| List-I (Hybridization) | List-II (Orientation in Space) |
|---|---|
| (A) sp3 | (I) Trigonal bipyramidal |
| (B) dsp2 | (II) Octahedral |
| (C) sp3d | (III) Tetrahedral |
| (D) sp3d2 | (IV) Square planar |
To determine the correct match between the hybridization states in List-I and their corresponding orientations in List-II, we need to understand the geometry associated with each type of hybridization:
By analyzing the hybridizations and corresponding orientations, we can match List-I and List-II as follows:
| List-I (Hybridization) | List-II (Orientation in Space) |
|---|---|
| (A) sp3 | (III) Tetrahedral |
| (B) dsp2 | (IV) Square planar |
| (C) sp3d | (I) Trigonal bipyramidal |
| (D) sp3d2 | (II) Octahedral |
Therefore, the correct answer is A-III, B-IV, C-I, D-II.
\[\text{sp}^3 \rightarrow \text{Tetrahedral}\]
\[\text{dsp}^2 \rightarrow \text{Square planar}\]
\[\text{sp}^3\text{d} \rightarrow \text{Trigonal bipyramidal}\]
\[\text{sp}^3\text{d}^2 \rightarrow \text{Octahedral}\]
Among SO₃, NF₃, NH₃, XeF₂, CIF$_3$, and SF₆, the hybridization of the molecule with non-zero dipole moment and one or more lone-pairs of electrons on the central atom is:
Given below are two statements: 
Statement (II): Structure III is most stable, as the orbitals having the lone pairs are axial, where the $ \ell p - \beta p $ repulsion is minimum. In light of the above statements, choose the most appropriate answer from the options given below:
Match list-I with list-II and choose the correct option.
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}