topic 1: periodicity Flashcards
classification
An element is classified as s, p, d or f block according to its position in the Periodic Table, which is determined by its proton number.
periodicity
Periodicity is the characteristics of elements showing a REPEATING pattern and recurring variation with increasing atomic number.
trends for period 3
-melting point
-atomic radius
-ionisation energy
what is the structure and melting point for Na, Mg and Al?
-Na, Mg and Al exist as GIANT METALLIC STRUCTURES
-They increase due to MORE OUTER SHELL ELECTRONS available
-as the CHARGE on the ion INCREASES and these join the ‘sea’ of delocalised electrons so a bigger charge on the ion produces a BIGGER ATTRACTION which increases the melting point.
what is the structure and melting point for Si?
Si exists as GIANT COVALENT MOLECULE (macromolecular structure) and appears at the peak of the graph as it has a high boiling points due to STRONG COVALENT BONDING
what is the structure and melting point for P, S and Cl?
-P, S and Cl exist as SIMPLE MOLECULES and Ar is a MONATOMIC SUBSTANCES
-These elements’ standard states formula of P4, S8, Cl2 and Ar
-These substances have WEAK VAN DER WAAL FORCES between molecules which results in a LOW MELTING POINT
-The size of the van der waals depends on the molecules involved
-Sulfur forms BIGGER MOLECULES as it is S8, therefore having STRONGER VAN DER WAALS FORCES and therefore MORE ENERGY is needed to overcome them
trend in atomic radius in period 3
ATOMIC RADIUS DECREASES ACROSS PERIOD 3
-As we move along the period we are adding protons to the nucleus and electrons to the third shell.
-The charge on the nucleus is increasing from 11+ to 17+.
-This increased nuclear charge pulls the electrons in closer to the nucleus
(There are no additional electron shells to provide more shielding).
-so the size of atom decreases across the period
trends in first ionisation energy in period 3
-GENERALLY INCREASE ACROSS A PERIOD - with exceptions of group 2-3 and 5-6
-Higher nuclear charge, smaller atomic radius.
-Outer electrons are MORE ATTRACTED to the nucleus.
-Shielding is CONSTANT
-DECREASE DOWN A GROUP
-Larger atomic radius
-Increase in shielding as MORE ENERGY LEVELS are added.
-Outer electron is LESS ATTRACTED to the nucleus and is therefore easier to lose
exception in the dip between group 2 and 3
-Dip between group 2 and 3:
-Going from S orbital into P orbital
-Higher energy shell
-This is further away from the nucleus, so increased shielding
expection in the dip between group 5 and 6:
-In group 5, p subshell has ONE ELECTRON in each orbital, NO REPULSION
-In group 6, p subshell has one orbital with 2 ELECTRONS IN
-The electrons in the same orbital REPEL each other, so LESS ENERGY is needed to remove the outer electron
