Question

What is the name of the ionic compound with the formula $Ni{{(P{{O}_{4}})}_{2}}$ ?

Answer 1

Hint : We know that nickel phosphate is an inorganic compound with the formula $Ni{{(P{{O}_{4}})}_{2}}$ . It is a mint green paramagnetic solid that is insoluble in water and also occurs as the mineral arupite. It features octahedral $Ni$ centers, which are bound to water and phosphate.

Complete Step By Step Answer:
As the name suggests, the given compound nickel phosphate comprises a cationic group and an anionic group of nonmetals. We know that any compound containing a cation and an anion is considered to be an ionic compound. In ionic compounds, we have to take into account the charges of each of the elements present in the compound. Now let us identify the formula for the given compound i.e. nickels phosphate. Nickel phosphate possesses two polyatomic ions (i.e. group of nonmetals (nickel and phosphate in this case) so we'll use a table of names for the common polyatomic ions, along with the Periodic Table. Rules to write formula of a ternary ionic compound are listed below:
Using the periodic table, write down the symbol for the element along with its charge. In the present case, we have polyatomic ions which are not available in periodic tables.
Using the Common ion table, identify the symbol as well as charge of the polyatomic ions
In the present case, polyatomic ion nickel is written as $Ni$ having a charge $2+$ and phosphate is written as $P{{O}_{4}}$ having a charge $-3.$
Check if the charges are balanced; In the present case it becomes $N{{i}^{2+}}P{{O}_{4}}^{-3}$ . The charges are not balanced as summation of charges is non-zero. $(+2)+(-3)=-1\ne 0$
If unbalanced charges are there, add subscripts following the crisscross method such that net charge for the compound is zero. In the present case, we will add subscripts to each polyatomic ion such that the net charge becomes zero. So it becomes ${{\left( N{{i}^{2+}} \right)}_{3}}{{\left( P{{O}_{4}}^{-3} \right)}_{2}}$ and now the net charge is balanced i.e. $\left[ 3\times (+2) \right]+\left[ 2\times (-3) \right]\Rightarrow 6-6\Rightarrow 0=0.$
Therefore, the name of the ionic compound with the formula $Ni{{(P{{O}_{4}})}_{2}}$ is ${{\left( N{{i}^{2+}} \right)}_{3}}{{\left( P{{O}_{4}}^{-3} \right)}_{2}}.$

Note :
While writing the chemical formula of an ionic compound, never write the subscript ' $1$ '. Always remember that we can also have two polyatomic ions in the same compound as it is in the present case. In this case, we always have to find and write both of the names as identified from the Common Ion Table instead of periodic table.