The hexahydrate nitrate of a fifth-period metal element $\mathbf{M}$ is mixed with $\mathrm{Cr(NO_3)_3\cdot 9H_2O}$ in stoichiometric ratio and gradually heated in an oxygen stream to produce compound $\mathbf{X}$ and release a reddish-brown mixed gas (Reaction 1). Further studies have shown that $\mathbf{X}$ is a paramagnetic solid with the chemical formula $\mathbf{M}\mathrm{CrO_4}$; 0.2786 g of solid $\mathbf{X}$ is dissolved in 0.10 mol/L $\mathrm {H_2SO_4}$ without gas evolution (Reaction 2); the resulting solution is titrated to the endpoint with 23.76 mL of 0.1144 mol/L $\mathrm { F e } ^ { 2 + }$ solution (Reaction 3). In the above process, the oxidation state of the metal element $\mathbf{M}$ remains unchanged.

Let the atomic number of $\mathbf{M}$ be a, the valence of $\mathrm{Cr}$ in $\mathbf{X}$ be b, and the absolute values of the difference between the number of reactant molecules and the number of product molecules in reactions 1, 2, and 3 (Reaction 1 is a chemical equation, 2 and 3 are ionic equations with the simplest integer ratio coefficients) be c, d, and e, respectively. Then the value of $\log_{min\{c,d,e\}}\frac{max\{c,d,e\}+b}{a}$ is (an error within 0.01 is considered correct):
将第五周期金属元素$\mathbf{M}$的六水合硝酸盐与$\mathrm{Cr(NO_3)_3\cdot 9H_2O}$按计量比混合，并在氧气流中逐步加热，可制得化合物 $\mathbf{X}$ 并放出红棕色混合气体(反应1)。进一步研究表明，$\mathbf{X}$是一种顺磁性固体，化学式为 $\mathbf{M}\mathrm{CrO_4}$ ；将0.2786 g固体$\mathbf{X}$用0.10 mol/L $\mathrm {H_2SO_4}$ 溶解，无气体放出(反应2)；所得溶液用0.1144 mol/L $\mathrm { F e } ^ { 2 + }$ 溶液滴定至终点(反应3)，消耗23.76 mL。以上过程中，金属元素$\mathbf{M}$的氧化态始终未发生变化。

记$\mathbf{M}$的原子序数为a，$\mathbf{X}$中 $\mathrm{Cr}$的化合价为b，反应1，2，3（反应1为化学方程式，2，3均为离子方程式且系数均为最简整数比）反应物的分子数与生成物的分子数之差的绝对值分别为c，d，e，则$\log_{min\{c,d,e\}}\frac{max\{c,d,e\}+b}{a}$的值为（误差在0.01内视为正确）：
All other options are incorrect
-0.63
-0.44
-0.37
-0.11
0.23
0.41
0.64
0.80
1.03
1.32
其他选项均不正确
-0.63
-0.44
-0.37
-0.11
0.23
0.41
0.64
0.80
1.03
1.32
The number of moles of electrons transferred from 0.2786 g of solid $\mathbf{X}$ can be calculated from the titration results:

$n(e^-)=0.1144~mol/L\times23.76~mL=2.718\times10^{-3}~mol$

<Checkpoint>[The number of moles of electrons transferred from 0.2786 g of solid $\mathbf{X}$ is $n(e^-)=2.718\times10^{-3}~mol$](1)</Checkpoint>

Based on reactivity, it is known that $\mathrm{Cr}$ will be oxidized to a valence state greater than +3. Therefore, we will discuss the valence states of $\mathrm{Cr}$ above +3 separately:

If it is +4, then $n(\mathbf{X})=2.718\times10^{-3}~mol$

$M(\mathbf{X})=102.5~g/mol$

Unreasonable

If it is +6, then $n(\mathbf{X})=9.06\times10^{-4}~mol$

$M(\mathbf{M})=191.5~g/mol$

Not a fifth-period element.

If it is +5, then $n(\mathbf{X})=1.359\times10^{-3}~mol$

$M(\mathbf{M})=89.0~g/mol$

Is element $\mathrm{Y}$, reasonable.

<Checkpoint>[The valence of $\mathrm{Cr}$ in $\mathbf{X}$ is +5](1)</Checkpoint>

<Checkpoint>[$\mathbf{M}$ is $\mathrm{Y}$](1)</Checkpoint>

Therefore $a=39,b=5$

After knowing the substance, it is easy to write the equations for the three reactions:

Reaction 1:

$\mathrm{Y(NO_3)_3\cdot 6H_2O+Cr(NO_3)_3\cdot 9H_2O\to YCrO_4+6NO_2+O_2+15H_2O}$

<Checkpoint>[$\mathrm{Y(NO_3)_3\cdot 6H_2O+Cr(NO_3)_3\cdot 9H_2O\to YCrO_4+6NO_2+O_2+15H_2O}$](1)</Checkpoint>

Reaction 2:

$\mathrm{3YCrO_4+10H^+\to 3Y^{3+}+Cr_2O_7^{2-}+Cr^{3+}+5H_2O}$

<Checkpoint>[$\mathrm{3YCrO_4+10H^+\to 3Y^{3+}+Cr_2O_7^{2-}+Cr^{3+}+5H_2O}$](1)</Checkpoint>

Reaction 3:

$\mathrm{Cr_2O_7^{2-}+6Fe^{2+}+14H^+\to 2Cr^{3+}+6Fe^{3+}+7H_2O}$

<Checkpoint>[$\mathrm{Cr_2O_7^{2-}+6Fe^{2+}+14H^+\to 2Cr^{3+}+6Fe^{3+}+7H_2O}$](1)</Checkpoint>

Therefore $c=21,d=3,e=6$

<Checkpoint>[$a=39$](0.2)</Checkpoint>

<Checkpoint>[$b=5$](0.2)</Checkpoint>

<Checkpoint>[$c=21$](0.2)</Checkpoint>

<Checkpoint>[$d=3$](0.2)</Checkpoint>

<Checkpoint>[$e=6$](0.2)</Checkpoint>

$\log_{min\{c,d,e\}}\frac{max\{c,d,e\}+b}{a}=\log_3\frac{21+5}{39}=-0.37$
通过滴定结果可以计算出0.2786 g固体$\mathbf{X}$转移的电子的摩尔数：

$n(e^-)=0.1144~mol/L\times23.76~mL=2.718\times10^{-3}~mol$

<Checkpoint>[0.2786 g固体$\mathbf{X}$转移的电子的摩尔数$n(e^-)=2.718\times10^{-3}~mol$](1)</Checkpoint>

根据反应性可以知道$\mathrm{Cr}$会被氧化为大于+3的价态，因此对$\mathrm{Cr}$在+3价以上的价态分别进行讨论：

若为+4价，则$n(\mathbf{X})=2.718\times10^{-3}~mol$

$M(\mathbf{X})=102.5~g/mol$

不合理

若为+6价，则$n(\mathbf{X})=9.06\times10^{-4}~mol$

$M(\mathbf{M})=191.5~g/mol$

不是第五周期元素。

若为+5价，则$n(\mathbf{X})=1.359\times10^{-3}~mol$

$M(\mathbf{M})=89.0~g/mol$

为元素$\mathrm{Y}$，合理。

<Checkpoint>[$\mathbf{X}$中$\mathrm{Cr}$的化合价为+5](1)</Checkpoint>

<Checkpoint>[$\mathbf{M}$为$\mathrm{Y}$](1)</Checkpoint>

因此$a=39,b=5$

知道物质之后就很容易写出三个反应的方程式：

反应1：

$\mathrm{Y(NO_3)_3\cdot 6H_2O+Cr(NO_3)_3\cdot 9H_2O\to YCrO_4+6NO_2+O_2+15H_2O}$

<Checkpoint>[$\mathrm{Y(NO_3)_3\cdot 6H_2O+Cr(NO_3)_3\cdot 9H_2O\to YCrO_4+6NO_2+O_2+15H_2O}$](1)</Checkpoint>

反应2：

$\mathrm{3YCrO_4+10H^+\to 3Y^{3+}+Cr_2O_7^{2-}+Cr^{3+}+5H_2O}$

<Checkpoint>[$\mathrm{3YCrO_4+10H^+\to 3Y^{3+}+Cr_2O_7^{2-}+Cr^{3+}+5H_2O}$](1)</Checkpoint>

反应3：

$\mathrm{Cr_2O_7^{2-}+6Fe^{2+}+14H^+\to 2Cr^{3+}+6Fe^{3+}+7H_2O}$

<Checkpoint>[$\mathrm{Cr_2O_7^{2-}+6Fe^{2+}+14H^+\to 2Cr^{3+}+6Fe^{3+}+7H_2O}$](1)</Checkpoint>

因此$c=21,d=3,e=6$

<Checkpoint>[$a=39$](0.2)</Checkpoint>

<Checkpoint>[$b=5$](0.2)</Checkpoint>

<Checkpoint>[$c=21$](0.2)</Checkpoint>

<Checkpoint>[$d=3$](0.2)</Checkpoint>

<Checkpoint>[$e=6$](0.2)</Checkpoint>

$\log_{min\{c,d,e\}}\frac{max\{c,d,e\}+b}{a}=\log_3\frac{21+5}{39}=-0.37$