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 |
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $
Electron Configuration is referred to as the distribution of electrons in an atom's orbitals. An electron in an atom is defined by a set of four quantum numbers (n), the most important of which defines the main energy level known as a shell. The filling of electrons into different subshells, also known as orbitals (s, p, d, f) in an atom. The position of an element in the periodic table is determined by the quantum numbers of the last orbital filled.