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
O
O S O
O
FOUR O ATOMS HAVE
THE CHARGE IS -2
6…. 1 X 6 = 6
6…. 4 X 6 = 24
= 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
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,
O
O S O
O
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.
STEP 4: USE THE VSEPR CHART TO DETERMINE THE GEOMETRY
THERE ARE 4 DOMAINS, 4 BONDS ON THE CENTRAL ATOM…TETRAHEDRAL
-1
MOST STABLE MOLECULES
WILL:
1)
2)
3)
4)
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.
O
O S O
O
S ATOM: F.C.= 6-(0 + 4)= +2
-1
A FORMAL CHARGE OF +2 IS LARGE.
-1
-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)
-1
O
O S O
O
S ATOM: F.C.= 6-(0 + 5)= +1
-1
F.C.= 6-(4 + 2)= 0
ALL OXYGEN ATOMS OF THIS
CONFIGURATION ARE 0
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)
-1
O
O S O
O
S ATOM: F.C.= 6-(0 + 6)= +0
F.C.= 6-(4 + 2)= 0
F.C.= 6-(4 + 2)= 0
ALL OXYGEN ATOMS OF THIS
CONFIGURATION ARE 0
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)
0
O
O S O
O
S ATOM: F.C.= 6-(0 + 7)= -1
F.C.= 6-(4 + 2)= 0
-1
F.C.= 6-(4 + 2)= 0
ALL OXYGEN ATOMS OF THIS
CONFIGURATION ARE 0
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
POLARITY: USE IONIC CHARACTER AND ELECTRONEGATIVITY.
-ASSIGN ELECTRONEGATIVITIES FROM TABLE S (OR RELATIVE POSITION ON PERIODIC
TABLE).
O
O S O
O
3.4
2.6
3.4
3.4
3.4
THE IONIC CHARATER IS /2.6 – 3.4/ = 0.8 (0 TO 1.7 IONIC CHARACTER IS COVALENT,
ABOVE 0 IS POLAR, (0.0-0.4 IS WEAKLY POLAR)
POLARITY(CONTINUED): USE IONIC CHARACTER AND ELECTRONEGATIVITY.
DRAW DIPOLE ARROWS POINTNG TO THE MORE ELECTRONEGATIVE ATOM IN EACH
BOND.
O
O S O
O
3.4
ð3.4
2.6
ð+
ð-
3.4
ð-
ð- 3.4
PLACE THE NEGATIVE POLE SYMBOL (ð-) NEAR THE MOST ELECTRONEGATIVE ATOM,
AND THE POSITIVE POLE SYMBOL (ð+) NEAR THE LEAST ELECTRONEGATIVE ELEMNT IN
EACH BOND.
GEOMETRY: THE CENTRAL ATOM HAS 4 ELECTRON DOMAINS, ALL BONDS. ACCORDING
TO VSEPR THIS IS THE TETRAHEDRAL FAMILY AND IS A TETRAHEDRON. THIS IS A
SYMMETRICAL GEOMETRY WHEN ALL LIGANDS ARE IDENTICAL AND THEREFORE
CANNOT BE POLAR (SNAP..SYMETRICAL NONPOLAR ASYMETRICAL POLAR.)
O
O S O
O
3.4
ð3.4
2.6
ð+
ð-
3.4
ð-
ð- 3.4
NOTE: FOR VSEPR MULTIPLE BONDS COUT AS ONE DOMAIN.THIS IS AN EXAMPLE OF A
NON POLAR MOLECULE WITH POLAR BONDS.