Calculating total capacitance of a circuit
The capacitor C4 does not appear in the equivalent capacitance of the circuit. This is because the circuit is symmetrical around the two ends of the capacitor and there is no reason for the voltage to be different across these points. This would imply a zero potential difference (and charge) across the capacitor which can thus be removed without affecting the overall circuit.
But you can still apply KVL and charge equations to arrive at the same result:
Suppose the battery transfers a charge Q into the capacitors and the charge distribution across the capacitors looks as in the figure above:
Applying KVL across C2-C5-C6-C3 loop, we get:
$$\frac{q1}{C} + \frac{q2}{C} = \frac{Q-q1}{C} + \frac{Q-q2}{C}$$
$$\implies q1 + q2 = Q$$
Another KVL across C2-C1-C3 loop gives:
$$-\frac{q1}{C} - \frac{q1-q2}{C1} + \frac{Q-q1}{C} = 0$$
$$\implies (\frac{1}{C1} + \frac{1}{C})(q1-q2) = 0$$
Solving these equations we get,
$$q1 = q2$$
Thus, there is no charge across the C6 capacitor concluded before.
It is now pretty straight forward to solve for the equivalent capacitance now.