NEET-UG 2005

Plank constant (h) = $\frac{E\lambda }{c}$ = $\frac{\left[M{L}^2{T}^{-2}\right]\left[L\right]}{\left[L{T}^{-1}\right]}$ = [ML²T^-1]

Moment of inertia (I) = mr² = [ML²]

$\therefore$ $\frac{h}{I}=$ $\frac{\left[M{L}^2{T}^{-1}\right]}{\left[M{L}^2\right]}$ = [T^-1] = frequency

x = ae^$-$$\alpha$t + be^$\beta$t

$\frac{dx}{dt}$ = - $\alpha$ae^$-$$\alpha$t + b$\beta$e^$\beta$t

v = - $\alpha$ae^$-$$\alpha$t + b$\beta$e^$\beta$t

Velocity will go on increasing with time.

At height $\frac{h}{2}$, speed is = 10 m/s

and at height h, speed = 0

Using formula,

v² = u² - 2gh

0 = (10)² - 2(10)$\frac{h}{2}$

$\Rightarrow$ h = $\frac{100}{10}$ = 10 m

For two vectors to be perpendicular to each other
$\overrightarrow{A}.\overrightarrow{B}=0$

$\left(2\hat{i}+3\hat{j}+8\hat{k}\right).\left(4\hat{j}-4\hat{i}+\alpha \hat{k}\right)=0$

–8 + 12 + 8$\alpha$ = 0

$\alpha =-\frac{4}{8}=-\frac{1}{2}$

Let $\overrightarrow{A}\times \overrightarrow{B}=\overrightarrow{C}$
The cross product of $\overrightarrow{B}$ and $\overrightarrow{A}$ is perpendicular to the plane containing $\overrightarrow{A}$ and $\overrightarrow{B}$
i.e. perpendicular to $\overrightarrow{B}$

If a dot product of this cross product and $\overrightarrow{A}$ is taken, as
the cross product is perpendicular to $\overrightarrow{A}$, $\overrightarrow{C}$ $\times$ $\overrightarrow{A}$ = 0

Therefore product of $\left(\overrightarrow{A}\times \overrightarrow{B}\right)\times \overrightarrow{A}=0$

Velocity of A relative to B is given by
$\overrightarrow{v_{A/B}}=\overrightarrow{v_A}-\overrightarrow{v_B}=\overrightarrow{v}-\overrightarrow{v_1}$    ...(i)

By taking x-components of equation (i), we get
$0=v\sin \theta -v_1\Rightarrow \sin \theta =\frac{v_1}{v}$    ...(ii)

By taking Y-components of equation (i), we get
$v_y=v\cos \theta$    ...(iii)

Time taken by boy at A to catch the boy at B is given by

t = $\frac{\mathrm{Relative}displacementalongY-axis}{\mathrm{Relative}velocityalongY-axis}$

= $\frac{a}{v\cos \theta }=\frac{a}{v.\sqrt{1-{\sin }^2\theta }}=\frac{a}{v.\sqrt{1-{\left(\frac{v_1}{v}\right)}^2}}$
[From equation (i)]

= $\frac{a}{v.\sqrt{\frac{v^2-v_1^2}{v^2}}}=\frac{a}{\sqrt{v^2-v_1^2}}$

= $\sqrt{\frac{a^2}{v^2-v_1^2}}$

$a_r={\omega }^{2}R$ & $a_t=\frac{dv}{dt}=0$

or, ar= (2$\pi$n)2R = 4${\pi }^2$n²R²
= $4{\pi }^2{\left(\frac{22}{44}\right)}^2{\left(1\right)}^2$

a_net = a_r = ${\pi }^2$ms^–2 and direction along the radius towards the centre.

Work done = area under F-x graph

= area of trapezium OABC

= $\frac{1}{2}\left(3+6\right)\left(3\right)=13.5J$

As per law of conservation of linear momentum

m₁v₁ + m₂v₂ = 0

$\Rightarrow v_2=\left(-\frac{m_1}{m_2}\right)v_1$

$\Rightarrow v_2=\left(-\frac{18}{12}\right)6=-9m/s$

Now, kinetic energy of second piece

$K.E=\frac{1}{2}\left(m{v}_2^2\right)$

$=\frac{1}{2}\times 12\times {\left(-9\right)}^2$

$\therefore$ K.E = 486 J

$K.E.=\frac{1}{2}I{\omega }^2$

$\therefore \frac{1}{2}{I}_1{\omega }_1^2=\frac{1}{2}.2I_1{\omega }_2^2$

$\frac{{\omega }_1^2}{{\omega }_2^2}=\frac{2}{1}\Rightarrow \frac{{\omega }_1}{{\omega }_2}=\frac{\sqrt{2}}{1}$

It is noted that net work done by frictional force when drum rolls down without slipping is zero, so W_net = 0

W_trans + W_rotational = 0

ΔK_trans + ΔK_rotational = 0

ΔK_trans = – ΔK_rotational

Hence, frictional force converts translational energy to rotational energy.

Moment of Inertia of uniform circular disc with radius ‘R’ and mass ‘M’ about axis passing through C.M. and normal to disc will be :

I_CM = (1/2) MR²

Moment of Inertia about axis tangential to disc :

I_T = I_CM + MR²

I_T = (1/2) MR² + MR²

= (3/2) MR²

K.E. of satellite moving in an orbit around the earth is

$K=\frac{1}{2}m{v}^2=\frac{1}{2}m{\left(\sqrt{\frac{GM}{r}}\right)}^2=\frac{GMm}{2r}$

P.E. of satellite and earth system is

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

$\Rightarrow \frac{K}{U}=\frac{\frac{GMm}{2r}}{\frac{GMm}{r}}=\frac{1}{2}$

We know that

$g=\frac{GM}{R^2}=\frac{G\left(\frac{4}{3}\pi R^3\right)}{R^2}=\frac{4}{3}\pi GR\rho$

$\frac{g^{\prime }}{g}=\frac{R^{\prime }}{R}=\frac{3R}{R}=3$

$\therefore$ g' = 3g

Heat conducted

$=\frac{KA\left(T_1-T_2\right)t}{l}=\frac{K\pi r^2\left(T_1-T_2\right)t}{l}$

The rod with the maximum ratio of A/$l$ will conduct most. Here the rod with r = 2r₀ and $l=l_0$ will conduct most.

For a process to be reversible, it must be quasi-static. For quasi static process, all changes take place infinitely slowly. Isothermal process occur very slowly so it is quasi-static and hence it is reversible.

$1-\frac{T_2}{T_1}=1-\frac{Q_2}{Q_1}$

$\Rightarrow 1-\frac{400}{500}=1-\frac{Q_2}{6\times {10}^4}$

$\Rightarrow \frac{4}{5}=\frac{Q_2}{6\times {10}^4}\Rightarrow Q_2=4.8\times 10^{4}cal$

Net heat converted into work

$=6.0\times 10^4-4.8\times 10^4=1.2\times 10^{4}cal$

In circular motion of a particle with constant speed, particle repeats its motion after a regular interval of time but does not oscillate about a fixed point. So, motion of particle is periodic but not simple harmonic.

d₁ = 2 m, d₂ = 3 m

$Intensity\propto \frac{1}{{\left(distance\right)}^2}$

$I_1\propto \frac{1}{2^2}$ and $I_2\propto \frac{1}{3^2}$

$\therefore \frac{I_1}{I_2}=\frac{9}{4}$

We know that potential energy of discrete system of charges is given by

$U=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q_1{q}_2}{r_{12}}+\frac{q_2{q}_3}{r_{23}}+\frac{q_3{q}_1}{r_{31}}\right)$

According to question,

$U_{initial}=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q_1{q}_2}{0.3}+\frac{q_2{q}_3}{0.5}+\frac{q_3{q}_1}{0.4}\right)$

$U_{final}=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q_1{q}_2}{0.3}+\frac{q_2{q}_3}{0.1}+\frac{q_3{q}_1}{0.4}\right)$

$U_{final}-U_{initial}=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q_1{q}_2}{0.1}-\frac{q_2{q}_3}{0.5}\right)$

$=\frac{1}{4\pi {\epsilon }_0}\left[10q_2{q}_3-2q_2{q}_3\right]=\frac{q_3}{4\pi {\epsilon }_0}\left(8q_2\right)$

Work done is equal to zero because the
potential of A and B are the same = $\frac{1}{4\pi {\epsilon }_0}\frac{q}{a}$


No work is done if a particle does not change its potential energy.
i.e. initial potential energy = final potential energy.

It is a balanced Wheatstone bridge. Hence resistance 4$\Omega$ can be eliminated.

$\therefore$ $R_{eq}=\frac{6\times 9}{6+9}=\frac{18}{5}$

$\therefore i=\frac{V}{R_{eq}}=\frac{5V}{18}$

P = i²R or 1 = 25 × R

$R=\frac{1}{25}=0.04\Omega$

$R_{eq}=\frac{R_1{R}_2}{R_1+R_2}=\frac{\left(\frac{R}{2}.\frac{R}{2}\right)}{\frac{R}{2}+\frac{R}{2}}$

$\therefore$ $R_{eq}=\frac{R}{4}$

From Kirchhoff’s law, I × 2 + I × 1 = 18 – 12

Current in the circuit,

$I=\frac{V}{R}=\frac{6}{2}=2A$

Voltage drop across $2\Omega$,

V₁ = 2 × 2 = 4 V

Voltmeter reading = 18 – 4 = 14 V.

Equivalent capacitance for three capacitors (C₁, C₂ & C₃) in series is given by

$\frac{1}{C_{eq.}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$

$=\frac{C_2{C}_3+C_3{C}_1+C_1{C}_2}{C_1{C}_2{C}_3}$

$\Rightarrow C_{eq.}=\frac{C_1{C}_2{C}_3}{C_2{C}_3+C_3{C}_1+C_1{C}_2}$

$\Rightarrow C_{eq.}=\frac{2\left(2C\right)\left(3C\right)}{C\left(2C\right)+\left(2C\right)\left(3C\right)+\left(3C\right)C}=\frac{6}{11}C$

$\Rightarrow$ Charge on capacitors (C₁, C₂ & C₃) in series

$=C_{eq}V=\frac{6C}{11}V$

Charge on capacitor $C_4=C_{4}V=4CV$

$\frac{Chargeon{C}_2}{Chargeon{C}_4}=\frac{\frac{6C}{11}V}{4CV}=\frac{6}{11}\times \frac{1}{4}=\frac{3}{22}$

The direction of $\overrightarrow{B}$ is due to i is acting inwards i.e. into the paper along $\left(-\hat{k}\right)$.

$\therefore$ The magnetic force

$\overrightarrow{F}=Q\left(\overrightarrow{v}\times \overrightarrow{B}\right)=Q\left(\overrightarrow{v}\hat{i}\right)\times B\left(-\hat{k}\right)=QvB\hat{j}$

So, $\overrightarrow{F}$ is along Oy direction.

The magnetic field produce by moving electron in circular path

B = $\frac{{\mu }_{0}i}{2r}$

and i = $\frac{q}{2\pi r}\times v$

$\therefore$ B = $\frac{{\mu }_{0}qv}{4\pi r^2}$

$\Rightarrow$ r $\propto$ $\sqrt{\frac{v}{B}}$

The magnetic dipole moment of diamagnetic material is zero as each of its pair of electrons have opposite spins, i.e., $\mu$_d = 0.

Paramagnetic substances have dipole moment $>$ 0, i.e. $\mu$_p $\ne$ 0, because of excess of electrons in its molecules spinning in the same direction.

Ferro-magnetic substances are very strong magnets and they also have permanent magnetic moment, i.e. $\mu$_f $\ne$ 0.

The current flowing clockwise in the equilateral triangle has a magnetic field in the direction $\hat{k}$.

$\tau$ = BiNAsin$\theta$ = BiAsin90°

$\therefore$ A = $\frac{\tau }{iB}$

Area of equilateral triangle

A = $\frac{1}{2}\left(l\right)\sqrt{l^2-{\left(\frac{l}{2}\right)}^2}$ = $\frac{\sqrt{3}}{4}{l}^2$

$\Rightarrow$ $\frac{\sqrt{3}}{4}{l}^2=\frac{\tau }{iB}$

$\Rightarrow$ $l=2{\left(\frac{\tau }{\sqrt{3}Bi}\right)}^{\frac{1}{2}}$

Work done due to a charge W = QV

tan$\varphi$ = $\frac{X_C-X_L}{R}$

$\Rightarrow$ $\tan \left(\frac{\pi }{4}\right)=\frac{\frac{1}{\omega C}-\omega L}{R}$

$\Rightarrow$ R = $\frac{1}{\omega C}-\omega L$

$\Rightarrow$ R + 2$\pi$fL = $\frac{1}{2\pi fC}$

$\Rightarrow$ C = $\frac{1}{2\pi f\left(R+2\pi fL\right)}$

${\lambda }_m>{\lambda }_v>{\lambda }_x$

In spectrum X-rays has minimum wavelength and microwave has maximum wavelength.

The angular resolution,

$\Delta$$\theta$ = $\frac{1.22\lambda }{D}$

= $\frac{1.22\times 5000\times {10}^{-8}}{0.1}$

= 6.1 $\times$ 10^-4

$\therefore$ Order = 10^-4

$\therefore$ (E_C – E_B) + (E_B – E_A) = (E_C – E_A)

$\Rightarrow$ $\frac{hc}{{\lambda }_3}=\frac{hc}{{\lambda }_1}+\frac{hc}{{\lambda }_2}$

$\Rightarrow$ ${\lambda }_3=\frac{{\lambda }_1{\lambda }_2}{{\lambda }_1+{\lambda }_2}$

For nuclei having A $>$ 56 binding energy per nucleon gradually decreases.

Binding energy per nucleon for fission products is higher relative to Binding energy per nucleon for parent nucleus, i.e., more masses are lost and are obtained as kinetic energy of fission products. So, the given ratio $<$ 1.

Isotones means number of neutron remains same.

Apply Bohr's Atomic model for H-atom

K.E = -T.E = +3.4 eV

Energy released in given reaction

= BE of products – BE of reactants = c – (a+ b) = c – a – b

2h${\nu }_0$ = h${\nu }_0$ + $\frac{1}{2}m{\left(4\times {10}^6\right)}^2$

5h${\nu }_0$ = h${\nu }_0$ + $\frac{1}{2}m{v}^2$

$\Rightarrow$ $4\times \frac{1}{2}m{\left(4\times {10}^6\right)}^2$ = $\frac{1}{2}m{v}^2$

$\Rightarrow$ v = 8 $\times$ 10⁶ m/s

Energy of photon = $\frac{12400}{4100}$ = 3 eV

$\therefore$ A and B will emit photoelectrons.

Electromotive force depends on intensity of light that falls on it and does not depend on the frequency.

Lattice constant for (f.c.c.)

= a = interatomic spacing $\times$ $\sqrt{2}$ = 3.59 $\stackrel{\circ }{A}$

Band gap of carbon is 5.5 eV while that of silicon is 1.1 eV

$\therefore$ (E_g)_C $>$ (E_g)_Si

Zener diode is used for stabilisation while p-n junction diode is used for rectification.

Number of donors is more because electrons from –ve terminal of the cell pushes (enters) the n side and decreases the number of uncompensated pentavalent ion due to which potential barrier is reduced. The neutralised pentavalent atom are again in position to donate electrons.

Option (a) is correct as in conductor conduction and valance band overlap and the conduction band is partially filled

Option (b) is correct as insulators have energy gap of 5 - 10 eV.

Option (c) is incorrect as resistivity decreases with increase in temperature.

Option (d) is correct as with increase in temperature, more and more electrons jump to conduction band, hence conductivity increases.

One molal solution means one mole solute present in 1 kg (1000 g) solvent.

$\therefore$ mole of solute = 1

Mole of solvent (H₂O) = $\frac{1000}{18}$

X_solute = $\frac{1}{1+\frac{1000}{18}}$ = 0.018

E_n = -K${\left(\frac{Z}{n}\right)}^2$

Z = 1; n = 2

E₂ = $\frac{-K\times 1}{4}$$\Rightarrow$ E₂ = -328 kJ mol^-1;

K = 4 $\times$ 328

E₄ = $\frac{-K\times 1}{16}$ $\Rightarrow$ E₄ = -4 $\times$ 328 $\frac{1}{16}$

= -82 kJ mol^-1

NO_(g) + O_2(g) ⇌ NO_2(g) ,    K₁

Reverse the above equation

NO_2(g) ⇌ NO_(g) + O_2(g),     $\frac{1}{K_1}$

Multiply the above equation by 2, we get

NO_2(g) ⇌ NO_(g) + O_{2(g),    ${\left(\frac{1}{K_1}\right)}^2=K_2$}

	BOH	⇌	B+	+	OH-
Initially	C		0		0
At equilibbrium	C - C$\alpha$		C$\alpha$		C$\alpha$

[OH^-] = C$\alpha$

$\Rightarrow$ [OH^-] = $\sqrt{K_{b}C}$

$\Rightarrow$ [OH^-] = $\sqrt{1\times {10}^{-12}\times {10}^{-2}}$

= 1.0 $\times$ 10^$-$7 mol L^$-$1

H₂S ⇌ H⁺ + HS^–

HCl ⇌ H⁺ + Cl^–

In presence of HCl this ionization of H₂S is suppressed due to the presence of extra H⁺ ions from HCl and produces less amount of sulphide ions due to common ion effect, thus HCl decreases the solubility of H₂S which is sufficient to precipitate II^nd group radicals.

$\Delta$T_f = K_fm ......(1)

$\Delta$T_b = K_bm ......(2)

$\Rightarrow$ $\frac{\Delta T_f}{\Delta T_b}=\frac{K_f}{K_b}$ .....(3)

$\Delta$T_b = T₂ – T₁ = 100.18 – 100 = 0.18

k_f for water = 1.86 K kg mol^–1

k_b for water = 0.512 K kg mol^–1

$\frac{\Delta T_f}{0.18}=\frac{1.86}{0.512}$

$\Rightarrow$ $\Delta T_f$ = 0.654

$\Delta$T_b = T₂ – T₁

$\Rightarrow$ 0.654 = 0 – T₂

$\Rightarrow$ T₂ = $-$ 0.654^oC

As the ratio of pentane to hexane = 1 : 4

$\therefore$ Mole fraction of pentane = 1/5

Mole fraction of hexane = 4/5

Total vapour pressure

= $\left(\frac{1}{5}\times 440+\frac{4}{5}\times 120\right)$

= 184 mm of Hg

$\therefore$ Vapour pressure of pentane in mixture

= (Vapour pressure of mixture pentane $\times$ Mole fraction of in vapour phase)

$\Rightarrow$ 88 = 184 × mole fraction of pentane in vapour phase

$\Rightarrow$ Mole fraction of pentane in vapour phase

= $\frac{88}{184}$ = 0.478

Hence total vapour pressure

= [(Mole fraction of P) × (Vapour pressure of P)] + [(Mole fraction of Q) × Vapour pressure of Q)]

= $\left(\frac{3}{5}\times 80+\frac{2}{5}\times 60\right)$

= 48 + 24 = 72 torr

$\Delta$G = $\Delta$H – T$\Delta$S

For spontaneous reaction, $\Delta$G has to be negative.

Among the given options, it is positive only when T$\Delta$S > $\Delta$H and both $\Delta$H and $\Delta$S are +ve .

$\Delta$G = $\Delta$H – T$\Delta$S

For a reaction to become spontaneous, $\Delta$G must be negative.

For case (a) exothermic and increasing disorder

For exothermic reaction,

$\Delta$H = –ve and increasing disorder,

$\Delta$S = +ve

$\Delta$G = –ve –T(+ve)

Thus, $\Delta$G is negative for all temperature range

For case (b) exothermic and decreasing disorder For exothermic reaction, $\Delta$H = –ve and for decreasing disorder,

$\Delta$S = –ve

$\Rightarrow$ $\Delta$G = –ve – T(–ve)

Thus, $\Delta$G is not negative for all temperature range.

For case (c) endothermic and increasing disorder. For endothermic reaction, $\Delta$H = +ve and increasing disorder,

$\Delta$S = +ve

Thus, $\Delta$G is not negative for all temperature range.

For case (d) endothermic and decreasing disorder

For endothermic reaction,

$\Delta$H = +ve and decreasing disorder,

$\Delta$S = –ve

$\Delta$G = +ve – T (–ve)

Thus, $\Delta$G is positive for all temperature range.

We know that enthalpy of neutralization of a strong acid and a strong base is –57.33 kJ mol^–1.

Here, MgO is the weak base and HCl is a strong acid thus, a small amount of energy is used in the ionization of MgO thus, the heat of neutralization decreases. Therefore, enthalpy of neutralization is less than – 57.33 kJ mol^–1

2Al₂O₃ + 3C $\to$ 4Al + 3CO₂

From the above equation,

3 mol × 12 g mol^–1 = 36 g of carbon is consumed

to give 4 mol × 27 g mol^–1 = 108 g of Al

So, 108 g of Al is produced by 36 g of carbon

$\therefore$ 270000 g of Al is produced by

= $\frac{36}{108}\times 270000$ of C

= 90000 g of C

= 90 kg of C

Faraday second law of electrolysis

$\frac{M_{Al}}{M_H}=\frac{E_{Al}}{E_H}$

$\Rightarrow$ $\frac{4.5}{M_H}=\frac{\frac{27}{3}}{1}$

$\Rightarrow$M_H = 0.5 g

We know, Volume of 2 g H₂ at STP = 22.4 L

$\therefore$ Volume of 0.5 g H₂ at STP

= $\frac{22.4\times 0.5}{2}$ = 5.6 L

A + B $\to$ Product

Rate $\propto$ [A]^x [B]^y .......(1)

The rate of the reaction decreases by a factor of 4 if the concentration of reactant B is doubled.

$\frac{r}{4}$ $\propto$ [A]^x [2B]^y ......(2)

From equation (1) and (2), we get

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

$\Rightarrow$ y = -2

$\therefore$ Order of this reaction with respect to reactant B is -2.

Given [A] = 0.01 M

Rate = 2.0 × 10^–5 mol L^–1 S^–1

For a first order reaction

Rate = k[A]

k = $\frac{2\times {10}^{-5}}{0.01}=2\times 10^{-3}$

$\Rightarrow$ $t_{1/2}=\frac{0.693}{2\times {10}^{-3}}=347\sec$

When IIA group element (Ra) emits one $\alpha$-particle its group no. decreases by two unit. i.e., go into zero group (Gr. 16) But as it is radioactive thus due to successive emission last product is Pb i.e., (Gr.14).

In given unit cell it is shared equally by six faces of different unit cells.

Dodecyl trimethyl ammonium chloride is a cationic surfactant and forms micelles in aqueous solution above critical micelle concentration (CMC).

The molar enthalpy change accompanying the addition of an electron to an atom (or ion) is known as electron gain enthalpy.
Generally in increases on moving from left to right in a period and in a group it decreases as the size increases.
Exception: Because of the small size of F, electron-electron repulsion present in its relatively compact 2p-subshell, do not easily allow the addition of an extra electron. On the other hand, Cl because of its comparatively bigger size than F, allows the addition of an extra electron more easily.
$\underset{-1.48}{O}<\underset{-2.0}{S}<\underset{-3.6}{F}<\underset{-3.8}{\mathrm{Cl}}$

As difference of electronegativity increases % iconic charecter increases and covalent charecter decreases i.e, electronegativity difference decreases covalent charecter increases. Further greater the charge on the cation more will be its covalent character. Be has maximum (+2) charge.

Hydrogen bonding in H₂O > C₂H₅OH > CH₃OH

Hence, H₂O has maximum surface tension.

Bond lengths of O $-$ O in O₂ is 1.21 Å, in H₂O₂ is 1.48 Å and in O₃ is 1.28Å. Therefore, correct order of the O $-$ O bond length is H₂O₂ > O₃ > O₂ or O₂ < O₃ < H₂O₂

IF₃ has three bond pairs and two lone pairs, thus, it has sp³d hybridization.



PCl₃ has three bond pairs and one lone pair thus, it has sp³ hybridization.



NH₃ has three bond pairs and one lone pair thus, it has sp³ hybridisation.


BF₃ has three bond pairs and no lone pair thus, it has sp² hybridisation.

As the oxidation number of the central atom increases, strength of acid also increases.

Basic character decreases down the group Thus Na₂O is most basic.



We know that more the basic compound more is the pH.

Hence, pH₁ > pH₂ > pH₃ > pH₄

For third ionization enthalpy electronic configuration of

V²⁺ (21) : [Ar]₁₈ 3d³4s⁰

Cr²⁺ (22) : [Ar]₁₈ 3d⁴4s⁰

Mn²⁺ (23) : [Ar]₁₈ 3d⁵4s⁰

Fe²⁺ (24) : [Ar]₁₈ 3d⁵4s¹

Mn has most stable configuration due to half filled d-orbital. Hence 3rd ionization energy will be highest for Mn.

If the transition metal ion has unpaired electron then it shows colour.

Sc³⁺ : [Ar]₁₈ 3d⁰4s⁰

Fe²⁺ : [Ar]₁₈ 3d⁵4s¹

Ti³⁺ : [Ar]₁₈ 3d¹4s⁰

Mn²⁺ : [Ar]₁₈ 3d⁵4s⁰

Hence, Sc³⁺ do not contain unpaired electron and hence it will not undergo d – d transition and do not show colour.

The chemical reactions involved in this case are as follows :

KI + H₂O $\to$ KOH + HI

2KMnO₄ + 2KOH → 2K₂MnO₄ + H₂O + [O]

Thus, for overall reaction, two moles of KMnO₄ are reduced by two moles of KOH or KI. Hence, one mole of KI is required to reduce one mole of KMnO₄.

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.

cis-[Co(en)₂Cl₂]⁺ show optical isomerism because it forms a non-superimposable mirror image.
While trans-form contains plain of symmetry thus optically inactive.

In [Co(NH₃)₆]³⁺

Co⁺³ = [₁₈Ar] 3d⁶4s⁰

[Co(NH₃)₆]³⁺ is an inner orbital complex as well as diamagnetic in behaviour (due to absence of unpaired electrons).

Sublimation would be the best method because it is applicable for those organic compounds which pass directly from solid to vapour state on heating and vice versa on cooling. In these compounds naphthalene is volatile and benzoic acid is nonvolatile due to the formation of dimer via intermolecular hydrogen bonding.

3^o C is more stable due to the stabilization of the charge by three methyl groups (or inductive effect). It can also be explained on the basis of hyperconjugation.

Pair of optical isomerism and geometrical isomerism are able to exhibit the phenomenon of stereoisomerism because both type of isomers differ only in their orientation in space.

Greater the number of alkyl groups(R) attached to the double bonded carbon atoms more stable is the alkene. Further the relative rates of hydrogenation decrease with increase of steric hindrance.
Thus, the stability of alkene will be


Hence alkene which will react faster with H₂ is that which is most unstable.