Electronegativity & Intermolecular Forces EXAM QUESTIONS Flashcards
Explain what is meant by a ‘polar’ molecule. [1 mark]
A molecule with a difference in electronegativities which forms polar bonds.
The shape must be assymetrical.
Explain what is meant by a ‘non-polar’ molecule [1 mark]
A molecule with atoms with the same/similar electronegativities.
OR
A symmetric molecule that contains a difference in electronegativities, but the opposite direction and equal magnitude of polar bonds cancel out.
Explain what is meant by ‘electronegativity’ [2 marks]
The ability of an atom to attract a SHAIRED PAIR of electrons (1)
in a COVALENT bond. (1)
State and explain the trend in electronegativity values across period 3 from sodium to chlorine. [3 marks]
Trend/electronegativity increases (1)
Nuclear charge (number of protons) increases (1)
Electrons pulled closer to nucleus - decreases size of atom despite these atoms having the SAME number of shells. (1)
The BILLION dollar question of the century…
Are we breaking covalent bonds or intermolecular forces?
> IN SIMPLE COVALENT MOLECULES =
breaking intermolecular forces (not the bonds)
> IN GIANT COVALENT STRUCTURES =
breaking covalent bonds (because diamond is not a molecule, and are only held together by bonds)
Explain, in terms of electronegativity, how an ionic bond forms between sodium and iodine. [2 marks]
Iodine is more electronegative than sodium so…
it attracts the shared pair of electrons so much that it gains the electron to form an anion. (1)
The sodium atom has lost electrons and forms a cation. (1)
State the 3 factors affecting electronegativity [1 mark]
Nuclear charge
Atomic radius
Shielding
Explain which atom, fluorine or chlorine is more electronegative and why. [3 marks]
Fluorine (1)
Chlorine has a higher nuclear charge. (1)
However, F has a lower shielding effect, meaning electron pair in covalent bond are more attracted to nucleus. (1)
Explain why atomic radius decreases across a period. [4 marks]
Outer electrons are in the same shell for these atoms. (1)
Across period, more protons in nucleus, higher nuclear charge. (1)
So stronger attraction between nucleus and valence electrons. (1)
So valence electrons are pulled closer to nucleus. (1)
Therefore, atomic radius decreases.
Explain why the atomic radius increases down the group. [2 marks]
Outer electrons are in a shell further away from nucleus. (1)
So weaker attraction between nucleus and valence electrons. (1)
Compare the electronegativity of carbon and oxygen. [4 marks]
Oxygen is more electronegative than carbon. (1)
There is same level of electron shielding in O and C, there is 1 full shell of electrons between nucleus and bonding pair of electrons. (1)
However, Oxygen (8p+) has greater nuclear charge than carbon (6p+) (1)
and valence electrons are pulled closer to nucleus - smaller size.
State the Valence Shell Electron Pair Repulsion Theory (VSEPR) [1 mark]
Electron pairs ARRANGE themselves to MINIMISE REPULSION effects from one another.
What is the geometry of methane? (CH₄)
Draw the shape.
State any bond angles. [ 5 marks]
ELECTRON DENSITY:
There are 4 areas of ELECTRON DENSITY around CENTRAL carbon atom. (1)
BP/LP:
There are 0 lone pairs and 4 bonding pairs around the central atom. (1)
Electrons REPEL to be as far apart as possible and lone pairs repel MORE than bonding pairs. (1)
SHAPE:
Methane therefore has a TETRAHEDRAL shape. (1)
BOND ANGLE:
The bond ANGLES between the bonded pairs are 109.5° (1)
What is the success criteria for answering this type of question :
What is the geometry of ________ ? [5 marks]
1) State no. of areas of electron density around central atom.
2) State no. of lone & bonding pairs on central atom.
3) STATE “Electrons repel to be as far apart as possible, and lone pairs repel more than bonding pairs”.
4) State name of shape.
5) State bond angle of shape.
Describe the effect of lone pairs on the shape of a molecule. [2 marks]
Electrons repel as far apart as possible, and lone pairs repel more compared to bonded pairs, and decreases the bond angle by 2.5° each. (1)
There are E.G. 2 lone pairs, therefore the bond angle 109.5° decreases by 2.5° each, to 104.5° . (1)