A.
[Pt(Cl2)(NH3)2]The formula is most likely intended to be
[PtCl2(NH3)2], which has two chloride ligands and two ammonia ligands.
Find the oxidation state of Platinum (Pt):
Let the oxidation state of Pt be
x. The charge on each chloride ligand (
Cl−) is -1, and ammonia (
NH3) is a neutral ligand (charge 0). The overall complex is neutral.
x+2(−1)+2(0)=0x−2=0x=+2So, we have Platinum in the +2 oxidation state, which is
Pt2+.
Determine the electronic configuration:
The atomic number of Pt is 78. Its ground-state electronic configuration is
[Xe]4f145d96s1.
For
Pt2+, we remove two electrons (one from
6s and one from
5d).
The electronic configuration of
Pt2+ is
[Xe]4f145d8.
Determine the hybridization and shape:
The coordination number is 4. For a
d8 configuration, particularly for metals in the 4d and 5d series like Pt, the ligands always cause the electrons to pair up, leaving one
d-orbital empty. This leads to
dsp2 hybridization.
The orbital diagram for
Pt2+ in the complex will be:
5d6s6pThe empty
5d,
6s, and two
6p orbitals hybridize to form four
dsp2 hybrid orbitals.
The geometry corresponding to
dsp2 hybridization is Square planar.
Therefore, A matches with III.
B.
[Co(NH3)6]Cl3The complex ion here is
[Co(NH3)6]3+.
Find the oxidation state of Cobalt (Co):
Let the oxidation state of Co be
x. Ammonia (
NH3) is a neutral ligand. The overall charge on the complex ion is +3.
x+6(0)=+3x=+3So, we have Cobalt in the +3 oxidation state, which is
Co3+.
Determine the electronic configuration:
The atomic number of Co is 27. Its ground-state electronic configuration is
[Ar]3d74s2.
For
Co3+, we remove three electrons (two from
4s and one from
3d).
The electronic configuration of
Co3+ is
[Ar]3d6.
Determine the hybridization and shape:
The coordination number is 6. The ligand is
NH3, which is a strong-field ligand. It will cause the pairing of the
3d electrons.
The electrons in the
3d orbitals of
Co3+ will rearrange from
[↑↓][↑][↑][↑][↑] to
[↑↓][↑↓][↑↓][][].
This leaves two
3d orbitals vacant.
3d4s4pHybridization involves two
3d, one
4s, and three
4p orbitals, leading to
d2sp3 hybridization. This is an inner orbital complex.
The geometry corresponding to
d2sp3 hybridization is Octahedral.
Therefore, B matches with I.
C.
[NiCl4]2−Find the oxidation state of Nickel (Ni):
Let the oxidation state of Ni be
x. The charge on each chloride ligand (
Cl−) is -1. The overall charge on the complex ion is -2.
x+4(−1)=−2x−4=−2x=+2So, we have Nickel in the +2 oxidation state, which is
Ni2+.
Determine the electronic configuration:
The atomic number of Ni is 28. Its ground-state electronic configuration is
[Ar]3d84s2.
For
Ni2+, we remove two electrons from the
4s orbital.
The electronic configuration of
Ni2+ is
[Ar]3d8.
Determine the hybridization and shape:
The coordination number is 4. The ligand
Cl− is a weak-field ligand. It does not cause the pairing of
3d electrons.
The orbital diagram for
Ni2+ is:
3d4s4pSince no inner
d-orbitals are empty, the hybridization will involve the outer orbitals: one
4s and three
4p orbitals. This gives
sp3 hybridization.
The geometry corresponding to
sp3 hybridization is Tetrahedral.
Therefore, C matches with IV.
D.
[Fe(CO)5]Find the oxidation state of Iron (Fe):
The ligand is carbonyl (
CO), which is a neutral ligand. The overall complex is neutral.
Therefore, the oxidation state of Fe is 0.
Determine the electronic configuration:
The atomic number of Fe is 26. Its ground-state electronic configuration is
[Ar]3d64s2. It has 8 valence electrons.
Determine the hybridization and shape:
The coordination number is 5. The ligand
CO is a strong-field ligand. It causes the pairing of all valence electrons (from both
3d and
4s) into the
3d orbitals.
The 8 valence electrons will pair up in the
3d orbitals, resulting in a
3d8 configuration within the complex.
3d4s4pFor bonding with five
CO ligands, one empty
3d, one
4s, and three
4p orbitals are used for hybridization. This gives
dsp3 hybridization.
The geometry corresponding to
dsp3 hybridization is Trigonal bipyramidal.
Therefore, D matches with II.
Conclusion:
Let's summarise our findings:
A.
[Pt(Cl2)(NH3)2] is Square planar (III).
B.
[Co(NH3)6]Cl3 is Octahedral (I).
C.
[NiCl4]2− is Tetrahedral (IV).
D.
[Fe(CO)5] is Trigonal bipyramidal (II).
Matching these results with the given options, we get:
A-III, B-I, C-IV, D-II.
This corresponds to Option B.
The correct answer is Option B.
