Chemistry · Structure Of Atom · NEET
Think of an electron as a cloud that is thick near the nucleus and thin far away. A boundary surface diagram is one surface (like a skin) drawn in space where the probability density |psi|^2 has the SAME constant value at every point on that skin. NCERT picks the skin that traps about 90% of the electron cloud inside it. The SHAPE of that skin is what we call the shape of the orbital. So the diagram is basically the outer boundary of the region where the electron mostly lives.
Because the electron cloud never truly ends. |psi|^2 gets smaller and smaller as you go away from the nucleus but it never becomes exactly zero, even very far out. So a 100% boundary surface would have to be infinitely large, which is useless to draw. NCERT says a 100% surface is not possible. We pick a surface (about 90% probability) that is big enough to show the real shape but still has a definite size. This is a very common NEET trap.
For an s-orbital, the chance of finding the electron at a fixed distance r is the SAME in every direction (up, down, left, right). Equal in all directions means the boundary surface is a sphere. For a p-orbital, the wave function has a sign (+ and -) and the probability is high along one axis and ZERO in the plane passing through the nucleus. Two separate high-probability regions (lobes) sit on either side of that plane, giving the dumbbell (two-lobe) shape.
They answer different questions. A radial probability plot (like 1s vs 2s curves) tells you HOW FAR from the nucleus the electron is likely to be. A boundary surface diagram tells you the SHAPE and DIRECTION in 3D space. So one is about distance, the other is about shape. NEET can mix these up in a single question, so keep them separate.
No. All s-orbitals (1s, 2s, 3s, 4s) are spherical. Higher n only makes the sphere BIGGER, so 4s > 3s > 2s > 1s in size, and the electron sits further from the nucleus. The boundary surface diagram normally does not show the inner nodes (the empty shells inside); it only shows the overall outer shape.
Each p-orbital has 2 lobes (dumbbell) pointing along one axis: px, py, pz. There are 3 p-orbitals. Four of the five d-orbitals (dxy, dyz, dxz, dx^2-y^2) have 4 lobes (double dumbbell / cloverleaf), and dz^2 looks different: a dumbbell along the z-axis with a doughnut (ring) around its middle. All five d-orbitals have the same energy.
Which of the following pairs of d-orbitals will have electron density (lobes) along the axes?
Which one is a wrong statement?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is a surface drawn in space on which the probability density |psi|^2 is constant, and which encloses about 90% of the chance of finding the electron; its shape represents the shape of the orbital.
s-orbital is spherical, each p-orbital is dumbbell-shaped (2 lobes) along an axis, and d-orbitals are mostly cloverleaf (4 lobes) except d(z^2) which is a dumbbell with a ring around the middle.
Because |psi|^2 never becomes exactly zero even far from the nucleus, so a 100% surface would be infinitely large. We use about 90% probability instead to get a definite, useful shape.
No. 1s, 2s, 3s and 4s are all spherical. Only the size increases with n, so 4s > 3s > 2s > 1s. The shape stays a sphere.
On the plane that passes through the nucleus and separates the two lobes (this plane is a nodal plane). The electron is never found on this plane.
No. The boundary surface shows the 3D shape and orientation, while the radial probability curve shows how the chance of finding the electron changes with distance r from the nucleus.
