Ncert Solutions Chemistry Class 11th

This is a short answer type question as classified in NCERT Exemplar

The rate-determining step involved in the reaction is 

CH₃-CH=CH₂ + HX → CH₃-CH⁺-CH₃ + X^-

So the rate-determining step depends on the bond energy of HX i.e. higher the bond energy lesser would be reactivity.

Thus the order of reactivity of these halogen acids is

HI>HBr>HCl

Radius of orbit of H like species = (0.529 / Z) n²Å = (52.9 / Z) n² pm

r₁ = 1.3225 nm = 1322.5 pm = (52.9 / Z) n₁²

r₂ = 211.6 pm = 211.6 pm = (52.9 / Z) n₂²

∴ r₁ / r₂ = 1322.5 / 211.6

=>n₁² /n₂² = 6.25

=> n₁/n₂= (6.25)^1/2 = 2.5

=> n₁ = 2.5 n₂

=> 10 n₁= 25 n₂

=> 2 n₁= 5 n₂

If n₁ = 2, then n₂ = 5. That means transition occurs from 5^th orbit to 2^nd orbit. This means that the transition belongs to Balmer series.

Now, wave number? = (1.097 x 10⁷ m^-1) x (1/2² – 1/5²) = 1.097 x 10⁷ x 21/100 m^-1 = 23.037 x 10⁵ m^-1

λ = 1/? = 1/ 23.037 x 10⁵ m^-1

= 434 x 10^-9 m = 434 nm

This transition belongs to visible region of the spectrum of light.

Given, ν = (3.29 x 10¹⁵ Hz) (1/3² – 1/n²)

= (3.29 x 10¹⁵ Hz) (1/3² – 1/n²)

(3 x 10⁸ ms^-1) / (1.285 x 10^-6 m) = (3.29 x 10¹⁵ Hz) (1/3² – 1/n²)

2.3346 x 10¹⁴ = (3.29 x 10¹⁵ Hz) (1/3² – 1/n²)

2.3346 / 32.9 = 1/3² – 1/n²

0.071 = 1/9 – 1/n²

1/n² = 1/9 – 0.071= 0.111 – 0.071 = 0.04

n² = 1/ 0.04 = 25

=> n = 5

For n = 5, Paschen series lies in infrared region of the spectrum.

This is a short answer type question as classified in NCERT Exemplar

The rotation around C-C single bond is possible but it is not completely free due to the torsional strain which is about 1^-20 KJ mol^-1.

λ = 150 pm, v = 1.5 x 10⁷ m s^-1

Kinetic energy, K.E. = ½ mv2 = ½ x 9.1 x 10^-31 kg x (1.5 x 10⁷ ms^-1)²

= [ (9.1 x 1.5 x 1.5) / 2] x 10^{-31 +14}

= 10.2375 x 10^-17 J = 1.02375 x 10^-16 J

K.E = hc/ λ = (6.626 x 10^-34 kg m² s^-1) / (3 x 10⁸ ms^-1) / (1.5 x 10^-10 m)

= [ (6.626 x 3) x 10^-34+8+10] / 1.5  

= 13.252 x 10^-16 J

We know, E = W₀ + K.E.

W₀ = E – K.E.  = (13.252 – 1.024) x 10^-16 J

= 12.228 x 10^-16 J

= 12.228 10^-16 / 1.602 x 10^-19

= 7.63 x 10³ eV

λ = 256.7 nm = 256.7 x 10^-9 m

K.E. = 0.35 eV

E = hc/ λ = (6.626 x 10^-34Js) / (3 x 10⁸ ms^-1) / (256.7 x 10^-9 m)

= (6.626 x 3 x 10^-17) J/ 256.7

= (6.626 x 3 x 10^-17) / (256.7 x 1.602 x 10^-19) eV

E = 4.83 eV

The potential applied to silver gets converted into kinetic energy of the photoelectron.

So, Kinetic energy, K.E= 0.35 V

=> K.E= 0.35 eV          

E = W₀ + K.E.

=> W₀ = E – K.E.

= 4.83 eV – 0.35 eV = 4.48 eV.

This is a short answer type question as classified in NCERT Exemplar

Yes, it will execute geometrical isomerism as the product formed can be either cis-2-butene or trans-2-butene.

W₀ = 1.9 eV = 1.9 x 1.602 x 10^-19 J

Threshold frequency, v₀ = W₀ / h = (1.9 x 1.602 x 10^-19 J) / (6.626 x 10^-34Js)

= 0.459 x 10¹⁵ s^-1 = 4.59 x 10¹⁴ s^-1

Threshold wavelength? ₀ = c / v₀ = (3 x 10⁸ ms^-1) / (4.59 x 10¹⁴ s^-1)

= 0.6536 x 10^-6 m = 653.6 nm? 654 nm

Kinetic energy, E = E₀ + ½ mv²

(1/2 mv²) = E – E₀ = hc [ (1/? ) – (1/? ₀)]

= (6.626 x 10^-34Js) x (3 x 10⁸ ms^-1) / (10^-9) x [ (1/500) – (1/654)]

= 6.626 x 3 x 154 x 10-34+8+9) / (500 x 654)

= 9.36 x 10^-20 J

Velocity, v = [ (2 x 9.36 x 10^-20) / m]^1/2

= [ (2 x 9.36 x 10^-20) kg m² s^-2 / 9.1 x 10^-31 kg]^1/2

= (2.057 x 10¹¹ m²s^-2)^1/2= (20.57 x 10¹⁰ m²s^-2)^1/2

= 4.5356 x 10⁵ ms^-1

λ₁ = 589 nm = 589 x 10^-9 m

ν₁ = c / λ₁ = (3 x 10⁸ ms^-1) / (589 x 10^-9 m) = 5.0934 x 10¹⁴ s^-1

λ₂ = 589.6 nm = 589.6 x 10^-9 m

ν₂ = c / λ₂ = (3 x 10⁸ ms^-1) / (589.6 x 10^-9 m) = 5.0882 x 10¹⁴ s^-1

ΔE = E₁ – E₂ = h [ν₁ – ν₂]

= (6.626 x 10^-34Js) x [ (5.0934 x 10¹⁴ s^-1) – (5.0882 x 10¹⁴ s^-1)

= 3.31 x 10^-22 J