Step 1: Determine the number of unpaired electrons for each complex using Crystal Field Theory.
\( [Co(NH_3)_6]^{3+} \) (\( d^6 \), strong field): 0 unpaired electrons (diamagnetic)
\( [Co(C_2O_4)_3]^{3-} \) (\( d^6 \), weak field): 4 unpaired electrons (paramagnetic)
\( [MnCl_6]^{3-} \) (\( d^4 \), weak field): 4 unpaired electrons (paramagnetic)
\( [Mn(CN)_6]^{3-} \) (\( d^4 \), strong field): 2 unpaired electrons (paramagnetic)
\( [CoF_6]^{3-} \) (\( d^6 \), weak field): 4 unpaired electrons (paramagnetic)
\( [Fe(CN)_6]^{3-} \) (\( d^5 \), strong field): 1 unpaired electron (paramagnetic)
\( [FeF_6]^{3-} \) (\( d^5 \), weak field): 5 unpaired electrons (paramagnetic)
Step 2: Identify the number of unpaired electrons for each paramagnetic species.
The paramagnetic species have 4, 4, 2, 4, 1, and 5 unpaired electrons respectively.
Step 3: Count the number of paramagnetic species that share the same number of unpaired electrons.
One unpaired electron: \( [Fe(CN)_6]^{3-} \) (1 species)
Two unpaired electrons: \( [Mn(CN)_6]^{3-} \) (1 species)
Four unpaired electrons: \( [Co(C_2O_4)_3]^{3-} \), \( [MnCl_6]^{3-} \), \( [CoF_6]^{3-} \) (3 species)
Five unpaired electrons: \( [FeF_6]^{3-} \) (1 species)
The maximum number of paramagnetic species with the same number of unpaired electrons is 3 (all having 4 unpaired electrons).
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,