B{left( {OH} right)₃} + NaOH to NaB{O₂} + Naleft[ {B{{left( {OH} right)}₄}} right] + {H₂}O &

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p-Block Elements - I

normal

$B{\left( {OH} \right)_3} + NaOH \to NaB{O_2} + Na\left[ {B{{\left( {OH} \right)}_4}} \right] + {H_2}O$

How can this reaction is made to proceed in forward direction?

addition of cis, $2$ diol

addition of borax

addition of trans $1,\, 2$ diol

addition of $Na_2HPO_4$

Solution

$H_3BO_3$ acts are weak monobasic Lewis acid.

$(i)$ $B(OH)_3 + NaOH \to Na [B(OH)_4]$ On addition of cis $1, 2$ diol in $H_3BO_3$ solution, acidic strength of $H_3BO_3$ increases due to chelation effect.

Standard 11

Chemistry

$BCl_3$ does not exist as dimer but $BH_3$ exist as dimer $(B_2H_6)$ becuase

easy

The bond dissociation energy of $B- F$ in $BF_3$ is $646 \,kJ\, mol^{-1}$ whereas that of $C – F$ in $CF_4$ is $515\, kJ\, mol^{-1}.$ The correct reason for higher $B- F$ bond dissociation energy as compared to that of $C-F$ is

medium

(AIEEE-2008)

Give structure of gaseous $AlCl_3$.

easy

Given below are two statements :

Statement $I:$ Upon heating a borax bead dipped in cupric sulphate in a luminous flame, the colour of the bead becomes green.

Statement $II:$ The green colour observed is due to the formation of copper(I) metaborate.

In the light of the above statements, choose the most appropriate answer from the options given below :

medium

(JEE MAIN-2023)

Give suitable laboratory preparation for diborane.

easy

$B{\left( {OH} \right)_3} + NaOH \to NaB{O_2} + Na\left[ {B{{\left( {OH} \right)}_4}} \right] + {H_2}O$ How can this reaction is made to proceed in forward direction?

Solution

Similar Questions

$BCl_3$ does not exist as dimer but $BH_3$ exist as dimer $(B_2H_6)$ becuase

The bond dissociation energy of $B- F$ in $BF_3$ is $646 \,kJ\, mol^{-1}$ whereas that of $C – F$ in $CF_4$ is $515\, kJ\, mol^{-1}.$ The correct reason for higher $B- F$ bond dissociation energy as compared to that of $C-F$ is

Give structure of gaseous $AlCl_3$.

Give suitable laboratory preparation for diborane.

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$B{\left( {OH} \right)_3} + NaOH \to NaB{O_2} + Na\left[ {B{{\left( {OH} \right)}_4}} \right] + {H_2}O$

How can this reaction is made to proceed in forward direction?