The bond formed by complete transfer of electrons is
A Ionic bond
B Covalent bond
C Metallic bond
D Hydrogen bond
In ionic bonding, one atom transfers electron(s) to another forming cations and anions held by electrostatic attraction.
Which factor increases ionic character in a bond
A Small electronegativity difference
B Large electronegativity difference
C Equal electronegativity
D High bond length only
Greater electronegativity difference increases electron transfer tendency, raising ionic character.
According to Fajan’s rule, covalent character increases when
A Cation is large and anion is small
B Cation is small and highly charged
C Anion is small and non-polarizable
D Both ions are large
Small, highly charged cations strongly polarize anions, increasing covalent character.
Which compound has maximum covalent character
A NaCl
B KCl
C MgCl₂
D AlCl₃
Al³⁺ is small and highly charged, causing strong polarization of Cl⁻ → more covalent.
The shape of CH₄ molecule is
A Square planar
B Trigonal planar
C Tetrahedral
D Linear
Carbon in CH₄ is sp³ hybridized; four equivalent bonds arranged tetrahedrally (109.5°).
Hybridization of central atom in BF₃ is
A sp
B sp²
C sp³
D dsp²
Boron forms 3 σ bonds and has no lone pair → trigonal planar geometry → sp².
Shape of NH₃ molecule is
A Tetrahedral
B Trigonal pyramidal
C Trigonal planar
D Linear
N is sp³ with one lone pair; electron geometry tetrahedral, molecular shape trigonal pyramidal.
Bond angle in water (H₂O) is about
A 180°
B 120°
C 109.5°
D 104.5°
O is sp³ with two lone pairs causing stronger repulsion, compressing angle from 109.5° to ~104.5°.
Lone pair–lone pair repulsion is
A Least
B Equal to bond pair–bond pair
C Maximum
D Zero
Lone pairs occupy more space and repel strongly: LP–LP > LP–BP > BP–BP.
The molecular shape of CO₂ is
A Bent
B Tetrahedral
C Trigonal planar
D Linear
Central carbon has two electron domains (two double bonds) → sp hybridization → linear geometry (180°).
Which molecule has sp hybridization on central atom
A NH₃
B BF₃
C BeCl₂
D CH₄
Be forms 2 σ bonds, no lone pair → 2 electron domains → sp hybridization → linear.
Which species is trigonal bipyramidal
A SF₄
B PF₅
C XeF₂
D IF₇
PF₅ has 5 bond pairs, 0 lone pairs → trigonal bipyramidal (sp³d).
Shape of SF₄ is
A Tetrahedral
B Seesaw
C Square planar
D Linear
SF₄ has 4 bond pairs + 1 lone pair (AX₄E) → trigonal bipyramidal electron geometry → seesaw shape.
Shape of XeF₂ is
A Linear
B Bent
C Trigonal planar
D Octahedral
XeF₂ has 2 bond pairs + 3 lone pairs (AX₂E₃). Lone pairs occupy equatorial positions → linear F–Xe–F.
Shape of XeF₄ is
A Tetrahedral
B Square planar
C Trigonal pyramidal
D Seesaw
XeF₄ has 4 bond pairs + 2 lone pairs (AX₄E₂); lone pairs opposite → square planar.
Hybridization in PF₅ is
A sp²
B sp³
C sp³d
D sp³d²
Five electron domains → trigonal bipyramidal → sp³d.
Which has square planar shape
A SF₆
B XeF₄
C NH₃
D PF₅
AX₄E₂ arrangement gives square planar molecular geometry.
The bond order of O₂ molecule (MOT) is
A 1
B 2
C 2.5
D 3
For O₂, bond order = (bonding e⁻ − antibonding e⁻)/2 = 2 → double bond.
The bond order of O₂⁺ is
A 1.5
B 2
C 2.5
D 3
Removing one electron from antibonding π* increases bond order by 0.5 from 2 to 2.5.
Which species has the highest bond order
A O₂
B O₂⁻
C O₂⁺
D O₂²⁻
O₂⁺ has fewer antibonding electrons than O₂, giving higher bond order (2.5).
O₂ is paramagnetic because
A It has no electrons
B It has two unpaired electrons
C It has only paired electrons
D It has ionic bonding
In MOT, O₂ has two unpaired electrons in π* antibonding orbitals → paramagnetism.
Which molecule is diamagnetic
A O₂
B NO
C N₂
D O₂⁻
N₂ has all electrons paired in MO configuration → diamagnetic.
The bond order of N₂ molecule is
A 1
B 2
C 3
D 4
N₂ has bond order 3 (triple bond) from MO filling.
The correct order of bond length is
A N₂ < O₂ < O₂⁻
B O₂⁻ < O₂ < N₂
C O₂ < N₂ < O₂⁻
D N₂ < O₂⁻ < O₂
Higher bond order → shorter bond. N₂ (3) shortest, O₂ (2) next, O₂⁻ (1.5) longest.
The bond order of He₂ is
A 1
B 0
C 0.5
D 2
He₂ has equal bonding and antibonding electrons → bond order 0 → unstable.
The geometry of SF₆ is
A Trigonal bipyramidal
B Octahedral
C Square planar
D Tetrahedral
SF₆ has 6 bond pairs and no lone pair → octahedral (sp³d²).
Hybridization of central atom in SF₆ is
A sp²
B sp³
C sp³d
D sp³d²
Six electron domains correspond to octahedral arrangement → sp³d².
Which statement is correct about σ and π bonds
A π bond is stronger than σ bond always
B σ bond is formed by sidewise overlap
C σ bond allows free rotation more than π bond
D π bond is formed by head-on overlap
σ bonds are cylindrically symmetric; π bonds restrict rotation due to sidewise overlap.
A coordinate bond is formed when
A Both atoms share one electron each
B One atom donates both bonding electrons
C Electrons are fully transferred
D Protons are shared
In coordinate (dative) bond, both electrons come from donor atom.
In NH₄⁺, the bond between N and H formed during ion formation is
A Ionic bond
B Coordinate bond
C Metallic bond
D Hydrogen bond
NH₃ donates lone pair to H⁺ to form NH₄⁺ via coordinate bond (later all N–H become equivalent).
According to Werner’s theory, the ionizable ions are present in
A Primary valency sphere
B Secondary valency sphere
C Coordination sphere only
D Inner sphere only
Primary valency (oxidation state) is ionizable and satisfied by negative ions outside brackets.
Secondary valency of a metal corresponds to
A Atomic number
B Coordination number
C Mass number
D Valence electrons only
Werner’s secondary valency equals number of ligands directly bonded to metal (coordination number).
The coordination number of Co in [Co(NH₃)₆]³⁺ is
A 3
B 4
C 6
D 8
Six NH₃ ligands are directly attached to Co → coordination number 6.
Which ligand is bidentate
A NH₃
B H₂O
C en (ethylenediamine)
D Cl⁻
en has two donor nitrogen atoms, can attach at two sites → bidentate.
A ligand that can donate two pairs of electrons is called
A Monodentate
B Bidentate
C Ambidentate
D Bridging
Bidentate ligands coordinate through two donor atoms, donating two lone pairs.
Which is an ambidentate ligand
A NH₃
B CN⁻
C H₂O
D CO
CN⁻ can coordinate via C or N donor atom → ambidentate.
Oxidation state of Fe in [Fe(CN)₆]⁴⁻ is
A +1
B +2
C +3
D +4
Let oxidation state = x. x + 6(−1) = −4 → x − 6 = −4 → x = +2.
Oxidation state of Co in [Co(NH₃)₅Cl]Cl₂ is
A +1
B +2
C +3
D +4
Two Cl⁻ outside = −2, complex must be +2. Inside: NH₃ neutral, one Cl⁻ = −1. So Co + (−1) = +2 → Co = +3.
IUPAC name of [Co(NH₃)₆]Cl₃ is
A Hexaamminecobalt(III) chloride
B Hexaamminecobalt(II) chloride
C Trichloride hexaamminecobalt
D Cobalt hexaammine trichloride
Complex cation [Co(NH₃)₆]³⁺ → cobalt(III). Counter ions are 3 chloride → “chloride”.
The geometry of coordination number 6 complexes is commonly
A Linear
B Trigonal planar
C Octahedral
D Tetrahedral
Coordination number 6 generally forms octahedral structures (most stable arrangement).
Which complex shows geometrical isomerism
A [Co(NH₃)₆]³⁺
B [Pt(NH₃)₂Cl₂]
C [Zn(NH₃)₄]²⁺
D [Ag(NH₃)₂]⁺
Square planar complexes of type MA₂B₂ show cis/trans isomerism.
[Pt(NH₃)₂Cl₂] has how many geometrical isomers
A 1
B 2
C 3
D 4
Square planar MA₂B₂ gives cis and trans forms → 2 geometrical isomers.
Optical isomerism is shown by
A [Co(NH₃)₆]³⁺
B [Co(en)₃]³⁺
C [Pt(NH₃)₂Cl₂]
D [Ni(CO)₄]
Octahedral complexes with three bidentate ligands are chiral and show optical isomerism (Δ and Λ forms).
Coordination number of Ni in [Ni(CO)₄] is
A 2
B 4
C 6
D 8
Four CO ligands coordinate to Ni → coordination number 4.
The geometry of [Ni(CO)₄] is
A Square planar
B Tetrahedral
C Octahedral
D Linear
Ni(0) with strong-field CO forms sp³ tetrahedral complex.
In CFT, splitting of d-orbitals in octahedral field produces
A Two orbitals lower, three higher
B Three orbitals lower, two higher
C All five equal
D Four lower, one higher
Octahedral splitting: t₂g (dxy, dyz, dxz) lower; e_g (dz², dx²−y²) higher.
In tetrahedral field, the d-orbital splitting is
A Same as octahedral
B Opposite of octahedral and smaller magnitude
C No splitting occurs
D Only one orbital splits
In tetrahedral, e set lower and t₂ higher; splitting Δt is smaller (~4/9 of Δo).
Strong field ligands generally cause
A High spin complexes always
B Low spin complexes in octahedral cases
C No pairing of electrons
D No effect on splitting
Strong field ligands increase Δo, making pairing favorable → low spin.
Which ligand is strongest field among these
A I⁻
B F⁻
C H₂O
D CN⁻
Spectrochemical series places CN⁻ among strong field ligands producing large splitting.
Which complex is most likely low spin (octahedral)
A [FeF₆]³⁻
B [Fe(H₂O)₆]³⁺
C [Fe(CN)₆]³⁻
D [CoF₆]³⁻
CN⁻ is strong field ligand → large Δo → electron pairing → low spin complex.