| 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 |
The question requires matching elements with their respective electronic configurations. Let's analyze each element's correct electronic configuration to identify the correct match.
From the analysis above, the correct matching is:
| List-I (Element) | List-II (Electronic Configuration) |
|---|---|
| A. N | III. [He] 2s2 2p3 |
| B. S | II. [Ne] 3s2 3p4 |
| C. Br | I. [Ar] 3d10 4s2 4p5 |
| D. Kr | IV. [Ar] 3d10 4s2 4p6 |
Thus, the correct answer is: A-III, B-II, C-I, D-IV.
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).
The figures below show:
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Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}