1. We are asked for the order of standard reduction potentials E^_{M^{2+}/M} with negative sign for the successive elements:

$\text{Cr, Mn, Fe, Co}$

This means we compare the magnitudes of their negative standard electrode potentials for:

$M^{2+} + 2e^- \rightarrow M$

2. Standard values are approximately:

• $E^\circ_{\text{Cr}^{2+}/\text{Cr}} \approx -0.91\,\text{V}$
• $E^\circ_{\text{Mn}^{2+}/\text{Mn}} \approx -1.18\,\text{V}$
• $E^\circ_{\text{Fe}^{2+}/\text{Fe}} \approx -0.44\,\text{V}$
• $E^\circ_{\text{Co}^{2+}/\text{Co}} \approx -0.28\,\text{V}$

3. Since the question says "with negative sign", we compare them as negative quantities. The more negative the value, the greater it is in this ordering sense of negativity magnitude:

$-1.18 < -0.91 < -0.44 < -0.28$

So in terms of magnitude of negative sign:

$\text{Mn} > \text{Cr} > \text{Fe} > \text{Co}$

4. Checking options:

• A: $\text{Mn} > \text{Cr} > \text{Fe} > \text{Co}$ ✅
• B: $\text{Cr} > \text{Fe} > \text{Mn} > \text{Co}$ ❌
• C: $\text{Fe} > \text{Mn} > \text{Cr} > \text{Co}$ ❌
• D: $\text{Cr} > \text{Mn} > \text{Fe} > \text{Co}$ ❌

Therefore, the correct order is:

$\boxed{\text{Mn} > \text{Cr} > \text{Fe} > \text{Co}}$