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A metal, M forms chlorides in +2 and +4 oxidation states. Which of the following statements about these chlorides is correct?

  • Option 1)

    MCl_{2}  is more volatile than MCl_{4}

  • Option 2)

    MCl_{2}  is more soluble in anhydrous ethanol than  MCl_{4}

  • Option 3)

    MCl_{2}  is more ionic than  MCl_{4}

  • Option 4)

    MCl_{2}  is more easily hydrolysed than MCl_{4}

 

Answers (1)

As we have learned

Nature of boron halides -

All boron halides are covalent and Lewis acids.

Lewis acid character of boron trihalides obey the order BI_{3}> BBr_{3}> BCl_{3}> BF_{3}

They have plane triangular geometry (sp2 hybridization)

- wherein

Due to small size and high electronegativity

 

 In MCl_2   the oxidation state of M is +2 and in MCl_4 , The oxidation state pf M is +4 

Higher the oxidation state , smaller the size , Which results in greater polarizing power hence greater the covalent character 

So , MCl_4 is more covalent and MCl_2 is more iconic 

 

 

 

 

 

 


Option 1)

MCl_{2}  is more volatile than MCl_{4}

Option 2)

MCl_{2}  is more soluble in anhydrous ethanol than  MCl_{4}

Option 3)

MCl_{2}  is more ionic than  MCl_{4}

Option 4)

MCl_{2}  is more easily hydrolysed than MCl_{4}

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