Question

Which compound has a tetrahedral shape?
A. $SC{l_2} \\ $
B. $Al{I_3} \\ $
C. $BeB{r_2} \\ $
D. $SiC{l_4} $

Answer 1

Hint: Tetrahedral is one of the orientations of the hybrid orbital in space or electronic geometry around the central atom. It has a hybridization of $s{p^3}$ that is it has 25% s character and 75% p character involved.

Complete step by step solution:
Tetrahedral hybridization takes place when the compound has four electron pairs in its outer shell and the angle between its hybrid orbitals is approximately 109 degrees. As we can observe that all the above compounds are halides.
In this case $SiC{l_4}$ has a hybridization of $s{p^3}$ that is because silicon in excited state have four unpaired electron in 3s and 3p orbital where each of them has one-one unpaired electron that gets paired up by one electron from each chlorine attaining the $s{p^3}$ hybridization. Hence this results in tetrahedral shape. Note that there is no lone pair hence there will be no distortion in the shape of the compound and will be tetrahedral only. Mostly all tetrahalides are colourless volatile liquids. Colour is associated with electrons being promoted from one energy level to another, and absorbing or emitting the energy difference between the two levels. This is common between transition elements where there are unfilled d orbitals.
Dihalide of beryllium cannot have a tetrahedral shape because in the excited state when it has two unpaired electrons in 2s and 2p orbitals which even on pairing due two electrons from bromine gives sp hybridization. $SC{l_2}$ has an asymmetric distribution of electrons which is why distortion gives a bent geometric shape.

Note:
It is only tetrahalide that these compounds can have a tetrahedral shape. There is only one tetrahalide $SiC{l_4}$ in the options. $SiC{l_4}$ is commercially important as well. Small amounts are used to make pure Si for transistors