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Standard XII

Chemistry

Trend in Ionisation Enthalpy

Two metals ...

Question

Two metals $M$ and $N$ have the following values of ionization energies. Find energy required for formation of $M^{+}$ from $M$ , $M^{+ 2}$ from $M^{+}$ , and total energy required to form $M^{+ 2}$ from $M$ . Do similar calculations for metal $N$ .

$I E_{1}$ $I E_{2}$
$M$
496
4560
$N$
738
1450

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Solution

IE $_{1}$ is the energy required to form M $^{+}$ from M= 496 units.
IE $_{2}$ is the energy required to form M $^{+ 2}$ from M $^{+}$ =4560 units.
Energy required to form M $^{+ 2}$ from M is IE $_{1}$ +IE $_{2}$ = 496 +4560 =5056 units.
Similarly, to formN $^{+}$ from N,IE $_{1}$ is the energy required =738 units.
To form N $^{+ 2}$ from N $^{+}$ , IE $_{2}$ is the energy required =1450 units.
Total energy required to form N $^{+ 2}$ from N is IE $_{1}$ + IE $_{2}$
=738+1450 = 2188 units.

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Two metals M and N have the following values of ionization energies. Find energy required for formation of M+ from M, M+2 from M+, and total energy required to form M+2 from M. Do similar calculations for metal N.IE1 IE2M4964560N7381450

Two metals $M$ and $N$ have the following values of ionization energies. Find energy required for formation of $M^{+}$ from $M$ , $M^{+ 2}$ from $M^{+}$ , and total energy required to form $M^{+ 2}$ from $M$ . Do similar calculations for metal $N$ .

$I E_{1}$ $I E_{2}$
$M$
496
4560
$N$
738
1450