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A metal exists as an oxide with formula Mo.96 O . Metal M can exist as M 2 and M 3 in it's oxide Mo.96 O . The percentage of M 3 in the oxide is nearly ? a) 8.3%. b) 4.6%. c) 5%. d) 9.6.?
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A metal exists as an oxide with formula Mo.96 O . Metal M can exist as...
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A metal exists as an oxide with formula Mo.96 O . Metal M can exist as...
Solution:

Given: Metal oxide formula = Mo.96O

Metal M can exist as M2+ and M3+ in its oxide Mo.96O.

Let the percentage of M3+ in the oxide be x%.

Since the oxide is neutral, the sum of the charges of metal and oxygen should be equal to zero.

The atomic weight of Mo is 96, which means the oxide should contain 96g of Mo and 32g of O.

Let us assume that all the Mo in the oxide is in the form of M3+.

Then the formula for the oxide is M3+O2-

The weight of M3+ in the oxide will be 96g.

The weight of O2- in the oxide will be 32g.

If M exists as M2+ and M3+ in the oxide, then let us assume that x% of Mo exists as M3+ and (100-x)% of Mo exists as M2+.

Then the weight of M3+ in the oxide will be 0.96x g.

The weight of M2+ in the oxide will be 0.96(100-x) g.

Since the sum of the weights of M3+ and M2+ should be equal to 96g, we can write:

0.96x + 0.96(100-x) = 96

Solving this equation, we get x = 4.6%.

Therefore, the percentage of M3+ in the oxide is nearly 4.6%.

Answer: b) 4.6%.
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Read the passage given below and answer the following questions:The transition metals when exposed to oxygen at low and intermediate temperatures form thin, protective oxide films of up to some thousands of Angstroms in thickness. Transition metal oxides lie between the extremes of ionic and covalent binary compounds formed by elements from the left or right side of the periodic table. They range from metallic to semiconducting and deviate by both large and small degrees from stoichiometry. Since d-electron bonding levels are involved, the cations-exist in various valence states and hence give rise to a large number of oxides. The crystal structures are often classified by considering a cubic or hexagonal close-packed lattice of one set of ions with the other set of ions filling the octahedral or tetrahedral interstices. The actual oxide structures, however, generally show departures from such regular arrays due in part to distortions caused by packing of ions of different size and to ligand field effects. These distortions depend not only on the number of d-electrons but also on the valence and the position of the transition metal in a period or group.In the following questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices on the basis of the above passage.Assertion (

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A metal exists as an oxide with formula Mo.96 O . Metal M can exist as M 2 and M 3 in it's oxide Mo.96 O . The percentage of M 3 in the oxide is nearly ? a) 8.3%. b) 4.6%. c) 5%. d) 9.6.?
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A metal exists as an oxide with formula Mo.96 O . Metal M can exist as M 2 and M 3 in it's oxide Mo.96 O . The percentage of M 3 in the oxide is nearly ? a) 8.3%. b) 4.6%. c) 5%. d) 9.6.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A metal exists as an oxide with formula Mo.96 O . Metal M can exist as M 2 and M 3 in it's oxide Mo.96 O . The percentage of M 3 in the oxide is nearly ? a) 8.3%. b) 4.6%. c) 5%. d) 9.6.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A metal exists as an oxide with formula Mo.96 O . Metal M can exist as M 2 and M 3 in it's oxide Mo.96 O . The percentage of M 3 in the oxide is nearly ? a) 8.3%. b) 4.6%. c) 5%. d) 9.6.?.
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