The strength of a reducing agent is determined by the ability of its ion to donate electrons. The element Eu (Europium) in its \( \text{Eu}^{2+} \) state can easily lose an electron to form \( \text{Eu}^{3+} \), making it a strong reducing agent. The reduction process is as follows:
\(\text{Eu}^{2+} \rightarrow \text{Eu}^{3+} + e^-\)
This shows that \( \text{Eu}^{2+} \) acts as a strong reducing agent.
Thus, the correct answer is 3, which corresponds to \( \text{Eu}^{2+} \).
The Correct Answer is: $\text{Eu}^{2+}$
Sc Ti V Cr Mn Fe Co Ni Cu Zn
Y Zr Nb Mo Tc 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.
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: