Question 3.31
NCERT Class XI Chemistry
The first (∆iH1) and the second (∆iH2) ionization enthalpies (in kJ mol-1) and the (∆egH) electron gain enthalpy (in kJ mol-1) of a few elements are given below [see book]: (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). the metal which can form a predominantly stable covalent halide of the formula MX (X=halogen)?
Solution in Detail
(video solution below this)
$\displaystyle \underline{\underline{\text{Part (a) - least reactive}}}$
The least reactive element will have high ionization enthalpy and less negative or positive electron gain enthalpy. Element (V) has the highest IE 2372 units and positive EGE +40. Hence Element (V) is the answer!
$\displaystyle \underline{\underline{\text{Part (b) most reactive metal}}}$
Element II has the least first IE. It can easily lose an electron, hence it is a metal and the most reactive!
$\displaystyle \underline{\underline{\text{Part (c) most reactive non-metal}}}$
Element III has high ionization enthalpy so it is a non-metal. At the same time it has the most negative electron gain enthalpy of -328 units. Hence it is the most reactive non-metal!
$\displaystyle \underline{\underline{\text{Part (d) least reactive non-metal}}}$
Element V is the least reactive, and since it has a high first IE, it is a non-metal.
Alternately, the positive EGE indicates that element V is a noble gas, and therefore a non-metal.
Hence, element V is the least reactive non-metal!
NOTE: if noble gases are not treated as non-metals, then element IV is a non-metal because of high negative EGE, and it is less reactive because its first IE is not high, but rather low.
$\displaystyle \underline{\underline{\text{Part (e) binary halide MX}_2}}$
the metal which can form a stable binary halide is the one that will easily lose both its valence electrons.
Element VI has first and second IE quite close, and low, and since its EGE is less negative, it is a metal that can easily lose 2 electrons to form a binary halide!
$\displaystyle \underline{\underline{\text{Part (e) covalent halide MX}}}$
Element I has a moderately high first IE, very high second IE - which leads us to conclude that it has only one valence electron. At the same time it has a moderately high negative EGE. It can share its valence electron, and yet keep the electron cloud attracted towards itself to give it a covalent character!
Video Explanation
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