JEE Advanced 2017 · Paper 1 · Chemistry Q13

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ChemistrySingle correct+3 / -1Question 31 of 54

Passage

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

The wave function, $\psi_{n,l,m_l}$ is a mathematical function whose value depends upon spherical polar coordinates $(r, \theta, \phi)$ of the electron and characterized by the quantum numbers $n$ , $l$ and $m_l$ . Here $r$ is distance from nucleus, $\theta$ is colatitude and $\phi$ is azimuth. In the mathematical functions given in the Table, $Z$ is atomic number and $a_o$ is Bohr radius.

Column 1	Column 2	Column 3
(I) 1s orbital	(i) $\psi_{n,l,m_l} \propto \left(\frac{Z}{a_o}\right)^{3/2} e^{-\left(\frac{Zr}{a_o}\right)}$	(P) [plot of $\psi_{n,l,m_l}(r)$ against $r/a_0$ ]
(II) 2s orbital	(ii) One radial node	(Q) Probability density at nucleus $\propto \frac{1}{a_o^3}$
(III) $2p_z$ orbital	(iii) $\psi_{n,l,m_l} \propto \left(\frac{Z}{a_o}\right)^{5/2} r e^{-\left(\frac{Zr}{2a_o}\right)} \cos\theta$	(R) Probability density is maximum at nucleus
(IV) $3d_{z^2}$ orbital	(iv) $xy$ -plane is a nodal plane	(S) Energy needed to excite electron from $n = 2$ state to $n = 4$ state is $\frac{27}{32}$ times the energy needed to excite electron from $n = 2$ state to $n = 6$ state

The three-column table for Q.31-33. Column 3 entry (P) is a plot of ψ(r) against r/a₀: the curve starts high at r = 0, falls steeply, crosses zero once (one radial node), dips to a shallow negative minimum and rises back towards zero.

For the given orbital in Column 1, the only CORRECT combination for any hydrogen-like species is

(I) (ii) (S)

(IV) (iv) (R)

(II) (ii) (P)

(III) (iii) (P)

Answer and worked solution

Official answer: C

For any hydrogen-like species, the 2s orbital (II) has $n-l-1=1$ radial node (ii).

Its radial function starts positive at $r=0$ , crosses zero once, dips negative and returns to zero. That is the plot (P).

(C)

Topics:Orbitals and their shapes Quantum numbers

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JEE Advanced 2017 · Paper 1, Chemistry question 13