IIT-JEE 2010 Paper 2 Question Paper with Solutions

The fundamental mode in a pipe closed at one end and the second harmonic in a string are shown in the figure.

Fundamental frequency of a hollow pipe closed at one end is

${\upsilon _p} = {v \over {4L}}$

where, v = speed of sound, L = length of a pipe

Frequency of second harmonic of a string is

${\upsilon _s} = {2 \over {2l}}\sqrt {{T \over \mu }} = {1 \over l}\sqrt {{T \over \mu }}$

where, T = tension of the string, m = mass per unit length of the string, l = length of the string

${\upsilon _s} = {1 \over l}\sqrt {{T \over {{m \over l}}}} = {1 \over l}\sqrt {{{Tl} \over m}} = \sqrt {{T \over {ml}}}$ ...... (i)

where m is the mass of the string

According to given problem, ${\upsilon _p} = {\upsilon _s}$

$\therefore$ ${v \over {4L}} = \sqrt {{T \over {ml}}}$ or $m = {{16{L^2}T} \over {{v^2}l}}$

Substituting the given values, we get

$m = {{16 \times {{(0.8)}^2} \times (50)} \over {{{(320)}^2} \times 0.5}} = 0.01$ kg = 10 gram

The activity of a radioactive substance, having a decay constant $\lambda$ and number of nuclei N at time t, is given by

$A = \left| {dN/dt} \right| = \lambda N = \lambda {N_0}{e^{ - \lambda t}}$ ..... (1)

Take logarithm on both sides of equation (1) to get

$\ln \left| {dN/dt} \right| = \ln (\lambda {N_0}) - \lambda t$ ...... (2)

Thus, the graph between t and $\left| {dN/dt} \right|$ is a straight line with slope $- \lambda$.

Slope $= - \lambda = {{3 - 4} \over {6 - 4}}$ (From graph) or $\lambda = {1 \over 2}$ year^$-$1

Half life ${T_{1/2}} = {{0.693} \over \lambda } = 2 \times 0.693$ years = 1.386 years

4.16 years is approximately 3 half-lives

Nuclei will decay by a factor of 2³ = 8

$\therefore$ p = 8

The t - F diagram is a straight line passing through (0, 4) and (3, 0). The equation of this straight line is

$F = - {4 \over 3}t + 4$.

Newton's second law gives acceleration of the block as

$a = F/m = - {2 \over 3}t + 2$.

Integrate to get the velocity v at 4.5 s

$v = \int_0^{4.5} {a\,dt = \int_0^{4.5} {\left( { - {2 \over 3}t + 2} \right)dt = 2.25} }$ m/s.

Thus, kinetic energy of the block at 4.5 s is

$K = {1 \over 2}m{v^2} = 5.06$ J.

First refraction through lens

Here, u = $-$30 cm, f = +15 cm

Using ${1 \over v} - {1 \over u} = {1 \over f}$

${1 \over v} - {1 \over { - 30}} = {1 \over { + 15}}$ or ${1 \over v} = {1 \over {15}} - {1 \over {30}}$

v = +30 cm

$\therefore$ Image I₁ is real and formed at a distance of 30 cm on the right side of the lens.

Distance of image I₁ from the mirror is

30 cm $-$ 10 cm = 20 cm

The image I₁ acts as an virtual object for the mirror. The mirror forms an image I₂ at a distance of 20 cm in front of it.

The image I₂ acts as an object for the lens.

Second refraction through lens

Here, u = +10 cm, f = +15 cm

${1 \over v} - {1 \over { + 10}} = {1 \over { + 15}}$ or ${1 \over v} = {1 \over {15}} + {1 \over {10}}$ $\Rightarrow$ v = +6 cm

The final image I₃ is real and at a distance of 16 cm from the mirror.

Consider a small element of length dl on the drop-dropper interface. The force on this element is $dF = T\,dl$, and it makes an angle $\theta$ from the horizontal. Resolve $dF$ in the horizontal and the vertical directions to get, $d{F_h} = T\,dl\cos \theta$ and $d{F_v} = T\,dl\sin \theta$. By symmetry, the force $d{F_h}$ on two diametrically opposite elements on circular interface is equal and opposite. Thus, total force in the horizontal direction, ${F_h} = \int {d{F_h} = 0}$. Integrate the vertical component over the circular interface to get ${F_v} = T(2\pi r)\sin \theta = 2\pi rT(r/R) = 2\pi {r^2}T/R$.

In Vernier scale, 16 mm is divided into 20 parts, that is,

1 VSD = 16/20 mm.

Least count = 1 MSD $-$ 1 VSD

$LC = \left( {1 - {{16} \over {20}}} \right)$ mm = 0.22 mm

To find the quantized rotational energy of a diatomic molecule, we can use Bohr's quantization condition for angular momentum. Bohr's quantization condition states that the angular momentum $L$ of an electron in a hydrogen atom is quantized and given by:

$L = n \hbar$

where $n$ is a positive integer (known as the quantum number) and $\hbar$ is the reduced Planck's constant, given by $\hbar = {\hbar \over {2\pi}}$.

For a diatomic molecule, we extend this concept to its rotational motion. A rigid diatomic molecule can be approximated as a rigid rotor with a moment of inertia $I$. The angular momentum $L$ for the molecule in the nth level is given by:

$L = n \hbar$

where $n$ is the quantum number (note that $n = 0$ is not allowed).

The rotational kinetic energy for a rotating body with moment of inertia $I$ is given by:

$E = {1 \over 2} I \omega^2$

where $\omega$ is the angular velocity. The angular momentum $L$ is related to the angular velocity $\omega$ by:

$L = I \omega$

Solving for $\omega$, we get:

$\omega = {L \over I}$

Substituting this back into the expression for energy, we get:

$E = {1 \over 2} I \left( {L \over I} \right)^2 = {L^2 \over 2I}$

Using Bohr's quantization condition $L = n \hbar$, the energy expression becomes:

$E = {{(n \hbar)^2} \over {2I}}$

Substituting $\hbar = {\hbar \over {2\pi}}$, we obtain:

$E = {n^2 \over {2I}} \left( {{\hbar^2} \over {4\pi^2}} \right) = n^2 \left( {{\hbar^2} \over {8\pi^2 I}} \right)$

Therefore, the quantized rotational energy of the diatomic molecule in the nth level is:

$E_n = n^2 \left( {{\hbar^2} \over {8\pi^2 I}} \right)$

So, the correct option is:

Option D

$n^2 \left( {{{h^2} \over {8\pi^2 I}}} \right)$

The diatomic gas has five degrees of freedom i.e., f = 5. Thus, the internal energy per mole, specific heat at constant volume, specific heat at constant pressure, and the ratio of specific heats for a diatomic gas, are given by

$U = (f/2)RT = (5/2)RT$, ${C_v} = dU/dT = 5R/2$, ${C_p} = {C_v} + R = 7R/2$, $\gamma = {C_p}/{C_v} = 7/5$.

For an adiabatic process, $T{V^{\gamma - 1}}$ = constant

${T_i}V_i^{\gamma - 1} = {T_f}V_f^{\gamma - 1}$

Substituting the given values, we get

${T_i}V_i^{\gamma - 1} = a{T_i}{\left( {{{{V_i}} \over {32}}} \right)^{\gamma - 1}} \Rightarrow a = {32^{\gamma - 1}}$

For diatomic gas, $\gamma = {7 \over 5}$

$a = {32^{{7 \over 5} - 1}} = {32^{2/5}} = {2^2} = 4$

Electrostatic pressure is

$\left( {{{{\sigma ^2}} \over {2{\varepsilon _0}}}} \right)$ N/m²

The force is

$F = {{{\sigma ^2}} \over {2{\varepsilon _0}}} \times \pi {R^2}$

Therefore, $F \propto {{{\sigma ^2}{R^2}} \over {{\varepsilon _0}}}$

The drop falls down when the weight of the drop becomes larger than the surface tension. Hence,

$mg = {{2\pi {r^2}T} \over R}$

That is,

${4 \over 3}\pi {R^3}\rho g = {{2\pi {r^2}T} \over R}$

where

r = 5 $\times$ 10^$-$4 m ; $\rho$ = 10³ kg m^$-$3 ; g = 10 m/s² ; T = 0.11 Nm^$-$1

Therefore,

${R^4} = {{3{r^2}t} \over {2\rho g}} = {{3 \times {{(5 \times {{10}^{ - 4}})}^2} \times 0.11} \over {2 \times {{10}^3} \times 10}} = 4.125 \times {10^{ - 12}}$ m⁴

$\Rightarrow R = 1.425 \times {10^{ - 3}}$ m

The moment of inertia of CO molecule is

I = $\mu$r² ....... (i)

where, $\mu$ = reduced mass of the CO molecule, r = distance between C and O or bond length

The reduced mass $\mu$ of the CO molecule is

$\mu = {{{m_1}{m_2}} \over {{m_1} + {m_2}}} = \left[ {{{(12)(16)} \over {12 + 16}}} \right] \times {5 \over 3} \times {10^{ - 27}}$ kg

But $I = 1.87 \times {10^{ - 46}}$ kg m² (From the above question)

From equation (i), we get

${r^2} = {I \over \mu }$

Substituting the values of I and $\mu$ in above equation, we get

${r^2} = {{1.87 \times {{10}^{ - 46}}} \over {\left[ {{{12 \times 16} \over {28}} \times {5 \over 3} \times {{10}^{ - 27}}} \right]}}$

or ${r^2} = {{1.87 \times {{10}^{ - 46}} \times 28 \times 3} \over {12 \times 16 \times 5 \times {{10}^{ - 27}}}}$ or $r = 1.3 \times {10^{ - 10}}$ m

From the figure shown here, we have

$\tan {i_c} = {R \over t}$

$\sin {i_c} = {R \over {\sqrt {{R^2} + {t^2}} }} = {1 \over \mu } = {3 \over 5}$

$25{R^2} = 9{R^2} + 9{t^2}$

$16{R^2} = 9{t^2} \Rightarrow R = {{3t} \over 4} = {{3 \times 8} \over 4} = 6$ cm

Focal length of a convex mirror, $f = {R \over 2} = {{20} \over 2}$ m = 10 m

For first object, ${v_1} = + {{25} \over 3}$ m, $f = + 10$ m

Using mirror formula ${1 \over v} + {1 \over u} = {1 \over f}$

$\therefore$ ${1 \over {(25/3)}} + {1 \over {{u_1}}} = {1 \over {10}}$ or ${1 \over {{u_1}}} = {1 \over {10}} - {3 \over {25}}$

or ${u_1} = - 50$ m

For second object,

${v_2} = + {{50} \over 7}$ m, $f = + 10$ m

$\therefore$ ${1 \over {{v_2}}} + {1 \over {{u_2}}} = {1 \over f}$

${1 \over {(50/7)}} + {1 \over {{u_2}}} = {1 \over {10}}$ or ${1 \over {{u_2}}} = {1 \over {10}} - {7 \over {50}}$ or ${u_2} = - 25$ m

Speed of the object $= {{25} \over {30}}$ m s^$-$1

$= {{25} \over {30}} \times {{18} \over 5}$ km h^$-$1 = 3 km h^$-$1

The energy of photon is equal to the energy difference between the ground level and first excited level.

$hv = {E_2} - {E_1}$

$hv = {{(4 - 1){h^2}} \over {8{\pi ^2}I}} \Rightarrow I = {{3h} \over {8{\pi ^2}v}}$

$I = {{3 \times 2\pi \times {{10}^{ - 34}}} \over {[(8{\pi ^2}4)/\pi ] \times {{10}^{11}}}} = {3 \over {16}} \times {10^{ - 45}}$ kg-m² = 1.87 $\times$ 10^$-$46 kg-m²

The forces acting on the oil drop are its weight, buoyant force, and electrostatic force. The buoyant force on the oil drop is very small as compared to other two forces. Thus, the weight of the spherical oil drop is balanced by the electrostatic force

$qE = mg$

$qE = {4 \over 3}\pi {r^3}\rho g$ ..... (i)

where, r = radius of oil drop, $\rho$ = density of oil, q = charge on the oil drop

IN the absence of electric field,

IN equilibrium,

Viscous force on the drop = Weight of the drop

$6\pi \eta r{v_T} = mg$

$6\pi \eta r{v_T} = {4 \over 3}\pi {r^3}\rho g$ ...... (ii)

where, v_T = terminal velocity, $\eta$ = coefficient of viscosity of air

or ${r^2} = {{18 \times \eta \times {v_T}} \over {4 \times \rho \times g}}$

Substituting the given values, we get

${r^2} = {{18 \times 1.8 \times {{10}^{ - 5}} \times 2 \times {{10}^{ - 3}}} \over {4 \times 900 \times 9.8}}$ or $r = {3 \over 7} \times {10^{ - 5}}$ m

From equation (i), we get $q = {{4\pi {r^3}\rho g} \over {3E}}$

Substituting the given values, we get

$q = {{4 \times \pi \times {{\left( {{3 \over 7} \times {{10}^{ - 5}}} \right)}^3} \times 900 \times 9.8} \over {3 \times \left( {{{81\pi } \over 7} \times {{10}^5}} \right)}}$

$= {{4 \times \pi \times 7 \times 3 \times 3 \times 3 \times {{10}^{ - 15}} \times 900 \times 9.8} \over {3 \times 81\pi \times {{10}^5} \times 7 \times 7 \times 7}} = 8 \times {10^{ - 19}}C$

We have

Surface energy = Tension $\times$ Area

That is, E_surface = T $\times$ 4$\pi$R²

where T = 0.11 N/m and R = 1.4 $\times$ 10^$-$4 m. Therefore, the surface energy is

${E_{surface}} = T \times 4\pi {R^2} = 0.11 \times 4 \times 3.14 \times {(1.4 \times {10^{ - 3}})^2} = 2.7 \times {10^{ - 6}}J$

The equivalent resistance of the two parallel resistors is

$R = {{(2\,M\Omega )(2\,M\Omega )} \over {(2\,M\Omega ) + (2\,M\Omega )}} = 1M\Omega$

The equivalent capacitance of the two parallel capacitors is

$C = 2\mu F + 2\mu F = 4\mu F$

This corresponding equivalent diagram is as shown in the figure.

The voltage across the equivalent capacitor is same as the voltage across the individual capacitors (parallel combination). Thus, we need to find time t at which the voltage across C become 4 V in the equivalent circuit (charging of a capacitor). The voltage across C at time t is

$V = {V_0}\left[ {1 - {e^{ - t/(RC)}}} \right]$,

which simplifies to

$t = RC\ln \left( {{{{V_0}} \over {{V_0} - V}}} \right)$.

Substitute ${V_0} = 10$ V, $V = 4$ V, $R = 1 \times {10^6}\Omega$ and $C = 4 \times {10^{ - 6}}F$ to get

$t = 4\ln (5/3) = 4(\ln 5 - \ln 3) = 2\,s$.

If ${𝜇}_1<{𝜇}_2$ then the ray bends towards the normal after refraction at ${𝜇}_1-{𝜇}_2$ interface. Incident rays parallel to the optic axis bend towards the optic axis by the convex lens and away from the optic axis by the concave lens. If ${𝜇}_1>{𝜇}_2$ then the ray bends away from the normal. If ${𝜇}_2={𝜇}_3$ then the ray goes straight without bending at the ${𝜇}_2-{𝜇}_3$ interface. If ${𝜇}_2>{𝜇}_3$ then the ray bends away from the normal.

In circuit $\left(𝑃\right)$, under steady state, capacitor will act like an infinite impedance and inductor will act like a zero impedance. Thus, $𝐼=0,{𝑉}_1=0$, and ${𝑉}_2=𝑉$.

In circuit $\left(𝑄\right)$, inductor will act as zero resistance in steady state giving us $𝐼=𝑉/𝑅=𝑉/2,{𝑉}_1=0$, and ${𝑉}_2=𝑉$.

In circuit $\left(𝑅\right)$, the inductive reactance ${𝑋}_{𝐿}$ and impedance $𝑍$ are

$\begin{array}{cc}{𝑋}_{𝐿}& =𝜔𝐿=2𝜋𝜈𝐿=1.88\Omega ,\text{ and }\\ \\ 𝑍& =\sqrt{\surd {𝑋}_2^{𝐿}+{𝑅}^2}=2.75\Omega \end{array}$

Thus, $𝐼=𝑉/𝑍\ne 0,{𝑉}_1={𝑋}_{𝐿}𝐼=1.88𝐼,{𝑉}_2=𝑅𝐼=2𝐼$, and ${𝑉}_2>{𝑉}_1$.

In circuit (S), inductive reactance ${𝑋}_{𝐿}$, capacitive reactance ${𝑋}_{𝐶}$, and impedance $𝑍$ are

$\begin{array}{cc}{𝑋}_{𝐿}& =1.88\Omega \\ \\ {𝑋}_{𝐶}& =1/\left(𝜔𝐶\right)=1061\Omega ,\text{ and }\\ \\ 𝑍& ={𝑋}_{𝐶}-{𝑋}_{𝐿}=1059\Omega \end{array}$

Thus the current in the circuit $𝐼=𝑉/𝑍\ne 0,{𝑉}_1=$ ${𝑋}_{𝐿}𝐼=1.88𝐼,{𝑉}_2={𝑋}_{𝐶}𝐼=1061𝐼$, and ${𝑉}_2>{𝑉}_1$.

In circuit $\left(𝑇\right),{𝑋}_{𝐶},𝑅$, and $𝑍$ are

$\begin{array}{cc}{𝑋}_{𝐶}& =1/\left(𝜔𝐶\right)=1061\Omega \\ \\ 𝑅& =1000\Omega ,\text{ and }\\ \\ 𝑍& =\sqrt{\surd {𝑅}^2+{𝑋}_2^{𝐶}}=1458\Omega \end{array}$

Thus, $𝐼=𝑉/𝑍\ne 0,{𝑉}_1=𝑅𝐼=1000𝐼,{𝑉}_2={𝑋}_{𝐶}𝐼=$ $1061𝐼$, and ${𝑉}_2>{𝑉}_1$.

To identify the orbital angular momentum quantum number, $l$, for the state $S_2$ of a hydrogen-like species such as $Li^{2+}$, we can refer to the given information and the known equations for the energy levels of hydrogen-like atoms. Hydrogen-like atoms or ions have only one electron and their energy in a particular state is given by the equation:

$E_n = -\frac{Z^2}{n^2} E_0$

where:

• $E_n$ is the energy of the electron in the nth energy level,
• $Z$ is the atomic number of the species (for $Li$, $Z = 3$),
• $n$ is the principal quantum number,
• $E_0$ is the energy of the ground state of hydrogen ($-13.6$ eV).

The problem states that $S_2$ has energy equal to the ground state energy of the hydrogen atom. The ground state of hydrogen corresponds to $n = 1$ and $E_0 = -13.6$ eV. However, for $Li^{2+}$, with $Z = 3$, the same energy level could be attained at a different value of $n$ since the $Z^2$ factor magnifies the energy levels with increasing atomic number. Let's calculate the principal quantum number, $n$, for the $Li^{2+}$ ion that would give it an energy equal to $E_0$:

For $Li^{2+}$ ion to have the ground state energy of a hydrogen atom, we can set the energies equal and solve for $n$:

$-\frac{Z^2}{n^2}E_0 = E_0$

Substituting $Z = 3$ and simplifying:

$-\frac{9}{n^2} = 1$

From which we find, $n^2 = 9$ and thus $n = 3$.

Furthermore, the problem states that $S_1$ has one radial node and upon absorbing light, it transitions to $S_2$ which also has one radial node. Radial nodes are related to the principal quantum number $n$ and the angular quantum number $l$ by the formula:

Number of radial nodes = $n - l - 1$

Given that $S_2$ has one radial node, we can plug $n = 3$ into the radial node formula to solve for $l$:

$1 = 3 - l - 1$

This simplifies to:

$l = 3 - 2$

$l = 1$

Therefore, the orbital angular momentum quantum number of the state $S_2$ is 1, which corresponds to option B.

A diprotic acid is an acid that contains within its molecular structure two hydrogen atoms per molecule capable of dissociating (i.e., ionisable protons) in water.

$H_2{\mathrm{SO}}_4,H_2{\mathrm{CO}}_3,H_2{S}_2{O}_7,H_2{\mathrm{CrO}}_4,H_3{\mathrm{PO}}_3,H_2{\mathrm{SO}}_3$

Structure of each of the compounds is given below :

To address this question, we'll consider each element listed to determine if it exhibits only one non-zero oxidation state. This requires looking into common oxidation states that each of these elements can attain.

Oxygen (O): Oxygen typically shows an oxidation state of -2, except in peroxides where it has -1 and in compounds like OF2 (oxygen difluoride) where it exhibits +2. Therefore, it shows more than one non-zero oxidation state.

Chlorine (Cl): Chlorine commonly exhibits oxidation states of -1, +1, +3, +5, and +7 depending on the compound it is part of. Thus, it shows multiple non-zero oxidation states.

Fluorine (F): Fluorine is the most electronegative element and almost exclusively exhibits an oxidation state of -1 in all of its compounds. Hence, it shows only one non-zero oxidation state.

Nitrogen (N): Nitrogen can exhibit a variety of oxidation states ranging from -3 to +5. Therefore, nitrogen too shows multiple non-zero oxidation states.

Phosphorus (P): Phosphorus commonly exhibits oxidation states of -3, +3, and +5. So, phosphorus also has multiple oxidation states.

Tin (Sn): Tin typically shows two common oxidation states: +2 (Sn(II)) and +4 (Sn(IV)). Therefore, it has more than one non-zero oxidation state.

Thallium (Tl): Thallium primarily shows +1 and occasionally +3 oxidation states. Hence, thallium also exhibits more than one non-zero oxidation state.

Sodium (Na): Sodium almost always exhibits a +1 oxidation state in its compounds. Therefore, it shows only one non-zero oxidation state.

Titanium (Ti): Titanium can exhibit multiple oxidation states; common ones include +2, +3, and +4. Therefore, it has multiple non-zero oxidation states.

From the elements listed: Fluorine (F) and Sodium (Na) are the only ones that show only one non-zero oxidation state. Thus, among the given elements, 2 of them (F and Na) show just one non-zero oxidation state.

For a hydrogen-like ion, the energy levels can be given by the formula:

$E_n = -\frac{Z^2}{n^2} E_0$

where $E_n$ is the energy of the nth level, $Z$ is the atomic number (for Li$^{2+}$, $Z=3$), $n$ is the principal quantum number, and $E_0$ is the ground state energy of the hydrogen atom ($-13.6 \, eV$).

Given that state $S_2$ has energy equal to the ground state energy of the hydrogen atom and one radial node, we identify that $S_2$ corresponds to $n=2$ for a hydrogen atom. This is because for hydrogen-like species, the number of radial nodes is given by $n-1$, where $n$ is the principal quantum number. The ground state ($n=1$) has 0 nodes, the first excited state ($n=2$) has 1 radial node, etc.

Since the energy of $S_2$ is equal to the ground state energy of the hydrogen atom, we can directly compare the energies. For hydrogen ($Z=1$), the ground state energy ($n=1$) is:

$E = -E_0$

For the Li$^{2+}$ ion in state $S_2$ (which we established is equivalent to $n=2$ in terms of energy for hydrogen), we use the formula $E_n = -\frac{Z^2}{n^2} E_0$. Since $Z=3$ for Li$^{2+}$, and given that $S_2$ has the energy equivalent to the ground state of hydrogen ($E_0$), we solve for the energy ratio rather than the specific energy of $S_2$.

We then look at $S_1$, which we know must be the ground state for Li$^{2+}$ since it is the state before $S_2$ and has one radial node (indicating $n=2$ for $S_1$).

Thus, for $S_1$, which actually corresponds to $n=2$ for Li$^{2+}$, the energy in units of the hydrogen atom ground state energy is:

$E_{S_1} = -\frac{Z^2}{n^2} E_0 = -\frac{3^2}{2^2}E_0 = -\frac{9}{4}E_0$

Now, to express this in units of the hydrogen atom ground state energy ($-E_0$):

$\frac{E_{S_1}}{E_0} = -\frac{9}{4} = -2.25$

So, considering the provided options and the fact that energy levels are usually considered in positive values when comparing magnitudes, the correct answer is:

Option C: 2.25.

Molecular orbital configuration of B₂ (10 electrons) is

${𝜎}_{1{𝑠}^2}{𝜎}_{\ast }^{1{𝑠}^2}{𝜎}_{2{𝑠}^2}{𝜎}_{\ast }^{2{𝑠}^2}{𝜋}_{2{𝑝}_1^{𝑥}}={𝜋}_{2{𝑝}_1^{𝑦}}$

Here in B₂, 2 unpaired electrons present.

Since the Hund's rule is violated, two electrons are placed in $𝜋2{𝑝}_{𝑥}$ molecular orbital, so

$B_2\left(10\right):{𝜎}_{1{𝑠}^2}{𝜎}_{\ast }^{1{𝑠}^2}{𝜎}_{2{𝑠}^2}{𝜎}_{\ast }^{2{𝑠}^2}{𝜋}_{2{𝑝}_2^{𝑥}}$

Thus, bond order $=\frac{6-4}{2}=1$.

As there are no unpaired electrons, the nature is diamagnetic.

${\mathrm{OSF}}_2$ has $𝑠{𝑝}^3$ hybridisation. According to VSEPR theory, we can conclude that the shape of ${\mathrm{OSF}}_2$ is a trigonal pyramid.

Sulphur has a non-bonding pair of electrons and three sulphur-oxygen ( $S-O)$ bond pairs. ${\mathrm{SO}}_2$ is $𝑠{𝑝}^2$ hybridised with three bond pairs (two sigma and one pi) and one lone pair. It is replace trigonal by bent.

${\mathrm{BrF}}_3$ is $𝑠{𝑝}^3𝑑$ hybridized with three bond pair and two lone pairs. It is T shaped.

${\mathrm{SiO}}_{2-}^3$ is $𝑠{𝑝}^2$ hybridised with four bond pairs (three sigma and one pi) and no lone pair. It has trigonal planar shape.

For hydrogen-like species, the energy levels depend on the principal quantum number $n$ and are inversely proportional to the square of $n$, and they can be characterized by their nuclear charge $Z$. The energy for a hydrogen-like ion can be expressed as:

$E_n = -\dfrac{Z^2R_H}{n^2}$

where $R_H$ is the Rydberg constant for hydrogen, $Z$ is the atomic number (for Li²⁺, $Z=3$), and $n$ is the principal quantum number associated with the energy level.

Given that S₂ has energy equal to the ground state energy of the hydrogen atom ($E_n$ for hydrogen when $n=1$ and $Z=1$ is $E_1 = -R_H$), and given that S₂ has one radial node. The radial node information tells us about the principal quantum number $n$, as the number of radial nodes is given by $n-l-1$, where $l$ is the azimuthal quantum number.

The ground state of hydrogen corresponds to $n=1$. For the given species Li²⁺ to have the same energy as the ground state of the hydrogen atom but for state S₂, we can use the energy relation. Since the energy is specified to be the same as hydrogen's ground state, let's set up the equality according to the equation given and solve for $n$ specific to the condition (Li²⁺ is considered here, but the condition is about energy equivalence).

Given that S₂ has one radial node, it cannot be the ground state energy level for Li²⁺ (which would directly correspond to $n=1$), it implies a different $n$. For one radial node, the condition $n-l-1 = 1$ must be met.

The options presented are:

• 1s ($n=1$, $l=0$) - No radial nodes
• 2s ($n=2$, $l=0$) - One radial node
• 2p ($n=2$, $l=1$) - No radial nodes, due to the different $l$
• 3s ($n=3$, $l=0$) - Two radial nodes

S₁ is described as having one radial node. Based on the rule for the number of radial nodes ($n-l-1$), the only states that fits this condition directly from the options provided are $2s$ (since for a $2s$ orbital, $n=2$, $l=0$, yielding $2-0-1=1$ radial node).

Hence, the correct option for S₁ is:

Option B: 2s

$\text{ The products obtained on nitration of }𝐏,𝐐\text{ and }𝐒\text{ are }$

Here strongly activating and ortho-, para directing -OH group determines the site for electrophilic substitution.

Here $-{\mathrm{OCH}}_3$ is a stronger activator than $-{\mathrm{CH}}_3$ and both are ortho-, para directing.

Here, the substitution takes place on the activated ring (with substituent ${\mathrm{PhCOO}}^{-}$ ) at the sterically unhindered para-position.

The diagonal is given as 4r, where r is the radius of atom or sphere forming close packed structure.

$\begin{array}{cc}\text{ Diagonal }& =\sqrt{\surd L^2+L^2}=\sqrt{\surd 2L^2}\\ \\ 4𝑟& =\sqrt{\surd 2L^2}\text{ or }L=\frac{4𝑟}{\sqrt{\surd 2}}=2\sqrt{\surd 2}𝑟\\ \\ \text{ or }\text{ Total area }& =L^2\\ \\ (2\sqrt{\surd 2}𝑟{)}^2& =8{𝑟}^2\end{array}$

Number of spheres inside the square is

$1+4\left(\frac{1}{4}\right)=2$

Area of each sphere $=𝜋{𝑟}^2$

Total area of spheres $=2\times 𝜋{𝑟}^2$

$\begin{array}{cc}\text{ Packing fraction }& =\frac{\text{ Total area of spheres }}{\text{ Total area }}\\ \\ & =\frac{2\times 𝜋{𝑟}^2}{8{𝑟}^2}=\frac{𝜋}{4}=0.785\end{array}$

So, the percentage fraction is $78.5\%$.

$\begin{array}{cc}& {\left[{\mathrm{NiCl}}_4\right]}^{2-}\\ \\ & \text{ Oxidation state of }{\left[{\mathrm{NiCl}}_4\right]}^{2-}=+2\\ \\ & \mathrm{Ni}\left(28\right)=\left[\mathrm{Ar}\right]3{𝑑}^{8}4{𝑠}^2\\ \\ & {\mathrm{Ni}}^{2+}=\left[\mathrm{Ar}\right]3{𝑑}^8\end{array}$

Among the given complexes of Ni , only ${\mathrm{Cl}}^{-}$is a weak ligand and does not cause pairing of electrons to take place.

Thus, the spin only magnetic moment, with two unpaired electrons $(𝑛=2)$ configuration is

$\sqrt{\surd 𝑛(𝑛+2)}=\sqrt{\surd 2(2+2)}=2.82\mathrm{BM}=2.82\mathrm{BM}$

The reaction taking place is Hoffmann bromamide degradation followed by benzoylation.

For calculating work done, we need to calculate the area under curve for solid and dotted lines.

Let ' ${𝑤}_{𝑑}$ and ' $𝑤$ ' be work done along the dotted and solid path respectively.

$\begin{array}{cc}& W_{𝑑}=\text{ Area ABCD }+\text{ Area EFGC + Area FGIH }\\ \\ & {𝑤}_{𝑑}=4\times 1.5+1\times 1+2.5\times 2/3\\ \\ & =8.65\end{array}$