Consider the following reaction
`Na_(3)PO_(4)+(NH_(4))_(2)MoO_(4)+HNO_(3)(dil) to 'X'`
Then calculate total number of atoms of 15th group elements which are `sp^(3)` hybridized in compound 'X'.

Explanation

To solve the given problem, we need to follow these steps:

### Step 1: Identify the Reaction
We start with the reaction:
\[ \text{Na}_3\text{PO}_4 + (\text{NH}_4)_2\text{MoO}_4 + \text{HNO}_3 \text{(dil)} \rightarrow X \]

### Step 2: Determine the Product 'X'
The product 'X' formed from the reaction is:
\[ X = (\text{NH}_4)_3\text{PO}_4 \cdot 12\text{MoO}_3 \]
This compound is known as ammonium phosphomolybdate.

### Step 3: Identify the 15th Group Elements
In compound 'X', the relevant 15th group elements are:
- Nitrogen (N) from the ammonium ion \((\text{NH}_4)^+\)
- Phosphorus (P) from the phosphate ion \((\text{PO}_4)^{3-}\)

### Step 4: Determine Hybridization of Nitrogen
For the ammonium ion \((\text{NH}_4)^+\):
- Valency of nitrogen = 5 (as it belongs to group 15)
- Number of attached atoms = 4 (4 hydrogen atoms)
- Charge = -1 (since it is a cation, we subtract 1)

Using the formula for hybridization:
\[ \text{Hybridization} = \frac{(\text{Valency} + \text{Attached Atoms} - \text{Positive Charge})}{2} \]
\[ \text{Hybridization} = \frac{(5 + 4 - 1)}{2} = \frac{8}{2} = 4 \]
Thus, nitrogen in \((\text{NH}_4)^+\) is \(sp^3\) hybridized.

### Step 5: Determine Hybridization of Phosphorus
For the phosphate ion \((\text{PO}_4)^{3-}\):
- Valency of phosphorus = 5
- Number of attached atoms = 4 (4 oxygen atoms)
- Charge = 3 (since it is an anion, we add 3)

Using the same hybridization formula:
\[ \text{Hybridization} = \frac{(5 + 4 + 3)}{2} = \frac{12}{2} = 6 \]
However, since we are interested in the hybridization of the central atom, we consider the effective valency:
\[ \text{Hybridization} = \frac{(5 + 3)}{2} = \frac{8}{2} = 4 \]
Thus, phosphorus in \((\text{PO}_4)^{3-}\) is also \(sp^3\) hybridized.

### Step 6: Count the Total Number of Atoms
Now, we need to count the total number of atoms of 15th group elements that are \(sp^3\) hybridized in compound 'X':
- Nitrogen: 3 atoms (from \((\text{NH}_4)^+\))
- Phosphorus: 1 atom (from \((\text{PO}_4)^{3-}\))

Total number of \(sp^3\) hybridized atoms:
\[ \text{Total} = 3 \text{ (N)} + 1 \text{ (P)} = 4 \]

### Final Answer
The total number of atoms of 15th group elements which are \(sp^3\) hybridized in compound 'X' is **4**.

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