Low spin tetrahedral complexes are not known.
Low spin tetrahedral complexes are not known.
Show Hint
Solution and Explanation
Low spin tetrahedral complexes are not known because tetrahedral complexes typically have a weaker ligand field splitting energy compared to octahedral complexes. In a tetrahedral field, the splitting energy (\(\Delta_T\)) is smaller, which does not create a significant enough energy difference between the d-orbitals to cause pairing of electrons. Therefore, in a tetrahedral complex, the electrons tend to occupy all available orbitals singly (high-spin state), and low-spin tetrahedral complexes are not stable.
Top Questions on Coordination chemistry
Werner’s coordination theory in 1893 was the first attempt to explain the bonding in coordination complexes. It must be remembered that this theory was put forward before the electron had been discovered by J.J. Thomson in 1897, and before the electronic theory of valency. Werner did not have any of the modern instrumental techniques and all his studies were made using simple experimental techniques. Werner was able to explain the nature of bonding in complexes and he concluded that in complexes, the metal shows two different sorts of valency: primary and secondary. Primary valences are normally ionisable whereas secondary valences are non-ionisable.
- CBSE CLASS XII - 2025
- Chemistry
- Coordination chemistry
- Write the name and structure of a hexadentate ligand.
- CBSE CLASS XII - 2025
- Chemistry
- Coordination chemistry
- Why is [Ni(CN)4]2- square planar while [Ni(CO)4] is tetrahedral? [Atomic number : Ni = 28]
- CBSE CLASS XII - 2025
- Chemistry
- Coordination chemistry
- Based on Valence Bond Theory, explain the geometry and magnetic character of [NiCl$_4$]$^{2-$. [Atomic number: Ni = 28]}
- CBSE CLASS XII - 2025
- Chemistry
- Coordination chemistry
- Explain why in tetrahedral coordination entities, low-spin configurations are rarely observed.
- CBSE CLASS XII - 2025
- Chemistry
- Coordination chemistry
Questions Asked in CBSE CLASS XII exam
- Calculate the emf of the following cell at 25°C : \[ \text{Zn(s)} | \text{Zn}^{2+} (0.1 \, \text{M}) \parallel \text{H}^+ (0.01 \, \text{M}) | \text{Hg}(l) | \text{Hg}_2^{2+} (1 \, \text{bar}), \, \text{Pt(s)} \] \[ \left[ E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \, \text{V}, \, E^\circ_{\text{H}_2/\text{H}^+} = 0.00 \, \text{V}, \, \log 10 = 1 \right] \]
- CBSE CLASS XII - 2025
- Electrochemistry
- The side of an equilateral triangle is increasing at the rate of 3 cm/s. At what rate is its area increasing when the side of the triangle is 15 cm?
- CBSE CLASS XII - 2025
- Differentiation
- If a random variable \( X \) has the probability distribution:
\[ P(X = x) = \begin{cases} k, & \text{if } x = 0 \\ 2k, & \text{if } x = 1 \text{ or } 2 \\ 0, & \text{otherwise} \end{cases} \] then the value of \( k \) is- CBSE CLASS XII - 2025
- Probability
- If $P(A) = \frac{1}{7}$, $P(B) = \frac{5}{7}$ and $P(A \cap B) = \frac{4}{7}$, then $P(A | B)$ is:
- CBSE CLASS XII - 2025
- Probability
- Apoorv Ltd. acquired building worth Rs 15,50,000, Machinery worth Rs 11,40,000 and Furniture worth Rs 1,10,000 from Dhruv Ltd. and took over its liabilities of Rs 2,00,000 for a purchase consideration of Rs 25,00,000. Apoorv Ltd. paid the purchase consideration by issuing 12% debentures of Rs 100 each at a premium of 25%.
Pass the necessary journal entries in the books of Apoorv Ltd. for the above transactions.- CBSE CLASS XII - 2025
- Debentures