JEE Challenger
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ChemMedAdv 2017 • Paper 1

Match Quantum Numbers and Orbitals for Hydrogen Like Species

Comprehension Passage

Answer Q.31, Q.32 and Q.33 by appropriately matching the information provided in the three columns of the table below.

The wave function, ψn,l,ml\psi_{n,l,m_l}, is a mathematical expression depending upon spherical polar coordinates (r,θ,ϕ)(r, \theta, \phi) of the electron, characterized by quantum numbers nn, ll, and mlm_l. Here rr denotes the radial distance from the nucleus, θ\theta represents colatitude, and ϕ\phi is the azimuth. In the given formulas, ZZ represents the atomic number and aoa_o is the Bohr radius.

Column 1Column 2Column 3(I) 1s orbital(i) ψn,l,ml∝(Zao)32e−(Zrao)(P) Wavefunction plot vs r/ao(II) 2s orbital(ii) One radial node(Q) Probability density at nucleus ∝1ao3(III) 2pz orbital(iii) ψn,l,ml∝(Zao)52re−(Zr2ao)cos⁡θ(R) Probability density is maximum at nucleus(IV) 3dz2 orbital(iv) xy-plane is a nodal plane(S) Energy needed to excite electron from n=2 to n=4 state is 2732 times that from n=2 to n=6\begin{array}{|l|l|l|} \hline \text{Column 1} & \text{Column 2} & \text{Column 3} \\ \hline \text{(I) 1s orbital} & (i)\ \psi_{n,l,m_l} \propto \left(\frac{Z}{a_o}\right)^{\frac{3}{2}} e^{-\left(\frac{Zr}{a_o}\right)} & \text{(P) Wavefunction plot vs } r/a_o \\ \hline \text{(II) 2s orbital} & \text{(ii) One radial node} & \text{(Q) Probability density at nucleus } \propto \frac{1}{a_o^3} \\ \hline \text{(III) } 2p_z \text{ orbital} & (iii)\ \psi_{n,l,m_l} \propto \left(\frac{Z}{a_o}\right)^{\frac{5}{2}} r e^{-\left(\frac{Zr}{2a_o}\right)} \cos\theta & \text{(R) Probability density is maximum at nucleus} \\ \hline \text{(IV) } 3d_{z^2} \text{ orbital} & \text{(iv) } xy\text{-plane is a nodal plane} & \text{(S) Energy needed to excite electron from } n = 2 \text{ to } n = 4 \text{ state is } \frac{27}{32} \text{ times that from } n = 2 \text{ to } n = 6 \\ \hline \end{array}

For the specified orbital in Column 1, identify the only CORRECT combination valid for any hydrogen-like species:

Question Diagram 1

Options

A

(I) (ii) (S)

B

(IV) (iv) (R)

C

(II) (ii) (P)

Correct
D

(III) (iii) (P)

Topics & Concepts

Structure of AtomQuantum NumbersBohr ModelElectronic Configuration

Step-by-Step Solution

To determine the correct combination for Question 31, let me analyze each of the given options step-by-step:

  1. Option (A): (I) (ii) (S)

    • Column 1 (I) corresponds to the 1s orbital (n=1,l=0n=1, l=0).
    • Column 2 (ii) states "One radial node".
    • The number of radial nodes for an orbital is calculated using the formula: Radial nodes=n−l−1\text{Radial nodes} = n - l - 1
    • For a 1s orbital: Radial nodes=1−0−1=0\text{Radial nodes} = 1 - 0 - 1 = 0
    • Since the 1s orbital has 00 radial nodes, statement (ii) is incorrect for (I). Thus, Option A is incorrect.

  2. Option (B): (IV) (iv) (R)

    • Column 1 (IV) corresponds to the 3dz23d_{z^2} orbital.
    • Column 2 (iv) states "xyxy-plane is a nodal plane".
    • The 3dz23d_{z^2} orbital has two conical nodal surfaces centered along the z-axis, not nodal planes. The xyxy-plane is a nodal plane for orbitals like 3dxy3d_{xy} or 2pz2p_z, but not for 3dz23d_{z^2}. Thus, Option B is incorrect.

  3. Option (C): (II) (ii) (P)

    • Column 1 (II) corresponds to the 2s orbital (n=2,l=0n=2, l=0).
    • Column 2 (ii) states "One radial node". Radial nodes=2−0−1=1\text{Radial nodes} = 2 - 0 - 1 = 1 This is correct.
    • Column 3 (P) shows the plot of the wave function ψn,l,ml\psi_{n,l,m_l} vs ra0\frac{r}{a_0}.
    • The radial wave function for a 2s orbital is proportional to: ψ2s(r)∝(2−Zra0)e−Zr2a0\psi_{2s}(r) \propto \left(2 - \frac{Zr}{a_0}\right) e^{-\frac{Zr}{2a_0}} At r=0r = 0, ψ2s(0)>0\psi_{2s}(0) > 0. As rr increases, ψ2s\psi_{2s} decreases, crosses zero at r=2a0Zr = \frac{2a_0}{Z} (representing 1 radial node), becomes negative, reaches a local minimum, and then asymptotically approaches zero as r→∞r \to \infty. This precisely matches graph (P).
    • Therefore, the combination (II) (ii) (P) is correct.

  4. Option (D): (III) (iii) (P)

    • Column 1 (III) corresponds to the 2pz2p_z orbital (n=2,l=1n=2, l=1).
    • For any orbital with angular momentum quantum number l>0l > 0 (such as p,d,fp, d, f orbitals), the wave function value at the nucleus (r=0r = 0) is zero (ψ(0)=0\psi(0) = 0).
    • Graph (P) shows a non-zero positive value of the wave function at r=0r = 0, which is characteristic of an ss orbital (l=0l=0). Thus, graph (P) does not represent a 2pz2p_z orbital. Option D is incorrect.

Conclusion: The only correct combination valid for any hydrogen-like species is (C) (II) (ii) (P).

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