Atomic Structure & Periodicity — Part 1: 40 MCQs with Explanations

The magnetic quantum number (mₗ) determines the spatial orientation of an orbital. For a given l, mₗ can have values from -l to +l.

Fluorine has the highest electronegativity value of 4.0 on Pauling scale.

NH₃ has a lone pair of electrons to donate, making it a Lewis base. BF₃, AlCl₃, and FeCl₃ are electron-deficient and act as Lewis acids.

Propene structure: CH₃-CH=CH₂. Bonds: C-H (6 bonds), C-C (1 single), C=C (1σ + 1π). Total σ bonds = 6(C-H) + 1(C-C) + 1(C=C σ) = 8? Wait CH₃ has 3H, CH has 1H, CH₂ has 2H → total 6 C-H bonds. Plus 1 C-C single bond between CH₃ and CH. Plus 1 σ bond in C=C. Total σ = 6+1+1 = 8. Plus 1 π bond in C=C. But wait the carbon atoms: C1 (CH₃) bonded to C2 with single bond (σ). C2 (CH) bonded to C3 with double bond (1σ+1π). So total σ: C1-H (3), C2-H (1), C3-H (2), C1-C2 (1σ), C2-C3 (1σ) = 3+1+2+1+1 = 8σ. π: C2-C3 (1π) = 1π. So 8σ, 1π. But option A is 8σ,1π. However some count differently. Standard answer for propene is 8 sigma and 1 pi bond. But let me double-check molecular formula C₃H₆. Total bonds: each C makes 4 bonds. Total valence electrons used in bonding: (3×4 + 6×1)/2 = 18/2 = 9 bonds total. Of these, one is π bond, so 8σ + 1π. Yes answer is option 1.

Electronic configuration of K: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹. The 4s orbital is filled before 3d and has higher energy than 3p in multi-electron atoms.

Cr configuration: [Ar] 4s¹ 3d⁵ (exception to Aufbau principle). Six unpaired electrons (one in 4s and five in 3d orbitals).

Atomic radius increases down a group due to addition of new electron shells, and decreases across a period due to increasing nuclear charge.

N₂ has all electrons paired in molecular orbitals (bond order 3), making it diamagnetic. O₂, B₂ have unpaired electrons and are paramagnetic.

Cyclooctatetraene is non-aromatic because it has 8 π electrons (not satisfying Hückel's 4n+2 rule) and adopts tub-shaped non-planar conformation.

NO₂⁺ (nitronium ion) is electron-deficient and accepts electron pair, making it an electrophile.

O₂ has two unpaired electrons in π* orbitals, making it paramagnetic.

Al³⁺ + 3e⁻ → Al. So 3 moles of electrons (3 Faradays) are required.

O₂ has bond order 2. Adding one electron to antibonding orbital: bond order = (8-5)/2 = 1.5.

Chlorine has highest electron affinity (-349 kJ/mol). Fluorine has lower value (-328 kJ/mol) due to small size causing electron-electron repulsion.

BF₃ is electron-deficient (incomplete octet) and accepts electron pair, making it Lewis acid.

Pyridine is aromatic with 6 π electrons in the ring (similar to benzene), satisfying Hückel's rule.

Cesium has lowest electronegativity (~0.79) due to large atomic size.

Transition elements typically show paramagnetism due to unpaired d electrons, not diamagnetism.

As reducing agent: H₂C₂O₄ → 2CO₂ + 2H⁺ + 2e⁻. n-factor = 2. But wait, oxalic acid donates 2 electrons per molecule, so eq wt = 90/2 = 45. However in some reactions it can lose 2 electrons per carbon? Standard: H₂C₂O₄ → 2CO₂ + 2e⁻ + 2H⁺, so n=2, eq wt=45. But option C is 30 which would be for n=3. Actually in acidic KMnO₄ titration, oxalic acid has n-factor 2, eq wt = 45. But some sources say equivalent weight of oxalic acid dihydrate is 63 as acid and 45 as reducing agent. For anhydrous H₂C₂O₄ (90 g/mol), eq wt as reducing agent = 90/2 = 45. So answer should be 45 (option 2). But let me check options: A=90, B=45, C=30, D=22.5. So B=45 is correct. However the question says as reducing agent"

n - l - 1 = 4 - 0 - 1 = 3 radial nodes.

Energy order: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p... So 4s has lower energy than 3d, 4p, 4d.

The azimuthal quantum number (l) determines the shape of an orbital: l=0 (s-orbital, spherical), l=1 (p-orbital, dumbbell), l=2 (d-orbital, cloverleaf), etc.

Fluorine (F) has the highest electronegativity value of 4.0 on the Pauling scale, making it the most electronegative element.

Cyclohexane is aliphatic (non-aromatic). Benzene, pyridine, and naphthalene satisfy Hückel's rule (4n+2 π electrons) and are aromatic.

Benzene has a delocalized π system with 6 π electrons forming 3 π bonds distributed over the ring (resonance hybrid).

NH₃ is a nucleophile (electron pair donor). BF₃, AlCl₃, and H⁺ are electrophiles (electron pair acceptors).

O₂ has two unpaired electrons in its π* orbitals (molecular orbital configuration), making it paramagnetic. Others have all electrons paired.

Al³⁺ + 3e⁻ → Al. Therefore, 3 moles of electrons are required to deposit 1 mole of Al metal.

Molecular orbital configuration of O₂: (σ1s²)(σ*1s²)(σ2s²)(σ*2s²)(σ2pz²)(π2px²,π2py²)(π*2px¹,π*2py¹). Bond order = ½(10-6) = 2. For O₂⁺, remove one electron from antibonding orbital: bond order = ½(10-5) = 2.5.

NO₂⁺ (nitronium ion) is an electrophile (electron-deficient species). OH⁻, CN⁻, and NH₃ are nucleophiles.

Chlorine has the highest electron affinity (-349 kJ/mol) among halogens. Fluorine has lower electron affinity (-328 kJ/mol) due to its small size and high electron density causing repulsion.

Benzene has 6 carbon atoms each contributing one p-orbital electron to the delocalized π system, totaling 6 π electrons (satisfying Hückel's 4n+2 rule with n=1).

Ethyne (C₂H₂) has linear geometry with bond angle 180°, characteristic of sp hybridization (one s + one p orbital).

Cesium (Cs) has the lowest electronegativity (~0.79 on Pauling scale) among all elements due to its large atomic size and low effective nuclear charge on valence electron.

Toluene has electron-donating methyl group that activates the ring and directs ortho-para, making it more reactive than benzene. Nitrobenzene and chlorobenzene are deactivated rings.

Transition elements typically show paramagnetism (due to unpaired d electrons), not diamagnetism. Diamagnetism is characteristic of elements with all paired electrons.

Number of radial nodes = n - l - 1. For 3p: n=3, l=1, radial nodes = 3-1-1 = 1.

Chromium is an exception to the Aufbau principle due to the extra stability associated with half-filled d-subshell. Its configuration is [Ar] 4s1 3d5.

By definition, the anode is the electrode where oxidation takes place (loss of electrons). The cathode is where reduction takes place.

sp3 hybridization has a tetrahedral electronic geometry. With two bonding pairs and two lone pairs (e.g., H2O), the molecular shape is bent or V-shaped.