Question

The number of correct statement from the following is _________ .
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 diagram 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

Answer 1

Correct Answer: 3

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

• Correct. The 1s orbital has a spherically symmetrical distribution, and the probability density is highest at the nucleus, decreasing as the distance from the nucleus increases.

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

• Incorrect. For the 2s orbital, the probability density is maximum at the nucleus, decreases to zero at a radial node, increases again to a smaller maximum, and finally decreases asymptotically to zero. It does not simply increase to a maximum before decreasing.
  

C. Boundary surface diagrams of the orbitals enclose a region of 100% probability of finding the electron.

• Incorrect. Boundary surface diagrams typically enclose a region where there is a high (usually 90%) probability of finding the electron, not 100%. Electrons can still be found outside this region with a small probability.

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

• Correct. The number of angular nodes is equal to the azimuthal quantum number (\( \ell \)). For p orbitals, \( \ell = 1 \), and for d orbitals, \( \ell = 2 \).

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

• Correct. p-orbitals have a nodal plane passing through the nucleus, making the probability density at the nucleus zero.

Final Answer:

The number of correct statements is 3 (A, D, and E).