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: \[ 5 \]
The monomer (X) involved in the synthesis of Nylon 6,6 gives positive carbylamine test. If 10 moles of X are analyzed using Dumas method, the amount (in grams) of nitrogen gas evolved is ____. Use: Atomic mass of N (in amu) = 14
The correct match of the group reagents in List-I for precipitating the metal ion given in List-II from solutions is:
List-I | List-II |
---|---|
(P) Passing H2S in the presence of NH4OH | (1) Cu2+ |
(Q) (NH4)2CO3 in the presence of NH4OH | (2) Al3+ |
(R) NH4OH in the presence of NH4Cl | (3) Mn2+ |
(S) Passing H2S in the presence of dilute HCl | (4) Ba2+ (5) Mg2+ |
Match List I with List II:
Choose the correct answer from the options given below:
Let $ S $ denote the locus of the point of intersection of the pair of lines $$ 4x - 3y = 12\alpha,\quad 4\alpha x + 3\alpha y = 12, $$ where $ \alpha $ varies over the set of non-zero real numbers. Let $ T $ be the tangent to $ S $ passing through the points $ (p, 0) $ and $ (0, q) $, $ q > 0 $, and parallel to the line $ 4x - \frac{3}{\sqrt{2}} y = 0 $.
Then the value of $ pq $ is