =A𝑘[\widehat{𝑖}\cos (𝑘𝑡)-\widehat{𝑗}\sin (𝑘𝑡\left)\right]\end{array}$

By applying dot product between $\overrightarrow{𝐹}\&\overrightarrow{𝑃}$ we have

$\begin{array}{cc}& \overrightarrow{F}\cdot \overrightarrow{P}=0\\ \\ & \Rightarrow \left|\overrightarrow{𝐹}\right|\left|\overrightarrow{𝑃}\right|\cos 𝜃=0\\ \\ & \Rightarrow \cos 𝜃=0\\ \\ & \text{Hence,}𝜃={90}^{\circ }\end{array}$

Wave produced by tuning Fork in a vertical plane as shown in above figure.

Therefore, fringes of tuning Fork is kept in vertical plane.

By applying conservation of Energy, we have,

$\frac{1}{2}𝑚{𝑣}^2+\frac{1}{2}I{𝜔}^2=𝑚𝑔{ℎ}_{max}$ ..... (i)

For pure rolling,

$𝑣=𝑅𝜔$

$\Rightarrow 𝜔=𝑣/R$ ..... (ii)

Putting eq. (ii) in eq. (i), we get

$\Rightarrow \frac{1}{2}𝑚{𝑣}^2+\frac{1}{2}I\left(\frac{{𝑣}^2}{{𝑅}^2}\right)=𝑚𝑔\times \frac{3{𝑣}^2}{4𝑔}$ $(\because {ℎ}_{max}\times \frac{3{𝑣}^2}{4𝑔})$

$\Rightarrow I=\frac{1}{2}{\mathrm{MR}}^2$

Hence, this is a required formula of moment of Inertia of Disc. Therefore, the object a disc.

$2^{\text{nd }}$ Method : For pure rolling, the velocity of point of contact is zero i.e., $𝑣=0$, So, $\Delta 𝑥$ (displacement) is zero. Then work done by the frictional force should be

$𝑊={𝑓}_0\Delta 𝑥=0$

If work is done by dissipative force, work done should be zero. It means that work done of system is zero. So energy should be conserved. Initially, the object has both translation as well as rotational motion. If it attains '${ℎ}_{max}$' kinetic energy changes into potential energy.

From energy conservation principle,

$K.E_{\text{translational }}+K.E_{\text{rotational }}=$ Potential Energy

$\Rightarrow \frac{1}{2}𝑚{𝑣}^2+\frac{1}{2}I{𝜔}^2=𝑚𝑔\times \frac{3{𝑣}^2}{4𝑔}$

$\Rightarrow 𝑚{𝑣}^2+I{𝜔}_2=\frac{3{𝑣}^2}{2}𝑚$

$\Rightarrow I{𝜔}^2=\frac{1}{2}𝑚{𝑣}^2$

$\Rightarrow I=\frac{1}{2}\frac{𝑚{𝑣}^2}{{𝜔}^2}$ ..... (i)

For No slipping,

$\Rightarrow$ $\begin{array}{cc}& 𝑣=𝜔𝑅\\ & 𝜔=\frac{𝑣}{𝑅}\end{array}$ ..... (ii)

Put eq (ii) in eq (i)

Hence, $I=\frac{1}{2}{\mathrm{MR}}^2$

$I=\frac{1}{2}M\frac{{𝑣}^2}{{𝑣}^2}\times R^2$

this is moment of inertia of disc.

So, object is a Disc.

Pressure is acting on area $𝐴𝐵𝐶𝐷$, we have

Pressure Force $={\int }_{ℎ}^0(P_0+𝜌𝑔𝑥)\times 𝑑𝑥\times 2R$

Force due to pressure (push)

$\begin{array}{cc}& ={\left[(P_0+𝜌𝑔\frac{{𝑥}^2}{2})2R\right]}_{ℎ}^0\\ \\ & =(P_0ℎ+𝜌𝑔{ℎ}^2/2)2R\\ \\ & =2P_0ℎR+𝜌𝑔{ℎ}^{2}R\end{array}$

Surface tension force (push) $=𝑇\times 2𝑅$

$\therefore$ Net Force $=2P_0ℎR+𝜌𝑔{ℎ}^{2}R-2\mathrm{TR}$

When cavity is not present:

Electric field inside the solid sphere

$\stackrel{\to \leftarrow }{E_1}=\frac{𝜌}{3{𝜀}_0}\overrightarrow{{𝑟}_1}$ ..... (i)

Solid sphere after making cavity inside solid sphere,

$\stackrel{\to \leftarrow }{E_2}=\frac{𝜌}{3{𝜀}_0}\overrightarrow{{𝑟}_2}$ ...... (ii)

${\overrightarrow{𝐸}}_{𝑛𝑒𝑡}=\stackrel{\to \leftarrow }{{𝐸}_1}-\stackrel{\to \leftarrow }{{𝐸}_2}=\frac{𝜌}{3{𝜀}_0}\overrightarrow{{𝑟}_1}-\frac{𝜌}{3{𝜀}_0}\overrightarrow{{𝑟}_2}$

$=\frac{𝜌}{3{𝜀}_0}(\overrightarrow{{𝑟}_1}-\overrightarrow{{𝑟}_2})$ ...... (iii)

From eq (iii), we get

$\therefore E_{\mathrm{net}}=\frac{𝜌}{3{𝜀}_0}\left(\overrightarrow{𝑟}\right)$

Hence, electric field inside the emptied space (cavity) is non-zero and uniform.

Consider that,

$P=(-𝑎,0,0)$ and $Q=(0,𝑎,0)$ points lie on the equatorial plane of dipole.

The charges makes an electric dipole.

Therefore, potential at $P=$ Potential at $Q=0$

Work done $\left(W\right)=𝑞(V_P-V_Q)=0$

For $𝑎<𝑥<2𝑎$

$\overrightarrow{B}=B_0\widehat{𝑗}$

The initial velocity is $\overrightarrow{𝑣}=𝑣\cdot \widehat{𝑗}$

From the above diagram it is clear that force on the particle is toward $+𝑧$-axis (by applying Fleming's left hand rule) at $𝑥=𝑎$, which shifts the particle as shown in the x-z plane

For $2𝑎<𝑥<3𝑎$

$\overrightarrow{B}=-B_0\widehat{𝑗}$

The direction of velocity is shown at $𝑥=2𝑎$.

Again by using Fleming's left hand rule, we get direction of force.

The cut off wavelength is given by

${𝜆}_0=\frac{ℎ𝑐}{𝑒𝑣}$ ..... (i)

According to de-Broglie equation

$𝜆=\frac{ℎ}{𝑝}=\frac{ℎ}{\sqrt{\surd 2𝑚𝑒V}}$

$\Rightarrow {𝜆}^2=\frac{{ℎ}^2}{2𝑚𝑒V}$

$\Rightarrow V=\frac{{ℎ}^2}{2𝑚𝑒{𝜆}^2}$ ..... (ii)

From (i) and (ii)

${𝜆}_0=\frac{ℎ𝑐\times 2𝑚𝑒{𝜆}^2}{𝑒{ℎ}^2}=2𝑚𝑐{𝜆}^2$

Statement 1 : For velocity of centre of mass remains constant, net force acting on body must be zero. Therefore the statement 1 is false.

Statement 2 : The linear momentum of an isolated system remains constant. The statement is true.

Statement 1 : Cloth can be pulled out without dislodging the dishes from the table because of inertia. Therefore, statement 1 true.

Statement 2 : This is Newton's third law and hence true. But statement 2 is not a correct explanation of statement 1 .

As shown in the figure, the horizontal component of the magnetic field interacts with the current in the conducting ring, which produces an average force in the upwards direction (by using Fleming's left hand rule).

Statement 1 :

Total translational kinetic energy $=\frac{3}{2}𝑛\mathrm{RT}$

where,

n = no. of moles

R = gas constant

T = Temperature

but, PV = nRT

$\therefore$ K.E. = 1.5 PV

Statement 2 : Molecules of a gas collide with each other and the velocities of the molecules change due to collision.

But statement 2 is not a correct explanation of statement 1 .

${𝑣}_{\mathrm{SA}}=(340+20)m/s=360m/s$

${𝑣}_{\mathrm{SB}}=(340-30)m/s=310m/s$

where, speed of sound in air = 340 m/s

Hence, the speed of sound depends on the frame of reference of observer.

Since all the passengers in train A are moving with a velocity of $20m/s$ therefore the distribution of sound intensity of the whistle by the passengers in the train $𝐴$ is uniform.

Trick : Total frequency $=(1120-800)\mathrm{Hz}=320\mathrm{Hz}$

There is no relative motion between source and observer.

Therefore, all the passengers will hear the sound of same intensity.

Frequency $\left({𝑓}^{\prime }\right)={𝑓}_1\left(\frac{𝑣-{𝑣}_0}{𝑣-{𝑣}_{𝑠}}\right)=800\left[\frac{340-30}{340-20}\right]$

$=800\times \frac{31}{32}$ ..... (i)

where, $\left(\frac{𝑣-{𝑣}_0}{𝑣-{𝑣}_S}\right)=\frac{31}{32}$ ..... (ii)

Frequency $\left({𝑓}^{\prime }\right)={𝑓}_2\left(\frac{𝑣-{𝑣}_0}{𝑣-{𝑣}_S}\right)=1120\times \frac{31}{32}$ ..... (iii)

$\therefore {𝑓}^{\prime }-{𝑓}^{\prime }=(1120-800)\frac{31}{32}=320\times \frac{31}{32}$

$=310\mathrm{Hz}$

Since the wave fronts are parallel in each medium, the beam is also a parallel beam travelling in the direction perpendicular to the wave fronts. Since the beam would bend inside medium 2 when travelling from medium 1 , the second medium has the higher refractive index.

Trick : For plane wave fronts the beam of light is parallel.

Since point c and d are on same wave front.

Therefore, ${𝜙}_{𝑑}={𝜙}_{𝑐}$

Similarly, ${𝜙}_{𝑒}={𝜙}_{𝑓}$

$\therefore {𝜙}_{𝑑}-{𝜙}_{𝑓}={𝜙}_{𝑐}-{𝜙}_{𝑓}$

The gap between consecutive wave fronts in medium 2 is less than that in medium 1. Therefore, wavelength of light in medium 2 in less than that in medium 1. Therefore, speed of light is more in medium 1 and less in medium 2.

Reason:

In above figure shown that S.H.M

Therefore, ${𝑓}_{𝑠}=𝐾𝑥$

Given, $𝑣={𝑐}_1\sqrt{\surd {𝑐}_2-{𝑥}^2}$ ..... (i)

For a simple harmonic motion,

$𝑣=𝜔\sqrt{\surd {𝐴}^2-{𝑥}^2}$ ..... (ii)

On comparing eq. (i) & (ii) therefore it shows the S.H.M of particle.

A $\to$ P

(b) Given,

$𝑣=-𝑘𝑥$

At $𝑥=0,𝑣=0$

$𝑥=2,𝑣=-2𝑘$

$𝑥=-5,𝑣=+5𝑘$

The above consideration show that particle is coming from minus x-axis.

Velocity is decreasing with time.

Graph:

Therefore, no change in direction, velocity keeps on decreasing. So that kinetic energy is decreasing with time.

B $\to {𝑞}_1𝑟$

(C) From the reference frame of the observer in the elevator a constant force acts on the object. Hence the motion of the object will remain simple harmonic.

Therefore, option 1 is correct.

Velocity of object

$𝑣=2\sqrt{\surd 𝐺\frac{{𝑀}_{𝑒}}{{𝑅}_{𝑒}}}=\sqrt{\surd 2}\sqrt{\surd 2𝐺\frac{{𝑀}_{𝑒}}{{𝑅}_{𝑒}}}=\sqrt{\surd 2}{𝑣}_{𝑒}$

Since, $𝑣>{𝑣}_{𝑒}$ (where ${𝑣}_{𝑒}$ is escape velocity) object will move out of earth.

Hence, option (Q) and (R) are correct.

We can find the direction of magnetic field at $𝑃$ due to a wire by using thumb rule, and the direction of force on a wire due to other is calculated by the direction of cross product of direction of current and direction of magnetic field, $\overrightarrow{𝑓}=𝑞(\overrightarrow{𝑙}\times \overrightarrow{𝐵})$

(A) From thumb rule, the magnetic field at point $P$ due to both wires is in opposite direction. Since $P$ is at midpoint, net magnetic field at $𝑃$ is zero. The direction of force on one wire is towards another (from above equation), hence they attract each other. Thus, option $\left(Q\right)$ are $\left(R\right)$ are correct.

(B) Direction of magnetic field at $P$ is same due to both the loop. Hence option $\left(P\right)$ is correct. If both the loop is very close they might act as single (like solenoid), so we cannot predict loop will attract nor repel.

(C) Similar to case $𝐴$, net $𝐵$ at point $P$ is zero. And direction of force is towards another wire which means attraction. Hence, option (Q) and (R) are correct.

(D) Magnetic field at the center of current carrying loop is given by $B=\frac{{\propto }_0𝑖}{2𝑅}$

Current in the two loop is in opposite direction so magnetic field is in opposite direction due to both the loops. Radius of two loops is different, so it will have non zero field at point $𝑃$.

(B) Steam engine is a device which converts heat energy into mechanical energy and heat is supplied into the engine through the medium of steam. Thus energy conversion is right answer here.

(C) The incandescent light bulb works on incandescence, which is the emission of light radiations caused by heating the filament.

(D) Fuse wire works on the principle of heat and melting, when excess current passes in the circuit, due to heat produced by the excess current fuse melts and prevent the damage.

The formula of Mohr's salt is ${\mathrm{FeSO}}_4({\mathrm{NH}}_4{)}_2{\mathrm{SO}}_4$. It is a mixture of ferrous sulfate and ammonium sulfate. Ferrous sulfate is oxidized by potassium dichromate to ferric sulphate.

Reaction is as follows:

$K_2{\mathrm{Cr}}_2{O}_7+6{\mathrm{FeSO}}_4+7H_2{\mathrm{SO}}_4\to K_2{\mathrm{SO}}_4+{\mathrm{Cr}}_2({\mathrm{SO}}_4{)}_3+3{\mathrm{Fe}}_2({\mathrm{SO}}_4{)}_3+7H_{2}O$

1 mole of potassium dichromate reacts with 6 moles of ferrous sulphate. Thus, number of moles of Mohr's salt required per mole of dichromate is 6.

Statement‑1:

“Band gap in germanium is small.”

✅ True.

Germanium is a semiconductor with an energy band gap of about 0.66 eV at 300 K, which is much smaller than that of silicon (~1.1 eV).

Statement‑2:

“The energy gap of each germanium atomic energy level is infinitesimally small.”

❌ False.

In a single germanium atom, the energy levels are discrete with finite energy differences between them — not infinitesimally small.

In a solid, when many atoms come together, each discrete atomic level splits into closely spaced energy states due to interactions between atoms, forming continuous bands; the gap between valence and conduction bands (the band gap) is small but finite.

So, the “infinitesimally small” part is incorrect.

✅ Correct Option: C

Statement‑1 is True, Statement‑2 is False.

The rate law expression is

$R=𝑘\left[G{]}^{𝑎}\right[H{]}^{𝑏}$ ...... (i)

The rate of reaction increases by a factor of 2 on doubling the concentration of '$G$'.

The rate law expression becomes

$R^{\prime }=2R=𝑘2^{𝑎}\left[G{]}^{𝑎}\right[H{]}^{𝑏}$ ..... (ii)

Divided equation (ii) with eqn. (i)

$\begin{array}{cc}\frac{R^{\prime }}{R}& =\frac{2R}{R}=\frac{𝑘2^{𝑎}\left[G{]}^{𝑎}\right[H{]}^{𝑏}}{𝑘2\left[G{]}^{𝑎}\right[H{]}^{𝑏}}\\ 2& =2^{𝑎}\\ 𝑎& =1\end{array}$

The rate of reaction increases by a factor of 8 on doubling the concentration of $G$ and '$H$'.

The rate law expression becomes

Rate $=R^{\prime \prime }𝑘2^{𝑎}\left[G{]}^{𝑎}{2}^{𝑏}\right[H{]}^{𝑏}$ ..... (iii)

Divide equation (iii) with equation (i)

$\begin{array}{cc}\frac{8R}{R}& =\frac{𝑘2^{𝑎}\left[G{]}^{𝑎}{2}^{𝑏}\right[H{]}^{𝑏}}{𝑘\left[G{]}^{𝑎}\right[H{]}^{𝑏}}\end{array}$

$8=2^{𝑎}{2}^{𝑏}$

or $𝑏=2$

The overall order of reaction is

$𝑎+𝑏=1+2=3$

(A) ${\mathrm{Mn}}^{+}=3{𝑑}^{5}4{𝑠}^1$; in presence of CO, effective configuration = $3{𝑑}^{6}4{𝑠}^0$

Three lone pair of electrons in metal 3d orbitals for back bonding with vacant ($𝜋$) pi* orbital of C in CO.

(B) ${\mathrm{Fe}}^0=3{𝑑}^{6}4{𝑠}^2$; in presence of CO, effective configuration = $3{𝑑}^8$

Four lone pair for back bonding with CO.

(C) ${\mathrm{Cr}}^0=3{𝑑}^{5}4{𝑠}^1$; effective configuration = $3{𝑑}^6$

Three lone pair for back bonding with CO.

(D) $V^{-}=3{𝑑}^{4}4{𝑠}^2$; effective configuration = $3{𝑑}^6$

Three lone pair for back bonding with CO.

Since, Vanadium has a negative charge, it is present in lower oxidation state and can easily back donate the electrons from metal $𝑑$ orbitals to ($𝜋$) pi* orbital of carbonyl ligand. Hence, it forms a much stronger bond with carbon or carbonyl thereby weakening the carbon oxygen bond.

A solution of metal ion (Hg${}^{2+}$) when treated with KI gives a red precipitate which dissolves in excess KI to give a colourless solution. Moreover, the solution of metal ion on treatment with a solution of cobalt (II) thiocyanate gives rise to a deep blue crystalline precipitate. The metal ion is Hg${}^{2+}$.

${\mathrm{Hg}}_2+\mathrm{KI}\to \mathrm{Hg}{I}_2$ (red ppt)

${\mathrm{HgI}}_2+\mathrm{KI}\to K_2{\mathrm{HgI}}_4$

$\mathrm{Co}(\mathrm{SCN}{)}_2+K_2{\mathrm{HgI}}_4=\mathrm{Co}[\mathrm{Hg}\left(\mathrm{SCN}{)}_4\right]$ (blue crystalline precipitates)

Its open structure is

If we add halogens on alkenes it is mostly anti addition. Anti addition of ${\mathrm{Br}}_2$ on $𝑡𝑟𝑎𝑛𝑠$ alkene gives meso compound.

So, here only 1 stereoisomer is formed.

$𝑡𝑟𝑎𝑛𝑠$-2-butene is

Bromination of $𝑡𝑟𝑎𝑛𝑠$-2-butene

Among the given resonance structures, the least stable resonance structure is given by option (A) as the two positive charges lie adjacent to each other and cause maximum repulsion.

(A)

Same charges are present at nearest position making the molecule less stable.

Given:

A positron is emitted from:

${}_{23}^{11}\text{Na}$

Step 1: Recall what happens in positron emission.

A positron is a ${𝛽}^{+}$ particle, i.e. ${}_0^{+1}𝑒$.

In positron emission:

$𝑝\to 𝑛+{𝑒}^{+}+{𝜈}_{𝑒}$

So inside the nucleus:

• Number of protons decreases by 1,

• Number of neutrons increases by 1,

• The mass number (A) remains the same.

Step 2: Apply this to sodium-23.

${}_{23}^{11}\text{Na}\to {}_{23}^{10}\text{X}+{𝑒}^{+}+{𝜈}_{𝑒}$

So the new nucleus has:

• Atomic number $𝑍=10$

• Mass number $𝐴=23$

The element with $𝑍=10$ is Neon (Ne).

Hence, the daughter nucleus is ${}_{23}^{10}\text{Ne}$.

Step 3: Find the required ratio.

They ask for the ratio of atomic mass (mass number) to atomic number of the resulting nuclide:

$\frac{𝐴}{𝑍}=\frac{23}{10}$

✅ Final Answer:

$\overline{)\frac{23}{10}}$

Option C: $\frac{23}{10}$

Process :

${\text{H}}_2\text{O}(𝑙,1\text{bar},373\text{K})\to {\text{H}}_2\text{O}(𝑔,1\text{bar},373\text{K})$

Step 1. Identify what temperature and pressure mean

At 373 K and 1 bar, this is the normal boiling point of water (approximately, at 1 atm ≈ 1.013 bar).

So, at this condition, liquid water and water vapor are in equilibrium.

Step 2. Thermodynamic definitions

At equilibrium between two phases at a given $𝑇$ and $𝑝$:

$\Delta 𝐺=0$

because the two phases have equal molar Gibbs energies.

So for a reversible phase change at equilibrium:

$\overline{)\Delta 𝐺=0}$

Step 3. Entropy change

When liquid water vaporizes,

• Disorder increases,

• Number of microstates increases,

so

$\overline{)\Delta 𝑆>0}$

(positive entropy change).

✅ Therefore, the correct set is:

$\overline{)\Delta 𝐺=0,\Delta 𝑆=+\text{ve}}$

✅ Correct option: A

Statement-1 :

Alkali metals dissolve in liquid ammonia to give blue solutions.

✅ True

Explanation: When alkali metals (Li, Na, K, etc.) are dissolved in liquid ammonia, they form deep blue conducting solutions at low concentrations.

Reaction:

$M\left(s\right)\stackrel{\mathrm{liq}.{\mathrm{NH}}_3}{\to \leftarrow }{M}_{+}^{\left(\mathrm{solv}\right)}+e_{-}^{\left(\mathrm{solv}\right)}$

The blue color arises from solvated electrons (${𝑒}_{-}^{\left(𝑁{𝐻}_3\right)}$), which absorb visible light.

Statement-2 :

Alkali metals in liquid ammonia give solvated species of the type ${\left[M{\left({\mathrm{NH}}_3\right)}_{𝑛}\right]}^{+}(M=\text{alkali metals})$.

✅ True

During dissolution, the metal forms metal cations solvated by ammonia molecules, i.e., $M^{+}$ surrounded by several ${\mathrm{NH}}_3$ molecules.

$M^{+}+e^{-}\to \left[M\right({\mathrm{NH}}_3{)}_n{]}^{+}+e_{-}^{\left(\mathrm{solv}\right)}$

But:

The blue color (Statement-1) is not due to $[M({\mathrm{NH}}_3{)}_n{]}^{+}$, but due to the solvated electrons (${𝑒}_{-}^{\left(𝑁{𝐻}_3\right)}$).

Therefore, although both statements are true, Statement-2 does not explain the blue color.

✅ Final Answer :

Option B — Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.

Fehling's solution contains copper (II) ions complexed with tartrate ions in sodium hydroxide solution. Complexing the copper (II) ions with tartrate ions prevents precipitation of copper (II) hydroxide.

The reducing group of glucose $(-\mathrm{CHO})$ is oxidised by the ${\mathrm{Cu}}^{2+}$ ions in Fehling's to ${\mathrm{Cu}}^{+}$ (red copper oxide).

Red solution is produced due to formation of cuprous oxide. ${\mathrm{Cu}}_{2}O$ is produced instead of $\mathrm{CuO}$.

Hence, statement- 1 is true but statement- 2 is incorrect.

A molecule is said to be chiral if it cannot be super imposed on its mirror image by any combination of rotations or translation. A chiral molecule exists as two stereoisomers that are mirror images of each other called enantiomers. Chirality is based on molecular symmetry. A chiral molecule cannot have an improper axis of rotation which includes planes of symmetry and inversion center.

Statement- 1 is true as molecules that are not superimposable on their mirror images are a symmetric and hence chiral. Such molecules are known as enantiomers.

Statement- 2: All chiral molecules don’t have to have a stereogenic center. A molecule can be chiral even if it does not have a stereocenter. A molecule’s chirality depends entirely on whether it is a symmetrical. Example-Biphenyls do not have a chiral carbon but they cannot rotate freely due to steric crowding. Hence, they are a symmetrical. Therefore, we can conclude that Statement-2 is false.

The $E^o$ value for $I_2/I^{-}$ half cell is much lower than that for ${\mathrm{Cl}}_2/{\mathrm{Cl}}^{-}$ half cell. Therefore, ${\mathrm{Cl}}_2$ has a higher tendency to get reduced.

It can therefore oxidise $I^{-}$ ion to $I_2$.

The cell emf corresponding to this reaction's

$\begin{array}{cc}{E}_o^{\text{cell }}& =E_o^{\text{cathode }}-E_o^{\text{anode }}=1.36V-0.54V\\ & =0.82V\end{array}$

Positive value of $E_o^{\text{cell }}$ shows that this reaction is spontaneous in the direction as written.

Stability of ${\mathrm{Fe}}^{3+}$ and ${\mathrm{Mn}}^{3+}$ in Acidic Solution

${\mathrm{Fe}}^{3+}$ is stable in acid, but ${\mathrm{Mn}}^{3+}$ is not. Let's see why.

Standard Electrode Potentials

The standard electrode potential, $E^o$, is a measure of how easily an ion can be reduced. For ${\mathrm{Mn}}^{3+}\to {\mathrm{Mn}}^{2+}$, $E^o$ is a high positive value. This means ${\mathrm{Mn}}^{3+}$ is eager to gain electrons and become ${\mathrm{Mn}}^{2+}$.

Comparing Potentials

The half-cell for $O_2,H^{+}/H_{2}O$ has a lower $E^o$ value than for ${\mathrm{Mn}}^{3+}$. So, ${\mathrm{Mn}}^{3+}$ can take electrons from water, causing water to split into $O_2$ and $H^{+}$.

The Reaction and Its Cell Potential

Here is the reaction:

$\begin{array}{c}4{\mathrm{Mn}}^{3+}+2H_{2}O\to 4{\mathrm{Mn}}^{2+}+4H^{+}+O_2\\ E_o^{\text{cell}}=E_o^{{\mathrm{Mn}}^{3+}/\mathrm{Mn}}-E_o^{H_{2}O,H^{+}/O_2}\\ =1.50V-1.23V\\ =0.27V\end{array}$

The $E_o^{\text{cell}}$ is positive (0.27 V), which means the reaction happens easily (is spontaneous).

Conclusion

Since ${\mathrm{Mn}}^{3+}$ reacts with water and changes into ${\mathrm{Mn}}^{2+}$, it is not stable in acid. In contrast, ${\mathrm{Fe}}^{3+}$ does not do this and is stable.

Aniline contains nitrogen in addition to carbon and hydrogen. During fusion with Na metal, the following reaction takes place

$C+N+\mathrm{Na}\stackrel{\text{ Fuse }}{⟶}\mathrm{NaCN}$

During the test, following reactions take place.

$6\mathrm{NaCN}+{\mathrm{FeSO}}_4\to {\mathrm{Na}}_4\left[\mathrm{Fe}\right(\mathrm{CN}{)}_6]+{\mathrm{Na}}_2{\mathrm{SO}}_4{\mathrm{Na}}_4\left[\mathrm{Fe}\right(\mathrm{CN}{)}_6]+2{\mathrm{Fe}}_2{\left({\mathrm{SO}}_4\right)}_3\to$ $\underset{Pr\mathrm{ussian}\mathrm{Blue}}{F{e}_4{\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}_3}$ $+6{\mathrm{Na}}_2{\mathrm{SO}}_4$

The reagent used in the reaction is aq $\mathrm{NaOH}+{\mathrm{CHCl}}_3$

Reaction is

Mechanism :

In this reaction electrophile involved is dichloro carbene (${\mathrm{CCl}}_2$) as it contains a vacant orbital.

When the same molecule is oxidised or reduced, then the reaction is called disproportionation reaction.

The reaction in which oxidation and reduction takes place simultaneously are called redox reactions. Thus,

(A)

(B)

(C)

(D)

(A)

(B)

(C) CN${}^{-}$ is negatively charged, so it is a nucleophile

${\mathrm{CN}}^{-}+{\mathrm{AgNO}}_3\to \mathrm{AgCN}\downarrow +{\mathrm{NO}}_{-}^3$