why do electrons occupy the 4s orbital first before the 3d orbital?
4s orbital is at a lower energy level than the 3d orbital
explain the electronic configuration of Cr
instead of 3d^4 4s^2 is 3d^5 4s^1
by the time Cr is reached, 3d and 4s orbitals are about equal in energy
by having one electron each in the 3d and 4s orbital, inter electronic repulsion is minimised,
hence 3d^5 4s^1 more stable
explain the electronic configuration of Cu
instead of 3d^9 4s^2 is 3d^10 4s^1
the fully filled 3d subshell is unusually stable due to symmetrical charge distribution around the metal centre,
hence 3d^10 4s^1 more stable
why does atomic radius decrease across a period
across period,
increasing protons so greater nuclear charge but e are added to the valence shell so SE remains constant, hence ENC increases valence e more tightly held, smaller in size
why does atomic radius increase down a group
down group,
- no. of e shells increases
- distance between nucleus and valence e increases
- shielding effect increases
- despite increasing nuclear charge,
- electrostatic attraction between nucleus and valence e decreases, resulting in increase in size of e cloud
why cations smaller than atoms
cations have 1 less occupied e shell
- distance between nucleus and valence e decreases
why anions bigger than atoms
anions have more electrons, inter electronic repulsion increases
explain ionic radius of isoelectronic ions across a period
decreases across period, (Na+ to Si4+, P3- to Cl-)
- no. of protons increases, hence nuclear charge increases
- valence e experience same shielding effect
- ENC increases
- electrostatic attraction between nucleus and valence e increases, resulting in decrease in size of e cloud
define first ionisation energy
the energy required to remove 1 mole of electrons from 1 mole of gaseous M atoms to form 1 mole of gaseous M+ ions
is ionisation energy endo or exo
endo, they are alw positive as they absorb energy to overcome attraction between electron and the nucleus
explain general trend of IE across a period (1/radius)
generally increase across period,
- no. of e shells remain the same
- no. of protons increases hence nuclear charge increases
- shielding effect constant, ENC increases
- electrostatic attraction between nucleus and valence e increases, hence requiring more energy to remove valence e
explain the 2 irregularities of IE across period
- grp 2 & grp 13 (Al
explain general trend of IE down a grp
decrease down a grp,
- no. of e shells increases
- distance between valence e and nucleus increases
- shielding effect increases
- despite increasing nuclear charge, electrostatic attraction between nucleus and valence e decreases, hence less energy required to remove valence e
explain trend in successive IE of an element
increases,
once the first e removed from neutral atom, each successive e removed from an ion of increasing positive charge which attracts the e more strongly
for graphs of IE
highest IE: grp 1
lowest IE:grp 2
from there you count the group
define EN
is the ability of an atom to attract shared electrons towards itself
explain trend of EN across a period (1/radius)
increases,
- no. of e shells remain the same
- no. of protons increases hence nuclear charge increases
- shielding effect constant, ENC increases
- electrostatic attraction between nucleus and shared e increases
explain trend of EN down a grp (1/radius)
decreases,
- no. of e shells increases
- distance between nucleus and shared e increases
- shielding effect increases
- despite increasing nuclear charge, electrostatic attraction between nucleus and shared e decreases
