NEET Practice Questions (MCQs) with Answers & Solutions

The complex $[Cr(NH_3)_6]Cl_3$ involves $d^2sp^3$ hybridization, making it an inner orbital complex (using inner d-orbitals). The shape of the complex is octahedral. It is paramagnetic due to the presence of unpaired electrons. The complex also gives a white precipitate with silver nitrate solution due to the presence of chloride ions. Therefore, the incorrect statement is that it is an outer orbital complex.

The magnetic moment of a complex can be calculated using the formula $ ext{Magnetic Moment} = ext{BM} = ext{n(n+2)}$, where $n$ is the number of unpaired electrons. For $[NiCl_4]^{2-}$, Ni is in the +2 oxidation state with a configuration of $3d^8$. In a tetrahedral field, this results in 2 unpaired electrons. Thus, the magnetic moment is $ ext{BM} = ext{2(2+2)} = 2.82$ BM.

The field strength of ligands is determined by their ability to split the d-orbitals of the central metal ion. In the spectrochemical series, the increasing order of field strength for the given ligands is: $NH_3 < en < CN^- < CO$. This order is based on the fact that $CO$ is a strong field ligand, followed by $CN^-$, then $en$ (ethylenediamine), and $NH_3$ is the weakest field ligand among the given options.

The magnetic moment is related to the number of unpaired electrons in the complex. The formula for spin-only magnetic moment (μ) is given by: $ ext{μ} = ext{BM} = ext{n(n+2)}^{1/2}$, where n is the number of unpaired electrons. For [NiCl_4]^{2-}, Ni(II) has a 3d^8 configuration with 2 unpaired electrons, leading to a magnetic moment of approximately 2.82 BM.

In $Cr(CO)_6$, the chromium is in the zero oxidation state (Cr^0) and has a 3d^6 4s^0 configuration. The strong field ligand CO causes pairing of all the electrons in the d-orbitals, resulting in no unpaired electrons. Therefore, the spin-only magnetic moment is 0 BM.

Potassium ferrocyanide, K4[Fe(CN)6], has an octahedral geometry around the Fe2+ ion. The six cyanide (CN-) ligands are arranged in an octahedral fashion around the central iron ion.

In an octahedral structure, the pair of d-orbitals involved in $d^2sp^3$ hybridization are $d_{x^2-y^2}$ and $d_{z^2}$. These orbitals point directly along the axes and are capable of forming strong directional bonds with the surrounding ligands.

$[Fe(CN)_6]^{4-}$ is diamagnetic because the Fe2+ ion in this complex has a low-spin configuration due to the strong field ligand (cyanide, CN-). This results in paired electrons in all the d-orbitals, leading to no unpaired electrons and hence diamagnetic behavior.

The magnitude of \( \\Delta_0 \\) (crystal field splitting energy) depends on the field strength of the ligands. In the given complexes, the ligands are CN^-, C_2O_4^{2-}, H_2O, and NH_3. Among these, CN^- is a strong field ligand and causes the maximum splitting. Therefore, \( [Co(CN)_6]^{3-} \\) will have the highest \( \\Delta_0 \\).

For the complex \( [Co(NH_3)_6]^{3+} \\), Co is in the +3 oxidation state (d^6 configuration). NH_3 is a strong field ligand causing pairing of electrons, resulting in 0 unpaired electrons. For \( [CoF_6]^{3-} \\), Co is also in the +3 oxidation state (d^6 configuration) but F^- is a weak field ligand, so it does not cause pairing of electrons, resulting in 4 unpaired electrons.