- A$Fe (III)$
- B$Mn (II)$
- C$Cr (I)$
- ✓$P (0)$
All other elements are $d-block$ elements therefore has valance electron in $3 d$ subshell.
Electronic configuration of $Cr ( I )$ is $1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^5$.
Electronic configuration of $F (III)$ is $1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^5$
Electronic configuration of $Mn (II)$ is $1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^5$.
So, all these have valence electrons in $3 d$ - subshell.
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$\begin{array}{|l|l|l|} \hline P_0\,\,(mmHg) & 250 & 300 \\ \hline t_{1/2}\,\,(minutes) & 135 & 112.5 \\ \hline \end{array}$
The order of reaction is -
$(A)$ $2 CO ( g )+ O _2( g ) \rightarrow 2 CO _2( g ) \quad \Delta H _1^\theta=- x\,kJ\,mol { }^{-1}$
$(B)$ $C$ (graphite) $+ O _2$ (g) $\rightarrow CO _2$ (g) $\Delta H _2^\theta=- y\,kJ\,mol -1$ The $\Delta H ^\theta$ for the reaction $......$.$C ($ graphite $)+\frac{1}{2} O _2( g ) \rightarrow CO ( g )$ is

