Boron fluoride exists as $BF_3$ but boron hydride doesn't exist as $BH_3$ . Give reason. In which form, does it exist ? Explain its structure.
Boron fluoride exists as $BF_3$ but boron hydride doesn't exist as $BH_3$ . Give reason. In which form, does it exist ? Explain its structure.
In $\mathrm{BF}_{3}$, lone pair of fluorine gives back support of electrons to boron atom by $p \pi-p \pi$ back bonding. This delocalization of electrons reduces the deficiency of electrons and thus reduces Lewis acidic character and increases stability of $\mathrm{BF}_{3}$.
In $\mathrm{BH}_{3}$, there is no lone pair of electrons on $\mathrm{H}$ atom, therefore, $\mathrm{BH}_{3}$ dimerizes to give $\mathrm{B}_{2} \mathrm{H}_{6}$. $4$ terminal $\mathrm{H}$ atoms and$2$ boron atoms lie in one plane and above and below the plane. Sothere are two bridging $\mathrm{H}$-atoms.
Similar Questions
Draw the structures of $BCl_3. NH_3$ and $AlCl_3$ (dimer).
Draw the structures of $BCl_3. NH_3$ and $AlCl_3$ (dimer).
Inorganic graphite is:
Inorganic graphite is:
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.
Assertion : Both $Be$ and $Al$ can form complexes such as $BeF_4^{2-}$ and $AlF_6^{3-}$ respectively, $BeF_6^{3-}$ is not formed.
Reason : In case of $Be$, no vacant $d-$ orbitals are present in its outermost shell.
Assertion : Both $Be$ and $Al$ can form complexes such as $BeF_4^{2-}$ and $AlF_6^{3-}$ respectively, $BeF_6^{3-}$ is not formed.
Reason : In case of $Be$, no vacant $d-$ orbitals are present in its outermost shell.
- [AIIMS 2015]
In the extraction of aluminium, bauxite is dissolved in cryolite because
In the extraction of aluminium, bauxite is dissolved in cryolite because