NEET-UG 2017

Dimension of
$\frac{e^2}{4\pi {\epsilon }_0}$ = [ F$\times$ d² ] = [ML³T^-2]

Dimension of G = [M^-1L³T^-2],

Dimension of c = [LT^-1]

Now assume dimension of length is related as,

L = [c]^x[G]^y[$\frac{e^2}{4\pi {\epsilon }_0}$]^z

$\therefore$ [L¹] = [ML³T^-2]^z [M^-1L³T^-2]^y [LT^-1]^x

Comparing both sides and solving we get,

x = -2, y = $\frac{1}{2}$, z = $\frac{1}{2}$

$\therefore$ L = $\frac{1}{c^2}{\left[G\frac{e^2}{4\pi {\epsilon }_0}\right]}^{1/2}$

Velocity of preeti with respect to elevator v₁ = $\frac{d}{t_1}$

Velocity of elevator with respect to ground v₂ = $\frac{d}{t_2}$

$\therefore$ Net velocity of preeti on moving escalator with respect to the ground

v = v₁ + v₂

$\frac{d}{t}$ = $\frac{d}{t_1}$ + $\frac{d}{t_2}$

$\frac{1}{t}$ = $\frac{1}{t_1}$ + $\frac{1}{t_2}$

$\therefore$ t = $\frac{t_1{t}_2}{t_1+t_2}$

Here t is the time taken by preeti to walk up on the moving escalator.

x = 5t $-$ 2t² and y = 10t

$\frac{dx}{dt}$ = 5 - 4t, $\frac{dy}{dt}$ = 10

$\therefore$ v_x = 5 - 4t, v_y = 10

$\frac{d{v}_x}{dt}$ = - 4, $\frac{d{v}_y}{dt}$ = 0

$\therefore$ $a_x$ = - 4, $a_y$ = 0

$\overrightarrow{a}=a_x\hat{i}+a_y\hat{j}$ = $-4\hat{j}$ m/s²

$\therefore$ Acceleration of the particle at t = 2 s is - 4 m/s²

Centripetal force $\left(\frac{m{v}^2}{l}\right)$ is provided by tension so net force on the particle will be equal to tension T.

Before cutting the string
kx = T + 3 mg ...(i)
T = mg ...(ii)
$\Rightarrow$ kx = 4mg

After cutting the string T = 0
$a_A=\frac{4mg-3mg}{3m}$


$a_A=\frac{g}{3}\uparrow$ and $a_B=\frac{mg}{m}=g\downarrow$

From work-energy theorem,
W_g + W_a = $\Delta$K.E

or mgh + W_a = $\frac{1}{2}m{v}^2-0$

$10^{-3}\times 10\times 10^3+W_a=\frac{1}{2}\times 10^{-3}\times {\left(50\right)}^2$

$\Rightarrow W_a=-8.75J$

which is the work done due to air resistance

Work done due to gravity = mgh

= 10^–3 × 10 × 10³ = 10 J

Centre of gravity of a body is the point at which the total gravitational torque on body is zero. Centre of mass and centre of gravity coincides only for symmetrical bodies.
Hence statements (1) and (2) are incorrect.
A couple of a body produces rotational motion only.
Hence statement (3) is incorrect.
Mechanical advantage greater than one means that the system will require a force that is less than the load in order to move it.
Hence statement (4) is correct.

Given, mass of cylinder m = 3kg
R = 40 cm = 0.4 m
F = 30 N ; $\alpha$ = ? As we know, torque $\tau$ = Ia

F × R = MR²$\alpha$

$\alpha$ = $\frac{F\times R}{M{R}^2}$

$\alpha =\frac{30\times \left(0.4\right)}{3\times {\left(0.4\right)}^2}$

$\Rightarrow \alpha =25rad/s^2$

According to the problem,

I${\omega }_1$ + I${\omega }_2$ = 2I${\omega }_0$

${\omega }_0$ = $\frac{\left({\omega }_1-{\omega }_2\right)}{2}$

$K_i=\frac{1}{2}I\left({{\omega }_1}^2+{{\omega }_2}^2\right)$

$K_f=\frac{1}{4}{\left({\omega }_1+{\omega }_2\right)}^2$

Loss $\Delta K=I\left[\frac{{{\omega }_1}^2}{2}+\frac{{{\omega }_2}^2}{2}-\frac{{{\omega }_1}^2}{4}-\frac{{{\omega }_2}^2}{4}-\frac{2{\omega }_1{\omega }_2}{4}\right]$

$=I\left[\frac{{{\omega }_1}^2}{4}+\frac{{{\omega }_2}^2}{4}-\frac{2{\omega }_1{\omega }_2}{4}\right]$

$=\frac{I}{4}{\left[{\omega }_1-{\omega }_2\right]}^2$

Above earth surface

$g_h=g\left(1-\frac{2h}{R_e}\right)$

Below earth surface

$g_d=g\left(1-\frac{d}{R_e}\right)$

According to question, g_h = g_d

$g\left(1-\frac{2h}{R_e}\right)$ = $g\left(1-\frac{d}{R_e}\right)$

Clearly, d = 2h = 2 km

Since two astronauts are floating in gravitational free space. The only force acting on the two astronauts is the gravitational pull of their masses, $F=\frac{G{m}_1{m}_2}{r^2}$

which is attractive in nature.
Hence they move towards each other.

Bulk modulus is given by

$B=\frac{P}{\left(\frac{\Delta V}{V}\right)}\Rightarrow \frac{\Delta V}{V}=\frac{P}{B}$

$3\frac{\Delta R}{R}=\frac{P}{B}$   (here, $\frac{\Delta R}{R}$ = fractional decreases in radius)

$\Rightarrow \frac{\Delta R}{R}=\frac{P}{3B}$

Equivalent thermal conductivity of the composite rod in parallel combination will be,

$K=\frac{K_1{A}_1+K_2{A}_2}{A_1+A_2}=\frac{K_1+K_2}{2}$

Here, $h_{oil}\times {\rho }_{oil}\times g=h_{water}\times {\rho }_{water}\times g$

${\rho }_{0}g\times 140\times 10^{-3}={\rho }_{w}g\times 130\times 10^{-3}$

${\rho }_{oil}=\frac{130}{140}\times 10^3\approx 928kg/m^3$
$\left[\because {\rho }_w=1kg{m}^{-3}\right]$

Given r₁ = 12 cm , r₂ = 6 cm

T₁ = 500 K and T₂ = 2 × 500 = 1000 K

P₁ = 450 watt

Rate of power loss $P\propto r^2{T}^4$

$\frac{P_1}{P_2}=\frac{r_1^2{T}_1^4}{r_2^2{T}_2^4}$

$P_2=P_1\frac{r_1^2{T}_1^4}{r_2^2{T}_2^4}$

Solving we get, P₂ = 1800 watt

Internal energy of the system is given by

$U=\frac{f}{2}nRT$

Degree of freedom
F_diatomic = 5
f_monoatomic = 3

and, number of moles n(O₂) = 2
n(Ar) = 4

$U_{total}=\frac{5}{2}\left(2\right)RT+\frac{3}{2}\left(4\right)RT=11RT$

Process I volume is constant hence, it is isochoric

In process IV, pressure is constant hence, it is isobaric.

Given, efficiency of engine, $\eta =\frac{1}{10}$

work done on system W = 10J

Coefficient of performance of refrigerator

$\beta =\frac{Q_2}{W}=\frac{1-\eta }{\eta }=\frac{1-\frac{1}{10}}{\frac{1}{10}}=\frac{\frac{9}{10}}{\frac{1}{10}}=9$

Energy absorbed from reservoir

$Q_2=\beta w$

$Q_2=9\times 10=90J$

Let us assume, the length of spring be l.

When we cut the spring into ratio of length 1 : 2 : 3, we get three springs of lengths $\frac{l}{6},\frac{2l}{6}$ and $\frac{3l}{6}$ with force constant,

$\therefore$ $k_1=\frac{kl}{l_1}=\frac{kl}{l/6}=6k$

$k_2=\frac{kl}{l_2}=\frac{kl}{2l/6}=3k$

$k_3=\frac{kl}{l_3}=\frac{kl}{3l/6}=2k$

When connected in series,

$\frac{1}{k^{\prime }}=\frac{1}{6k}+\frac{1}{3k}+\frac{1}{2k}=\frac{1+2+3}{6k}=\frac{1}{k}$

$\therefore k^{\prime }=k$

When connected in parallel,

k'' = 6k + 3k + 2k = 11k

$\frac{k^{\prime }}{k^{\prime \prime }}=\frac{k}{11k}=\frac{1}{11}$

Given, Amplitude A = 3 cm
When particle is at x = 2 cm
According to question, magnitude of velocity = acceleration

$\omega \sqrt{A^2-x^2}=x{\omega }^2$

$\sqrt{{\left(3\right)}^2-{\left(2\right)}^2}=2\left(\frac{2\pi }{T}\right)$

$\sqrt{5}=\frac{4\pi }{T}\Rightarrow T=\frac{4\pi }{\sqrt{5}}$

As we known from Doppler's Effect

$f_{apprent}=f_0\left[\frac{v+v_0}{v-v_s}\right]=400\left[\frac{340+16.5}{340-22}\right]$

$\therefore$ f_apprent = 448 Hz

The difference of frequencies of closed pipe
$\frac{2v}{4l}=260-220=40Hz$

So, the fundamental frequency $f^{\prime }=\frac{v}{4l}=20Hz$

As the regions are of equipotentials, so Work done W = q$\Delta$V

$\Delta$V is same in all the cases hence work - done will also be same in all the cases.

According to question, the net electrostatic force (F_E) = gravitational force (F_G)

F_E = F_G

$\Rightarrow \frac{1}{4\pi {\epsilon }_0}\frac{\Delta e^2}{d^2}=\frac{G{m}^2}{d^2}$

$\Rightarrow \Delta e=m\sqrt{\frac{G}{K}}\left(\frac{1}{4\pi {\epsilon }_0}=k=9\times {10}^9\right)$

$=1.67\times 10^{-27}\sqrt{\frac{6.67\times {10}^{-11}}{9\times {10}^9}}$

$\Delta e\approx 1.436\times 10^{-37}C$

A potentiometer is an accurate and versatile device to make electrical measurements of E.M.F. because the method involves zero deflection without any current in galvanometer.

Resistance $R=\rho l/A$

$l^{\prime }=nl$ and $A^{\prime }=\frac{A}{n},R^{\prime }=\frac{\rho l}{A^{\prime }}$

$R^{\prime }=\frac{\rho nl}{A/n}=\frac{\rho l}{A}{n}^2$

$R^{\prime }=R{n}^2$

When the capacitor is charged by a battery of potential V, then energy stored in the capacitor,

$U_i=\frac{1}{2}C{V}^2$   ...(i)

When the battery is removed and another identical uncharged capacitor is connected in parallel


Common potential, $V^{\prime }=\frac{CV}{C+C}=\frac{V}{2}$

Then the energy stored in the capacitor,

$U_f=\frac{1}{2}\left(2C\right){\left(\frac{V}{2}\right)}^2=\frac{1}{4}C{V}^2$   ...(ii)

$\therefore$ From eqns. (i) and (ii)

$U_f=\frac{U_i}{2}$

that means the total electrostatic energy of resulting system will decreases by a factor of 2.

Since same current flowing through both the wires

i_i = i₂ = i

So F₁ = $\frac{{\mu }_0{i}^2}{2\pi d}$ = F₂

$\therefore$ Magnitude of force per unit length on the middle wire 'B'

F_net = $\sqrt{F_1^2+F_2^2}$ = $\frac{{\mu }_0{i}^2}{\sqrt{2}\pi d}$

Given: n = 250, l = 2.1 cm, W = 1.25 cm, i = 85 μA, B = 0.85 T

Now W = MB (cos θ₁ – cos θ₂) = 2MB = 2(iNA)B

= 2 × 85 × 10^–6 × 250 × 2.1 × 1.25 × 10^–4 × 0.85

= 9.1 × 10^–6 J

If $\theta$₁ and $\theta$₂ are opparent angles of dip.

Let $\alpha$ be the angle which one of the plane make with the magnetic meridian.

For magnetic meridian and plane (1)

tan $\theta$₁ = $\frac{\tan \theta }{\cos \alpha }$

$\Rightarrow$ cos $\alpha$ = $\frac{\tan \theta }{\tan {\theta }_1}$ .....(1)

For magnetic meridian and plane (2)

tan $\theta$₁ = $\frac{\tan \theta }{\cos \left(90^{\circ }-\alpha \right)}$

$\Rightarrow$ $\sin \alpha =\frac{\tan \theta }{\tan {\theta }_2}$ ....(2)

(1)² + (2)²

1 = $\frac{{\tan }^2\theta }{{\tan }^2{\theta }_1}+\frac{{\tan }^2\theta }{{\tan }^2{\theta }_2}$

$\therefore$ cot²$\theta$ = cot²$$\theta$$₁ + cot²$$\theta$$₂

q = $-\frac{d\varphi }{R}$

= $-\frac{NA\left(B_2-B_1\right)}{R}$

= $\frac{NA{\mu }_{0}n\left(i_1-i_2\right)}{R}$

= $\frac{100\times \pi \times {10}^{-4}\times 4\pi \times {10}^{-7}\times 2\times {10}^4\times 4}{10{\pi }^2}$

= 32 $\times$ 10^-6 C = 32 $\mu$C

At t = 0, there will be no current flowing through an inductor as it resists the change in current, whereas current will slowly build in it. Also, as switch is closed, inductor will give an infinite resistance where capacitor shows zero resistance, so, current through battery, will be

I = $\frac{E}{\left(R/2\right)}=\frac{2E}{R}=\frac{2\times 18}{9}$ = 4 A

The rms value of magnetic field

B_rms = $\frac{E_{rms}}{c}$ = $\frac{6}{\left(3\times {10}^8\right)}$ = 2 × 10^{– 8} T

Now peak value of magnetic field

B_peak = $\sqrt{2}{B}_{rms}$ = 2.83 $\times$ 10^$-$8 T

The condition for dispersion without deviation is given as

|($\mu$ – 1)|A = |($\mu$' – 1)| A'

$\Rightarrow$ (1.42 – 1) × 10^o = (1.7 – 1)A'

$\Rightarrow$ 4.2 = 0.7A' or A' = 6^o

As the mirror is rotated θ, the reflected ray rotates 2θ.

Using trigonometry, tan 2θ = y/x

θ being small, 2θ = y/x or θ = y/2x

Position of 8^th bright fringe in medium

x = $\frac{8{\lambda }_{m}D}{\mu d}$

Position of 5^th dark fringe in air,

x' = $\frac{\left(5-\frac{1}{2}\right){\lambda }_{air}D}{d}$ = $\frac{4.5{\lambda }_{air}D}{d}$

According to question

$\frac{8{\lambda }_{m}D}{\mu d}$ = $\frac{4.5{\lambda }_{air}D}{d}$

$\Rightarrow$ $\mu$ = 1.78

Apply Malus' Law,

Intensity of light from first polaroid P₁, I₁ = $\frac{I_0}{2}$

Intensity of light from second polaroid P₃,

I₂ = $\frac{I_0}{2}{\cos }^{2}45^{\circ }$ = $\frac{I_0}{2}\times \frac{1}{2}$

Intensity of transmitted light through P₂,

I₃ = $\frac{I_0}{4}{\cos }^{2}45^{\circ }$ = $\frac{I_0}{4}\times \frac{1}{2}$ = $\frac{I_0}{8}$

The resolving power of an optical microscope,

RP = $\frac{2\mu \sin \theta }{\lambda }$

$\Rightarrow$ RP $\propto$ $\frac{1}{\lambda }$

Now, $\frac{R{P}_1}{R{P}_2}=\frac{{\lambda }_2}{{\lambda }_1}$

= $\frac{6000}{4000}$ = $\frac{3}{2}$

Ratio of number of nuclei

$\frac{N_1}{N_2}=\frac{e^{-8\lambda t}}{e^{-\lambda t}}$ = $e^{-7\lambda t}$

According to question,

$e^{-7\lambda t}$ = $\frac{1}{e}$

$\Rightarrow$ t = $\frac{1}{7\lambda }$

In case of Balmer series :

n₁ = $\infty$, n₂ = 2

$\therefore$ $\frac{1}{{\lambda }_B}=Rc\left(\frac{1}{2^2}-\frac{1}{{\infty }^2}\right)$ = $\frac{Rc}{4}$

In case of Lyman series :

n₁ = $\infty$, n₂ = 1

$\therefore$ $\frac{1}{{\lambda }_L}=Rc\left(\frac{1}{1^2}-\frac{1}{{\infty }^2}\right)$ = Rc

$\therefore$ $\frac{{\lambda }_B}{{\lambda }_L}$ = $\frac{4}{1}$ = 4

Wavelength, $\lambda$ = $\frac{h}{p}$

$\Rightarrow$ $\lambda$ = $\frac{h}{\sqrt{2mE}}$

$\Rightarrow$ $\lambda$ = $\frac{h}{\sqrt{2m\times \frac{3}{2}kT}}$

$\Rightarrow$ $\lambda$ = $\frac{h}{\sqrt{3mkT}}$

Minimum energy, E₀ = $\frac{12375}{3250}$ eV

$\Rightarrow$ E₀ = 3.81 eV

Ultraviolet wavelength, $\lambda$ = 2536 $\stackrel{o}{A}$

$\therefore$ Imparted energy, E = $\frac{12375}{2536}$ = 4.88 eV

Now, max kinetic energy, $\frac{1}{2}m{v}^2$ = E - E₀

$\Rightarrow$ v = $\sqrt{\frac{2\times 1.07\times 1.6\times {10}^{-19}}{9.1\times {10}^{-31}}}$

$\Rightarrow$ v = 0.61 × 10⁶ m/s = 6.0 × 10⁵ m/s

Given: V_i = 3 V, R_C = 3 k$\Omega$, R_B = 2 k$\Omega$, $\beta$ = 100

Voltage gain of the CE amplifier

A_V = $\beta \frac{R_c}{R_b}$ = 100$\left(\frac{3}{2}\right)$ = 150

Power gain = A_V$\beta$ = 150(100) = 15000

A diode is said to be forward biased if p-side is at higher potential than n-side of p-n junction.

y₁ = $\overline{A+B}$

y₂ = $\overline{y_1+y_1}$

= $A+B$

y = $\overline{y_2}$ = $\overline{A+B}$

So it is NOR gate.

Molarity =

Number of moles of solute

Volume of solution (in L)

Molality =

Number of moles of solute

Mass of solvent (in kg)

Mole fraction =

Number of moles of component

Total number of moles of all components

Weight percentage =

Weight of a component

Total weight of solution

$\times$ 100

KEY Concept : Those units which are volume related will be affected by the change in temperature.

The definition of molarity is - Number of moles of solute present in one litre of solution. From definition you can see it is depends on volume which increases with increasing temperature and decreases with decreasing temperature.

All the other three options (Molality, Mole fraction, Weight fraction of solute) are mass related units and temperature has no effect on mass.

In the case of hydrogen type atoms, energy depends on the principal quantum number only. Therefore 2-s orbital will have energy equal to 2-p orbital.

2NH₃ $\rightleftharpoons$ N₂ + 3H₂;     $\frac{1}{K_1}$

N₂ + O₂ $\rightleftharpoons$ 2NO;     K₂

3H₂ + $\frac{3}{2}$O₂ $\rightleftharpoons$ 3H₂O;     (K₃)³

By adding all equations, we get

2NH₃ + $\frac{5}{2}$ O₂ $\rightleftharpoons$ 2NO + 3H₂O

$\therefore$ K = $\frac{K_2\times {\left(K_3\right)}^3}{K_1}$

Initially, P₁ = 0.4 atm

V₁ = 20 L

T₁ = 400 K

Max. pressure of CO₂ = Pressure of CO₂ at equilibrium

For reaction,

SrCO₃(s) ⇌ SrO(s) + CO₂

K_p = P_CO2 = 1.6 atm = maximum pressure of CO₂ volume of container at this stage.

V₂ = $\frac{nRT}{P_2}$ ......(i)

Since, reaction was not at equilibrium earlier and also container is sealed.

$\therefore$ n = constant

n = $\frac{P_1{V}_1}{RT}$ .....(ii)

From (i) and (ii)

V₂ = $\frac{P_1{V}_1}{RT}\times \frac{RT}{P_2}$ = $\frac{0.4\times 20}{1.6}$ = 5 L

Ag₂C₂O_4(s) $\to$ 2Ag+ + C₂O₄ ^2–
      S                  2S           S

K_sp = [Ag⁺] ² [C₂O₄ ^2–] = [2S]² [S]

[Ag+] = 2S = 2.2 × 10^–4

or S = 1.1 × 10^–4 M

$\therefore$ K_sp = [2.2 × 10^–4 ]² [1.1 × 10^–4] = 5.3 × 10^–12

The value of molal depression constant, K_f is constant for a particular solvent, thus, it will be unchanged when molality of the dilute solution is doubled.

For a spontaneous reaction,

$\Delta$G < 0 i.e. $\Delta$H - T$\Delta$S < 0

$T>\frac{\Delta H}{\Delta S}$

$T>\left(\frac{35.5\times 1000}{83.6}=424.6\approx 425K\right)$

$\therefore T>425K$

w = - P_ext$\Delta$V = -2.5(4.50 - 2.50)

$\Rightarrow$ - 5 L atm = - 5 $\times$ 1.01.325 J = - 506.625 J

$\Delta$U = q + w

As, the container is insulted, thus q = 0

Hence, $\Delta$U = w = -506.625 J

E_cell = E^o - $\frac{0.059}{n}\log \frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$

E₁ = E^o - $\frac{0.059}{2}\log \frac{0.01}{1}$

= E^o - $\frac{0.059}{2}\left(-2\right)\log 10$

E₁ = E^o + 0.059

E₂ = E^o - $\frac{0.059}{2}\log \frac{1}{0.01}$

$\Rightarrow$ E₂ = E^o - $\frac{0.059}{2}\log 100$

$\Rightarrow$ E₂ = E^o - 0.059

$\therefore$ E₁ > E₂

Slow step is the rate determining step.

Rate = k[X][Y₂] ......(i)

Equilibrium constant for fast step,

K = $\frac{{\left[X\right]}^2}{\left[X_2\right]}$

$\Rightarrow$ [X] = $\sqrt{K\left[X_2\right]}$

By substituting [X] in equation (i), we get

Rate = k$\sqrt{K\left[X_2\right]}$[Y₂] = k' ${\left[X_2\right]}^{\frac{1}{2}}$[Y₂]

$\therefore$ Order of reaction = $\frac{1}{2}+1$ = 1.5

For a first order reaction,

k = $\frac{2.303}{t}\log \frac{{\left[A\right]}_0}{{\left[A\right]}_t}$

$\Rightarrow$ 10^-2 = $\frac{2.303}{t}\log \frac{20}{5}$

$\Rightarrow$ t = 138.6 sec

Frenkel defect is favoured in those ionic compounds in which there is large difference in the size of cations and anions.

Non-stoichiometric defects due to metal deficiency is shown by FexO where x = 0.93 to 0.96.

A catalyst accelerates the forward and backward reaction to the same extent thus, the equilibrium constant does not change. Catalyst only speed up the reactions to attain equilibrium faster and does not alter equilibrium constant. Equilibrium constant varies only when temperature condition of reaction gets changed.

The electronic configuration of th element with Z = 114 is [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p². Hence it belongs to the carbon family which has the same outer electronic configuaration.

CN^- and CO have same no. of electrons and have same bond order equal to 3.

BCl₃-Trigonal planar, sp²-hybridised, 120^o angle

Zn being more reactive displaces Ag and Au.

4M(s) + 8NaCN(aq) + 2H₂O(aq) + O₂(g) $\to$
          4Na[M(CN)₂](aq) + OH(aq)

2Na[M(CN)₂](aq) + Zn(s) $\to$ Na₂[Zn(CN)₄](aq) +2M(s)

Here M = Ag or Au

This method is known as Mac-Arthur forest cyanide process.

Li⁺ being smallest, has maximum charge density.

So, Li⁺ is most heavily hydrated among all alkali metal ions. Effective size of Li⁺ in aqueous solution is therefore, largest. So, moves slowest under electric field.

XX' $\to$ Linear (e.g. ClF, BrF)

XX₃' $\to$ T-Shape (e.g. ClF₃, BrF₃)

XX₅' $\to$ Square pyramidal (e.g. BrF₅, IF₅)

XX₇' $\to$ Pentagonal bipyramidal (e.g. IF₇)

The inertness of s-subshell electrons towards bond formation is known as inert pair effect. This effect increases down the group thus, for Sn, +4 oxidation state is more stable, whereas, for Pb, +2 oxidation state is more stable, i.e., Sn²⁺ is reducing while Pb⁴⁺ is oxidising

S₄O₆^2– and S₂O₃^2– have S—S bond.

HgCl₂ + 4I^– (aq) $\to$ HgI₄^2– (aq) + 2Cl^–(aq)

I₂(s) + I^– (aq) $\to$ I₃^–(aq)

Acidified KMnO₄ is a strong oxidizing agent thus, among the given option which readily undergoes oxidation with KMnO₄ will decolourise it. CO₂, NO₂ and P₂O₅ are already in their highest oxidation state while SO₂ can further oxidize with KMnO₄ to give sulphate ions.

2MnO₄^– + 5SO₂ + 2H₂O $\to$ 2Mn²⁺ + 5SO₄^2– + 4H⁺

Minimum or comparable energy gap between 5f, 6d and 7s subshell makes electron excitation easier, hence there is a greater range of oxidation states in actinoids.

₂₅Mn = [Ar] 3d⁵4s²

Mn⁺³ = [Ar] 3d⁴

CN^– is a strong field ligand thus, it causes pairing of electrons in 3d-orbital.
As, coordination number of Mn = 6, so it will form an octahedral complex.