Question:

Which of the following statements are incorrect about boundary surface diagram of orbital?} (i) A boundary surface is drawn in space for an orbital on which the value of probability density $|\Psi|^2$ is constant.} (ii) The boundary surface for $|\Psi|^2$ and $|\Psi|$ are not identical.} (iii) The density of the dots represents the probability density of finding the electron in the region.} (iv) In $1s$ and $2s$ orbitals, the probability of finding electron at a given distance is not equal in all the directions.} (v) The $2p$ orbitals in the boundary surface diagram are not spherical.

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All s-orbitals are spherical, while p-orbitals are dumbbell shaped.
Updated On: Apr 24, 2026
  • (i), (iv)
  • (iii), (iv)
  • (ii), (iv)
  • (ii), (iii)
  • (i), (ii)
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The Correct Option is C

Solution and Explanation

Concept: Chemistry - Atomic Orbitals and Boundary Surface Diagrams.
Step 1: Check statement (i). True. Boundary surface is drawn where probability density remains constant.
Step 2: Check statement (ii). False. Since $|\Psi|^2$ and $|\Psi|$ represent same shape region, surfaces are identical in shape.
Step 3: Check statement (iii). True. Dot density indicates probability of electron presence.
Step 4: Check statement (iv). False. $s$ orbitals are spherical, so probability at given distance is same in all directions.
Step 5: Check statement (v). True. $2p$ orbitals are dumbbell shaped, not spherical.
Step 6: Incorrect statements. [ \boxed{(ii),(iv)} ]
Hence, correct option is (C).
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