NEET-UG 2003

Let the time taken by the ball to reach the top most point T.

$\therefore$ v = u - gT

v = 0 at top most point.

T = $\frac{u}{g}$

Velocity of the ball after (T - t) second

v = u - g(T - t)

= u - gT + gt = u - g$\frac{u}{g}$ + gt

$\therefore$ v = gt

Distance covered during the last t seconds of its ascent is

s = (gt)t - $\frac{1}{2}g{t}^2$ = $\frac{1}{2}g{t}^2$

When more than two balls means minimum three balls are in the sky then time of flight for the first ball should be more than 4 s.

T > 4 s

$\frac{2u}{g}$ > 4

$\Rightarrow$ u > 19.6 m/s

Given : $\left(\overrightarrow{F_1}+\overrightarrow{F_2}\right)\perp \left(\overrightarrow{F_1}-\overrightarrow{F_2}\right)$

$\left(\overrightarrow{F_1}+\overrightarrow{F_2}\right).\left(\overrightarrow{F_1}-\overrightarrow{F_2}\right)=0$

$F_1^2-F_2^2-\overrightarrow{F_1}.\overrightarrow{F_2}+\overrightarrow{F_2}.\overrightarrow{F_1}=0$

$\Rightarrow F_1^2=F_2^2$   (i.e. F₁, F₂ are equal to each other in magnitude.)

T = Tension caused in string by monkey

= m (g + a)

$T\le 25\times 10\Rightarrow 20\left(10+a\right)\le 250$

$\Rightarrow 10+a\le 12.5$

$\Rightarrow a\le 2.5$

Reading of the scale = Apparent wt. of the man = m(g + a)

= 80 (10 + 5) = 1200 N

W_mg = $\Delta$K

$\Rightarrow$ -mg$\ell$ = ½ mv² – ½ mu²
or, mv² = m(u² – 2g$\ell$]
or, $v=\sqrt{u^2-2g\ell }\hat{j}$

$\overrightarrow{u}=u\hat{i}$

$\therefore \overrightarrow{v}-\overrightarrow{u}=\sqrt{u^2-2g\ell }\hat{j}-u\hat{i}$

$\therefore \left|\overrightarrow{v}-\overrightarrow{u}\right|={\left[\left(u^2-2g\ell \right)+u^2\right]}^{\frac{1}{2}}$

$=\sqrt{2\left(u^2-g\ell \right)}$

If k be the spring constant, then

$U=\frac{1}{2}\times k\times {\left(2\right)}^2=2k$

$U_{final}=\frac{1}{2}\times k\times {\left(10\right)}^2=50k$

$\Rightarrow \frac{U}{U_{final}}=\frac{2k}{50k}=\frac{1}{25}$

$\Rightarrow U_{final}=25U$

From conservation law of momentum, before collision and after collision linear momentum (p) will be same. That is,

initial momentum = final momentum.

$\Rightarrow$ 0 = m₁v₁ – m₂v₂ $\Rightarrow$ m₁v₁ = m₂v₂

p₁ = p₂

Now, $E=\frac{p^2}{2m}$

$\therefore$ $\frac{E_1}{E_2}=\frac{p_1^2}{2m_1}\times \frac{2m_2}{p_2^2}$

$\Rightarrow \frac{E_1}{E_2}=\frac{m_2}{m_1}\left[p_1=p_2\right]$

The total energy at A = the total energy at B

$\Rightarrow \frac{1}{2}m{u}^2=\frac{1}{2}m{v}^2+mgl$

$\Rightarrow v=\sqrt{u^2-2gl}$

The change in magnitude of velocity = $\sqrt{u^2+v^2}$

$=\sqrt{2\left(u^2-gl\right)}$

Rotational energy =$\frac{1}{2}I{\left(\omega \right)}^2=\frac{1}{2}\left(m{K}^2\right){\omega }^2$

Linear kinetic energy = $\frac{1}{2}m{\omega }^2{R}^2$

$\therefore$ Required fraction

$=\frac{\frac{1}{2}\left(m{K}^2\right){\omega }^2}{\frac{1}{2}\left(m{K}^2\right){\omega }^2+\frac{1}{2}m{\omega }^2{R}^2}=\frac{K^2}{K^2+R^2}$

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

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

$=\frac{1}{4}m{v}^2+\frac{1}{2}m{v}^2=\frac{3}{4}m{v}^2$

Now, gain in K.E. = Loss in P.E.

$\frac{3}{4}m{v}^2=mgh\Rightarrow v=\sqrt{\left(\frac{4}{3}\right)gh}$

Applying conservation law of angular momentum, $I_1{\omega }_1=I_2{\omega }_2$

$I_2=\left(M{r}^2\right)+4\left(m\right)\left(r^2\right)=\left(M+4m\right)r^2$

(Taking ${\omega }_1=\omega$ and ${\omega }_1={\omega }_1$)

$\Rightarrow M{r}^2\omega =\left(M+4m\right)r^2{\omega }_1$

$\Rightarrow {\omega }_1=\frac{M\omega }{M+4m}$

Gravitational force is independent of medium, Hence, this will remain same.

The velocity of the mass while reaching the surface of both the planets will be same.

i.e., $\sqrt{2g^{\prime }{h}^{\prime }}=\sqrt{2gh}$

$\sqrt{2\times g\times h^{\prime }}=\sqrt{2\times 9g\times 2}$

$\Rightarrow 2h^{\prime }=36$

$\Rightarrow h^{\prime }=18m$

Gravitational potential energy on earth’s surface = $-\frac{GMm}{R}$, where M and R are the mass and radius of the earth respectively, m is the mass of the body and G is the universal gravitational constant.

Gravitational potential energy at a height h = 3R

$=-\frac{GMm}{R+h}=-\frac{GMm}{R+3R}=-\frac{GMm}{4R}$

$\therefore$ Change in potential energy

$=-\frac{GMm}{4R}-\left(-\frac{GMm}{R}\right)$

$=-\frac{GMm}{4R}+\frac{GMm}{R}$

$=\frac{3}{4}\frac{GMm}{R}$ $=\frac{3}{4}mgR$

In series, equivalent thermal conductivity

$K_{eq}=\frac{2K_1{K}_2}{K_1+K_2}$

$\Rightarrow K_{eq}=\frac{2\times K\times 2K}{K+2K}=\frac{4}{3}K$

Efficiency of Carnot engine

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

$\frac{W}{6}=1-\frac{400}{500}=\frac{1}{5}\Rightarrow W=\frac{6}{5}=1.2kcal$

Let k be the force constant of spring. If k' is the force constant of each part, then

$\frac{1}{l}=\frac{4}{k^{\prime }}\Rightarrow k^{\prime }=4k$

$\therefore$ Time period
= $2\pi \sqrt{\frac{m}{4k}}=\frac{1}{2}\times 2\pi \sqrt{\frac{m}{k}}=\frac{T}{2}$

The resonance wave becomes very sharp when damping force is small.

Potential energy of particle performing SHM varies parabolically in such a way that at mean position it becomes zero and maximum at extreme position.

Potential energy of simple harmonic oscillator = $\frac{1}{2}m{\omega }^2{y}^2$

for $y=\frac{a}{2},P.E=\frac{1}{2}m{\omega }^2\frac{a^2}{4}$

$\Rightarrow P.E=\frac{1}{4}\left(\frac{1}{2}m{\omega }^2{a}^2\right)=\frac{E}{4}$

In simple harmonic motion velocity
$=A\omega \sin \left(\omega t+\pi /2\right)$



acceleration $=A{\omega }^2\sin \left(\omega t+\pi \right)$ from this we can easily find out that when v is maximum, then $a$ is zero.

Apparent frequency, $f^{\prime }=\frac{v+v_0}{v}f$

$=\frac{v+(1/5)v}{v}f=1.2f$

Wavelength does not change by motion of observer.

The total flux through the cube ${\varphi }_{total}=\frac{q}{{\epsilon }_0}$

$\therefore$ The electric flux through any face
${\varphi }_{face}=\frac{q}{6{\epsilon }_0}=\frac{4\pi q}{6\left(4\pi {\epsilon }_0\right)}$

Fuse wire should have high resistance and low melting point.

Let R₁ and R₂ be the resistance of the two coils and V be the voltage supplied.

Effective resistance of two coils in parallel = $\frac{R_1{R}_2}{R_1+R_2}$

Let H be the heat required to begin boiling in kettle.

Then H = Power $\times$ time = $\frac{V^2{t}_1}{R_1}=\frac{V^2{t}_2}{R_2}$

For parallel combination, H = $\frac{V^2\left(R_1+R_2\right)t_p}{R_1{R}_2}$

$\Rightarrow \frac{1}{t_p}=\left(\frac{t_2+t_1}{t_2{t}_1}\right)$

$\therefore t_p=\frac{t_2{t}_1}{t_2+t_1}=\frac{10\times 40}{10+40}=8$ minute

$R=\frac{V^2}{P}=\frac{220\times 220}{100}=484\Omega$

In series, R_eq = 484 + 484 = 968 $\Omega$

$\therefore P_{eq}=\frac{V^2}{968}=\frac{220\times 220}{968}=50watt$

In parallel, $R_{eq}=242\Omega$

$\therefore P_{eq}=\frac{V^2}{242}=\frac{220\times 220}{242}=200watt$

In balance Wheatstone bridge, the galvanometer arm can be neglected so equivalent resistance = R.

For series,
$C^{\prime }=\frac{C_1\times C_2}{C_1+C_2}=\frac{4\times 4}{4+4}=2\mu F$

For parallel
$C_{eq}=C^{\prime }+C_3=2+4=6\mu F$

If a moving charged particle is subjected to a perpendicular uniform magnetic field, then according to F = qvB sin $\theta$, it will experience a maximum force which will provide the centripetal force to particle and it will describe a circular path with uniform speed.

Magnetic field induction at point inside the solenoid of length l, having n turns per unit length carrying current i is given by

B = $\mu$₀ni

If i $\to$ doubled, n $\to$ halved then B $\to$ remains same.

The time period of a bar magnet in a magnetic field is given by

T = $2\pi \sqrt{\frac{I}{MB}}=2\pi \sqrt{\frac{m{k}^2}{MB}}\propto \sqrt{m}$

$\therefore$ $\frac{T_1}{T_2}=\frac{\sqrt{m_1}}{\sqrt{m_2}}$ = $\frac{T_1}{T_2}=\frac{\sqrt{m_1}}{\sqrt{m_2}}=\frac{\sqrt{m}}{\sqrt{4m}}=\frac{1}{2}$

$\Rightarrow$ T₂ = 2T₁ = 2T.

According to Curie’s law, $\chi$ $\propto$ $\frac{1}{T}$

A diamagnetic material in a magnetic field moves from stronger to the weaker parts of the field.

Every body at all time, at all temperatures emit radiation (except at T = 0), which fall in the infrared region.

Electromagnetic waves are a form of energy waves that have both an electric and magnetic field such as Radio Waves, Microwaves, Infrared, Visible light, Ultraviolet, X-rays, Gamma rays. So $\beta$-rays are not electromagnetic waves, but it is a beam of fast electrons.

We know from Lens maker's formula

$\frac{1}{f}=\left(\mu -1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Initially, Here R₁ = R and R₂ = –R by convection.

$\therefore$ $\frac{1}{f}=\left(\mu -1\right)\frac{2}{R}$

$\Rightarrow$ $\frac{1}{2f}=\left(\mu -1\right)\frac{1}{R}$

If we cut the lens along XOX' then the two halves of the lens will be having the same radii of curvature and so, focal length f' = f .

But when we cut it along YOY' then, we will have

R₁ = R but R₂ = $\infty$

$\therefore$ $\frac{1}{f^{\prime \prime }}=\left(\mu -1\right)\left(\frac{1}{R}-\frac{1}{\infty }\right)$ = $\left(\mu -1\right)\frac{1}{R}$ = $\frac{1}{2f}$

$\Rightarrow$ f'' = 2f

$\frac{1}{f}=\left(\mu -1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Where $\mu =\frac{{\mu }_{convexlens}}{{\mu }_{liquid}}$ = 1

$\therefore$ $\frac{1}{f}=\left(1-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=0$

$\Rightarrow$ f = $\infty$

Mass number = atomic number + no. of neutrons

For hydrogen, number of neutrons = 0

So, mass number = Atomic number.

Hence mass number is sometimes equal to atomic number.

Mass defect($\Delta$m) = 2M_P + 2M_N – M_He

$\Rightarrow$ $\Delta$m = 2 × 1.0074 + 2 × 1.0087 – 4.0015

$\Rightarrow$ $\Delta$m = 0.0307

Now as per relation, E = $\Delta$m × 931 MeV

= 0.0307 × 931 = 28.4 MeV

Using the relation for mean life.

Given : t = 2t = 2$\left(\frac{1}{\lambda }\right)$

Then from M = M₀e^-$\lambda$t = 10e^{-$\lambda$$\left(\frac{1}{\lambda }\right)$} = $10{\left(\frac{1}{e}\right)}^2$ = 1.35 g

As V $\propto$ r³

$\Rightarrow$ V_atom = $\frac{r_{atom}^3}{r_{nucleus}^3}$ $\times$ V_nucleus

= $\frac{{\left({10}^{-10}\right)}^3}{{\left({10}^{-15}\right)}^3}$ $\times$ V_nucleus

= 10¹⁵ $\times$ V_nucleus

As a result of fusion, enormous amount of heat is liberated which is the main cause of source of solar energy.

The charge on hydrogen nucleus q₁ = +e

charge on electron, q₂ = –e

Coulomb force, F = $K\frac{\left(+e\right)\left(-e\right)}{r^2}$

= $-\frac{K{e}^2}{r^3}\overrightarrow{r}$

= $-\frac{K{e}^2}{r^2}\hat{r}$

It represents $\beta$-decay.

As Emission of electron (e^– ) + antineutrino ($\overline{\nu }$) $\Rightarrow$ $\beta$-decay.

As Radius of first orbit, r $\propto$ $\frac{1}{Z}$

for doubly ionized lithium Z (= 3) will be maximum, hence for doubly ionized lithium, r will be minimum.

For a light source of power P watt, the intensity at a distance d is given by

I = $\frac{P}{4\pi d^2}$

$\therefore$ Intensity $\propto$ $\frac{1}{{\left(dis\tan ce\right)}^2}$

${\left(\frac{e}{m}\right)}_{proton}=\frac{1}{1837}{\left(\frac{e}{m}\right)}_{electron}$

$\therefore$ ${\left(\frac{e}{m}\right)}_{proton}\ll {\left(\frac{e}{m}\right)}_{electron}$

When the collector is positive and emitter is negative with respect to base, it causes the forward biasing for each junction, which causes conduction of current.

In reverse bias, the size of the depletion region increases thereby increasing the potential barrier.

In full wave rectifier the fundamental frequency in ripple is twice of input frequency.

Barrier potential does not depend in diode design while barrier potential depends upon temperature, doping density and forward biasing.

$v$ = c/$\lambda$

$\lambda$ = $\frac{c}{v}$ = $\frac{3\times {10}^8}{8\times {10}^{15}}$ = 37.5 $\times$ 10^-9 m

= 37.5 nm $\approx$ $4\times 10^1$ nm

SO₃^2– : oxidation state of ‘S’ is +4

S₂O₄^2– : oxidation state of ‘S’ is +3.

S₂O₆^2– : oxidation state of ‘S’ is +5.

So, the order is S₂O${}_4^{2-}$ < SO${}_3^{2-}$ < S₂O${}_6^{2-}$.

Balanced chemical equation for Haber's process is as follows :

3H₂ + N₂ $\to$ 2NH₃

It is given that only 50% of the expected product is formed hence only 10 litre of NH₃ is formed.

Therefore, composition of gaseous mixture at the end is as follows :

N₂ used = 5 litres

N₂ left = 30 L – 5 L = 25 L

H₂ used = 15 litres,

H₂ legt = 30 L – 15 L = 15 L

NH₃ = 10 L

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

For any reaction to process in forward direction the reaction quotient (Q) must be less than equilibrium constant K_C.

Q < K_c

AgI	⇌	Ag+	+	I-
s		s		s

K_sp = s²

$\Rightarrow$ 1.0 × 10^–16 = s²

$\Rightarrow$ s = 1.0 × 10^–8 mol L^–1

$\therefore$ [Ag⁺] = 1.0 × 10^–8 mol L^–1

Also, in 10^–4 N KI solution,

[I^–1] = (10^–4 + 1.0 × 10^–8) mol L^–1

$\Rightarrow$ [I^–1] = (10^–4) mol L^–1

[As 1.0 × 10^–8 mol L^–1 << 1.0 × 10^–4 mol L^–1]

$\therefore$ K_sp of AgI = [Ag⁺][I^–]

= (1.0 × 10^–8)(10^–4)

= 1.0 × 10^–12 mol L^–1

H₂O(s) → H₂O(l)

$\Delta$H = 6.0 kJ mol^–1

T = 0 + 273 K = 273 K; $\Delta$S = ?

$\Delta$S = $\frac{\Delta H}{T}$ = $\frac{6000}{273}$ = 21.98 JK^–1 mol^–1

Heat of formation, $\Delta$H^o_f of a substance is the amount of heat absorbed or released when one mole of this substance is formed directly from its constituent elements.

In option (a), one mole of XeF₄ is formed from its constituent elements i.e., Xe and F₂ thus, the equation has equal value of $\Delta$H^o_r and $\Delta$H^o_f .

In option (b), the constituent atoms should be carbon and oxygen only but the reactant used is CO thus,

$\Delta$H^o_r $\ne$ $\Delta$H^o_f

In option (c), the reactant used is O₃ which is again not in its element form thus,

$\Delta$H^o_r $\ne$ $\Delta$H^o_f

In option (d), two products are formed thus

$\Delta$H^o_r $\ne$ $\Delta$H^o_f

For ideal solution,

$\Delta$H_soln = $\Delta$H₁ + $\Delta$H₂ + $\Delta$H₃

C_(graphite) $\to$ C_(diamond)

Volume of graphite = $\frac{12}{2.25}=5.33$ cm³ mol^–1

Volume of diamond = $\frac{12}{3.31}=3.63$ cm³ mol^–1

$\Delta$V = V_graphite – V_diamond

= 1.70 cm³ mol^–1

= 1.70 × 10^–3 L mol^–1

So, $\Delta$G^o = P$\Delta$V

$\Rightarrow$ 1895 J mol^–1 = P(1.70 × 10^–3 L mol^–1)

$\Rightarrow$ 1114.70 × 10³ J/L = P

$\Rightarrow$ $\frac{1114.7\times {10}^3}{101.33}$ = P

$\Rightarrow$ P = 11000.69 atm × 1.013 × 10⁵ Pa

= 11143.69 × 10⁵ Pa

= 11.14 × 10⁸ Pa

As we know, q = nC$\Delta$T

q = 1.0 kJ = 1000 J

C = 75 JK^–1 mol^–1

m = 100 g ⇒ Number of moles = $\frac{100}{18}$ g mol^–1

1000 = $\frac{100}{18}$ $\times$ 75 $\times$ $\Delta$T

$\Rightarrow$ $\Delta$T = $\frac{10\times 18}{75}$ K

$\Rightarrow$ $\Delta$T = 2.4 K

$\Delta$H = $\Delta$E + $\Delta$n_g RT

$\Delta$n_g = 3 – (1 + 5) = – 3

$\Rightarrow$ $\Delta$H = $\Delta$E + (– 3RT)

$\Rightarrow$ $\Delta$H – $\Delta$E = – 3RT

4/3Al + O₂ $\to$ 2/3Al₂O₃,   $\Delta$G = $-$ 827 kJ mol^$-$1

For 1 mol of Al, n = 3

$\therefore$ For $\frac{4}{3}$ mol of Al, n = $3\times \frac{4}{3}=4$

As $\Delta$G = - nFE^o

$\Rightarrow$ – 827 × 10³ J = – 4 × E° × 96500

$\Rightarrow$ E^o = $\frac{827\times {10}^3}{4\times 96500}$ = 2.14 V

Cell reaction is,

Zn + Cu²⁺ $\to$ Zn²⁺ + Cu

E_cell = E^o_cell - $\frac{RT}{nF}\ln \frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$

Greater the factor $\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$, less is the EMF.

Hence E₁ > E₂

For first order reaction

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

$\Rightarrow$ k = $\frac{2.303}{1}\log \frac{0.8}{0.2}$ = 2.303 log 4 ....(1)

Let t₁ hour is required for changing the concentration of A from 0.9 mole to 0.675 mole of B.

Remaining mole of A = 0.9 – 0.675 = 0.225

$\therefore$ k = $\frac{2.303}{t_1}\log \frac{0.9}{0.225}$ .....(2)

From equation (1) and (2)

$\frac{2.303}{t_1}\log \frac{0.9}{0.225}$ = 2.303 log 4

$\Rightarrow$ $t_1$ = 1 hr

Arrhenius equation k = $A{e}^{-\frac{E_a}{RT}}$

$\Rightarrow$ log k = log A - $\frac{E_a}{2.303RT}$

Comparing it with equation of straight line i.e.,

y = mx + C

On plotting log k vs $\frac{1}{T}$, we get a straight line, the slope indicates the value of activation energy.

The energy of activation of reverse reaction is less than or more than energy of activation E_$$a$$ of forward reaction

$\Delta H={\left(E_a\right)}_F-{\left(E_a\right)}_R$

Because it depends upon the nature of reaction.

If ${\left(E_a\right)}_F>{\left(E_a\right)}_R$, reaction is endothermic.

or ${\left(E_a\right)}_F<{\left(E_a\right)}_R$, reaction is exothermic.

As r = k[A]ⁿ

if n = 0

r = k[A]⁰

or r = k thus for zero order reactions rate is equal to the rate constant.

Given t_1/2 = 12.3 years

Initial amount (N₀) = 32 mg

Total time = 49.2 years

No. of half lives(n) = $\frac{T}{t_{1/2}}=\frac{49.2}{12.3}=4$

N = N₀${\left(\frac{1}{2}\right)}^n$ = $32{\left(\frac{1}{2}\right)}^4$ = 2 mg

Molar volume from pyknometric density

= $\frac{M}{2.165\times {10}^3}$ m³

Molar volume from X-ray density

= $\frac{M}{2.178\times {10}^3}$ m³

Volume unoccupied

= $\frac{M}{{10}^3}\left(\frac{1}{2.165}-\frac{1}{2.178}\right)$ m³

Fraction unoccupied

= $\frac{\left(\frac{0.013M\times {10}^{-3}}{2.165\times 2.178}\right)}{\left(\frac{M\times {10}^{-3}}{2.165}\right)}$

= 5.96 $\times$ 10^-3

The speed of the reaction increases because adsorption lowers the activation energy of the reaction.

Amongst isoelectronic ions, ionic radii of anions is more than that of cations. Further size of the anion increases with increase in -ve charge and size of cation decreases with increase in +ve charge. Hence, the correct order is

O^2- > F^- > Na⁺ > Mg²⁺ > Al³⁺

As sigma bond is stronger than the $\pi$(pi) bond, so it must be having higher bond energy than $\pi$(pi) bond.

In H₂O₂ , the O–H groups are not in the same plane. So it has non-planar structure. It has a half-opened book structure in which the two O–H groups lie on the two pages of the book. The angle between two pages of the book is 94^o and H–O–O bond angle is 97^o.

F is more electronegative than Cl therefore HF bond is stronger than HCl and hence proton is not given off easily and hence HF is a weakest acid.

B(OH)₃ does not provide H⁺ ions in water instead it accepts OH^– ion and hence it is Lewis acid.

B(OH)₃ + H₂O ⇌ [B(OH)₄]^- + H⁺

In lanthanide series there is a regular decrease in the atomic as well as ionic radii of trivalent ions (M³⁺) as the atomic number increases. Although the atomic radii do show some irregularities but ionic radii decreases from La(103 pm) to Lu (86pm).

The basic character of the transition metal monoxide is
TiO > VO > CrO > FeO because basic character of oxides decrease with increase in atomic number. Oxides of transitional metals in low oxidation state i.e., + 2 and + 3 are generally basic except Cr₂O₃.

The catalytic activity of transitional metals is due to their variable oxidation states.

$\pi$-bonded organometallic compound includes organometallic compounds of alkenes, alkynes and some other carbon containing compounds having electrons in their p orbitals.

Thus, the number of unpaired electrons = 4.