I am aware that during addition/subtraction in the accumulator, if a 9-bit value is resulted the carry bit is set, but what if this is used:
mov A,-200
-200 is a 9-bit value in 2's compliment form. what's going to happen if I move it to accumulator? Is the extra bit going to drop off or go set the carry?
A is an 8-bit register.
But -200 is not an eight-bit value. When clipped to 8 bits, it's the value 56 - and 56 is well within range.
In this case, an assembler could decide to convert -200 to +56. Or it could decide to produce a warning/error that -200 doesn't make sense in an 8-bit assign.
The first place that you must always look for a question like that is the Manual for the tool in question.
Keil's assembler manual is here: http://www.keil.com/support/man/docs/a51
And the specific page about instructions is here: http://www.keil.com/support/man/docs/a51/a51_wp_instructions.htm
Linking to: http://www.keil.com/support/man/docs/is51/is51_overview.htm
The MOV instruction is described here: http://www.keil.com/support/man/docs/is51/is51_mov.htm
As Per says, the immediate operand is 8 bits - there is no MOV instruction which takes a 9-bit immediate operand!
Or, to be pedantic,
there is no MOV instruction with an 8-bit target which takes a 9-bit immediate operand!
I can think of no 8051 instruction that takes a 9-bit immediate operand.
MOV DPTR, #immediate
Takes a 16-bit immediate operand - so it could accommodate a 9-bit value.
But you're right - there's nothing that takes an actual 9-bit operand.