A) 2
B) 1
C) 4
D) 3
Correct Answer: D
Solution :
\[{{M}_{2}}{{O}_{x}}\xrightarrow{\text{Reduction}}M\] Eq. of \[{{M}_{2}}{{O}_{x}}=Eq.\]of Metal \[\frac{\text{Weight}\,\text{of}\,{{\text{M}}_{\text{2}}}{{\text{O}}_{\text{x}}}}{\text{Equivalent}\,\text{weight}\,\text{of}\,{{\text{M}}_{\text{2}}}{{\text{O}}_{\text{x}}}}=\] \[\frac{\text{Weight}\,\text{of}\,\text{metal}}{\text{Equivalent}\,\text{weight}\,\text{of}\,\text{metal}}\,\text{=}\] \[\frac{4}{\frac{2\times 56+x\times 16}{2x}}=\frac{2.8}{\frac{56}{x}}\] (1) On solving we get, \[\frac{4}{56+8x}=\frac{2.8}{56}\] \[\frac{1}{14+2x}=\frac{1}{20}\] \[2x=6\] \[x=3.\] Hence, the oxide is\[{{M}_{2}}{{O}_{3}}.\] Hence, the correct option is (d).
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