How to determine bond angles?
Surfed all over the internet and couldn’t find a reasonable method to determine bond angles in chemical molecules? Well, you have reached the right spot because in this article we have tried to address all your questions regarding bond angles in covalently bonded molecules.
So, without any further delay, let’s start reading.
What is a bond angle?
A bond angle is simply defined as the geometric angle formed between two adjacent covalent bonds that share a common atom at the center.
According to the Valence Shell Electron Pair Repulsion (VSEPR) theory of chemical bonding, covalently bonded molecules consist of two different types of electron pairs i.e., a bond pair and a lone pair of electrons. 
The electron pairs stay furthest away from each other to minimize their repulsive effect. This controls the shape of the molecule which in turn controls the bond angles present in this molecule.
So, now we know that the bond angle in a molecule can be determined using the VSEPR theory. Let’s see how.
What is the VSEPR theory?
The Lewis structure is a very useful way of representing the structure of covalently bonded molecules. However, it does not give any information about the geometry and shape of a molecule. This information can be obtained from the VSEPR theory.
Also check –
A molecule consists of two different geometries i.e., an electronic geometry and a molecular geometry. The electronic geometry represents the 3dimensional structure of the molecule. It depends on the total number of electron pairs present around the central atom in the molecule.
The molecular geometry on the other hand determines the shape of the molecule. It takes into account the different numbers of bond pairs and lone pairs around the central atom. Bond angles help molecules maintain their specific shapes and molecular geometries.
The VSEPR notation uses an alphabet A to represent the central atom to which other atoms X are covalently bonded while E denotes a lone pair present on the central atom.
Determining bond angles using VSEPR theory (AXE method)
Linear geometry
If the central atom (A) in a molecule is bonded to two other atoms (X) and there is no lone pair (E) present on it then the molecule occupies a linear geometry and shape.
It is represented by the VSEPR notation AX_{2}. The central atom is sp hybridized. Its linear shape signifies that all the bonded atoms lie on a straight line thus they form a mutual bond angle of 180°.
Trigonal planar geometry
If the central atom (A) is bonded to three other atoms (X) and there is no lone pair on the central atom, then the molecule occupies a trigonal planar geometry and shape.
The VSEPR notation for this molecule is AX_{3}. The central atom is sp^{2} hybridized. Three X around the central A atom form an equilateral triangle. A complete circle is made up of 360°, when it is divided into three equal parts (360/3 = 120) then each AX bond angle in a trigonal planar shape has a 120° bond angle.
In AX_{2}E type molecules, one of the three bonded atoms in the trigonal planar molecule gets replaced by a lone pair of electrons (E). This leads to lonepair bondpair repulsions which in turn decreases the XAX bond angle. The lonepair bondpair repulsions are significantly greater than bondpair bondpair repulsions.
Thus, although the molecule has a trigonal planar electronic geometry, but its molecular geometry changes and it occupies a bent shape to minimize the repulsive effect. The bond angle decreases from the ideal 120° to approximately 118°.
As a general rule of thumb, for each X replaced by a lone pair (E), the bond angle gets reduced by 2°. So, you need to subtract 2° from the ideal bond angle to obtain the bond angle for a specific shape.
Tetrahedral geometry
AX_{4}type molecules have a tetrahedral geometry and shape. Four atoms (X) are bonded to the central atom (A) like the four vertices of a tetrahedron. The central atom is sp^{3} hybridized. It has an XAX bond angle of 109.5°.
You should also note that a greater p character in the hybrid orbitals is another factor contributing to a reduced bond angle.
Bond angle decreases in the order : linear (sp)> trigonal planar (sp^{2}) > tetrahedral (sp^{3}).
If one X atom gets replaced by a lone pair (E), it forms AX_{3}Etype molecules. The lone pair is placed at the apex of the molecule to minimize lonepair bondpair repulsions.
The XAX bond angle decreases (109.5° 2° = 107.5° ). The molecule occupies a trigonal pyramidal shape and molecular geometry. It has a triangular base and a pyramid at the top of it.
If two X atoms get replaced by two lone pairs, AX_{2}E_{2}type molecules are formed. The repulsive effect is further increased as lonepair lonepair repulsions > lonepair bondpair repulsions > bondpair bondpair repulsions.
To accommodate the strong electronic repulsions, both the lone pairs are placed as far apart from each other as possible. Thus, the molecule occupies a bent shape, and the XAX bond angle decreases further to 104.5°.
Trigonal bipyramidal geometry
AX_{5}type molecules have a trigonal bipyramidal electronic geometry. The central atom is sp^{3}d hybridized. The central atom (A) is surrounded by five X atoms on the sides. It forms a triangular base and two pyramids above and below the triangle.
The bonded atoms form three different bond angles i.e., 120° at the triangular base, 180° along the XAX straight line, and 90° where the XA atoms lie at a right angle to each other.
In the trigonal bipyramidal geometry, the two axial X atoms are held fixed while equatorial X atoms can be removed and replaced with lone pairs. One X atom replaced by an E forms an AX_{4}Etype molecule. Such molecules occupy an irregular seesaw shape to minimize their electronic repulsions.
So, they do not really have definite bond angle values. Different types of bond angles can be present at different positions in the seesaw shape. The bond angles are usually > 90° and <180°.
Two equatorial X replaced with E in AX_{3}E_{2}type molecules makes the molecule occupy a Tshape with a bond angle < 180° while all three equatorial X replaced by lone pairs forms linear AX_{2}E_{3}type molecules with a bond angle =180°.
Octahedral geometry
AX_{6}type molecules possess an octahedral electronic geometry and shape. The central atom is sp^{3}d^{2} hybridized. The six X atoms lie around the central A atom in an eight vertices arrangement i.e., an octahedron.
It forms two different bond angles i.e., 180° at the center while the peripheral XAX atoms lie at right angles (90°) to each other.
The four X atoms forming a square base are held fixed while the atom at the top and bottom of the octahedron can be removed and replaced with lone pairs. The top X atom removed and replaced with a lone pair (AX_{5}E) lends the molecule a square pyramidal shape and geometry.
The XAX bond forms a 90° bond angle at the square base while the bond angles at other positions on the molecule are < 90°.
Both the top and bottom X atoms replaced with lone pairs make the molecule occupy a square planar shape and a 90° bond angle. The lone pairs are placed opposite each other to minimize their repulsive effect.
What factors affect bond angles?
Overall, we have identified four main factors that affect the bond angles present in a molecule namely:
 Molecular geometry or shape.
 Lonepair lonepair and lonepair bondpair repulsions are present in the molecule.
 Hybridization of the central atom. The bond angle decreases as the p character increases.
 Electronegativity of the central atom. The bond angle decreases by decreasing the electronegativity of the central atom.
Chart of bond angles
The following chart will help you in determining the bond angles for different molecules having varying shapes/molecular geometries according to the VSEPR concept.
VSEPR notation  Bond pairs  Lone pairs  Ideal electronic geometry  Molecular geometry  Bond angles  Examples 
AX_{2}  2  0  Linear  Linear  180°  BeCl_{2}, CO_{2} 
AX_{3}  3  0  Trigonal planar  Trigonal planar  120°  BF_{3}, SO_{3} 
AX_{2}E  2  1  Trigonal planar  Bent  118°  SO_{2}, SnCl_{2} 
AX_{4}  4  0  Tetrahedral  Tetrahedral  109.5°  CH_{4}, CCl_{4} 
AX_{3}E  3  1  Tetrahedral  Trigonal pyramidal  107.5°  NH_{3}, PCl_{3} 
AX_{2}E_{2}  2  2  Tetrahedral  Bent  104.5°  H_{2}O, N_{2}O 
AX_{5}  5  0  Trigonal bipyramidal  Trigonal bipyramidal  90°, 120°, 180°  PCl_{5, }PF_{5} 
AX_{4}E  4  1  Trigonal bipyramidal  Seesaw  101.6°, 173°, 187°  SF_{4}, TeCl_{4} 
AX_{3}E_{2}  3  2  Trigonal bipyramidal  Tshape  175°  ClF_{3}, BrF_{3} 
AX_{2}E_{3}  2  3  Trigonal bipyramidal  Linear  180°  XeF_{2} 
AX_{6}  6  0  Octahedral  Octahedral  90°, 180°  SF_{6}, PCl_{6}^{–} 
AX_{5}E  5  1  Octahedral  Square pyramidal  90°, <90°  BrF_{5}, IF_{5} 
AX_{4}E_{2}  4  2  Octahedral  Square Planar  90°  XeF_{4}, ICl_{4}^{–} 
FAQ
What is the Bond angle? 
A bond angle is simply defined as the geometric angle formed between two adjacent covalent bonds that share a common atom at the center. 
How can you find the bond angle? 
The bond angle can easily find by using the VSEPR theory –

What is the molecular geometry and bond angle of the water (H_{2}O)? 
According to the VSEPR theory, H_{2}O is an AX_{2}E_{2}type molecule. The central oxygen (O) atom belongs to Group VI A of the Periodic Table. It has a total of 6 valence electrons. 2 bond pairs and 2 lone pairs around oxygen make the H_{2}O molecule occupy a bent molecular geometry and shape with a bond angle of 104.5°. 
Why is the bond angle of H_{2}S less than that of H_{2}O although both have bent shapes? 
Both H_{2}S and H_{2}O are AX_{2}E_{2}type molecules with 2 bond pairs and 2 lone pairs around the central atom. Both the molecules have a bent molecular shape and geometry. The HOH bond angle in H_{2}O is 104.5° while the HSH bond angle in H_{2}S is 92.1° because the sulfur (S) atom is less electronegative as compared to oxygen. The bondpair bondpair repulsions between HS bonds are reduced so bonds come closer, consequently, the bond angle decreases. 
Why is the bond angle in NF_{3} less than that in NH_{3}? 
The bond angle in NF_{3 }is 101.9° while that in NH_{3} is 107.5° although both have a trigonal pyramidal shape with 3 bond pairs and 1 lone pair around the central nitrogen (N) atom. This is because the F atoms in NF_{3 }are highly electronegative. The bonded electron pairs are furthest away from the central N atom in NF_{3.} Consequently, there is less distortion present in the molecule. The bondpair bondpair repulsions between NF atoms decrease which leads to a reduced bond angle. 
Why do bond angles in CH_{4}, NH_{3, }and H_{2}O differ? 
The bond angles in these four molecules differ based on their different molecular geometries/shapes. CH_{4} has an ideal tetrahedral electronic geometry. There are 4 bond pairs and no lone pair around the central carbon atom. The H atoms arrange symmetrically around the central atom and form a mutual bond angle of 109.5°. The ideal tetrahedral geometry gets distorted when a bond pair is replaced by a lone pair as in NH_{3, }so it occupies a trigonal pyramidal geometry with a bond angle of 107.5°. Two bond pairs replaced by lone pairs further distort the shape. Lonepair lonepair repulsions exist which makes H_{2}O occupy a bent shape and molecular geometry with a bond angle of 104.5°. 
Short tricks for bond angle comparison between different molecules? 
The bond angle is:

Summary
 A bond angle is the geometric angle between two adjacent bonds joined to a mutual atom at the center.
 Bond angles contribute to the overall shape of the molecule.
 A molecule may have a different molecular geometry or shape from its ideal electronic geometry as per VSEPR theory.
 The molecule occupies a shape that demonstrates minimum repulsive effect between its different electronic regions.
 An ideal bond angle is a maximum angle at which the electronic repulsions are minimized.
 The bond angles help differentiate between linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral electronic and molecular geometries.
 Greater the distortion present in a molecule, the lower the bond angle between its covalently bonded atoms.
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