'X' is the number of acidic oxides among $$VO_2$$, $$V_2O_3$$, $$CrO_3$$, $$V_2O_5$$ and $$Mn_2O_7$$. The primary valency of cobalt in $$[Co(H_2NCH_2CH_2NH_3)_3]_2(SO_4)_3$$ is Y. The value of X + Y is :

For acidic‐basic behaviour of transition metal oxides, the general trend is: basic nature at low oxidation state $$\rightarrow$$ amphoteric $$\rightarrow$$ acidic at the highest oxidation state. We first write the oxidation state of the metal in each oxide.

$$VO_2:\; V+2(-2)=0 \Rightarrow V^{+4}$$
$$V_2O_3:\;2V+3(-2)=0 \Rightarrow V^{+3}$$
$$CrO_3:\; Cr+3(-2)=0 \Rightarrow Cr^{+6}$$
$$V_2O_5:\;2V+5(-2)=0 \Rightarrow V^{+5}$$
$$Mn_2O_7:\;2Mn+7(-2)=0 \Rightarrow Mn^{+7}$$

Nature of these oxides:
• $$V^{+3}$$ oxide ($$V_2O_3$$) is basic.
• $$V^{+4}$$ oxide ($$VO_2$$) is amphoteric.
• $$V^{+5}$$ oxide ($$V_2O_5$$) is also amphoteric (it reacts both with acids and with alkalies).
• $$Cr^{+6}$$ oxide ($$CrO_3$$) is the anhydride of $$H_2CrO_4$$ and is strongly acidic.
• $$Mn^{+7}$$ oxide ($$Mn_2O_7$$) is the anhydride of $$HMnO_4$$ and is strongly acidic.

Thus only $$CrO_3$$ and $$Mn_2O_7$$ are counted as acidic. Therefore
$$X = 2$$

Primary valency (Werner) = oxidation state of the metal.

The compound is $$[Co(H_2NCH_2CH_2NH_3)_3]_2(SO_4)_3$$.
Each ligand $$(H_2NCH_2CH_2NH_3)$$ is treated as neutral for the charge calculation (only the unprotonated $$NH_2$$ donates the lone pair; the protonated $$NH_3^+$$ part does not donate and its positive charge is balanced internally by the non-coordinating N).
Let the oxidation state of cobalt be $$y$$.

Charge balance for one complex ion:
$$y + 3(0)=\text{charge of }[CoL_3]^{n+}$$

Overall compound is electrically neutral, so

$$2\bigl(\text{charge of }[CoL_3]^{n+}\bigr) + 3(-2)=0 \; -(1)$$

From $$(1)$$: $$2n - 6 = 0 \Rightarrow n = +3$$. Hence $$y = +3$$.

Therefore the primary valency of cobalt is
$$Y = 3$$

Finally, $$X+Y = 2 + 3 = 5$$.

Option A is correct.