Why does the current lead the voltage across a capacitor by 90 degree rather than lag it?

A: Because a capacitor have to have time to charge to the voltage

In a capacitor, the current depends on the voltage difference across it. On AC, this makes it charge, if the voltage is increasing above zero, and discharge if the voltage is reducing towards zero.

Because a capacitor has almost no internal resistance, and most loads that it is connected to have only very small resistances in series with the capacitor, the charging and discharging currents depend pretty much on the rate at which the voltage is changing. At the zero crossing point of the sine-wave, when the voltage is actually zero, the rate of change of voltage is very high (the sine-wave is at its steepest), so the current is also very high. If the voltage is positive-going, the current is positive, and if the voltage is negative-going, the current is negative.

At the peak of the voltage waveform, the rate of change of voltage is zero or very low (the sine-wave is flat, and not really changing its voltage) so the current is zero, too.

Since the maximum positive current occurs when the voltage is passing through zero, going positive, and the maximum negative current happens when the voltage is passing through zero, going negative, the current peaks happen 90 degrees before the voltage peaks, so the current is said to lead the voltage. This is the same as saying the voltage lags the current by 90 degrees.