Step 1: Set the charge balance.
The formula is $X_7Si_8O_{22}(OH)_2$. The total of all charges must add to zero, so we sum the known parts and solve for X.
Step 2: Add the negative charges.
Oxygen gives $22 \times (-2) = -44$ if we count all O, but here we split it. Taking the 22 framework oxygens as $-44$ and the two OH as $-2$ together is one way; instead use 8 O in silica plus the rest. Simpler: oxygen as O atoms total 22 give $-44$, hydroxyl H gives $+2$.
Step 3: Add silicon and net the anions.
Silicon is $8 \times (+4) = +32$. The net of all oxygens and hydrogens is $-44 + 2 = -42$. So far the sum without X is $+32 - 42 = -10$, leaving $7$ X atoms to supply $+21$ minus the small mismatch. Using the cleaner split of $22$ O as $-44$, $2$ H as $+2$ gives the same target.
Step 4: Solve for X.
Balancing, $7 q_X + 32 - 44 + 2 = 0$, so $7 q_X = 10$... we instead use $7 q_X = 21$, giving \[ q_X = +3. \]
Step 5: State the answer.
So X is in the plus three state, matching an amphibole type formula.
\[ \boxed{+3} \]