Question

One mole of atoms of an element M absorb 576 kJ energy for the formation of $M^{+}$ ions. The total amount of energy required for one mole of M atoms to form tripositive ions is 5140 kJ.$2^{nd}$ and $3^{rd}IP$ values of the metal are in 2 : 3 ratio. Calculate $I{P}_2$ and $I{P}_3$ in $kJmo{l}^{-1}$ respectively.

• A. $1825,2738$
• B. $2738,1825$
• C. $1453,2765$
• D. $2765,1453$

Answer 1

The correct option is A $1825,2738$

$M\to M^{+}Energyabsorbed$
$=576kJmol{e}^{-1}$
$I{P}_1=576kJmol{e}^{-1}$
$M\to M^{+3}Energyabsorbed$
$=5140kJmol{e}^{-1}$
$M\to M^{+3}Energyabsorbed$
$=5140-576=4564kJmol{e}^{-1}$
$RatioofI{P}_2$ & $I{P}_3,=2:3$
$I{P}_2+I{P}_3=4564kJ$
$I{P}_2=\frac{4564}{5}\times 2=1825kJmol{e}^{-1}$
$I{P}_3=\frac{4564}{5}\times 3=2738kJmol{e}^{-1}$.