Classification of Elements and Periodicity in Properties
Q1
Errorless
Latest Trend
Numerical
Arrange the bonds in order of increasing ionic character in the molecules. LiF, $K_{2}O$ , $N_{2}$ , $SO_{2}$ and $ClF_{3}$ [a) $ClF_{3} < N_{2} < SO_{2} < K_{2}O < LiF$ (b) $LiF < K_{2}O < ClF_{3} < SO_{2} < N_{2}$ (c) $N_{2} < SO_{2} < ClF_{3} < K_{2}O < LiF$ (d) $N_{2} < ClF_{3} < SO_{2} < K_{2}O < LiF$
Correct Answer: C
Explanation:
Increasing order of ionic character $N_{2} < SO_{2} < ClF_{3} < K_{2}O < LiF$ Ionic character depends upon difference of electronegativity (bond polarity).
Q2
Errorless
JEE Section
Numerical
The $1^{\mathrm{st}}$ , $2^{\mathrm{nd}}$ and the $3^{\mathrm{rd}}$ ionization enthalpies $I_{1}, I_{2}$ and $I_{3}$ , of four atoms with atomic numbers $n, n+1, n+2$ and $n+3$ , where $n < 10$ , are tabulated below. What is the value of $n$
| Atomic number | Ionization Enthalpy (kJ/mol) | ||
| $I_1$ | $I_2$ | $I_3$ | |
| n | 1681 | 3374 | 6050 |
| n+1 | 2081 | 3952 | 6122 |
| n+2 | 496 | 4562 | 6910 |
| n+3 | 738 | 1451 | 7733 |
Correct Answer: 9
Explanation:
| Atomic number | Ionization Enthalpy (kJ/mol) | ||
| $I_1$ | $I_2$ | $I_3$ | |
| n | 1681 | 3374 | 6050 |
| n+1 | 2081 | 3952 | 6122 |
| n+2 | 496 | 4562 | 6910 |
| n+3 | 738 | 1451 | 7733 |
By observing the values of $I_1, I_2$ , and $I_3$ for atomic number $(n + 2)$ , it is observed that $I_2 > > I_1$ .
This indicates that number of valence shell electrons is 1 and atomic number $(n+2)$ should be an alkali metal.
Also for atomic number $(n+3)$ , $I_{3} >> I_{2}$ .
This indicates that it will be an alkaline earth metal which suggests that atomic number $(n+1)$ should be a noble gas and atomic number $(n)$ should belong to Halogen family. Since n < 10; hence n = 9 (F atom)
Note : n = 1 (H atom) cannot be the answer because it does not have $I_{2}$ and $I_{3}$ values.