The first (ΔiH₁) and the second (ΔiH₂) ionization enthalpies (in kJ mol⁻¹) and th (ΔegH) electron gain enthalpy (in kJ mol⁻¹) of a few elements are given below:Elements III IIIIVV VIΔH₁520419 168110082372738ΔH₂ 730030513374184652511451ΔegH-60-48-328-295+48-40Which of the above elements is likely to be : (a) the least reactive element. (b) the most reactive element. (c) the most reactive metal. 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)?
Answers
Answer:
a. helium is least reacting element.
b. flourine is the most reactive metal
c. sodium is the most reactive metal
d. helium is the least reacting non metal
Answer:
Explanation:
a) The least reactive component is probable to be Element V. This is because it has the lowest enthalpy for the first ionization (ΔH1) and a favorable enthalpy for the benefit of electrons (ΔH).
b) Element II is likely to be the most reactive metal as it has the smallest enthalpy of the first ionization (ΔH1) and a small enthalpy of the adverse electron gain (ΔH).
c) Element III is probably the most reactive non-metal because it has a high first enthalpy of ionization (ΔiH1) but less than the noble gas elements and the highest enthalpy of negative electrons (Δe.gh.).
d) Element V is likely to be the least reactive non-metal as it has a very high first enthalpy of ionization (ΔH2) and a positive enthalpy of electron gain (ΔH).
e) Element VI has a small enthalpy of adverse electron gain (element VI). It's a metal, therefore. It also has the smallest second enthalpy of ionization (ΔiH2). It can therefore form a stable MX2 (X= halogen) binary halide.
f) Element I has the low energy for first ionization and elevated energy for second ionization. Thus, the formula MX (X= halogen) can form a predominantly stable covalent halide.