O O S O O STEP 1(CALCULATE VALENCE ELECTRONS IN MOLECULAR ORBITAL. RENDER THE LEWIS DOT STRUCTURE OF THE SO4 2- POLYATOMIC ION, THE BONDS ARE COVALENT. STEP 1(CALCULATE VALENCE ELECTRONS IN MOLECULAR ORBITAL. ONE S HAS 6…. 1 X 6 = 6 FOUR O ATOMS HAVE 6…. 4 X 6 = 24 THE CHARGE IS -2 = 2 32 (TOTAL) STEP 2, CONNECT THE LIGANDS (THE OXYGENS) WITH SINGLE BONDS INITIALLY, THEN DRAW THE STRUCTURE WITH THE CENTRAL ATOM BEING THE LEAST ELECTRONEGATIVE NON HYDROGEN. O O S O O

O O S O O STEP 2 CONTINUED: SUBTRACT 2 ELECTRONS FOR EACH SINGLE BOND FROM THE TOTAL VALENCE ELECTRONS. 4 BONDS AT 2 EACH IS 8 ELECTRONS. 32 – 8 = 24 ELECTRONS REMIN TO BE PLACED

STEP 3. ADD 6 MORE ELECTRONS TO COMPLETE THE OCTET OF EACH OXYGEN, STEP 3 CONTINUED: SUBTRACT THE VALENCE ELECTRONS FROM THE RUNNING TOTAL. 24 –(4 X 6), 24-24=0 ALL YOUR ELECTONS ARE PLACED AS OF NOW.

O O S O O STEP 4: USE THE VSEPR CHART TO DETERMINE THE GEOMETRY THERE ARE 4 DOMAINS, 4 BONDS ON THE CENTRAL ATOM…TETRAHEDRAL O -1 MOST STABLE MOLECULES WILL: HAVE ATOMS WITH THE SMALLEST FORMAL CHARGES, NEAT 0. NEGATIVE FORMAL CHARGES RESIDE ON THE MOST ELECTRONEGETIVE ATOMS. ATOMS DIRECTLY BONDED DO NOT HAVE THE SAME SIGN OF FORMAL CHARGE. THE SUM OF THE FORMAL CHARGE ADDS UP TO THE IONIC CHARGE. S ATOM: F.C.= 6-(0 + 4)= +2 A FORMAL CHARGE OF +2 IS LARGE. O S O -1 -1 O -1 F.C.= 6-(6 + 1)= -1 ALL OXYGENS OF THIS CONFIGURATION ARE -1

STEP 5: BEGIN TO BACK BOND TO FORM DOUBLE BONDS, USE FORMAL CHARGE TO CHOSE THE BEST RENDITION. FORMAL CHARGE = VALENCE ELECTRONS – (UNBONDED ELECTRONS – BONDS) O -1 S ATOM: F.C.= 6-(0 + 5)= +1 O S O -1 F.C.= 6-(4 + 2)= 0 ALL OXYGEN ATOMS OF THIS CONFIGURATION ARE 0 O THIS SEEMS MORE STABE THAN THE ONE ON THE LAST SLIDE, THE S ATOM HAS A SMALLER FORMAL CHARGE MAGNITUDE AND ONE OF THE OZYGENS IS 0. -1 F.C.= 6-(6 + 1)= -1 ALL OXYGENS OF THIS CONFIGURATION ARE -1

STEP 5: BEGIN TO BACK BOND TO FORM DOUBLE BONDS, USE FORMAL CHARGE TO CHOSE THE BEST RENDITION. FORMAL CHARGE = VALENCE ELECTRONS – (UNBONDED ELECTRONS – BONDS) O -1 S ATOM: F.C.= 6-(0 + 6)= +0 O S O F.C.= 6-(4 + 2)= 0 F.C.= 6-(4 + 2)= 0 ALL OXYGEN ATOMS OF THIS CONFIGURATION ARE 0 O THIS SEEMS MORE STABE THAN THE EITHER OF THE FORMER STRUCTURES, THE S ATOM IS NOW 0, AND 2 OF THE OXYGENS ARE 0. -1 F.C.= 6-(6 + 1)= -1 ALL OXYGENS OF THIS CONFIGURATION ARE -1

STEP 5: BEGIN TO BACK BOND TO FORM DOUBLE BONDS, USE FORMAL CHARGE TO CHOSE THE BEST RENDITION. REMEMBER…MORE BONDS MORE STABLE. FORMAL CHARGE = VALENCE ELECTRONS – (UNBONDED ELECTRONS – BONDS) O S ATOM: F.C.= 6-(0 + 7)= -1 O S O F.C.= 6-(4 + 2)= 0 -1 F.C.= 6-(4 + 2)= 0 ALL OXYGEN ATOMS OF THIS CONFIGURATION ARE 0 O THIS ONE IS REJECTED DUE TO DIRECTLY BONDED ATOMS HAVING THE SAME SIGN OF FORMAL CHARGE. THE S ATOM, LEAST ELECTRONEGATIVE HAS A – FORMAL CHARGE…UNSTABLE. -1 F.C.= 6-(6 + 1)= -1 ALL OXYGENS OF THIS CONFIGURATION ARE -1