Question

Observe the following reactions: \[ \text{Dimer} \xrightarrow{\text{vapour phase}} MCl_2 \xrightarrow{1200\,K} \text{linear monomer} \] What is \(M\)?

• A. Ca
• B. Sr
• C. Mg
• D. Be

Hint:

Beryllium compounds often show anomalous behavior because of the small size and high charge density of \(Be^{2+}\). In the vapour phase, \(BeCl_2\) exists as a bridged dimer \((Be_2Cl_4)\) and at high temperatures forms a linear monomer \((Cl-Be-Cl)\).

Answer 1

The Correct Option is D

Beryllium is the smallest member of group 2 and its $Be^{2+}$ ion is unusually small and strongly polarising, so beryllium chloride behaves very differently from the chlorides of the heavier alkaline earth metals, which are essentially ionic.

In the vapour phase at moderate temperature, $BeCl_2$ exists as a dimer held together by two bridging chlorine atoms, each beryllium completing its octet by sharing a chlorine lone pair with its neighbour.

When heated further to around 1200 K, there is enough thermal energy to break these weak bridging bonds, and the dimer falls apart into two monomeric $BeCl_2$ units. In the monomer, beryllium bonds to just two chlorines with no bridging, so it adopts $sp$ hybridisation and a perfectly linear $Cl-Be-Cl$ geometry.

Magnesium, calcium and strontium chlorides do not show this dimer to monomer transition because their bonding is largely ionic from the start. So the metal M in this transformation is beryllium, and the correct choice is option (4).