List-I (Element) | List-II (Electronic Configuration) |
---|---|
A. N | I. [Ar] 3d10 4s2 4p5 |
B. S | II. [Ne] 3s2 3p4 |
C. Br | III. [He] 2s2 2p3 |
D. Kr | IV. [Ar] 3d10 4s2 4p6 |
Let us match the electronic configurations:
A. N (Nitrogen):} Nitrogen has the electronic configuration $1s^2 2s^2 2p^3$, which corresponds to configuration III.
B. S (Sulfur):} Sulfur has the electronic configuration $[\text{Ne}] 3s^2 3p^4$, which corresponds to configuration II.
C. Br (Bromine):} Bromine has the electronic configuration $[\text{Ar}] 3d^{10} 4s^2 4p^5$, which corresponds to configuration I.
D. Kr (Krypton):} Krypton has the electronic configuration $[\text{Ar}] 3d^{10} 4s^2 4p^6$, which corresponds to configuration IV.
Thus, the correct match is A-III, B-II, C-I, D-IV. Hence, the answer is (2).
If the system of equations \[ x + 2y - 3z = 2, \quad 2x + \lambda y + 5z = 5, \quad 14x + 3y + \mu z = 33 \] has infinitely many solutions, then \( \lambda + \mu \) is equal to:}
The equilibrium constant for decomposition of $ H_2O $ (g) $ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \quad (\Delta G^\circ = 92.34 \, \text{kJ mol}^{-1}) $ is $ 8.0 \times 10^{-3} $ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation ($ \alpha $) of water is _____ $\times 10^{-2}$ (nearest integer value). [Assume $ \alpha $ is negligible with respect to 1]