The atomic number of Einsteinium (Es) is 99. The electronic configuration of an element can be determined by following the Aufbau principle, which fills the orbitals in the order of increasing energy.
The configuration for Einsteinium is $[\text{Rn}]5f^{11}6d^07s^2$, where:
$[\text{Rn}]$ represents the electron configuration of the nearest noble gas, Radon (86).
The $5f^{11}$ represents the 11 electrons in the 5f subshell.
The $6d^0$ indicates no electrons in the 6d subshell.
The $7s^2$ indicates the 2 electrons in the 7s subshell.
Thus, the correct answer is (2).
Which of the following is the correct electronic configuration for \( \text{Oxygen (O)} \)?
Correct statements for an element with atomic number 9 are
A. There can be 5 electrons for which $ m_s = +\frac{1}{2} $ and 4 electrons for which $ m_s = -\frac{1}{2} $
B. There is only one electron in $ p_z $ orbital.
C. The last electron goes to orbital with $ n = 2 $ and $ l = 1 $.
D. The sum of angular nodes of all the atomic orbitals is 1.
Choose the correct answer from the options given below:
The energy of an electron in first Bohr orbit of H-atom is $-13.6$ eV. The magnitude of energy value of electron in the first excited state of Be$^{3+}$ is _____ eV (nearest integer value)
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} $
20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Nearest Integer value) (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol$^{-1}$)