The question pertains to the oxidation states of lanthanoids, specifically focusing on why cerium can exhibit a +4 oxidation state despite +3 being the most common. Let's analyze the options and arrive at the correct conclusion.
The electronic configuration of cerium (Ce), which has an atomic number of 58, is \([Xe] \, 4f^1 \, 5d^1 \, 6s^2\). Typically, lanthanoids are known for losing electrons to achieve a stable configuration, often leading to a +3 oxidation state. However, a special case occurs with cerium.
Upon losing three electrons, the typical +3 oxidation state configuration for cerium is \([Xe] \, 4f^1\). However, cerium can also lose one additional electron to achieve the +4 oxidation state, leading to the configuration \([Xe] \, 4f^0\). The stability of this configuration is due to the completely emptied 4f subshell, which is energetically favorable.
Let's evaluate the options:
After losing one more electron, it acquires \(4f^{14}\) electronic configuration.
- - This is incorrect. Cerium does not reach a \(4f^{14}\) configuration, as such a configuration involves filling rather than emptying the 4f subshell.
Its atomic number is 61.
- - This is incorrect. Cerium's atomic number is 58, not 61.
After losing one more electron, it acquires \(4f^0\) electronic configuration.
- - This is correct. The removal of one additional electron results in a stable, fully emptied 4f subshell.
Its nearest inert gas is Radon.
- - This is incorrect. The nearest inert gas prior to cerium is Xenon, not Radon.
Thus, the correct answer is the option stating that cerium acquires the \(4f^0\) electronic configuration upon losing one more electron, making its +4 oxidation state stable.
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Important Questions from Inorganic Chemistry
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A. The complex exhibits geometrical isomerism.
B. The complex is white in colour.
C. The calculated spin-only magnetic moment of the complex is 2.84 BM.
D. The calculated CFSE (Crystal Field Stabilization Energy) of Ni in this complex is $-0.8 \, \Delta_0$.
E. The geometrical arrangement of ligands in this complex is similar to that in $\text{Ni(CO)}_4$.
Choose the correct answer from the options given below : - Given below are two statements :
Statement I : The first ionization enthalpy of Cr is lower than that of Mn.
Statement II : The second and third ionization enthalpies of Cr are higher than those of Mn.
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Statement I : Elements 'X' and 'Y' are the most and least electronegative elements, respectively among N, As, Sb and P. The nature of the oxides $\text{X}_2\text{O}_3$ and $\text{Y}_2\text{O}_3$ is acidic and amphoteric, respectively.
Statement II : $\text{BCl}_3$ is covalent in nature and gets hydrolysed in water. It produces $[\text{B(OH)}_4]^-$ and $[\text{B(H}_2\text{O)}_6]^{3+}$ in aqueous medium.
In the light of the above statements, choose the correct answer from the options given below :
