Correct match is :
Correct match is :
- A
Ordinary form of borax $:$ $Na_2B_4O_7 · 5H_2O$
- B
Colemanite $:$ $Ca_2B_6O_{11}·5H_2O$
- C
Boronatrocalcite $:$ $2Mg_3B_8O_{15}·MgCl_2$
- D
Octahedral form of borac $:$ $Na_2B_4O_7·10H_2O$
Similar Questions
Aluminum dissolves in mineral acids and aqueous alkalis and thus shows amphoteric character. A piece of aluminum foil is treated with dilute hydrochloric acid or dilute sodium hydroxide solution in a test tube and on bringing a burning match stick near the mouth of the test tube, a pop sound indicates the evolution of hydrogen gas. The same activity when performed with concentrated nitric acid, reaction doesn't proceed. Explain the reason.
Aluminum dissolves in mineral acids and aqueous alkalis and thus shows amphoteric character. A piece of aluminum foil is treated with dilute hydrochloric acid or dilute sodium hydroxide solution in a test tube and on bringing a burning match stick near the mouth of the test tube, a pop sound indicates the evolution of hydrogen gas. The same activity when performed with concentrated nitric acid, reaction doesn't proceed. Explain the reason.
A neutral molecule $XF_3$ has a zero dipole moment. The element $X$ is most likely
A neutral molecule $XF_3$ has a zero dipole moment. The element $X$ is most likely
- [AIIMS 2016]
The structure of diborane contains :-
The structure of diborane contains :-
Which of the following is an electron deficient molecule?
Which of the following is an electron deficient molecule?
The reaction of $\mathrm{H}_{3} \mathrm{N}_{3} \mathrm{B}_{3} \mathrm{Cl}_{3}$ $(A)$ with $\mathrm{LiBH}_{4}$ in tetrahydrofurane gives inorganic benzene $(B)$. Further, the reaction of $(A)$ with $(C)$ leads to $\mathrm{H}_{3} \mathrm{N}_{3} \mathrm{B}_{3}(\mathrm{Me})_{3}$. than Compounds $(\mathrm{B})$ and $(\mathrm{C})$ respectively, are
The reaction of $\mathrm{H}_{3} \mathrm{N}_{3} \mathrm{B}_{3} \mathrm{Cl}_{3}$ $(A)$ with $\mathrm{LiBH}_{4}$ in tetrahydrofurane gives inorganic benzene $(B)$. Further, the reaction of $(A)$ with $(C)$ leads to $\mathrm{H}_{3} \mathrm{N}_{3} \mathrm{B}_{3}(\mathrm{Me})_{3}$. than Compounds $(\mathrm{B})$ and $(\mathrm{C})$ respectively, are
- [JEE MAIN 2020]