Question

# Justify the following :(i) $$K_2CO_3$$ cannot be prepared by Solvays process(ii) Be and Mg do not impart colour to the flame (iii) Boron forms covalent compounds in +3 state(iv) CO is highly poisonous

Solution

## Solvay process cannot be used to prepare potassium carbonate. This is because unlike sodium bicarbonate, potassium bicarbonate is fairly soluble in water and does not precipitate out.On heating an alkali earth metal or its salt, the electrons are excited easily to higher energy levels because of the absorption of energy. When these electrons return to their ground states, they emit extra energy in form of radiations which fall in the visible region thereby imparting a characteristic colour to the flame. Be and Mg do not impart any colour to the fame i.e. they do not give the flame test. This is due to their very small size. Boron because of its small size and a high sum of first three ionization enthalpies, does not lose its 3 Valence electrons to form $$B^{3+}$$ ions. It does not form ionic compounds. Instead, boron always forms covalent compounds by sharing its valence electron.Carbon monoxide, on the other hand, binds very strongly to the iron in haemoglobin. Once carbon monoxide attaches, it is very difficult to release. So if we breathe in carbon monoxide, it sticks to our haemoglobin and takes up all of the oxygen binding sites. Eventually, blood loses all of its ability to transport oxygen, and you suffocate.Chemistry

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