NEET-UG 2006

Given, v = at + $\frac{b}{t+c}$

Dimension of at = Dimension of velocity v

$\therefore$ [at] = [LT^-1]

$\Rightarrow$ [a] = $\frac{\left[L{T}^{-1}\right]}{\left[T\right]}$ = [LT^-2]

As c is added to t,

$\therefore$ [c] = [T]

$\frac{\left[b\right]}{\left[T\right]}$ = [LT^-1]

$\therefore$ [b] = [L]

Given x = 40 + 12t $-$ t³

v = $\frac{dx}{dt}$ = 12 - 3t²

When particle comes to rest then v = 0.

$\therefore$ 12 - 3t² = 0

$\Rightarrow$ t = 2 s

Distance travelled before coming to rest,

s = $\underset{0}{\overset{2}{\int }}vdt$ = $\underset{0}{\overset{2}{\int }}\left(12-3t^2\right)dt$

= ${\left[12t-\frac{3t^3}{3}\right]}_0^2$

= 24 - 8 = 16 m

Distance covered in one circular loop = 2$\pi$r = 2$\times$3.14$\times$100 = 628 m

Average speed = $\frac{628}{62.8}=10$ m/s

Net displacement in one loop = 0

$\therefore$ Average velocity = $\frac{0}{time}$ = 0

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

For body A, 16 = $\frac{1}{2}g{t}_1^2$

For body B, 25 = $\frac{1}{2}g{t}_2^2$

$\therefore$ $\frac{16}{25}=\frac{t_1^2}{t_2^2}$

$\Rightarrow$ $\frac{t_1}{t_2}=\frac{4}{5}$

${\left|\overrightarrow{A}+\overrightarrow{B}\right|}^2={\left|\overrightarrow{A}-\overrightarrow{B}\right|}^2$

$={\left|\overrightarrow{A}\right|}^2+{\left|\overrightarrow{B}\right|}^2+2\overrightarrow{A}\overrightarrow{B}=A^2+B^2+2AB\cos \theta$

$={\left|\overrightarrow{A}-\overrightarrow{B}\right|}^2={\left|\overrightarrow{A}\right|}^2+{\left|\overrightarrow{B}\right|}^2-2\overrightarrow{A}.\overrightarrow{B}$

$=A^2+B^2-2AB\cos \theta$

So, A² + B² + 2AB cos $\theta$

= A² + B² – 2AB cos $\theta$

4AB cos$\theta$ = 0 $\Rightarrow$ cos$\theta$ = 0

$\theta$ = 90º

So, angle between A & B is 90º.

(45º – $\theta$) & (45º + $\theta$) are complementary angles as 45º – $\theta$ + 45º + $\theta$ = 90º. We know that if angle of projection of two projectiles make complementary angles, their ranges are equal. In this case also, the range will be same. So the ratio is 1 : 1.

Acceleration = d²s/dt²
= 2/3 m/s²

Now force acting on body = mass × acceleration

F = 3 × (2/3) = 2 Newton

Hence displacement in 2 secs
= (1/3) × 2 × 2 = 4/3 m

Finally, work done = 2 × 4/3 = 8/3 J

Potential energy of a spring

= $\frac{1}{2}$ $\times$ force constant $\times$ (extension)²

$\therefore$ Potential energy $\propto$ (extension)²

or, $\frac{U_1}{U_2}={\left(\frac{x_1}{x_2}\right)}^2$ or, $\frac{U_1}{U_2}={\left(\frac{2}{8}\right)}^2$

$\Rightarrow \frac{U_1}{U_2}=\frac{1}{16}$

$\Rightarrow U_2=16U_1=16U\left(\because U_1=U\right)$

Loss in potential energy = mgh

= 2 × 10 × 10 = 200 J.

Gain in kinetic energy = work done = 300 J

$\therefore$ Work done against friction = 300 – 200 = 100 J

Resolving the velocities in vertical and horizontal directions, resolved parts of first velocity



v cos$\theta$ perpendicular to the wall and v sinq parallel to the wall. In the second case, they are –v sin$\theta$ & v cos$\theta$ respectively. Here, –ve sign is because direction is opposite to the earlier ones. So we see a net change in velocity perpendicular to way
= v sin$\theta$ – (–v sin$\theta$) = 2v sin$\theta$

This change has occured in 0.25 sec, so, rate of

change of velocity = $\frac{2v\sin \theta }{0.25}$

$=\frac{2\times 12\times \sin 30^{\circ }}{0.25}\Rightarrow \frac{24\times 1}{2\times 0.25}=48$

Thus, acceleration a = 48 m/sec²

Force applied = m $\times$ a = 0.5 × 48 = 24 N

Torque about A,

$\tau =mg\times \frac{l}{2}=\frac{mgl}{2}$

$\tau =I\alpha$

$\therefore$ Angular acceleration,
$\alpha =\frac{\tau }{I}=\frac{mgl/2}{m{l}^2/3}=\frac{3g}{2l}$

The centre of the tube will be at length L/2. So radius r = L/2.

The force exerted by the liquid at the other end = centrifugal force

Centrifugal force
= $Mr{\omega }^2=M\left(\frac{L}{2}\right){\omega }^2=\frac{ML{\omega }^2}{2}$

Moment of inertia of a uniform circular disc about an axis through its centre and perpendicular to its plane is $I_C=\frac{1}{2}M{R}^2$ by the theorem of parallel axes.

$\therefore$ Moment of inertia of a uniform circular disc about an axis touching the disc at its diameter and normal to the disc is I.

$I=I_C+M{h}^2=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$

Escape velocity of the body from the surface of earth is v = $\sqrt{2gR}$

For escape velocity of the body from the platform potential energy + kinetic energy = 0

$-\frac{GMm}{2R}+\frac{1}{2}m{v}^2=0$

$\Rightarrow f{v}_{escape}=\sqrt{\frac{GM}{R^2}.R}=\sqrt{gR}=fv$

From the surface of the earth, $v_{escape}=\sqrt{2gR}$

$\therefore f{v}_{escape}=\frac{v_{escape}}{\sqrt{2}}$, $\therefore$ $f=\frac{1}{\sqrt{2}}$

Applying Wein's displacement law,

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

$5000\mathring{\mathbf{A}}\times \left(1227+273\right)=\left(2227+273\right)\times {\lambda }_m$

${\lambda }_m=\frac{5000\times 1500}{1000}=3000\mathring{A}$

We know that efficiency of Carnot Engine

$=\frac{T_1-T_2}{T_1}$

where, T₁ is temp. of source & T₂ is temp. of sink

$\therefore 0.40=\frac{T_1-300}{T_1}\Rightarrow T_1-300=0.40T_1$

$0.6T_1=300\Rightarrow T_1=\frac{300}{0.6}=\frac{3000}{6}=500K$

Now efficiency to be increased by 50%

$\therefore$ $0.60=\frac{T_1-300}{T_1}\Rightarrow T_1-300=0.6T_1$

$0.4T_1=300\Rightarrow T_1=\frac{300}{0.4}=\frac{300\times 10}{4}=750$

Increase in temp = 750 – 500 = 250 K

$C_P=\frac{7}{2}R;C_V=C_P-R$

$=\frac{7}{2}R-R=\frac{5}{2}R$

$\frac{C_P}{C_V}=\frac{7/2R}{5/2R}=\frac{7}{5}$

Let l be the length of block immersed in liquid as shown in the figure. When the block is floating,



$\therefore$ mg = Al$\rho$m

If the block is given vertical displacement y then the effective restoring force is

$F=-\left[A\left(l+y\right)\rho g-mg\right]$

$=-\left[A\left(l+y\right)\rho g-Al\rho g\right]=-Al\rho gy$

Restoring force = $-\left[Al\rho g\right]y$. As this F is directed towards its equilibrium position of block, so if the block is left free, it will execute simple harmonic motion.

Here inertia factor = mass of block = m
Spring factor = $A\rho g$

$\therefore$ Time period = $T=2\pi \sqrt{\frac{m}{A\rho g}}$   i.e. $T\propto \frac{1}{\sqrt{A}}$

Y₁ = 4 sin500$\pi$t, Y₂ = 2 sin506$\pi$t

${\omega }_1=500\pi$, ${\omega }_1=2\pi \upsilon$,
${\upsilon }_1=250,{\upsilon }_2=253$

$\upsilon ={\upsilon }_2-{\upsilon }_1=253-250=3$ beats/s

$\therefore$ Number of beats per minute = 3 × 60 = 180

Reverberation time is defined as the time during which the intensity of sound in the auditorium becomes one millionth of the initial intensity.

Sabine has shown that standard reverberation time for an auditorium is given by the formula

$T_R=K\frac{V}{\alpha S}$

Here, V is volume of the auditorium, S is the surface area.

$\therefore T_R=\frac{K.V}{\alpha S}=1$ (given)

$\Rightarrow T_R=\frac{K.l^3}{6\alpha l^2}=\frac{K}{6\alpha }l$ $T_R=\frac{K.l^3}{6\alpha l^2}=\frac{K}{6\alpha }l$
(Assuming auditorium to be cubic in shape)

$\therefore T_R\propto l$

If dimension is doubled, reverberation time t will be doubled.
So, New T_R = 2 sec.

Frequencies of sound waves are
$\frac{330}{5}$ & $\frac{330}{5.5}$

i.e., 66 Hz and 60 Hz

Frequencies of beat = 66 – 60 = 6 per second

Light waves are electromagnetic waves. Light waves are transverse in nature and do not require a medium to travel, hence they can travel in vacuum. Sound waves are longitudinal waves and require a medium to travel. They do not travel in vacuum.

$y\left(x,t\right)=8.0\sin \left(0.5\pi x-4\pi t-\frac{\pi }{4}\right)$

Compare with a standard wave equation,

$y=a\sin \left(\frac{2\pi x}{\lambda }-\frac{2\pi t}{T}+\varphi \right)$

we get $\frac{2\pi }{\lambda }=0.5\pi \Rightarrow \lambda =\frac{2\pi }{0.5\pi }=4m$

$\frac{2\pi }{T}=4\pi \Rightarrow T=\frac{2\pi }{4\pi }=\frac{1}{2}\sec$

$\upsilon =1/T=2Hz$.

Wave velocity, $v=\lambda \upsilon =4\times 2=8$ m/sec.

Work done in deflecting a dipole through an angle $\theta$ is given by

$W=\underset{0}{\overset{\theta }{\int }}pE\sin \theta d\theta =pE\left(1-\cos \theta \right)$

Since $\theta$ = 90°

$\therefore$ W = pE(1 – cos90°) $\Rightarrow$ W = pE

Electric flux, ${\varphi }_E=\int \overrightarrow{E}.d\overrightarrow{S}$

$=\int EdS\cos \theta =\int EdS\cos {90}^o=0$

The lines are parallel to the surface.

Power = V . I = I²R

$i_2=\sqrt{\frac{Power}{R}}=\sqrt{\frac{2}{8}}$

= $\sqrt{\frac{1}{4}}=\frac{1}{2}A$

Potential over $8\Omega =R{i}_2=8\times \frac{1}{2}=4V$

This is the potential over parallel branch. So,

$i_1=\frac{4}{4}=1A$

Power of $3\Omega ={i_1}^{2}R=1\times 1\times 3=3W$

Kirchhoff’s first law of electrical circuit is based on conservation of charge and Kirchhoff’s second law of electrical circuit is based on conservation of energy.

Kirchhoff’s law has to be applied to the whole loop.


while $i=\frac{2E}{\left(r_1+r_2+R\right)}$

If through one section (here the first battery) has zero potential difference, current cannot flow. The question could have been modified.

The statement that when the circuit is closed, the potential difference across the first cell is zero implies that in a series circuit, one part cannot conduct current which is wrong. Kirchhoff’s law is violated.

Assuming that ir₁ = E as given in the question paper, some students could have found that R = r₁ – r₂. They have to be given marks.

Current will flow from B to A


Potential drop over the resistance CA will be more due to higher value of resistance. So potential at A will be less as compared with at B. Hence, current will flow from B to A.

If we increase the distance between the plates its capacity decreases resulting in higher potential as we know Q = CV. Since Q is constant (battery has been disconnected), on decreasing C, V will increase.

Let r₁ and r₂ are the radius of coil 1 and 2. If B₁ and B₂ are magnetic induction at their centre, then

$B_1=\frac{{\mu }_0{I}_1}{2r_1}$ and $B_2=\frac{{\mu }_0{I}_2}{2r_2}$

Since B₁ = B₂ ; and r₁ = 2r₂ therefore I₁ = 2I₂.

Again if R₁ and R₂ are resistance of the coil 1 and 2 then R₁ = 2R₂ (as R $\propto$ length = 2$\pi$r) and if V₁ and V₂ are the potential difference across them respectively, then

$\frac{V_1}{V_2}=\frac{I_1{R}_1}{I_2{R}_2}=\frac{\left(2I_2\right)\left(2R_2\right)}{I_2{R}_2}=4$

Force on a particle moving with velocity $\overrightarrow{v}$ in a magnetic field $\overrightarrow{B}$ is

$\overrightarrow{F}=q\left(\overrightarrow{v}\times \overrightarrow{B}\right)$

$\Rightarrow$ F = qvBsin$\theta$

(i) When $\theta$ = 0°, F = qvBsin0° = 0

(ii) When $\theta$ = 90°, F = qvBsin90° = qvB

(iii) When $\theta$ = 180°, F = qvBsin180° = 0

If angle between v and B is either zero or 180º, then value of F will be zero.

Curie temperature is the temperature above which ferromagnetic material becomes paramagnetic material.

Mutual inductance is given as,

M = K$\sqrt{L_1\times L_2}$

Now given, K = 1, L₁ = 2 mH, L₂ = 8 mH

$\therefore$ M = $1\times \sqrt{2\times 8}$ = 4 mH

We know that frequency of electrical oscillation in L.C. circuit is

f = $\frac{1}{2\pi }\sqrt{\frac{1}{LC}}$

$\therefore$ $\frac{f_1}{f_2}={\left(\frac{L_2{C}_2}{L_1{C}_1}\right)}^{1/2}$

= ${\left(\frac{2L\times 4C}{L\times C}\right)}^{1/2}={\left(8\right)}^{1/2}$

$\therefore$ $\frac{f_1}{f_2}=2\sqrt{2}$

$\Rightarrow$ $f_2=\frac{f_1}{2\sqrt{2}}$

$\Rightarrow$ $f_2=\frac{f}{2\sqrt{2}}$ [As f₁ = f]

The core of a transformer is laminated to minimise the energy losses due to eddy currents.

Impedance of series LCR is

Z = $\sqrt{R^2+{\left(X_L-X_C\right)}^2}$

= $\sqrt{8^2+{\left(31-25\right)}^2}$

= $\sqrt{64+36}$ = 10 $\Omega$

Power factor, cos$\varphi$ = $\frac{R}{Z}=\frac{8}{10}=0.8$

Power of combination in dioptres

P = P₁ + P₂

= $\frac{100}{f_1}$ + $\frac{100}{f_2}$

= $\frac{100}{25}-\frac{100}{25}$ = 0

Apparent depth = $\frac{realdepth}{\mu }$ = $\frac{3}{1.5}$ = 2 cm

As image appears to be raised by 1 cm, therefore, microscope must be moved upwards by 1 cm.

For berillium, R₁ = R₀ (9)^1/3

For germanium, R₂ = R₀ A^1/3

$\frac{R_1}{R_2}$ = $\frac{{\left(9\right)}^{1/3}}{A^{1/3}}$

$\Rightarrow$ $\frac{1}{2}$ = $\frac{{\left(9\right)}^{1/3}}{A^{1/3}}$

$\Rightarrow$ $\frac{1}{8}=\frac{9}{A}$

$\Rightarrow$ A = 72

Energy required for exciting the hydrogen atom in the ground state to orbit n is given by

E = E_n - E₁

$\Rightarrow$ 12.1 = $-\frac{13.6}{n^2}$ - $\left(-\frac{13.6}{1^2}\right)$ = 3

$\Rightarrow$ n² = $\frac{13.6}{1.5}$ = 9

$\Rightarrow$ n = 3

Number of spectral lines emitted

= $\frac{n\left(n-1\right)}{2}$ = $\frac{3\times 2}{2}$

R₁ = R₀e^{-$\lambda$t₁} and R₂ = R₀e^{-$\lambda$t₂}

$\therefore$ $\frac{R_1}{R_2}$ = $\frac{R_0{e}^{-\lambda t_1}}{R_0{e}^{-\lambda t_2}}$ = $e^{-\lambda \left(t_1-t_2\right)}$

$\Rightarrow$ R₁ = R₂e^{$-$$\lambda$(t₁$-$t₂)}

${}_1^{2}H$ + ${}_1^{2}H$ $\to$ ${}_2^{4}He$ + energy

Energy released = 28 – 2 × 2.2 = 23.6 MeV

The photoelectric current is directly proportional to the intensity of illumination. Therefore a change in the intensity of the incident radiation will change the photocurrent also.

According to Einstein’s photoelectric equation,

K = h$\nu$ – E₀ ... (i)

and K' = h(2$\nu$) – E₀ ... (ii)

= 2h$\nu$ – E₀ = h$\nu$ + h$\nu$ – E₀

K'= h$\nu$ + K [using (i)]

In a discharge tube ionization of enclosed gas is produced due to collisions between negative electrons and neutral atoms/molecules.

Momentum of photon p = E/c

$\therefore$ p = $\frac{1\times {10}^6\times 1.6\times {10}^{19}}{3\times {10}^8}$

= 5 $\times$ 10^$-$22

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

Current gain, $\beta$ = $\frac{\Delta I_C}{\Delta I_B}$

= $\frac{\left(10-5\right)\times {10}^{-3}}{\left(150-100\right)\times {10}^{-6}}$ = 100

The truth table corresponding to waveform is given by

$\begin{array}{ccc}A& B& C\\ 1& 1& 1\\ 0& 1& 0\\ 1& 0& 0\\ 0& 0& 0\end{array}$

$\therefore$ The given logic circuit gate is AND gate.

$\Delta x.m\Delta v=h/4\pi$

0.1 $\times$ 10^-10 $\times$ 9.11 $\times$ 10^-31 $\times$ $\Delta v$ = $\frac{6.626\times {10}^{-34}}{4\times 3.143}$

$\therefore$ $\Delta v$ = $\frac{6.626\times {10}^{-34}}{4\times 3.143\times 0.1\times {10}^{-10}\times 9.11\times {10}^{-31}}$

= 5.79 $\times$ 10⁶ m s^-1

Magnetic quantum number describes the orientation.

Option C is incorrect as the value of K_P given is wrong. It should have been

K_P = $\frac{P_{C{O}_2}}{P_{C{H}_4}\times {\left[P_{O_2}\right]}^2}$

HNO₂ is a weak acid and NaNO₂ is its salt thus these two constitute to form an acidic buffer.

In HCl solution [H⁺] = 10^-8 M

Also from water, [H⁺] = 10^-7 M

$\therefore$ Total [H⁺] = 10^-8 M + 10^-7 M

= 1.1 ×10^–7 M

$\simeq$ 1.0525 $\times$ 10^$-$7 M

A solution of acetone in ethanol shows positive deviation from Raoult's law. It is because ethanol molecules are strongly hydrogen bonded. When acetone is added, these molecules break the hydrogen bonds and ethanol becomes more volatile. Therefore its vapour pressure is increased.

The correct answer is Option D: from both sides of semipermeable membrane with unequal flow rates.

In osmosis, water (or any solvent) moves through a semipermeable membrane from a region of lower solute concentration to a region of higher solute concentration. This semipermeable membrane allows the passage of solvent molecules but not solute molecules. The primary aim of this process is to equalize the solute concentrations on both sides of the membrane. However, it is essential to note that while water molecules predominantly move from the area of lower concentration to the area of higher concentration, there is still some movement of water from the higher concentration to the lower concentration side as well. This bidirectional movement occurs because water molecules are constantly in motion, but the net flow results in a higher volume of water moving to the side with higher solute concentration.

This process does not occur with equal flow rates because the driving force of osmosis is to reduce the concentration gradient between the two sides. Therefore, the flow rate is predominantly higher from the lower solute concentration side to the higher solute concentration side until an equilibrium is reached, at which point the net flow of water equalizes, but does not necessarily stop. This imbalance in flow rates demonstrates why Option D is the most accurate description of osmotic flow through a semipermeable membrane.

Molality of non-electrolyte solute

= $\frac{1}{250\times 0.0512}$ = 0.0781 m

$\Delta$T_f = k_f m

= 5.12 × 0.0781 = 0.4 K

For isotonic solution,

osmotic pressure of urea = osmotic pressure of nonvolatile solute

$\frac{10}{60\times 1000}$ = $\frac{5}{m\times 100}$

$\Rightarrow$ m = 300 g mol^–1

$\Delta$G_system < 0, process is spontaneous.

$\Delta$G_system = 0, process is in equilibrium.

$\Delta$G_system > 0, process is not spontaneous.

We know that

$\Delta$H = $\Delta$E + $\Delta$n_gRT

In the reaction, H₂(g) + Br₂(g) $\to$ 2HBr(g)

$\Delta$n_g = 2 - (1 + 1) = 0

So, $\Delta$H = $\Delta$E for this reaction.

$\Delta$G = $\Delta$H – T$\Delta$S

Now, at equilibrium $\Delta$G = 0

0 = $\Delta$H – T$\Delta$S

$\Rightarrow$ 0 = 30000 –T (105)

$\Rightarrow$ T = $\frac{30000}{105}$ = 285.7 K

The resonance energy provides extra stability to the benzene molecule so it has to over come for hydrogenation to take place.

So $\Delta$H = – 358.5 – (–150.4) = –208.1 kJ

From the given expression:

At anode : A $\to$ A⁺ + e^–

At cathode : B⁺ + e^– $\to$ B

Overall reaction is : A + B⁺ $\to$ A⁺ + B

At anode : Fe $\to$ Fe²⁺ + 2e^-; E^o = -0.441 V

At cathode : Fe³⁺ + e^– $\to$ Fe²⁺; E^o = 0.771 V

Fe + 2Fe³⁺ $\to$ 3Fe²⁺; E^o = ?

To get the above equation, (ii) × 2 – (i)

2Fe³⁺ + 2e^– $\to$ 2Fe²⁺; E^o = 0.771 V

Fe $\to$ Fe²⁺ + 2e^-; E^o = - 0.441 V
------------------------------------------------
Fe + 2Fe³⁺ $\to$ 3Fe²⁺

E^o = 0.771 + 0.441 = 1.212 V

Given,

2A + B $\to$ 3C + D

Rate of reaction =

$-\frac{1}{2}\frac{d\left[A\right]}{dt}=-\frac{d\left[B\right]}{dt}$ = $\frac{1}{3}\frac{d\left[C\right]}{dt}=\frac{d\left[C\right]}{dt}$

N_2(g) + 3H_2(g) $\to$ 2NH_3(g)

Rate = $\frac{-d\left[N_2\right]}{dt}=-\frac{1}{3}\frac{d\left[H_2\right]}{dt}=\frac{1}{2}\frac{d\left[N{H}_3\right]}{dt}$

$\Rightarrow$$\frac{d\left[N{H}_3\right]}{dt}=-\frac{2}{3}\frac{d\left[H_2\right]}{dt}$

F-centres are the sites where anions are missing and instead electrons are present. They are responsible for colours.

Density of CsBr = $\frac{Z\times M}{V\times N_A}$

= $\frac{1\times 213}{{\left(436.6\times {10}^{-10}\right)}^3\times 6.023\times {10}^{23}}$

= 4.25 g/cm³

According to Freundlich adsorption isotherm, in the median range of pressure

       $\frac{x}{m}\propto P^{\frac{1}{n}}$

$\Rightarrow$  $\frac{x}{m}$ = kP${}^{\frac{1}{n}}$

taking log both sides, we get,

log$\frac{x}{m}$ = logk + $\frac{1}{n}$ logP

Here in graph between log$\frac{x}{m}$ and logP, slope is $\frac{1}{n}$ and intercepts = log k.

Bond dissociation energy is kcal/mol of F - F, Cl - Cl, Br - Br and I - I are 38, 57, 45.5 and 35.6 respectively.

Bond dissociation energy of fluorine is less because of its small size and repulsion between electrons of two atoms.

$\underset{38}{F-F}\underset{57}{Cl-Cl}\underset{45.5}{Br-Br}\underset{35.6}{I-I}$
X - X bond dissociation energy (kcal/mol)

The lower value of bond dissociation energy of fluorine is due to the high inter-electronic repulsion between non-bonding electrons in the 2p-orbitals of fluorine. As a result F - F bond is weaker in comparison to Cl - Cl and Br - Br bonds.

SiCl₄, NH₄⁺, SO₄^2- and PO₄^3- ions are the examples of molecules/ions which are of AB₄ type and have tetrahedral structure. SCl₄ is AB₄(lone pair) types species. Although the arrangement of five sp³d hybrid orbitals in space is trigonal bipyramidal, due to the presense of one lone pair of electron in the basal hybrid orbital, the shape of AB₄(lone pair) species gets distorted and becomes distorted tetrahedral or see-saw.

AB₅ molecule has generally trigonal bipyramidal structure. Sometimes due to presence of lone pair of electrons the structure gets distorted.

The dipole moment of NF₃ is 0.24 D and of NH₃ is 1.48 D. In NF₃, the dipole moment vector are aligned in opposite direction to that of the dipole moment vector of lone pair on N–atom which partly cancel out. On the other hand, N–H bonds in NH₃ are in the same direction of dipole moment of the lone pair on N-atom which adds up the new resultant dipole moment.

Due to sp hybridisation of N⁺, NO₂⁺ ion has linear shape.

Ionic radii of alkali metals in water follows the order
Li⁺ > Na⁺ > K⁺ > Rb⁺ > Cs⁺

Thus in aqueous solution due to larger ionic radius Li⁺ has lowest mobility and hence the correct order of ionic mobility is

Rb⁺ > K⁺ > Na⁺ > Li⁺

SeO₂ $\to$ acidic oxide

Al₂O₃ $\to$ amphoteric,

Sb₂O₃ $\to$ amphoteric

Bi₂O₃ $\to$ basic oxide.

The Cl - F (Cl - F_b) bond length is equal to 1.60 $\stackrel{o}{A}$ while each of two axial Cl - F (Cl - F_a) bond length is equal to 1.70 $\stackrel{o}{A}$.

Among the three given hybrid orbitals, sp hybrid orbital is most electronegative. Contribution of s in sp hybrid orbital is maximum (50%) so this orbital is closer to nucleus. Naturally it will have greater tendency to pull electron towards it. Hence it becomes more electronegative and sp³ becomes least electronegative as it has only 25% s-character.

Copper sulphate on treatment with excess of KCN forms complex K₃[Cu(CN)₄] or [Cu(CN)₄]^3–

2CuSO₄ + 10KCN $\to$ 2K₃[Cu(CN)₄] + 2K₂SO₄ + (CN)₂

The 5f-orbitals extend into space beyond the 6s and 6p-orbitals and participate in bonding. This is in direct contrast to the lanthanides where the 4f-orbitals are buried deep inside in the atom, totally shielded by outer orbitals and thus unable to take part in bonding.

For Ni⁺² = [Ar] 3d⁸4s⁰ (2 Unpaired electrons)

Ti⁺³ = [Ar] 3d⁸4s⁰ (1 unpaired electron)

Sc⁺³ = [Ar] (0 unpaired electron)

Cu⁺ = [Ar] 3d¹⁰ (0 unpaired electron)

Hence, Ni⁺² and Ti³⁺ are coloured in aqueous solution due to the presence of unpaired electrons in d-subshell.

Ionization isomerism arises when the coordination compounds give different ions in solution.

Co(NH₃)₄(NO₂)₂]Cl ⇌ Co(NH₃)₄(NO₂)₂]⁺ + Cl^-

Co(NH₃)₄(NO₂)₂]Cl and Co(NH₃)₄(ONO)₂]Cl are linkage isomers as NO₂^– is linked through N or through O.

Octahedral complexes of the type MA₄B₂ exhibit geometrical isomerism.

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

$\Rightarrow$ n = 3

Thus, number of unpaired electrons in d-orbitals subshells of chromium (Cr = 24) = 3

$\therefore$ Configuration of Cr = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³

In 3d³ the distribution of electrons

$3d_{xy}^1,3d_{yz}^1,3d_{xz}^1$, $3d_{x^2-y^2}^0,3d_{z^2}^0$

Due to absence of asymmetric carbon atom.

All alcohols follow the general formula as C_nH_2n+2O.

In case of unsymmetric ethers when treated with HI then I^– attaches to smaller and less complex alkyl group as it follows S_N2 mechanism

CH₃ $-$ O $-$ CH(CH₃)₂ + HI $\to$ CH₃I + (CH₃)₂CHOH