A radioactive substance `.)(88)X^(228)` (IIA) emits `3alpha` and
`3beta`-particles to form 'Y'. O which group of long form of the periodic table
does 'Y' belong?

Options

A. IVA
B. VA
C. VIA
D. VIIA

Explanation

To solve the problem, we need to determine the group of the periodic table that the element 'Y' belongs to after the radioactive substance \( (^{228}_{88}X) \) emits 3 alpha particles and 3 beta particles.

### Step-by-Step Solution:

1. **Identify the Initial Element:**
The initial radioactive substance is \( (^{228}_{88}X) \). Here, the mass number is 228 and the atomic number is 88.

2. **Calculate the Effect of Alpha Decay:**
Each alpha particle emission decreases the mass number by 4 and the atomic number by 2. Since 3 alpha particles are emitted:
- Mass number decrease: \( 228 - (3 \times 4) = 228 - 12 = 216 \)
- Atomic number decrease: \( 88 - (3 \times 2) = 88 - 6 = 82 \)

After 3 alpha emissions, we have:
\[ Y = (^{216}_{82}Y) \]

3. **Calculate the Effect of Beta Decay:**
Each beta particle emission increases the atomic number by 1 (as a neutron is converted to a proton). Since 3 beta particles are emitted:
- Atomic number increase: \( 82 + (3 \times 1) = 82 + 3 = 85 \)
- The mass number remains the same (216).

After 3 beta emissions, we have:
\[ Y = (^{216}_{85}Y) \]

4. **Identify the Element:**
The element with atomic number 85 is Astatine (At).

5. **Determine the Group of the Periodic Table:**
Astatine (At) belongs to Group 17 of the periodic table, which is also known as the halogens. In the long form of the periodic table, this is referred to as Group 7A.

### Final Answer:
The element 'Y' belongs to Group 7A of the long form of the periodic table.