Although both CO_(2) and H_(2)O are triatomic molecular , the shape of H_(2)O molecule is bent while that of CO_(2) is liner. Explain this on the bases of dipole moment.

Although both CO_(2) and H_(2)O are triatomic molecular , the shape of

1.	Although both CO_(2) and H_(2)O are triatomic molecular , the shape of H_(2)O molecule is bent while that of CO_(2) is liner. Explain this on the bases of dipole moment.
Answer» Solution :ALWAYS, POLARITY of molecule = `({:("The vecotr sum "),("of the dipole"),("moments of "),("various bonds"):})` Shape of `CO_(2) `: the shpae of `CO_(2)` is linear so its become nonpolar. Practically `CO_(2)` is nonpolar. `(mu = 0 )` Polarity of molecule = Vectro addition of Dipole moment of all bands inmolecule `CO_(2)` is linear molecule so it is nonpolar experiment `CO_(2)` is NON polar `(mu = 0 ) CO_(2)` is nonpolar it is indirect thus it shape must be like this vector addition of DIPOLAR becomes zero. In `CO_(2) " two C" OVERSET(rarr)(_)` Obonds are same but if the are in opposite direction than nullify the effect of each other . `O overset(larr)(_) " C " overset(rarr)(_) O equiv [""^(-delta)O -overset(-2 delta)(C) - O^(-delta) ]^(0)` `therefore mu = 0 "so , "CO_(2)` is non polar `H_(2)O : H_(2) O ` is angular `mu` = 1.85 D so it is polar . IN `H_(2) `O .O. is more electronegative than H. `therefore H - O` bond is polar . `H^(+) o O^(-delta) and H overset(rarr)(_)` son bond has dipole . The sharp of `H_(2)`O should be like this the net dipole not become zero. So the arrangement of bond must be angular . Bond angle is `1.4.5^(@) " net " mu = 1.85 D `