Question

Part A What is the value of the bond angles in CCl4? Enter the bond angle of the molecule. Part B What is the value of the smallest bond angle in ClF4- ? Enter the smallest bond angle of the molecule. Part C What is the value of the bond angles in COH2? Enter the bond angle of the molecule. Part D What is the value of the bond angle in ICl2- ? Enter the bond angle of the molecule.

Answer 1

Concepts and reason

The concept used to solve this problem is based on the Valence Shell Electron Pair Repulsion theory.

The VSEPR theory is used to predict the shape of the molecule with the help of number of electrons pair surrounding the central metal atom.

By predicting the shape of molecule, bond can be predicted.

Fundamentals

For predicting the shape of molecule by using VSEPR theory, it is necessary to draw Lewis structure of the molecule.

Lewis structure can be drawn by counting the total number of valence electrons in the molecule. The rules for drawing Lewis structure are as follow.

First write the structure of the compound by placing bonding atoms next to each other. The least electronegative atom occupies the central position.

Count the total number of valence electrons in the molecule. add the number of negative charge to the total number of valence electrons and subtract the number of positive charge from total number of valence electrons.

Complete the octets of the atoms bonded to central atom.

(1)

The atomic number of chlorine and carbonare17 and6. The number of valence electrons presentare 7 and 4 respectively. But there are 4chlorine atoms. So, total number of valence electrons in chlorine are 28. Thus, total number of valence electrons are 32. These electrons are distributed in such a way that every atom has complete octet.

The Lewis structure of ${\rm{CC}}{{\rm{l}}_4}$ is drawn as follows:

Now, according to the VSEPR theory, the molecule with central atom bonded to four electron groups has the tetrahedral shape. For tetrahedral shape, the bond angles between atoms are ${109.5^ \circ }$ .

(2)

The atomic number of chlorine and fluorine are 17 and 9. The number of valence electrons present in both atoms are 7. But there are 4 fluorine atoms. So, total number of valence electrons in fluorine are 28. Also, there is a negative charge. Thus, total number of valence electrons are 36. These electrons are distributed in such a way that every atom has complete octet.

The Lewis structure of ${\rm{ClF}}_4^ -$ is drawn as follows:

Now, according to the VSEPR theory, the molecule with central atom bonded to six electron groups has the square planar shape. For square planar shape, the bond angles between atoms are ${90^ \circ }$ .

(3)

The atomic number of carbon and oxygen are 6 and 8. The number of valence electrons present are 4 and 6 respectively.Also, there are two hydrogen atoms. Thus, total number of valence electrons are 12. These electrons are distributed in such a way that every atom has complete octet.

The Lewis structure of ${\rm{CO}}{{\rm{H}}_2}$ is drawn as follows:

Now, according to the VSEPR theory, the molecule with central atom bonded to three electron groups has the trigonal planar shape. For trigonal planar shape, the bond angles between atoms are ${120^ \circ }$ .

(4)

The number of valence electrons present in iodine and chlorine are 7 and 7 respectively. But there are two chlorine atoms. Thus, total number of valence electrons are 22. These electrons are distributed in such a way that every atom has complete octet.

The Lewis structure of ${\rm{ICl}}_2^ -$ is drawn as follows:

Now, according to the VSEPR theory, the molecule with central atom bonded to five electron groups has the linear shape. For linear shape, the bond angles between atoms are ${180^ \circ }$ .

Ans: Part 1

The bond angle of ${\rm{CC}}{{\rm{l}}_4}$ molecule is ${109.5^ \circ }$ .

Part 2

The bond angle of ${\rm{ClF}}_4^ -$ molecule is ${90^ \circ }$ .

Part 3

The bond angle of ${\rm{CO}}{{\rm{H}}_2}$ molecule is ${120^ \circ }$ .

Part 4

The bond angle of ${\rm{ICl}}_2^ -$ molecule is ${180^ \circ }$ .