Class 11th

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

The electron donating group increases the reactivity in electrophilic substitution reaction whereas the electron withdrawing group decreases the reactivity in electrophilic substitution.

Thus, the order of reactivity

ν = 4.37 x 10⁵ m s^-1, m = 0.1 kg

As per de Brogile's equation,

λ= m/v = (6.626 x 10^-34 kg m² s^-1) / (0.1 kg) x (4.37 x 10⁵ m s^-1)

=6.626/0.437  x 10^-34-5 m

= 1.516 x 10^-38 m

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

ν = 2.19 x 10⁶ m s^-1

As per de Brogile's equation,

λ= m/v = (6.626 x 10^-34 kg m² s^-1) / (9.1 x 10^-31 kg) x (2.19 x 10⁶ m s^-1)

6.626/91 * 2.19=  x 10^-34+25 m = 0.33243 x 10^-9 m = 332.43 pm

λ = 800 pm = 800 x 10^-12 m

m = 1.675 x 10^-27 kg

As per de Brogile's equation,

ν =h/mλ  = (6.626 x 10^-34 kg m² s^-1) / (1.675 x 10^-27 kg) x (800 x 10^-12 m)

=6.626/1.675 * 8  x 10^-34+27+10

= 0.494 x 10³ ms^-1

= 494 ms^-1

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.