Q 1 :

The number of radial node/s for 3p orbital is:             [2024]

  • 3

     

  • 2

     

  • 1

     

  • 4

     

(3)         

              Number of radial nodes = n-l-1.

               For 3p orbital, n=3, l=1

               Number of radial nodes = 3 - 1 - 1 = 1 

 



Q 2 :

Which one of the following about an electron occupying the 1s orbital in a hydrogen atom is incorrect? (Bohr’s radius is represented by a0)           [2025]

  • The probability density of finding the electron is maximum at the nucleus.

     

  • The electron can be found at a distance 2a0 from the nucleus.

     

  • The 1s orbital is spherically symmetrical.

     

  • The total energy of the electron is maximum when it is at a distance a0 from the nucleus.

     

(4)

|ψ|2  max at r=0 (nucleus)

|ψ|20 at any point

E=-13.6Z2n2,  Emax=0 at n=



Q 3 :

The wave function (Ψ) of 2s is given by

Ψ2s=122π(1a0)1/2(2-ra0)e-r/2a0

At r=r0, radial node is formed. Thus, r0 in terms of a0.                         [2023]

  • r0=a02

     

  • r0=2a0

     

  • r0=4a0

     

  • r0=a0

     

(2)

At node Ψ2s2=0, Ψ2s=0

So, (2-ra0)=0

2-r0a0=0  r0=2a0



Q 4 :

The number of correct statements from the following is __________.              [2023]

(A) For 1s orbital, the probability density is maximum at the nucleus

(B) For 2s orbital, the probability density first increases to maximum and then decreases sharply to zero.

(C) Boundary surface diagrams of the orbitals encloses a region of 100% probability of finding the electron.

(D) p and d-orbitals have 1 and 2 angular nodes respectively

(E) Probability density of p-orbital is zero at the nucleus



(4)