Correct option is (1) -x
En = -RH (Z2 / n2) J
For He+ (n = 1), En = -x = -RH (22 / 12) = -4RH
∴ RH = x / 4
For Be3+ (n = 2), En = -RH (Z2 / n2) J
= - (x / 4) × (4 × 4 / 2 × 2) = -x J
List-I ( Ions ) | List-II ( No. of unpaired electrons ) | ||
A | Zn$^{2+}$ | (I) | 0 |
B | Cu$^{2+}$ | (II) | 4 |
C | Ni$^{2+}$ | (III) | 1 |
D | Fe$^{2+}$ | (IV) | 2 |
List I | List II | ||
A | Down’s syndrome | I | 11th chormosome |
B | α-Thalassemia | II | ‘X’ chromosome |
C | β-Thalassemia | III | 21st chromosome |
D | Klinefelter’s syndrome | IV | 16th chromosome |