To determine the number of radial nodes in a 3p orbital, we need to understand the concept of radial nodes and how they are calculated for atomic orbitals.
Understanding Radial Nodes:
Radial nodes are regions where the probability of finding an electron is zero. They can be calculated using the formula:
\(N_{\text{radial}} = n - \ell - 1\)
where:
For a 3p orbital:
Substituting these values into the formula gives:
\(N_{\text{radial}} = 3 - 1 - 1 = 1\)
Thus, the number of radial nodes in a 3p orbital is 1.
Conclusion:
The correct answer is 1, which is consistent with the calculation based on quantum numbers. The remaining options (4, 2, 3) are incorrect because they do not align with the formula for radial nodes.
The number of radial nodes for an orbital is given by the formula:
Number of radial nodes = n − l − 1
where: - n is the principal quantum number. - l is the azimuthal quantum number
(orbital angular momentum quantum number).
For a 3p orbital: - n = 3 - l = 1 (since it is a p-orbital)
Applying the formula:
Number of radial nodes = 3 − 1 − 1 = 1
Conclusion: The number of radial nodes for the 3p orbital is 1.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)