JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

**Simulator**

**Previous Years JEE Advance Questions**

**Paragraph for questions 1 to 3**

The hydrogen-like species $\mathbf{L} \mathbf{i}^{2+}$ is in a spherically symmetric state $\mathrm{S}_{1}$ with one radial node. Upon absorbing light the ion undergoes transition to a state $\mathrm{S}_{2}$ The state $\mathrm{S}_{2}$ has one radial node and its energy is equal to the ground state energy of the hydrogen atom.

(A) 1s (B) 2s (C) 2p (D) 3s

**[JEE 2010]**

**Sol.**(B)

$\mathrm{S}_{1}=2 \mathrm{s}$

$\mathrm{S}_{2}=3 \mathrm{p}$

(A) 0.75 (B) 1.50 (C) 2.25 (D) 4.50

**[JEE 2010]**

**Sol.**(C)

$\mathrm{E}=13.6 \times \frac{3}{4}$

(A) 0 (B) 1 (C) 2 (D) 3

**[JEE 2010]**

**Sol.**(B)

**[JEE 2011]**

**Sol.**9

**[JEE 2011]**

**Sol.**4

$(\mathrm{A}) \frac{\mathrm{h}^{2}}{4 \pi^{2} \mathrm{ma}_{0}^{2}}$

$(\mathrm{B}) \frac{\mathrm{h}^{2}}{16 \pi^{2} \mathrm{ma}_{0}^{2}}$

(C) $\frac{\mathrm{h}^{2}}{32 \pi^{2} \mathrm{ma}_{0}^{2}}$

(D) $\frac{\mathrm{h}^{2}}{32 \pi^{2} \mathrm{ma}_{0}^{2}}$

**[JEE 2012]**

**Sol.**(C)

**[JEE 2013]**

**Sol.**(5)

_{=}$4,\left|\mathrm{m}_{\ell}\right|=1$ and $\mathrm{m}_{\mathrm{s}}=-\frac{1}{2} \mathrm{is}$

**[JEE 2014]**

**Sol.**(6)

**[JEE Adv. 2017]**

**Sol.**(B)

**Answer Q.10, Q.11 and Q.12 by appropriately matching the information given in the three columns of the following table.**

The wave function n,* l *, m1 is a mathematical function whose value depends upon spherical polar coordinates $(\mathrm{r}, \theta, \phi)$ of the electron and characterized by the quantum numbers n, l and $\mathrm{m}_{1}$. Here r is distance from nucleus, is colatitude and is azimuth. In the mathematical functions given in the Table, Z is atomic number $\mathrm{a}_{0}$

is Bohr radius.

hydrogen – like species is :

(A) (IV) (iv) (R)

(B) (II) (ii) (P)

(C) (III) (iii) (P)

(D) (I) (ii) (S)

**[JEE – Adv. 2017]**

**Sol.**(B)

(A) (IV) (iv) (R) incorrect, because, $\mathrm{d}_{\mathrm{z}^{2}}$ has no nodal plane.

(B) (II) (ii) (P) correct, because 2s orbtial has 1 radial node.

(C) (III) (iii) (P) incorrect, because probability density for 2p at nucleus is zero.

(D) (I) (ii) (S) incorrect, because 1s orbital has no radial node.

(A) (II) (ii) (Q)

(B) (I) (i) (S)

(C) (I) (i) (R)

(D) (I) (iii) (R)

**[JEE – Adv. 2017]**

**Sol.**(D)

The option (D) is incorrect because in the wave function of 1s orbital , no angular function should be present.

(A) (I) (iv) (R)

(B) (I) (i) (P)

(C) (II) (i) (Q)

(D) (I) (i) (S)

**[JEE – Adv. 2017]**

**Sol.**(D)

We have to select only correct combination hence, the option (D) is correct.

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