Question

The formula of hydrogen phosphate of a metal M is $MHP{O_4}$ . Give the formula of metal chloride.
A.MCl
B.$MC{I_3}$
C.$MC{I_2}$
D.${M_2}CI$

Answer 1

Hint: It is given that the formula of hydrogen phosphate of a metal M is $MHP{O_4}$ So, the valency of $HP{O_4}$ is -2. M replaced two hydrogen atoms of hydrogen phosphate. So, the valency of metal M is +2 (divalent). And the valency of chlorine is 1 as it is monovalent. Hence, the formula of the metal chloride is $MC{I_2}$ .

Complete step by step answer:
Given, $MHP{O_4}$
The given compound is the hydrogen phosphate of a metal M
 $MHP{O_4}$ is derived from the acid ${H_3}P{O_4}$ .
Valency of $HP{O_4}$ is -2. Therefore, the valency of metal attached to $HP{O_4}$ will have the valency of +2 i.e. it is divalent $(M{2^ + })$ .
Also, the metal M replaced two hydrogen atoms of hydrogen phosphate. So, its valency is 2.
When the metal M combines with Chlorine (Cl) whose valency is 1, then the reaction will be as follows:
 ${M^{2 + }} + 2CI \to MC{I_2}$
Hence, the formula of the metal chloride is $MC{I_2}$ as the valency of M is 2 and the valency of Cl is 1.

Therefore, the correct answer is option (C).

Note: In $MC{I_2}$ , M is the symbol for the metal that combines with Chlorine (Cl). The 2 after Cl means the valency of the metal has to be 2. We can now replace it with any metal from Group 2 like Calcium (Ca), Magnesium (Mg), Barium (Ba) etc. or with any element from the transition group which show variable valency like Iron (Fe) in +2 valency (ferrous state), Lead (Pb) etc.