A good answer might be:

No. Because the 16-bit pattern is zero-extended to 32 bits, the resulting pattern is never a two's-complement negative integer.

OR Immediate used with Non-zero Operands

Here is a description of the ori instruction used as a bitwise OR instruction:

ori  d,s,const     # register d <-- bitwise OR of const
                   #                with the contents of register $s
                   #                const is 16-bits, so
                   #                0x0000 ... const ... 0xFFFF

The three operands of the instruction must appear in the correct order, and const must be within the specified range. If the immediate operand in a source instruction is less than sixteen bits (such as 0x2) the assembler expands it to sixteen. If it is more than sixteen bits the assembler writes an error message.

The const part of the assembly language instruction can be a positive decimal or a hexadecimal constant. The assembler translates the constant into a 16-bit pattern in the machine instruction. For example, the following two assembly language instructions translate into the same machine language instruction:

ori  $5,$4,0x10
ori  $5,$4,16

QUESTION 5:

Is the following instruction correct?

ori  $0,$9,0x32