The crystal field stabilization energy (CFSE) for [CoCl₆]^4– is 18000 cm^–1.
The CFSE for [CoCl₄]^2–  will be:

1.	$6000$ $c{m}^{-1}$	2.	$16000$ $c{m}^{-1}$
3.	$18000$ $c{m}^{-1}$	4.	$8000$ $c{m}^{-1}$

Zero magnetic moment of the octahedral complex ${\mathrm{K}}_2\left[{\mathrm{NiF}}_6\right]$ is due to: 

1. Low spin ${\mathrm{d}}^6$ Ni(IV) complex

2. Low spin ${\mathrm{d}}^8$ Ni(ll) complex

3. High spin ${\mathrm{d}}^8$ Ni(ll) complex

4. High spin ${\mathrm{d}}^6$ Ni(IV) complex

According to Crystal Field Theory (CFT), the compound, among the following, that forms a colored complex is:

1. [Ti(H₂O)₆]⁴⁺
2. [Cr(NH₃)₆]³⁺
3. K[VF₆]
4. (NH₄)₂[TiCl₆]

The hybridizations of [Ni(CO)₄] and [Cr(H₂O)₆]³⁺, respectively, are : 

1. sp³ and d³sp²

2. dsp² and d²sp³

3. sp³ and d²sp³

4. dsp²  and sp³d²

A metal ion has ${\mathrm{d}}^6$ configuration. In octahedral complex entity of low spin, the number of electrons present in ${\mathrm{e}}_{\mathrm{g}}$ set of orbitals will be :

1. Six (6)
2. Four (4)
3. Zero (0)
4. Two (2)

Spin-only magnetic moments of [Fe(NH₃)₆]³⁺ and [FeF₆]^3-  in BM are, respectively :

The number of geometrical isomers of [CrCl₂en(NH₃)₂] are:
[en-ethylenediamine]

Find the geometry and the number of unpaired electrons present in the complex ion [MnBr₄]²⁻.

2.335 g of compound X (empirical formal CoH₁₂N₄Cl₃) upon treatment with excess AgNO₃ solution produces 1.435 g of a white precipitate. Determine the primary and secondary valences of cobalt in compound X, respectively.

[Given Atomic mass: Co = 59, Cl = 35.5, Ag = 108]

The total number of geometrical isomers possible for an octahedral complex of $\left[{\mathrm{MA}}_2{\mathrm{B}}_2{\mathrm{C}}_2\right]$ is:
(M = transition metal; A, B and C are monodentate ligands)