Class 11th

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New answer posted

7 months ago

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A
alok kumar singh

Contributor-Level 10

The nuclear charge experienced by electrons depends on the distance between the nucleus and orbital. The greater is this distance, the lesser is the effective nuclear charge. Among all the p orbitals, 4p orbital lies the farthest from the nucleus and thus experiences the lowest effective nuclear charge because of the maximum magnitude of screening or shielding effect.

New answer posted

7 months ago

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P
Payal Gupta

Contributor-Level 10

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

Due to the +R effect halogens are ortho and para directing groups.

New answer posted

7 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

The electrons may be assigned to the following orbitals :
(i) 4d
(ii) 3d
(iii) 4p
(iv) 3d
(v) 3p
(vi) 4p.
The increasing order of energy is :
(v) < (ii) = (iv) < (vi) = (iii) < (i)

New answer posted

7 months ago

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A
alok kumar singh

Contributor-Level 10

Since actual momentum is smaller than the uncertainty in measuring momentum, therefore, the momentum of electron cannot be defined

New answer posted

7 months ago

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Payal Gupta

Contributor-Level 10

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

New answer posted

7 months ago

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A
alok kumar singh

Contributor-Level 10

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

New question posted

7 months ago

0 Follower 2 Views

New answer posted

7 months ago

0 Follower 1 View

P
Payal Gupta

Contributor-Level 10

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

New answer posted

7 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

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

New answer posted

7 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

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

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