Q.Choose the correct option.
p-orbitals are....... in shape.

For a one-electron system, the Schrödinger equation's solution in spherical polar coordinates factorises as $\Psi(r,\theta,\varphi)=R(r)\cdot f(\theta)\cdot g(\varphi)$, letting the probability distribution be studied separately as a function of distance from the nucleus (the radial distribution) and as a function of direction (the angular distribution).

Radial distribution and nodes. Plotting the raw $R(r)^2$ against $r$ is misleading, since it ignores how much physical space exists at each distance; correctly weighting by the volume of a thin spherical shell, $dV=4\pi r^2\,dr$, gives the true radial probability curve, $4\pi r^2\cdot R(r)^2$ versus $r$. For hydrogen's 1s orbital this curve peaks exactly at the Bohr radius (0.529 Å), the electron's single most probable distance from the nucleus. For higher orbitals the curve can touch zero at one or more specific radii called radial nodes - spherical shells where the electron is essentially never found - and the general counting rule is

Angular distribution and orbital shape. For an s orbital ($l=0$), the angular part works out to a constant, independent of direction - giving the familiar perfect sphere shape (with $(n-1)$ concentric radial nodes for higher s orbitals, but always spherically symmetric overall). A p orbital ($l=1$, three orientations $p_x,p_y,p_z$) has a two-lobed "dumbbell" shape with one nodal plane. A d orbital ($l=2$, five orientations) has a "clover leaf" shape (with $d_{z^2}$ looking different: a dumbbell plus a ring) and two nodal planes. An f orbital ($l=3$, seven orientations) has a more complex multi-lobed shape with three nodal planes. In general, the number of angular nodes for any orbital equals its azimuthal quantum number, $l$, so the total node count (radial plus angular) for any orbital works out to $(n-1)$.