Ncert Exercises & Solutions

(a) Element V is likely to be the least reactive element. This is because it has the highest

first ionization enthalpy ( $_{∆ i H_{1}}$ ) and a positive electron gain enthalpy (∆ $_{e g}$ H).

(b) Element II is likely to be the most reactive metal as it has the lowest first ionization

enthalpy ( $_{∆ i H_{1}}$ ) and a low negative electron gain enthalpy (∆ $_{e g}$ H).

enthalpy( $_{∆ i H_{1}}$ ) and the highest negative electron gain enthalpy (∆ $_{e g}$ H).

(d) Element V is likely to be the least reactive non–metal since it has a very high first

ionization enthalpy $(∆_{i H^{2}})$ and a positive electron gain enthalpy (∆ $_{e g}$ H).

(e)Element VI has a low negative electron gain enthalpy (∆ $_{e g}$ H). Thus, it is a metal.

Further, it has the lowest second ionization enthalpy $(∆_{i H^{2}})$ Hence, it can form a stable

binary halide of the formula MX $_{2}$ (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).

3.32 Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements.

(a) Lithium and oxygen

(b) Magnesium and nitrogen

(d) Silicon and oxygen

(e) Phosphorus and fluorine

(f) Element 71 and fluorine