MCQ
Which electronic arrangement shows ground state of an element
- A

- B

- C

- ✓






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Reason : Low spin complexes have lesser number of unpaired electrons.
| Rate constant | Activation energy | |
| Step $1$ | $k_1$ | $E_{a1} = 180\ kJ/mol$ |
| Step $2$ | $k_2$ | $E_{a2} = 80\ kJ/mol$ |
| Step $3$ | $k_3$ | $E_{a3} = 50\ kJ/mol$ |
If overall rate constant, $k = {\left( {\frac{{{k_1}{k_2}}}{{{k_3}}}} \right)^{2/3}}$ ,then overall activation energy of the reaction will be .......... $ kJ/mol$
$\mathop {{H_2}(Pt)}\limits_{1\,atm} \left| {\begin{array}{*{20}{c}}
{H_{aq}^ + }\\
{pH = 4}
\end{array}} \right|\left| {\begin{array}{*{20}{c}}
{A{g^ + }}\\
{xM}
\end{array}} \right|\begin{array}{*{20}{c}}
{Ag}\\
{}
\end{array}$
What is the value of $x$ ? ........... $\mathrm{M}$
( give : $E_{A{g^ + }|Ag}^o = + 0.8$ $\frac{{2.303\,RT}}{F} = 0.06 $)