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.
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Previous Years IIT/JEE Advance Questions
[JEE 2009]
$B \rightarrow Q, R, S, T$ $C \rightarrow P, Q, R$ $D \rightarrow P, Q, R, S$
(A) $\mathrm{N}_{2} \mathrm{O}$ (B) $\mathrm{N}_{2} \mathrm{O}_{3}$ (C) $\mathrm{N}_{2} \mathrm{O}_{4}$ (D) $\mathrm{N}_{2} \mathrm{O}_{5}$ [JEE 2009]
$2 \mathrm{X}+\mathrm{B}_{2} \mathrm{H}_{6} \longrightarrow\left[\mathrm{BH}_{2}(\mathrm{X})_{2}\right]^{+}\left[\mathrm{BH}_{4}\right]^{-}$ the amine(s) X is (are) (A) $\mathrm{NH}_{3}$ (B) $\mathrm{CH}_{3} \mathrm{NH}_{2}$ (C) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}$ $(\mathrm{D})\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}$ [JEE 2009]
[JEE 2009]
(A) $\mathrm{SO}_{3}$ (B) $\operatorname{BrF}_{3}$ (C) $\mathrm{SiO}_{3}^{2-}$ (D) $\mathrm{OSF}_{2}$ [JEE 2010]
(A) 1 and diamagnetic (B) 0 and diamagnetic (C) 1 and paramagnetic (D) 0 and paramagnetic [JEE 2010]
If Hund’s rule is violated in Boron then unpaired electron of 
gets paired up and boron becomes diamagnetic.
[JEE 2010]
According to VSEPR theory all bond angle are $<90 \%$
[JEE 2010]
[JEE 2010]
state is O, Cl, F, N, P, Sn, Tl, Na, Ti [JEE 2010]
Na & F
[JEE 2011]
$\mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6}$
(A) $\mathrm{HNO}_{3}, \mathrm{NO}, \mathrm{NH}_{4} \mathrm{Cl}, \mathrm{N}_{2}$ (B) $\mathrm{HNO}_{3}, \mathrm{NO}, \mathrm{N}_{2}, \mathrm{NH}_{4} \mathrm{Cl}$ (C) $\mathrm{HNO}_{3}, \mathrm{NH}_{4} \mathrm{Cl}, \mathrm{NO}, \mathrm{N}_{2}$ (D) $\mathrm{NO}, \mathrm{HNO}_{3}, \mathrm{NH}_{4} \mathrm{Cl}, \mathrm{N}_{2}$ [JEE 2012]
(A) sp and $\mathrm{sp}^{3}$ (B) sp and $\mathrm{sp}^{2}$ (C) only $\mathrm{sp}^{2}$ (D) $\mathrm{sp}^{2}$ and $\mathrm{sp}^{3}$ [JEE 2012]
(A) Trigonal bipyramidal (B) Square planar (C) tetrahedral (D) see-saw [JEE 2012]
$\mathrm{sp}^{3} \mathrm{d} \quad$ See-Saw shape
[JEE Adv. 2013]
(A) $\mathrm{Be}_{2}$ (B) $\mathrm{B}_{2}$ (C) $\mathrm{C}_{2}$ (D) $\mathrm{N}_{2}$ [JEE Adv. 2014]
If 2s–2p is not operative than $\mathrm{C}_{2}$ becomes paramagnetic.
[JEE Adv. 2014]
(A) $\mathrm{BrF}_{5}$ (B) $\mathrm{ClF}_{3}$ (C) $\mathrm{XeF}_{4}$ (D) SF $_{4}$ [JEE – ADV. 2016]
Hence Ans. (B,C)
(A)Tetranuclear$\left[\mathrm{B}_{4}\mathrm{O}_{5}(\mathrm{OH})_{4}\right]^2-}$ unit (B) All boron atoms in the same plane (C) Equal number of $\operatorname{sp}^{2}$ and $\mathrm{sp}^{3}$ hybridized boron atoms (D) One terminal hydroxide per boron atom [JEE – ADV. 2016]
(A)Having $\left[\mathrm{B}_{4} \mathrm{O}_{5}(\mathrm{OH})_{4}\right]^{2-}$ tetranuclear (boron) unit (B) All boron atoms not in same plane (C) Two boron are $\mathrm{Sp}^{2}$ hybridised and two boron are sp3 hybridised (D) One terminal hydroxide per boron atom is present.
(A) $\mathrm{C}_{2}^{2-}$ is expected to be diamagnetic (B) $\mathrm{O}_{2}^{2+}$ is expected to have a longer bond length than O2 (C) $\mathrm{N}_{2}^{+}$ and $\mathrm{N}_{2}$– have the same bond order (D) $\mathrm{He}_{2}^{+}$ has the same energy as two isolated He atoms [JEE – ADV. 2016]
In the MO of $\mathrm{C}_{2}^{2-}$ there is no unpaired electron hence it is diamagnatic (B) Bond order of $\mathrm{O}_{2}^{2+}$ is 3 and $\mathrm{O}_{2}$ is 2 therefore bond length of $\mathrm{O}_{2}$ is greater than $\mathrm{O}_{2}^{2+}$ (C) The molecular orbital energy configuration of $\mathrm{N}_{2}^{+}$ is (D) $\mathrm{He}_{2}^{+}$ has less energy as compare to two isolated He atoms
(A) $\mathrm{HClO}_{4}$ is more acidic than HClO because of the resonance stabilization of its anion (B) $\mathrm{HClO}_{4}$ is formed in the reaction between $\mathrm{Cl}_{2}$ and $\mathrm{H}_{2} \mathrm{O}$ (C) The central atom in Both $\mathrm{HClO}_{4}$ and HClO is $\mathrm{sp}^{3}$ hybridized (D) The conjugate base of $\mathrm{HClO}_{4}$ is weaker base than $\mathrm{H}_{2} \mathrm{O}$ [JEE – Adv. 2017]
A) the physical state of $\mathrm{x}_{2}$ at room temperature changes from gas to solid down the group (B) decrease in HOMO-LUMO gap down the group (C) decrease in (D) decrease in ionization energy down the group [JEE – Adv. 2017]
down the group
_{2},\mathrm{B}_{2},\mathrm{C}_{2}, \mathrm{N}_{2}, \mathrm{O}_{2}^{-}$ the number of diamagnetic species is – (Atomic number) : H = 1, He = 2, Li = 3, Be = 4, B = 5, C = 6, N = 7, O = 8 , f = 9) [JEE – Adv. 2017]
If existence of $\mathrm{Be}_{2}$ is considered in atomic form or very weak bonded higher energetic species having zero bond order then it is diamagnetic , then answer will be 6. But if existence of molecular form of $\mathrm{Be}_{2}$ is not considered then magnetic property can’t be predicted then answer will be 5.
$\left[\mathrm{TeBr}_{6}\right]^{2-},\left[\mathrm{BrF}_{2}\right]^{+}, \mathrm{SNF}_{3}$ and $\left[\mathrm{XeF}_{3}\right]^{-}$ [Atomic number : N = 7, F = 9, S = 16, Br = 35, Te = 52, Xe = 54] [JEE – Adv. 2017]
(A) $\mathrm{H}_{3} \mathrm{PO}_{4}>\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}>\mathrm{H}_{3} \mathrm{PO}_{3}>\mathrm{H}_{3} \mathrm{PO}_{2}$ (B) $\mathrm{H}_{3} \mathrm{PO}_{3}>\mathrm{H}_{3} \mathrm{PO}_{2}>\mathrm{H}_{3} \mathrm{PO}_{4}>\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}$ (C) $\mathrm{H}_{3} \mathrm{PO}_{2}>\mathrm{H}_{3} \mathrm{PO}_{3}>\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}>\mathrm{H}_{3} \mathrm{PO}_{4}$ (D) $\mathrm{H}_{3} \mathrm{PO}_{4}>\mathrm{H}_{3} \mathrm{PO}_{2}>\mathrm{H}_{3} \mathrm{PO}_{3}>\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}$ [JEE – Adv. 2017]
(A) $\mathrm{Cr}_{2} \mathrm{O}_{3}, \mathrm{CrO}, \mathrm{SnO}, \mathrm{PbO}$ (B) $\mathrm{NO}, \mathrm{B}_{2} \mathrm{O}_{3}, \mathrm{PbO}, \mathrm{SnO}_{2}$ (C) $\mathrm{Cr}_{2} \mathrm{O}_{3}, \mathrm{BeO}, \mathrm{SnO}, \mathrm{SnO}_{2}$ (D) $\mathrm{ZnO}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{PbO}, \mathrm{PbO}_{2}$ [JEE – Adv. 2017]
(A) $\mathrm{Al}\left(\mathrm{CH}_{3}\right)_{3}$ has the three-centre two-electron bonds in its dimeric structure (B) $\mathrm{AlCl}_{3}$ has the three-centre two-electron bonds in its dimeric structure (C) BH3 has the three-centre two-electron bonds in its dimeric structure (D) The Lewis acidity of $\mathrm{BCl}_{3}$ is greater than that of $\mathrm{AlCl}_{3}$ [JEE – Adv. 2017]
(A) (B) (C) (D) Lewis acidic strength decreases down the group. The decrease in acid strength occurs because as size increases, the attraction between the incoming electron pair and the nucleus weakens. Hence Lewis acidic strength of $\mathrm{BCl}_{3}$ is more than $\mathrm{AlCl}_{3}$.
(A) $\mathrm{Bi}_{2} \mathrm{O}_{5}$ is more basic than $\mathrm{N}_{2} \mathrm{O}_{5}$ (B) $\mathrm{NF}_{3}$ is more covalent than $\mathrm{BiF}_{3}$ (C) PH3 boils at lower temperature than $\mathrm{NH}_{3}$ (D) The N–N single bond is stronger than the P–P single bond [JEE – Adv. 2018]
(D) Due to small size in N–N single bond l.p. – l.p. repulsion is more than P–P single bond therefore N–N single bond is weaker than the P–P single bond.
