Under which conditions of temperature and pressure does a real gas behave most like an ideal gas
1 answer:
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Answer:
Forms ionic bonds, insoluble, doesn't participate in single or double displacement reactions, non-reactive, high heat of combustion
Explanation:
Beryllium is a metal, since it belongs to group 2A, the alkaline earth metals. It has a total of 2 valence electrons.
Phosphorus, on the other hand, belongs to group 5A and has a total of 5 valence electrons.
We have a compound which has a metal in it, therefore, it's an ionic compound. Beryllium, our metal, loses its 2 electrons to gain an octet and phosphorus, our nonmetal, should gain 3 electrons to have an octet. The oxidation states are +2 and -3 respectively. This means we need 3 beryllium cations and 2 phosphide anions in our formula .
Beryllium phosphide would be expected to be insoluble, as only beryllium chloride, fluoride, nitrate, phosphate and sulfate are soluble substances, while the remaining ones are expected to be insoluble.
Due to its insolubility, beryllium phosphide would not participate in any ionic reactions, such as single displacement or double displacement.
Since it's insoluble, we expect this compound to be chemically stable and not reactive. This implies that if we wanted to burn it, the heat of combustion would be very high, as a lot of energy would be needed to be supplied in an endothermic reaction in order to burn it.