Class 11th

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

The benzene here would undergo Friedel-Crafts alkylation reaction where carbocation is formed as an intermediate, thus secondary carbocation is formed as an intermediate due to its greater stability than that of the primary carbocation.

The final product obtained is 

As per de Brogile's equation,

λ = h / mv = (6.626 x 10^-34 kg m² s^-1) / (9.1 x 10^-31 kg) x (1.6 x 10⁶ ms^-1)

= 0.455 x 10^-9 m = 0.455 nm = 455 pm.

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.