AP/gen chem atomic structures and properties

Mass percent formula

X = ((# of atoms)(element atomic mass))/formula weight of compound

Empirical formula steps to solve

Assume 100 gram sample and convert grams to moles

Divide all moles by the smallest mole count of the elements

Coulombs Law equation

F = k(q1q2)/d²

F=force of attraction

K= constant

q = magnitude of charge with a particle - electrons and protons

d = distance

Shielding effect

Electrons farthest away from the nucleus is partially shielded by the inner core electrons due to repulsion

This reduces the electrostatic attraction between the nucleus and the outer electrons

First ionization energy

The minimum amount of energy that is required to remove an outermost, least tightly held, electron from an atom in gas state

First ionization energy periodic trends

As shells count decreases the more energy is require

As valance electrons increase the more energy is required

Electron configuration d’s and f’s knowledge

The d’s are always one shell down, and the f’s are always two shells down

ex: (U) 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s²4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s² 5f³ 6d¹

Effective nuclear charge

Z(eff) = Z - sigma

Z(eff) - the charge experienced by an electron

Z - the actual nuclear charge (atomic number of element)

Sigma - shielding constant (0<sigma<Z)

Repulsive forces caused by shielding effect reduce the effective nuclear charged by outer electrons