Which bond is formed due to overlap of orbitals along internuclear axis
A Pi bond
B Sigma bond
C Hydrogen bond
D Ionic bond
Sigma (σ) bond is formed by head-on (axial) overlap of orbitals along the internuclear axis.
Which overlap results in formation of a π bond
A s–s overlap
B s–p overlap
C p–p sidewise overlap
D p–s overlap
π bonds are formed by lateral (sidewise) overlap of p orbitals above and below the internuclear axis.
Which bond restricts free rotation around bond axis
A Single bond
B Sigma bond
C Pi bond
D Ionic bond
π bonds involve sidewise overlap; rotation breaks overlap, so rotation is restricted.
Hybridization in ethene (C₂H₄) is
A sp
B sp²
C sp³
D sp³d
Each carbon forms three σ bonds and one π bond → sp² hybridization.
The bond angle in trigonal planar geometry is approximately
A 90°
B 104.5°
C 109.5°
D 120°
sp² hybridization gives trigonal planar geometry with 120° bond angles.
Which molecule shows sp hybridization
A CO₂
B NH₃
C CH₄
D BF₃
CO₂ has two electron domains → sp hybridization → linear shape.
Which species has bent molecular shape
A CO₂
B BeCl₂
C SO₂
D BF₃
SO₂ has one lone pair on sulfur → bent shape despite trigonal planar electron geometry.
According to VSEPR theory, shape of PCl₅ is
A Tetrahedral
B Trigonal pyramidal
C Trigonal bipyramidal
D Octahedral
PCl₅ has 5 bond pairs, no lone pairs → trigonal bipyramidal.
Which molecule has T-shaped geometry
A SF₄
B ClF₃
C XeF₂
D PF₅
ClF₃ has 3 bond pairs + 2 lone pairs (AX₃E₂) → T-shaped geometry.
Shape of IF₇ molecule is
A Octahedral
B Pentagonal bipyramidal
C Square planar
D Trigonal bipyramidal
IF₇ has 7 bond pairs and no lone pair → pentagonal bipyramidal.
Hybridization involved in IF₇ is
A sp³d
B sp³d²
C sp³d³
D dsp²
Seven electron domains require sp³d³ hybridization.
Which molecule has zero dipole moment
A NH₃
B H₂O
C CO₂
D SO₂
CO₂ is linear and symmetrical, so bond dipoles cancel out.
The bond order of NO molecule is
A 1
B 1.5
C 2
D 2.5
NO has odd number of electrons; bond order calculated from MOT is 1.5.
Which species is paramagnetic
A N₂
B O₂
C CO
D H₂
O₂ has two unpaired electrons in antibonding orbitals.
The bond order of O₂²⁻ is
A 2
B 1.5
C 1
D 0
Adding two electrons to O₂ fills antibonding orbitals, reducing bond order to 1.
Which species is most stable
A He₂
B O₂
C N₂
D Ne₂
N₂ has highest bond order (3), indicating maximum stability.
The magnetic behavior of N₂ is
A Paramagnetic
B Diamagnetic
C Ferromagnetic
D Antiferromagnetic
All electrons in N₂ are paired.
Which molecular orbital has highest energy
A σ1s
B π2p
C π2p
D σ2p
Antibonding σ* orbitals are highest in energy.
Which statement about molecular orbitals is correct
A They belong to individual atoms
B They are formed by atomic orbital overlap
C They cannot extend over whole molecule
D They contain only paired electrons
Molecular orbitals are formed by linear combination of atomic orbitals.
Which of the following has maximum bond length
A N₂
B O₂
C O₂⁻
D O₂²⁻
Bond length increases as bond order decreases; O₂²⁻ has lowest bond order.
Which ligand causes maximum splitting of d-orbitals
A F⁻
B H₂O
C NH₃
D CN⁻
CN⁻ is a strong field ligand (spectrochemical series).
The coordination number of metal in [Cr(en)₃]³⁺ is
A 3
B 4
C 6
D 8
en is bidentate; 3 en ligands attach via 6 donor atoms.
Which complex shows optical isomerism
A [Co(NH₃)₆]³⁺
B [CoCl₆]³⁻
C [Co(en)₃]³⁺
D [Pt(NH₃)₂Cl₂]
Three bidentate ligands in octahedral complex produce chirality.
Which complex does NOT show geometrical isomerism
A [Pt(NH₃)₂Cl₂]
B [Co(NH₃)₄Cl₂]⁺
C [Ni(CO)₄]
D [Cr(en)₂Cl₂]⁺
Tetrahedral complexes do not show geometrical isomerism.
Oxidation state of Cr in [Cr(H₂O)₆]Cl₃ is
A +1
B +2
C +3
D +6
Three Cl⁻ outside → complex ion has +3 charge → Cr = +3.
The ligand which is neutral is
A CN⁻
B Cl⁻
C NH₃
D OH⁻
NH₃ is a neutral ligand donating lone pair.
The geometry of coordination number 4 complex with strong field ligands is
A Tetrahedral
B Square planar
C Octahedral
D Linear
d⁸ metal ions with strong field ligands form square planar complexes.
Which metal commonly forms square planar complexes
A Zn²⁺
B Ni²⁺
C Pt²⁺
D Fe²⁺
Pt²⁺ (d⁸) strongly favors square planar geometry.
The splitting of d-orbitals in octahedral field is represented by
A Δt
B Δo
C Δ
D δ
Δo denotes crystal field splitting energy in octahedral complexes.
In octahedral field, which d-orbitals experience greater repulsion
A dxy, dyz, dxz
B dx²–y², dz²
C All equally
D None
These orbitals point directly at ligands → higher energy (eg set).
High spin complexes are formed with
A Strong field ligands
B Weak field ligands
C Neutral ligands only
D π-acceptor ligands
Weak field ligands cause small splitting → electrons remain unpaired.
Which complex is paramagnetic
A [Fe(CN)₆]⁴⁻
B [Fe(CN)₆]³⁻
C [Co(NH₃)₆]³⁺
D [Zn(NH₃)₄]²⁺
Fe³⁺ (d⁵) with CN⁻ still has unpaired electrons → paramagnetic.
Which ligand is π-acceptor
A NH₃
B H₂O
C CO
D OH⁻
CO accepts electron density back via π-bonding (back bonding).
The denticity of EDTA is
A 2
B 4
C 5
D 6
EDTA coordinates through six donor atoms → hexadentate.
The complex [Co(NH₃)₄Cl₂]⁺ shows
A Only ionization isomerism
B Only geometrical isomerism
C Only optical isomerism
D No isomerism
MA₄B₂ octahedral complexes show cis–trans isomerism.
Which compound violates Werner’s theory
A [Co(NH₃)₆]Cl₃
B [Co(NH₃)₅Cl]Cl₂
C [Co(NH₃)₄Cl₂]Cl
D [Co(NH₃)₃Cl₃]
This compound shows both ionization and geometrical isomerism, challenging early Werner assumptions.
The ligand field splitting is minimum in
A Octahedral
B Square planar
C Tetrahedral
D Linear
Δt is smaller than Δo; tetrahedral complexes have weakest splitting.
Which coordination number is most common
A 2
B 4
C 6
D 8
Octahedral coordination (CN=6) is most stable and common.
Which complex is diamagnetic
A [Fe(H₂O)₆]²⁺
B [Mn(H₂O)₆]²⁺
C [Zn(NH₃)₄]²⁺
D [FeF₆]³⁻
Zn²⁺ is d¹⁰ → all electrons paired.
The number of unpaired electrons in high-spin Fe³⁺ (d⁵) is
A 1
B 2
C 3
D 5
High spin d⁵ → all five electrons unpaired.
Which ligand forms chelate ring
A NH₃
B H₂O
C en
D Cl⁻
en binds through two donor atoms forming a ring with metal (chelation).
Chelate effect refers to
A Formation of colored complexes
B Increased stability of chelated complexes
C Decrease in coordination number
D Weak bonding
Multidentate ligands form more stable complexes due to entropy gain.
Which complex shows linkage isomerism
A [Co(NH₃)₅NO₂]²⁺
B [Co(en)₃]³⁺
C [Pt(NH₃)₂Cl₂]
D [Ni(CO)₄]
NO₂⁻ can coordinate via N or O → linkage isomerism.
The IUPAC name of [Cr(H₂O)₄Cl₂]Cl is
A Tetraaquadichlorochromium(II) chloride
B Tetraaquadichlorochromium(III) chloride
C Dichlorotetraaquachromium(III) chloride
D Dichlorotetraaquachromium(II) chloride
Oxidation state of Cr = +3; ligands named alphabetically.
Which complex does NOT show optical isomerism
A [Co(en)₃]³⁺
B [Cr(en)₃]³⁺
C [Co(NH₃)₆]³⁺
D [Fe(en)₃]²⁺
All ligands identical and monodentate → achiral.
Which ligand is weakest field
A CN⁻
B NH₃
C H₂O
D I⁻
I⁻ lies at weak-field end of spectrochemical series.
Which metal ion cannot form coordination compound
A Fe²⁺
B Cu²⁺
C Zn²⁺
D Na⁺
Alkali metal ions rarely form stable coordination complexes.
Which complex has coordination number 2
A [Ag(NH₃)₂]⁺
B [Zn(NH₃)₄]²⁺
C [Ni(CO)₄]
D [Co(NH₃)₆]³⁺
Two NH₃ ligands attached → coordination number 2.
The geometry of coordination number 2 complex is usually
A Linear
B Bent
C Tetrahedral
D Square planar
CN=2 complexes (Ag⁺, Au⁺) usually adopt linear geometry.
Which statement about coordination compounds is correct
A All are colored
B All are paramagnetic
C Transition metals form them due to vacant orbitals
D Only ionic bonding is involved
Presence of vacant d-orbitals allows acceptance of electron pairs from ligands.