Question 31: The first (ΔiH1) and the second (ΔiH) ionization enthalpies (in kJ mol–1) and the (ΔegH) electron gain enthalpy (in kJ mol–1) of a few elements are given below:
Elements |
ΔiH1 |
ΔiH |
ΔegH |
I |
520 |
7300 |
-60 |
II |
419 |
3051 |
-48 |
III |
1681 |
3374 |
-328 |
IV |
1008 |
1846 |
-295 |
V |
2372 |
5251 |
+48 |
VI |
738 |
1451 |
-40 |
Which of the above elements is likely to be :
(a) the least reactive element.
(b) the most reactive metal.
(c) the most reactive non-metal.
(d) the least reactive non-metal.
(e) the metal which can form a stable binary halide of the formula MX2, (X=halogen).
(f) the metal which can form a predominantly stable covalent halide of the formula MX (X=halogen)?
Answer:
(a) Element V is likely to be the least reactive element. This is because it has the highest first ionization enthalpy (ΔiH1) and a positive electron gain enthalpy (ΔegH). It is a noble gas.
(b) Element II is likely to be the most reactive metal as it has the lowest first ionization enthalpy (ΔiH1) and a low negative electron gain enthalpy (ΔegH).The with lowest first ionization enthalpy is likely to be a reactive metal.
(c) Element III is likely to be the most reactive non–metal as it has a high first ionization enthalpy (ΔiH1) but less than the noble gas elements and the highest negative electron gain enthalpy (ΔegH).
(d) Element V is likely to be the least reactive non–metal since it has a very high first ionization enthalpy (ΔiH2) and a positive electron gain enthalpy (ΔegH).
(e) Element VI has a low negative electron gain enthalpy (ΔegH). Thus, it is a metal. Further, it has the lowest second ionization enthalpy (ΔiH2). Hence, it can form a stable binary halide of the formula MX2 (X=halogen).
(f) Element I has low first ionization energy and high second ionization energy. Therefore, it can form a predominantly stable covalent halide of the formula MX (X=halogen).
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