IO4- lewis structure is notation of IO4- valence electrons in line and dot form also known as lewis dot structure. Let us learn IO4- lewis structure and characteristics.
IO4- lewis structure composed of two atoms 1 iodine and 4 oxygen atoms. Iodine is centrally placed and bonding to 4 O atoms shown by line (4 covalent bonds) and lone pair electrons placed on 4 O atoms shown by dots. IO4- lewis structure is closed in a square bracket due to negative charge on it.
IO4- is the chemical formula of periodate ion, which is an ester or salt of periodic acid composed of mainly two elements iodine and oxygen which are non-metals. IO4- i.e. periodate is an inorganic mono – valent negative ion (anion). Let us have some discussion its structure, characteristics, shape, polarity and other properties.
How to draw IO4- lewis structure?
Lewis structure is basically a simplified method to show valence electrons of a molecule forming a simple structure. Following are some steps to draw IO4- lewis structure.
- Valence electrons: Count valence electrons of IO4- lewis structure and select the least electronegative atom as a central atom i.e. iodine is less electronegative than oxygen.
- Lone pair electrons and octet rule: Make single bonds between all I and O atoms of IO4- lewis structure, place remaining valence electrons on bonding atoms i.e. O atoms. These non-bonding electrons are lone pair electrons note them how much they are in the structure. Apply octet rule and check whether the atoms has complete or incomplete octets.
- Formal charge and shape: Evaluate the formal charge of IO4- lewis structure with given formula and predict the shape, hybridization and bond angle of it.
IO4- valence electrons
Valence electron are the electrons on outermost valence shell of an element which they gain or lose to form a nearest noble gas configuration. Let us discuss this.
IO4- lewis structure contains 32 valence electrons. I and O atoms of IO4- ion has 7 and 6 valence electrons respectively, as they belongs to 17th and 16th group of periodic table. Thus the valence shell electronic configuration of I is 5s2, 5p5 and O is 2s2, 2p4.
Let’s calculate overall valence electrons present on IO4- lewis structure.
- Valence electrons on iodine and four oxygen = 7 + ( 6 x 4 ) = 31
- Add one extra electron due to -1 charge on IO4- ion = 01
- Total valence electrons on IO4- lewis structure = 31 + 1 = 32
- To find overall electron pairs divide valence electrons by 2 = 32 / 2 = 16
Therefore, the IO4- lewis structure has thirty two valence electrons and sixteen electron pairs.
IO4- lewis structure lone pairs
Lone pair electrons are those unshared pair of valence electrons which are present in any molecule or ion. Let us have a discussion of lone pairs in IO4- lewis structure.
The IO4- lewis structure composed of 12 lone pair electrons. It has total 32 valence electrons. Eight valence electrons are being bond electron pairs as they form covalent bonds within I and O atoms. The remaining 24 non-bonding electrons are being 12 lone pair electrons.
Each oxygen atom of IO4- has 6 non – bonding electrons. Thus, each oxygen atom has three pairs of lone electron pairs. As IO4- have four oxygen atom, therefore 4 (O) x 3 (L P) = 12 lone pair electrons on IO4- ion..
IO4- lewis structure octet rule
Octet rule defines the presence of eight electrons on any atom which shows the complete or incomplete octet. Below is brief discussion on IO4- lewis structure octet rule.
IO4- lewis structure satisfying octet rule as the I and O atoms has complete octets. IO4- ion have 32 valence electrons, 8 bonding electrons around central I atom and 2 bonding and 6 non-bonding electrons around 4 O atoms. Thus, I and 4 O atoms have complete octets in IO4- lewis structure.
The central I atom has 8 electrons (bonding) due to which it has complete octet. All 4 O atoms has 8 electrons (6 non-bonding / 2 bonding) due to which they have complete octets.
IO4- lewis structure formal charge
Formal charge is the positive or negative charge present on any ion or molecule. Here see more discussion on IO4- formal charge.
Formal charge = (valence electrons – non-bonding electrons – ½ bonding electrons)
|IO4- atoms||Valence electrons||Non – bonding electrons||Bonding electrons||Formal charge|
|Iodine||7||0||8||(7 – 0 – 8 / 2) = + 3|
|Oxygen||6||6||2||(6 – 6 – 2 / 2) = -1|
IO4- lewis structure resonance
Resonance structure are 2 or more forms of same molecule with same atomic position but different electronic position and bonds. Take a look on IO4- resonance structure.
|Structure:||Central Iodine F C||Upper oxygen F C||Lower oxygen F C||Right oxygen F C||Left oxygen F C|
Resonance structure can be possible if the molecule has multiple bonds, formal charge and lone electron pairs. All the 4 resonance structures of IO4- ion shows stable form. There is movement of electrons on 3 O atoms to form 3 I=O double bonds reducing the formal charge on I and 3 O to zero.
IO4- lewis structure shape
Molecular shape and electron geometry of any lewis structure is determined by VSEPR theorys given module. Let us discuss IO4- lewis structure shape.
The shape and geometry of IO4- lewis structure is tetrahedral. The IO4- lewis structure is composed of five atoms having one iodine atom and four oxygen atoms. Iodine is placed centrally due to its low electronegativity than oxygen atoms.
Shape of IO4- ion
These medial iodine atom have zero lone electron pairs and attached with 4 O atoms around it. Thus, the IO4- lewis structure follows the AX4 generic formula of in which A is central atom and X is bonding atoms to central atom. AX4 generic formula shows tetrahedral shape of IO4- ion.
Know more about Hydrophobic Examples
IO4- lewis structure angle
Bond angle is a particular angle within the bonds of any molecular structure. Let us discuss the bond angle in IO4- lewis structure below:
IO4- lewis structure has 109.5 degree bond angle within O-I-O bonds of IO4- ion. 4 I-O bonds within central iodine and 4 O atoms. As the central atom does not contains any lone pair electron it has tetrahedral shape. Hence it has 109.5 degree bond angle.
The bond angles can be predicted with the VSEPR theory as per the structures. As the IO4- lewis structure has the tetrahedral geometry as it is composed of 5 atoms, 1 central I atom which is get joined with 4 O atoms. VSEPR theory prediction says it has 109.5 degree angle.
The mixing and recasting of atomic orbitals to form a new hybrid orbitals is known as hybridization. Let us have some discussion on IO4- lewis structure hybridization.
The IO4- lewis structure has sp3 or tetrahedral hybridization. Hybridization can be determine with steric number of central atom. IO4- ion has steric number 4 and tetrahedral shape or geometry.
When 1 ‘s’ and 3 ‘p’ orbitals of same atomic shell of element overlaps with each other to form 4 new equivalent hybrid orbitals is known as tetrahedral or sp3 hybridization. IO4- ion comes under AX4 generic formula of VSEPR theory due to which it has tetrahedral molecular shape and electron geometry.
IO4- oxidation number
The atoms having total electron to which they donate or accept with other atoms to form chemical bonds is oxidation number. Below the discussion in IO4- oxidation number.
The oxidation number of IO4- ion is +7. The oxidation number of iodine atom is basically -1 as per the rules but it has exception if these halogen atom gets attached to oxygen or fluorine atoms. So, in IO4- ion iodine cannot have -1 oxidation number but has calculated oxidation number +7.
Why the IO4- have + 7 oxidation number?
In ionic compounds like IO4- all the oxidation numbers should be summed up to the charge on the ion.
The iodine of IO4- ion has calculated oxidation number value +7, because the iodine atom is a non – metal and thus comes under 17th periodic table group. It means the iodine atom always requires one electron for completion of octet and form a stable state.
How the IO4- have + 7 oxidation number?
IO4- is an ionic compound and as per oxidation rules we have to sum up charges on it.
To evaluate the oxidation number of iodine in IO4- ion, we have to arrange all the charges on RHS (right hand side) which equals to zero. Oxygen atom has – 2 oxidation number and the IO4- ion contains four oxygen atoms. Thus, the oxidation number of iodine is:
Let us consider the oxidation number of iodine in IO4- = x,
Four oxygen atoms have oxidation number = – 2,
Therefore, x + 4 (- 2) = – 1
x – 8 = – 1
x = – 1 + 8
x = + 7
So, the iodine atom of IO4- ion has + 7 oxidation number.
Is IO4- polar or nonpolar
Non-polar compounds are those which do not have dipole or zero dipole due to symmetrical arrangement of atoms in its structure. Here, we can see below.
IO4- (periodate) is a non – polar ion, because it has equal distribution of electrons and zero dipole moment. The IO4- ion has tetrahedral geometry and also has symmetrical arrangements of atoms in its structure.
Why IO4- is non – polar?
As per the VSEPR theory, lewis structure having central atom with no lone pair electrons and it gets joined with similar atoms then it is a non – polar compound.
In IO4- lewis structure, the middle iodine atom do not have any lone pair of electrons and also it get attached with similar atoms i.e. four oxygen atoms around it. Hence, the IO4- ion is a non – polar ion in nature.
How IO4- is non – polar?
IO4- ion has all the same O atoms around central I atom. Also all O atoms has equal electrons pairs, though it is non-polar.
IO4- ion is non-polar as it has equivalent sharing of electrons within the I and O atoms due to which the dipole moment cancels out each other of both atoms and IO4- ion has zero dipole.
Is IO4- a polyatomic ion?
Polyatomic ions are those which have more than one atom and bonded covalent to each other and also it has some charge present on it. Describing here Is IO4- polyatomic.
IO4- is a polyatomic ion. From its structure it is clear that, there are more than one atoms are present in IO4- ion i.e 5 (poly) atoms 1 I and 4 O.
Why IO4- a polyatomic ion?
Poly means many or more. Polyatomic ion is charge containing molecule having more than one atom in its molecule.
IO4- ion is a polyatomic ion because it has total five atoms present on it which are have covalent bonded with each other. Also the atoms of the IO4- ion have some charge like it has +3 charge on central iodine atom and -1 charge on bonding four oxygen atoms.
IO4- lewis structure has tetrahedral shape and geometry with sp3 hybridization and 109.5 degree bond angle. It is a polyatomic ion and non – polar in nature.