Find the metallic and ferromagnetic substance.

We begin by recalling two key definitions. A metallic substance conducts electricity because it possesses delocalised (mobile) electrons. A ferromagnetic substance shows spontaneous alignment of magnetic moments even in the absence of an external field and therefore exhibits strong, permanent magnetism.

To decide which oxide satisfies both conditions, we must examine the electronic configuration of the metal ion in each compound and the type of magnetic ordering it gives rise to.

First, we note the common oxidation state of the metal in each dioxide:

Because oxygen is almost always $$\left( -2 \right)$$ in binary oxides, the charge balance condition $$\text{metal charge} + 2(-2) = 0$$ leads to

$$\text{metal charge}=+4.$$

Hence, in every oxide listed, the metal is in the $$+4$$ oxidation state. We now inspect the corresponding $$3d$$ electron count and magnetic behaviour one by one.

1. For $$VO_2$$ we have the electron configuration

$$V: [Ar]\;3d^3\,4s^2 \;\Rightarrow\; V^{4+}: [Ar]\;3d^1.$$

This single $$d$$ electron produces only a small magnetic moment. Experimentally, $$VO_2$$ undergoes a metal-insulator transition: above $$\approx 68^{\circ}{\rm C}$$ it is metallic, but at room temperature it is insulating and merely paramagnetic. Thus it is not ferromagnetic at ordinary conditions.

2. For $$MnO_2$$:

$$Mn: [Ar]\;3d^5\,4s^2 \;\Rightarrow\; Mn^{4+}: [Ar]\;3d^3.$$

Three unpaired $$d$$ electrons give a magnetic moment, but in the crystal these moments couple antiferromagnetically, cancelling each other. The oxide is a semiconductor and not metallic. Therefore, it is not ferromagnetic and not metallic.

3. For $$TiO_2$$:

$$Ti: [Ar]\;3d^2\,4s^2 \;\Rightarrow\; Ti^{4+}: [Ar]\;3d^0.$$

No $$d$$ electrons remain, so $$TiO_2$$ is diamagnetic and a wide-band-gap insulator. Hence it is neither metallic nor ferromagnetic.

4. For $$CrO_2$$:

$$Cr: [Ar]\;3d^5\,4s^1 \;\Rightarrow\; Cr^{4+}: [Ar]\;3d^2.$$

The two $$d$$ electrons of $$Cr^{4+}$$ occupy the $$t_{2g}$$ orbitals in such a way that one electron becomes itinerant (delocalised) while the other remains localised. According to band-structure calculations and experimental measurements, this produces a so-called half-metallic ferromagnet:

$$\text{Delocalised }e^- \; \Longrightarrow \; \text{metallic conduction}$$

$$\text{Parallel alignment of spins} \; \Longrightarrow \; \text{ferromagnetism}$$

Indeed, $$CrO_2$$ is widely used as a magnetic pigment in audio and video recording tapes precisely because it is both a good electrical conductor and strongly ferromagnetic at room temperature.

Comparing all four oxides, only $$CrO_2$$ simultaneously exhibits substantial electrical conductivity and robust ferromagnetic ordering.

Hence, the correct answer is Option 4.