In an isolated transition metal atom the five d orbitals have the same energy. When a transition metal ion forms a tetrahedral complex the d orbitals are split into two groups of different energies.

Question 3 [Transition Metals]

In an isolated transition metal atom the five d orbitals have the same energy. When a transition metal ion forms a tetrahedral complex the d orbitals are split into two groups of different energies.

Reference: Past Exam Paper – 9701 March 2016 Paper 42 Q3

Solution:

a) [electron configuration in Transition Metals]

Co [Ar] 4s² 3d⁷         Co²⁺ [Ar] 3d⁷

(recall, in formation of a cation, electrons are first lost from 4s subshell than 3d. This is due to 4s subshell being at a lower and more accessible energy level)

b) [coloured compounds in Transition Metals]

Splitting happens in ratio of 3:2 :: higher energy level: lower energy level.

(this was a trick question, most commonly octahedral shapes are asked; in octahedral complexes, 2 orbitals are promoted to a higher energy level while 3 remain on lower level. It is the other way around for tetrahedral complexes.)

c)

i) [transitional metal complexes in Transition Metals]

[Co(H₂O)₃(Cl)₃]^-

 Few points to keep in mind:

-         Even though question states chloride ions, we place no charge on the individual ligand- in both the diagram and formula. Therefore [Co(H₂O)₃(Cl^-)₃]^- is incorrect.

-         Calculation for overall charge must always be done, it is NOT always the charge on the transition metal ion.  Question states Cobalt(II) therefore net charge= +2+3(0)+3(-1)= -1 [water is a neutral ligand, chloride ions carry unit negative charge]

ii) [isomerism in Transition Metals]  

When there are 2 different types of ligands, 3 each, the isomers are a special case of “fac-mer” isomerism.

Isomer 1 is a FAC isomer. With respect to water ligands, the plane created (highlighted in pink) does NOT pass through the central metal ion. In case of isomer 2(MER isomer), the plane created contains the central metal ions. These are the only 2 possible isomers. All other versions will be identical to either of these and will be penalised in the marking scheme.

d)i) [Pt(NH₃)₂(Cl)₂] [transitional metal complexes in Transition Metals]

      - Pt has charge +2 as stated in the question

      - NH₃ molecules are neutral ligands; charge=0

      - Chloride ions carry -1 charge each.

      - Therefore net charge= +2+2(-1)=0

ii) [isomerism in Transition Metals]

Please keep in mind, lone pair in NH3 is located on N atom, Therefore bonding must be between N atom and Pt atom. Inversion of formula is necessary when ligand is attached on left hand side(as seen in diagram 2)

iii) Cis-platin; It is able to bind to (the guanine bases) in DNA strands and this prevent DNA replication.

(this is a memory based question. Trans-platin is ineffective in the body.)

e)

i) [stability constants in Transition Metals]

points to note:

-         Brackets are very important especially when dealing with complexes in K_stab

-         Sometimes double square brackets are required ( as seen in the image); The inner square bracket is part of the formulaic representation while the outer square bracket is the standard notation for “concentration”.

-         Remember to write the net charge of the complex after the first inner bracket.

-         Any stoichiometric coefficient in the equation is transposed to become the power/exponent of its corresponding species.

ii) since value is in order of 10¹³, Kstab value is very large. (=> numerator value is much greater than denominator=>tetrammine complex is much more stable)

Solutions provided by Kashish Varshney, India