Problem 76 - A Complete Resource Book in Chemistry for JEE Main

Question

The complex ion which has no d electrons in the central metal atom is (Atomic number \(\mathrm{Cr}=24\), \(\mathrm{Mn}=25, \mathrm{Fe}=26, \mathrm{Co}=27)\) (a) \(\left[\mathrm{MnO}_{4}\right]\) (b) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) (c) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3}\) (d) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\)

Step-by-Step Solution

Answer

The complex ion \([\mathrm{MnO}_{4}]^{-}\) has no d electrons in the central metal atom.

Step 1: Determine the Oxidation State of Each Metal In Complexes

To find out the oxidation state of the central metal atom in the complex ions, first note that complex ions have ligands that donate electron pairs. We need to assume typical ligand charges and solve for the metal's oxidation state. For example, oxygen usually has a -2 charge, ammonia is a neutral ligand, and cyanide has a -1 charge.

Step 2: Calculate Oxidation States

(a) In \([\mathrm{MnO}_{4}]\), the oxidation state of \(\mathrm{Mn}+4(-2) = -1\). Therefore, \(\mathrm{Mn} = +7\).(b) In \([\mathrm{Co}(\mathrm{NH}_{3})_{6}]^{3+}\), \(\mathrm{Co}+0 = 3\), so \(\mathrm{Co} = +3\).(c) In \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\), \(\mathrm{Fe}+6(-1)=-3\). Thus, \(\mathrm{Fe} = +3\).(d) In \([\mathrm{Cr}(\mathrm{H}_{2}\mathrm{O})_{6}]^{3+}\), \(\mathrm{Cr}+0 = 3\), so \(\mathrm{Cr} = +3\).

Step 3: Determine the "d" Electron Configuration of Each Oxidation State

Calculate the number of "d" electrons remaining for each metal's oxidation state.(a) For \(\mathrm{Mn}^{+7}\), the electron configuration starts from \(\mathrm{Mn}\) which is \([\mathrm{Ar}]3d^5 4s^2\). Removing 7 electrons results in no electrons in the 3d subshell, \([\text{0 in } 3d]\).(b) \(\mathrm{Co}^{+3}\): Starts from \([\mathrm{Ar}]3d^7 4s^2\), so \(\mathrm{Co}^{3+}\) is \([\mathrm{Ar}]3d^6\) - 6 d-electrons.(c) \(\mathrm{Fe}^{+3}\): Starts from \([\mathrm{Ar}]3d^6 4s^2\), so \(\mathrm{Fe}^{3+}\) is \([\mathrm{Ar}]3d^5\) - 5 d-electrons.(d) \(\mathrm{Cr}^{+3}\): Starts from \([\mathrm{Ar}]3d^5 4s^1\), so \(\mathrm{Cr}^{3+}\) is \([\mathrm{Ar}]3d^3\) - 3 d-electrons.

Step 4: Compare the Electron Configurations of Central Atoms

From these results, we see that only \(\mathrm{Mn}^{+7}\) in \([\mathrm{MnO}_{4}]^{-}\) has zero d-electrons. All others have d-electrons remaining - \(\mathrm{Co}^{3+}\) has 6, \(\mathrm{Fe}^{3+}\) has 5, and \(\mathrm{Cr}^{3+}\) has 3 d-electrons.

Key Concepts

Oxidation States

Understanding oxidation states is crucial in coordination chemistry, particularly for transition metals. Oxidation state is essentially the charge of a metal ion when bonded with ligands, and it determines how many electrons a metal atom can accept or donate. Transition metals often have multiple oxidation states due to their ability to lose different numbers of d electrons. For example:

• In \([\text{MnO}_4]^-\), manganese has an oxidation state of \(+7\) because each oxygen atom carries a \(-2\) charge, contributing to a total charge of \(-8\). Thus, manganese has to be \(+7\) to balance and meet the complex's overall \(-1\) charge.
• Cobalt in \([\text{Co}(\text{NH}_3)_6]^{3+} \) is in the \(+3\) oxidation state, ammonia being a neutral ligand that does not alter the oxidation state.
• In \([\text{Fe}(\text{CN})_6]^{3-} \), iron is \(+3\) since each cyanide acts as a \(-1\) charged ligand.
• Chromium in \([\text{Cr}(\text{H}_2\text{O})_6]^{3+} \) also has an oxidation state of \(+3\), with water as a neutral ligand.

Knowing these oxidation states allows us to decipher electron configurations and reactivity patterns in coordination complexes.

Transition Metals

Transition metals are situated in the d-block of the periodic table and are known for their unique properties, which include their variable oxidation states, formation of colored complexes, and catalytic abilities.
All transition metals have:

• Partially filled d-orbitals, which allow them to form a wide array of complex ions by accepting or donating electrons to/from surrounding ligands.
• Exhibit different physical properties such as high melting points and magnetic properties due to the presence of unpaired electrons.
• Can participate in redox reactions due to their ability to change oxidation states easily.

The flexibility in oxidation states, seen in metals like manganese (\(+7\)), iron (\(+3\)) and cobalt (\(+3\)), gives transition metals their distinctiveness and practical import in technological and biological systems.
These traits make them exceptional for forming complex ions like \([\text{MnO}_4]^-\), well-known for vivid colors and versatility in oxidation-reduction reactions.

Electron Configuration

Electron configuration for transition metals involves understanding how electrons fill the orbitals. The filling order follows the principle that electrons are added to orbitals in order of increasing energy levels, specifically in the pattern:

• "4s" is filled before "3d" due to its lower energy
• d-orbitals can be partially filled due to their similar energy levels

For clarity, consider:

• Manganese (Mn) with an atomic number of 25 has a base configuration of \([\text{Ar}]3d^5 4s^2\). When it forms \([\text{MnO}_4]^-\), it loses all 7 outer electrons ending up with \(0\) d-electrons, a rare case for d-block elements.
• Cobalt (Co) in \(+3\) state maintains an electron configuration of \([\text{Ar}]3d^6\), losing its 4s and some 3d electrons.
• Iron (\text{Fe}^{3+}) similarly sheds electrons to reach \([\text{Ar}]3d^5\).

These configurations explain why some ions appear more stable or exhibit particular magnetic properties. Electrons in these configurations directly influence the chemical behavior and formation of coordination compounds, vital for studies in chemistry and applications in materials science.