To determine the number of complexes with no electrons in the \(t_2\) orbital, analyze each complex and its oxidation state, electronic configuration, and crystal field splitting.
Step 1: Analyze each complex
1.\(\text{TiCl}_4\):
- Oxidation state of Ti: \(+4 \, (\text{Ti}^{4+})\).
- Electronic configuration of \(\text{Ti}^{4+}\): \(3d^0\).
- No electrons in \(t_2\) orbitals.
2.\([\text{MnO}_4]^-\):
- Oxidation state of Mn: \(+7 \, (\text{Mn}^{7+})\).
- Electronic configuration of \(\text{Mn}^{7+}\): \(3d^0\).
- No electrons in \(t_2\) orbitals.
3.\([\text{FeO}_4]^{2-}\):
- Oxidation state of Fe: \(+6 \, (\text{Fe}^{6+})\).
- Electronic configuration of \(\text{Fe}^{6+}\): \(3d^0\).
- No electrons in \(t_2\) orbitals.
4.\([\text{FeCl}_4]^-\):
- Oxidation state of Fe: \(+2 \, (\text{Fe}^{2+})\).
- Electronic configuration of \(\text{Fe}^{2+}\): \(3d^6\).
- \(t_2\) orbitals are populated with electrons (\(t_2^3e^3\)).
5.\([\text{CoCl}_4]^{2-}\):
- Oxidation state of Co: \(+2 \, (\text{Co}^{2+})\).
- Electronic configuration of \(\text{Co}^{2+}\): \(3d^7\).
- \(t_2\) orbitals are populated with electrons (\(t_2^4e^3\)).
Step 2: Count complexes with no \(t_2\) electrons
- \(\text{TiCl}_4\), \([\text{MnO}_4]^-\), and \([\text{FeO}_4]^{2-}\) have no electrons in \(t_2\) orbitals.
- \([\text{FeCl}_4]^-\) and \([\text{CoCl}_4]^{2-}\) have electrons in \(t_2\) orbitals.
Conclusion:
The number of complexes with no electrons in the \(t_2\) orbital is:
\[3 \, (\text{TiCl}_4, \, [\text{MnO}_4]^-, \, [\text{FeO}_4]^{2-}).\]
Final Answer: (1).
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
| List I (Substances) | List II (Element Present) |
| (A) Ziegler catalyst | (I) Rhodium |
| (B) Blood Pigment | (II) Cobalt |
| (C) Wilkinson catalyst | (III) Iron |
| (D) Vitamin B12 | (IV) Titanium |
| List-I (Complex ion) | List-II (Spin only magnetic moment in B.M.) |
|---|---|
| (A) [Cr(NH$_3$)$_6$]$^{3+}$ | (I) 4.90 |
| (B) [NiCl$_4$]$^{2-}$ | (II) 3.87 |
| (C) [CoF$_6$]$^{3-}$ | (III) 0.0 |
| (D) [Ni(CN)$_4$]$^{2-}$ | (IV) 2.83 |
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,