The $1^{\text {st }}, 2^{\text {nd }}$, and the $3^{\text {rd }}$ ionization enthalpies, $I_{1}, I_{2}$, and $I_{3}$, of four atoms with atomic numbers $n, n+1$, $n+2$, and $n+3$, where π < 10, are tabulated below. What is the value of n?Atomic Number Ionization Enthalpy (kJ/mol) I1
I2
I3
n 1681 3374 6050 n+1 2081 3952 6122 n+2 469 4562 6910 n+3 738 1451 7733
| Atomic Number | Ionization Enthalpy (kJ/mol) | ||
I1 | I2 | I3 | |
| n | 1681 | 3374 | 6050 |
| n+1 | 2081 | 3952 | 6122 |
| n+2 | 469 | 4562 | 6910 |
| n+3 | 738 | 1451 | 7733 |
Correct Answer: 9
Solution and Explanation
Since I.E. of Na is 495.8 kJ molβ1 , so (n + 2) should be 11.
n + 2 = 11, So n = 9
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