Assertion $(A)$: $[B(OH_2)_6]^{3+}$ and $[B(OH)_4]^{-}$ form octahedral and tetrahedral structures.
Reason $(R)$: Being electron deficient, boron readily reacts with Lewis bases like $H_2O$ and $OH^{-}$.
The correct option among the following is
Reason $(R)$: Being electron deficient, boron readily reacts with Lewis bases like $H_2O$ and $OH^{-}$.
The correct option among the following is
- A$A$ is true, $R$ is true and $R$ is the correct explanation for $A$
- B$A$ is true, $R$ is true but $R$ is not the correct explanation for $A$
- C$A$ is true but $R$ is false
- D$A$ is false but $R$ is true
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Similar Questions
On the basis of reactions given below identify $X, Y, Z$.
$Boron \xleftarrow{Mg \, or \, Fe} X \xleftarrow{\Delta} B(OH)_3 + \text{Gas-}Z$
$B(OH)_3 \xleftarrow{H_2O} Y \xrightarrow{2NH_3/\Delta} \text{Inorganic benzene}$
$Boron \xleftarrow{Mg \, or \, Fe} X \xleftarrow{\Delta} B(OH)_3 + \text{Gas-}Z$
$B(OH)_3 \xleftarrow{H_2O} Y \xrightarrow{2NH_3/\Delta} \text{Inorganic benzene}$
The dissolution of $Al(OH)_3$ by a solution of $NaOH$ results in the formation of
In the reaction $B(OH)_{3} + 2 H_{2}O \rightarrow [B(OH)_{4}]^{-} + H_{3}O^{+}$,$B(OH)_{3}$ functions as:
Write balanced equations for :
$(i) \, BF_{3} + LiH \to $
$(ii) \, B_{2}H_{6} + H_{2}O \to $
$(iii) \, NaH + B_{2}H_{6} \to $
$(iv) \, H_{3}BO_{3} \xrightarrow{\Delta }$
$(v) \, Al + NaOH \to $
$(vi) \, B_{2}H_{6} + NH_{3} \to $
$(i) \, BF_{3} + LiH \to $
$(ii) \, B_{2}H_{6} + H_{2}O \to $
$(iii) \, NaH + B_{2}H_{6} \to $
$(iv) \, H_{3}BO_{3} \xrightarrow{\Delta }$
$(v) \, Al + NaOH \to $
$(vi) \, B_{2}H_{6} + NH_{3} \to $
Medium
View Solution$Na_2B_4O_7 \cdot 10H_2O \xrightarrow{\Delta} 2NaBO_2 + B_2O_3 (A) + 10H_2O$. $(A) + MnO \xrightarrow{\Delta} Mn(BO_2)_2 (B)$. $(A)$ and $(B)$ are
Difficult
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