The pair of compounds having metals in their highest oxidation state is:

To decide which pair shows the metals in their highest possible oxidation state, we must first calculate the oxidation number of the metal in every compound listed.

Recall the rule: The algebraic sum of the oxidation numbers of all atoms in a neutral compound is zero, while in an ion it equals the ionic charge. Oxygen almost always has $$-2$$, chlorine almost always has $$-1$$, and the cyanide ion $$CN^-$$ carries a charge of $$-1$$. We now apply this rule to every species.

Option A  $$Mn_2O_7$$ and $$CrO_2Cl_2$$

For $$Mn_2O_7$$ (a neutral molecule):

Let the oxidation number of each manganese atom be $$x$$.

We have two manganese atoms, so their total contribution is $$2x$$.

There are seven oxygens, each at $$-2$$, giving $$7(-2)=-14$$.

The compound is neutral, so the total is zero:

$$2x + (-14) = 0 \;\Longrightarrow\; 2x = +14 \;\Longrightarrow\; x = +7.$$

Thus manganese is in the $$+7$$ state, the highest oxidation state known for Mn.

For $$CrO_2Cl_2$$ (chromyl chloride, also neutral):

Let chromium have oxidation number $$y$$.

Two oxygens contribute $$2(-2)=-4$$, and two chlorines contribute $$2(-1)=-2$$.

Total charge being zero, we write

$$y + (-4) + (-2) = 0 \;\Longrightarrow\; y = +6.$$

Chromium is therefore $$+6$$. Chromium cannot go beyond $$+6$$ in stable compounds, so this is also its highest oxidation state.

Option B  $$[Fe(CN)_6]^{3-}$$ and $$[Cu(CN)_4]^{2-}$$

For $$[Fe(CN)_6]^{3-}$$:

Let Fe be $$z$$. Six $$CN^-$$ ligands give $$6(-1)=-6$$. The complex ion has an overall charge of $$-3$$, so:

$$z + (-6) = -3 \;\Longrightarrow\; z = +3.$$

Iron can exhibit up to $$+6$$ in ferrate $$FeO_4^{2-}$$, so $$+3$$ is not its highest.

For $$[Cu(CN)_4]^{2-}$$:

Let Cu be $$w$$. Four $$CN^-$$ ligands give $$4(-1)=-4$$. The ion’s charge is $$-2$$:

$$w + (-4) = -2 \;\Longrightarrow\; w = +2.$$

Copper can reach $$+3$$ in compounds like $$KCuO_2$$, so $$+2$$ is not its highest.

Option C  $$[NiCl_4]^{2-}$$ and $$[CoCl_4]^{2-}$$

For $$[NiCl_4]^{2-}$$:

Let Ni be $$a$$. Four chlorides: $$4(-1)=-4$$. Overall charge $$-2$$:

$$a + (-4) = -2 \;\Longrightarrow\; a = +2.$$

Nickel can exhibit $$+3$$ and $$+4$$ in other species, so $$+2$$ is not the highest.

For $$[CoCl_4]^{2-}$$:

Let Co be $$b$$. Four chlorides: $$4(-1)=-4$$. Overall charge $$-2$$:

$$b + (-4) = -2 \;\Longrightarrow\; b = +2.$$

Cobalt easily attains $$+3$$ in many compounds; $$+2$$ is therefore not its highest.

Option D  $$[FeCl_4]^{-}$$ and $$Co_2O_3$$

For $$[FeCl_4]^{-}$$:

Let Fe be $$c$$. Four chlorides: $$4(-1)=-4$$. Overall charge $$-1$$:

$$c + (-4) = -1 \;\Longrightarrow\; c = +3.$$

Again, iron can reach $$+6$$, so $$+3$$ is not the highest.

For $$Co_2O_3$$ (neutral):

Let Co be $$d$$. Three oxygens: $$3(-2)=-6$$. The molecule is neutral:

$$2d + (-6) = 0 \;\Longrightarrow\; 2d = +6 \;\Longrightarrow\; d = +3.$$

But cobalt is known up to $$+3$$ only in its stable compounds, so while $$+3$$ is Co's usual maximum, we have already rejected the pair because iron is not at its maximum.

Summarising, only Option A provides both metals in their respective highest oxidation states: manganese at $$+7$$ and chromium at $$+6$$.

Hence, the correct answer is Option A.