Colour of the bead in borax bead test is mainly due to the formation of
Colour of the bead in borax bead test is mainly due to the formation of
- A
metal oxides
- B
boron oxide
- C
metal metaborates
- D
elemental boron
Similar Questions
The Lewis acid character of boron tri halides follows the order:
The Lewis acid character of boron tri halides follows the order:
- [JEE MAIN 2023]
Which one of the following is the correct statement ?
Which one of the following is the correct statement ?
- [AIEEE 2008]
Identify the statement that is not correct as far as structure of diborane is concerned
Identify the statement that is not correct as far as structure of diborane is concerned
Identify the correct statement for $B _{2} H _{6}$ from those given below.
$(A)$ In $B _{2} H _{6}$, all $B - H$ bonds are equivalent.
$(B)$ In $B _{2} H _{6}$ there are four $3-$centre$-2-$electron bonds.
$(C)$ $B _{2} H _{6}$ is a Lewis acid.
$(D)$ $B _{2} H _{6}$ can be synthesized form both $BF _{3}$ and $NaBH _{4}$.
$(E)$ $B _{2} H _{6}$ is a planar molecule.
Choose the most appropriate answer from the options given below..... .
Identify the correct statement for $B _{2} H _{6}$ from those given below.
$(A)$ In $B _{2} H _{6}$, all $B - H$ bonds are equivalent.
$(B)$ In $B _{2} H _{6}$ there are four $3-$centre$-2-$electron bonds.
$(C)$ $B _{2} H _{6}$ is a Lewis acid.
$(D)$ $B _{2} H _{6}$ can be synthesized form both $BF _{3}$ and $NaBH _{4}$.
$(E)$ $B _{2} H _{6}$ is a planar molecule.
Choose the most appropriate answer from the options given below..... .
- [JEE MAIN 2022]
When metal $X$ is treated with sodium hydroxide, a white precipitate $(A)$ is obtained, which is soluble in excess of $NaOH$ to give soluble complex $(B)$. Compound $(A)$ is soluble in dilute $HCl$ to form compound $(C)$. The compound $(A)$ when heated strongly gives $(D)$, which is used to extract metal. Identify $(X), (A), (B), (C)$ and $(D)$. Write suitable equations to support their identities.
When metal $X$ is treated with sodium hydroxide, a white precipitate $(A)$ is obtained, which is soluble in excess of $NaOH$ to give soluble complex $(B)$. Compound $(A)$ is soluble in dilute $HCl$ to form compound $(C)$. The compound $(A)$ when heated strongly gives $(D)$, which is used to extract metal. Identify $(X), (A), (B), (C)$ and $(D)$. Write suitable equations to support their identities.