Chemical Bonding - JEE Advanced Previous Year Questions with Solutions
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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 IIT/JEE Advance Questions 
[JEE 2009] 
gets paired up and boron becomes diamagnetic. 
According to VSEPR theory all bond angle are $<90 \%$ 
[JEE 2010] 
$\mathrm{sp}^{3} \mathrm{d} \quad$ See-Saw shape 
[JEE Adv. 2014] 
Hence Ans. (B,C) 
(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. 
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 
down the group (D) decrease in ionization energy down the group [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. 
(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 $\mathrm{AlCl}_{3}$. 
(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.
Frequently Asked Questions
Find answers to common questions.
What is the bond order of He₂⁺ and is it stable?
He₂⁺ has a bond order of 0.5, calculated as (2 bonding electrons − 1 antibonding electron) ÷ 2 = 0.5. It is a stable species because bond order is greater than zero, and it has lower energy than two isolated helium atoms. This directly contradicts a common misconception tested in JEE Advanced 2016.
Is Molecular Orbital Theory more important than VSEPR for JEE Advanced?
Both are equally important, but MOT has appeared slightly more frequently (7 questions vs. 5 for VSEPR in this dataset). MOT questions are also harder to guess, so mastering it gives a higher score advantage. Do not skip VSEPR — shape-based questions appear every 2–3 years
How many questions from Chemical Bonding appear in JEE Advanced each year?
Chemical Bonding typically contributes 2–4 questions per year in JEE Advanced, making it one of the most consistently tested chapters in Inorganic Chemistry. Over the 2009–2018 period analysed here, the chapter contributed an average of 3 questions per paper.
Can I skip Chemical Bonding if I am weak at it and focus on other chapters?
Skipping Chemical Bonding is not advisable. It connects directly to Coordination Chemistry, Acid-Base Chemistry, and Molecular Structure — topics that together account for 20–25% of JEE Advanced Chemistry. A weak foundation here creates compounding errors. Spend 2 focused weeks on it using PYQs plus NCERT theory.
How do I calculate lone pairs on central atoms quickly for integer-type questions?
Use the formula: Lone pairs on central atom = (Valence electrons of central atom − Electrons used in bonding) ÷ 2. For [XeF₃]⁻ (Xe: 8 valence e⁻, 3 bonds = 6 e⁻ used, plus 1 extra from negative charge = 9 total) → (9 − 6) ÷ 2 = 1.5 → 3 lone pairs on Xe. Practice this for all species in the 2017 question.
Does JEE Advanced ask questions based on 2s–2p mixing in MOT?
Yes. JEE Advanced 2014 explicitly asked what happens when 2s–2p mixing is NOT operative. Without this mixing, the σ2p orbital lies below π2p, which changes the magnetic character of C₂ from diamagnetic (normal) to paramagnetic. Always note which condition the question specifies.