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).
Based on Heisenberg's uncertainty principle, the uncertainty in the velocity of the electron to be found within an atomic nucleus of diameter \( 10^{-15} \, \text{m} \) is \( \dots \dots \times 10^9 \, \text{ms}^{-1} \) (nearest integer). \[ \text{[Given: mass of electron} = 9.1 \times 10^{-31} \, \text{kg, Planck's constant (} h \text{)} = 6.626 \times 10^{-34} \, \text{Js]} \] \[ \text{(Value of } \pi = 3.14) \]