Electronic – Why not use only a MOV as a snubber for relay control of AC motor

You can place the RC either at the B side or the A side. When components are placed in series the order of them doesn't matter for the working.

About the diodes. When you switch off the relay it will cause a (possibly large) negative voltage on the FET's drain, and a flyback diode is used to limit that voltage to a 0.7 V diode drop. So the diode(s) don't serve to protect the coil, but the FET. Using the zeners will allow this voltage to go to -5.7 V or -15.7 V if you'd use the 15 V zeners. There's no reason for taking risks here, even if the FET can handle -30 V. So I would just use a rectifier or signal diode, or even better a Schottky diode.

edit re your comment
You can indeed use a zener (combined with a common diode, D1 doesn't have to be a zener) to decrease switch-off time, and Tyco also mentions it in this application note, but I don't read it as if they insist on it. The scope images in the first link show a dramatic decrease in switch-off time, but that measures the time between deactivating the relay and the first opening of the contact, not the time between first opening and the return to the rest position, which will change much less.

edit re the 6 V relay and the RC circuit
Like I says in this answer you can operate a relay below its rated voltage, and since its operate voltage is 4.2 V the 6 V version of your relay can also be used at 5 V. If you use a series resistor not higher than 9 Ω you'll have that 4.2 V, and then you don't need the capacitor (keep an eye on the tolerance for the 5 V!). If you want to go lower you're on your own; the datasheet doesn't give a must hold voltage. But let's say this would be 3 V. Then you can use a series resistor of 32 Ω and you'll need the capacitor to get the relay activated.

Operate time is maximum 15 ms (which is long), so as the capacitor charges the relay voltage shouldn't go below 4.2 V until 15 ms after switching on.

Now we have to calculate the RC time for that. R is the parallel of the relay's coil resistance and the series resistance (that's Thévenin's fault), so that's 19.3 Ω. Then

\$ 3 V + 2 V \cdot e^{\dfrac{- 0.015 ms}{19.3 \Omega \text{ C}}} = 4.2 V \$

Solving for \$\text{C}\$ gives us 1500 µF minimum.

Re switching off:
You can't violate Q = CV, it's the Law. Your clamping voltage is 3.3 V + 0.7 V = 4 V. That means that when you switch the FET off the low side of the capacitor momentarily will be pulled to -4 V, and quickly rise again to 0 V. The high side is 2 V higher, and will simply follow that 4 V drop while the capacitor discharges through the parallel resistor. The capacitor won't even notice the drop. The discharge time constant is 1500 µF \$\times\$ 32 Ω = 48 ms, then the capacitor will discharge to 20 mV (1% of its initial value) in 220 ms.

The 62 mA won't charge nor discharge the capacitor. We often apply Kirchhoff's Current Law (KCL) to nodes, but it also applies to regions:

Draw a boundary around C1 and R1, and you'll see there's only one path to the outer world since the way to the FET is cut off. Since the total current has to be zero there can't be any current through that unique connection. The coil has to take care of the 62 mA on its own, and it does so by using the loop formed by the zeners.

The values used will often be OK.
A larger than usual motor inductance may cause problems.

The snubber's job is to protect the switch contacts from inductive turnoff transients from the motor. Stopping the transient at source (across the motor) or at destination (across the contacts) both work. Arguably, having it at the switch is better as it deals with the energy that will do damage, as opposed to energy that may do damage, so it is more focused and it also then deals with other spikes that may happen along.

If you look at your circuit you'll note that in both cases the snubber connects from the motor-switch connection point to one leg of the mains. If the mains impedance is low at the spike frequency (-ies) then both are about equivalent.

The circuit current continues instantaneously at switch off. If it all flows through the snubber then it will pass through the 120 ohm resistor, so the voltage spike will initially will be \$V=IR = 10\mathrm{A} \times 120\mathrm{\Omega} = 1200\mathrm{V}\$. While that is a lot it is usually within the switch break capability (or else), and there are usually other impedances present which will also help to damp it.

The snubbing current will flow only until the capacitor charges to the driving voltage. If the motor inductance is large the capacitor may charge to a higher or much higher voltage.

The capacitor needs to be large enough to not be charged to the point where current decays through charging of the cap before the resistor dissipates the energy. To be sure that the component values present will do the job, you need to know motor inductance.

Energy in inductor is \$E=\frac{1}{2}LI^2\$

Capacitor will "ring" with an energy of \$E=\frac{1}{2}CV^2\$

The resistor needs to dissipate this energy.

Energy = \begin{align}\frac{1}{2}Li^2 &= \frac{1}{2}CV^2 \\ \\ \Rightarrow V &= \sqrt{\frac{Li^2}{C}} \end{align}

Then there is some \$L/R\$ time constant as well and ...

You can start to calculate this (if you know L) or simulate it, but in most cases the values shown are OK for typical equipment.

Place a scope across the contacts. What peak V do you see (use a suitable probe!). Do the contacts spark? They shouldn't.

Note that increasing C improves snubbing action but also increases losses from the mains in normal operation. Note also that a capacitor across a mains switch may be frowned at in some contexts.

Added:

Dario said: One problem with placing the RS across the switch is that now you have some current in the circuit in the switched off mode. ...

User_long_gone responded: I'm absolutely certain that the 4-5 MILLIAMPS of current flowing through a 0.1 microfarad capacitor at 60 Hz will present no problem to a motor circuit. Wasteful of energy? It's less than 1/2 watt.

It's worth noting that

1. The snubber across the motor may not bother the motor itself but may well severely bother anyone silly enough to think that the switch being off means that the circuit is "safe" or "dead". If the switch is in the phase/live lead the motor side of the switch may be near ground due to relative impedances. But there is no certainty that this will always be the way the connection is made - even if regulations say that it should be.

2 "Even" 1/2 a Watt of pointlessly wasted energy in an appliance is frowned on in modern scenarios.