Question:

The first\((\triangle_ iH_1)\) and the second \((\triangle_iH)\) ionization enthalpies (in \(kJ\; mol ^{-1}\) ) and the \((\triangle_{eg}H)\) electron gain enthalpy (in \(kJ \;mol^{ -1}\) ) of a few elements are given below: 
Elements                       \(\triangle_iH\)                \(\triangle_iH\)                \(\triangle_{eg}H\) 
   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 : 
  1. the least reactive element.
  2. the most reactive metal.
  3. the most reactive non-metal.

Updated On: Nov 9, 2023
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Solution and Explanation

(a) Element \(V\) is likely to be the least reactive element. This is because it has the highest first ionization enthalpy \((\triangle_iH_1)\) and a positive electron gain enthalpy \((\triangle_{eg}H)\).


(b) Element \(II\) is likely to be the most reactive metal as it has the lowest first ionization enthalpy \((\triangle_iH_1)\) and a low negative electron gain enthalpy \((\triangle_{eg}H)\).


(c) Element \(III\) is likely to be the most reactive non-metal as it has a high first ionization enthalpy \((\triangle_iH_1)\) and the highest negative electron gain enthalpy \((\triangle_{eg}H)\).


(d) Element \(V\) is likely to be the least reactive non-metal since it has a very high first ionization enthalpy \((\triangle_iH_2)\) and a positive electron gain enthalpy \((\triangle_{eg}H)\).


(e) Element \(VI\) has a low negative electron gain enthalpy \((\triangle_{eg}H)\). Thus, it is a metal. Further, it has the lowest second ionization enthalpy \((\triangle_iH_2)\). Hence, it can form a stable binary halide of the formula \(MX_2\) (\(X\)= halogen).


(f) Element \(V\) has the highest 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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Concepts Used:

Classification of Elements & Periodicity in Properties

Since many elements were being discovered in the 19th century and the study of these elements individually was proving difficult, classification of elements was made necessary.

Classification by Johann Dobereiner - German chemist Johann Dobereiner classified certain elements on the basis of their similar properties in the groups of continuing - three elements each. These groups were called ‘triads’. In every triad, the atomic weight of the middle element was equal to the average of the atomic weights of the first and third elements. 

Newlands Law of Octaves - The elements were arranged in increasing order of their atomic weights and found that every 8th element shows similarity with the 1st element. 

Mendeleev’s Periodic Table - The arrangement of all 63 elements in rows or columns in order of their atomic weight was made by Mendeleev. He left some space for corresponding elements in his periodic table which were not even discovered till then. Although he predicted the properties of those elements through his periodic classification of elements. 

Modern Periodic Law - The properties of the elements of the modern periodic law are periodic functions of their atomic numbers.