NEET-UG 2013

Given P $=\frac{a^3{b}^2}{cd}$

% error in P :

$\frac{\Delta P}{P}\times 100$ = $\left[3\frac{\Delta a}{a}+2\frac{\Delta b}{b}+\frac{\Delta c}{c}+\frac{\Delta d}{d}\right]\times 100$

                  = 3 $\times$ 1% + 2 $\times$ 2% + 3% + 4% = 14%

Here h = $\frac{1}{2}g{t}^2$

$\therefore$ h₁ = $\frac{1}{2}g{\left(5\right)}^2$ = 125

h₁ + h₂ = $\frac{1}{2}g{\left(10\right)}^2$ = 500

$\Rightarrow$ h₂ = 375

h₁ + h₂ + h₃ = $\frac{1}{2}g{\left(15\right)}^2$ = 1125

$\Rightarrow$ h₃ = 625

$\therefore$ h₂ = 3h₁ and h₃ = 5h₁

or $h_1$ = $\frac{h_2}{3}$ = $\frac{h_3}{5}$

At point B X component of velocity remains unchanged while Y component reverses its direction.
$\therefore$ The velocity of the projectile at point B is $2\hat{i}-3\hat{j}$ m/s.

For upper half of inclined plane

v² = u² + 2a S/2 = 2 (g sin $\theta$) S/2 = gS sin $\theta$

For lower half of inclined plane

0 = u² + 2 g (sin $\theta$ – $\mu$ cos $\theta$) S/2

$\Rightarrow -gS\sin \theta =gS\left(\sin \theta -\mu \cos \theta \right)$

$\Rightarrow 2\sin \theta =\mu \cos \theta$

$\Rightarrow \mu =\frac{2\sin \theta }{\cos \theta }=2\tan \theta$

From figure

F = 6 mg,

As speed is constant, acceleration a = 0

$\therefore$ 6 mg = 6ma = 0, F = 6 mg

$\therefore$ T = 5 mg , T' = 3 mg

T'' = 0 F_net on block of mass 2 m

= T – T' – 2 mg = 0

ALTERNATE :
$\because$ v = constant so, a = 0, Hence, F_net = ma = 0

Here, $\overrightarrow{F}=\left(3\hat{i}+\hat{j}\right)N$

Initial position, $\overrightarrow{r_1}=\left(2\hat{i}+\hat{k}\right)m$

Final position, $\overrightarrow{r_2}=\left(4\hat{i}+3\hat{j}-\hat{k}\right)m$

Displacement, $\overrightarrow{r}=\overrightarrow{r_2}-\overrightarrow{r_1}$

$\overrightarrow{r}=\left(4\hat{i}+3\hat{j}-\hat{k}\right)m-\left(2\hat{i}+\hat{k}\right)m=2\hat{i}+3\hat{j}-2\hat{k}m$

Work done, $W=\overrightarrow{F}.\overrightarrow{r}=\left(3\hat{i}+\hat{j}\right).\left(2\hat{i}+3\hat{j}-2\hat{k}\right)=6+3=9J$

The situation is as shown in the figure.


According to law of conservation of linear momentum

${\overrightarrow{p}}_1+{\overrightarrow{p}}_2+{\overrightarrow{p}}_3=0$
$\therefore$ ${\overrightarrow{p}}_3=-\left({\overrightarrow{p}}_1+{\overrightarrow{p}}_2\right)$

Here, ${\overrightarrow{p}}_1=(1kg)(12m{s}^{-1})\hat{i}=12\hat{i}kgm{s}^{-1}$

${\overrightarrow{p}}_2=(2kg)(8m{s}^{-1})\hat{j}=16\hat{i}kgm{s}^{-1}$

$\therefore {\overrightarrow{p}}_3=-\left(12\hat{i}+16\hat{j}\right)kgm{s}^{-1}$

The magnitude of p₃ is

$p_3=\sqrt{{\left(12\right)}^2+{\left(16\right)}^2}=20kgm{s}^{-1}$

$\therefore m_3=\frac{p_3}{v_3}=\frac{20kgm{s}^{-1}}{4m{s}^{-1}}=5kg$

The kinetic energy of the rolling object is converted into potential energy at height

$h=\left(\frac{3v^2}{4g}\right)$

So by the law of conservation of mechanical energy, we have

$\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2=Mgh$  $\left(\because \omega =\frac{v}{R}\right)$

$\frac{1}{2}M{v}^2+\frac{1}{2}I{\left(\frac{v}{R}\right)}^2=Mg\left(\frac{3v^2}{4g}\right)$

$\frac{1}{2}I\frac{v^2}{R^2}=\frac{3}{4}M{v}^2-\frac{1}{2}M{v}^2$

$\frac{1}{2}I\frac{v^2}{R^2}=\frac{1}{4}M{v}^2$

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

Hence, the object is disc.

When the string is cut, the rod will rotate about P. Let $\alpha$ be initial angular acceleration of the rod.Then

Torque, $\tau =I\alpha =\frac{M{L}^2}{3}\alpha$      ...(i)

(Moment of inertia of the rod about one end = $\frac{M{L}^2}{3}$)

Also, $\tau =Mg\frac{L}{2}$ ...(ii)

Equating (i) and (ii), we get

$Mg\frac{L}{2}=\frac{M{L}^2}{3}\alpha \Rightarrow \alpha =\frac{3g}{2L}$

Gravitational potential V = $\frac{-Gm}{r}$

$V_0=-\frac{G\times 2}{1}-\frac{G\times 2}{2}-\frac{G\times 2}{4}-\frac{G\times 2}{8}$

= $-2G\left[1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...\infty \right]$

$=-2G\times \frac{1}{1-\frac{1}{2}}$

$=-2G\times \frac{1}{\frac{1}{2}}=-4G$

Gravitational potential energy at any point at a distance r from the centre of the earth is

$U=-\frac{GMm}{r}$

where M and m be masses of the earth and the body respectively.

At the surface of the earth, r = R

$U_i=-\frac{GMm}{R}$

At a height h from the surface,

r = R + h = R + 2R = 3R    ( h = 2R (Given))

$\therefore$ $U_f=-\frac{GMm}{3R}$

Change in potential energy,

$\Delta U=U_f-U_i$

$=-\frac{GMm}{3R}-\left(-\frac{GMm}{R}\right)=\frac{GMm}{R}\left(1-\frac{1}{3}\right)$

$=\frac{2}{3}\frac{GMm}{R}=\frac{2}{3}mgR$  $\left(\because g=\frac{GM}{R^2}\right)$

The wettability of a surface by a liquid depends primarily on angle of contact between the surface and the liquid.

In case of ideal gas

C_p – C_v = R

but $\gamma$ = C_p/C_v

So, $C_v=\frac{R}{\gamma -1}$

$F=\frac{YA}{L}\times \ell$

So, extension, $\ell \propto \frac{L}{A}\propto \frac{L}{D^2}$
[$\therefore$ F and Y are constant]

${\ell }_1\propto \frac{100}{1^2}\propto 100$ and ${\ell }_2\propto \frac{200}{2^2}\propto 50$

${\ell }_3\propto \frac{300}{3^2}\propto \frac{100}{3}$ and ${\ell }_4\propto \frac{50}{\frac{1}{4}}\propto 200$

The ratio of $\frac{L}{D^2}$ is maximum for case (length = 50 cm, diameter = 0.5 mm)

According to Wien’s displacement law

${\lambda }_{m}T$ = constant

${\lambda }_m=\frac{constant}{T}$

So when a piece of iron is heated, ${\lambda }_m$ decreases i.e. with rise in temperature the maximum intensity of radiation emitted gets shifted towards the shorter wavelengths. So the colour of the heated object will change that of longer wavelength (red) to that of shorter (reddish yellow) and when the temperature is sufficiently high and all wavelengths are emitted, the colour will become white.

W_net = Area of triangle ABC

$=\frac{1}{2}AC\times BC$

$=\frac{1}{2}\times 5\times 10^{-3}\times 4\times 10^5=1000J$

According to ideal gas equation PV = nRT

$\Rightarrow V=\frac{nRT}{P}$

For an isobaric process, P = constant and V $\propto$ T
Therefore, V – T graph is a straight line passing through origin. Slope of this line is inversely proportional to P.
In the given figure,

(Slope)₂ > (Slope)₁   $\therefore$ P₂ < P₁

According to question $P\propto T^3$

But as we know for an adiabatic process the pressure $P\propto T^{\frac{\gamma }{\gamma -1}}$

So, $\frac{\gamma }{\gamma -1}=3\Rightarrow \gamma =\frac{3}{2}$

$\therefore$ $\frac{C_p}{C_v}=\frac{3}{2}$

From first law of thermodynamics

$\Delta Q=\Delta U+\Delta W=\frac{3}{2}.\frac{1}{4}R\left(T_2-T_1\right)+0$

$=\frac{3}{8}{N}_a{K}_B\left(T_2-T_1\right)$    $\left[\because K=\frac{R}{N}\right]$

Let $\upsilon$ be frequency of the unknown source. As it gives 4 beats per second when sounded with a source of frequency 250 Hz,

$\therefore \upsilon =250\pm 4=246Hzor254Hz$

Second harmonic of this unknown source = 492 Hz or 508 Hz which gives 5 beats per second, when sounded with a source of frequency 513 Hz.

Therefore unknown frequency, $\upsilon$ = 254 Hz.

Pressure change will be minimum at both ends.

The standard equation of a wave travelling along +ve x-direction is given by

$\mathrm{y}=\mathrm{Asin}(\mathrm{kx}-\omega \mathrm{t})$

where A = Amplitude of the wave
k = angular wave number
$\omega$ = angular frequency of the wave

Given: A = 1 m, $\lambda =2\pi m,\upsilon =\frac{1}{\pi }Hz$

As $k=\frac{2\pi }{\lambda }=\frac{2\pi }{2\pi }=1$

$\omega =2\pi \upsilon =2\pi \times \frac{1}{\pi }=2$

$\therefore$ The equation of the given wave is

y = 1 sin (1x – 2t) = sin(x – 2t)

In the direction of electric field, electric potential decreases.

$\therefore$ V_B > V_C > V_A

From figure,
$\tan \theta =\frac{F_e}{mg}\Rightarrow \frac{r/2}{y}=\frac{\frac{k{q}^2}{r^2}}{mg}$
   $\left[\because F=\frac{k{q}^2}{r^2}fromcoulomb’slaw\right]$

$\Rightarrow r^3\propto y\Rightarrow r{}^{\prime 3}\propto \frac{y}{2}\Rightarrow \frac{r^{\prime }}{r}=\frac{1}{2^{1/3}}$

$\Rightarrow r^{\prime }=\frac{r}{\text{root}3\text{of}2}$

From the question, total resistance of Wheatstone bridge is:
(40) × (120)/(40 + 120) = 30$\Omega$

Now the current through the cell is:
= 7V/(5+30)$\Omega$ = (1/5)A = 0.2A

Resistance of a wire,
$R=\rho \frac{l}{A}=4\Omega$   ...(i)

When wire is stretched twice, its new length be $l^{\prime }$. Then

$l^{\prime }=2l$

On stretching volume of the wire remains constant.

$\therefore lA=l^{\prime }{A}^{\prime }$ where A' is the new cross-sectional area

$\Rightarrow A^{\prime }=\frac{l}{l^{\prime }}A=\frac{l}{2l}A=\frac{A}{2}$

$\therefore$ Resistance of the stretched wire is

$R^{\prime }=\rho \frac{l^{\prime }}{A^{\prime }}=\rho \frac{2l}{\left(A/2\right)}=4\rho \frac{l}{A}$

$=4\left(4\Omega \right)=16\Omega$

$I=\frac{\epsilon }{R+r}$

$\Rightarrow IR+Ir=\epsilon$

Here,
$R=10\Omega ,r=?,\epsilon =2.1V,I=0.2A$

$\therefore$ 0.2 × 10 + 0.2 × r = 2.1

2 + 0.2r = 2.1

0.2r = 0.1 $\Rightarrow$ r = $\frac{1}{2}$ = 0.5 $\Omega$

When moves with an acceleration a₀ towards west, electric field

$E=\frac{F}{q}=\frac{m{a}_0}{e}\left(West\right)$

When moves with an acceleration 3a₀ towards east, magnetic field

$B=\frac{2m{a}_0}{e{v}_0}\left(downward\right)$

If the magnetic pole has a strength ‘m’ then,

M = m $\times$ $l$

When the bar magnet is bent in the form of an arc as shown in figure then

l = $\frac{\pi }{3}\times r=\frac{\pi r}{3}$

$\Rightarrow$ r = $\frac{3l}{\pi }$

New magnetic dipole moment

M' = m $\times$ 2rsin 30^o

= m $\times$ 2 $\times$ $\frac{3l}{\pi }\times \frac{1}{2}$ = $\frac{3ml}{\pi }=\frac{3M}{\pi }$

A current loop in a magnetic field is in equilibrium in two orientations one is stable and another unstable.

When a current loop is placed in a magnetic field it experiences a torque. It is given by

$\overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}$ = MB sin$\theta$

where, $\overrightarrow{M}$ is the magnetic moment of the loop and $\overrightarrow{B}$ is the magnetic field.

If $\theta$ = 0^o $\Rightarrow$ $\tau$ = 0 (stable equilibrium)

If $\theta$ = 180^o $\Rightarrow$ $\tau$ = 0 (unstable equilibrium)

This is the case of periodic EMI



From graph, it is clear that direction is changing twice in 1 cycle.

If a bulb B and AC source are connected in series with self-inductance coil, then in such case the brightness of the bulb decreases when an iron rod is inserted in the coil which further increases impedance of the circuit.

The natural frequency of vibration of the water molecules in the food item should be same as that of the frequency of microwaves.

Equivalent focal length is $\frac{1}{f_{eq}}=\frac{1}{f_1}+\frac{1}{f_2}$

$\Rightarrow$ $\frac{1}{f_{eq}}$ = $\left({\mu }_1-1\right)\left(\frac{1}{\infty }-\frac{1}{-R}\right)$ + $\left({\mu }_2-1\right)\left(\frac{1}{-R}-\frac{1}{\infty }\right)$

$\Rightarrow$ f_eq = $\frac{R}{\left({\mu }_1-{\mu }_2\right)}$

P_cornea = + 40 D

P_e = + 20 D

Total power of combination = 40 + 20 = 60 D

Focal length of combination = $\frac{1}{60}\times 100$ cm = $\frac{5}{3}$ cm

For minimum converging state of eye lens,

u = -$\infty$ v = ? f = $\frac{5}{3}$

From lens formula,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\Rightarrow$ v = $\frac{5}{3}$ cm

Distance between retina and cornea-eye lens

= $\frac{5}{3}$ cm = 1.67 cm

From the question, n₁$$\lambda$$₁ = n₂$$\lambda$$₂

So, $\frac{n_1}{n_2}=\frac{{\lambda }_2}{{\lambda }_1}$ = $\frac{10000}{12000}=\frac{5}{6}$

Hence, minima n₁ and n₂ are 5 and 6.

X_min = $\frac{n_1{\lambda }_{1}D}{d}=\frac{5\left(12000\times {10}^{-10}\right)2}{2\times {10}^{-3}}$

= 6 × 10^–3 m = 6 mm

We see that wavelength of Lyman series,

n_f = 3, n_i = 4

$\frac{1}{{\lambda }_L}=R\left[\frac{1}{3^2}-\frac{1}{4^2}\right]$ = $\frac{7R}{144}$

We see that wavelength of Balmer series :

n_f = 2, n_i = 3

$\frac{1}{{\lambda }_B}=R\left[\frac{1}{2^2}-\frac{1}{3^2}\right]$ = $\frac{5R}{36}$

Now ratio of longest wavelengths corresponds to Lyman and Balmer series:

$\frac{{\lambda }_L}{{\lambda }_B}=\frac{5}{36}\times \frac{144}{7}$ = $\frac{5}{27}$

Mass defect $\Delta$m = 0.02866 a.m.u.

Energy = 0.02866 × 931 = 26.7 MeV

As ₁H² + ₁H² $$\to$$ ₂He⁴

Energy liberated per a.m.u = $\frac{13.35}{2}$ MeV = 6.675 MeV

Initial number of atoms of X is N₀

and Initial number of atoms of Y is 0

Number of atoms after time t, for X is N and for Y is N₀ - N

According to the question,

$\frac{N}{N_0-N}$ = $\frac{1}{7}$

$\Rightarrow$ $\frac{N}{N_0}=\frac{1}{8}$

As $\frac{N}{N_0}={\left(\frac{1}{2}\right)}^n$ where n is the no. of half lives

$\therefore$ ${\left(\frac{1}{2}\right)}^n=\frac{1}{8}$

$\Rightarrow$ n = 3

t = nT_1/2 = 3 × 20 = 60 years

Hence, the age of rock is 60 years.

As speed of an electron increases.

Its de-Broglie wavelength decreases

$\lambda =\frac{h}{mv}$

and angular width for central maxima is $\omega$ = $\frac{2\lambda }{d}$

$\omega \propto \lambda \propto \frac{1}{v}$

Wavelength of an electron of energy E is

${\lambda }_e=\frac{h}{\sqrt{2m_{e}E}}$

$\Rightarrow$ ${\lambda }_e^2=\frac{h^2}{2m_{e}E}$ .....(1)

Wavelength of a photon of same energy E is

${\lambda }_p=\frac{hc}{E}$

$\Rightarrow$ $E=\frac{hc}{{\lambda }_p}$ .....(2)

Equating (1) and (2), we get

$\frac{hc}{{\lambda }_p}=\frac{h^2}{2m_e{\lambda }_e^2}$

$\Rightarrow$ ${\lambda }_p=\frac{2m_{e}c}{h}{\lambda }_e^2$

$\Rightarrow$ ${\lambda }_p\propto {{\lambda }_e}^2$

Work function, $\varphi$ = h$\nu$

According to Einstein’s photoelectric equation

$\frac{1}{2}m{v}_{\max }^2$ = h(2$\nu$) - h$\nu$

$\Rightarrow$ $\frac{1}{2}m{v}_{\max }^2=h\nu$

$\Rightarrow$ $v_{\max }^2=\frac{2h\nu }{m}$

$\Rightarrow$ v_max = $\sqrt{\frac{2h\upsilon }{m}}$

The output of the given logic circuit is

X = $\overline{\overline{A.B}}$ = A.B

In n-type semiconductor, electrons are majority charge carriers and holes are minority charge carriers and pentavalent atoms are dopants.

Formula for trans conductance $g$_m is

$g$_m = $\frac{\beta }{r_i}$

Formula for voltage gain A,

A = $\beta$$\frac{R_L}{r_i}$

$\Rightarrow$ A = $g$_mR_L

Given A = G

$\therefore$ G = $g$_mR_L

$\Rightarrow$ G $\propto$ $g$_m

$\Rightarrow$ $\frac{G_2}{G_1}=\frac{g_{m_2}}{g_{m_1}}$

$\Rightarrow$ G₂ = $\frac{0.02}{0.03}$ $\times$ G = $\frac{2}{3}G$

$2=\frac{m}{63\times 0.25}$

$\Rightarrow$ m = 2 × 63 × 0.25 = 31.5 g

Now, if concentrated HNO₃ is 100% then it requires 31.5 g. But the original solution of HNO₃ is 70% concentrated.

Hence, the mass of HNO₃ required to produce 2.0 M solution

= $\frac{100}{75}\times 31.5$

= 45 g of HNO₃

Moles of urea = $\frac{6.02\times {10}^{20}}{6.02\times {10}^{23}}=0.001$

Concentration of solution = $\frac{0.001}{100}\times 1000$ = 0.01 M

The orbitals associated with n = 3, l = 1 is 3p. One orbital (with m = -1) of 3p-subshell can only accomodate maximum 2 electrons.

The electron is more tightly bound in the smallest allowed orbit.

c = $\nu \lambda$

$\lambda$ = $\frac{c}{\nu }$

= $\frac{3\times {10}^{17}}{6\times {10}^{15}}$ = 50 nm

The compressibility factor is the term which is measured for a gas to study its deviation from the ideal behaviour. Compressibility factor,

$Z=\frac{PV}{nRT}$

Greater is the difference in the value of Z from 1, greater is the deviation of the gas from ideal behaviour.

Among the given molecules, NH₃ is easily liquefiable gas which deviates from ideal behaviour to the maximum extent.

KOH is the base thus, it gives OH^– ions thus it cannot remove OH^– ions from reaction mixture but it adds on the concentration of OH^– ions.

So, an acid must be added but if a strong acid is added to the reaction mixture then in acidic condition the MnO₄ ^– formed reduces to give Mn²⁺ thus, HCl which is a strong acid and SO₂ which on treating with water forms a strong H₂SO₄ cannot be used for this purpose.

Thus, CO₂ which forms H₂CO₃ a weak acid reacts to remove OH^– but not that much acidic that MnO₄ ^– undergo reduction. Thus, CO₂ is used for this reaction for completion.

BF₃ acts as Lewis acid. It is electron pair acceptor.

For a general reaction,

ΔH = Activation energy of forward reaction – Activation energy of backward reaction.

As, both the energies of activation have same value thus, ΔH = 0.

$\Delta$G is not equal to zero because if it is so the reaction must be in equilibrium which is not in this case

At anode :Zn²⁺_(aq) + 2e^$$-$$ $\to$ Zn_(s);  E^o = $-$0.76 V

At cathode : Ag₂O_(s) + H₂O_(l) + 2e^$$-$$ $\to$ 2Ag_(s) + 2OH^$$-$$_(aq), E^o = 0.34 V

E°_cell = E°_cathode – E°_anode

= 0.34 – (–0.76) = 1.10 V

Oxidation half reaction is

H₂ $\to$ 2H⁺ + 2e^–

If pH = 10, [H⁺] = 10^–10

We know, Nernst Equation

E_cell = E°_cell - $\frac{0.0591}{2}\log \frac{{\left[H^{+}\right]}^2}{P_{H_2}}$

For hydrogen electrode E°_cell = 0

E_cell = - $\frac{0.0591}{2}\log \frac{{\left({10}^{-10}\right)}^2}{1}$

= 0.0591log10¹⁰

E_cell = 0.591 V

Degree of dissociation($\alpha$) = $\frac{9.54}{238}$ = 0.04008 = 4.008%

$\log \frac{k_2}{k_1}=\frac{E_a}{2.303R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

Initial temperature, T₁ = 20 + 273 = 293 K

Final temperature, T₂ = 35 + 273 = 308 K

R = 8.314 JK^–1 mol^–1

As rate becomes double on raising temperature

$\therefore$ r₂ = 2r₁

As rate constant, k $\infty$ r

k₂ = 2k₁

$\therefore$ $\log 2=\frac{E_a}{2.303\times 8.314}\left(\frac{1}{293}-\frac{1}{308}\right)$

$\Rightarrow$ $0.301=\frac{E_a}{19.147}\times \frac{15}{293\times 308}$

$\Rightarrow$ E_$$a$$ = 34673 J mol^–1 = 34.7 kJ mol^–1

for fcc Z = 4.

d = $\frac{ZM}{N_A{a}^3}$

$\Rightarrow$ M = $\frac{d{N}_A{a}^3}{Z}$

= $\frac{2.72\times 6.023\times {10}^{23}\times {\left(404\times {10}^{-10}\right)}^3}{4}$

= 27 g mol^–1

Total number of carbon atoms per unit cell

= $8\times \frac{1}{8}+6\times \frac{1}{2}+4$ = 8

The given molecules SiF₄, XeF₄, and BF₃ are symmetric molecules that is why due to cancellation of bond dipoles they are non-polar even though they contain polar bonds.

But in SiF₄ there are four Si—F bonds and one lone pair due to which its structure is unsymmetrical. Hence, it is a polar molecule.

HCl is polar ($\mu$ $\ne$ 0) and He is non-polar ($\mu$ = 0) gives dipole-induced dipole interaction.

Water is a polar compound so the salt which is more polar or having more ionic character will be more soluble in it. According to Fajan's rule, ionic character of compound increases with increase in the size of cation. Now, among the given compounds, the size of cations are in the order as follows :

Na⁺ > Zn²⁺ > Cu²⁺

Thus, the order of ionic character and the solubility order in aqueous medium is as follows :

Na₂S > ZnS > CuS

$O_2^{-}$ superoxide has one unpaired electron.

Boron hydrides are electron deficient compound because boron atom has three half-filled orbitals in excited state.

SO₂ gas is liberated when any sulphide ore is roasted.

2M₂S + 3O₂ $\to$ 2M₂O + 2SO₂

SO₂ is a colourless gas with chocking smell of burnt sulphur.

HClO₄ with highest oxidation number and its conjugate base is resonance stabilised, hence it is most acidic.

Boron nitride (BN) is known as inorganic graphite. The most stable form is hexagonal one. It has layered structure similar to graphite.

(NH₄)₂Cr₂O₇ $\frac{\text{buildrel}\Delta }{⟶}$ N₂ + Cr₂O₃ + 4H₂O

Zn(ClO₃)₂ $\frac{\text{buildrel}\Delta }{⟶}$ ZnCl₂ + 3O₂

KClO₃ $\frac{\text{buildrel}\Delta }{⟶}$ KCl + $\frac{3}{2}$O₂

2K₂Cr₂O₇ $\frac{\text{buildrel}\Delta }{⟶}$ 2K₂CrO₄ + Cr₂O₃ + $\frac{3}{2}$ O₂

H₂O does not contain any $\pi$-bond.

SiO${}_4^{4-}$ orthosilicate is basic unit of silicates.

Since Me₃SiCl contains only one Cl, therefore it can’t form high molecular mass silicon polymer. It can form only dimer.

Ion with no unpaired electron is diamagnetic in nature.

Ce²⁺ = [Xe] 4f² → 2 unpaired electrons

Sm²⁺ = [Xe] 4f⁶ → 6 unpaired electrons

Eu²⁺ = [Xe] 4f⁷ → 7 unpaired electrons

Yb²⁺ = [Xe] 4f¹⁴ → No unpaired electron

Interstitial compounds are generally chemically inert.

For $\mu$ = 1.73 B.M. n will be :

1.73 = $\sqrt{n\left(n+2\right)}$

$\Rightarrow$ n = 1

In [Cu(NH₃)₄]²⁺. Cu⁺² = [Ar]3d⁹

NH₃ = Strong field ligand

Here, unpaired electron gets excited to higher energy level but it still remains unpaired.

AgNO₃ + [Cr(H₂O)₄Cl₂]Cl $\to$ AgCl

Excess mole = MV

= 0.01 $\times$ $\frac{100}{1000}$

= 0.001

In benzyl free radical it has 6 p orbitals, each containing one unpaired electron, in a six membered cyclic structure is in accordance with Huckel rule of aromaticity.

IUPAC name of the structure is

3-Ethyl-2-hydroxy-4-methylhex-3-en-5-ynoic acid