chemistry quiz
Draw the Lewis structure of CO2 and predict its shape and bond angle.
linear and 180°; 2 Accept crosses, lines or dots as electron pairs.
Describe the bonding in iron and explain the electrical conductivity and malleability of the metal.
metallic (bonding); positive ions/cations and delocalized/sea of electrons; electrostatic attraction between the two; Award [2 max] for description of bonding Conductivity: electrons delocalized/free to move; Malleability: atoms/ions/cations can move without breaking bonds / atoms/ions/ cations can slide past each other;
Describe the structure and bonding in silicon dioxide.
network/giant lattice / macromolecular / repeating tetrahedral units; Bonding: (single) covalent (bonds);
Describe the structure and bonding in SiO2.
network/giant structure / macromolecular; each Si atom bonded covalently to 4 oxygen atoms and each O atom bonded covalently to 2 Si atoms / single covalent bonds;
Predict and explain how the hardness and electrical conductivity of C60 fullerene would compare with that of diamond and graphite.
softer than diamond/harder than graphite; as C60 molecules can move over each other; conducts better than diamond/worse than graphite; as C60 has less delocalisation (of the unpaired bonding electrons) than graphite;
Draw the Lewis structure of NH3, state its shape and deduce and explain the H-N-H bond angle in NH3.
trigonal/triangular pyramidal; (~)107° / less than 109.5°; Do not allow ECF. LP-BP repulsion > BP-BP repulsion / one lone pair and three bond pairs / lone pairs/non-bonding pairs repel more than bonding-pairs;
Predict the shape and bond angles for the following species: (i) CO2 (ii) CO3 2- (iii) BF4
(CO2) linear; 180° (ii) (CO3 2- ) triangular planar; 120° (iii) (BF4 - ) tetrahedral; 109
Explain the term electronegativity.
(relative) measure of an atom's attraction for electrons; in a bond;
State the H-N-H bond angle in an ammonia molecule.
107°;
State the H-N-H bond angle in an ammonium ion.
109.5
Ammonia, NH3, is a weak base. (i) Draw the Lewis structure of ammonia and state the shape of the molecule and its bond angles. (ii) The conjugate acid of ammonia is the ammonium ion, NH4 + . Draw the Lewis structure of the ammonium ion and deduce its shape and bond angles.
Accept any combination of dots/crosses and lines to represent electron pairs. (trigonal/triangular) pyramid; Allow 3D representation using wedges and dotted bonds of trigonal pyramidal molecule. 107°; 3 Accept any angle between 105o and 108.5o. No ECF for shape based on incorrect Lewis structure. (ii) Charge needed for mark. tetrahedral; Allow a 3D representation using wedges and dotted bonds of tetrahedral molecule. 109.5°/109°/109° 28';
Predict and explain which of the following compounds consist of molecules: NaCl, BF3, CaCl2, N2O, P4O6, FeS and CBr4.
BF3, N2O, P4O6 and CBr4; Non-metals only / small difference in electronegativity values of the elements;
Predict and explain which of the following compounds consist of molecules: NaCl, BF3, CaCl2, N2O, P4O6, FeS and CBr4.
BF3, N2O, P4O6 and CBr4; Non-metals only/small difference in electronegativity values of the elements;
Describe and compare three features of the structure and bonding in the three allotropes of carbon: diamond, graphite and C60 fullerene.
Bonding Graphite and C60 fullerene: covalent bonds and van der Waals'/London/ dispersion forces; Diamond: covalent bonds (and van der Waals'/London/dispersion forces); Delocalized electrons Graphite and C60 fullerene: delocalized electrons; Diamond: no delocalized electrons; Structure Diamond: network/giant structure / macromolecular / three-dimensional structure and Graphite: layered structure / two-dimensional structure / planar; C60 fullerene: consists of molecules / spheres made of atoms arranged in hexagons/pentagons; Bond angles Graphite: 120° and Diamond: 109°; C60 fullerene: bond angles between 109-120°; Allow Graphite: sp 2 and Diamond: sp 3 . Allow C60 fullerene: sp 2 and sp 3 . Number of atoms each carbon is bonded to Graphite and C60 fullerene: each C atom attached to 3 others; Diamond: each C atom attached to 4 atoms / tetrahedral arrangement of C (atoms);
Using Table 7 of the Data Booklet, predict and explain which of the bonds O-H, O-N or N-H would be most polar.
C2H4: is non-polar and explanation either using a diagram or in words, involving no net dipole moment;
Draw Lewis (electron dot) structures for CO2 and H2S showing all valence electrons.
CO2 is linear; two charge centres or bonds and no lone pairs (around C); H2S is bent/v-shaped/angular; two bond pairs, two lone pairs (around S); 4 (iii) CO2 is non-polar, H2S is polar; bond polarities cancel CO2 but not in H2S;
Predict and explain whether each species is polar.
H3O + : is polar and explanation either using a diagram or in words, involving the net dipole moment;
Use the VSEPR theory to deduce the shape of H3O + and C2H4. For each species, draw the Lewis structure, name the shape, and state the value of the bond angle(s). Predict and explain whether each species is polar.
H3O + Trigonal pyramidal; 109 s polar and explanation either using a diagram or in words, involving the net dipole moment C2H4 Trigonal planar; 120°; H3O + : ; C2H4: is non-polar and explanation either using a diagram or in words, involving no net dipole moment;
Use the VSEPR theory to deduce the shape of H3O + and C2H4. For each species, draw the Lewis structure, name the shape, and state the value of the bond angle(s).
H3O + Trigonal/triangular pyramidal; Allow values in the range 106° to 109.5°; C2H4 Trigonal/triangular planar; Allow values of approximately 120°;
Explain why the H-N-H bond angle of NH3 is different from the H-N-H bond angle of NH4 + ; referring to both species in your answer.
NH4 + has four bonding pairs (around central atom so is a regular tetrahedron); NH3 has three bonding pairs (of electrons) and one non-bonding pair; non-bonding pairs (of electrons) exert a greater repulsive force;
Using Table 7 of the Data Booklet, predict and explain which of the bonds O-H, O-N or N-H would be most polar.
O-H is most polar; O-H has greatest difference between electronegativities/calculation showing values of 1.4, 0.5 and 0.9 respectively;
Deduce whether or not each molecule is polar, giving a reason for your answer.
SCl2 is polar; C2Cl2 is non-polar; 3 No net dipole movement for C2Cl2 but angular SCl2 has a resultant dipole
Use the VSEPR theory to predict and explain the shape and the bond angle of each of the molecules SCl2 and C2Cl2
SCl2 two bonding pairs, two non-bonding pairs; angular/bent/non-linear/V-shaped; . 90° < angle < 107°; C2Cl2 two charge centres around each C; linear; angle = 180°;
State the type of bonding in the compound SiCl4. Draw the Lewis structure for this compound.
Si—Cl bonds are covalent;
Draw a diagram to show hydrogen bonding between two molecules of NH3. The diagram should include any dipoles and/or lone pairs of electrons
a
Explain why the ammonia molecule is polar.
asymmetrical
Draw the Lewis structure of NCl3. Predict, giving a reason, the Cl - N - Cl bond angle in NCl3.
bond angle: 107°-109° greater repulsion between lone pair and bonding pairs
Predict and explain the order of the melting point for propanol, butane and propanone with reference to their intermolecular forces.
butane < propanone < propanol; butane has van der Waals' forces; . propanone has dipole-dipole attractions; propanol has (the stronger) H-bonding;
Outline the principles of the valence shell electron pair repulsion (VSEPR) theory.
electron pairs/charge centres (in valence shell) of central atom repel each other; to positions of minimum energy/repulsion/maximum stability; pairs forming a double or triple bond act as a single bond; non-bonding pairs repel more than bonding pairs
Diamond, graphite and C60 fullerene are three allotropes of carbon. (i) Describe the structure of each allotrope. (ii) Compare the bonding in diamond and graphite.
Diamond 3D array/network involving tetrahedral carbons / each carbon atom joined to four others; Graphite layer structure involving trigonal (triangular) planar carbons / with each carbon atom joined to three others / with hexagonal (six-membered) rings of carbon atoms; C60 fullerene truncated icosahedrons; Accept carbon atoms form a "ball" with 32 faces, of which 12 are pentagons and 20 are hexagons, exactly like a soccer ball. Do not accept soccer ball alone.
Diamond, graphite and C60 fullerene are three allotropes of carbon. (i) Describe the structure of each allotrope.
Diamond 3D array/network involving tetrahedral carbons/each carbon atom joined to four others; Graphite layer structure involving trigonal (triangular) planar carbons/with each carbon atom joined to three others/with hexagonal (six-membered) rings of carbon atoms; C60 fullerene truncated icosahedrons; Accept carbon atoms form a 'ball' with 32 faces, of which 12 are pentagons and 20 are hexagons, exactly like a soccer ball. Do not accept soccer ball alone.
Compare and explain the hardness and electrical conductivity of diamond and graphite.
Diamond Graphite poor/non-conductor good conductor no delocalized electrons delocalized electrons hard soft rigid structure layers can slide
State the structures of and the bonding in diamond and graphite.
Diamond giant molecular/macromolecular/3-D covalent bonds only; Graphite covalent bonds and van der Waals' forces layer structure;
Compare the bonding in diamond and graphite.
Diamond: covalent bonds (only); Graphite: covalent bonds and the separated layers held together by (weak) London/van der Waals'/dispersion forces;
(i) Outline the principles of the valence shell electron pair repulsion (VSEPR) theory.
Find number of electron pairs/charge centres in (valence shell of) central atom; electron pairs/charge centres (in valence shell) of central atom repel each other; to positions of minimum energy/repulsion / maximum stability; pairs forming a double or triple bond act as a single bond; non-bonding pairs repel more than bonding pairs
