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(i) Explain the observations from the Ellingham diagram. (ii) Write a short note on anamolous properties of the first element of p-block. |
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Answer» Solution :(i) 1. For most of the metal oxide formation, the slope is positive. It can be explained as follows. Oxygen gas is consumed during the formation of metal oxides which results in the decrease in RANDOMNESS. Hence, `DeltaS` becomes negative and it makes the term, `TDeltaS` positive in the straight line equation. 2. The graph for the formation of carbon monoxide is a straight line with negative slope. In this case `DeltaS` is positive as 2 moles of CO gas is formed by the consumption of one mole of oxygen gas. It indicates that CO is more stable at higher temperature. 3. As the temperature increases, generally `DeltaG` value for the formation of the metaloxide become less negative and becomes zero at a particular temmperature. Below this temperature, `DeltaG` is negative and the oxide is stable and above this temperature`DeltaG` is positive. This general trend suggests that metal oxides become less stable at higher temperature and their decomposition becomes easier. 4. There is a sudden change in the slope at a particulartemperature for some metal oxides like MgO, HgO. This is due to the phase transition (melting or evaporation). (ii) In p-block elements the first member of each group differs from the other elements of the corresponding group. The following FACTORS are responsible for this anomalous behaviour. Small size of the first member. High ionisation enthalpy and high ELECTRONEGATIVITY. Absence of d-orbitals in their VALANCE shell. The first member of the group-13, boron is a metalloid while others are reactive metals. Moreover, boron shows diagonal RELATIONSHIP with silicon of group 14. The oxides of boron and silicon are similar in their acidic nature. |
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