To determine which ion is isoelectronic with CO, we first need to understand what "isoelectronic" means. Two species are isoelectronic if they have the same number of electrons.

Carbon monoxide (CO) has a total of 10 electrons:

Carbon (C) has 6 electrons and Oxygen (O) has 8 electrons. Therefore:

$$6 + 8 = 14\, \text{electrons}$$

Now, let's examine each option:

Option A: CN^-

Cyanide ion (CN^-) has:

Carbon (C) with 6 electrons, Nitrogen (N) with 7 electrons, and one additional electron because of the negative charge:

$$6 + 7 + 1 = 14\, \text{electrons}$$

Option B: NO₂

Nitrogen dioxide (NO₂) has:

Nitrogen (N) with 7 electrons and two Oxygen (O) each with 8 electrons:

$$7 + 8 \times 2 = 23\, \text{electrons}$$

Option C: SO₂

Sulfur dioxide (SO₂) has:

Sulfur (S) with 16 electrons and two Oxygen (O) each with 8 electrons:

$$16 + 8 \times 2 = 32\, \text{electrons}$$

Option D: ClO₂

Chlorine dioxide (ClO₂) has:

Chlorine (Cl) with 17 electrons and two Oxygen (O) each with 8 electrons:

$$17 + 8 \times 2 = 33\, \text{electrons}$$

From the above calculations, we can conclude that the ion CN^- (option A) has the same number of electrons as CO and is therefore isoelectronic with it. So, the correct answer is:

Option A: CN^-