Question

(i) (I) Write IUPAC name of the given complexes :

\( \mathrm{[Ag(CN)_2]} \) \( \mathrm{[Ag(NH_3)_2]} \)

(II) What is the coordination number of the central metal ion in the following complexes ?

\( \mathrm{[Fe(C_2O_4)_3]^{3-}} \) and \( \mathrm{[Cr(NH_3)_3Cl_3]} \)

(III) Draw a geometrical isomer of the complex \( \mathrm{[CoCl_2(en)_2]^+} \), which is optically active.

(ii) Using valence bond theory, explain the hybridization of \( \mathrm{[Ni(CO)_4]} \) and its magnetic behaviour.

[Atomic number : Ni = 28]

Hint:

Name the ligands first, then the metal. Oxalate is bidentate, so the coordination number is 6. Only the cis isomer is chiral. Ni(0) in Ni(CO)4 is sp3 and diamagnetic.

Answer 1

Step 1: Part (i)(I): Naming rules:
In a name, ligands come first, then the metal, with the oxidation state in Roman numerals.
If the complex ion is an anion, the metal name ends in -ate. Cyanide is written cyanido and ammonia is written ammine.
The charges are not printed, so the usual ions \( \mathrm{[Ag(CN)_2]^-} \) and \( \mathrm{[Ag(NH_3)_2]^+} \) are named.

Step 2: Part (i)(I): The two names:
In \( \mathrm{[Ag(CN)_2]^-} \) silver is +1 and the ion is an anion, so silver becomes argentate.
In \( \mathrm{[Ag(NH_3)_2]^+} \) silver is +1 and the ion is a cation, so the name keeps silver.
\[ \mathrm{[Ag(CN)_2]^-}\ :\ \text{dicyanidoargentate(I)} \] \[ \mathrm{[Ag(NH_3)_2]^+}\ :\ \text{diamminesilver(I)} \]

Step 3: Part (i)(II): Coordination numbers:
Coordination number is the number of donor atoms attached to the metal.
Oxalate \( \mathrm{C_2O_4^{2-}} \) is bidentate. Three oxalates give \( 3 \times 2 = 6 \), so Fe has coordination number 6.
In \( \mathrm{[Cr(NH_3)_3Cl_3]} \) there are three NH3 and three Cl, each unidentate. So Cr has \( 3 + 3 = 6 \).

Step 4: Part (i)(III): Optically active isomer:
\( \mathrm{[CoCl_2(en)_2]^+} \) is octahedral and has trans and cis forms.
The trans form has a plane of symmetry, so it is optically inactive.
The cis form, with the two Cl on adjacent corners (90 degrees apart) and the two en rings each spanning two adjacent positions, has no plane of symmetry.
Its mirror image cannot be superimposed on it, so cis-\( \mathrm{[CoCl_2(en)_2]^+} \) is optically active and exists as d and l forms.

Step 5: Part (ii): Hybridization of \( \mathrm{[Ni(CO)_4]} \):
Nickel is in the zero oxidation state. Ni(0) is \( 3d^8 4s^2 \).
CO is a strong field ligand. It shifts the two 4s electrons into 3d and pairs them, giving \( 3d^{10} \).
The 4s and three 4p orbitals are empty and mix to form four \( sp^3 \) hybrid orbitals. Each takes one lone pair from a CO molecule.
The shape is tetrahedral.

Step 6: Part (ii): Magnetic behaviour:
All ten d electrons are paired. No unpaired electron is present, so the complex is diamagnetic.

Final Answer:
(i)(I) Dicyanidoargentate(I); diamminesilver(I). (II) 6 and 6. (III) The cis isomer is optically active. (ii) \( \mathrm{[Ni(CO)_4]} \) is \( sp^3 \) hybridised, tetrahedral and diamagnetic.\[ \boxed{sp^3,\ \text{diamagnetic}} \]