
The $ {\text{MS}}{{\text{O}}_{\text{4}}} $ is the formula of sulphate of Metal M. Write down the formula of its:
-Hydroxide
-Chlorite
-chloride
-Nitrate
-Nitrite
-Peroxide
-Chromate
-Phosphate
Answer
538.2k+ views
Hint: The number of electrons gained or lost by an tom to complete its outermost octet is known as its valency. Valency is also called the combining capacity of an atom of a particular element. Valency is also discussed as the combining of the number of atoms with single valency.
Complete step by step answer:
Valency: The combining capacity for an atom of an element is known as valency or we can say the number of electrons gained or loosed by an atom to complete its outermost octet is known as its valency.
In this problem to find out the chemical compounds of metal M first we have to find out its combining capacity or its valency. From the sulphate of metal $ {\text{MS}}{{\text{O}}_{\text{4}}} $ we can see that the valency of metal is 2.
Therefore if we separate the metal we will get $ {{\text{M}}^{{\text{ + 2}}}} $ .
So,
The hydroxide of the metal is $ {\text{M(OH}}{{\text{)}}_{\text{2}}} $
The chlorite of the metal is $ {\text{M(Cl}}{{\text{O}}_{\text{2}}}{{\text{)}}_{\text{2}}} $ . [ $ {\text{Cl}}{{\text{O}}_{\text{2}}}^ - $ is the chlorite ion]
The chloride of the metal is $ {\text{MC}}{{\text{l}}_{\text{2}}} $
The Nitrate of the metal is $ {\text{M(N}}{{\text{O}}_3}{{\text{)}}_{\text{2}}} $ [ $ {\text{N}}{{\text{O}}_3}^ - $ is the Nitrate ion]
The Nitride of the metal is $ {\text{M(N}}{{\text{O}}_{\text{2}}}{{\text{)}}_{\text{2}}} $ [ $ {\text{N}}{{\text{O}}_2}^ - $ is the Nitrite ion]
The peroxide of the metal is $ {\text{M}}{{\text{O}}_{\text{2}}} $ [ $ {{\text{O}}^ - } $ is the peroxide ion]
The chromate of the metal is $ {\text{M(Cr}}{{\text{O}}_4}) $ [ $ {\text{Cr}}{{\text{O}}_4}^{ - 2} $ is the chromate ion]
The phosphate of the metal is $ {{\text{M}}_{\text{3}}}{{\text{(P}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{2}}} $ [ $ {\text{P}}{{\text{O}}_4}^{ - 3} $ is the phosphate ion]
Note:
According to modern definitions the defining statement for valency has been changed. Now valency is defined as the number of hydrogen atoms that can combine with the atom of an element is considered as its valency.
Complete step by step answer:
Valency: The combining capacity for an atom of an element is known as valency or we can say the number of electrons gained or loosed by an atom to complete its outermost octet is known as its valency.
In this problem to find out the chemical compounds of metal M first we have to find out its combining capacity or its valency. From the sulphate of metal $ {\text{MS}}{{\text{O}}_{\text{4}}} $ we can see that the valency of metal is 2.
Therefore if we separate the metal we will get $ {{\text{M}}^{{\text{ + 2}}}} $ .
So,
The hydroxide of the metal is $ {\text{M(OH}}{{\text{)}}_{\text{2}}} $
The chlorite of the metal is $ {\text{M(Cl}}{{\text{O}}_{\text{2}}}{{\text{)}}_{\text{2}}} $ . [ $ {\text{Cl}}{{\text{O}}_{\text{2}}}^ - $ is the chlorite ion]
The chloride of the metal is $ {\text{MC}}{{\text{l}}_{\text{2}}} $
The Nitrate of the metal is $ {\text{M(N}}{{\text{O}}_3}{{\text{)}}_{\text{2}}} $ [ $ {\text{N}}{{\text{O}}_3}^ - $ is the Nitrate ion]
The Nitride of the metal is $ {\text{M(N}}{{\text{O}}_{\text{2}}}{{\text{)}}_{\text{2}}} $ [ $ {\text{N}}{{\text{O}}_2}^ - $ is the Nitrite ion]
The peroxide of the metal is $ {\text{M}}{{\text{O}}_{\text{2}}} $ [ $ {{\text{O}}^ - } $ is the peroxide ion]
The chromate of the metal is $ {\text{M(Cr}}{{\text{O}}_4}) $ [ $ {\text{Cr}}{{\text{O}}_4}^{ - 2} $ is the chromate ion]
The phosphate of the metal is $ {{\text{M}}_{\text{3}}}{{\text{(P}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{2}}} $ [ $ {\text{P}}{{\text{O}}_4}^{ - 3} $ is the phosphate ion]
Note:
According to modern definitions the defining statement for valency has been changed. Now valency is defined as the number of hydrogen atoms that can combine with the atom of an element is considered as its valency.
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