Identify the two feedback paths — that is what distinguishes every circuit in the option list.

Positive feedback: the R₁–R₂ divider returns a fraction of the output to the non-inverting input. This gives hysteresis: the comparator threshold is

\(\pm\beta V_Z, \qquad \beta=\dfrac{R_1}{R_1+R_2}\)

where the back-to-back zeners clamp the output at ±V_Z.

Negative feedback through a capacitor: R_f and C form a timing branch on the inverting input, so that node cannot follow the output instantly — it charges exponentially towards it.

How the circuit behaves. Suppose the output is at +V_Z. The capacitor charges through R_f towards that level. The moment its voltage passes the upper threshold +βV_Z, the comparator flips to −V_Z; the capacitor then discharges and recharges towards the negative rail until it crosses −βV_Z, and the output flips again. The cycle repeats indefinitely with period

\(T=2R_fC\ln\!\left(\dfrac{1+\beta}{1-\beta}\right)\)

The key point: nothing triggers it. There is no external input at all — the circuit oscillates on its own, producing a square wave at the output and an exponential (roughly triangular) waveform on the capacitor. A self-starting oscillator with two quasi-stable states is a free-running (astable) multivibrator.

Why the other three are wrong.

A monostable has one stable state and needs an external trigger pulse to produce a single timed output; the giveaway is a trigger input and usually a clamping diode, neither of which is present.

A comparator has an input signal applied to one terminal and no feedback at all.

A Schmitt trigger does have the positive-feedback divider, but its inverting input receives an external signal; here that input is occupied by the timing capacitor, which is exactly what converts a Schmitt trigger into an astable.

Hence, the circuit is a free running multivibrator.