Practicing Success

Target Exam

CUET

Subject

Chemistry

Chapter

Inorganic: D and F Block Elements

Question:

Read the following passage and answer the next five questions based on it:

Sc Ti V Cr Mn Fe Co Ni Cu Zn
Y Zr Nb Mo Te Ru Rh Pd Ag Cd
La Hf Ta W Re Os Ir Pt Au Hg

In any transition series, as we move from left to right the \(d-\)orbitals are progressively filled and their properties vary accordingly

Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr

The above are the two series of f-block elements in which the chemical properties won't change much. The 5f-series elements are radioactive in nature and mostly are artificially synthesized in laboratories and thus much is not known about their chemical properties.

Which of the following pair of compounds exhibits same colour in aqueous solution?

Options:

\(FeCl_2, CuCl_2\)

\(VOCl_2, CuCl_2\)

\(VOCl_2, FeCl_2\)

\(VOCl_2, MnCl_2\)

Correct Answer:

\(VOCl_2, CuCl_2\)

Explanation:

The correct answer is option 2. \(VOCl_2, CuCl_2\).

In transition element series, colour is observed due to d-d transition of d orbital electrons so compounds with same no. of unpair electron will show same colour

Transition metal complexes often exhibit color due to d-d transitions, where electrons in the d orbitals absorb visible light and transition to higher energy levels. Compounds with the same number of unpaired electrons in their d orbitals tend to exhibit similar or related colors. Out of the given compounds \(VOCl_2\) and  \(CuCl_2\) will have the same number of unpaired electrons.

In the case of \(VOCl_2\) vanadium is in +IV oxidation state (\(V^{4+}\)) and in case of \(CuCl_2\) copper is in +II oxidation state (\(Cu^{2+}\)). Vanadium(IV) (\(V^{4+}\)) has one unpaired electron in the \(d\) orbital, making it paramagnetic and potentially exhibiting color due to d-d transitions. Copper(II) (\(Cu^{2+}\)) also has one unpaired electron in the \(d\) orbital, making it paramagnetic and capable of showing color due to d-d transitions.

Therefore, both \(V^{4+}\) and \(Cu^{2+}\) ions can exhibit similar colors in their complexes due to the presence of one unpaired electron in their respective d orbitals, as suggested. However, it's essential to consider the ligands surrounding the metal ion and their influence on the color of the complex as well. Different ligands can lead to variations in color despite the same metal ion having the same number of unpaired electrons.