H2O is polar, whereas BeF2 is not because

Question

H2O is polar, whereas BeF2 is not because (A) electronegativity of F is greater than that of O (B) H2O involves H-bonding, whereas BeF2 is not a discrete molecule (C) H2O involves H-bonding, whereas BeF2 is a discrete molecule (D) H2O is linear and BeF2 is angular.

Tardigrade Admin · Accepted Answer

Because of linear shape, dipole moments cancel each other in BeF 2( F leftarrows Be leftarrows F ) and thus, it is non-polar, whereas H 2 O is V-shaped and hence, it is polar.

Q. $H_{2} O$ is polar, whereas $B e F_{2}$ is not because

electronegativity of F is greater than that of O

$H_{2} O$ involves H-bonding, whereas $B e F_{2}$ is not a discrete molecule

$H_{2} O$ involves H-bonding, whereas $B e F_{2}$ is a discrete molecule

$H_{2} O$ is linear and $B e F_{2}$ is angular.

Because of linear shape, dipole moments cancel each other in $B e F_{2} (F ⇄ B e ⇄ F)$ and thus, it is non-polar, whereas $H_{2} O$ is $V$ -shaped and hence, it is polar.

Q. H2​O is polar, whereas BeF2​ is not because

electronegativity of F is greater than that of O

H2​O involves H-bonding, whereas BeF2​ is not a discrete molecule

H2​O involves H-bonding, whereas BeF2​ is a discrete molecule

H2​O is linear and BeF2​ is angular.

Because of linear shape, dipole moments cancel each other in BeF2​(F⇄Be⇄F) and thus, it is non-polar, whereas H2​O is V-shaped and hence, it is polar.

Q. $H_{2} O$ is polar, whereas $B e F_{2}$ is not because

$H_{2} O$ involves H-bonding, whereas $B e F_{2}$ is not a discrete molecule

$H_{2} O$ involves H-bonding, whereas $B e F_{2}$ is a discrete molecule

$H_{2} O$ is linear and $B e F_{2}$ is angular.

Because of linear shape, dipole moments cancel each other in $B e F_{2} (F ⇄ B e ⇄ F)$ and thus, it is non-polar, whereas $H_{2} O$ is $V$ -shaped and hence, it is polar.