Question

The first $\left(\right.\left(\Delta \right)_{i}H_{1}\left.\right)$ and second $\left(\right.\left(\Delta \right)_{i}H_{2}\left.\right)$ ionization enthalpies (in kJ $mol^{- 1}$ ) and the electron gain enthalpy $\left(\right.\left(\Delta \right)_{e g}H\left.\right)$ (in kJ $mol^{- 1}$ ) of the elements I, II, III, IV and V are given below:

Element	$\Delta _{i}H_{1}$	$\Delta _{i}H_{2}$	$\Delta _{e g}H$
I	$520$	$7300$	$-60$
II	$419$	$3051$	$-48$
III	$1681$	$3374$	$-328$
IV	$1008$	$1846$	$-295$
V	$2372$	$5251$	$+48$

The most reactive metal and the least reactive non-metal of these are respectively

• A. V and II
• B. II and V
• C. V and III
• D. IV and V

Answer 1

Answer: (B)

$\Delta H_{1}=520$
$\Delta H_{2}=7300$
$\Delta H_{2}$ is $10$ times greater than $\Delta H_{1}$ mean first $e^{-}$ remove from inner shell {Ionisation energy decreases the number of shells increases because $Z_{e f f}$ decreases hence, removal of electron is easy so, $I.E.\downarrow$
III $\Delta _{i}H_{1} < \Delta _{i}H_{2}$
IV $\Delta _{i}H_{1} < \Delta _{i}H_{2}$
V $\Delta _{i}H_{1} < \Delta _{i}H_{2}$
$\Delta H_{e f f}H=+ve$
$\Delta H_{e f f}H=+ve$ Means it belongs to inert gas & these are belongs to non-metal and very less reactive
So, II most reactive metal.
V is least reactive non-metal.