Question

Which molecule is not linear?
A) \[{\rm{Be}}{{\rm{F}}_{\rm{2}}}\]
B) \[{\rm{Be}}{{\rm{H}}_{\rm{2}}}\]
C) \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\]
D) \[{{\rm{H}}_{\rm{2}}}{\rm{O}}\]

Answer 1

Hint: The shape of a molecule is determined from the hybridization of the central atom in a compound and the count of bond pairs and lone pairs in a compound. Here, we will use a formula to find out hybridization of these compounds.

Formula used:
\[H = \dfrac{{V + X - C + A}}{2}\]; Here, H stands for hybridization, V is count of valence electrons, X is the count of monovalent groups bonded to the central atom, C stands for cationic charge and A stands for charge of anion.

Complete Step by Step Answer:
Let’s find out the hybridization of each compound.
For \[{\rm{Be}}{{\rm{F}}_{\rm{2}}}\], Count of valence electrons=2, count of monovalent atoms=2
Therefore, \[H = \dfrac{{2 + 2}}{2} = 2\]
The value of H is 2, this indicates that hybridization of the Be atom is sp. So, this molecule has a linear shape.
Now,

For \[{\rm{Be}}{{\rm{H}}_{\rm{2}}}\], Count of valence electrons=2, count of monovalent atoms=2
Therefore, \[H = \dfrac{{2 + 2}}{2} = 2\]
The value of H is 2, this indicates that hybridization of the Be atom is sp. So, this molecule also has a linear shape.

For\[{\rm{C}}{{\rm{O}}_{\rm{2}}}\], Count of valence electrons=4, count of monovalent atoms=0
Therefore, \[H = \dfrac{{4 + 0}}{2} = 2\]
The value of H is 2, this indicates that hybridization of the C atom is sp. So, this molecule also has a linear shape.

For \[{{\rm{H}}_{\rm{2}}}{\rm{O}}\], the valence electrons of O atom=6, Count of monovalent atoms=2
Therefore, \[H = \dfrac{{6 + 2}}{2} = 4\]
The value of H is 4, this indicates that hybridization of the C atom is \[s{p^3}\] . So, this molecule does not have a linear shape but angular.
Therefore, option D is right.

Note: The VSEPR theory also guides in the prediction of shapes of molecules by giving some rules. These rules are based on the count of bond pairs and lone pairs surrounding the central atom in a compound.