Consider the lattice energies of the following group 2a compounds: beh2, 3205 kj/mol; mgh2, 2791 kj/mol; cah2, 2410 kj/mol; srh2, 2250 kj/mol; bah2, 2121 kj/mol. part a what is the oxidation number of h in these compounds?

(-1) is the oxidation number of H in these compounds.

Explanation:

Oxidation state is defined as the number that is given to an atom when it looses or gains electron. It is always written as a superscript.

When an atom looses electron, it will attain a positive oxidation state.When an atom gains electron, it will attain a negative oxidation state.

Elements of group-2 have fixed oxidation state that +2.

Oxidation of Hydrogen atom in all :

Oxidation state of each metal in the given compounds = +2

Let the oxidation state of hydrogen atom be = x

Since, these given compounds are neutral that is overall charge on the compound is 0.

(-1) is the oxidation number of H in these compounds.

The velocity would be 2 

Density= mass/volume so 5.1 divided by 4.7 is  5.1 / 4.7 =1.08510638298round to the correct number of sig figs