Question

Which of the following statements are incorrect ?

• A. In $B_2{H}_6\left(s\right)$ there are four $2c-2e^{-}$ bonds
• B. $(C{H}_3{)}_3\ddot{N}$ and $(Si{H}_3{)}_3\ddot{N}$ are pyramidal
• C. $A{l}_{2}C{l}_6\left(s\right)$ has banana bonding
• D. In $(Si{H}_3{)}_3\ddot{P}$ there is back bonding

Answer 1

Answer: (B), (D)

Only statement A is correct. All terminal $B-H$ bonds are 2c-2e bonds.
In ${\left(Si{H}_3\right)}_{3}N$, $Si$ has vacant d-orbital which forms $p\pi -d\pi$ bonding. So there is no back bonding in ${\left(Si{H}_3\right)}_{3}N$.
${\left(C{H}_3\right)}_{3}N$ has $s{p}^3$ hybridization on nitrogen with pyramidal shape due to $p\pi -p\pi$ bonding whereas in ${\left(Si{H}_3\right)}_{3}N$, $Si$ has vacant d orbitals which forms $p\pi -d\pi$ bonding. So, ${\left(Si{H}_3\right)}_{3}N$ has planar structure whereas ${\left(C{H}_3\right)}_{3}N$ has pyramidal structure.
$A{l}_{2}C{l}_6$ is a dimer of $AlC{l}_3$ in which $Cl$ forms coordinate bond with $Al$ as shown in figure.