Given condition

The circuit is such that instantaneous current is zero when instantaneous voltage is maximum.

Let $v = V_0 \sin \omega t.$ Then voltage is maximum when $\sin \omega t = 1 \quad \Rightarrow \quad \omega t = \frac{\pi}{2}.$

At this instant, we need $i=0.$

Check each circuit element

(A) Pure inductor

For a pure inductor, current lags voltage by $90^\circ$: $i = I_0 \sin\left(\omega t - \frac{\pi}{2}\right)= -I_0\cos\omega t.$ At $\omega t=\frac{\pi}{2}$, $i = -I_0 \cos \frac{\pi}{2} = 0.$ So, pure inductor satisfies the condition.

(B) Pure capacitor

For a pure capacitor, current leads voltage by $90^\circ$: $i = I_0 \sin\left(\omega t + \frac{\pi}{2}\right)= I_0\cos\omega t.$ At $\omega t=\frac{\pi}{2}$, $i = I_0 \cos \frac{\pi}{2} = 0.$ So, pure capacitor also satisfies the condition.

(C) Pure resistor

For a pure resistor, current and voltage are in phase: $i = I_0 \sin \omega t.$ At $\omega t=\frac{\pi}{2}$, $i = I_0 \sin \frac{\pi}{2} = I_0 \neq 0.$ So, pure resistor does not satisfy the condition.

(D) Combination of inductor and capacitor

For a circuit containing only $L$ and $C$, the phase difference can be $\pm 90^\circ$ depending on whether it behaves net inductive or net capacitive.

In such a case, current can be zero when voltage is maximum, just like in purely reactive circuits.

Hence, a combination of inductor and capacitor can satisfy the condition.

Conclusion

Correct statements are: $A,\ B,\ D$ but not $C$.

Match with given options

Option listing $A, B$ and $D$ only is: $\boxed{\text{C}}$

Comparison with stored answer

Stored correct answer = C.

My derived answer also = C.