Question

Choose the correct sequence on the basis of number of unpaired electrons present in the given field:
A. $Cr(+II)$ in weak field
B. $Co(+II)$ in strong field
C. $Ni(+II)$ in strong field
D. $Pt(+II)$ in weak field
E. $Ag(+II)$ in strong field
Choose the correct answer from the options given below :

• A. C = D < B = E < A
• B. B = E < A < C = D
• C. B < C < D = A < E
• D. A < B < C < D = E

Hint:

Key exceptions to memorize: 4d (Pd, Ru, Ag) and 5d (Pt, Ir, Au) transition metals essentially never form high-spin complexes. Furthermore, $d^8$ metals in a strong field (like $Ni^{2+}$ with $CN^-$) or heavy $d^8$ metals (like $Pt^{2+}, Pd^{2+}$) predominantly form square planar complexes, which are completely diamagnetic (0 unpaired electrons).

Answer 1

The Correct Option is A

Step 1: Concept:
The question requires determining the number of unpaired d-electrons in various transition metal complex ions, taking into account their oxidation state and the strength of the ligand field (Crystal Field Theory), and then ordering them.

Step 2: Key Formula or Approach:
1. Determine the oxidation state and $d^n$ configuration for the metal ion.
2. Apply Crystal Field Theory based on geometry and field strength:
- Strong field ligands cause large splitting ($\Delta_o$), forcing electrons to pair up in lower energy orbitals (low-spin).
- Weak field ligands cause small splitting, allowing electrons to occupy higher energy orbitals before pairing (high-spin).
- Remember that 4d and 5d metals essentially always form low-spin complexes, regardless of ligand strength, because of their larger spatial orbital extent. Furthermore, $d^8$ configurations for strong fields or heavy metals frequently form square planar, diamagnetic ($n_{unpaired} = 0$) complexes.

Step 3: Step-by-step Explanation:

• A. $Cr(+II)$ in weak field: Cr is $[Ar] 4s^1 3d^5$. Cr(II) is $d^4$. In a weak octahedral field, it forms a high-spin complex: $t_{2g}^3 e_g^1$. This gives 4 unpaired electrons.

• B. $Co(+II)$ in strong field: Co is $[Ar] 4s^2 3d^7$. Co(II) is $d^7$. In a strong octahedral field, it forms a low-spin complex: $t_{2g}^6 e_g^1$. This gives 1 unpaired electron.

• C. $Ni(+II)$ in strong field: Ni is $[Ar] 4s^2 3d^8$. Ni(II) is $d^8$. In a strong field, Ni(II) typically forms a square planar complex, where the $d_{x^2-y^2}$ orbital is very high in energy and empty, leading to a paired configuration. Thus, it is diamagnetic with 0 unpaired electrons.

• D. $Pt(+II)$ in weak field: Pt is a 5d transition metal. Its compounds always experience a very large crystal field splitting, effectively acting as "strong field" cases regardless of the ligand. $Pt(+II)$ is $d^8$ and almost exclusively forms square planar complexes that are diamagnetic. Thus, it has 0 unpaired electrons.

• E. $Ag(+II)$ in strong field: Ag is $[Kr] 5s^1 4d^{10}$. Ag(II) is $d^9$. In any ligand field (octahedral or square planar), a $d^9$ configuration must have exactly 1 unpaired electron.
Summary of Unpaired Electrons:
C ($0$) = D ($0$) $<$ B ($1$) = E ($1$) $<$ A ($4$).

Step 4: Final Answer:
This specific ordering directly matches option (A).