Question

Which of the following compound(s) terminal $(2c-2e^{-})$ bonds and bridge bonds are lying in same plane?

• A. $I_{2}C{l}_6$
• B. $F{e}_{2}C{l}_4$
• C. Solid $BeC{l}_2$
• D. $G{a}_2{H}_6$

Answer 1

Answer: (A), (B)

The correct options are
A $I_{2}C{l}_6$
B $F{e}_{2}C{l}_4$

a) $IC{l}_3$ dimerizes to give $I_{2}C{l}_6$ in which two $Cl$ atoms are bridged and donate their lone pair to $I$ atoms resulting in a $3C-4e^{-}$ bond known as bridged bonding, whereas four $Cl$ are bonded to $I$ via $2C-2e^{-}$ bond known as terminal bonds. So each $I$ has two lone pairs and four $\sigma$ bond pairs, i.e. $s{p}^3{d}^2$ hybridised and has octahedral geometry around it. Therefore, $Cl$ atoms along with $I$ lies in the same plane.
b) In $F{e}_{2}C{l}_4$, two $Cl$ atoms are bonded with $Fe$ via $2C-2e^{-}$ bond which are called terminal $Cl$, while the other two $Cl$ are bonded with $Fe$ via $3C-4e^{-}$ bond in which the $Cl$ atom donates its lone pair to the vacant $d$ orbitals of $Fe$ also known as bridged atoms. Terminal as well as bridged atoms lies in the same plane of $Fe$.c) In solid state, $BeC{l}_2$ polymerises to give $(BeC{l}_2{)}_n$. In polymeric chain each $Be$ is tetrahedrally surrounded by $Cl$ atoms, two bridged $Cl$ atoms lies in same plane of $Be$ and other two bridged $Cl$ atoms lies above and below this plane i.e perpendicularly. So, two $Cl$ falls in same plane of $Be$ and other two $Cl$ atoms are above and below this plane.
d) In $G{a}_2{H}_6$ each $Ga$ is $s{p}^3$ hybridized and is tetravalent. So, two$H$lies in the same plane and other two atoms lies above and below the plane as shown.