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.

Answer 1

(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 $(\triangle_{eg}H)$.

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(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 $(\triangle_{eg}H)$.

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(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 $(\triangle_{eg}H)$.

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(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 $(\triangle_{eg}H)$.

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(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).

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(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).