Respuesta :
Answer:
H-O-H polar
O-C-O nonpolar
H-C-N polar
Explanation:
Looking up the electronegativities of the atoms involved in this question, we have:
Atom Electronegativity
H 2.2
C 2.55
N 3.04
O 3.44
All of the atoms differ in electronegativity resulting in individual dipole moments in H-O, O-C, H-C and C-N bonds. To find if the molecules will be polar we need to consider the structure of the compound to see if there is a resultant dipole moment.
In H-O-H, we have 2 lone pairs of electrons around the central oxygen atom which push the angle H-O-H of the ideal tetrahedral structure to be smaller than 109.5 º resulting in an overall dipole moment making it polar.
In O-C-O, we have two dipole moments that exactly cancel each other in the linear molecule since the central carbon atom does not have lone pairs of electrons since it has 2 double bonds. Therefore the molecule is nonpolar.
In H-C-N, again we have have a central carbon atom without lone pairs of electrons and the shape of the molecule is linear. But, now we have that the dipole moment in C-N is stronger than the H-C dipole because of the difference in electronegativity of nitrogen compared to hydrogen. The molecule has an overall dipole moment and it is polar.
Answer:
H-O-H is polar
O-C-O is non polar
H-C-N is polar
Explanation:
- H-O-H
The electronegativity value of H is 2.1 and the Electronegavity value for O is 3.5 which makes the electronegativity difference of 1.4. O has 2 lone pair of electrons which makes the shape non- linear and polar.
- O-C-O
The bond between C and O is a double bond. The Electronegativity of O is 3.5 while the Electronegativity value for C is 2.5 which gives the electronegativity difference of 1.0. There are no lone pairs on carbon and the dipole or partial charges on carbon and oxygen cancels out making O-C-O non-polar.
- H-C-N
The Electronegativity value of H is 2.1, electronegativity value for C is 2.5 and the electronegativity value for N is 3.0 which makes the molecule to develop partial positive charge on H and partial negative charge on the C-N making the compound polar.