हिंदी

F-Be-F is a liner molecule but H-O-H is angular. Explain.

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प्रश्न

F-Be-F is a liner molecule but H-O-H is angular. Explain.

संक्षेप में उत्तर
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उत्तर

  1. In the BeF2 molecule, the central beryllium atom undergoes sp hybridization giving rise to two sp hybridized orbitals placed diagonally opposite with an angle of 180°. Thus, F-Be-F is a linear molecule.
  2. In the H2O molecule, the central oxygen atom undergoes sp3 hybridization giving rise to four sp3 hybridized orbitals directed towards four corners of a tetrahedron. There are two lone pairs of electrons in two of the sp3 hybrid orbitals of oxygen. The lone pair-lone pair repulsion distorts the structure. Hence, H-O-H is angular or V-shaped.
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अध्याय 5: Chemical Bonding - Exercises [पृष्ठ ७९]

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बालभारती Chemistry [English] Standard 11 Maharashtra State Board
अध्याय 5 Chemical Bonding
Exercises | Q 3. (K) | पृष्ठ ७९

संबंधित प्रश्न

Select and write the most appropriate alternatives from the given choices.

The angle between the two covalent bonds is minimum in:


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Write the geometry and bond angle present in ammonia.


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Define Bond Length


Predict the shape and bond angles in the following molecule:

CF4


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H2S


Complete the flow chart.

Molecular Formula Structural Formula Shape/ Geometry Bond angle
BeCl2     180°
  O=C=O Linear  
C2H2      

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Which of following bonds has maximum bond length?


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Consider the reactions.

\[\ce{C_{(s)} + 2H2_{(g)} -> CH4_{(g)}}\], ΔH = −x kcal

\[\ce{C_{(g)} + 4H_{(g)} -> CH4_{(g)}}\], ΔH = −x1 kcal

\[\ce{CH4_{(g)} -> CH3_{(g)} + H_{(g)}}\], ΔH = +y kcal

The average C-H bond enthalpy is:


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(Nb = bonding electrons, Na = antibonding electrons)


Consider the ions/molecule:

\[\ce{O^+_2, O_2, O^-_2, O^{2-}_2}\]

For increasing bond order the correct option is:


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