Question

COIT0402786 

Rank the following compounds in order of the length of the carbon-halogen bond. Give the number 1 to the shortest bond and the number 4 to the longest bond. 

C01T041173 

What type(s) of orbital overlap Is(are) indicated on the following structure:

CO1T0406878 

Rank the following compounds in order of the length of the carbon-halogen bond. Give the number 1 to the shortest bond and the number 4 to the longest bond 

Answer 1

Answer of question 1 :  Fluorine is the most electronegative element in the periodic table and electronegativity order decrease from fluorine to iodine i.e. fluorine is more elctronegative than chlorine, bromine and iodine. Due to high electronegativity and smaller size of fluorine the bond length of CH3F is smaller i.e. number 1 given to CH3F. After that number 2 given to CH3Cl and number 3 given to CH3I. As we moved from Fluorine to Iodine in periodic table size increases and bond length also increases.

Answer of question 2 : The orbital overlap shown in below is sp3-sp3 overlap because both carbons are sp3 hybridised so they utilised there sp3 orbitals for bonding.

т I H -I 0-I HC- -C- C C= I I H sp3-sp

Answer of question 3 : As we move from Fluorine to Iodine in periodic table size increases and bond length also increases. Based on this the shortest bond length to CH3CH2CH2F i.e. rank 1, rank 2 given to CH3CH2CH2Cl, Rank 3 given to CH3CH2CH2Br and rank 4 given to CH3CH2CH2I.