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 \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).