Step 1: Analyze each compound to identify tetrahedral geometry.
1. \([Co(CN)_4]^{4-}\): CN is a strong ligand. Due to pairing, it undergoes \(sp^3\) hybridization, forming a tetrahedral structure.
2. \([Co(CO)_3(NO)]\): This forms a trigonal planar geometry due to the coordination environment.
3. \(XeF_4\): Square planar geometry, not tetrahedral.
4. \([PCl_4]^+\): Tetrahedral geometry due to \(sp^3\) hybridization.
5. \([PdCl_4]^{2-}\): Square planar geometry.
6. \([ICl_4]^-\): Square planar geometry.
7. \([Cu(CN)_4]^{3-}\): CN being a strong ligand leads to \(sp^3\) hybridization, hence tetrahedral.
8. \(P_4\): Tetrahedral geometry due to its molecular structure.
Step 2: Count the species with tetrahedral geometry.
Tetrahedral species: \([Co(CN)_4]^{4-}\), \([PCl_4]^+\), \([Cu(CN)_4]^{3-}\), and \(P_4\).
Step 3: Final answer. The total number of tetrahedral species is: \[ 5 \]
In the group analysis of cations, Ba$^{2+}$ & Ca$^{2+}$ are precipitated respectively as
Let $ P(x_1, y_1) $ and $ Q(x_2, y_2) $ be two distinct points on the ellipse $$ \frac{x^2}{9} + \frac{y^2}{4} = 1 $$ such that $ y_1 > 0 $, and $ y_2 > 0 $. Let $ C $ denote the circle $ x^2 + y^2 = 9 $, and $ M $ be the point $ (3, 0) $. Suppose the line $ x = x_1 $ intersects $ C $ at $ R $, and the line $ x = x_2 $ intersects $ C $ at $ S $, such that the $ y $-coordinates of $ R $ and $ S $ are positive. Let $ \angle ROM = \frac{\pi}{6} $ and $ \angle SOM = \frac{\pi}{3} $, where $ O $ denotes the origin $ (0, 0) $. Let $ |XY| $ denote the length of the line segment $ XY $. Then which of the following statements is (are) TRUE?