Is Berrylium Capable of Showing an Oxidation State of +6?

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Discussion Overview

The discussion revolves around the oxidation state of beryllium in a proposed chemical reaction involving beryllium hydroxide and hydrochloric acid. Participants explore the possibility of beryllium exhibiting an oxidation state of +6 and the implications of ligand theory in this context.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant presents a reaction suggesting that beryllium has an oxidation state of +6, questioning how this is possible given its electron configuration.
  • Another participant emphasizes the importance of checking if the equation is balanced, implying that an unbalanced equation is incorrect.
  • Some participants speculate that hydroxide ions may donate electrons to beryllium's empty p orbitals, relating this to ligand theory.
  • There are multiple mentions of the equation being unbalanced, with suggestions that the correct form involves water as a ligand.
  • Participants discuss the number of water molecules required in the complex, with some arguing for three based on balancing, while others suggest four for stability.
  • One participant acknowledges a misunderstanding regarding the existence of the proposed beryllium compound and corrects their earlier claim about the reaction.
  • Another participant notes that the coordination entity involves Be2+, which influences the hybridization of its orbitals to accommodate ligands.

Areas of Agreement / Disagreement

Participants generally agree that the original equation is unbalanced and that the proposed beryllium compound as written does not exist. However, there remains disagreement on the correct number of water molecules and the implications of ligand theory, with no consensus on the oxidation state of +6.

Contextual Notes

Limitations include the unresolved nature of the original equation's validity, the dependence on definitions of oxidation states, and the lack of clarity on the correct stoichiometry of the reaction.

mooncrater
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Homework Statement


While studying alkaline Earth metals of the s block elements I found this equation :
Be (OH)2+2HCl +2H2O →[Be (OH)4]Cl2

It is clear that in the complex, the oxidation state of Berrylium is +6. But how is even that possible?

Homework Equations


None

The Attempt at a Solution


We know that Be has only 2 electrons in its s orbital and moreover has a total of only 4 electrons, so how is even that possible?
 
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Wouldn't hurt to check if the equation is balanced. If it is not, it is wrong.
 
Well it's an amazing question.
Don't know about you but find that reaction in my NCERT book.
The only reason I can guess is OH donating its electrons to empty p orbitals of Berylium and therefore Berylium having that additional six electrons.
It is related to ligand theory.
 
Borek said:
Wouldn't hurt to check if the equation is balanced. If it is not, it is wrong.
Yeah it is not balanced. Don't know why in my book it is given as same as the mooncrater's equation.
Can you, mooncrater give the reference of taking the reaction?
 
Mistake is quite obvious here.

Hint: water is a good ligand.
 
Borek said:
Mistake is quite obvious here.

Hint: water is a good ligand.
Is that intended for me? Well I am not a expert in ligand theory. This thread brought an interest in me and I wanted to be in discussion.
So then my reason is incorrect for OH donating electrons to empty p orbitals of Berylium?
 
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You don't have to be an expert to understand that the reaction that is not balanced is wrong. Beryllium compound - as written - doesn't exist. And I gave you both a hint to what does the real compound look like.
 
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Yeah, the equation should be,
## Be(OH)_2 + 2HCl + H_2O → [Be(H_2O)_3] Cl_2 ##
Thanks Borek Sir.
The question remains for mooncrater that how in book that he is reading, wrong equation is written?

As in my book also same equation is given and our teacher ask to rote memorise it without proper logic.

Though I have passed my 11th grade, I never realized it.
Once again thanks Borek Sir
 
Why Be(H2O)32+?
 
  • #10
Borek said:
Why Be(H2O)32+?
Because H2O has a zero charge and Berylium would get +2 charge, which is its normal state.
 
  • #11
But why 3 water molecules? Why not 2 or 4?
 
  • #12
For balancing purpose. Otherwise reaction would not be balanced.
Like the way, you are making sure that I get every concept of this question. :smile:
 
  • #13
It is from NCERT 11TH
 
  • #14
As I was suspecting.
 
  • #15
Raghav Gupta said:
For balancing purpose. Otherwise reaction would not be balanced.

Be(OH)2 + 2HCl + 75H2O → [Be(H2O)77]Cl2

Balanced?

Google for beryllium aqua complex.
 
  • #16
Main reason for not balancing it:
It is in our course book.
Now it's purely impossible for the complex to exist.
According to me [Be (H2O)3]Cl2(as Raghav said)
should exist as Be has 2 electrons in its outermost orbital and needs 6 more for inert gas configuration. For that 3 H2Ogroups are needed.
Is it correct?
 
  • #17
Borek said:
Be(OH)2 + 2HCl + 75H2O → [Be(H2O)77]Cl2

Balanced?

Google for beryllium aqua complex.
Searched it and got that Be is more stable when 4 molecules of H2O are bonded to it as Ligands.
Reaction should be
Be(OH)2 + 2HCl + 2H2O → [Be(H2O)4]Cl2

mooncrater I realized that I was wrong.
The compound should be [Be(H2O)4]Cl2.
The reason is that , the coordination entity is having Be2+ not simply Be
So 2s orbital , 2py,2px, 2pz orbitals are empty which hybridise to accommodate lone pairs of 4 water Ligands.
 
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  • #18
Okay I got it now. Thanks.
 
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