
The formula of metal phosphate is $$MHP{O_3}$$ , the formula of metal hypophosphite is:
A. $$M{H_2}P{O_2}$$
B. $$M{({H_2}P{O_2})_2}$$
C. $$M{({H_2}P{O_2})_3}$$
D. $${M_2}{({H_2}P{O_2})_2}$$
Answer
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Hint:In the given formula of the metal phosphide we see that the M represents the metal in the formula and the $$HP{O_3}$$ represents the phosphite anion. It is also known as meta phosphoric acid. As we know that the valency of the phosphite ion is 2- so now let us calculate the valency of the metal attached to it to find the answer of this question.
Complete step-by-step answer: The $$MHP{O_3}$$ has been derived from the $${H_3}P{O_4}$$ acid. The anion given to us in the metal phosphite is phosphite anion which has the valency of 2-. So let us assume that the valency of M that is metal be x. Now being a neutral molecule the sum of oxidation states or the valency of the molecule will be equal to zero. So now see the following calculations:
$$\eqalign{
& x + ( - 2) = 0 \cr
& x = 2 \cr} $$
So the valency of metal M is +2. This is so because it has the tendency to replace 2 hydrogen atoms of hydrogen phosphate.
Now let me tell you the formula of the hypophosphite ion and that is $${H_2}P{O_2}^ - $$. So the hypophosphite has the valency or charge of 1- on it. Here the metal has the valency of +2 and the other has the valency of 1-. On reciting them together the formula which we will get is $$M{({H_2}P{O_2})_2}$$. Look at the following equation for better understanding:
$${M^{2 + }} + 2{H_2}P{O_2} \to M{({H_2}P{O_2})_2}$$
So the formula will be $$M{({H_2}P{O_2})_2}$$ as the valency of metal is +2 and hypophosphite is 1.
So the correct answer is option B.
Note: In $$M{({H_2}P{O_2})_2}$$,M is the symbol of metal which combines with the hypophosphite. The number 2 after the hypophosphite means or indicates the valency of metal which is equal to 2. Here two molecules of hypophosphite reacts with the metal of divalent valency to form the given compound. If the metal valency is changed then the formula would have been changed.
Complete step-by-step answer: The $$MHP{O_3}$$ has been derived from the $${H_3}P{O_4}$$ acid. The anion given to us in the metal phosphite is phosphite anion which has the valency of 2-. So let us assume that the valency of M that is metal be x. Now being a neutral molecule the sum of oxidation states or the valency of the molecule will be equal to zero. So now see the following calculations:
$$\eqalign{
& x + ( - 2) = 0 \cr
& x = 2 \cr} $$
So the valency of metal M is +2. This is so because it has the tendency to replace 2 hydrogen atoms of hydrogen phosphate.
Now let me tell you the formula of the hypophosphite ion and that is $${H_2}P{O_2}^ - $$. So the hypophosphite has the valency or charge of 1- on it. Here the metal has the valency of +2 and the other has the valency of 1-. On reciting them together the formula which we will get is $$M{({H_2}P{O_2})_2}$$. Look at the following equation for better understanding:
$${M^{2 + }} + 2{H_2}P{O_2} \to M{({H_2}P{O_2})_2}$$
So the formula will be $$M{({H_2}P{O_2})_2}$$ as the valency of metal is +2 and hypophosphite is 1.
So the correct answer is option B.
Note: In $$M{({H_2}P{O_2})_2}$$,M is the symbol of metal which combines with the hypophosphite. The number 2 after the hypophosphite means or indicates the valency of metal which is equal to 2. Here two molecules of hypophosphite reacts with the metal of divalent valency to form the given compound. If the metal valency is changed then the formula would have been changed.
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