JEE Chemistry 2023 Question with Solution

Given: Five statements about orbitals, probability density, boundary surface diagrams, and angular nodes are given.

Find: The number of correct statements.

Evaluate each statement using the standard properties of atomic orbitals.

A. For $1s$ orbital, the probability density is maximum at the nucleus. This statement is correct. The $1s$ orbital is spherically symmetric, and its probability density is highest at the nucleus and decreases with increasing distance.

B. For $2s$ orbital, the probability density first increases to maximum and then decreases sharply to zero. This statement is incorrect. For the $2s$ orbital, the probability density is maximum at the nucleus, then decreases to zero at a radial node, then increases again to a smaller maximum, and finally decreases asymptotically to zero.

C. Boundary surface diagrams of the orbitals enclose a region of $100\%$ probability of finding the electron. This statement is incorrect. Boundary surface diagrams usually represent a region containing about $90\%$ probability of finding the electron, not $100\%$.

D. $p$ and $d$-orbitals have $1$ and $2$ angular nodes respectively. This statement is correct. The number of angular nodes is equal to the azimuthal quantum number $\ell$. Thus, for $p$ orbitals, $\ell = 1$ and for $d$ orbitals, $\ell = 2$.

E. Probability density of $p$-orbital is zero at the nucleus. This statement is correct. A $p$-orbital has angular node behavior through the nucleus, so the probability density at the nucleus is zero.

Therefore, the correct statements are A, D, and E.

Hence, the number of correct statements is $3$.

Given: Statements about orbital probability density and nodes.

Find: Count how many are true.

Use these facts:

1. $s$ orbitals can have non-zero probability density at the nucleus.
2. $p$ orbitals have zero probability density at the nucleus.
3. Number of angular nodes equals $\ell$.
4. Boundary surface diagrams do not represent $100\%$ probability.

Applying these facts:

• A is true.
• B is false because the $2s$ orbital has a radial node and two maxima pattern in probability density.
• C is false because the enclosed probability is high but not complete.
• D is true.
• E is true.

So, total correct statements $= 3$.

Therefore, the answer is $3$.