Let's analyze the molecular orbital configuration of CO and its ions.

The molecular orbital configuration of CO (14 electrons) is: $({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\sigma }_{2p}{)}^2({\pi }_{2p}{)}^4({\sigma }_{2p}{)}^2$. The highest occupied molecular orbital (HOMO) is ${\sigma }_{2p}$ and the lowest unoccupied molecular orbital (LUMO) is ${\pi }_{2p}^{\ast }$.

Option A: The electron is being added to ${\pi }^{\ast }$ orbital to get ${\mathrm{CO}}^{-}$from $\mathrm{CO}$
When an electron is added to CO to form ${\mathrm{CO}}^{-}$, it enters the LUMO, which is the ${\pi }_{2p}^{\ast }$ (antibonding) orbital. This statement is correct.

Option B: The electron is being removed from ${\pi }^{\ast }$ orbital to get ${\mathrm{CO}}^{+}$from $\mathrm{CO}$
When an electron is removed from CO to form ${\mathrm{CO}}^{+}$, it is removed from the HOMO, which is the ${\sigma }_{2p}$ (bonding) orbital, not the ${\pi }_{2p}^{\ast }$ (antibonding) orbital. Therefore, this statement is incorrect.

Option C: In the formation of ${\mathrm{CO}}^{-}$from $\mathrm{CO}$, bond length increases
For CO, bond order = (number of bonding electrons - number of antibonding electrons) / 2 = (10 - 4) / 2 = 3.
For ${\mathrm{CO}}^{-}$, an electron is added to the ${\pi }_{2p}^{\ast }$ (antibonding) orbital. So, bond order = (10 - 5) / 2 = 2.5. Since the bond order decreases from 3 to 2.5, the bond length increases. This statement is correct.

Option D: In the formation of ${\mathrm{CO}}^{+}$from $\mathrm{CO}$, bond length decreases
For CO, bond order = 3.
For ${\mathrm{CO}}^{+}$, an electron is removed from the ${\sigma }_{2p}$ (bonding) orbital. So, bond order = (9 - 4) / 2 = 2.5. Since the bond order decreases from 3 to 2.5, the bond length increases, not decreases. This statement is incorrect. However, the question asks for the incorrect statement among the given options, and option B is definitively incorrect based on the removal of an electron from an antibonding orbital when it should be from a bonding orbital. Let's re-evaluate D. If an electron is removed from a bonding orbital, the bond order decreases, and bond length increases. So D is also incorrect. There might be an error in the question or options provided, as both B and D appear incorrect. However, typically, when forming ${\mathrm{CO}}^{+}$, the electron is removed from the highest energy bonding orbital. Let's re-check the MO diagram for CO. The HOMO is ${\sigma }_{2p}$. Removing an electron from a bonding orbital decreases bond order, thus increasing bond length. So, statement D is indeed incorrect. Given the options, and assuming only one incorrect statement, let's re-examine the MO diagram. For CO, the order of MOs is ${\sigma }_{2s}<{\sigma }_{2s}^{\ast }<{\pi }_{2p}<{\sigma }_{2p}<{\pi }_{2p}^{\ast }<{\sigma }_{2p}^{\ast }$. The HOMO is ${\sigma }_{2p}$. So, removing an electron from CO to form ${\mathrm{CO}}^{+}$ means removing it from ${\sigma }_{2p}$. This is a bonding orbital. Removing an electron from a bonding orbital decreases the bond order, which increases the bond length. Therefore, statement D, which says bond length decreases, is incorrect.

Comparing B and D:
Statement B: Electron removed from ${\pi }^{\ast }$ orbital to get ${\mathrm{CO}}^{+}$. This is incorrect because the electron is removed from the ${\sigma }_{2p}$ (bonding) orbital.
Statement D: In the formation of ${\mathrm{CO}}^{+}$ from $\mathrm{CO}$, bond length decreases. This is incorrect because removing an electron from a bonding orbital decreases bond order (from 3 to 2.5), which increases bond length.

Both B and D are incorrect statements. However, in multiple-choice questions, we often look for the most fundamentally incorrect statement or the one that directly contradicts the MO theory. The removal of an electron from ${\pi }^{\ast }$ for ${\mathrm{CO}}^{+}$ (Statement B) is a direct misidentification of the HOMO. The consequence on bond length (Statement D) is a derived property. If we assume the standard MO diagram for CO, the HOMO is ${\sigma }_{2p}$. Therefore, removing an electron from ${\sigma }_{2p}$ (a bonding orbital) decreases the bond order from 3 to 2.5, leading to an increase in bond length. Thus, statement D is incorrect. Statement B is also incorrect because the electron is removed from a bonding orbital, not an antibonding orbital.

Let's re-evaluate the MO diagram for CO. The correct order of MOs for CO is $({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\pi }_{2p}{)}^4({\sigma }_{2p}{)}^2$. The HOMO is ${\sigma }_{2p}$. The LUMO is ${\pi }_{2p}^{\ast }$.

A) ${\mathrm{CO}}^{-}$: Electron added to LUMO (${\pi }_{2p}^{\ast }$). Correct.
B) ${\mathrm{CO}}^{+}$: Electron removed from HOMO (${\sigma }_{2p}$). Statement says removed from ${\pi }^{\ast }$. Incorrect.
C) ${\mathrm{CO}}^{-}$: Bond order of CO = 3. Bond order of ${\mathrm{CO}}^{-}$ = (10-5)/2 = 2.5. Bond order decreases, so bond length increases. Correct.
D) ${\mathrm{CO}}^{+}$: Bond order of CO = 3. Bond order of ${\mathrm{CO}}^{+}$ = (9-4)/2 = 2.5. Bond order decreases, so bond length increases. Statement says bond length decreases. Incorrect.

Since both B and D are incorrect, there might be an issue with the question or options. However, if we must choose one, statement B makes