NEET Practice Questions (MCQs) with Answers & Solutions

To determine the oxidation state of the central metal atom in various complexes, we need to consider the charges of the ligands and the overall charge of the complex. In $[Cr(CO)_6]$, carbon monoxide (CO) is a neutral ligand, meaning it has an oxidation state of 0. Therefore, the oxidation state of chromium in $[Cr(CO)_6]$ is 0. This makes $[Cr(CO)_6]$ the correct answer where the oxidation state of the metal is zero.

To find the oxidation state of nickel in the complex compound $K_4[Ni(CN)_4]$, we can use the following steps:

1. Let the oxidation state of nickel (Ni) be x.
2. Potassium (K) has an oxidation state of +1, and there are 4 potassium ions, so the total oxidation state contribution from potassium is +4.
3. Cyanide (CN) is a ligand with an oxidation state of -1, and there are 4 cyanide ions, so the total oxidation state contribution from cyanide is -4.
4. The overall charge of the complex ion $[Ni(CN)_4]$ is -4 (since it needs to balance the +4 charge from potassium to make the compound neutral).

So, we have the equation:

$$4(+1) + x + 4(-1) = 0$$

Simplifying this, we get:

$$4 + x - 4 = 0$$

$$x = 0$$

Therefore, the oxidation state of nickel in $K_4[Ni(CN)_4]$ is 0.

The highest possible oxidation states for transition metals are generally found by considering their position in the periodic table. Chromium (Cr) can reach an oxidation state of +6, and manganese (Mn) can reach an oxidation state of +7.

In the compound $CrO_2Cl_2$, chromium is in the +6 oxidation state because oxygen has a -2 oxidation state and chlorine has a -1 oxidation state. For the overall charge to be neutral, Cr must be +6.

In the compound $MnO_4^-$, manganese is in the +7 oxidation state because oxygen has a -2 oxidation state and there are four oxygen atoms contributing a total of -8. For the overall charge to be -1, Mn must be +7.

Thus, both $CrO_2Cl_2$ and $MnO_4^-$ have metals in their highest possible oxidation states.

To determine the number of unpaired electrons in the complex ion $[CoF_6]^{3-}$, we follow these steps:

1. The atomic number of cobalt (Co) is 27, so its electron configuration is $[Ar] 3d^7 4s^2$.
2. In the $[CoF_6]^{3-}$ complex, cobalt is in the +3 oxidation state. This means cobalt loses three electrons, resulting in the electron configuration $[Ar] 3d^6$.
3. Fluoride (F-) is a weak field ligand and does not cause pairing of electrons.
4. In a weak field, the $3d$ orbitals remain unpaired as much as possible.

Thus, the electron configuration for $Co^{3+}$ in a weak field is $t_{2g}^4 e_g^2$, where there are 4 electrons in the $t_{2g}$ orbitals and 2 electrons in the $e_g$ orbitals.

Among the 6 electrons in the $d$ orbitals, 4 will remain unpaired.

Therefore, the number of unpaired electrons in $[CoF_6]^{3-}$ is 4.

The compound $[Co(NH_3)_5NO_2]Cl_2$ does not show geometrical isomerism because it contains only one bidentate ligand (NO2), and the positions of the ligands around the central metal ion do not allow for geometrical isomers. Geometrical isomerism typically occurs in complexes with at least two different types of ligands that can occupy different spatial arrangements around the central metal ion.

The complex $[Cr(NH_3)_6][Co(CN)_6]$ can exhibit coordination isomerism. In coordination isomerism, the composition of the coordination entities (complex ions) is interchangeable between the cation and anion. Thus, $[Cr(NH_3)_6][Co(CN)_6]$ can have its metal ions swap places to form $[Co(NH_3)_6][Cr(CN)_6]$.

The compounds $[Co(NH_3)_5NO_2]Cl_2$ and $[Co(NH_3)_5(ONO)]Cl_2$ are related to each other as linkage isomers. Linkage isomerism occurs when a ligand can coordinate to the metal ion through two different atoms. In this case, the nitrite ligand (NO2) can bind through the nitrogen atom (NO2) or the oxygen atom (ONO).

The compounds $[Co(NH_3)_5Br]SO_4$ and $[Co(NH_3)_5SO_4]Br$ are examples of ionisation isomerism. This type of isomerism occurs when the counter ion in a complex salt is itself a potential ligand and can displace the ligand within the coordination sphere to give a different ionisable species.

$[Co(NH_3)_5 Br]SO_4$ exhibits ionisation isomerism. Ionisation isomerism occurs when an ionisable counter ion such as SO_4^{2-} or Br^- is swapped with a ligand in the coordination sphere, leading to different compounds that ionise to give different ions in solution.

Among the given options, $[Fe(H_2O)_6]^{+2}$ has the highest paramagnetism. This is because it has the maximum number of unpaired electrons compared to the other ions listed. Iron(II) in $[Fe(H_2O)_6]^{+2}$ has 4 unpaired electrons (d^6 configuration), making it the most paramagnetic.