IB DP Chemistry Topic 13.2 Coloured complexes HL Paper 1

Question

Which process is responsible for the colour of a transition metal complex?

A.     The absorption of light when electrons move between s orbitals and d orbitals

B.     The emission of light when electrons move between s orbitals and d orbitals

C.     The absorption of light when electrons move between different d orbitals

D.     The emission of light when electrons move between different d orbitals

▶️Answer/Explanation

C

The striking colors exhibited by transition-metal complexes are caused by excitation of an electron from a lower-energy d orbital to a higher-energy d orbital, which is called a d–d transition. Light is absorbed when an electron jumps to a higher energy orbit.

Question

Which electron transitions are responsible for the colours of transition metal compounds?

A.     Between d orbitals and s orbitals

B.     Among the attached ligands

C.     From the metal ion to the attached ligands

D.     Between d orbitals

▶️Answer/Explanation

D

The striking colors exhibited by transition-metal complexes are caused by excitation of an electron from a lower-energy d orbital to a higher-energy d orbital, which is called a d–d transition. Light is absorbed when an electron jumps to a higher energy orbit.

Question

Ligands can form dative covalent bonds with metal ions to form complex ions. Which of the following can act as a ligand?

I.     \({\text{C}}{{\text{l}}^ – }\)

II.     \({\text{N}}{{\text{H}}_{\text{3}}}\)

III.     \({{\text{H}}_{\text{2}}}{\text{O}}\)

A.     I and II only

B.     I and III only

C.     II and III only

D.     I, II and III

▶️Answer/Explanation

D

Ligand, is any atom or molecule attached to a central atom, usually a metallic element, in a coordination or complex compound. The atoms and molecules used as ligands are almost always those that are capable of functioning as the electron-pair donor in the electron-pair bond (a coordinate covalent bond) formed with the metal atom. Examples of common ligands are the neutral molecules water (H2O), ammonia (NH3), and carbon monoxide (CO) and the anions cyanide (CN), chloride (Cl), and hydroxide (OH).

Question

Which complex is colourless in solution?

A.     \({\text{[Fe(}}{{\text{H}}_2}{\text{O}}{{\text{)}}_6}{\text{]C}}{{\text{l}}_2}\)

B.     \({\text{[Ni(N}}{{\text{H}}_3}{{\text{)}}_6}{\text{]C}}{{\text{l}}_2}\)

C.     \({\text{[Zn(}}{{\text{H}}_2}{\text{O}}{{\text{)}}_6}{\text{](N}}{{\text{O}}_3}{{\text{)}}_2}\)

D.     \({{\text{K}}_3}[{\text{Co(CN}}{{\text{)}}_6}]\)

▶️Answer/Explanation

C

Electron configuration for zinc = [Ar] 3d10 4s2 – with all orbitals full up to the 4p orbitals which are empty.

The definition of a transition element is an element which has one or more vacancies in d orbitals, therefore zinc is NOT a transition element by the nature of all of its 3d orbitals being full.

Furthermore, the 3d orbitals are of lower energy than the 4s electrons which means that zinc has 2 (and 2 only) electrons available for ionic bonding. That is, none of the 3d electrons can be lost.

Hence, there is no d-d transition of electrons, hence it form colourless complex. 

Question

Which compound is likely to be colourless?

A.     \({\text{[Zn(}}{{\text{H}}_{\text{2}}}{\text{O}}{{\text{)}}_{\text{6}}}{\text{]C}}{{\text{l}}_{\text{2}}}\)

B.     \({{\text{[N}}{{\text{H}}_{\text{4}}}{\text{]}}_{\text{2}}}{\text{[Fe(}}{{\text{H}}_{\text{2}}}{\text{O}}{{\text{)}}_{\text{6}}}{\text{][S}}{{\text{O}}_{\text{4}}}{{\text{]}}_{\text{2}}}\)

C.     \({{\text{K}}_{\text{3}}}{\text{[Co(CN}}{{\text{)}}_{\text{6}}}{\text{]}}\)

D.     \({\text{[Ni(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{6}}}{\text{][B}}{{\text{F}}_{\text{4}}}{{\text{]}}_{\text{2}}}\)

▶️Answer/Explanation

A

Electron configuration for zinc = [Ar] 3d10 4s2 – with all orbitals full up to the 4p orbitals which are empty.

The definition of a transition element is an element which has one or more vacancies in d orbitals, therefore zinc is NOT a transition element by the nature of all of its 3d orbitals being full.

Furthermore, the 3d orbitals are of lower energy than the 4s electrons which means that zinc has 2 (and 2 only) electrons available for ionic bonding. That is, none of the 3d electrons can be lost.

Hence, there is no d-d transition of electrons, hence it form colourless complex. 

Question

Which ion is colourless?

A.     \({{\text{[Sc(}}{{\text{H}}_{\text{2}}}{\text{O}}{{\text{)}}_{\text{6}}}{\text{]}}^{3 + }}\)

B.     \({{\text{[Cr(}}{{\text{H}}_{\text{2}}}{\text{O}}{{\text{)}}_{\text{6}}}{\text{]}}^{3 + }}\)

C.     \({{\text{[Fe(}}{{\text{H}}_{\text{2}}}{\text{O}}{{\text{)}}_{\text{6}}}{\text{]}}^{3 + }}\)

D.     \({{\text{[Fe(CN}}{{\text{)}}_{\text{6}}}{\text{]}}^{3 – }}\)

▶️Answer/Explanation

A

Scandium full electron configuration will be 1s2 2s2 2p6 3s2 3p6 3d1 4s2.

The electron configuration of scandium ion(Sc3+) is 1s2 2s2 2p6 3s2 3p6. This electron configuration shows that scandium ion(Sc3+) has acquired the electron configuration of argon and it achieves an octave full stable electron configuration.

It has no 3d electrons. Hence no d-d transition is possible. Hence it will form colourless compound. 

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