The bonding in diborane $(B_2H_6)$ can be described by
- A$4$ two-centre-two-electron bonds and $2$ three-centre-two-electron bonds
- B$3$ two-centre-two-electron bonds and $3$ three-centre-two-electron bonds
- C$2$ two-centre-two-electron bonds and $4$ three-centre-two-electron bonds
- D$4$ two-centre-two-electron bonds and $4$ two-centre-two-electron bonds
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$BCl_3$ exists as a monomer,whereas $AlCl_3$ is dimerized through halogen bridging. Give reason. Explain the structure of the dimer of $AlCl_3$ also.
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View SolutionDiborane reacts with $HCl$ in the presence of $AlCl_3$ and liberates:
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Consider the following statements for diborane $(B_2H_6)$:
$1.$ Boron is approximately $sp^3$ hybridized.
$2.$ $B-H-B$ angle is $180^{\circ}$.
$3.$ There are two terminal $B-H$ bonds for each boron atom.
$4.$ There are only $12$ bonding electrons available.
Which of these statements are correct?
$1.$ Boron is approximately $sp^3$ hybridized.
$2.$ $B-H-B$ angle is $180^{\circ}$.
$3.$ There are two terminal $B-H$ bonds for each boron atom.
$4.$ There are only $12$ bonding electrons available.
Which of these statements are correct?
Sodium borate on reaction with $H_2SO_4$ (conc.) and $C_2H_5OH$ gives a compound '$A$' which burns with a green flame. The compound '$A$' is
Medium
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