Because of the inductive kick when the power is removed from the relay. The kick can be hundreds of volts, and any transistor or IC connected to it could be ruined.
The diode shorts this inductive kick to Vcc so it does not occur.
It's call "Fly back" voltage. A relay coils is an inductor and the current in an inductor does not change instantaneously. When you open the current path, the inductor will generate what ever voltage is required to keep the current flowing for as long as it takes to discharge the inductor magnetic field. When the switch driving the relay coil is a transistor, the transistor is destroyed by a huge voltage when it is turned off. The transistor will survive only if the transistor is big enough to absorb the energy stored in the inductor. The Diode provides a current path for voltage generated by the inductor because the fly back voltage across the coil is the reverse direction of the on voltage. Because the diode forward voltage is very small, the coil current will take longer to drop off, and the relay release, than it would for a higher fly back voltage.
When the transistor in the diagram is switched on, the supply voltage is applied across an inductance and the current through it increases linearly. As it does so, a magnetic field is created around the coil.
When the transistor is switched off, the coil is disconnected from the supply. The current through it is abruptly interrupted and is no longer able to maintain the field. The field which has been established therefore collapses very rapidly. In doing so it causes a voltage to be impressed across the coil and because of the speed at which the field collapses this voltage can be very high.
To short away counter EMF is generated by the relay coil while it is turning OFF. This EMF is quite high,capable to destroy parts like transistor that links to the coil.
To prevent reverse flow of current.
When a relay switches off, the electromagnet that controls it will have a collapsing magnetic field that can cause a reversing of the polarity. I.e a +5v supply will turn off, but a reverse -5V will come back. The reverse voltage could even be higher than the original voltage, which could damage your circuit. The diode creates a short circuit which will dissipate that voltage.
Why place a diode across the relay coil?
