NEET-UG 2000

Dimension of force = [MLT^-2]

Dimension of impulse = [MLT^-1]

So both of them have different dimensions.

Given, s = (3t³ + 7t² + 14t + 8)

$\frac{ds}{dt}$ = 9t² + 14t + 14

$\frac{d^{2}s}{d{t}^2}$ = 18t + 14

$a$ = 18t + 14

At t = 1, a = 18 $\times$ 1 + 14 = 32 m/s²

Vertical acceleration in both the cases is g, whereas horizontal velocity is constant.

Given, u₁ = u₂ = u, ${\theta }_1$ = 60º, ${\theta }_2$ = 30º

In 1^st case, we know that range
$R_1=\frac{u^2\sin 2\left(60^{\circ }\right)}{g}=\frac{u^2\sin 120^{\circ }}{g}$
$=\frac{u^2\sin \left(90^{\circ }+30^{\circ }\right)}{g}$
$=\frac{u^2\left(\cos 30^{\circ }\right)}{g}=\frac{u^2\sqrt{3}}{2g}$

In 2^nd case, when ${\theta }_2$ = 30° , then
$R_2=\frac{u^2\sin 60^{\circ }}{g}=\frac{u^2\sqrt{3}}{2g}$

$\Rightarrow R_1=R_2$

$v_{\mathrm{Resultant}}=\frac{1km}{1/4hr}=4km/hr$

$\therefore$ $v_{River}=\sqrt{5^2-4^2}=3km/hr$

Let T be the tension in the string.

$\therefore$ 10g – T = 10a     ....(i)

T – 5g = 5a             ....(ii)

Adding (i) and (ii),

$5g=15a\Rightarrow a=\frac{g}{3}m/s^2$

Change in momentum = 2 × 3 × 10 × sin60°

= $60\times \frac{\sqrt{3}}{2}$

Force = Change in momentum/Impact time

$=\frac{30\sqrt{3}}{0.2}=150\sqrt{3}N$

P = $\overrightarrow{F}.\overrightarrow{v}=\left(60\hat{i}+15\hat{j}-3\hat{k}\right).\left(2\hat{i}-4\hat{j}+5\hat{k}\right)$

= 120 – 60 – 15 = 45 watts.

Velocity after 5 sec,

v = u – gt = 100 – 10 × 5 = 50 m/s

By conservation of momentum

1 × 50 = 0.4 × (–25) + 0.6 × v'

60 = 0.6 × v' $\Rightarrow$ v' = 100 m/s upwards

Solid sphere reaches the bottom first because for solid cylinder $\frac{K^2}{R^2}=\frac{1}{2}$, and for hollow cylinder $\frac{K^2}{R^2}=1$

Acceleration down the inclined plane $\propto \frac{1}{\frac{K^2}{R^2}}$

Solid cylinder has greater acceleration, so it reaches the bottom first.

As effective distance of mass from BC is greater than the effective distance of mass from AB, therefore I₂ > I₁.

When a sphere is rotating in a vertical circle, it exerts the maximum outward pull when it is at the lowest point B.

Therefore, tension at B is maximum = $\frac{m{v}^2}{R}$

So, the string breaks at point B.

$F_{surface}=G\frac{Mm}{R_e^2}$

$F_{R_e/2}=G\frac{Mm}{{\left(R_e+R_e/2\right)}^2}=\frac{4}{9}\times F_{surface}$

$=\frac{4}{9}\times 72=32N$

$v_{earth}=\sqrt{\frac{2G{M}_e}{R_e}}$

$v_{planet}=\sqrt{\frac{2G{M}_p}{R_p}}=\sqrt{\frac{2G{M}_e}{R_e/4}}=\sqrt{\frac{8G{M}_e}{R_e}}$

$\frac{v_{planet}}{v_{earth}}=\sqrt{\frac{8G{M}_e}{R_e}}\times \sqrt{\frac{R_e}{2G{M}_e}}=2$

$\therefore$ $v_{planet}=2\times v_{earth}$

= 2 $\times$ 11.2 = 22.4 km/s

Gravitational force is required for convection

According to Wein's displacement law,

${\lambda }_{m}T$ = 2.88 × 10^–3

When T = 2000 K,

${\lambda }_m$(2000) = 2.88 × 10^–3   ....(1)

When T = 3000 K,

${\lambda }_m^{{}^{\prime }}$ (3000) = 2.88 × 10^–3    ....(2)

Dividing (1) by (2),

$\frac{2}{3}\frac{{\lambda }_m}{{\lambda }_m^{{}^{\prime }}}=1$

$\Rightarrow \frac{{\lambda }_m}{{\lambda }_m^{{}^{\prime }}}=\frac{3}{2}$

$\Rightarrow {\lambda }_m^{{}^{\prime }}=\frac{2}{3}{\lambda }_m$

Initially the efficiency of the engine was $\frac{1}{6}$ which increases to $\frac{1}{3}$ when the sink temperature reduces by 62º C.

$\eta =\frac{1}{6}=1-\frac{T_2}{T_1}$ when T₂ = sink temperature , T₁ = source temperature

$\Rightarrow T_2=\frac{5}{6}{T}_1$

Secondly,
$\frac{1}{3}=1-\frac{T_2-62}{T_1}=1-\frac{T_2}{T_1}+\frac{62}{T_1}$

$=1-\frac{5}{6}+\frac{62}{T_1}$

$\Rightarrow T_1=62\times 6=372K=372-273=99^{o}C$

& $T_2=\frac{5}{6}\times 372=310K=310-273=37^{o}C$

$\gamma =1+\frac{2}{f}$

where f is the degree of freedom

$\therefore$ $\frac{2}{f}=\gamma -1\Rightarrow f=\frac{2}{\gamma -1}$

$f_A=2f_B$

$\Rightarrow \frac{1}{2\pi }\sqrt{\frac{g}{l_A}}=2\times \frac{1}{2\pi }\sqrt{\frac{g}{l_B}}$

$\Rightarrow \frac{l}{l_A}=4\times \frac{l}{l_B}$

$\Rightarrow l_A=\frac{l_B}{4}$ , which does not depend on mass.

In $\Delta OAC,\cos \theta =A/l$

$\Rightarrow OA=lcos\theta$

$\therefore$ $AB=l\left(1-\cos \theta \right)=h$

At point, C the velocity of bob = 0. The vertical acceleration = g

$\therefore$ $v^2=2gh$

$\Rightarrow v=\sqrt{2gl\left(1-\cos \theta \right)}$

As $n\propto \left(1/l\right)$

$l=l_1+l_2+l_3$

$\therefore \frac{1}{n}=\frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3}$

Given : x = acos($\omega$t + $\delta$)
and y = acos($\omega$t + $\alpha$)     ...(i)

where, $\delta =\alpha +\pi /2$

$\therefore \mathrm{x}=\mathrm{acos}(\omega \mathrm{t}+\alpha +\pi /2)$

= –asin($\omega$t + $\alpha$) ...(ii)

Given the two waves are acting in perpendicular direction with the same frequency and phase difference $\pi$/2.

From equations (i) and (ii),
x² + y² = a²
which represents the equation of a circle.

$f^{\prime }=\frac{v-u}{v}f;f^{\prime \prime }=\frac{v+u}{v}f$

Number of beats = $f^{\prime \prime }-f^{\prime }=\frac{2u}{\lambda }$

Using Gauss’s law, the total electric flux through a closed surface is

$$\Phi =\frac{q_{\text{enclosed}}}{{\epsilon }_0}$$

Here, the charge $Q$ is placed at a corner of the cube, not at the centre.

To apply Gauss’s law easily, imagine $8$ identical cubes joined together so that the charge $Q$ comes at the common corner and becomes the centre of the larger cube.

Now for the larger cube:

• the charge enclosed = $Q$

• so total flux through the larger cube is

$${\Phi }_{\text{large cube}}=\frac{Q}{{\epsilon }_0}$$

Since the larger cube is made of $8$ identical small cubes, by symmetry the flux is equally shared by them.

So, flux through one small cube is

$${\Phi }_{\text{one cube}}=\frac{1}{8}\cdot \frac{Q}{{\epsilon }_0}=\frac{Q}{8{\epsilon }_0}$$

Hence, the electric flux through all the six faces of the given cube is

$$\boxed{\frac{Q}{8{\epsilon }_0}}$$

So, the correct answer is:

$$\boxed{\text{Option B}}$$

Considering symmetric elements each of length dl at A and B, we note that electric fields perpendicular to PO are cancelled and those along PO are added. The electric field due to an element of length $dl(=ad\theta )$ along PO.

$dE=\frac{1}{4\pi {\epsilon }_0}\frac{dq}{a^2}\cos \theta$
$\left(\because dl=ad\theta \right)$

$=\frac{1}{4\pi {\epsilon }_0}\frac{\lambda dl}{a^2}\cos \theta$

$=\frac{1}{4\pi {\epsilon }_0}\frac{\lambda \left(ad\theta \right)}{a^2}\cos \theta$ Net electric field at O

$E={\int }_{-\pi /2}^{\pi /2}dE$

$=2{\int }_O^{\pi /2}\frac{1}{4\pi {\epsilon }_0}\frac{\lambda \left(ad\theta \right)}{a^2}\cos \theta$

$=2.\frac{1}{4\pi {\epsilon }_0}\frac{\lambda }{a}{\left[\sin \theta \right]}_o^{\pi /2}$

$=2.\frac{1}{4\pi {\epsilon }_0}\frac{\lambda }{a}.1=\frac{\lambda }{2\pi {\epsilon }_{0}a}$

When we measure the emf of a cell by the potentiometer then no current draws in the circuit in zero-deflection condition i.e., cell is in open circuit. Thus, in this condition the actual value of a cell is found. In this way potentiometer is equivalent to an ideal voltmeter of infinite resistance.

Note: The emf by the potentiometer is measured from null method in which zero deflection position is found on the wire.

$P=\frac{V^2}{R}$

$\Rightarrow R\propto \frac{1}{P}$

$\therefore R_{40}>R_{100}$

$\frac{V-E}{r}=I\Rightarrow \frac{V-12}{5\times {10}^{-2}}=60$

$\Rightarrow V=15V$

This is a balanced Wheatstone’s bridge so no current flows through the 7 $\Omega$ resistor.

$\therefore$ $\frac{1}{R_{eq}}=\frac{1}{4+3}+\frac{1}{6+8}$

$\Rightarrow R_{eq}=\frac{14}{3}\Omega$

Let q be the charge on each capacitor.

$\therefore$ Energy stored, $U=\frac{1}{2}C{V}^2=\frac{1}{2}\frac{q^2}{C}$

Now, when battery is disconnected and another capacitor of same capacity is connected in parallel to the first capacitor, then voltage across each capacitor, $V=\frac{q}{2C}$

$\therefore$ Energy stored = $\frac{1}{2}C{\left(\frac{q}{2C}\right)}^2=\frac{1}{4}.\frac{1}{2}\frac{q^2}{C}=\frac{1}{4}U.$

Magnetic field due to 5A = $\frac{5{\mu }_0}{2\pi \times 2.5}$ = $\frac{2{\mu }_0}{2\pi }\otimes$

Magnetic field due to 2.5A = $\frac{2.5{\mu }_0}{2\pi \times 2.5}$ = $\frac{{\mu }_0}{2\pi }\odot$

Resultant Magnetic field = $\frac{2{\mu }_0}{2\pi }-\frac{{\mu }_0}{2\pi }=\frac{{\mu }_0}{2\pi }\otimes$

The frequency of radiations of waves are :

From one side, $\frac{t-x}{5}=1.5$

From other side, $\frac{x}{2}$ = 1.5 $\Rightarrow$ x = 3

$\therefore$ $\frac{t-3}{5}=1.5$ $\Rightarrow$ t = 10.5 cm

The minimum mirror length should be half of the height of man.

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

= $\left(1.5-1\right)\left(\frac{1}{\infty }-\frac{1}{-10}\right)$ = $0.5\left(\frac{1}{10}\right)$

$\Rightarrow$ f = 20 cm

When plane surface is silvered, F = $\frac{f}{2}=\frac{20}{2}$ = 10 cm

Rainbow is formed due to combination of internal reflection and dispersion.

$\nu$ $\propto$ $\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$

Maximum frequency of emission happens when n₁ = 1 and n₂ = 2

Atomic hydrogen is unstable and it has life period of a fraction of a second.

The relation is :

T_1/2 = $\frac{\ln 2}{\lambda }$

Jump to second orbit leads to Balmer series. The jump from 4th orbit shall give rise to second line of Balmer series.

According to liquid drop model of nucleus, an excited nucleus breaks into lighter nuclei just like an excited drop breaks into tiny drops.

The particle X is neutrino.

Thomson evaluated the mass of electron infirectly with help of charge.

By photoelectric effect, Einstein proved E = h$\upsilon$.

A diode is said to be forward biased if p-type semiconductor of p-n junction is at positive potential with respect to n-type semiconductor of p-n junction. It is so for circuit (a).

$\beta$ = $\frac{I_C}{I_B}$ = $\frac{I_C}{I_E-I_C}$ = $\frac{\frac{I_C}{I_E}}{1-\frac{I_C}{I_E}}$ = $\frac{\alpha }{1-\alpha }$

In an Ionic crystal, the cations and anions are arranged in alternate form.

BaCO₃ $\to$ BaO + CO₂

197.34 g $\to$ 22.4 L at N.T.P

9.85 g $\to$ $\frac{22.4}{197.34}\times 9.85=1.118L$

In A₃(BC₄)₂, (+2) $\times$ 3 + 2[+5+4(-2)]

$\Rightarrow$ +6 + 10 -16 = 0

$\therefore$ Oxidation numbers of A, B, C are + 2, +5 and $-$2 respectively.

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

$\Rightarrow \lambda$ = $\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{3.03\times {10}^{-19}}$

= 656 nm

No. of electrons CO = CN^- = NO⁺ = $C_2^{2-}$ = 14, So these are isoelectronics.

Average molar kinetic energy = $\frac{3}{2}nRT$

As temperature is same hence average kinetic energy of CO and N₂ will be same.

Equilibrium constant of a reaction is independent of the concentration of species involved in the reaction but dependent only on the temperature.

K_p = P_CO2

As solids do not exert pressure, so their partial pressure is taken as unity.

CH₃COOH is weak acid while NaOH is strong base, so one equivalent of NaOH can not be neutralized with one equivalent of CH₃COOH. Hence the solution of one equivalent of each does not have pH value as 7. Its pH will be towards basic side as NaOH is a strong base hence conc. of OH^– will be more than the conc. of H⁺.

NH₃ after losing a proton (H⁺) gives NH₂^–. So, NH₃ is the conjugate acid of NH₂^–.

Note : Conjugate acid-base pair differ only by a proton.

Molecular masses of polymers are best determined by osmotic pressure method . Firstly because other colligative properties give so low values that they cannot be measured accurately and secondly, osmotic pressure measurements can be made at room temperature and do not require heating which may change the nature of the polymer.

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

= $\frac{2930}{300}$

= 9.77 J/mol-K

2Zn + O₂ $\to$ 2ZnO;  $\Delta$G^o = $-$ 616 J ....(1)
2ZnS $\to$ 2Zn + O₂;  $\Delta$G^o = $+$ 293 J.....(2)
S₂ + 2O₂ $\to$ 2SO₂;   $\Delta$G^o = $-$408 J .....(3)

$\Delta$G^o for the reaction can be obtained by adding (1), (2) and (3)

$\therefore$ $\Delta$G^o = 293 - 616 - 408 = -731 J

As we know, $\Delta$H = $\Delta$E + $\Delta$n_g RT

Now, $\Delta$n_g = Number of gaseous moles of products – number of gaseous moles of reactions

= 2 – 3 = – 1

$\Rightarrow$ $\Delta$H = $\Delta$E + (–1) RT

$\Delta$H = $\Delta$E – RT

For spontaneous reaction $\Delta$G^o = – ve and

E^o_cell = + ve

as $\Delta$G^o = – nFE^o_cell

where, n = number of electrons taking part

E^o = emf of cell

F = Faraday constant

Cu²⁺ + 2e^– $\to$ Cu; E°₁ = 0.34 V .....(1)

Cu²⁺ + e^– $\to$ Cu⁺ ; E°₂ = 0.15 V.....(2)

Cu⁺ + e^– $\to$ Cu; E°₃ = ? .....(3)

$\therefore$ $\Delta$G^o₁ = -2$\times$0.34$\times$F

and $\Delta$G^o₂ = -1$\times$0.15$\times$F

and $\Delta$G^o₃ = -1$\times$E°₃$\times$F

Also, $\Delta$G^o₁ = $\Delta$G^o₂ + $\Delta$G^o₃

$\Rightarrow$ -0.68F = -0.15F - E°₃ $\times$ F

$\Rightarrow$ E°₃ = 0 68 - 0 15 = 0 53 V

$\therefore$ E°_cell = 0.53 - 0.15 = 0.38 V

According to Kohlrausch’s law, the equivalent conductance of BaCl₂ at infinite dilution,

$\lambda$_$$\infty$$ of BaCl₂ = $\frac{1}{2}$$\lambda$_$$\infty$$ of Ba²⁺ + $\lambda$_$$\infty$$ of Cl^-

$\Rightarrow$ $\lambda$_$$\infty$$ of BaCl₂ = $\frac{1}{2}\times 127+76$ = 139 5

$-\frac{1}{3}$$\frac{d\left[B{r}^{-}\right]}{dt}=+\frac{1}{5}\frac{d\left[B{r}_2\right]}{dt}$

$\Rightarrow$ $\frac{d\left[B{r}^{-}\right]}{dt}=-\frac{3}{5}\frac{d\left[B{r}_2\right]}{dt}$

Enzymes act like catalyst in biochemical reactions. Presence of an enzyme increases the rate of reaction by lowering the activation energy of the reactant.

No. of half life’s gone = $\frac{18}{3}$ = 6

N₀ = 300 g

N_t = N₀ $\times$ ${\left(\frac{1}{2}\right)}^n$

= $300\times {\left(\frac{1}{2}\right)}^6$

= $300\times \frac{1}{64}=4.68g$

We know that electrostatic force is that binds the oppositely charged ions which are formed by transfer of electron from one atom to another is called ionic bond. We also know that cation and anion are oppositely charged particles therefore they form ionic bond in crystal.

A atoms are at 8 corners of the cube

$\therefore$ No. of A atoms per unit cell = 8 $\times$ $\frac{1}{8}$ = 1

B atoms are at the face centre of six faces.

$\therefore$ No. of B atoms per unit cell = 6 $\times$ $\frac{1}{2}$ = 3

Thus, formula of compound = AB₃

By adding electrolytes the colloidal particles are precipitated. The electrolytes neutralise the charge of colloids leading to their coagulation and thus destroy the colloid.

Bond energy generally decreases on moving from top to bottom along a group. It happens due to fact that size increases on moving down the group and thus, the two nuclei are far apart and less capable of holding the two atoms together.

But in case of F₂ molecule due to more repulsion in between non-bonding electron pair (2p) of two fluorines (due to small size of F-atom) in comparison to non-bonding electron pair (3p) in chlorine, the bond energy of F₂ is less than Cl₂.

BE(F₂) = 158.5 kJ/mole and

BE(Cl₂)= 242.6 kJ/mole

$N_2$ has 14 electrons.

Moleculer orbital configuration of $N_2$

= ${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\pi }_{2p_x^2}={\pi }_{2p_y^2}{\sigma }_{2p_z^2}$

$\therefore$ N_b = 10

N_a = 4

$\therefore$ BO = $\frac{1}{2}\left[10-4\right]$ = 3

N₂⁺ has 13 electrons.

Moleculer orbital configuration of N₂⁺

= ${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\pi }_{2p_x^2}={\pi }_{2p_y^2}{\sigma }_{2p_z^1}$

$\therefore$ N_b = 9

N_a = 4

$\therefore$ BO = $\frac{1}{2}\left[9-4\right]$ = 2.5

As the bond order in N₂ is more than N₂⁺ so the dissociation energy of N₂ is higher than N₂⁺.

Those species which have unpaired electrons are called paramagnetic species.

And those species which have no unpaired electrons are called diamagnetic species.

(A) N₂⁺ has 13 electrons.

Moleculer orbital configuration of N₂⁺

= ${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\pi }_{2p_x^2}={\pi }_{2p_y^2}{\sigma }_{2p_z^1}$

Here 1 unpaired electrons present, so it is paramagnetic.

(B) Molecular orbital configuration of $O_2^{-}$ (17 electrons) is

${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\sigma }_{2p_z^2}{\pi }_{2p_x^2}={\pi }_{2p_y^2}{\pi }_{2p_x^2}^{\ast }={\pi }_{2p_y^1}^{\ast }$

Here 1 unpaired electrons present, so it is paramagnetic.

(A) CO has 14 electrons.

Moleculer orbital configuration of CO

= ${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\pi }_{2p_x^2}={\pi }_{2p_y^2}{\sigma }_{2p_z^2}$

Here 0 unpaired electrons present, so it is diamagnetic.

(D) Molecular orbital configuration of NO (15 electrons) is

${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\sigma }_{2p_z^2}{\pi }_{2p_x^2}={\pi }_{2p_y^2}{\pi }_{2p_x^1}^{\ast }={\pi }_{2p_y^o}^{\ast }$

Here 1 unpaired electrons present, so it is paramagnetic.

In PO₄^3–, P atom has vacant d-orbitals, thus it can form p$\pi$ - d$\pi$ bond. ‘N’ and ‘C’ have no vacant ‘d’ orbital in their valence shell, so they cannot form such bond.

XeF₄ is sp³d² hybridised and it is square planar :

In BCl₃ the central atom ‘B’ is sp² hybridised and contains only ‘six’ electrons in its valence shell. Therefore it is electron deficient.

Mn : [Ar] 3d⁵4s²

Shows +2, +3, +4, +5, +6 & +7 oxidation states.

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

Sc³⁺ : [Ar]₁₈ 3d⁰4s⁰

Ti³⁺ : [Ar]₁₈ 3d¹4s⁰

Cr⁴⁺ : [Ar]₁₈ 3d²4s⁰

V³⁺ : [Ar]₁₈ 3d²4s⁰

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

K₃[Cu(CN)₂] = 3(+1)+x+2(-1) = 0

$\Rightarrow$ x = -1

$\therefore$ the oxidation no. of 'Cu' is -1 (-ve), so this complex is unstable so it is not formed.

In Fe(CO)₅, the Fe-atom is dsp³ hybridised and has shape of trigonal bipyramidal.

₂₆Fe = [Ar]3d⁶4s²

Complex [Co(en)₂Cl₂]Cl show 4 different isomerism.

The staggered form of ethane has the following structure and the dihedral angle is 60^o, which means ‘H’ atoms are at an angle of 60^o to each other.

Due to restricted rotation about double bond, 2-butene shows geometrical isomerism.

This is most stable as the repulsion between two methyl groups is least.

Among —CH₃, —OCH₃ and CF₃,

CH₃ and —OCH₃ are electron donating groups. Hence, they activate the benzene ring. In these the order of activation is —OCH₃ $>$ —CH₃ while —CF₃ group deactivates the benzene nucleus. So, it shows lower rate of electrophilic substitution on benzene ring. Thus, order of electrophilic substitution is :

Propene adds to diborane (B₂H₆) giving an addition product. The addition compound on oxidation gives 1-propanol. Here addition of water takes place according to anti-Markownikoff’s rule.

O-atom is more electronegative than C-atom, therefore O-atom bears partial –ve charge and C-atom to which it is attached bear partial +ve charge.

Here H₂O is continuously removed from reaction so that reaction always proceeds to product formation i.e., Ethyl benzonate.

HF form linear polymer due to H-bonding.

H - F .... H - F .... H - F .... H - F

Due to high electronegativity value of 'F' atom it forms effective hydrogen bonds.