Covalent Bond Model
Basics of Chemical Bond
CHM 001
Oluwakemi A. Oloba-Whenu, Ph. D
147A Faculty of Science
Covalent Bond
• Formed by sharing electrons
• More common
• A bond where electrons from each atom are shared
•
•
•
•
Each covalent bond has 2 electrons that are shared.
Only the Valence electrons are involved in these covalent bonds.
Between nonmetallic elements of similar electronegativity.
The atoms involved will share sufficient numbers of electrons in order to
achieve a octet or noble gas electron configuration (that is, eight valence
electrons).
• Examples; O2, CO2, C2H6, H2O, SiC
Covalent Bonding
• There are several electrostatic
interactions in these bonds:
– attractions between electrons and nuclei,
– repulsions between electrons, and
– repulsions between nuclei.
• For a bond to form, the attractions must
be greater than the repulsions.
Covalent Bonding
• There are several electrostatic
interactions in these bonds:
– attractions between electrons and nuclei,
– repulsions between electrons, and
– repulsions between nuclei.
• For a bond to form, the attractions must
be greater than the repulsions.
Formation
• between atoms of the same element
N2, O2, diamond,graphite
• between atoms of different elements
CO2, SO2 on the RHS of
the table
• when one of the elements is in the middle of the table; CCl4, SiCl4
• with head-of-the-group elements with high ionization energies;
BeCl2
Covalent bond and Lewis structure
• The shared electrons in H2 spend part of the time in the region around each
atom.
• In this sense, each atom in H2 has a helium configuration
• The formation of a bond between H and Cl to give an HCl molecule can be
represented in a similar way.
: :
: :
. + .Cl :
: :
H
H Cl
• Thus, hydrogen has two valence electrons about it (as in He) and Cl has eight
valence electrons about it (as in Ar)
Covalent bonding
• Covalently bound species are different from their ionic counterparts
• exist as individual, discrete species (vs. 3-D crystal lattice structure for
ionic)
• tend to exhibit much lower melting and boiling points (vs. ionic)
HYDROGEN
H
H
both atoms need one electron
to complete their outer shell
DOT AND
CROSS
DIAGRAM
H H
H
H
atoms share a pair of electrons
to form a single covalent bond
H
H
FLUORINE
F
F
both atoms need one electron
to complete their outer shell
DOT AND
CROSS
DIAGRAM
F F
F
F
atoms share a pair of electrons
to form a single covalent bond
F
F
FLUORINE
H
F
both atoms need one electron
to complete their outer shell
DOT AND
CROSS
DIAGRAM
H F
H
F
atoms share a pair of electrons
to form a single covalent bond
H
F
Covalent bond and Lewis structure
• An electron pair that is
shared is called a bonding
pair or bond pair
• An electron pair localized
on an atom a lone pair (an
electron pair that is not
shared).
METHANE
H
H
H
H
C
H
atom needs four
electrons to complete
its outer shell
DOT AND
CROSS
DIAGRAM
H
each atom needs one
electron to complete
its outer shell
C
H
H
Carbon shares all 4 of its
electrons to form 4 single
covalent bonds
H
H
H C H
H C H
H
H
AMMONIA
H
H
H
N
H
atom needs three
electrons to complete
its outer shell
N
H
H
each atom needs one
electron to complete
its outer shell
Nitrogen can only share 3 of
its 5 electrons otherwise it will
exceed the maximum of 8
A LONE PAIR REMAINS
H N H
H
H
N
H
H
WATER
H
H
O
H
atom needs two
electrons to complete
its outer shell
O
H
each atom needs one
electron to complete
its outer shell
Oxygen can only share 2 of its
6 electrons otherwise it will
exceed the maximum of 8
TWO LONE PAIRS REMAIN
H O
H
H
O
H
Multiple bond
• A shared pair is a single bond
and is represented by a line
• If more than an electron pair is
shared then there is more
than one bond
• Two pair are called double
bonds
O
O
O O
each atom needs
two electrons to
complete its outer
shell
O
each oxygen
shares 2 of its
electrons to form a
O
DOUBLE
COVALENT BOND
C
H
:
H
C
:::
• If more than an electron pair is
shared then there is more
than one bond
• Two pair are called double
bonds
• Three pairs single bonds
H
: :C
C
:
Multiple bond
H
H
H
or
H
C
C
H
H or H
H
C
C
H
Multiple Bonds
single bond – 2 electrons (1 pair of electrons) are
shared between 2 atoms
double bond – 4 electrons (2 pairs of electrons) are
shared between 2 atoms
triple bond – 6 electrons (3 pairs of electrons) are
shared between 2 atoms
Multiple
bond
• The more the
number of electron
pairs the shorter
and stronger the
bond
• Bond Order Number of electron
pairs that are
shared between
two atoms
Exercise
• Write Lewis structures for the following covalent molecules
• then write their structural formulae using — to represent a shared
pair of electrons:
• (i) H2
(ii) CH4
• (iii) SiCl4
(iv) OCl2
(vii) PCl3
Polar and non-polar covalent bonds
• When the atoms are alike, as in the H-H bond of H2 , the bonding
electrons are shared equally (a nonpolar covalent bond).
• When the two atoms are of different elements, the bonding electrons
need not be shared equally, resulting in a “polar” bond.
Polar covalent bond
• A polar covalent bond is one in which
the bonding electrons spend more time
near one of the two atoms involved
• It is as a result of unequal sharing of
electron pair
• Fluorine pulls harder on the shared
electrons than hydrogen does.
• Therefore, the fluorine end has more
electron density than the hydrogen end.
• Due to electronegativity difference
• The end with the larger electron density
gets a partial negative charge and the end
that is electron deficient gets a partial
positive charge.
Polar covalent bond-Electronegativity
• Electronegativity is a measure of the
ability of an atom in a molecule to draw
bonding electrons to itself.
• In general, electronegativity increases
from the lower-left corner to the
upper-right corner of the periodic
table.
• The current electronegativity scale,
developed by Linus Pauling, assigns a
value of 4.0 to fluorine and a value of
0.7 to cesium.
Polar covalent bond
Electronegativity
values have no
units
Arrange the following in order of
increasing electronegativity: Na, F, O, K,
Al, Si, Mg
Polar Covalent Bonds – dipole moment
• When two atoms share electrons unequally, a bond dipole
results.
• Recall that un equal sharing results in partial charges
• A bond dipole is a measure of separation of electronic
charge between two atoms bonded together.
• The size of a dipole is measured by its dipole moment ( 𝜇)
• The dipole moment, , produced by two equal but opposite
charges separated by a distance, 𝑟, is calculated as
𝜇 = 𝑄𝑟
• It is a quantitative measure of the magnitude of a
dipole
• It is measured in debyes (D).
• It increases as the magnitude of Q increases and as r
increases
Polar Covalent Bonds-dipole moment
• The greater the
difference in
electronegativity,
the more polar is
the bond.
• The higher the
dipole moment
Electronegativity and types of bond
• The absolute value of the difference in electronegativity of two bonded
atoms gives a rough measure of the polarity of the bond.
• When this difference is small (less than 0.5), the bond is nonpolar.
• When this difference is large (greater than 0.5), the bond is considered
polar.
• If the difference exceeds approximately 1.8, sharing of electrons is no
longer possible and the bond becomes ionic.
Rank the following in order of
increasing bond polarity:
H─F H─Br F─F
a)H─F < H─Br < F─F < Na─Cl
b)F─F < H─F < H─Br < Na─Cl
c)H─Br < H─F < F─F < Na─Cl
d)F─F < H─Br < H─F < Na─Cl
e)Na─Cl < H─F < H─Br < F─F
Na─Cl
Exercise
• Work out the differences in electronegativities between
the elements in the following covalent bonds, rank them
in order of increasing polarity, and show the direction of
each dipole:
• (i) Be—H
(ii) O—H
• (iii) C—H
(iv) C—Cl
• (v) C—O.
POLAR MOLECULES
Occurrence
HYDROGEN CHLORIDE
NET DIPOLE - POLAR
not all molecules containing polar bonds are polar overall
if bond dipoles ‘cancel each other’ the molecule isn’t polar
if there is a ‘net dipole’ the molecule will be polar
TETRACHLOROMETHANE
WATER
NET DIPOLE - POLAR
NON-POLAR
Lewis structure for molecules
• The Lewis electron-dot formula of a covalent compound is a simple
two-dimensional representation of the positions of electrons in a
molecule.
• Bonding electron pairs are indicated by either two dots or a dash.
• In addition, these formulas show the positions of lone pairs of
electrons
Drawing Lewis Structures
1.
2.
3.
4.
5.
Sum the valence electrons from all atoms. Add one electron to the total for each negative
charge for an anion and subtract one electron from the total for each positive charge for a
cation.
Write the symbols for the atoms, show which atoms are attached to which, and connect them
with a single bond. (Chemical formulas are often written in the order in which the atoms are
connected in the molecule or ion). For example, the formula HCN indicates that the carbon
atom is bonded to the H and to the N. In many polyatomic molecules and ions, the central
atom is usually written first, as in and SF4. Remember in most cases, that the central atom is
generally less electronegative than the atoms surrounding it.
Complete the octets around all the atoms bonded to the central atom. Note: a hydrogen atom
has only a single pair of electrons around it.
Place any leftover electrons on the central atom, even if doing so results in more than an octet
of electrons around the atom.
If there are not enough electrons to give the central atom an octet, try multiple bonds. Use one
or more of the unshared pairs of electrons on the atoms bonded to the central atom to form
double or triple bonds. Halogens and hydrogen do not take part in multiple bond
Drawing Lewis Structures
1. Consider the structure of PCl3,
phosphorus trichloride
2. Total of 5(𝑃) + 3 × 7 𝐶𝑙 = 26
• Write the symbols for the atoms, show
which atoms are attached to which,
and connect them with a single
(Heavier element or The least
electronegative atom is often the
central element)
Drawing Lewis Structures
1. Consider the structure of PCl3,
phosphorus trichloride
2. Total of 5(𝑃) + 3 × 7 𝐶𝑙 = 26
3. Write the symbols for the atoms, show
which atoms are attached to which, and
connect them with a single
4. Complete the octets around all the atoms
bonded to the central atom.
5. Place any leftover electrons on the central
atom, even if doing so results in more
than an octet of electrons around the
atom.
Writing Lewis Dot Formulas
• Consider the covalent compound 𝑆𝐶𝑙2.
• Total electrons: 6 𝑆 + 2 × 7 𝐶𝑙 = 20𝑒 −
• Sketch structure with single bonds
• 4 electrons gone for bonds, 20 − 4 = 16𝑒 − 𝑙𝑒𝑓𝑡
• Complete octet of terminal atoms 12 gone 16 − 12 = 4𝑒 −
• Leftovers on central atom
Writing Lewis Dot Formulas
• Try drawing Lewis dot formulas for the following
covalent compound 𝐶𝑂𝐶𝑙2.
• Total electrons: 4 C + 6 𝑂 + 2 × 7 𝐶𝑙 =
24𝑒 −
• Sketch structure with single bonds
• 6 electrons gone for bonds, 24 − 6 = 18𝑒 − 𝑙𝑒𝑓𝑡
• Complete octet of terminal atoms;18 gone;18 −
18 = 0𝑒 −
• No Leftovers, central atom octet not complete
• One of the oxygen or chlorines must share a lone
pair
Writing Lewis Dot Formulas
• 𝐻𝐶𝑁.
• Total electrons: 1 𝐻 + 4 𝐶 + 5 𝑁 = 10𝑒 −
• Sketch structure with single bonds
• 4 electrons gone for bonds, 10 − 4 = 6𝑒 − 𝑙𝑒𝑓𝑡
• Complete octet of terminal atoms; 6 gone; 6 −
6 = 0𝑒 − (Hydrogen can only take two electrons)
• No Leftovers, central atom octet not complete
• Nitrogen shares two lone pairs
Draw the Lewis structure for the
−
𝐵𝑟𝑂3 ion.
• Total electrons: 7 𝐵𝑟 + 3 × 6 𝑂 + 1 𝑓𝑟𝑜𝑚 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑐ℎ𝑎𝑟𝑔𝑒 =
26𝑒 −
• Sketch (For oxyanions 𝐵𝑟𝑂3− , Cl𝑂3− , 𝑆𝑂42− the oxygen atoms surround
the central nonmetal atom) 26 − 6 = 20𝑒 −
• Put eight electrons around the outer atoms 20 − 18 = 2𝑒 −
• Add excess electrons to central atom
• How many valence electrons should appear in the Lewis structure for
CH2Cl2?
• (b) Draw the Lewis structure
• Draw the Lewis structure for (a) NO+ ion, (b) C2H4 (c) ClO2-(d) PO43-
Formal Charge and Alternative Lewis
Structures
• In certain instances, more than one feasible Lewis
structure can be illustrated for a molecule. For
example, consider the 𝐶𝑂2 molecule
• In such cases, to determine the dominant structure
formal charge is used
• The formal charge of any atom in a molecule is the
charge the atom would have if all the atoms in the
molecule had the same electronegativity
Formal Charge and Lewis Structures
• The formal charge of an atom is determined by subtracting the number
of electrons in its “domain” from its group number.
“domain”
electrons
1 e- 4 e-
5 e-
H
C
N:
I
IV
V
or
1 e-
4 e-
H
N C:
I
V
5 e-
group
number
IV
– The number of electrons in an atom’s “domain” is
determined by counting one electron for each
Copyright © Houghton Mifflin bond and two electrons for each lone pair.
Company.All rights reserved.
Presentation of Lecture Outlines, 9–42
Formal Charge and Lewis Structures
• FC = VE (neutral atom) - LE (atom in molecule) - 1/2BE (atom in molecule)
• VE =the number of valence electrons in the neutral atom
• LE = the number of lone pair electrons on the atom in the molecule
• BE= the number of bonding electrons on the atom in the molecule
“domain”
electrons
1 e- 4 e-
5 e-
H
C
N:
I
IV
V
or
1 e-
4 e-
H
N C:
I
V
5 e-
group
number
IV
– The number of electrons in an atom’s “domain” is
determined by counting one electron for each
Copyright © Houghton Mifflin bond and two electrons for each lone pair.
Company.All rights reserved.
Presentation of Lecture Outlines, 9–43
Formal Charge and Lewis Structures
• The most likely structure is the one with the least number of atoms carrying
formal charge. If they have the same number of atoms carrying formal charge,
choose the structure with the negative formal charge on the more
electronegative atom.
formal charge
H
C
0
0
N:
0
or
H
0
N C:
+1
-1
– In this case, the structure on the left is most likely
correct.
Copyright © Houghton Mifflin
Company.All rights reserved.
Presentation of Lecture Outlines, 9–44
Sample Exercise 8.9 Lewis Structures and Formal Charges
Three possible Lewis structures for the thiocyanate ion, NCS – , are
(a) Determine the formal charges in each structure. (b) Based on the formal charges, which Lewis structure
is the dominant one?
Solution
(a) Neutral N, C, and S atoms have five, four, and six valence electrons, respectively.
We can determine the formal charges in the three structures by using the rules we just
discussed:
As they must, the formal charges in all three structures sum to 1–, the overall charge of
the ion. (b) The dominant Lewis structure generally produces formal charges of the
smallest magnitude (guideline 1). That rules out the left structure as the dominant one.
Further, as discussed, N is more electronegative than C or S. Therefore, we expect any
negative formal charge to reside on the N atom (guideline 2). For these two reasons, the
middle Lewis structure is the dominant one for NCS–.
Formal Charges
Formal charge vs oxidation number vs partial
charges
Resonance Structures
• The structure of ozone, O3, can be represented by two
different Lewis electron-dot formulas.
• Both are identical and have equal weight i.e. they are
both equally valid
• Experimentally the two bonds have the same length
• Both structures are called resonance structures
• This is called delocalized bonding, in which a bonding
pair of electrons is spread over a number of atoms
• According to the resonance description, you describe
the electron structure of molecules with delocalized
bonding by drawing all of the possible electron-dot
formulas.
Sample Exercise 8.10 Resonance Structures
Which is predicted to have the shorter sulfur–oxygen bonds, SO3or SO32– ?
Solution
The sulfur atom has six valence electrons, as does oxygen. Thus, SO 3 contains 24 valence electrons. In writing
the Lewis structure, we see that three equivalent resonance structures can be drawn:
As with NO3– , the actual structure of SO3 is an equal blend of all three. Thus, each S—O bond length should be
about one-third of the way between the length of a single bond and the length of a double bond. That is, they
should be shorter than single bonds but not as short as double bonds.
The SO32– ion has 26 electrons, which leads to a dominant Lewis structure in which all the S—O bonds
are single:
Our analysis of the Lewis structures leads us to conclude that SO 3 should have the shorter S—O bonds and
SO32– the longer ones. This conclusion is correct: The experimentally measured S—O bond lengths are 1.42 Å
in SO3 and 1.51 Å in SO32–.
Structure of Nitrate ion
• NO3-
Resonance
• The organic compound
benzene, C6H6, has two
resonance structures.
• It is commonly depicted as
a hexagon with a circle
inside to signify the
delocalized electrons in
the ring.
Exceptions to the octet rule
• Although many molecules obey the octet rule, there are exceptions where
the central atom has more than eight electrons.
• There are three types of ions or molecules that do not follow the
octet rule:
• ions or molecules with an odd number of electrons,
• ions or molecules with less than an octet,
• ions or molecules with more than eight valence electrons (an expanded octet
• Generally, if a nonmetal is in the third period or greater it can
accommodate as many as twelve electrons, if it is the central atom.
• These elements have unfilled “d” subshells that can be used for
bonding.
Odd Number of Electrons
Though relatively rare and usually quite unstable and reactive,
there are ions and molecules with an odd number of electrons.
Fewer Than Eight Electrons
• Consider BF3:
• Giving boron a filled octet places a negative
charge on the boron and a positive charge on
fluorine.
• This would not be an accurate picture of the
distribution of electrons in BF3.
Fewer Than Eight Electrons
Therefore, structures that put a double bond
between boron and fluorine are much less
important than the one that leaves boron with
only 6 valence electrons.
Fewer Than Eight Electrons
The lesson is: If filling the octet of the central
atom results in a negative charge on the
central atom and a positive charge on the
more electronegative outer atom, don’t fill the
octet of the central atom.
More Than Eight Electrons
• The only way PCl5 can exist is if phosphorus
has 10 electrons around it.
• It is allowed to expand the octet of atoms on
the third row or below.
• Presumably d orbitals in these atoms participate in
bonding.
Exceptions to the Octet Rule
• In xenon tetrafluoride, XeF4, the xenon atom must accommodate two
extra lone pairs.
F:
: :
Xe
:
F:
:
:
Copyright © Houghton Mifflin
Company.All rights reserved.
:
: :
:F
:
:
:F
Presentation of Lecture Outlines, 9–58
More Than Eight Electrons
Even though we can draw a Lewis structure for the
phosphate ion that has only 8 electrons around the
central phosphorus, the better structure puts a
double bond between the phosphorus and one of
the oxygens.
More Than Eight Electrons
• This eliminates the charge on the phosphorus and
the charge on one of the oxygens.
• The lesson is: When the central atom is on the
third row or below and expanding its octet
eliminates some formal charges, do so.
Covalent Bond Strength
H = 242 kJ/mol
• The strength of a bond is usually measured by
determining how much energy is required to break
the bond.
• This is the bond enthalpy.
• The bond enthalpy for a Cl—Cl bond,
D(Cl—Cl), is 242 kJ/mol.
Average Bond
Enthalpies
• Average bond
enthalpies are positive,
because bond breaking
is an endothermic
process.
NOTE: These are average
bond enthalpies, not
absolute bond
enthalpies; the C—H
bonds in methane, CH4,
will be a bit different
than the
C—H bond in
chloroform, CHCl3.
Enthalpies of Reaction
• Yet another way to estimate H for a reaction is to
compare the bond enthalpies of bonds broken to the
bond enthalpies of the new bonds formed.
• In other words,
Hrxn = (bond enthalpies of bonds broken) −
(bond enthalpies of bonds formed)
Enthalpies of Reaction
CH4(g) + Cl2(g) 
CH3Cl(g) + HCl(g)
In this example, one C—H
bond and one Cl—Cl bond
are broken; one C—Cl and
one H—Cl bond are
formed.
CH4(g) + Cl2(g) 
CH3Cl(g) + HCl(g)
Enthalpies of Reaction
So,
Hrxn = [D(C—H) + D(Cl—Cl) 
[D(C—Cl) + D(H—Cl)
= [(413 kJ) + (242 kJ)] 
[(328 kJ) + (431 kJ)]
= (655 kJ)  (759 kJ)
= 104 kJ
Bond Enthalpy and Bond Length
• We can also measure an average bond length for
different bond types.
• As the number of bonds between two atoms
increases, the bond length decreases.
Enthalpy problem:
• Calculate the enthalpy of reaction
for:
• CH4 + O2 ---> CO2 + H2O
• Calculate the enthalpy of reaction for:
• CH4 + 3/2O2 ---> CO2 + 2H2O
• 4(C--H) + 3/2(O==O) - 2(C==O) - 4(OH)
• 4(413) + 3/2(495) - 2(800) - 4(463) = -563 kJ
• HC CH + O2 ------> CO2 + H2O
• HC CH + 5/2O2 ------> 2CO2 + H2O
• 1(CC) + 2CH + 5/2(O=O) - 4(C==O) - 2(OH)
• 1(834) + 2(413) +5/2(495) - 4(800) - 2(463) = -1229 kJ
Enthalpy problem:
• Calculate the enthalpy of reaction for:
• CH4 + 3/2O2 ---> CO2 + 2H2O
• 4(C--H) + 3/2(O==O) - 2(C==O) - 4(OH)
• 4(413) + 3/2(495) - 2(800) - 4(463) = -563 kJ
• HC CH + 5/2O2 ------> 2CO2 + H2O
• 1(CC) + 2CH + 5/2(O=O) - 4(C==O) - 2(OH)
• 1(834) + 2(413) +5/2(495) - 4(800) - 2(463) = -1229 kJ
Sample Exercise 8.12 Using Average Bond Enthalpies
Using data from Table 8.4, estimate H for the reaction
Solution
Analyze We are asked to estimate the enthalpy change for a chemical reaction by
using average bond enthalpies for the bonds broken and formed.
Plan In the reactants, we must break twelve C—H bonds and two C—C bonds in the
two molecules of C2H6 and seven O2 bonds in the seven O2 molecules. In the
products, we form eight C==O bonds (two in each CO2) and twelve O—H bonds
(two in each H2O).
Solve Using Equation 8.12 and data from Table 8.4, we have
H = [12D(C—H) + 2D(C—C) + 7D(O2)] – [8D(C==O) + 12D(O—H)]
= [12(413 kJ) + 2(348 kJ) + 7(495 kJ)] – [8(799 kJ) + 12(463 kJ)]
= 9117 kJ – 11948 k J
= –2831 kJ
More than one central atom
• Many atoms have more than 1 central atom
• You just deal with each separately
• Example:
• Lewis structure for acetate:
• CH3CO2-
More than 1 central atom
• The organic compound
benzene, C6H6, has two
resonance structures.
• It is commonly depicted as
a hexagon with a circle
inside to signify the
delocalized electrons in
the ring.
Properties of covalent compound
Bonding
Atoms are joined together within the molecule by covalent bonds.
Electrical
Don’t conduct electricity as they have no mobile ions or electrons
Solubility
Tend to be more soluble in organic solvents than in water;
some are hydrolysed
Boiling point
Low -
e.g.
intermolecular forces (van der Waals’ forces) are weak;
they increase as molecules get a larger surface area
CH4
-161°C
C2H6
- 88°C
C3H8
-42°C
as the intermolecular forces are weak, little energy is required to
to separate molecules from each other so boiling points are low
some boiling points are higher than expected for a given mass
because you can get additional forces of attraction
More Practice
• Draw lewis structures for:
• SO4-2, CO3-2, CHCl3, CN3H6+ (H’s are attached to the
N’s). SO2, PO33-, NO2-1, BrO3-, ClO4-,
DATIVE COVALENT (CO-ORDINATE) BONDING
A dative covalent bond differs from covalent bond only in its formation
Both electrons of the shared pair are provided by one species (donor) and it
shares the electrons with the acceptor
Donor species will have lone pairs in their outer shells
Acceptor species will be short of their “octet” or maximum.
Lewis base
Lewis acid
a lone pair donor
a lone pair acceptor
Ammonium ion, NH4+
The lone pair on N is used to share
with the hydrogen ion which needs
two electrons to fill its outer shell.
The N now has a +ive charge as
- it is now sharing rather than
owning two electrons.
Boron trifluoride-ammonia NH3BF3
Boron has an incomplete shell in BF3 and can accept a share of a pair of
electrons donated by ammonia. The B becomes -ive as it is now shares a
pair of electrons (i.e. it is up one electron) it didn’t have before.
Ionic versus Covalent
IONIC
COVALENT
Bonded Name
Salt
Molecule
Bonding Type
Transfer e-
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Types of Elements
Metal & Nonmetal
Nonmetals
Physical State
Solid
Solid, Liquid, or Gas
Melting Point
High (above 300ºC)
Low (below 300 ºC)
Solubility
Dissolves in Water
Varies
Conductivity
Good
Poor